Timeline for How to think about parabolic induction.
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2012 at 12:50 | comment | added | Emerton | Dear labirintas, The situation in the modular case will be similar to the situation for $p$-adic groups: if you work in the $K$-group (i.e. up to semisimplification) then you get equality. Regards, Matthew | |
| Dec 10, 2012 at 13:23 | answer | added | Wilberd van der Kallen | timeline score: 9 | |
| Dec 10, 2012 at 12:25 | answer | added | Alexander Braverman | timeline score: 3 | |
| Dec 10, 2012 at 9:37 | comment | added | labirintas | In the previous comment I consider representations of $GL_2(F_p)$ on $F_p$-vector spaces. | |
| Dec 10, 2012 at 9:36 | comment | added | labirintas | Let me just note that this fails if you consider modular representations. So if you induce a regular character $\chi$ from the upper triangular matrices in $GL_2(F_p)$ and from the lower triangular matrices, you get non-isomorphic representations, which have the same semi simplification. Regular means $\chi^s\neq \chi$, and $s$ is the matrix (0 & 1// 1 & 0). | |
| Dec 10, 2012 at 7:05 | answer | added | Sam Gunningham | timeline score: 6 | |
| Feb 25, 2011 at 5:19 | answer | added | Victor Ginzburg | timeline score: 7 | |
| Dec 21, 2009 at 16:10 | answer | added | David Bar Moshe | timeline score: 2 | |
| Dec 21, 2009 at 0:31 | comment | added | Kevin McGerty | I'm used to thinking of this in the context of finite groups of Lie type, where induction is exact (even in the modular case). Of course that's not true in other contexts, so I need to think about them... | |
| Dec 21, 2009 at 0:29 | history | edited | Kevin McGerty | CC BY-SA 2.5 |
Added some explanation of the "Mackey formula" argument.
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| Dec 19, 2009 at 17:15 | comment | added | David Treumann | What I said is not quite right. It can happen (and does happen here) that FG is exact while F is only left exact. The question I should have asked is: are the modules (R^i Ind_P^G)(Res_L^P M) independent of P? | |
| Dec 18, 2009 at 20:42 | comment | added | David Treumann | Parabolic induction has right-derived functors, because Ind_P^G does. Are these also independent of P? Also, can you say more about the Mackey formula argument? | |
| Dec 18, 2009 at 3:46 | answer | added | S Kitchen | timeline score: 1 | |
| Dec 17, 2009 at 21:54 | answer | added | Ben Webster♦ | timeline score: 3 | |
| Dec 17, 2009 at 14:11 | comment | added | Ben Webster♦ | Another way to say this is that a Levi in SL(n) is a choice of direct sum decomposition of your space, so up to conjugacy, they correspond to partitions. A parabolic is a choice of a flag, and so correspond to partitions with ordered parts, or compositions. As Kevin points out, there are many more compositions than partitions. | |
| Dec 17, 2009 at 13:09 | comment | added | Kevin McGerty | Yes surely: say in GL(3) I can take the stabilizer of a line and a complementary plane as the Levi L, then the parabolics corresponding to the stabilizer of the line or the plane both contain L but are not conjugate. In fact the number of (conjugacy classes of ) Levi subgroups is p(n) the number of partitions of n, while for parabolics it is $2^{n-1}$, so there are "a lot more" parabolics than Levis. | |
| Dec 17, 2009 at 10:24 | comment | added | Leonid Positselski | Is it really true that the same reductive subgroup can be a Levi subgroup of two nonconjugate parabolic subgroups? | |
| Dec 17, 2009 at 5:04 | history | asked | Kevin McGerty | CC BY-SA 2.5 |