Timeline for Restrictions of irreducible representations of algebraic groups to root SL_2-subgroups
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2016 at 20:41 | comment | added | Jim Humphreys | @David Stewart: It's not quite so easy (and like Premet's approach it requires extra care for some types if the characteristic is 2, 3). But see (6.1) in the AMS Memoir No. 365 (1987) by Seitz for the prime characteristic irreducibles. As he notes, Freudenthal's multiplicity formula indicates in characteristic 0 that classical weight multiplicities tend to grow quickly in most cases. | |
| Aug 18, 2016 at 10:40 | answer | added | Alexander Premet | timeline score: 5 | |
| Aug 16, 2016 at 10:21 | comment | added | Alexandre Borovik | Motivation for this question comes from model-theoretic algebra, more precisely from the theory of groups of finite Morley rank. In certain situations we have to deal with a definable action of a simple group $G$ of finite Morley rank on an abelian group $A$ with a much smaller simple definable subgroup $L < G$ acting on $[A,L]$ irreducibly (that is, without nontrivial proper $L$ -invariant subgroups) and with $[A,L] < A$. Notice, there is no vector space structure on $A$ yet, $A$ is just an abelian group. I realised that I do not know answer even in simplest algebraic group cases. | |
| Aug 16, 2016 at 10:10 | comment | added | Alexandre Borovik | Alas, I do not know the answer even in zero characteristics. | |
| Aug 11, 2016 at 21:32 | comment | added | David Stewart | There can't be many examples. In char $0$ you may as well assume $V$ is irreducible. Then you're looking for representations which restrict with the non-$0$ weight spaces only being $1$-dimensional. The only examples I can think of are $SL_2<SL_n$ and $SL_2<Sp_{2n}$ both on the natural rep. Without doing it, I would guess it should be quite easy? | |
| Aug 11, 2016 at 20:17 | comment | added | YCor | @AlexandreBorovik Allen asks you whether you know the answer to your question in characteristic zero | |
| Aug 11, 2016 at 17:18 | comment | added | Alexandre Borovik | Yes, of course, $V = [V,L] \oplus V^L$ (over algebraically closed field of characterstic $0$). | |
| Aug 11, 2016 at 16:29 | comment | added | Allen Knutson | So if $K=\mathbb C$, anyway, this is the unique $L$-invariant complement to the invariant vectors $V^L$? (If your question is easy in characteristic $0$, do say so; certainly I don't already know the answer there.) | |
| Aug 11, 2016 at 16:18 | comment | added | Alexandre Borovik | The commutator, $[V,L]$ is the span of vectors of the form $v - v^l$ for $v \in V$ and $l \in L$. | |
| Aug 11, 2016 at 15:17 | comment | added | Allen Knutson | What does $[V,L]$ mean here? | |
| Aug 11, 2016 at 15:10 | review | First posts | |||
| Aug 11, 2016 at 15:23 | |||||
| Aug 11, 2016 at 15:09 | history | asked | Alexandre Borovik | CC BY-SA 3.0 |