Timeline for Can the image of a Schur functor always be made an irreducible representation?
Current License: CC BY-SA 2.5
4 events
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| Jul 19, 2010 at 17:03 | comment | added | Theo Johnson-Freyd | Ah, so I was in error. It was late last night, after some wine, that I wrote my answer :) | |
| Jul 19, 2010 at 13:21 | comment | added | Jack Schmidt | Yeah, I think all this is saying is that the Frobenius–Schur indicator indicates whether the trivial rep is a summand of the 2nd symmetric power, the 2nd exterior power, or neither. O(n) reps means it is a summand of the 2nd symmetric power. It's pretty common for an irrep not to be realized in O(n). | |
| Jul 19, 2010 at 12:33 | comment | added | Jack Schmidt | It sounds like you mean that Sym^2(V) is reducible unless dim(V) = 1? SL(2,3) has some 2 dim reps with Sym^2 irreducible. This appears somewhat common. Checking a standard list of character tables, one sees there are examples for Sym^2 in dimensions 2,3,4,5,6,8,9,10,11,12,13,14,18,20,21,26,28,32,41,42,43,45,60,342,1333; examples for Sym^3 in dimensions 2,3,4,5,6,8,9,10,12,13,14,18,20,32; examples for Sym^4 in dimensions 2,4,6,12; and examples for Sym^5 in dimensions 2,4,6,12. There were no examples for Sym^6 as indicated by Guralnick and Thiep. Maybe this is a U(n) versus O(n) problem? | |
| Jul 19, 2010 at 4:30 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |