Timeline for Is there any Lefschetz-like principle for representations of finite groups?
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15 events
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| Feb 12, 2015 at 23:37 | comment | added | P Vanchinathan | Nice to get a comment from the master himself! Thanks for your explicit statement about divisibility that I did not know earlier. (Perhaps I did not study the textbooks carefully). As the divisibility proofs used the fact an algebraic integer that is rational is a usual integer I could not guess about representations in prime characteristic. | |
| Feb 12, 2015 at 19:38 | comment | added | Marty Isaacs | Although degrees of irreducible representations of finite groups over algebraically closed fields of prime characteristic p need not divide the group order, it is worth mentioning that this divisibility property does hold for solvable groups, and more generally, for p-solvable groups. Since a group with order not divisible by p is automatically p-solvable, this generalizes the fact that the divisibility property holds if p does not divide the group order. | |
| Feb 12, 2015 at 3:02 | review | Close votes | |||
| Feb 12, 2015 at 10:59 | |||||
| Feb 3, 2015 at 9:40 | comment | added | Qiaochu Yuan | There's no need to mention complex conjugation in the theory. The conjugate of a character $\chi(g)$, over any field, is $\chi(g^{-1})$. | |
| Feb 3, 2015 at 2:33 | comment | added | P Vanchinathan | Yes, I have seen that wonderful book. Remember reading from there that an irreducible complex character of degree > 1 vanishes somewhere. And the proof uses AM>GM inequality! | |
| Feb 2, 2015 at 18:30 | comment | added | Jim Humphreys | By the way, there's no need to abandon characters altogether (but they might not always give much information). AMS has reprinted the classic text by I.M. Issacs, where you should look at Chapter 9: ams.org/bookstore-getitem/item=CHEL-359-H | |
| Feb 2, 2015 at 16:57 | vote | accept | P Vanchinathan | ||
| Feb 2, 2015 at 16:55 | answer | added | Jay Taylor | timeline score: 3 | |
| Feb 2, 2015 at 16:48 | comment | added | P Vanchinathan | @Jay Taylor: Thanks for the reference. Somehow the massive size of Curtis and Reiner made me stay away from it. I'll now definitely look up there. You have now given a clear answer. Yes, Wedderburn's structure theorem gives a way avoiding complex characters. You can make this comment an answer. | |
| Feb 2, 2015 at 16:41 | comment | added | Jay Taylor | See for instance (27.21) on pg. 186 of Curtis and Reiner's fantastic book "Representation theory of finite groups and associative algebras". | |
| Feb 2, 2015 at 16:40 | comment | added | Jay Taylor | Character theory provides a very elegant tool when studying representations of finite groups but many statements can be proved without using characters. For instance, the identity you mention follows from the fact that the group algebra $\mathbb{K}G$ is semisimple when $\mathbb{K}$ is an algebraically closed field whose characteristic doesn't divide $|G|$. In particular $\mathbb{K}G$ is isomorphic to a direct sum of matrix algebras $M_{d_i}(\mathbb{K})$. Counting dimensions as $\mathbb{K}$-vector spaces gives you the numerical identity. This can be proved without character theory. | |
| Feb 2, 2015 at 16:19 | review | Close votes | |||
| Feb 3, 2015 at 21:37 | |||||
| Feb 2, 2015 at 16:12 | comment | added | P Vanchinathan | I am uncomfortable about using the property of complex conjugation. This seems to play a crucial role in the proof of Schur orthogonality relations. | |
| Feb 2, 2015 at 15:59 | comment | added | Jim Humphreys | Certainly any algebraically closed field of characteristic 0 will reproduce the classical results (in fact, a much smaller extension of $\mathbb{Q}$ suffices). But in prime characteristic $p$ dividing $|G|$ everything tends to change. It's true however that when $p$ doesn't divide $|G|$, you get back the same picture as in characteristic 0. All of this can be found in standard sources. | |
| Feb 2, 2015 at 15:54 | history | asked | P Vanchinathan | CC BY-SA 3.0 |