text:
\section{Introduction}
\label{sec:intro}
Let $G$ be a reductive algebraic group over an algebraically closed
field $k$.  A subgroup $H$ of $G$ is said to be $G$-completely
reducible ($G$-cr) if whenever $H$ is contained in a parabolic
subgroup $P$ of $G$, $H$ is contained in some Levi subgroup of $P$.
(If $G$ is non-connected, then we have to modify this definition
slightly; see Subsection \ref{subsec:noncon}.)  In the special case $G =
\GL_n(k)$, $H$ is $G$-completely reducible if and only if $H$ acts
completely reducibly on $k^n$.  Thus $G$-complete reducibility
generalizes the notion of semisimplicity from representation theory,
\cite{serre1.5, serre1, serre2}.

Much effort has been made to study $G$-completely reducible subgroups
of reductive groups.  There are applications to finite groups of Lie
type \cite{LMS}.  Ideas from the theory of $G$-complete reducibility
play an important part in the study by Liebeck and Seitz
\cite{liebeckseitz0} of the reductive subgroups of the exceptional
simple groups.

In characteristic zero a subgroup $H$ of $G$ is $G$-completely
reducible if and only if $H$ is reductive (cf.\ \cite[Lem.\ 2.6]{BMR}).
In this case, it is
trivial to show that the class of $G$-completely reducible subgroups
of $G$ is closed under certain natural constructions: for instance, if
$H,N$ are $G$-completely reducible subgroups of $G$ such that $H$
normalizes $N$, then $H$ and $N$ are reductive, so $HN$ is reductive,
and thus $HN$ is $G$-completely reducible.  The analogous result need
not hold in positive characteristic \cite[Examples 5.1 and 5.3]{BMR2}.
Nonetheless, results concerning $G$-completely reducible subgroups
which hold in
characteristic zero tend to hold ``generically'' in positive characteristic,
that is, if the
characteristic is large enough, or under certain extra natural
restrictions on the subgroups considered.  We give some other examples
below: see Theorem~\ref{thm:verygood}
and Theorem~\ref{thm:adjsc}.

The purpose of this paper is to explore the bounds of these generic
results, proving that they hold without the extra hypotheses or
finding counterexamples when the extra hypotheses are removed.
Characteristic $2$ is a particularly fertile ground for
counterexamples.  To prove positive results, we use geometric
techniques from our earlier works \cite{BMR} and \cite{BMR2}.
Especially important is the property of {\em separability} (see
\cite[Def.\ 3.27]{BMR}): the hypothesis that certain subgroups are
separable is required for several results involving $G$-complete
reducibility, e.g., \cite[Thm.\ 3.35, Thm.\ 3.46]{BMR}.

Let $\gg = \Lie G$ be the Lie algebra of $G$.
\begin{defn}
\label{def:sep}
 A subgroup $H$ of $G$ is said to be \emph{separable in $G$} if $\Lie
C_G(H) = \cc_\gg(H)$, that is, if the scheme-theoretic centralizer of $H$
in $G$ is smooth.
\end{defn}
Here $C_G(H)$ and $\cc_\gg(H)$
denote the (set-theoretic) centralizer of $H$ in $G$ and
in $\gg$ respectively.
Note that we always have
$\Lie C_G(H) \subseteq \cc_\gg(H)$.

The motivation for this terminology is as follows.  Given $n\in
\mathbb{N}$, we let $G$ act on $G^n$ by simultaneous conjugation:
\[
g\cdot (g_1,g_2,\ldots, g_n) = (gg_1g^{-1},gg_2g^{-1},\ldots,
gg_ng^{-1}).
\]
Suppose $H$ is the algebraic subgroup of $G$ generated by elements
$g_1, \ldots, g_n\in G$.  Then we say $H$ is \emph{(topologically)
finitely generated} by $g_1, \ldots, g_n$.  The orbit map $G \to
G\cdot(g_1,\ldots,g_n)$, $g\mapsto g\cdot (g_1,g_2,\ldots, g_n)$ is a
separable morphism of varieties if and only if
$H$ is separable in $G$ (see
\cite[Sec.~3.5]{BMR}).

There is an analogous notion of separability for subalgebras of $\gg$ ---
see Definition \ref{def:sepalg} --- which similarly is related to
the separability of orbit maps $G \to G\cdot(x_1,\ldots,x_n)$,
where $x_i \in \gg$ and $G$ acts diagonally on $\gg^n$ by simultaneous
adjoint action.

Our first main result concerning this notion of separability is as follows.

\begin{thm}
\label{thm:sep2}
Let $G$ be connected reductive and suppose that $\Char k$ is very good for
$G$.  Then any subgroup of $G$ is separable in $G$ and any subalgebra
of $\gg$ is separable in $\gg$.
\end{thm}

The proof (see Section \ref{sec:redpairs}) depends on Theorem \ref{thm:redpairs} below and the
existence of reductive pairs under the given characteristic
restrictions, where for a reductive subgroup $H$ of a reductive group $G$,
we say that $(G,H)$ is a \emph{reductive pair} if
the Lie algebra $\hh = \Lie H$ is an $H$-module direct
summand of $\gg$ (of course, this is automatically satisfied in
characteristic zero).
Special cases of Theorem \ref{thm:sep2}  are known due to work by
Liebeck--Seitz \cite[Thm.\ 3]{liebeckseitz0} and Lawther--Testerman
\cite[Thm.\ 2]{lawthertesterman}.

A prototype of the results we consider is the following fundamental
result of P.\ Slodowy \cite[Thm.~1]{slodowy}, which is obtained by applying a standard
tangent space argument due to R.W.\ Richardson
(cf.\ \cite[Sec.~3]{rich2} and \cite[Lem.~3.1]{rich0}).
Theorem \ref{thm:finorbit} has many applications: see \cite{BMR},
\cite{slodowy}, \cite[Sec.~3]{vinberg}, for example.  Some of the
results that follow from Theorem \ref{thm:finorbit} are still valid
even when the hypotheses of the theorem are not met, but one often has
to work much harder to obtain them (cf.\ \cite[Sec.~1]{martin1}).

\begin{thm}
\label{thm:finorbit}
Let $H$ be a reductive subgroup of a reductive group $G$.
Let $n\in \mathbb{N}$, let $(h_1,\ldots,h_n)\in H^n$ and let $K$ be
the algebraic subgroup of $H$ generated by $h_1,\ldots,h_n$.  Suppose
that $(G,H)$ is a reductive pair and that $K$ is separable in $G$.
Then
\begin{itemize}
\item[(a)] for all $(g_1,\ldots,g_n)\in G\cdot (h_1,\ldots,h_n)\cap H^n$, the $H$-orbit map $H\ra H\cdot (g_1,\ldots,g_n)$ is
separable.  In particular, $K$ is a separable subgroup of $H$;
\item[(b)] the intersection $G\cdot (h_1,\ldots,h_n)\cap H^n$ is a
finite union of $H$-conjugacy classes;
\item[(c)] each $H$-conjugacy class in $G\cdot (h_1,\ldots,h_n)\cap H^n$ is
closed in $G\cdot(h_1,\ldots,h_n)\cap H^n$.
\end{itemize}
\end{thm}

The following consequence \cite[Thm.\ 3.35]{BMR} of Theorem
\ref{thm:finorbit} gives an application to $G$-complete reducibility.
The second assertion follows from Theorem \ref{thm:finorbit}(c) and
the geometric characterization of $G$-complete reducibility given in
\cite{BMR} (see Theorem \ref{thm:gcrcrit} below); the first does not
appear explicitly in \cite[Thm.\ 3.35]{BMR}, but it follows
immediately from the proof together with Theorem
\ref{thm:finorbit}(a).

\begin{thm}
\label{thm:redpairs}
Suppose that $(G,H)$ is a reductive pair. Let $K$ be a
subgroup of $H$ such that $K$ is a separable subgroup of $G$.  Then
$K$ is separable in $H$. Moreover, if $K$ is $G$-completely
reducible, then it is also $H$-completely reducible.
\end{thm}

Theorem \ref{thm:finorbit}(b) can be used to give a simple proof that
a connected reductive group $G$ has only finitely many conjugacy
classes of unipotent elements in generic characteristic: one takes an
embedding of $G$ in some $\GL(V)$ such that $(\GL(V),G)$ is a
reductive pair then applies Theorem \ref{thm:finorbit}(b) (taking $n=1$) to deduce
the result for $G$ from the result for $\GL(V)$.  In small positive
characteristic there need not exist any such reductive pair $(\GL(V),G)$ (see
Subsection \ref{subsec:sep}).  Nonetheless Lusztig proved without any
restriction on the characteristic, using some very deep mathematics,
that $G$ has only finitely many unipotent conjugacy classes
\cite{lusztig}.  Guralnick recently extended this result to
non-connected reductive groups \cite[Thm.\ 3.3]{guralnick}; he then
used it to prove that Theorem
\ref{thm:finorbit}(b) still holds for an arbitrary reductive subgroup $H$ of $G$ if $n=1$ \cite[Thm.\
1.2]{guralnick}, even though Richardson's proof no longer goes
through.

In view of this, it is natural to ask whether Theorem
\ref{thm:finorbit}(b) holds in more generality.  In Example
\ref{exmp:orbitcountereg}, we prove the following result which shows
that Guralnick's conjugacy result \cite[Thm.\ 1.2]{guralnick} does not
extend to conjugacy classes of $n$-tuples for $n>1$.

\begin{thm}
\label{thm:countereg}
 There exists a reductive group $G$, a reductive subgroup $M$ of $G$
 and a pair $(m_1,m_2)\in M^2$ such that $(G,M)$ is a reductive
 pair but $G\cdot (m_1,m_2)\cap M^2$ is {\bf not} a finite union
 of $M$-conjugacy classes.
\end{thm}

Thus Theorem \ref{thm:finorbit}(b) is false in general without the
separability hypothesis.  We give an example (Proposition
\ref{prop:GcrnotMcr}) showing that the second assertion of Theorem
\ref{thm:redpairs} also fails without the separability hypothesis.
This implies that Theorem
\ref{thm:finorbit}(c) fails as well without the separability
hypothesis (see Remark \ref{rem:thm1.3(c)fails}).  \smallskip

The paper is split into several sections, as we now outline.  In
Section \ref{sec:prelims} we introduce the notation and preliminary
results needed for the rest of the exposition.  In particular, we
recall the formalism of R-parabolic subgroups from \cite{BMR}, which
allows us to consider reductive groups which are not connected; this
is very important for many of our results.

In Section \ref{sec:redpairs} we discuss the question of separability
in connection with reductive pairs.  In Proposition
\ref{prop:C_G(S)redpair} we give a construction for certain reductive
pairs, where the reductive subgroup of $G$ is of the form $C_G(S)$; here $S$ is a reductive group acting suitably on $G$.

In Section~\ref{sec:sep1} we investigate the connection between
the $G$-complete reducibility of a subgroup $H$ and the semisimplicity of the adjoint
representation of $H$ on $\gg$.  The
following result is the basis of our discussion \cite[Thm.\
3.46]{BMR}.

\begin{thm}
\label{thm:adjcr}
Let $H$ be a separable subgroup of $G$.  If $\gg$ is semisimple as an
$H$-module, then $H$ is $G$-completely reducible.
\end{thm}

One way of removing the separability hypothesis from
Theorem~\ref{thm:adjcr} is to combine Theorems \ref{thm:sep2} and
\ref{thm:adjcr}. This immediately gives the following result.

\begin{thm}
\label{thm:verygood}
Let $G$ be connected reductive and suppose that $\Char k$ is very good
for $G$. Let $H$ be a subgroup of $G$ such that $\gg$ is a semisimple
$H$-module.  Then $H$ is $G$-completely reducible.
\end{thm}

The assumption in Theorem \ref{thm:verygood}
that $\Char k$  is very good for $G$ is rather
restrictive. In Section~\ref{sec:sep1} we discuss to what extent we
can remove the separability hypothesis from Theorem~\ref{thm:adjcr}
with a weaker assumption on $\Char k$;
see Theorem~\ref{thm:allregss}, Theorem~\ref{thm:adjsc} --- which is our second main result ---
and Corollary~\ref{cor:adjc}.

In Section \ref{sec:orbitthm} we extend two general results of Richardson
concerning orbits of reductive groups on affine varieties
\cite[Thm.\ A, Thm.\ C]{rich0}, see Theorem \ref{thm:richC}.
We apply these extensions in turn to
questions of $G$-complete reducibility; in particular,
we discuss some results which examine the
relationship between $G$-complete
reducibility and $H$-complete reducibility of a subgroup $K$ of $H$,
where $H$ is a subgroup of $G$ of the form $H = C_G(S)$,
with $S$ a reductive group acting suitably on $G$:
see Proposition \ref{prop:SnormalizesK}.
The case when the group $S$ considered above is a subgroup of $G$
is discussed in Section \ref{sec:cent+norm}.

In Section \ref{sec:ex}, we provide an important collection of closely
related constructions.  They give counterexamples to several of our
results, including Theorems \ref{thm:finorbit} and
\ref{thm:redpairs}, under weakened hypotheses.
In particular, here we also prove Theorem \ref{thm:countereg}.

\section{Preliminaries}
\label{sec:prelims}

\subsection{Notation}
Throughout, we work over an algebraically closed field $k$;
we let $k^*$ denote the multiplicative
group of $k$.
All algebraic groups are assumed to be linear.  By a subgroup of an algebraic group we mean a closed subgroup
and by a homomorphism of algebraic groups we mean a homomorphism
of abstract groups that is also a morphism of
algebraic varieties.  Let $H$ be a linear algebraic group.  We denote
by $\overline{\langle S\rangle}$ the algebraic subgroup of $H$
generated by a subset $S$.
We let $DH$ denote the derived group $[H,H]$, $Z(H)$ the centre of $H$,
and $H^0$ the connected component of $H$ that contains $1$.  If $S$ is
a subset of $H$, then $C_H(S)$ is the centralizer of $S$ in $H$ and
$N_H(S)$ is the normalizer of $S$ in $H$.  In general we use
an upper-case roman letter $G,H,K$, etc., to denote an algebraic group
and the corresponding lower-case gothic letter $\gg,\hh,\kk$, etc., to
denote its Lie algebra. If $\hh$ is a Lie algebra and $S$ is a
subset of $\hh$, then $\cc_\hh(S)$ is the centralizer of $S$ in
$\hh$.  We denote the centre of $\hh$ by $\zz(\hh)$.

Let $\Ad: H \to \GL(\hh)$ denote the adjoint representation; then we
let $H_{\rm ad}$ denote the image of $H$ under this map and $\hh_{\rm
ad}$ denote $\Lie H_{\rm ad}$.  Note that $(H_{\rm ad})^0$ is the adjoint form of $D(H^0)$ \cite[V.24.1]{borel}.

For the set of cocharacters (one-parameter subgroups) of $H$ we write
$Y(H)$; the elements of $Y(H)$ are the homomorphisms from $k^*$ to
$H$.

The \emph{unipotent radical} of $H$ is denoted $R_u(H)$; it is the
maximal connected normal unipotent subgroup of $H$.  The algebraic
group $H$ is called \emph{reductive} if $R_u(H) = \{1\}$; note that we
do not insist that a reductive group is connected.  In particular, $H$
is reductive if it is simple as an algebraic group ($H$ is said to be
\emph{simple} if $H$ is connected and all proper normal subgroups of
$H$ are finite).  If $N$ is a normal subgroup of $H$, then $H$ is reductive if and only if $N$ and $H/N$ are. The algebraic group $H$ is called \emph{linearly
reductive} if all rational representations of $H$ are semisimple.

If $H$ acts on the affine variety $X$, then we denote by $X^H$ the
fixed point subvariety of $X$: that is, $X^H = \{x \in X \mid h\cdot x
= x \ \forall h \in H\}$.  If $S$ is a subset of $X$, then we denote the pointwise
stabilizer of $S$ in $H$ by $C_H(S)$; we write $C_H(x)$ instead of
$C_H(\{x\})$ for $x$ in $X$.
If $X = K$ is an algebraic group and $H$ acts on $K$ by
automorphisms, then we write $C_K(H)$ instead of $K^H$.
Then we also have an
induced linear action of $H$ on $\kk = \Lie K$; we write $\cc_\kk(H)$ instead
of $\kk^H$.

Throughout the paper $G$ denotes a reductive algebraic group, possibly
non-connected, with Lie algebra $\gg$.
A subgroup of $G$ normalized
by some maximal torus of $G$ is called a \emph{regular} subgroup
of $G$ (connected reductive regular subgroups of connected reductive
groups are often also referred to as \emph{subsystem subgroups}, e.g.,
see \cite{liebeckseitz}).

