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Suppose $A = Rep_k(G)$ and $B=Rep_k(H)$ are tannakian categories and $F: A\to B$ is an equivalence of abelian categories with $F(1_A) = 1_B$ (but not a $\otimes$-equivalence). What can I say about $G$ and $H$?

Question: Suppose $G$ is pro-unipotent. Are $G$ and $H$ isomorphic?

I have a vague feeling that there should some $H^1$ classifying deformations of the $\otimes$-structure on $A$ and that if $G$ is pro-unipotent this group should vanish so that $G \simeq H$. Is there any reference for such a thing?

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Perhaps this example is illuminating. Let $G$ and $H$ be the two non-abelian groups of order 8 (considered, if you like, as etale group schemes over $k$). As Tannakian categories $A$ and $B$ are of course different, but as abelian categories I think they're often exactly the same, because both $A$ and $B$ are just the finitely-generated modules over the group rings $k[G]$ and $k[H]$, so if $k$ is big enough to make these rings isomorphic (for example if all the representations of $G$ and $H$ are defined over $k$, e.g. if $k$ is the complexes) then $A$ and $B$ are equivalent... –  Kevin Buzzard Dec 2 '10 at 11:17
...(and even isomorphic!). So in this finite group setting, if $k$ is alg closed of char 0, it seems to me that pretty much all you can say is "the degrees of the irreducible characters of $G$ and $H$ coincide". In fact I don't even need groups of order 8: consider the two groups of order 4! –  Kevin Buzzard Dec 2 '10 at 11:18
See David Jordan's answer to mathoverflow.net/questions/38089/…. –  Martin Brandenburg Dec 2 '10 at 11:37
I wasn't suggesting that $A\simeq B$ as abelian categories implies $G=H$ in general obviously. I do think it is the case if $G$ is pro-unipotent. I edited the question to clarify it. –  YBL Dec 2 '10 at 17:09
I don't have a good sense for the (pro)unipotent case. At the other extreme are the semisimples, e.g. finite groups over algebraically closed characteristic zero. Then from the abelian category the only thing you know is that the group is semisimple, and how many simple objects it has. –  Theo Johnson-Freyd Dec 2 '10 at 17:19

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