Timeline for Character table does not determine group Vs Tannaka duality
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2010 at 18:38 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
edited body
|
| Jan 11, 2010 at 17:37 | comment | added | David E Speyer | That's what R(G) \otimes C can tell you. If you don't tensor with C, you can get a little more. For example, R(Z/2 x Z/2) is not isomorphic to R(Z/4) because, after tensoring with a field k of characteristic 2, the former becomes k[u,v]/<u^2, v^2> and the latter becomes k[u]/u^4. But, morally, I agree with this answer. | |
| Jan 11, 2010 at 17:19 | comment | added | Qiaochu Yuan | Whoops. I seem to have been secretly assuming that R(G) came with the dual basis on Hom(G, C), which is obviously wrong. Thanks! | |
| Jan 11, 2010 at 17:18 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
added 73 characters in body
|
| Jan 11, 2010 at 16:59 | comment | added | Mariano Suárez-Álvarez | How do you get the class sizes from the abstract ring $R(G)\subseteq\mathbb C^G$? It is a semisimple commmutative $\mathbb C$-algebra, so its only invariant is the dimension. | |
| Jan 11, 2010 at 16:55 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |