TITLE: Is the normalizer of a reductive subgroup reductive?
QUESTION [14 upvotes]: Let $G$ be a reductive algebraic group over an algebraically closed field (of characteristic zero if it matters) and $H \subset G$ a subgroup, also reductive.  Is the identity component of the normalizer of $H$ in $G$ always reductive?

REPLY [14 votes]: In positive characteristic, the normalizer of a (connected) reductive
subgroup of a (connected) reductive group is not in general reductive.
I communicated the example below in an emailed answer to a query in
April 2002 (it took me some searching in old emails to find it!)  At
the time I wrote something at the end like "I'm not aware of a good
reference" -- this remains true.
Let $k$ be alg. closed of positive characteristic. I will exhibit $H
\subset G$ reductive groups such that $C_G(H)$ is not reductive;
since $C_G(H)$ is normal in $N_G(H)$, also $N_G(H)$ is not reductive.
Here is the example. Let $n \ge 2$, and consider the adjoint
representation $$\operatorname{Ad}: \operatorname{GL}_n \to
\operatorname{GL}(\mathfrak{gl}_n)$$ where $\mathfrak{gl}_n$ is the Lie algebra of
$\operatorname{GL}_n$.
The image $H \simeq \operatorname{PGL}_n$ of $\operatorname{Ad}$ is a
reductive subgroup of $G=\operatorname{GL}(\mathfrak{gl}_n)$, and the
centralizer $C_G(H)$ identifies with the group of units of the
endomorphism ring $\operatorname{End}_H(\mathfrak{gl}_n)$.
If $(n,p) = 1$, $\mathfrak{gl}_n$ is a semisimple representation of
$\operatorname{GL}_n$ with two distinct irreducible factors, so that
$\operatorname{End}_H(L) = k \times k$. In this case $C_G(H)$ is
a 2 dimensional torus and hence is reductive.
If $(n,p) = p$, $\mathfrak{gl}_n$ is an indecomposable representation
with 3 composition factors. Thus the endomorphism ring of
$\mathfrak{gl}_n$ is a local ring. It turns out that this endomorphism
ring $\operatorname{End}_H(\mathfrak{gl}_n)$ still has dimension 2,
but it is now isomorphic to the ring $k[t]/\langle t^2 \rangle$. To exhibit a non-0
nilpotent $H$-endomorphism $t$ of $\mathfrak{gl}_n$, take the
following map: to a matrix $X$ in $\mathfrak{gl}_n$, assign the matrix
$t(X) = \operatorname{tr}(X).1$ where 1 denotes the $n\times n$
identity matrix, and $\operatorname{tr}()$ denotes the trace.  Since
$n$ is divisible by $p$, applying $t$ twice gives 0.
The unit group of this local ring is isomorphic to the product of a
$1$ dimensional torus and the additive group of the field; such a
group is not reductive. (Explicitly: the additive subgroup is precisely the set of
all automorphisms $1 + at$ of $\mathfrak{gl}_n$ with $a \in k$.)
More generally, let $H$ be any reductive group, let $V$ be a finite
dimensional $H$-module, and write $G = \operatorname{GL}(V)$. 
Suppose that $V$ is indecomposable, has
composition length 3, and that $\operatorname{soc}(V) \simeq V
/\operatorname{rad}(V)$. Then $\operatorname{End}_H(V) \simeq
k[t] / \langle t^2 \rangle$ 
and $C_G(H) =
\operatorname{End}_H(V)^\times$ is not reductive.
Here are a few more examples of such $H$ and $V$ (I'll write
$T(\mu)$ for the indecomposable tilting module with highest weight
$\mu$).

$H = \operatorname{Sp}(W)$, $V = \bigwedge^2 W$ ("exterior square"), when $p>2$ and
$\dim W \equiv 0 \pmod p$.
$H = \operatorname{SO}(W)$, $V = \operatorname{Sym}^2 W$ ("symmetric square"), when
$p>2$ and $\dim W \equiv 0 \pmod p$.
$H = \operatorname{SL}_2$, $V = T(n)$ for $p \le n \le 2p-2$.
$H = \operatorname{SL}_{\ell + 1}$, $V = T(\varpi_i +
  \varpi_\ell)$ for $1 \le i < \ell$, when $\ell + 2 - i \equiv 0 \pmod p$,
[This more-or-less interpolates the original example since when $i=1$ and $\ell + 1 \equiv 0 \pmod p$, $T(\varpi_1 +
  \varpi_\ell) \simeq \mathfrak{gl}_{\ell+1}$.]
$H = \operatorname{SO}_{2\ell}$, $V = T(\varpi_1 +
  \varpi_\ell)$ when  $\ell \equiv 0 \pmod{p}$.