Fix a maximal torus $T$ of $G$.
We write $X(T)$ for the character group of $T$.
Let $\Psi = \Psi(G,T) \subseteq X(T)$
denote the set of roots of $G$ with respect to
$T$. We write $\frakt = \Lie T$ for the Lie algebra of $T$.
If $\alpha\in \Psi$, then $U_\alpha$ denotes the
corresponding root subgroup of $G$ and $\uu_\alpha$ denotes the root space
${\rm Lie}\,U_\alpha$ of $\gg$.
Thus the root space decomposition of $\gg$ is
given by
\[
\gg = \frakt \oplus \bigoplus_{\alpha \in \Psi} \uu_\alpha.
\]
We denote by $G_\gamma$ the simple rank $1$
subgroup $\langle U_\gamma\cup U_{-\gamma}\rangle$ of $G$ and by $\gg_\gamma$
the Lie algebra of $G_\gamma$. Fix a Borel subgroup $B$ of $G$
containing $T$ and let $\Sigma = \Sigma(G, T)$ be the set of simple
roots of $\Psi$ defined by $B$. Then $\Psi^+ = \Psi(B,T)$ is the set of
positive roots of $G$.
For $\beta \in \Psi^+$ write $\beta =
\sum_{\alpha \in \Sigma} c_{\alpha\beta} \alpha$ with $c_{\alpha\beta}
\in \mathbb N_0$.  A prime $p$ is said to be \emph{good} for $G$ if it
does not divide any non-zero $c_{\alpha\beta}$, and
\emph{bad} otherwise.  A prime $p$ is good for $G$ if and only if it
is good for every simple factor of $G^0$ \cite{SS}; the bad primes for
the simple groups are $2$ for all groups except type $A_n$, $3$ for
the exceptional groups and $5$ for type $E_8$.  A prime $p$ is said to
be \emph{very good} for $G$, if $p$ is good for $G$ and $p$ does not
divide $n+1$ for any simple component of $G$ of type $A_n$. If $G$ is
simple and $\Char k$ is very good for $G$, then the Lie algebra $\gg$ is
simple \cite{St3}.

\begin{rem}
Separability of subgroups of $G$ and of subalgebras of $\gg$
(see Definition \ref{def:sepalg}) is automatic
in characteristic zero (cf.\ \cite[Thm.~13.4]{Hum}). Likewise,
the notion of $G$-complete reducibility is not
interesting in characteristic zero,
as a subgroup of $G$ is $G$-completely reducible if and only if
it is reductive (cf.\ \cite[Lem.\ 2.6]{BMR}); most of our results
and proofs become trivial in characteristic zero.
In the remainder of the paper
$p$ denotes the characteristic $\Char k$ of $k$ in case $\Char k>0$.
\end{rem}

Let $\gamma\in \Psi$.  We denote by $\gamma^\vee\in Y(G)$ the
corresponding coroot.  Then $\gamma^\vee$ is a homomorphism from $k^*$
to $G_\gamma$.  If $\alpha,\beta\in \Sigma$, then we have
$s_\alpha\cdot \beta= \beta- \langle \beta,\alpha^\vee\rangle \alpha$
\cite[Lem.~7.1.8]{spr2}, where $s_\alpha$ is the reflection
corresponding to $\alpha$ in the Weyl group of $G$.

In Section~\ref{sec:ex} we need the following well-known result
(which is implicit for instance in \cite[6.5]{Jantzen1}).
For convenience we include a proof.

\begin{lem}
\label{lem:sc}
Assume that $G$ is connected.  If the derived group $DG$ of $G$ is
simply connected, then the same holds for any Levi subgroup of any
parabolic subgroup of $G$.
\end{lem}

\begin{proof}
 We use the following characterization of simply connectedness of $DG$:
Let $T$ be a maximal torus of $G$ and let $\alpha_1,\ldots,\alpha_n\in X(T)$
be the choice of simple roots corresponding to some Borel subgroup $B$ containing $T$. Then $DG$ is simply connected if and only if
there exist characters $\chi_1,\ldots,\chi_n\in X(T)$ such that
$\inprod{\chi_i,\alpha_j\check}=\delta_{ij}$ for all $i,j\in\{1,\ldots,n\}$.
For $G$ semisimple this is clear. The general case follows from the fact
that the integer $\inprod{\chi,\alpha\check}$ only depends on the
restriction of $\chi$ to $T\cap DG$ and the fact that any character
of $T\cap DG$ can be lifted to a character of $T$
(\cite[Prop.~III.8.2(c)]{borel}).

 Let $P$ be a parabolic subgroup of $G$ and let $L$ be a Levi subgroup of $P$.  We may assume that $P$ contains $B$ and $L$ contains $T$.  The result now follows immediately from the following
well-known description of $L$
\cite[Prop.~IV.14.18]{borel}: the simple roots of $L$ can
be chosen from the set $\{\alpha_1,\ldots,\alpha_n\}$.
\end{proof}

The next result allows us in positive characteristic
to replace an algebraic group $S$ acting on $G$ by automorphisms
with a finite subgroup of $S$. It is a slight strengthening of \cite[Thm.~7]{BGM},
since it asserts the existence of a {\it countable} locally finite dense subgroup.
We use this lemma in the proof of
Proposition~\ref{prop:C_G(S)redpair}
and also in several places in Section~\ref{sec:orbitthm}.

\begin{lem}
\label{lem:finitedense}
Assume that $\Char k  > 0$.
Let $H$ be a linear algebraic group over $k$.
Then there exists an ascending sequence
$H_1\subseteq H_2\subseteq\cdots$ of finite
subgroups of $H$ whose union is dense in $H$.
\end{lem}

\begin{proof}
We proceed by induction on $\dim H$.
If $H$ is reductive, then the result follows
from \cite[Sec.~3]{martin1}.
Otherwise $Z:=Z(R_u(H))^0$ is a connected
unipotent normal subgroup of $H$ of dimension $\ge 1$.
By \cite[III.10.6(2)]{borel}, $Z$ contains a subgroup
isomorphic to the additive group $\mathbb{G}_a$.
Let $C$ be the subgroup of $Z$
generated by the subgroups of $Z$ that are
isomorphic to $\mathbb{G}_a$. Then $C$ is the additive group
of a non-zero finite-dimensional vector space over $k$,
by \cite[Thm.~5.4]{Hoch}.
Furthermore, $C$ is normal in $G$.
Clearly, $C$ has an ascending sequence $C_1\subseteq C_2\cdots$
of finite subgroups whose union is dense in $C$. By the
induction hypothesis, $M:=H/C$ also has an ascending sequence
$M_1\subseteq M_2\subseteq\cdots$ of finite subgroups whose
union is dense in $M$. Let $\pi:H\to M$ be the canonical
projection. For each $i\ge1$ let $H_i$ be a finitely generated
subgroup of $H$ such that $\pi(H_i)=M_i$. Without loss of
generality we may assume that the $H_i$ form an ascending
sequence of subgroups and that $H_i$ contains $C_i$. Since
$H_i$ is finitely generated and $H_i\cap C$ is of finite index
in $H_i$, we have that $H_i\cap C$ is finitely generated,
by \cite[Thm.~11.54]{Rot}.
Since $C$ is a vector space, this means that $H_i\cap C$ is
finite. But then $H_i$ is finite. Now let $H'$ be the closure of
the union of the $H_i$. Then $H'$ is a closed subgroup of $H$
containing $C$. Its image $\pi(H')$ is a closed subgroup of $M$
containing the $M_i$ and is therefore equal to $M$. Consequently, $H'=H$.
\end{proof}

\subsection{$G$-Complete Reducibility}
\label{subsec:noncon}
In \cite[Sec.~6]{BMR}, Serre's original notion of $G$-complete
reducibility is extended to include the case when $G$ is reductive but
not necessarily connected (so that $G^0$ is a connected reductive
group).  The crucial ingredient of this extension is the introduction
of so-called \emph{Richardson-parabolic subgroups} (\emph{R-parabolic
subgroups}) of $G$.  We briefly recall the main definitions here; for
more details on this formalism, see \cite[Sec.~6]{BMR}.

For a cocharacter $\lambda
\in Y(G)$, the \emph{R-parabolic subgroup} corresponding to $\lambda$
is defined by $P_\lambda := \{ g\in G \mid \underset{a\to 0}{\lim}\,
\lambda(a)g\lambda(a)\inverse \textrm{ exists}\}$.  Then $P_\lambda$
admits a Levi decomposition $P_\lambda = R_u(P_\lambda) \rtimes
L_\lambda$, where $L_\lambda = \{ g\in G \mid \underset{a\to
0}{\lim}\, \lambda(a)g\lambda(a)\inverse = g \} = C_G(\lambda(k^*))$.
We call $L_\lambda$
an \emph{R-Levi subgroup} of $P_\lambda$.  For an R-parabolic subgroup
$P$ of $G$, the different R-Levi subgroups of $P$ correspond in this
way to different choices of $\lambda \in Y(G)$ such that $P =
P_\lambda$; moreover, the R-Levi subgroups of $P$ are all conjugate
under the action of $R_u(P)$.  An R-parabolic subgroup $P$ is a
parabolic subgroup in the sense that $G/P$ is a complete variety; the
converse is true when $G$ is connected, but not in general
(\cite[Rem.\ 5.3]{martin1}).  The map $c_\lambda :P_\lambda \to
L_\lambda$ given by $c_\lambda(g) = \underset{a\to 0}{\lim}\,
\lambda(a)g\lambda(a)\inverse$ is a surjective homomorphism of
algebraic groups with kernel $R_u(P_\lambda)$; it coincides with the
usual projection $P_\lambda\ra L_\lambda$.

\begin{rem}
For a subgroup $H$ of $G$, there is a natural inclusion $Y(H)
\subseteq Y(G)$.  If $\lambda \in Y(H)$, and $H$ is reductive, we can
therefore associate to $\lambda$ an R-parabolic subgroup of $H$ as
well as an R-parabolic subgroup of $G$.  To avoid confusion, we
reserve the notation $P_\lambda$ for R-parabolic subgroups of $G$, and
distinguish the R-parabolic subgroups of $H$ by writing $P_\lambda(H)$
for $\lambda \in Y(H)$.  The notation $L_\lambda(H)$ has the obvious
meaning.  Note that $P_\lambda(H) = P_\lambda \cap H$,
$L_\lambda(H) = L_\lambda \cap H$ and
$R_u(P_\lambda(H))= R_u(P_\lambda)\cap H$ for $\lambda \in Y(H)$.
\end{rem}

\begin{rem}
\label{rem:P0G0}
If $\lambda\in Y(G)$ and $P_\lambda^0=G^0$,
then $P_\lambda$ is an R-Levi subgroup of itself:
for $R_u(P_\lambda)$, being connected, is contained in $G^0$, so
$R_u(P_\lambda)= R_u(P_\lambda)\cap G^0= R_u(P_\lambda(G^0))= R_u(P_\lambda\cap G^0)= R_u(G^0)= \{1\}$.
\end{rem}

\begin{defn}
\label{defn:gcr}
Suppose $H$ is a subgroup of $G$.  We say $H$ is \emph{$G$-completely
reducible} ($G$-cr for short) if whenever $H$ is contained in an
R-parabolic subgroup $P$ of $G$, then there exists an R-Levi subgroup
$L$ of $P$ with $H \subseteq L$.
\end{defn}

\begin{rem}
If $H$ is a $G$-completely reducible subgroup of $G$,
then $H$ is reductive (cf.\ \cite[Sec.~2.5 and Thm.~3.1]{BMR}).
\end{rem}

\begin{comment}
We give a useful characterisation of $G$-complete reducibility in
terms of the map $c_\lambda$ \cite[Lem.~2.17 and Thm.~3.1]{BMR}.

\begin{lem}
\label{lem:gcrclambda}
 Let $H$ be a subgroup of $G$.  Then $H$ is $G$-completely reducible
 if and only if for every $\lambda\in Y(G)$ such that $H\subseteq
 P_\lambda$, $c_\lambda(H)$ is $R_u(P_\lambda)$-conjugate to $H$.
\end{lem}
\end{comment}

Since all parabolic subgroups (respectively\ all Levi subgroups of
parabolic subgroups) of a connected reductive group are R-parabolic
subgroups (respectively\ R-Levi subgroups of R-parabolic subgroups),
Definition \ref{defn:gcr} coincides with Serre's original definition
for connected groups \cite{serre2}.
Sometimes we come across subgroups of $G$ which
are not contained in any proper R-parabolic subgroup of $G$; these
subgroups are trivially $G$-completely reducible.  Following Serre
again, we call these subgroups \emph{$G$-irreducible} ($G$-ir).

\begin{rem}
\begin{comment}
 If $H$ is a $G$-completely reducible subgroup of $G$ and $P$ is
minimal among R-parabolic subgroups of $G$ containing $H$, then $H$ is
$L$-irreducible, where $L$ is an R-Levi subgroup of $P$ containing
$H$.  This observation often allows us to restrict attention to
$G$-irreducible and $G$-indecomposable subgroups.
\end{comment}
Since R-Levi
subgroups of R-parabolic subgroups play an important r\^ole in many of
our proofs, for brevity we sometimes abuse language and refer to an
\emph{R-Levi subgroup of $G$}; by this we mean an R-Levi subgroup of
some R-parabolic subgroup of $G$.  Similarly, when $G$ is connected,
we may refer to \emph{a Levi subgroup of $G$}; this means a Levi
subgroup of some parabolic subgroup of $G$.
\end{rem}

A key result is the following \cite[Cor.\ 3.7]{BMR}, which gives a
geometric criterion for $G$-complete reducibility.

\begin{thm}
\label{thm:gcrcrit}
 Let $g_1,\ldots, g_n\in G$ and let $H= \overline{\langle \{g_1,\ldots,
 g_n\}\rangle}$.  Then $H$ is $G$-completely reducible if and only if
 the conjugacy class $G\cdot (g_1,\ldots, g_n)$ is closed in $G^n$.
\end{thm}

We frequently require results from \cite[Sec.~6.3]{BMR} for non-connected $G$,
though
we usually simply cite the relevant result in \cite{BMR} for connected $G$.

\subsection{Separability}
\label{subsec:sep}
In Section~\ref{sec:redpairs} we require the following analogue of
Definition~\ref{def:sep} for subalgebras of $\gg$.
Recall that
$C_G(\hh) = \{g \in G \mid \Ad g(x) = x \text{ for all } x \in \hh \}$.

\begin{defn}
\label{def:sepalg}
A subalgebra $\hh$ of $\gg$ is said to be \emph{separable in $\gg$} if
$\Lie C_G(\hh) = \cc_\gg(\hh)$.
\end{defn}

The above definition has the same motivation as in the group
case. Given $n\in \mathbb{N}$, we let $G$ act on $\gg^n$ by diagonal
adjoint action.  Suppose $\hh$ is the subalgebra of $\gg$ generated by
elements $x_1, \ldots, x_n\in \gg$. Then the orbit map $G \to
G\cdot(x_1,\ldots,x_n)$, $g\mapsto g\cdot (x_1, \ldots,x_n)$
is a separable morphism of varieties if and only if $\hh$ is separable in
$\gg$ (see \cite[II.6.7]{borel}).

As in the group case (\cite[Thm.~1]{slodowy}), it is straightforward
to generalize Richardson's tangent space arguments
\cite[Sec.~3]{rich2} and \cite[Lem.~3.1]{rich0}
to the action of $G$ on $\gg^n$. We
then obtain the following analogue of Theorem~\ref{thm:finorbit}.

\begin{thm}
\label{thm:finorbitalg}
Let $H$ be a reductive subgroup of $G$.
Let $n\in \mathbb{N}$, let $(x_1,\ldots,x_n)\in \hh^n$ and let $\kk$
be the subalgebra of $\hh$ generated by $x_1,\ldots,x_n$.  Suppose
that $(G,H)$ is a reductive pair and that $\kk$ is separable in $\gg$.
Then
\begin{itemize}
\item[(a)] all $H$-orbit maps in $G\cdot (x_1,\ldots,x_n)\cap \hh^n$
are separable. In particular, $\kk$ is separable in $\hh$;
\item[(b)] the intersection $G\cdot (x_1,\ldots,x_n)\cap \hh^n$ is a
finite union of $H$-conjugacy classes;
\item[(c)] each $H$-conjugacy class in $G\cdot (x_1,\ldots,x_n)\cap \hh^n$ is
closed in $G\cdot(x_1,\ldots,x_n)\cap \hh^n$.
\end{itemize}
\end{thm}

We immediately obtain the following analogue of the first
assertion of Theorem~\ref{thm:redpairs}.

\begin{cor}
\label{cor:redpairs}
Suppose that $(G,H)$ is a reductive pair.  Let $\kk$ be a subalgebra
of $\hh$ such that $\kk$ is separable in $\gg$.  Then $\kk$ is
separable in $\hh$.
\end{cor}

Because every subgroup of $\GL(V)$ is separable in $\GL(V)$ and every
subalgebra of $\gl(V)$ is separable in $\gl(V)$ (e.g., see \cite[Ex.\
3.28]{BMR}), Theorem~\ref{thm:redpairs} and
Corollary~\ref{cor:redpairs} imply the following.

\begin{cor}
\label{cor:glv:g}
If $(\GL(V),G)$ is a reductive pair, then every subgroup of $G$ is
separable in $G$ and every subalgebra of $\gg$ is separable in $\gg$.
\end{cor}

We deduce from this that not every $G$ can be embedded in some
$\GL(V)$ in such a way that $(\GL(V),G)$ is a reductive pair: this
applies, for instance, to $G=\SL_2(k)$ when $p=2$, because $H=G$ is
not a separable subgroup of $G$.  However, if $G$ is of a given Dynkin
type, then generically --- that is, for almost all values of $p$ ---
the conclusion of Corollary \ref{cor:glv:g} holds; for example, if $G$
is an exceptional simple group of adjoint type and $p$ is good for
$G$, then $(\GL(\gg),G)$ is a reductive pair (cf.\ Example
\ref{exmp:Serre}).

