Timeline for Is the normalizer of a reductive subgroup reductive?

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Nov 26, 2012 at 16:19	comment	added	Ben Wieland		Here's a slightly different construction: If $V$ is a representation with nonsplit quotient $W$, then $V\oplus W$ admits an equivariant shear from $W$ to $V$ that cannot be complemented without splitting $V$. So the centralizer of the group in $GL(V\oplus W)$ is not reductive, so the normalizer is not.
Nov 25, 2012 at 17:45	history	edited	George McNinch	CC BY-SA 3.0	added 1010 characters in body
Nov 25, 2012 at 14:39	comment	added	George McNinch		@Jim: I think I'm skeptical that "very good primes" are the issue. If I view $M=\operatorname{GL}(L)$ as a Levi factor of a parabolic subgroup of $G=\operatorname{GL}(L \oplus k)$ and take the subgroup $H$ of $M$ as in the construction of my answer, I believe that $C_G(H)$ is still not reductive.
Nov 25, 2012 at 14:26	comment	added	Jim Humphreys		@George: This kind of example looks convincing. It would confirm that for a given $G$ , the normalizer of some connected reductive subgroup $H$ fails to be reductive at least when the prime $p$ fails to be very good for $G$ . I wonder if it's possible that very good primes (say for both $G$ and $H$ ) always lead to reductive normalizers? It seems mysterious to have both outcomes possible for different choices of $p$ relative to $G, H$ . More examples and counterexamples might help to decide whether or not there is a systematic pattern.
Nov 25, 2012 at 13:40	history	answered	George McNinch	CC BY-SA 3.0