Skip to main content
12 events
when toggle format what by license comment
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Nov 27, 2013 at 16:23 answer added Jim Humphreys timeline score: 2
Nov 27, 2013 at 9:34 vote accept Nick Gill
Nov 27, 2013 at 9:28 comment added Nick Gill @JimHumphreys, Cline, Parshall & Scott have around 30 papers - I came across their earliest in the course of reading around this problem, but I'm not sure if this is the paper you are referring to?
Nov 27, 2013 at 9:04 answer added Derek Holt timeline score: 7
Nov 26, 2013 at 20:35 comment added Derek Holt @Jim Humphreys: This is addressed more to you as the expert on representations of groups of L:ie type in natural characteristic. is it not true that, if $|E|=q^d$ as in Qn 2, the induced module action would have to be either trivial or the natural module or its dual (possibly twisted by a field automorphism)? The basic modules for ${\rm SL}_d(q)$ are all defined over ${\mathbb F}_q$, and tensor products that could be written over smaller fields would have order larger than $q^d$. For the natural module, the only nonsplit extension is for ${\rm SL}_5(2)$.
Nov 26, 2013 at 19:07 comment added Jim Humphreys P.S. Concerning the notion of Levi factor, this arises more generally than the parabolic case when you have an algebraic group and try to split off a semisimple (or reductive) complement to its radical (or unipotent radical)
Nov 26, 2013 at 19:01 comment added Jim Humphreys @Nick: As the preceding comment points out, cohomology questions are involved here; so it's useful to consult the paper by Cline-Parshall-Scott, van der Kallen and their references: gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002093200
Nov 26, 2013 at 17:27 comment added user76758 The conjugation action of $G$ on its abelian normal subgroup $E$ factors through a $G/E$-action on $E$, so the given extension structure makes $E$ into an ${\rm{SL}}_d(q)$-module. If you specify this module structure then the set of isomorphism classes of such extensions is in bijection with ${\rm{H}}^2({\rm{SL}}_d(q),E)$, so you're asking to compute degree-2 cohomology for ${\rm{SL}}_d(q)$ acting on $\mathbf{F}_p$-vector spaces; sounds hard! The case $|E| = q^d$ isn't special, but natural cases of interest are $\mathbf{F}_q^d$ with usual action and ${\mathfrak{sl}}_d(q)$ with adjoint action.
Nov 26, 2013 at 17:16 comment added Nick Gill @JimHumphreys, Thanks for your comment, I'll look up George's work. Could you just clarify what you mean when you say `Levi factor'... Are you talking about a subgroup $L$ of a parabolic $P$ for which $P=Rad(P).L$ and which is minimal with respect to this property?
Nov 26, 2013 at 16:48 comment added Jim Humphreys Full answers to Q1, Q2 may not be known, but the issue here might be studied in the broader framework of existence of Levi factors in connected algebraic groups in prime characteristic which are neither solvable nor semisimple. See in particular the work of George McNinch indicated in his answer to an older question of mine: mathoverflow.net/questions/22118.
Nov 26, 2013 at 14:30 history asked Nick Gill CC BY-SA 3.0