\medskip

The final result of this section shows that a non-separable $G$-cr
subgroup $K$ of $G$ is, up to isogeny, a separable subgroup of a
regular subgroup of $G$.
Given a reductive group $M$, we let $\pi_M : M\ra M_{\rm ad}$ denote the natural morphism.

\begin{prop}
\label{prop:sepovergp}
Let $K$ be a $G$-completely reducible subgroup of $G$.  Then there
exists a reductive subgroup $M$ of $G$ containing a maximal torus of $G$
such that $K \subseteq M$, $K$ is $M$-irreducible and $\pi_M(K)$ is
separable in $M_{\rm ad}$.
\end{prop}

\begin{proof}
Since
$K$ is $G$-cr, we may assume by \cite[Cor.~3.5]{BMR}
that $K$ is $G$-ir after replacing $G$ by an R-Levi subgroup of $G$
that is minimal with respect to containing $K$.  By
\cite[Cor.~2.7(i)]{BMR}, $K$ is $M$-ir in any reductive subgroup $M$
of $G$ containing $K$.  If $\pi_G(K)$ is separable in $G_{\rm ad}$,
then we can take $M=G$, so suppose not.
By \cite[Prop.\ 3.39]{BMR}, there exists a
reductive subgroup $M'$ of $G_{\rm ad}$ containing a maximal torus of $G_{\rm
ad}$ such that $\pi_G(K)\subseteq M'$ and $M'$ is not separable in
$G_{\rm ad}$.  As $(G_{\rm ad})^0$ is of adjoint type, its
Lie algebra has trivial centre,
so any overgroup of $(G_{\rm ad})^0$ is separable in $G_{\rm ad}$.
This forces $M'$ to be of strictly smaller dimension than $G_{\rm
ad}$.  Let $M= \pi_G^{-1}(M')$, a subgroup of $G$ which is of strictly
smaller dimension than $G$ and contains a maximal torus of $G$.  Since
$M$ is an overgroup of the $G$-ir group $K$, $M$ is reductive.  The
result now follows by induction on $\dim G$.
\end{proof}

Proposition \ref{prop:sepovergp} is false if we do not assume that $K$
is $G$-completely reducible: see Example \ref{exmp:nonGcr} below.

\begin{comment}
\subsection{Kempf--Rousseau Theory}
\label{subsec:kempf}
Let $G$ act on an affine variety $X$ and suppose $x \in X$ is a point
such that $G\cdot x$ is not closed in $X$.  Let $\Lambda(x) = \{
\lambda \in Y(G) \mid \underset{a\to 0}{\lim}\, \lambda(a)\cdot x
\textrm{ exists}\}$.  A theorem of Kempf (\cite[Thm.\ 3.4]{kempf}, see
also \cite{rousseau}) allows us to attach to $x$ a so-called
\emph{optimal class} $\varnothing \ne \Omega(x) \subseteq \Lambda(x)$,
with the
following properties:
\begin{itemize}
\item[(i)] Let $\mathcal O$ denote the unique closed $G$-orbit in the
closure of $G\cdot x$.  Then $\underset{a\to 0}{\lim}\,
\lambda(a)\cdot x \in \mathcal O$ for all $\lambda \in \Omega(x)$.  In
particular, $\underset{a\to 0}{\lim}\, \lambda(a)\cdot x\not\in G\cdot
x$.
\item[(ii)] For all $\lambda, \mu \in \Omega(x)$, we have $P_\lambda =
P_\mu$.  Let $P$ denote the unique R-parabolic subgroup of $G$ so
defined.
\item[(iii)] If $g \in G$, then $\Omega(g\cdot x) = g\cdot \Omega(x)$
and $gP_\lambda g\inverse = P_{g\cdot \lambda}$, for $\lambda \in
\Omega(x)$.
\item[(iv)] $R_u(P)$ acts simply transitively on $\Omega(x)$: that is,
for each R-Levi subgroup $L$ of $P$, there exists one, and only one,
$\lambda \in \Omega(x)$ with $L = L_\lambda$. Hence $C_G(x) \subseteq P$.
\end{itemize}
Note that Kempf only considered the case of connected $G$, but the
extension to non-connected $G$ is straightforward (see \cite[Sec.~1]{martin2}).
The final statement in (iv) is proved as follows. Fix $\lambda\in\Omega(x)$.
Let $g\in C_G(x)$. Then $g\cdot \Omega(x)=\Omega(g\cdot x)=\Omega(x)$ by (iii).
So $g\cdot\lambda\in\Omega(x)$. By (iv), $g\cdot\lambda=u\cdot\lambda$ for some
$u\in R_u(P)$. So $u^{-1}g\in L_\lambda\subseteq P$ and, therefore, $g\in P$.

\end{comment}

\section{Reductive Pairs and Separability}
\label{sec:redpairs}
The notion of separability is central to many of the results in this
paper.  Theorems \ref{thm:finorbit} and \ref{thm:redpairs} both
illustrate the importance of reductive pairs in this context.  In this
section we elaborate on this theme.  For examples and constructions of
reductive pairs, we refer to \cite[Sec.~I.3]{slodowy} and \cite[Sec.~3.5]{BMR}.

Recall that an isogeny
is an epimorphism with finite kernel and that it
is called \emph{separable}
if its differential is an isomorphism.

\begin{lem}
\label{lem:sep}
Let $\varphi:G\to G'$ be a separable isogeny of reductive
groups, let $H$ be a subgroup of $G$ and let $\kk$ be a subalgebra of
$\gg$.  Then $H$ is separable in $G$ if and only if $\varphi(H)$ is
separable in $G'$ and $\kk$ is separable in $\gg$ if and only if
$d\varphi(\kk)$ is separable in $\gg'$.
\end{lem}
\begin{proof}
Since the differential $d\varphi$ of $\varphi$ is an isomorphism, it is
clear that
$d\varphi(\cc_\gg(\kk))=\cc_{\gg'}(d\varphi(\kk))$. Furthermore, we
have
 \begin{equation}
 \label{eqn:ad}
\Ad(\varphi(g))\circ d\varphi
=d\varphi\circ\Ad(g)\text{\quad for all } g\in G,
 \end{equation}
from which it follows that
$d\varphi(\cc_\gg(H))=\cc_{\gg'}(\varphi(H))$.

So to prove the first statement it suffices to show that
$\varphi(C_G(H))^0=C_{G'}(\varphi(H))^0$.  For this, in turn, it
suffices to show that $\varphi^{-1}(C_{G'}(\varphi(H)))^0\subseteq
C_G(H)$.  If $y\in H$, then the image of
$\varphi^{-1}(C_{G'}(\varphi(H)))^0$ under the morphism $x\mapsto
xyx^{-1}y^{-1}$ is an irreducible subset of the finite set $\ker \varphi$ which
contains $1$, so it must equal $\{1\}$.

To prove the second statement, it suffices to show that
$\varphi(C_G(\kk))=C_{G'}(d\varphi(\kk))$. This follows easily from
Eqn.\ \eqref{eqn:ad}.
\end{proof}

\begin{lem}
\label{lem:sepisog}
Let $G$ be connected and suppose that $p$ is very good for
$G$.  Then there exists a separable isogeny $S\times H\to G$, where
$S$ is a torus and $H$ is a product of simply connected simple groups.
\end{lem}

\begin{proof}
 Let $G_1,\ldots, G_r$ be the simple factors of $DG$ and let
 $\widetilde{G}_i$ be the simply connected cover of $G_i$ for each
 $i$.  Then $\Lie \widetilde{G}_i$ is simple for each $i$, by our
 hypothesis on $p$.  Set $S=Z(G)^0$ and $H= \widetilde{G}_1\times
 \cdots \times \widetilde{G}_r$.  It is easily checked that the
 multiplication map $S\times H\ra G$ is a separable isogeny.
\begin{comment}
By \cite{St2} there exists a simply connected semi-simple connected
group $H$ and a central isogeny $H\to DG$. Let $(X,\Phi)$ be the root
datum of $DG$ and let $X_r$ resp.\ $X_w$ be the root lattice resp.\
the weight lattice of $\Phi$. Then the kernel of the differential of
this isogeny is isomorphic to $\Hom(X_w/X,k)$. Now $X_w/X$ is a
quotient of $X_w/X_r$ and an easy inspection of \cite[Planches I -
IX]{bou} (one may assume for this that $\Phi$ is irreducible) shows
that a very good prime does not divide the order of $X_w/X_r$.  So the
differential of the isogeny has trivial kernel and the isogeny is
separable.

Now $H$ is the direct product of almost simple, simply connected
groups (see \cite{St2}) and the Lie algebras of these groups are
simple (see \cite{St3}, from our assumptions on the characteristic of
$k$ it follows that a basis of the root system of such a group gives
rise to a basis of the dual of the Lie algebra of a maximal torus), so
$\Lie DG$ is the direct sum of simple Lie algebras.  In particular,
$\Lie DG$ is perfect and centreless.  Since we always have $[\gg,\gg]
\subseteq \Lie DG$ (\cite[3.17]{borel}), it follows that $\Lie DG =
[\gg,\gg]$.  Furthermore we have $\zz(\gg) \cap [\gg,\gg] = \{0\}$,
since $[\gg,\gg]$ is centreless. The multiplication $Z(G)^0\times DG
\to G$ is a central isogeny and this isogeny is separable, since $\Lie
Z(G) \cap \Lie DG = \{0\}$.  Combining this with the isogeny $H\to DG$
we obtain the assertion of the lemma.
\end{comment}
\end{proof}

We are now in a position to prove Theorem \ref{thm:sep2}.

\begin{proof}[Proof of Theorem \ref{thm:sep2}]
By Lemmas~\ref{lem:sep} and \ref{lem:sepisog},
we may assume that $G =
S\times H_1\times \cdots \times H_r$, where $S$
is a torus and each $H_i$ is a simply connected
simple group. Put $H_0=S$ and let $\pi_i:G\to H_i$ be the projection.
For every subgroup $K$ of $G$ we have
$C_G(K)=\prod_{i=0}^rC_{H_i}(\pi_i(K))$ and for every subalgebra $\mm$ of $\gg$ we have
$\cc_\gg(\mm)=\bigoplus_{i=0}^r\cc_{\hh_i}(d\pi_i(\mm))$. Since $S$ is abelian, we may now
assume that $G$ is a simply connected simple group.

It now suffices to prove that there
exists a simple group $G'$, a separable isogeny
$\eta\colon G\ra G'$ and an embedding of $G'$ in
some $\GL(V')$ such that $(\GL(V'),G')$ is a reductive
pair: for then every subgroup of $G$ is separable in $G$
and every subalgebra of $\gg$ is separable in $\gg$, by
Corollary~\ref{cor:glv:g} and Lemma~\ref{lem:sep}.
For every simple group $K$ of the same Dynkin type as $G$,
the natural isogeny $G\to K$ is separable, since $p$ is very good for $G$.
So to complete the proof, it is enough to show
that for every Dynkin type, there exists a simple group
$K$ of this Dynkin type and an embedding of $K$ in some
$\GL(V)$ such that $(\GL(V),K)$ is a reductive pair.

If $K = \SO(V)$ or $K = \SP(V)$ and $p\neq 2$, then $(\GL(V),K)$ is a
reductive pair \cite[Lem.~5.1]{rich2}.  This deals with types $B_n$,
$C_n$ and $D_n$.  If $K = \SL(V)$, then, since $p$ is very good for
$G$, it follows that $(\GL(V),K)$ is a reductive pair: the scalar
matrices form a $K$-stable direct complement to $\Sl(V)$ in
$\gl(V)$. This deals with type $A_n$.  If $K$ is an adjoint simple
group of exceptional type and $p$ is good for $K$, then
$(\GL(\kk),K)$ is a reductive pair, thanks to \cite[\S 5]{rich2}.
This completes the proof.
\begin{comment}
 By Lemmas~\ref{lem:sep} and \ref{lem:sepisog}, we may assume that $G =
S\times H$, where $S$ is a torus and $H$ is a product of simply connected
simple groups.  If $(\GL(V),G)$ is a reductive pair,
then every subgroup of $G$ is separable in $G$ and every subalgebra of
$\gg$ is separable in $\gg$, by Corollary~\ref{cor:glv:g}.  According
to \cite[I.3(4),(5),(7)]{slodowy}, if each factor of a direct product
of algebraic groups forms a reductive pair with some general linear
group, then this also holds for the direct product itself.  Thanks to
Lemma~\ref{lem:sep}, it thus suffices to show that for each simple
factor of $H$ there is a separable isogeny to an algebraic group which
forms a reductive pair with some general linear group.  As the simple
factors of $H$ are simply connected, it suffices to find for each
irreducible root system a simple algebraic group with that root system
which forms a reductive pair with some general linear group.  In each case the
isogeny from the simply connected covering group is separable, since
$p$ is very good; see the proof of Lemma~\ref{lem:sepisog}.

If $G = \SO(V)$ or $G = \SP(V)$ and $p\neq 2$, then $(\GL(V),G)$ is a
reductive pair \cite[Lem.~5.1]{rich2}.  This deals with types $B_n$,
$C_n$ and $D_n$.  If $G = \SL(V)$, then, since $p$ is very good for
$G$, it follows that $(\GL(V),G)$ is a reductive pair: the scalar
matrices form a $G$-stable direct complement to $\Sl(V)$ in
$\gl(V)$. This deals with type $A_n$.  If $G$ is an adjoint simple
group of exceptional type and $p$ is good for $G$, then
$(\GL(\gg),G)$ is a reductive pair, thanks to \cite[\S 5]{rich2}.
\end{comment}
\end{proof}

\begin{cor}
\label{cor:sep2}
Let $G$ be connected and suppose that $p$ is very good for
$G$.  Let $g_1, \ldots, g_n \in G$ and let $x_1, \ldots,x_m\in\gg$.
Then the orbit maps $G \to G \cdot (g_1, \ldots, g_n)$ and $G \to G
\cdot (x_1, \ldots, x_m)$
are separable.
\end{cor}

\begin{proof}
 The orbit map $G \to G \cdot (g_1, \ldots, g_n)$ is separable if and
only if the algebraic subgroup of $G$ generated by $g_1, \ldots, g_n$
is separable in $G$ and the orbit map $G \to G\cdot(x_1,\ldots,x_m)$
is separable if and only if the subalgebra of $\gg$ generated by
$x_1,\ldots, x_m$ is separable in $\gg$ (see the comment
after Definition \ref{def:sep} and Subsection \ref{subsec:sep}),
so the corollary follows
immediately from Theorem \ref{thm:sep2}.
\end{proof}

\begin{rems}
\label{rems:redpair}
\begin{comment}
 \begin{itemize}
  \item[(i)] For $G$ simple of exceptional type and for simple
subgroups of $G$ and $p>7$, Theorem \ref{thm:sep2} is due to
case-by-case checks of Liebeck--Seitz \cite[Thm.\ 3]{liebeckseitz0}
and Lawther--Testerman \cite[Thm.\ 2]{lawthertesterman}.
  \item[(ii)] The restriction on $p$ in Theorem \ref{thm:sep2} is
necessary.  For instance, for $G = \SL(V)$ with $\dim V = p$, the
group $G$ is not separable in itself.  Also for $G$ simple of
exceptional type and $p$ a bad prime for $G$, the pair $(\GL(\gg), G)$
need no longer be a reductive pair.
For instance, in Proposition \ref{prop:LsepGnonsep} we provide an
example for $G$ of type $G_2$ and $p=2$ of a non-separable subgroup of
$G$ (cf.\ Corollary \ref{cor:glv:g}).
  \item[(iii)] The requirement that $G$ be connected in Theorem \ref{thm:sep2}
  is also necessary.  For instance, if $G=k^*\rtimes C_2$, where the
  non-trivial element $c$ of the cyclic group $C_2$ acts on $k^*$ by
  $c\cdot a = a^{-1}$, then $C_2$ is a non-separable subgroup of $G$.
  \item[(iv)] The case $n=1$ in Corollary \ref{cor:sep2} is a
  well-known fundamental result due to P.\ Slodowy, \cite[p.\ 38]{slodowy1}.
 \end{itemize}
\end{comment}
(i). Consider the class of reductive groups $G$ that have the property
that each subgroup of $G$ is separable in $G$ and each subalgebra of $\gg$
is separable in $\gg$. Lemma~\ref{lem:sep}, the proof of Theorem~\ref{thm:sep2}
and Corollary~\ref{cor:LGsep} below show that this class is closed under separable
isogenies (in both directions), direct products and centralizers of subgroups $S$ acting on $G$ by automorphisms
as in Proposition~\ref{prop:C_G(S)redpair}.
In particular our class contains the ``strongly standard'' reductive groups of
\cite[\S 2.4]{mcninch-test}.

(ii). For $G$ simple of exceptional type and for simple subgroups of
$G$ and $p>7$, Theorem \ref{thm:sep2} is due to case-by-case checks of
Liebeck--Seitz \cite[Thm.\ 3]{liebeckseitz0} and
Lawther--Testerman \cite[Thm.\ 2]{lawthertesterman}.

