Timeline for How ugly is the isomorphism R[GxH] = R[G] (X) R[H] for groups G, H?
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9 events
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| Apr 12, 2021 at 14:33 | comment | added | Maxime Ramzi | @XandiTuni : do you by any chance know of an example where the tensor product is not irreducible in characteristic $0$, if the field is not algebraically closed ? | |
| Jun 30, 2010 at 19:21 | comment | added | Xandi Tuni | @darij: If by "composed of" you mean "a direct sum of" then certainely not, so i am just telling that at least you get the irreducuble ones. But you get surjectivity on the level of Grothendieck rings, which do not see extensions. | |
| Jun 30, 2010 at 17:26 | comment | added | darij grinberg | Xandi: but thinking about the irreps is not enough in characteristic $p$, since not everything is composed of irreps. | |
| Jun 30, 2010 at 15:59 | comment | added | George McNinch | @Xandi: right, thanks! I had forgotten that (hadn't thought about these matters in a while...) | |
| Jun 30, 2010 at 15:55 | comment | added | Xandi Tuni | Thanks George no "probably" needed. The counting argument also works if the characteristic $p$ divides $|G|$: The number of irreducible representations equals the number of conjugacy classes in whose elements are of order prime to $p$. | |
| Jun 30, 2010 at 15:34 | comment | added | George McNinch | @darij: in beginning of previous comment, I should have written "probably refers to". | |
| Jun 30, 2010 at 14:12 | comment | added | George McNinch | @darij: "just count them" refers to the fact that for finite $G$, sum of squares of dims of irred. $kG$ modules $=|G|$ provided (i) $k$ is a splitting field for $G$ and (ii) $kG$ is semissimple ($\iff$ $|G|$ is invertible in $k$). I don't see how to use counting arg if $kG$ isn't ss, though Xandi's assertion remains true. Also: taking a comp. series for $G \times H$-module $V$, it seems that question of expressing $V$ as $\mathbf{Z}$-lin comb of $\otimes$s in Groth. group can be reduced to the case of irred. $V$ and thus to $\otimes$ of irreds, provided $\otimes$s of irreds remain irred. | |
| Jun 29, 2010 at 20:36 | comment | added | darij grinberg | What do you mean bby "just count them"? -- And I'm not asking whether the tensor products will remain irreducible. I'm asking whether any representation (I am not talking about irreduciblity) is, in the Grothendieck category (note that cancellation is possible if $k$ is not of char $0$, although already the char $0$ case is interesting enough) a $\mathbb Z$-linear combination of tensor products. | |
| Jun 29, 2010 at 18:52 | history | answered | Xandi Tuni | CC BY-SA 2.5 |