# Bimodule categories realized as internal bimodules

Let $\mathcal C$ be a finite tensor category, and $\mathcal M$ a finite left $\mathcal C$-module category. By a result of P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik (http://www-math.mit.edu/~etingof/tenscat1.pdf , Thm. 2.11.6), there is an algebra object $A \in \mathcal C$ and an equivalence of left $\mathcal C$-module categories $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$, where $\mathsf{Mod}\mbox{-}A(\mathcal C)$ is the category of right $A$-modules internal in $\mathcal C$. There is a similar result for right $\mathcal C$-module categories.

Now assume that $\mathcal M$ is a $\mathcal C$-$\mathcal C$-bimodule category, e.g. $\mathcal M$ is braided (then there is a canonical right/left $\mathcal{C}$-module structure if $\mathcal M$ was a left/right $\mathcal{C}$-module). (Of course, one can also consider the more general case of a $\mathcal C$-$\mathcal D$-bimodule category.) What is the corresponding result to the one above? Something straight-forward would be the following: If $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ as left module categories and $\mathcal M \simeq B\mbox{-}\mathsf{Mod}(\mathcal C)$ as right module categories, then $\mathcal M \simeq B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ as bimodule categories, where $B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ is the category of $B$-$A$-bimodules in $\mathcal C$. If this is correct, how can it be proven? Is there already a paper covering this?

(By results of Douglas, Schommer-Pries and Snyder, it at least follows that $\mathcal M \boxtimes_{\mathcal C} \mathcal M \simeq B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ as $\mathcal C$-bimodule categories. Does this already show that my claim above does not hold?)

Edit: Rephrasing my question, how can the equivalence $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ be made into an equivalence of bimodule categories, if $\mathcal M$ is a bimodule category?

The equivalence is unfortunately not correct: take for example the category $\mathcal{C}=Vec_G$ for some finite group $G$. Take $\mathcal{M}=Vec$ be the category of vector spaces. Then the forgetful functor from $\mathcal{C}$ to $Vec$ enriches $\mathcal{M}$ with the structure of a $\mathcal{C-C}$ bimodule category. If we write $A=KG$, the group algebra, then $\mathcal{M}$ is equivalent to the category of right (left) $A$-modules as a left (right) $\mathcal{C}$-module category. However, the category of $A-A$-bimodules in $\mathcal{C}$ is equivalent to the category $Rep-G$, which is not equivalent to $\mathcal{M}$. One possible approach to the problem is to consider $\mathcal{M}$ as a left module category over the category $\mathcal{D}:=\mathcal{C}\boxtimes\mathcal{C}^{op}$. The category $\mathcal{M}$ is then equivalent to the module category of some algebra in $\mathcal{D}$, which will contain $A\boxtimes B$ as a subalgebra.
• Thanks for the help, Ehud, that's a nice counterexample! Do I understand correctly that $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ (as left module categories) can not be understood as an equivalence of bimodule categories (e.g. by endowing $\mathsf{Mod}\mbox{-}A(\mathcal C)$ with a bimodule category structure using the braiding or something)? I can only think of the trivial extension $\mathcal M \simeq 1_{\mathcal{C}}\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ (as bimodule categories), where $1_{\mathcal{C}}$ is the tensor unit. – kolaka Feb 27 '17 at 8:56
• exactly. Consider for example the case where $G$ is abelian. In this case the category $Vec_A$ is braided. The braiding will give you a functor from $\mathcal{M}$ to the category of $A$-bimodules. This functor will not be an equivalence though, since there are many different ways to endow objects of $\mathcal{M}$ with a bimodule structure. – Ehud Meir Feb 27 '17 at 9:32
• OK, but my question was if $\mathsf{Mod}\mbox{-}A(\mathcal C)$ can be made into a $\mathcal C$-bimodule category, such that $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ becomes an equivalence of bimodule categories. Sorry if this question is stupid... – kolaka Feb 27 '17 at 9:38
• It isn't. I am not sure that I fully understand the question though: if you have the left module category $Mod-A(\mathcal{C})$, you can use the braiding in $\mathcal{C}$ to make it a bimodule category over $\mathcal{C}$. As we have seen, this will not necessarily be equivalent to the category of $A$-bimodules in $\mathcal{C}$. About your last question (in the edit): this might be a bit more complicated than how it first appears, because there might be many different ways to make $\mathcal{M}$ a bimodule category. – Ehud Meir Feb 27 '17 at 9:57
• Now that I read your question again: if $\mathcal{M}=Mod-A(\mathcal{C})$, then $\mathcal{M}\cong Mod-A(\mathcal{C})$ also as bimodule categories (you just define a bimodule category structure on $\mathcal{M}$ via this equivalence. – Ehud Meir Feb 27 '17 at 10:00