(iii). The restriction on $p$ in Theorem \ref{thm:sep2} is necessary.
For instance, for $G = \SL(V)$ with $\dim V = p$, the group $G$ is not
separable in itself.  Also for $G$ simple of exceptional type and $p$
a bad prime for $G$, the pair $(\GL(\gg), G)$ need no longer be a
reductive pair, so the proof breaks down:
for instance, in Proposition \ref{prop:LsepGnonsep} we provide an
example for $G$ of type $G_2$ and $p=2$ of a non-separable subgroup of
$G$ (cf.\ Corollary \ref{cor:glv:g}).

(iv). The requirement in Theorem \ref{thm:sep2} that $G$ be connected
is also necessary.  For instance, if $G=k^*\rtimes C_2$, where the
non-trivial element $c$ of the cyclic group $C_2$ acts on $k^*$ by
$c\cdot a = a^{-1}$, then $C_2$ is a
non-separable subgroup of $G$.

(v). The case $n=1$ in Corollary \ref{cor:sep2} is a well-known
fundamental result due to P.\ Slodowy, \cite[p38]{slodowy1}.

(vi). Serre has asked whether Theorem \ref{thm:sep2} holds for an arbitrary group subscheme $H$ of $G$;
Theorem \ref{thm:sep2} deals with the two special cases $H$ smooth and $H$ infinitesimal of height one.
\end{rems}

We finish the section with some further useful results on
separability and reductive pairs.

\begin{lem}
\label{lem:almostfaithful}
 Suppose $G$ is connected.
Let $S$ be an algebraic group
acting faithfully on $G$ by automorphisms.
Then the corresponding representation of $S$ on $\gg$ has finite kernel.
\end{lem}

\begin{proof}
 It is enough to prove that $S=\{1\}$ under the
extra hypotheses that $S$ is connected and $S$
acts trivially on $\gg$, so we assume this.
Let $s\in S$.  Let $B,B^-$ be any pair of
opposite Borel subgroups of $G$.  Since
$s\cdot B$ is also a Borel subgroup of $B$,
there exists $g\in G$ such that $s\cdot B= gBg^{-1}$.
Let $\bb= \Lie B$.  Then $\bb= s\cdot \bb= \Ad g(\bb)$,
so $g\in B$, by \cite[IV.14.1 Cor.~2]{borel}, so $s$
normalizes $B$.  Similarly $s$ normalizes $B^-$, so
$s$ normalizes the maximal torus $T=B\cap B^-$ of $G$.
Hence $s$ centralizes $T$, by the rigidity of tori
\cite[Prop.~III.8.10]{borel}. Since $B$ and $B^-$
were arbitrary, $s$ centralizes the set of semisimple elements of $G$,
which is dense in $G$ as $G$ is reductive (e.g., see \cite[Thm.~IV.12.3(2)]{borel}).
Thus $s$ centralizes $G$. But the
$S$-action on $G$ is faithful, so $S=\{1\}$ as required.
\end{proof}

\begin{prop}
\label{prop:C_G(S)redpair}
Let $S$ be
an algebraic group acting on $G$ by automorphisms.
Suppose that $S$ acts semisimply on $\gg$ and $\Lie C_G(S)=\cc_\gg(S)$.
Then
 \begin{itemize}
  \item[(a)] $C_G(S)$ is $G$-completely reducible
  and $(G,C_G(S))$ is a reductive pair.
  \item[(b)] If $S$ is a subgroup of $G$, then $N_G(S)$ is $G$-completely reducible and
  if further $S^0$ is central in $S$, then $(G,N_G(S))$ is a
  reductive pair.
 \end{itemize}
\end{prop}

\begin{proof}
 (a). Let $H$ be the union of the components of $G$ that meet $C_G(S)$.  Then $H$ is a finite-index subgroup of $G$, so $C_G(S)$ is $G$-cr if and only if it is $H$-cr, by \cite[Prop.~2.12]{BMR2}.  Hence we can assume that $G=H$.  Replacing $S$ by $S/C_S(G)$, we can also assume that $S$ acts faithfully on $G$.  Since $C_G(S)$ meets every component of $G$, it follows that $S$ acts faithfully on $G^0$.  The completely reducible representation $S\ra \GL(\gg)$ has finite kernel by Lemma \ref{lem:almostfaithful}, so $S$ is reductive.

If $\Char k=0$, then, as $S$ is reductive, it is linearly reductive.
So in this case all we have to show is that $C_G(S)$ is reductive.
Since $C_G(S)^0=C_{G^0}(S)^0$, this follows immediately from
\cite[Prop.~10.1.5]{rich0}.

Now assume that $\Char k = p > 0$.
\begin{comment}
Since $S$ is reductive, $S$ has an
$\overline{\mathbb{F}_p}$-structure, thanks to \cite[Prop.~3.2]{martin1}.
As pointed out in \cite[Notation~3.3]{martin1}, this implies that $S$ has an
increasing chain of finite subgroups $S_1\subseteq S_2\subseteq\cdots$
such that $\bigcup_{i\ge1}S_i$ is Zariski dense in $S$.  From this it
follows that for some $i\ge1$, $\gg$ is a semisimple $S_i$-module (cf.\ \cite[Lem.~2.10]{BMR}),
and we have $C_G(S_i) = C_G(S)$ and
$\cc_\gg(S_i) = \cc_\gg(S)$.
\end{comment}
By Lemma~\ref{lem:finitedense}, $S$ admits an
ascending chain of finite subgroups $S_1\subseteq S_2\subseteq\cdots$
such that
their union is Zariski dense in $S$. From this it
follows that for some $i\ge1$, $\gg$ is a semisimple $S_i$-module (that is, the image of $S_i$ in $\GL(\gg)$ is $\GL(\gg)$-cr --- cf.\ \cite[Lem.\ 2.10]{BMR}) and we have $C_G(S_i) = C_G(S)$ and $\cc_\gg(S_i) = \cc_\gg(S)$.
So, after replacing $S$ by $S_i$, we may assume that $S$ is finite.
Put $G' = G \rtimes  S$. Clearly, $G'$ is reductive.
Furthermore, $\gg' = \Lie G' = \gg$ is a semisimple $S$-module,
and $\Lie C_{G'}(S) = \cc_{\gg'}(S)$ by construction.
Therefore, $S$ is $G'$-cr, by Theorem \ref{thm:adjcr}.
Since $S$ is $G'$-cr, $C_{G'}(S)$ is $G'$-cr,
by \cite[Cor.\ 3.17]{BMR}.
So the normal subgroup $C_G(S)$ of $C_{G'}(S)$ is also $G'$-cr,
\cite[Thm.\ 3.10]{BMR}.
Now $C_G(S)$ is $G$-cr, by \cite[Prop.\ 2.12]{BMR2}.

\begin{comment}
We have a completely reducible representation from $S$ into
$\GL(\gg)$, so $C_{\GL(\gg)}(S)$ is $\GL(\gg)$-cr. Let $\pi_G\colon
G\ra \GL(\gg)$ be the canonical projection.  Then
$\pi_G(C_G(S))\subseteq C_{\GL(\gg)}(S)$, so $C_G(S)\subseteq
\pi_G^{-1}(C_{\GL(\gg)}(S))$.  Hence $C_G(S)^0$ is an extension of the
reductive group $C_{\GL(\gg)}(S)^0$ by a subgroup of the linearly
reductive group $Z(G^0)$.  This implies that $C_G(S)^0$ is reductive.
\end{comment}

Since $\gg$ is a semisimple $S$-module, it has a direct sum
decomposition into $S$-isotypic summands.
By hypothesis,  $\Lie C_G(S)=\cc_\gg(S)$,
so $\Lie C_G(S)$ is the trivial $S$-isotypic summand.  Hence
there is a unique $S$-stable complement to $\Lie C_G(S)$ in $\gg$,
namely, the sum $\mm$ of the non-trivial $S$-isotypic summands. The
uniqueness of $\mm$ implies that it is also $C_G(S)$-stable, so it
follows that $(G,C_G(S))$ is a reductive pair.

(b).
Now assume that $S$ is a subgroup of $G$.
Then $S$ is $G$-cr by Theorem \ref{thm:adjcr}, so $N_G(S)$ is
$G$-cr \cite[Thm.~3.14]{BMR}. Moreover, $N_G(S)/SC_G(S)$ is
finite by \cite[Lem.\ 6.8]{martin1}. Since $S^0\subseteq C_G(S)$ by
assumption, $N_G(S)/C_G(S)$ is finite. It follows that $\Lie
N_G(S)=\Lie C_G(S)$.  By the uniqueness of the subspace $\mm$
of $\gg$ from the proof of part (a) above, $\mm$ is also
$N_G(S)$-stable.  Hence $(G,N_G(S))$ is a reductive pair.
\end{proof}

\begin{comment}
\begin{prop}
\label{prop:C_G(S)redpair}
Let $S$ be a reductive algebraic group acting on $G$ by automorphisms.
Suppose that $S$ acts semisimply on $\gg$ and $\Lie C_G(S)=\cc_\gg(S)$.
Then
 \begin{itemize}
  \item[(a)] $C_G(S)$ is $G$-completely reducible
  and $(G,C_G(S))$ is a reductive pair.
  \item[(b)] If $S$ is a subgroup of $G$, then $N_G(S)$ is $G$-completely reducible and
  if further $S^0$ is central in $S$, then $(G,N_G(S))$ is a
  reductive pair.
 \end{itemize}
\end{prop}

\begin{proof}
(a). If $\Char k=0$, then, as $S$ is reductive it is linearly reductive.
So in this case all we have to show is that $C_G(S)$ is reductive.
Since $C_G(S)^0=C_{G^0}(S)^0$, this follows immediately from
\cite[Prop.~10.1.5]{rich0}.

Now assume that $\Char k = p > 0$.
Since $S$ is reductive, $S$ has an
$\overline{\mathbb{F}_p}$-structure, thanks to \cite[Prop.~3.2]{martin1}.
As pointed out in \cite[Notation~3.3]{martin1}, this implies that $S$ has an
increasing chain of finite subgroups $S_1\subseteq S_2\subseteq\cdots$
such that $\bigcup_{i\ge1}S_i$ is Zariski dense in $S$.  From this it
follows that for some $i\ge1$, $\gg$ is a semisimple $S_i$-module, and we have $C_G(S_i) = C_G(S)$ and
$\cc_\gg(S_i) = \cc_\gg(S)$. So,
after replacing $S$ by $S_i$, we may assume that $S$ is finite.
Put $G' = G \rtimes  S$. Clearly, $G'$ is reductive.
Furthermore, $\gg' = \Lie G' = \gg$ is a semisimple $S$-module,
and $\Lie C_{G'}(S) = \cc_{\gg'}(S)$ by construction.
Therefore, $S$ is $G'$-cr, by Theorem \ref{thm:adjcr}.
Since $S$ is $G'$-cr, $C_{G'}(S)$ is $G'$-cr by \cite[Cor.\ 3.17]{BMR}.
So the normal subgroup $C_G(S)$ of $C_{G'}(S)$ is also $G'$-cr,
\cite[Thm.\ 3.10]{BMR}.
Now $C_G(S)$ is $G$-cr, by \cite[Prop.\ 2.8]{BMR2}.

\begin{COMMENT}
We have a completely reducible representation from $S$ into
$\GL(\gg)$, so $C_{\GL(\gg)}(S)$ is $\GL(\gg)$-cr.  Let $\pi_G\colon
G\ra \GL(\gg)$ be the canonical projection.  Then
$\pi_G(C_G(S))\subseteq C_{\GL(\gg)}(S)$, so $C_G(S)\subseteq
\pi_G^{-1}(C_{\GL(\gg)}(S))$.  Hence $C_G(S)^0$ is an extension of the
reductive group $C_{\GL(\gg)}(S)^0$ by a subgroup of the linearly
reductive group $Z(G^0)$.  This implies that $C_G(S)^0$ is reductive.
\end{COMMENT}

Since $\gg$ is a semisimple $S$-module, it has a direct sum
decomposition into $S$-isotypic summands.
By hypothesis,  $\Lie C_G(S)=\cc_\gg(S)$,
so $\Lie C_G(S)$ is the trivial $S$-isotypic summand.  Hence
there is a unique $S$-stable complement to $\Lie C_G(S)$ in $\gg$,
namely, the sum $\mm$ of the non-trivial $S$-isotypic summands. The
uniqueness of $\mm$ implies that it is also $C_G(S)$-stable, so it
follows that $(G,C_G(S))$ is a reductive pair.

(b).
Now assume that $S$ is a subgroup of $G$.
Then $S$ is $G$-cr by Theorem \ref{thm:adjcr}, so $N_G(S)$ is
$G$-cr \cite[Thm.~3.14]{BMR}. Moreover, $N_G(S)/SC_G(S)$ is
finite by \cite[Lem.\ 6.8]{martin1}. Since $S^0\subseteq C_G(S)$ by
assumption, $N_G(S)/C_G(S)$ is finite. It follows that $\Lie
N_G(S)=\Lie C_G(S)$.  By the uniqueness of the subspace $\mm$
of $\gg$ from the proof of part (a) above, $\mm$ is also
$N_G(S)$-stable.  Hence $(G,N_G(S))$ is a reductive pair.
\end{proof}
\end{comment}

\begin{comment}
R:We can actually omit the assumption that $S$ is reductive. This works as follows.
If $\Char k=p>0$, then we apply the lemma on ascending sequences of
finite subgroups from the relative preprint and then the same arguments as in the proof work.
Now assume $\Char k=0$. We have to show that $C_G(S)$ is reductive.
By \cite[IV.14.11]{borel} we only have to show that $\Ad(C_G(S))$ is reductive.
Let $S'$ be the image of $S$ in $\GL(\gg)$. Then $S'$ is reductive, since the representation
$S'\to\GL(\gg)$ is faithful and $S'$ acts semi-simply on $\gg$ ($R_u(S^0)$ has to act semi-simply
by Clifford's theorem, but a unipotent algebraic group has a unique irreducible module: the trivial
$1$-dimensional module). So $C_{G_{\rm ad}}(S')$ is reductive by the same arguments as in the
proof. We now finish by showing that $\Ad(C_G(S))$ is normal in $C_{G_{\rm ad}}(S')$.
Let $g\in C_G(S)$ and let $h\in G$ such that $\Ad(h)\in C_{G_{\rm ad}}(S')$. Let $s\in S$. Then $(s\cdot h)h^{-1}\in\ker(\Ad)$. So $s\cdot h=zh$ for some $z\in Z(G)$. Since $S$ acts by automorphisms
we must have that $s\cdot h^{-1}=z^{-1}h^{-1}$. But then $s\cdot hgh^{-1}=hgh^{-1}$.
Since this holds for all $s\in S$, we see that $\Ad(h)\Ad(g)\Ad(h)^{-1}\in\Ad(C_G(S))$.
\end{comment}

The following is immediate by
Theorem \ref{thm:redpairs}, Corollary~\ref{cor:redpairs} and Proposition~\ref{prop:C_G(S)redpair}(a).
Note that it applies in particular when $S$ is a torus, so that $C_G(S)$ is an R-Levi subgroup
of $G$ \cite[Cor.~6.10]{BMR}.

\begin{cor}
\label{cor:LGsep}
Let $G$ and $S$ be as in the statement of Proposition \ref{prop:C_G(S)redpair}.
Then every subgroup of $C_G(S)$ which is separable in $G$
is separable in $C_G(S)$ and every subalgebra of $\cc_\gg(S)$
which is separable in $\gg$ is separable in $\cc_\gg(S)$.
\end{cor}

For a regular reductive subgroup $H$ of $G$ the next lemma gives
a useful criterion for $(G,H)$ to be a reductive pair
(cf.~\cite[Ex.~3.33]{BMR}).

\begin{lem}
\label{lem:redpair}
Let $T$ be a maximal torus of $G$ and let $H$ be a reductive subgroup
of $G$ containing $T$. Assume that $\Psi(H)=\Psi(H,T)$ is a closed
subsystem of $\Psi=\Psi(G,T)$. Then $(G,H)$ is a reductive pair.
\end{lem}

\begin{proof}
Let $\mm$ be the sum of the root spaces $\uu_\alpha$ with
$\alpha\notin\Psi(H)$. Then $\gg=\hh\oplus\mm$.
By the conjugacy of the maximal tori in $H^0$,
we have $H=H^0N_H(T)$. Since $N_H(T)$ permutes the root spaces in $\hh$
and also those outside $\hh$, $N_H(T)$ stabilizes $\mm$.
So all we have to show is that $\mm$ is stable
under the $U_\beta$, for $\beta\in\Psi(H)$.

