Timeline for Finite groups G with Rep(G) Grothendieck equivalent to a modular category
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2021 at 3:23 | vote | accept | Sebastien Palcoux | ||
| Oct 18, 2020 at 3:21 | comment | added | Victor Ostrik | @Sebastien Palcoux I think that the assumption of Saksonov's theorem is that you have isomorphism of the rings in characteristic $p$ (i.e. you take integral character ring and conjugacy class ring and tensor them with ${\mathbb Z}/p$). And in the paper of Andrus et al you have an isomorphism in characteristic zero, i.e. after tensoring with rational numbers. I don't think that these two sets of assumptions are related by implication in either direction. | |
| Oct 17, 2020 at 9:28 | comment | added | Sebastien Palcoux | @VictorOstrik There is something unclear: the paper of Andrus et al provides examples of non-abelian self-transpose p-groups (called stem groups). As you wrote, the property to be self-transpose should be much stronger than the (unbased) ring isomorphism between character ring and conjugacy class ring. So, according to Saksonov (mentioned above) these groups should be p-nilpotent with abelian Sylow p-subgroup, but they already are p-groups, so they should be abelian... Anyway, do you believe that Rep(G) is Grothendieck equivalent to a modular category if and only if G is self-transpose? | |
| Oct 16, 2020 at 22:25 | comment | added | Victor Ostrik | @Sebastien Palcoux I think that my "stronger condition" from the answer to Qiaochu is equivalent to the existence of isomorphism between character ring and conjugacy class ring which send irreducible characters to class sums divided by square roots of class sizes (so it is almost isomorphism of based rings). This is much stronger than just isomorphism of rings. Also this is equivalent to group $G$ being self-dual in the sense of Bannai (see the answer by A.Stasinski in Qiaochu's link). For example paper by Andrus et al in the same answer proves that such groups are nilpotent. | |
| Oct 16, 2020 at 20:02 | comment | added | Sebastien Palcoux | What about the groups with character ring equivalent to conjugacy class ring? According to this answer they are exactly the groups $p$-nilpotent with abelian Sylow $p$-subgroup. Perhaps this paper in Russian of Saksonov (cited as reference) contains a non-abelian example. I don’t know if the equivalence must keep the usual bases. | |
| Oct 16, 2020 at 19:00 | comment | added | Victor Ostrik | @Qiaochu Yuan Thanks for pointing out this example! The same arguments as in my answer give an apparently much stronger (but more difficult to verify) necessary condition: after normalization and permutation of columns as above the character table must become symmetric matrix. It would be interesting to check whether it works for the group of order 64 from your comment. | |
| Oct 16, 2020 at 17:49 | comment | added | Qiaochu Yuan | Apparently this was asked about on MO before! The smallest group with this property has order $64$, no non-abelian finite simple group has this property, and conjecturally all finite groups with this property are nilpotent: mathoverflow.net/questions/20374/… | |
| Oct 16, 2020 at 17:01 | history | answered | Victor Ostrik | CC BY-SA 4.0 |