Timeline for Tannakian categories equivalent as abelian categories
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2010 at 19:48 | comment | added | Ben Wieland | Here is a similar question that I believe is open: is a $p$-group determined by its mod $p$ group ring? | |
| Dec 2, 2010 at 17:28 | comment | added | Theo Johnson-Freyd | I'm a little worried about an argument about "deformations", since to my ear the word "deformation" means that you're only allowing those in a neighborhood. And my experience has been that often the different symmetric monoidal structures on a category are different by a lot. For example, consider the semisimple category (over $\mathbb C$) with four simples, one of which you call the "trivial representation / monoidal unit". Then there are two symmetric monoidal structures you can put on this, corresponding to $\mathbb Z/3$ and $S_3$. Do you consider these "deformations" of each other? | |
| Dec 2, 2010 at 17:20 | comment | added | Theo Johnson-Freyd | On the other hand, since you say that you know the trivial representation, then you do know all its Ext groups with itself. So this is something. | |
| Dec 2, 2010 at 17:19 | comment | added | Theo Johnson-Freyd | 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. | |
| Dec 2, 2010 at 17:09 | comment | added | AFK | 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. | |
| Dec 2, 2010 at 17:07 | history | edited | AFK | CC BY-SA 2.5 |
added 77 characters in body
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| Dec 2, 2010 at 11:37 | comment | added | Martin Brandenburg | See David Jordan's answer to mathoverflow.net/questions/38089/…. | |
| Dec 2, 2010 at 11:18 | comment | added | Kevin Buzzard | ...(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! | |
| Dec 2, 2010 at 11:17 | comment | added | Kevin Buzzard | 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... | |
| Dec 2, 2010 at 11:11 | history | asked | AFK | CC BY-SA 2.5 |