Let $\alpha\in\Psi\setminus\Psi(H)$ and $\beta\in\Psi(H)$.
If $\gamma = \alpha+i\beta$ is a root for some integer $i\ge0$,
then we must have $\gamma\notin\Psi(H)$, as otherwise
$\alpha = \gamma-i\beta\in\Psi(H)$,
because $\Psi(H)$ is closed and symmetric;
see \cite[Ch.~6 Prop.~23(iii)]{bou}.
Now let $u\in U_\beta$.
Then, by \cite[Lem.~5.2]{borel1},
$\Ad u(\uu_\alpha)\subseteq\bigoplus\uu_{\alpha+i\beta}$,
where the sum is over all integers $i\ge0$ such that $\alpha+i\beta$ is a root.
By the above, this sum is contained in $\mm$.
\end{proof}

\begin{rems}
\label{rem:C_G(S)redpair}
\begin{comment}
 \begin{itemize}
  \item[(i)] We can apply Proposition~\ref{prop:C_G(S)redpair} and the
second part of Theorem~\ref{thm:redpairs} to prove that if $S$ is a
linearly reductive subgroup of $G$ with $S^0$ central in $S$ and $H$
is a $G$-cr subgroup of $N_G(S)$ such that $H$ is separable in $G$,
then $H$ is $N_G(S)$-cr.  Here the separability condition cannot be
removed; see Example~\ref{exmp:KnotKS}.
  \item[(ii)] We observe that Slodowy's example \cite[I.3(7)]{slodowy}
  is a special case of Proposition \ref{prop:C_G(S)redpair}, namely
  when $G = \GL(V)$ and $C_G(S)$ is a Levi subgroup of $G$.
  \item[(iii)] The converse of Corollary~\ref{cor:LGsep} is false: see
  Proposition \ref{prop:LsepGnonsep}.
 \end{itemize}
\end{comment}

(i).
Note that if $S$ is a linearly reductive group acting on $G$ by
automorphisms, then the conditions of Proposition~\ref{prop:C_G(S)redpair}(a)
are satisfied, \cite[Lem.\ 4.1]{rich0}.

(ii).
We can apply Proposition~\ref{prop:C_G(S)redpair}(b) and Theorem~\ref{thm:redpairs} to prove that if $S$ is a linearly
reductive subgroup of $G$ with $S^0$ central in $S$ and $H$ is a
$G$-cr subgroup of $N_G(S)$ such that $H$ is separable in $G$, then
$H$ is $N_G(S)$-cr.  Here the separability condition cannot be
removed; see Example~\ref{exmp:KnotKS}.

(iii).
Note that the hypothesis on $S^0$ in Proposition
\ref{prop:C_G(S)redpair}(b) cannot be removed.  For example, suppose
$p=3$ and let $G= (k^*\times k^*\times k^*)\rtimes (C_3\times C_2)$, where the
cyclic group $C_3$ acts on $k^*\times k^*\times k^*$ by a cyclic permutation of the factors and the cyclic group $C_2= \{1,a\}$ acts on $k^*\times k^*\times k^*$ by $a\cdot (x,y,z)= (x^{-1},y^{-1},z^{-1})$.  Let $S$ be the linearly reductive subgroup
$\Delta C_2$, where $\Delta$ is the diagonal inside $k^*\times k^*\times k^*$.
Then $N_G(S)^0= \Delta$ and $C_3\subseteq N_G(S)$, so $(G,N_G(S))$ is not a
reductive pair because $\Lie \Delta$ does not admit a $C_3$-stable
complement in $\Lie (k^*\times k^*\times k^*)$.

(iv).
We observe that Slodowy's example \cite[I.3(7)]{slodowy} is a
special case of Proposition \ref{prop:C_G(S)redpair}(a), namely when $G
= \GL(V)$ and $C_G(S)$ is a Levi subgroup of $G$.

(v).
In the special case of Proposition \ref{prop:C_G(S)redpair} when
$S$ is linearly reductive and $G$ is connected, Richardson showed
in \cite[Prop.\ 10.1.5]{rich0} that $C_G(S)$ is reductive and, if $S$ is a subgroup of $G$,
$N_G(S)$ is reductive.

(vi).
The converse of Corollary~\ref{cor:LGsep} is false: see
Proposition \ref{prop:LsepGnonsep}.
\end{rems}

We give another application of Proposition \ref{prop:C_G(S)redpair}.

\begin{exmp}
\label{ex:d4g2}
Suppose that $p = 2$.  Let $G$ be simple of type $D_4$ and let $S$ be the
group of order $3$ generated by the triality graph automorphism of
$G$.  Then $K = C_G(S)^0$ is of type $G_2$.  Since $S$ is linearly
reductive, Proposition \ref{prop:C_G(S)redpair}(a) implies that $(G, K)$
is a reductive pair.  In Section \ref{sec:ex}, we construct a subgroup
$H$ of $K$ isomorphic to $S_3$ which is not separable in $K$, see
Proposition \ref{prop:LsepGnonsep}. It follows from Theorem
\ref{thm:redpairs} that $H$ is also non-separable in $G$.  In
addition, by Lemma \ref{lem:HGcr}(a), $H$ is $K$-cr and thus, thanks
to \cite[Cor.\ 3.21]{BMR}, $H$ is also $G$-cr.

This example also gives rise to a non-separable subgroup in the
exceptional group of type $F_4$ as follows.
Let $G'$ denote this group; then,
since $D_4$ is a closed subsystem of the root system of type $F_4$
(the $D_4$-subsystem consists of the long roots in the $F_4$ system),
Lemma \ref{lem:redpair} implies that $(G',G)$ is a reductive pair.
It follows from Theorem \ref{thm:redpairs}
that $H$ is also non-separable in the
group $G'$.
\end{exmp}

\section{The Adjoint Module and Complete Reducibility}
\label{sec:sep1}

In the proof of Theorem \ref{thm:adjcr}, the hypothesis of separability
is used only for a rather coarse dimension-counting argument, so it is natural
to ask whether it can be removed.
This is a more subtle problem than it at first appears.

\begin{comment}
\begin{exmp}[Serre]
\label{ex:serre}
Let $H$ be an exceptional simple group of adjoint type and
suppose $k$ is a field of positive characteristic
$p < 2h-2$, where $h$ denotes the Coxeter number of $H$.
Then, with the possible exception of type $G_2$ in characteristic $5$,
there exists a simple subgroup $K$ of $H$ of type $A_1$ such that
$K$ is $H$-irreducible, but $K$ does not act semisimply on $\hh$ --- that is, $K$ is not $G$-completely reducible for $G:= \GL(\hh)$.
We can choose $p$ and $H$ such that $p$ is good for $H$; then $(G,H)$ is a reductive
pair, since the Killing form on $\hh$ is non-degenerate \cite[\S 5]{rich2}.  It then follows from Theorem \ref{thm:redpairs} that $K$ is separable in $H$, since $K$ is separable in $G$.
\end{exmp}
\end{comment}

Our first result shows that
we can remove the separability assumption
from $H$ in Theorem \ref{thm:adjcr}
under extra hypotheses on $H$.

\begin{thm}
\label{thm:allregss}
Suppose that $H$ is a subgroup of $G$ such
that $H$ acts semisimply on ${\mf m}_{\rm ad}$
for every reductive subgroup $M$ of $G$ that contains $H$
and a maximal torus of $G$ (this includes the case $M = G$).
Then $H$ is $G$-completely reducible.
\end{thm}

\begin{proof}
Suppose $P$ is an $R$-parabolic subgroup of $G$ containing $H$,
let $T$ be a maximal torus of $P$
and let $\lambda\in Y(T)$ such that $P=P_\lambda$.  By Remark \ref{rem:P0G0}, we can assume that $P^0$ is proper in $G^0$.  This implies that $\lambda$ is non-central in $G^0$ \cite[Lem.~2.4]{BMR}.

First assume that $H$ acts semisimply on $\gg$ and
that $\zz(\gg)=\{0\}$. We then show
that there exists a reductive subgroup $M$ of $G$
such that $\dim M<\dim G$ and $M$ contains $H$,
$\lambda(k^*)$ and a maximal torus of $G$.
After that we prove the statement of the theorem
using induction on the dimension of $G$.

Put $\nn=\Lie R_u(P)$, $\frakl_\lambda=\Lie L_\lambda $
and $\mf{s}=\Lie\lambda(k^*)$.
Since $L_\lambda$ centralizes $\lambda(k^*)$ and normalizes $R_u(P)$,
$\lambda(k^*)R_u(P)$ is normalized by $L_\lambda$.
Therefore, $\lambda(k^*)R_u(P)$ is a normal subgroup of $P$.
So $\Lie (\lambda(k^*)R_u(P)) = \nn\oplus\mf{s}$
is an $H$-submodule of $\gg$, since $H$ is contained in $P$.
Now $\nn$ is an $H$-submodule of $\nn\oplus\mf{s}$, so,
since $\gg$ is a semisimple $H$-module,
there exists a $1$-dimensional $H$-submodule $\mf{s}_1$ of $\nn\oplus\mf{s}$
which is a direct $H$-complement to $\nn$.

As $L_\lambda$ acts trivially on $\zz(\frakl_\lambda)$,
$(\nn\oplus \zz(\mf{l}_\lambda))/\nn$
is a trivial $L_\lambda$-module. By \cite[Prop.~I.3.17]{borel},
we have $\Ad(u)(x)-x\in[\frakl,\nn]\subseteq\mf{n}$ for all
$u\in R_u(P)$ and all $x\in\pp$, so $(\nn\oplus \zz(\mf{l}_\lambda))/\nn$
is also a trivial $R_u(P)$-module and therefore a trivial $P$-module.
So $\mf{s}_1$ must be a trivial $H$-module.
Let $x\in\mf{s}_1$ be non-zero.
Clearly, $H$ fixes the nilpotent and semisimple parts of $x$.
By \cite[Thm.~III.10.6]{borel}, $\nn$ is the set of nilpotent
elements of $\nn\oplus\mf{s}$.
So $x$ has non-zero semisimple part and we may assume that $x$ is semisimple.
Thus we have found a non-zero semisimple element of $\Lie P$ that is fixed by $H$ and
$\lambda(k^*)$.
After conjugating $\lambda$ and $T$ by the same element of
$P$, we may assume that $x\in\Lie T$.

Put $M:=C_{G}(x)$.
This is a reductive subgroup of $G$
which contains $T$, $H$ and $\lambda(k^*)$.
Furthermore, $\dim M < \dim G$,
since otherwise $G^0\subseteq M$ which is impossible,
because $\cc_\gg(G^0) = \zz(\gg)=\{0\}$.
So $M$ has the required properties.

To prove the assertion of the theorem, we pass to the adjoint group.
Let $\pi_G:G\to G_{\rm ad}$ be the natural morphism and let
$\tilde\lambda=\pi_G\circ\lambda\in Y(G_{\rm ad})$.
By our hypothesis, $\pi_G(H)$ acts semisimply on $\gg_{\rm ad}$.
Note that $\tilde\lambda$ is non-trivial.
If we apply the above argument to $G_{\rm ad}$, $\pi_G(H)$ and
$\tilde\lambda$, we
get a reductive subgroup $M'$ of $G_{\rm ad}$ such that
$\dim M'<\dim G_{\rm ad}$ and $M'$ contains $\pi_G(T)$,
$\pi_G(H)$ and ${\tilde\lambda}(k^*)$.
But then $M:=\pi_G^{-1}(M')$ is a reductive subgroup of
$G$ such that $\dim M<\dim G$ and $M$ contains $T$, $H$ and
${\lambda}(k^*)$. Clearly, $M$ and $H$ satisfy the same assumptions as $G$
and $H$.
By the induction hypothesis, $H$ is $M$-cr, so there exists $u\in P_\lambda(M)$
such that $H\subseteq uL_\lambda(M)u^{-1}$.
But $R_u(P_\lambda(M))=R_u(P_\lambda)\cap M$ and similarly for $L_\lambda(M)$.
So $H$ is contained in the Levi subgroup $uL_\lambda u^{-1}$ of $P$.
\end{proof}

\begin{comment}
The following example indicates the sort of problem that can arise
without the assumptions on $H$ made in Theorem \ref{thm:allregss}.
\end{comment}

The following example indicates the sort of problem that can arise
without the assumptions on $H$ made in Theorem \ref{thm:allregss}.  First we need some terminology. If $G$ is
connected and simple of type $A_{n-1}$, then either $\Lie G
\cong\mf{sl}_n$, $\Lie G \cong\mf{pgl}_n$, or $p^2|n$ and $\Lie G$
is of {\it intermediate type}; see \cite[Table~1]{hogeweij}.  In the latter case, $\gg$ is the direct sum of its centre, which is $1$-dimensional, and its derived algebra, which is isomorphic to $\mf{psl}_n:=\mf{sl}_n/(k\cdot{\rm id})$.

\begin{exmp}
\label{ex:gvsgtilde}
Let $p$ be a prime, put $n=p^2$ and let $G$ be the simple algebraic
group of type $A_{n-1}$ whose character group is the lattice that is
strictly between the root lattice and the weight lattice
(the quotient group of the latter two lattices is a cyclic group of order $p^2$).  Then $\gg$ is of intermediate type.
Thus $\gg$ is the direct sum of two simple $G$-modules and is therefore
a semisimple $G$-module.  But the only proper non-zero ideal of $\gg_{\rm ad}\cong\mf{pgl}_n$ is its derived subalgebra, which is of dimension $n^2-2$
(\cite[Table~1]{hogeweij}).  Since every $G$-submodule of $\gg_{\rm ad}$ is an ideal, it follows that $G$ does not act semisimply on $\gg_{\rm ad}$.
\end{exmp}

\begin{comment}
 \begin{exmp}
\label{ex:gvsgtilde}
Let $p$ be a prime, put $n=p^2$ and let $G$ be the simple algebraic
group of type $A_{n-1}$ whose character group is the lattice that is
strictly between the root lattice and the weight lattice
(the quotient group of the latter two lattices is a cyclic group of order $p^2$).
Then $\gg$ is the direct sum of its centre, which is $1$-dimensional,
and its derived
algebra, which is isomorphic to $\mf{psl}_n$ (see \cite[Table~1]{hogeweij}).
Thus $\gg$ is the direct sum of two simple $G$-modules and is therefore
a semisimple $G$-module.
But the only proper non-zero ideal of $\gg_{\rm ad}\cong\mf{pgl}_n$
is its derived subalgebra, which is of dimension $n^2-2$
(\cite[Table~1]{hogeweij}).
Since every $G$-submodule of $\gg_{\rm ad}$ is an ideal,
it follows that $G$ does not act semisimply on $\gg_{\rm ad}$.
\end{exmp}

\end{comment}

\begin{rems}
\begin{comment}
 \begin{itemize}
  \item[(i)] Example \ref{ex:gvsgtilde} shows that it is possible for a subgroup
$H$ to act semisimply on $\gg$,
but not on $\gg_{\rm ad}$.
Thus the argument in the proof of Theorem \ref{thm:allregss}
breaks down in general.
  \item[(ii)] Suppose $G$ is semisimple (so $Z(G)$ is finite),
and $H$ is a subgroup of $G$ which acts semisimply on $\gg$, but
$H$ is not separable in $G$.
Observe that
$H$ might be ``trivially'' non-separable in $G$, in the sense that $\gg$
has non-zero centre so that $G$ is not even separable in itself.
One might hope to remove such possibilities by passing to the
adjoint form $G_{\rm ad}$ of $G$, but
Example \ref{ex:gvsgtilde} shows that if we do this, the image
of $H$ in $G_{\rm ad}$ may fail to act
semisimply on $\gg_{\rm ad}$.
 \end{itemize}
\end{comment}
  (i). Example \ref{ex:gvsgtilde} shows that it is possible for a subgroup
$H$ to act semisimply on $\gg$,
but not on $\gg_{\rm ad}$.
Thus the argument in the proof of Theorem \ref{thm:allregss}
cannot be used to extend Theorem \ref{thm:adjcr} to the non-separable case

  (ii). Suppose $G$ is semisimple (so $Z(G)$ is finite),
and $H$ is a subgroup of $G$ which acts semisimply on $\gg$, but
$H$ is not separable in $G$.
Observe that
$H$ might be ``trivially'' non-separable in $G$, in the sense that $\gg$
has non-zero centre so that $G$ is not even separable in itself.
One might hope to deal with such possibilities by passing to the
adjoint form $G_{\rm ad}$ of $G$, but
Example \ref{ex:gvsgtilde} shows that if we do this, the image
of $H$ in $G_{\rm ad}$ may fail to act
semisimply on $\gg_{\rm ad}$.
\end{rems}

\begin{comment}
{\tt Where should this next bit go????? What does it prove????}  In
the case that $H$ does not act semisimply on $\gg_{\rm ad}$, still
assuming that $\mf{s}\subseteq\zz(\gg)$, we convinced ourselves in the
situation that $G$ is almost simple that $H$ cannot lie in a Levi
subgroup of $P$.  Because by \cite{BMR} $\tilde H$ would lie in a Levi
subgroup $\tilde L$ of $\tilde P$.  Then we could write $\tilde
L=\tilde L_\lambda$ for some $\tilde\lambda\in Y(\tilde G)$ and then
$\tilde{\mf s}:=\Lie\tilde\lambda(k^\times)$ would be an $H$-module
complement in $\tilde\gg$ to $\pi(\gg)$ on which it acts semisimply.
This is a contradiction.

We now translate the above "proof" into the following algorithm that,
starting with a parabolic containing $H$, tries to put $H$ into a Levi
of that parabolic.

