Timeline for Bimodule categories realized as internal bimodules
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2017 at 14:07 | comment | added | Ehud Meir | Yes, this is the case. And you are welcome :) | |
| Feb 27, 2017 at 14:03 | vote | accept | kolaka | ||
| Feb 27, 2017 at 14:00 | comment | added | kolaka | Exactly, that was my question, thanks for your patience. So to wrap up, by using the braiding we can have an equivalence $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ as bimodule categories, but there is no such equivalence of $\mathcal M$ to a category of $A$-bimodules (or $B$-$A$-bimodules, if $\mathcal M \simeq B\mbox{-}\mathsf{Mod}(\mathcal C)$ as right module categories). | |
| Feb 27, 2017 at 10:00 | comment | added | Ehud Meir | 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. | |
| Feb 27, 2017 at 9:57 | comment | added | Ehud Meir | 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. | |
| Feb 27, 2017 at 9:38 | comment | added | kolaka | 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... | |
| Feb 27, 2017 at 9:32 | comment | added | Ehud Meir | 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. | |
| Feb 27, 2017 at 8:56 | comment | added | kolaka | 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. | |
| Feb 26, 2017 at 22:03 | history | answered | Ehud Meir | CC BY-SA 3.0 |