Let $G$ and $H$ be as above and let $P$ be a parabolic of $G$
    containing $H$.\\ Case 1. $P=G$. Take $L=G$.\\ Case 2.1. There
    exists a $\lambda\in Y(G)$ such that $P=P_\lambda$ and
    $\mf{s}:=\Lie\lambda(k^\times)$ is not central in $\gg$.  In this
    case we can find a semisimple element $x\in\gg$ such that $x$ is
    not central in $\gg$ and $M:=C_G(x)$ contains $H$.  Now we can
    replace $G$ and $P$ by $C_G(x)$ and $P_\lambda(M)$ and start the
    procedure again.\\ Case 2.2. For every $\lambda\in Y(G)$ such that
    $P=P_\lambda$ we have that $\mf{s}$ is central in $\gg$.\\ Case
    2.2.1. $H$ does not act completely reducibly on $\tilde\gg$.  Then
    we "know" that $H$ cannot lie in any Levi subgroup of $P$.\\ Case
    2.2.2. $H$ acts completely reducibly on $\tilde\gg$.  In this case
    we can find a proper regular reductive subgroup $M$

of $G$ such that $P=P_\lambda$ for some $\lambda\in Y(M)$.  Then we
replace $G$ and $P$ by $M$ and $P_\lambda(M)$ and start the procedure
again.

After termination of the algorithm we have either found a Levi
subgroup of $P$ that contains $H$ or we have found a regular reductive
subgroup $M$ of $G$ with $P=P_\lambda$ for some $\lambda\in Y(M)$ such
that $\mf{s}:=\Lie\lambda(k^\times)$ is central in $\gg$ and $H$ does
not act semisimply on $\tilde{\mf m}$.  In this case we would like to
conclude that $H$ is not contained in any Levi of $P$, but we can only
do this if $M$ is almost simple (counterexample where $M$ has two
simple factors).  In the first case we know that for all $M$ appearing
from Case 2.2.2 $H$ acts semisimply on $\tilde{\mf m}$.
\end{comment}

\begin{comment}
Our next result shows that we can also remove the separability
assumption in Theorem \ref{thm:adjcr} by strengthening the conditions
on $G$, rather than on $H$ (as in Theorem \ref{thm:allregss}).  In
order to state the result, we need some terminology. If $G$ is
connected and simple of type $A_{n-1}$, then either $\Lie G
\cong\mf{sl}_n$, $\Lie G \cong\mf{pgl}_n$, or $p^2|n$ and $\Lie G$
is of {\it intermediate type}; see \cite[Table~1]{hogeweij}.  In
contrast to Theorem \ref{thm:verygood}, this next result does not
impose any characteristic restrictions stemming from simple factors of
$G$ of type $A_n$.
\end{comment}

Our next result shows that we can also remove the separability
assumption in Theorem \ref{thm:adjcr} by strengthening the conditions
on $G$, rather than on $H$ (as in Theorem \ref{thm:allregss}).  In
contrast to Theorem \ref{thm:verygood}, this next result does not
impose any characteristic restrictions stemming from simple factors of
$G$ of type $A_n$.

\begin{thm}
\label{thm:adjsc}
Assume that $G$ is connected, $p$ is good for $G$, and no simple
factor of type $A_n$ of the derived group $DG$ of $G$ has Lie algebra
of intermediate type.  Let $H$ be a subgroup of $G$ which acts
semisimply on $\Lie DG$.  Then $H$ is $G$-completely reducible.
\end{thm}

\begin{proof}
Since $Z(G)^0$ acts trivially on $\gg$ and since it is contained in any Levi
subgroup of any parabolic subgroup of $G$, we may assume that $Z(G)^0\subseteq H$.
Then $H=Z(G)^0(H\cap DG)$.
Applying \cite[Lem.\ 2.12]{BMR} to the isogeny $Z(G)^0\times DG\to G$ and the projection $Z(G)^0\times DG\ra DG$,
we see that we may replace $G$ by $DG$ and $H$ by $H\cap DG$.

Let $G_1,G_2,\ldots, G_r$ be the simple factors of $G$ and let $\mu : \Pi_i G_i\to G$ be the isogeny given by multiplication.
Denote the projection $G\to G_i$ by $\pi_i$ and put $\gg_i=\Lie G_i$.
Note that $\gg_i$ is $H$-semisimple, since it is a $G$-submodule of $\gg$.
It is easily checked that for each $i$ the set of automorphisms of $\gg_i$ given by $H$
is the same as that given by $\pi_i(\mu^{-1}(H))$. So $\pi_i(\mu^{-1}(H))$ acts
semisimply on $\gg_i$ for each $i$. Thanks to \cite[Lem.\ 2.12]{BMR},
we may now assume that $G$ is simple.

First assume that $G$ is not of type $A$.
Then $H$ is $G$-cr by Theorem \ref{thm:verygood}.

Now assume that $G$ is of type $A_{n-1}$.
Since $\gg$ is not of
intermediate type, the isogenies $\varphi:\SL_n\to G$ and
$\psi:G\to\PGL_n$ cannot both be inseparable. So $\varphi^{-1}(H)$
acts semisimply on $\mf{sl}_n$ or $\psi(H)$ acts semisimply on
$\mf{pgl}_n$. By \cite[Lem.\ 2.12(ii)(b)]{BMR}, we may assume that
$G\cong\SL_n$ or that $G\cong\PGL_n$.

First assume that $G=\SL_n$.  We have $G_{\rm ad}=\PGL_n$ and
$\gg_{\rm ad}=\mf{pgl}_n$.  The trace form on $\gl_n$ is
non-degenerate and induces a non-degenerate $\GL_n$-invariant pairing between
$\mf{sl}_n$ and $\mf{pgl}_n$.  This shows that
$\mf{pgl}_n\cong\mf{sl}_n^*$ as $\GL_n$-modules and therefore that $H$
acts semisimply on $\gg_{\rm ad}$.  More generally, thanks to
\cite{BoSe}, if $M$ is any reductive subgroup of $G$ containing a
maximal torus of $G$, then $M^0$ is a Levi subgroup of $G$ (see also
\cite[Ex.\ Ch.\ VI \S 4.4]{bou}).  Regarding $G$ as a subgroup of
$\GL_n$, we can write $M^0= (\prod_{i=1}^r\GL_{n_i})\cap\SL_n$ and
$\mm= (\oplus_{i=1}^r\gl_{n_i})\cap\mf{sl}_n$ for some $n_i$.  Since
the restriction of the trace form of $\gl_n$ to
$\oplus_{i=1}^r\gl_{n_i}$ is non-degenerate --- it is the direct sum
of the trace forms of the $\gl_{n_i}$ --- the orthogonal complement of
$\mm $ in $\oplus_{i=1}^r\gl_{n_i}$ is 1-dimensional and is therefore
equal to $k\cdot {\rm id}_n$. So $\mm^*\cong
(\oplus_{i=1}^r\gl_{n_i})/(k\cdot{\rm id}_n)$ as $M$-modules.  Now
$\mm_{\rm ad}=\oplus_{i=1}^r\mf{pgl}_{n_i}$, which is isomorphic to a
quotient of $\mm^*=(\oplus_{i=1}^r\gl_{n_i})/(k\cdot {\rm id}_n)$.
\begin{comment}
\hfill\break BEGIN COMMENT\\ If $L =
(\GL_{n_1}\times\cdots\times\GL_{n_r})\cap\SL_n$ is a Levi of $\SL_n$,
then $\Lie L = \oplus_{i=1}^r\gl_{n_i}\cap\mf{sl}_n$. Since the
restriction of the trace form of $\gl_n$ to $\oplus_{i=1}^r\gl_{n_i}$
is non-degenerate (it is the direct sum of the trace forms of the
$\gl_{n_i}$), the orthogonal of $\Lie L $ in $\oplus_{i=1}^r\gl_{n_i}$
is one dimensional and therefore equal to $k{\rm id}_n$. So $\Lie
L^*\cong (\oplus_{i=1}^r\gl_{n_i})/k{\rm id}_n$ as
$N_{\GL_n}(\GL_{n_1}\times\cdots\times\GL_{n_r})$-modules.  Now $\Lie
\Ad(L)=\oplus_{i=1}^r\mf{pgl}_{n_i}$ which is clearly isomorphic to a
quotient of $(\oplus_{i=1}^r\gl_{n_i})/k{\rm id}_n$.\\ END COMMENT\\
\end{comment}
Thus if $H$ acts semisimply on $\gg$, then $H$ acts semisimply on
$\mm_{\rm ad}$ for any reductive subgroup $M$ which contains
$H$ and a maximal torus of $G$.  The result now follows from Theorem
\ref{thm:allregss}.

Finally, assume that $G$ is isomorphic to $\PGL_n$.  Let
$\varphi:\SL_n\to\PGL_n$ be the canonical projection.  Since
$\mf{pgl}_n\cong\mf{sl}_n^*$ as $\GL_n$-modules, $\varphi^{-1}(H)$
acts semisimply on $\mf{sl}_n$.  The desired result follows from the
previous case and \cite[Lem.\ 2.12(ii)(b)]{BMR}.
\end{proof}

We record two important special cases of Theorem \ref{thm:adjsc}.

\begin{cor}\label{cor:adjc}
Assume that $G$ is connected, $p$ is good for $G$, and $DG$ is either
adjoint or simply connected.  Let $H$ be a subgroup of $G$
which acts semisimply on $\Lie DG$. Then $H$ is $G$-completely
reducible.
\end{cor}

\begin{rems}
(i). Note that Theorem~\ref{thm:adjsc} applies in cases when Theorem~\ref{thm:adjcr}
does not. For example, let $G = \SL_2(k)$, where $k$
has characteristic $2$, let $T$ be a maximal torus of $G$, and set $H
= N_G(T)$.  Then $\gg$ is semisimple as an $H$-module, so Theorem
\ref{thm:adjsc} applies, but, as $H$ is not separable in $G$ (cf.\
\cite[Ex.\ 3.4(b)]{martin3}), Theorem \ref{thm:adjcr} does not apply.
We can also take $G=\GL_2(k)$, $k$ of characteristic $2$, and $H =
N_G(T)$, where $T$ is a maximal torus of $G$. Then $\Lie DG =
\mf{sl}_2$ is a semisimple $H$-module, but $\gg = \gl_2$ is not
semisimple as an $H$-module.

(ii). Let $H$ be a group and let $\rho:H\to\GL(V)$ be a finite-dimensional representation of $H$.  We have $V\otimes V^*\cong\gl(V)$
as $\GL(V)$-modules and therefore also as $H$-modules. Furthermore, $V$
is $H$-semisimple if and only if $\rho(H)$ is $\GL(V)$-cr. So, by
Theorem~\ref{thm:adjsc} (or Theorem \ref{thm:adjcr}), $V$ is
$H$-semisimple if $V\otimes V^*$ is $H$-semisimple. This result is a
special case of a theorem of Serre, see \cite[Thm.~3.3]{serre0}.

(iii). We do not know whether the assumptions other than $\gg$
being $H$-semisimple are necessary in Theorem~\ref{thm:adjsc}.
To show that the assumption on the simple type $A$ factors
can be removed, it would
suffice to show that, under the assumption that $p^2|n$, a subgroup $H$ of $\GL_n$ which acts
semisimply on $\mathfrak{psl}_n$
is $\GL_n$-cr --- that is, acts semisimply on the natural
module $k^n$.
\begin{comment}
Conversely, to show that the assumption on the simple type $A$
factors cannot be removed, it
suffices to show that the above assertion is false when $p^2|n$.
\end{comment}

(iv). It is easy to prove that $k^n$ is
$H$-semisimple if $\mf{pgl}_n$ (equivalently, $\mf{sl}_n$) is $H$-semisimple.
The arguments are straightforward modifications of the arguments at the
end of the proof of Weyl's theorem in \cite[Thm.~6.3]{Hum2} and work for an arbitrary field $k$.
So one can prove Theorem~\ref{thm:adjsc} without using Theorem~\ref{thm:allregss}.
\end{rems}

We finish this section with an example which illustrates that the
converse of Theorem \ref{thm:adjcr} is false.

\begin{exmp}
\label{exmp:Serre}
In \cite[Cor.\ 5.5]{serre2}, Serre
shows that $\gg$ is a semisimple $H$-module if and only if
$H$ is $G$-completely reducible, under the assumption that
$p > 2h-2$, where $h$ is the Coxeter number of $G$.
Theorem \ref{thm:verygood} improves the bound for the forward implication
of this equivalence.
However, Serre's bound is sharp for the reverse implication,
as the following example (due to some unpublished calculations of Serre)
shows, see also \cite[Rem.~3.43(iii)]{BMR}.

Let $H$ be an exceptional
simple group of adjoint type and suppose
$p \leq 2h-2$, where $h$ denotes the Coxeter number of $H$.
Then, with the possible exception of type $G_2$ in characteristic $5$,
Serre's construction yields a simple subgroup
$K$ of $H$ of type $A_1$ such that
$K$ is $H$-irreducible, but $K$ does not act
semisimply on $\hh$ --- that is, $K$ is not
$G$-completely reducible, where $G:=\GL(\hh)$.

This example also illustrates that the converse of the second
assertion of Theorem \ref{thm:redpairs} is not true in general,
even when $K$ is separable in $G$ and $(G,H)$
is a reductive pair.
For we can choose $p$ and $H$ above such that $p$ is
good for $H$; then $(\GL(\hh),H)$ is a reductive
pair (cf.\ \cite[Ex.\ 3.37]{BMR}), since the Killing form on $\hh$ is
non-degenerate \cite[\S 5]{rich2}.  It
then follows from Theorem \ref{thm:redpairs}
that $K$ is separable in $H$, since $K$ is separable in $\GL(\hh)$.
\end{exmp}

\section{Closed Orbits and Separability}
\label{sec:orbitthm}

In this section we generalize two results of R.W.~Richardson
\cite[Thm.\ A, Thm.\ C]{rich0},
which we are then able to apply to $G$-complete
reducibility.
We require two preliminary results,
which allow us to relax the hypotheses in the
original theorems.
Armed with these lemmas, it is quite straightforward to adapt
Richardson's proofs to our setting.

Our first result is an analogue of \cite[Prop.\ 6.1]{rich0},
which Richardson proves for $S$ linearly reductive
and $G$ connected, using non-commutative cohomology of
algebraic groups.
It should be noted that, although our result is more general
than Richardson's, in order to apply it to linearly reductive groups
one needs to know that they are always completely reducible in
any ambient reductive group (\cite[Lem.\ 2.6]{BMR}),
and the proof of this depends on the argument that Richardson uses.

\begin{lem}
\label{lem:Sstable}
Let $G$ be a normal subgroup of the reductive group $G'$ and
suppose that $S$ is a $G'$-completely reducible subgroup of $G'$.
Then for every R-parabolic subgroup $P$ of $G$ normalized by $S$,
there exists an R-Levi subgroup of $P$ normalized by $S$.
\end{lem}

\begin{proof}
Note that since $G$ is normal in $G'$, $G$ is reductive,
and $G^0$ is a connected reductive normal subgroup of $G'$.
We first show that it is enough to prove the result when $G$ is connected.
If $P$ is an R-parabolic subgroup of $G$ normalized by $S$,
then $P^0$ is a parabolic subgroup of $G^0$ normalized by $S$.
If $L$ is a Levi subgroup of $P^0$ normalized by $S$, then
$L = M^0$ for some R-Levi subgroup $M$ of $P$ \cite[Cor.~6.8]{BMR}.
Let $T$ be a maximal torus of $M$; then $T \subseteq L$.
Since $S$ normalizes $P$, it follows that $sMs\inverse$ is
an R-Levi subgroup of $P$ for any $s \in S$.
Moreover, since $S$ normalizes $L$, we see that
$T \subseteq L= sLs\inverse\subseteq sMs\inverse$.
This shows that $M = sMs\inverse$, for every $s \in S$,
by \cite[Cor.\ 6.5]{BMR},
and therefore $S$ normalizes $M$.
Thus, if the result is true for $G^0$, then it is true for $G$.

Now suppose $G$ is connected, and let $H = C_{G'}(G)^0$.
Since $G$ is normal in $G'$, we have $G'^0 = GH$, by
\cite[Lem.\ 6.8]{martin1}.  Let $P$ be a parabolic
subgroup of $G$ normalized by $S$.  Then $P\times H$
is a parabolic subgroup of $G\times H$, and it follows
from \cite[Prop.~IV.11.14(1)]{borel} applied to the
multiplication map $f\colon G\times H\ra G'^0$ that
$PH$ is a parabolic subgroup of $G'^0$.  As $S$
normalizes $P$, $S$ normalizes $PH$, so $S\subseteq N_{G'}(PH)$,
which is an R-parabolic subgroup of $G'$ by \cite[Prop.\ 5.4(a)]{martin1}.
Now $S$ is $G'$-cr by hypothesis, so $S$ is contained in some R-Levi
subgroup $M$ of $N_{G'}(PH)$, so $S$ normalizes $M^0$, which is a Levi
subgroup of $N_{G'}(PH)^0=PH$.  By \cite[Lem.~6.15(ii)]{BMR}, $f^{-1}(PH)$
is a parabolic subgroup of $G\times H$ with Levi subgroup $f^{-1}(M^0)$.

If $(g,h)\in f^{-1}(PH)$, then $gh=g'h'$ for some $g'\in P$
and some $h'\in H$,
so $(g')^{-1}g= h'h^{-1}\in G\cap H\subseteq Z(G)\subseteq P$, so $g\in P$.
It follows that $f^{-1}(PH)=P\times H$.  Hence $f^{-1}(M^0)= L\times H$
for some Levi subgroup $L$ of $P$ and hence $M^0=LH$.
A similar calculation shows that $L=M^0\cap G$, so $S$
normalizes $L$ and we are done.
\end{proof}

\begin{comment}
\begin{proof}
Note that since $G$ is normal in $G'$, $G$ is reductive,
and $G^0$ is a connected reductive normal subgroup of $G'$.
We first show that it is enough to prove the result when $G$ is connected.
If $P$ is an R-parabolic subgroup of $G$ normalized by $S$,
then $P^0$ is a parabolic subgroup of $G^0$ normalized by $S$.
If $L$ is a Levi subgroup of $P^0$ normalized by $S$, then
$L = M^0$ for some R-Levi subgroup $M$ of $P$ \cite[Cor.~6.8]{BMR}.
Let $T$ be a maximal torus of $M$; then $T \subseteq L$.
Since $S$ normalizes $P$, it follows that $sMs\inverse$ is
an R-Levi subgroup of $P$ for any $s \in S$.
Moreover, since $S$ normalizes $L$, we see that
$T \subseteq L= sLs\inverse\subseteq sMs\inverse$.
This shows that $M = sMs\inverse$, for every $s \in S$,
by \cite[Cor.\ 6.5]{BMR},
and therefore $S$ normalizes $M$.
Thus, if the result is true for $G^0$, then it is true for $G$.

Now suppose $G$ is connected, and let $H = C_{G'}(G)^0$.
Since $G$ is normal in $G'$, $G'^0 = GH$, by
\cite[Lem.\ 6.8]{martin1}.
Let $P$ be a parabolic subgroup of $G$.
Then $P = P_\lambda(G)$ for some $\lambda \in Y(G)$;
moreover, the image of $\lambda$ centralizes $H$,
since $H$ centralizes $G$,
so $PH = P_\lambda(G'^0)$ is a parabolic subgroup of $G'^0$,
and $H \subseteq L_\lambda(G'^0) = L_\lambda(G)H$.
In fact, since $H$ is centralized by $\lambda$,
$R_u(PH) = R_u(P) \subseteq G$ also centralizes $H$.
Thus $H$ is contained in
every Levi subgroup of $PH$, and if $M$ is
a Levi subgroup of $PH$, then $M = LH$, where
$L = M\cap G$ is a Levi subgroup of $P$.
It is also clear that $N_{G'}(P) \subseteq N_{G'}(PH)$,
since all of $G'$ normalizes $H$.
On the other hand, suppose $g \in N_{G'}(PH)$ and $x \in P$.
Then $gxg\inverse \in G \cap PH = P(G\cap H) \subseteq PZ(G) = P$,
since $Z(G) \subseteq P$.
Thus $N_{G'}(P) = N_{G'}(PH)$, which is an R-parabolic subgroup of $G'$,
by \cite[Prop.\ 5.4(a)]{martin1}.
Also note that $N_{G'}(PH)^0 = PH$, since $PH$ is a parabolic
subgroup of the connected reductive group $G'^0$, and is therefore equal
to its own normalizer in $G'^0$.

Finally, suppose $P$ is normalized by $S$.
Then $S \subseteq N_{G'}(P)$, and $S$ is $G'$-cr, so
there exists an R-Levi subgroup $M$ of $N_{G'}(P)$
with $S \subseteq M$.
Thus $S$ normalizes $M^0$, which is a Levi subgroup of $PH$.
But $M^0 \cap G$ is then a Levi subgroup of $P$ normalized by $S$,
as required.
\end{proof}

\end{comment}

\begin{lem}
\label{lem:richC}
Let $S$ be a group and let $f : V\to W$ be a surjective homomorphism of
finite-dimensional $S$-modules over some field. If $V/V^S$ has no trivial
composition factor, then $f(V^S)=W^S$.
\end{lem}

\begin{proof}
Since $f$ is surjective, it induces a surjective homomorphism
$V/V^S\to W/f(V^S)$ of $S$-modules. Since $V/V^S$ has no trivial
composition factor, neither does $W/f(V^S)$. In
particular, $W/f(V^S)$ does not have any non-zero $S$-fixed points.
So $W^S$ must be equal to $f(V^S)$.
\end{proof}

\begin{rem}
\label{rem:richC}
If $V$ is $S$-semisimple ---
in particular, if $S$ is a linearly reductive algebraic group --- then  $V/V^S$ has no
trivial composition factor, so Lemma \ref{lem:richC} applies.
\end{rem}

We can now state our generalization of Richardson's two results
\cite[Thm.\ A, Thm.\ C]{rich0}.
Observe that if $S$ is a subgroup of $G$, then
the condition on $S$ in Theorem \ref{thm:richC}(b)(ii)
is just that of separability of $S$ in $G$.

\begin{thm}
\label{thm:richC}
Let $G$ be a normal subgroup of the reductive group $G'$.  Let $S$ be
a subgroup of $G'$ such that $G'= GS$.  Let $G'$ act on a
variety $X$ and let $x\in X^S$.  Let $\mathcal O$ denote the unique
closed $G$-orbit in the closure of $G\cdot x$.
\begin{itemize}
\item[(a)] Suppose that $X$ is affine and $S$ is $G'$-completely reducible.  Then
$C_G(S)$ is $G$-completely reducible and there exists $\lambda\in
Y(C_G(S))$ such that $\underset{a\to 0}{\lim}\, \lambda(a)\cdot x$
exists and is a point of $\mathcal O$.  In particular, $G\cdot x$ is
closed if $C_G(S)\cdot x$ is closed.
\item[(b)] Suppose that
\begin{itemize}
\item[(i)] $\gg/\cc_\gg(S)$ does not have any trivial $S$-composition
factors;
\item[(ii)] $\cc_\gg(S)= \Lie C_G(S)$.
\end{itemize}
Then each
irreducible component of $G\cdot x \cap X^S$ is a single
$C_G(S)^0$-orbit. In particular, $C_G(S)\cdot x$ is closed if $G\cdot
x$ is closed.
 \end{itemize}
\end{thm}

\begin{proof}
(a). Suppose that $S$ is $G'$-cr. Then $C_G(S)$ is $G$-cr, by the same
arguments as in the proof of Proposition~\ref{prop:C_G(S)redpair}(a).
If $G\cdot x$ is closed, then we can take $\lambda$ to be the zero
cocharacter, so assume that $G\cdot x$ is not closed.  Then $G^0\cdot
x$ is not closed.  Moreover, the unique closed $G$-orbit $\mathcal{O}$
in the closure of $G\cdot x$ contains the unique closed $G^0$-orbit in
the closure of $G^0\cdot x$.  Thus, since $Y(G) = Y(G^0)$ and
$Y(C_G(S)) = Y(C_{G^0}(S))$, we may replace $G$ by $G^0$ and assume
that $G$ is connected.

We can now follow the first part of Richardson's original proof in
\cite[Sec.~8]{rich0} word for word to give the result, simply
replacing references to \cite[Prop.\ 6.1]{rich0} with our Lemma
\ref{lem:Sstable}.

\begin{comment}
(a).  We first observe that for $x \in X^S$, the unique closed
$G$-orbit $\mathcal O$ in the closure of $G\cdot x$ is equal to the
unique closed $G'$-orbit in the closure of $G'\cdot x$.  This follows
since $x \in X^S$ implies $G'\cdot x = G\cdot x$, hence
$\overline{G'\cdot x} = \overline{G\cdot x}$.  In particular,
$\overline{G\cdot x}$ is $S$-stable.  Let $y \in \mathcal O$; then for
any $s \in S$,
\[
s\cdot \mathcal O = \{ s\cdot(g\cdot y) \mid g \in G\} = \{sgs\inverse
\cdot (s\cdot y) \mid g \in G\} = G \cdot (s\cdot y)
\]
is a closed $G$-orbit in $\overline{G\cdot x}$, and thus equals
$\mathcal O$. So $\mathcal O$ is $S$-stable and, therefore, $\mathcal
O$ is the unique closed $G'$-orbit in the closure of $G'\cdot x$.

Suppose that $S$ is $G'$-cr.  Then $C_{G'}(S)$ is $G'$-cr by
\cite[Cor.\ 3.17]{BMR}.  So the normal subgroup $C_G(S)$ of
$C_{G'}(S)$ is also $G'$-cr, \cite[Thm.\ 3.10]{BMR}.  Now $C_G(S)$ is
$G$-cr by \cite[Prop.\ 3.8]{BMR2}.

If $G\cdot x$ is closed, then we can take $\lambda$ to be the zero
cocharacter, so assume that $G\cdot x$ is not closed.  Then $G'\cdot
x= G\cdot x$ is not closed.  Let $P = P(x)$ be the optimal
destabilizing parabolic subgroup of $G'$ corresponding to $x$ (see
Subsection \ref{subsec:kempf}).  Since $x \in X^S$, we have $S
\subseteq C_{G'}(x) \subseteq P$. Since $S$ is $G'$-cr, there exists
an R-Levi subgroup $L$ of $P$ with $S \subseteq L$.

Now $L = L_\mu$ for one and only one $\mu$ in the optimal class
$\Omega(x) \subseteq Y(G')$ associated to $x$. For any $s \in S$,
$s\cdot \Omega(x) = \Omega(s\cdot x) = \Omega(x)$, and hence $s \cdot
\mu \in \Omega(x)$, cf.\ \cite[Lem.\ 7.3]{rich0}.  But $L_{s\cdot \mu}
= s L_\mu s\inverse = L_\mu$, since $S\subseteq L$.  Thus we must have
$s\cdot \mu = \mu$ for all $s \in S$, so that $\mu \in Y(C_{G'}(S)) =
Y(C_{G'}(S)^0)$.  There exist commuting tori $T$, $A$ of $C_G(S)$,
$Z(S)$ respectively, there exist $\lambda\in Y(T)$, $\sigma \in Y(A)$
such that $\mu= \lambda+ \sigma$ (cf.\ \cite[Sec.~6.2]{BMR}).

{\tt R and M: still check!\\ R: We could probably make our life a lot
easier by assuming that $G'=G\rtimes S$.\\ It is of course not clear
that $S$ $G'$-cr implies $S$ $G\rtimes S$-cr.\\ R:Maybe we should
apply Kempf to $G$ rather than $G'$ as Richardson does and work with
an $S$-invariant length function. Does $S$ $G'$-cr imply $S$ $G$-cr in
the sense of Richardson: For every $S$-stable $R$-parabolic....?\\ The
assumption that $S$ is $G$-cr in this sense is much more natural.}\\

We have $\mu(a)\cdot x= \lambda(a)\cdot x$ for all $a\in k^*$.  Thus
$\underset{a\to 0}{\lim}\, \lambda(a)\cdot x= \underset{a\to
0}{\lim}\, \mu(a)\cdot x$ exists and is a point of $\mathcal O$.
\end{comment}

(b). Now assume that $\gg/\cc_\gg(S)$ does not have any trivial
$S$-composition factors and that $\cc_\gg(S)= \Lie C_G(S)$.  We can
write $G\cdot x$ as a finite union ${\mathcal
O}_1\cup\cdots\cup{\mathcal O}_n$ of $G^0$-orbits. Each ${\mathcal
O}_i$ is a $G$-translate of $G^0\cdot x$, from which we deduce that
the ${\mathcal O}_i$ are the irreducible components of $G\cdot x$. We
may assume that ${\mathcal O}_1,\ldots,{\mathcal O}_r$ $(r\le n)$, are
the $G^0$-orbits in $G\cdot x$ that meet $X^S$. Then $G\cdot x\cap
X^S=({\mathcal O}_1\cap X^S)\cup\cdots\cup({\mathcal O}_r\cap X^S)$, a
disjoint union of closed subsets of $G\cdot x$. This shows that, after
replacing $G'$ by $G^0S$, we may assume that $G$ is connected.

Note that $G\cdot x$ is $S$-stable. Put $Y=(G\cdot x)^S = G\cdot x\cap X^S$.
By the argument at the end of the proof of \cite[Thm.\ A]{rich0},
we may assume that the orbit map $G\to G\cdot x$ of $x$ is separable:
Let $y\in Y$.
Then the orbit map of $y$ is separable: that is, its differential
$\psi:\gg\to T_y(G\cdot y)$
is surjective. Since the orbit map of $y$ is $S$-equivariant,
$\psi$ is $S$-equivariant.
By Lemma~\ref{lem:richC}, we have $\psi(\cc_\gg(S))=(T_y(G\cdot y))^S$.
Clearly, $T_y(Y)\subseteq (T_y(G\cdot x))^S$,
so $T_y(Y)\subseteq\psi(\cc_\gg(S))$. Since $\cc_\gg(S)=\Lie C_G(S)$,
the Tangent Space Lemma \cite[Lem.~3.1]{rich0} gives the first assertion of (b).

It now follows that $C_G(S)^0\cdot x$, and therefore $C_G(S)\cdot x$,
is closed in $G\cdot x$.
So if $G\cdot x$ is closed, then $C_G(S)\cdot x$ is closed.
\end{proof}

\begin{rems}\label{rems:richC}
  (i). Assume that $S$ is a separable subgroup of
  $G$ which acts semisimply on $\gg$.
  Then $S$ is $G$-cr by Theorem \ref{thm:adjcr}.
  So parts (a) and (b) of Theorem~\ref{thm:richC} both apply.
  Thus in this case $C_G(S)\cdot x$ is closed if and
  only if $G\cdot x$ is closed.

  (ii). Suppose that $S$ is linearly reductive.  Then conditions (i)
and (ii) of Theorem \ref{thm:richC}(b) are satisfied (see
\cite[Lem.~4.1]{rich0}), and $S$ is $G'$-cr \cite[Lem.~2.6]{BMR};
the special case of
Theorem \ref{thm:richC} for $S$ linearly reductive is precisely
Richardson's original theorem \cite[Thm.\ C]{rich0} (clearly, the
hypothesis in Theorem \ref{thm:richC} that $G'=GS$ is harmless).

(iii). It follows from the proof of Theorem \ref{thm:richC}(b)
that \cite[Thm.~A]{rich0} holds with the hypothesis that $S$
acts semisimply on $\gg$ replaced by the weaker hypothesis of
Theorem \ref{thm:richC}(b)(ii).

(iv). The result in Theorem \ref{thm:richC}(a) that the $G$-orbit
is closed if the $C_G(S)$-orbit is closed is also a slight generalization of part of
\cite[Thm.\ 4.4]{bate}.

(v). Theorem \ref{thm:richC}(b) does not hold for an arbitrary
$G'$-completely reducible subgroup $S$.  In fact it does not even hold
when $S$ is a $G$-completely reducible subgroup of $G$: in
\cite[Prop.~3.9]{BMR2} it is shown that if $S$ is a $G$-completely
reducible subgroup of $G$ and $K$ is a subgroup of $C_G(S)$, then $K$
is $C_G(S)$-completely reducible if and only if $KS$ is $G$-completely
reducible.  Now \cite[Examples 5.1, 5.3, 5.5]{BMR2} give instances of
commuting $G$-completely reducible subgroups $K$ and $S$ such that
$KS$ is not $G$-completely reducible, whence $K$ is not
$C_G(S)$-completely reducible.  If the field $k$ is large enough (cf.\
\cite[Rem.~2.9]{BMR}), we can pick an $n$-tuple $(k_1, \ldots, k_n) \in
G^n$ topologically generating $K$ for some $n$.  Then by Theorem
\ref{thm:gcrcrit}, $G\cdot(k_1, \ldots, k_n)$ is closed in $G^n$, but
$C_G(S)\cdot(k_1, \ldots, k_n)$ is not.
\end{rems}

\vfill
\eject

\begin{comment}
{\tt R: Note that, because of the lemma on ascending sequences of
finite subgroups in the relative preprint, Corollary \ref{cor:SnormalizesK}
is valid without the assumption that $S$ is reductive
and Corollary \ref{cor:richC} is valid without the assumption that $S$ is
reductive in case char $=p>0$.}
\end{comment}

\begin{cor}
\label{cor:richC}
Let $G$, $G'$, $S$, $X$ and $x \in X^S$ be as in
Theorem~\ref{thm:richC}, and suppose $X$ is affine. Suppose that
\begin{itemize}
\item[(i)] the adjoint representation of $S$ on $\gg$ is semisimple;
\item[(ii)] $\cc_\gg(S)= \Lie C_G(S)$.
\end{itemize}
Then $S$ is reductive, $C_G(S)$ is $G$-completely reducible and $G \cdot x$ is closed if
and only if $C_G(S) \cdot x$ is closed.
\end{cor}

\begin{proof}
 Since $GS=G'$, the quotient $S/(S\cap G^0)$ is isomorphic to a finite-index subgroup of $G'/G^0$, and hence is reductive.  Now $S\cap G^0$ is normal in $S$, so $S\cap G^0$ acts semisimply on $\gg$ by Clifford's Theorem.  The kernel $N$ of the $(S\cap G^0)$-action on $\gg$ is a subgroup of the diagonalizable group $Z(G^0)$ and is therefore reductive. Since $N$ and $(S\cap G^0)/N$ are reductive, $S\cap G^0$ is reductive, and it follows that $S$ is reductive.

Suppose $\Char k=0$.  Note that, in this case, properties
(i) and (ii) and the $G'$-complete reducibility of $S$ follow from the fact that $S$ is reductive.  Therefore, the hypotheses of
Theorem~\ref{thm:richC}(a) and (b) are satisfied.

Now suppose $\Char k=p >0$.
By the same arguments as in the proof of
Proposition~\ref{prop:C_G(S)redpair}(a),
we may assume that $S$ is finite and deduce that then $S$ is $G'$-cr.
Thus, the hypotheses of Theorem~\ref{thm:richC}(a) and (b) are again
satisfied.
\end{proof}

\begin{comment}
\begin{cor}
\label{cor:richC}
Let $G$, $G'$, $S$, $X$ and $x \in X^S$ be as in
Theorem~\ref{thm:richC}. Suppose that
\begin{itemize}
\item[(i)] $S$ is reductive;
\item[(ii)] the adjoint representation of $S$ on $\gg$ is semisimple;
\item[(iii)] $\cc_\gg(S)= \Lie C_G(S)$.
\end{itemize}
Then $C_G(S)$ is $G$-completely reducible and $G \cdot x$ is closed if
and only if $C_G(S) \cdot x$ is closed.
\end{cor}

\begin{proof}
First suppose $\Char k=0$. Note that, in this case, properties (ii) and
(iii) follow from (i).  Furthermore, $S$ is $G'$-cr, since $S$ is
reductive and $\Char k=0$.  Therefore, the hypotheses of
Theorem~\ref{thm:richC}(a) and (b) are satisfied.

Now suppose $\Char k=p >0$.
By the same arguments as in the proof of
Proposition~\ref{prop:C_G(S)redpair}(a),
we may assume that $S$ is finite and deduce that then $S$ is $G'$-cr.
Thus, the hypotheses of Theorem~\ref{thm:richC}(a) and (b) are again
satisfied.
\end{proof}
\end{comment}

\begin{comment}
\begin{cor}
\label{cor:linred}
Let $G$ and $S$ satisfy the hypotheses of Corollary~\ref{cor:richC}.
Let $(k_1, \ldots, k_n) \in G^n$ such that $S$ permutes the $k_i$.
Then $G \cdot (k_1 , \ldots, k_n)$ is closed in $G^n$ if and only if
$C_G(S) \cdot (k_1 , \ldots, k_n)$ is closed in $G^n$.
\end{cor}

\begin{proof}
The symmetric group $S_n$ acts naturally on $G^n$, and the $G$-action
commutes with this action.  Set $X = G^n/S_n$ and let $\pi: G^n \to X$
be the natural map.  For any subgroup $M$ of $G$ and any $(g_1 ,
\ldots, g_n) \in G^n$, we have $M\cdot (g_1 , \ldots, g_n)$ is closed
in $G^n$ if and only if $M\cdot \pi((g_1 , \ldots, g_n))$ is closed in
$X$.
See \cite[\S 2]{BaRi} for more details and background on this type of
construction in Geometric Invariant Theory.

Now let $(k_1 , \ldots, k_n)$ be as in the statement of the corollary.
Since $S$ permutes the $k_i$, the point $x:= \pi((k_1 , \ldots, k_n))$
belongs to $X^S$.  By Corollary~\ref{cor:richC}, $G\cdot x$ is closed
if and only if $C_G(S)\cdot x$ is closed.  Thus $G\cdot (k_1 , \ldots,
k_n)$ is closed if and only if $C_G(S)\cdot (k_1 , \ldots, k_n)$ is
closed, as desired.
\end{proof}
\end{comment}

\begin{comment}
\begin{rem}
\label{rem:relative}
Note that if we let $K$ be the subgroup of $G$ topologically generated
by the $n$-tuple $(x_1,\ldots,x_n)$ in Corollary~\ref{cor:richC}, then
this result shows that $K$ is $G$-completely reducible if and only if
$C_G(S) \cdot (x_1,\ldots,x_n)$ is closed.  In other words, to
``test'' for $G$-complete reducibility of the subgroup $K$, one does
not need to consider every R-parabolic subgroup of $G$ containing $K$,
just those R-parabolic subgroups of $G$ which stem from cocharacters
evaluating in $C_G(S)$; we say that $K$ is \emph{relatively
$G$-completely reducible with respect to $C_G(S)$}.
\end{rem}
\end{comment}

Now we translate our results in terms of $G$-complete reducibility.

\begin{comment}
\begin{prop}
\label{prop:SnormalizesK}
Let $G$, $G'$ and $S$ be as in Theorem \ref{thm:richC}.  Let $K\subseteq H$ be
subgroups of $G$ such that $S$ normalizes $K$ and $H^0 = C_G(S)^0$.
\begin{itemize}
\item[(a)] Suppose that $S$ is $G'$-completely reducible.  Then
$H$ is reductive and $K$ is $G$-completely reducible if it is $H$-completely reducible.
\item[(b)] Suppose that
\begin{itemize}
\item[(i)] $\gg/\cc_\gg(S)$ does not have any trivial
$S$-composition factors;
\item[(ii)] $\cc_\gg(S)= \Lie C_G(S)$;
\item[(iii)] $C_G(S)$ is reductive;
\item[(iv)] $S$ is reductive.
\end{itemize}
Then $K$ is $H$-completely reducible if it is $G$-completely reducible.
\end{itemize}
\end{prop}
\end{comment}

\begin{prop}
\label{prop:SnormalizesK}
Let $G$ be a normal subgroup of the reductive group $G'$. Let $S$ be
a subgroup of $G'$ such that $G'= GS$. Let $K\subseteq H$ be
subgroups of $G$ such that $H^0 = C_G(S)^0$ and $S$ normalizes $K$.
\begin{itemize}
\item[(a)] Suppose that $S$ is $G'$-completely reducible.  Then
$H$ is reductive and $K$ is $G$-completely reducible if it is $H$-completely reducible.
\item[(b)] Suppose that
\begin{itemize}
\item[(i)] $\gg/\cc_\gg(S)$ does not have any trivial
$S$-composition factors;
\item[(ii)] $\cc_\gg(S)= \Lie C_G(S)$;
\item[(iii)] $C_G(S)$ is reductive.
\end{itemize}
Then $H$ is reductive and $K$ is $H$-completely reducible if it is $G$-completely reducible.
\end{itemize}
\end{prop}

\begin{proof}
In case (a), $C_G(S)$ is reductive by Theorem~\ref{thm:richC}(a) and in
case (b), this is true by assumption. Since $H^0 = C_G(S)^0$, it follows
that $H$ is reductive.

Clearly, we can now assume that $\Char k=p>0$.
By the argument in the proof of
Proposition~\ref{prop:C_G(S)redpair}(a),
we may also assume that $S$ is finite.  As
in \cite[Lem.~2.10]{BMR}, we can replace $K$ with a subgroup $K'$ that
is topologically generated by some $k_1, \ldots, k_n$ with the property
that for any $\lambda\in Y(G)$, we have $K\subseteq P_\lambda$ if and
only if $K'\subseteq P_\lambda$, and $K\subseteq L_\lambda$ if and
only if $K'\subseteq L_\lambda$.  Thus $K$ is $G$-cr (respectively
$H$-cr) if and only if $K'$ is $G$-cr (respectively $H$-cr).

Since $S$ is finite, we may assume, by replacing $(k_1, \ldots, k_n)$
with a larger tuple if necessary, that $S$
permutes the $k_i$ and therefore that $S$ also normalizes $K'$. Since
$H^0 = C_G(S)^0$, we have that $H \cdot (k_1, \ldots, k_n)$ is closed
if and only if $C_G(S) \cdot (k_1, \ldots, k_n)$ is closed.

Let $G'$ act on $G^n$ by simultaneous conjugation.
The symmetric group $S_n$ acts naturally on $G^n$, and the $G'$-action
commutes with this action.  Set $X = G^n/S_n$ and let $\pi: G^n \to X$
be the natural map; the fibres of $\pi$ are precisely the $S_n$-orbits (see \cite[Sec.~2]{BaRi}, for example).  For any subgroup $M$ of $G'$ and any $(g_1 ,
\ldots, g_n) \in G^n$, we have that $M\cdot (g_1 , \ldots, g_n)$ is closed
in $G^n$ if and only if $M\cdot \pi((g_1 , \ldots, g_n))$ is closed in
$X$. Put $x=\pi((k_1, \ldots, k_n))$. Then $x\in X^S$.
Assertions (a) and (b) now follow from
Theorem~\ref{thm:gcrcrit} and from Theorem~\ref{thm:richC}(a) and (b),
respectively.
\end{proof}

\begin{comment}
\begin{rem}
 The hypothesis that $S$ is reductive in Proposition \ref{prop:SnormalizesK} is unnecessary.  This is clear in characteristic 0 --- all that is needed is that $C_G(S)$ is reductive.  In characteristic $p>0$, it can be shown that {\bf any} algebraic group $S$ admits an ascending chain of finite subgroups $S_1\subseteq S_2\subseteq\cdots$
such that $\bigcup_{i\ge1}S_i$ is Zariski dense in $S$, so the argument of Proposition \ref{prop:C_G(S)redpair}(a) holds for non-reductive $S$ as well.
\end{rem}
\end{comment}

The next corollary is a generalization of \cite[Cor.\ 3.21]{BMR}.
Note that the hypotheses on $S$
in
Corollary \ref{cor:SnormalizesK} are satisfied if $S$
is linearly reductive.

\begin{cor}
\label{cor:SnormalizesK}
Let $S$ be
an algebraic group acting on $G$ by automorphisms.
Suppose that $S$ acts semisimply on $\gg$ and $\cc_\gg(S)=\Lie C_G(S)$.
Let $K\subseteq H$ be subgroups of $G$ such that $H^0 = C_G(S)^0$ and $S$ normalizes $K$.
Then $H$ is reductive and $K$ is $G$-completely reducible if and
only if $K$ is $H$-completely reducible.
\end{cor}

\begin{proof}
By Proposition~\ref{prop:C_G(S)redpair}(a), $C_G(S)$ is reductive.
So $H$ is also reductive.
We can now assume that $\Char k=p>0$.
By the argument in the proof of
Proposition~\ref{prop:C_G(S)redpair}(a),
we can assume that $S$ is finite. Then we put $G'=G\rtimes S$
and obtain, as in that proof, that $S$ is $G'$-cr and that $C_G(S)$ is $G$-cr.
Now the assumptions of Proposition~\ref{prop:SnormalizesK}(a) and (b) are satisfied.
\end{proof}

\begin{comment}
\begin{cor}
\label{cor:SnormalizesK}
Let $S$ be a reductive algebraic group acting on $G$ by automorphisms.
Suppose that $S$ acts semisimply on $\gg$ and $\cc_\gg(S)=\Lie C_G(S)$.
Let $K\subseteq H$ be subgroups of $G$ such that $S$ normalizes $K$
and $H^0 = C_G(S)^0$.
Then $H$ is reductive and $K$ is $G$-completely reducible if and
only if $K$ is $H$-completely reducible.
\end{cor}

\begin{proof}
By Proposition~\ref{prop:C_G(S)redpair}(a), $C_G(S)$ is reductive.
So $H$ is also reductive.
We can now assume that $\Char k=p>0$.
By the argument in the proof of
Proposition~\ref{prop:C_G(S)redpair}(a),
we may assume that $S$ is finite. Then we put $G'=G\rtimes S$
and obtain, as in that proof, that $S$ is $G'$-cr and that $C_G(S)$ is $G$-cr.
Now the assumptions of Proposition~\ref{prop:SnormalizesK}(a) and (b) are satisfied.
\end{proof}
\end{comment}

\section{Centralizers and Normalizers}
\label{sec:cent+norm}

\begin{comment}
In the previous section we looked at the transfer of
complete reducibility from $G$ to subgroups of $G$ of the form $C_G(S)$,
where $S$ is a group acting on $G$.
We now consider the situation that $S$ is a
subgroup of $G$ acting by inner automorphisms.
This additional requirement on $S$ allows us to
replace $C_G(S)$ by a subgroup $H$ that is closely related to
$C_G(S)$; our next result provides some information about what happens
when we just assume that $H$ contains $C_G(S)^0$, rather than $H^0 =
C_G(S)^0$, as in Proposition \ref{prop:SnormalizesK}.
The result also generalizes \cite[Prop.\ 3.9]{BMR2}, which
deals with the special case when $H=C_G(S)$; in fact, the argument we
use is very similar.
\end{comment}

In this section we continue the theme of Section \ref{sec:orbitthm}, looking at the special case when $S$ is a subgroup of $G$ acting on $G$ by inner automorphisms.  The extra restriction on $S$ allows us to consider subgroups $H$ sitting between $C_G(S)^0$ and $N_G(S)$.  The following result gives a criterion for $K$ to be $H$-cr; it generalizes \cite[Prop.\ 3.9]{BMR2} and the argument we use is very similar.

\begin{comment}
\begin{lem}
\label{lem:KN<=>K}
Suppose $S$ is a $G$-completely reducible subgroup of $G$, and suppose
$H$ is a subgroup of $G$ such that $C_G(S)^0 \subseteq H^0 \subseteq
N_G(S)$.  Then $HS$ is reductive and for any subgroup $K \subseteq H
\cap N_G(S)$, $K$ is $H$-completely reducible if and only if $KS$ is
$G$-completely reducible if and only if $KS$ is $N_G(S)$-completely
reducible if and only if $KS$ is $HS$-completely reducible.  If $H$ is
reductive, then $K$ is $H$-completely reducible if and only if $KS$ is
$G$-completely reducible.
\end{lem}
\end{comment}

\begin{comment}
{\tt R: I have now required that $H\subseteq N_G(S)$.
Note: If we only have $H^0\subseteq N_G(S)$, then why would we have $(HS)^0\subseteq N_G(S)$?
What does $HS$ mean in this situation? The closed subgroup generated by $H$ and $S$?}
\end{comment}

\begin{prop}
\label{prop:KN<=>K}
Suppose $S$ is a $G$-completely reducible subgroup of $G$, and suppose
$H$ is a subgroup of $G$ such that $C_G(S)^0 \subseteq H\subseteq
N_G(S)$.  Let $K$ be a subgroup of $H$.
Then:
 \begin{itemize}
  \item[(a)] $HS$ is reductive.
  \item[(b)] The following are equivalent:
 \begin{itemize}
\item[(i)] $KS$ is $G$-completely reducible;
\item[(ii)] $KS$ is $HS$-completely reducible;
\item[(iii)] $KS$ is $N_G(S)$-completely reducible;
\item[(iv)] $KS/S$ is $N_G(S)/S$-completely reducible.
 \end{itemize}
  \item[(c)] Suppose that $H$ is reductive.  Let $\psi$ be the
  canonical projection from $H$ to $H/C_G(S)^0$.  Then $K$ is
  $H$-completely reducible if and only if $KS/S$ is
  $N_G(S)/S$-completely reducible and $\psi(K)$ is
  $H/C_G(S)^0$-completely reducible.
 \end{itemize}
\end{prop}

\begin{proof}
(a). We have
  $C_G(S)^0S^0= N_G(S)^0$,
 by \cite[Lem.\ 6.8]{martin1}.
 As $C_G(S)^0\subseteq H$, we have $(HS)^0= N_G(S)^0$.
 Now $N_G(S)$ is reductive by \cite[Prop.\ 3.12]{BMR}, so $HS$ is reductive.

(b). By part (a), $(HS)^0 = N_G(S)^0$, so $HS$ is a finite-index subgroup of $N_G(S)$.
The equivalence of (ii) and (iii) now follows from \cite[Prop.\ 2.12]{BMR2}.
The equivalence of (i) and (iii) follows from \cite[Cor.\ 3.3]{BMR2},
and the equivalence of (iii) and (iv) from \cite[Thm.\ 3.4]{BMR2}.

(c). Now suppose that $H$ is reductive.
The subgroups $H\cap S$ and $(H\cap S)^0$ are normal in $H$.
Let $\pi$ be the canonical projection from $H$ to $H/(H\cap
S)^0$.
We have a homomorphism $\pi\times \psi$ from $H$ to $H/(H\cap
S)^0\times H/C_G(S)^0$.
If $h\in \ker (\pi\times \psi)$ then $h\in
(H\cap S)^0\cap C_G(S)^0\subseteq S^0\cap C_G(S)^0\subseteq Z(S^0)$,
so $\ker (\pi\times \psi)^0$ is a torus, so $\pi\times \psi$ is
non-degenerate.
By \cite[Lem.~2.12(i) and (ii)]{BMR}, $K$ is $H$-cr
if and only if $\pi(K)$ is $H/(H\cap S)^0$-cr and $\psi(K)$ is
$H/C_G(S)^0$-cr.
To finish the proof, it is enough to show that
$\pi(K)$ is $H/(H\cap S)^0$-cr if and only if $KS/S$ is $N_G(S)/S$-cr.
The natural map from $H/(H\cap S)^0\ra HS/S$ is an
isogeny, so $\pi(K)$ is $H/(H\cap S)^0$-cr if and only if $KS/S$ is
$HS/S$-cr, by \cite[Lem.~2.12(ii)]{BMR}.
As $HS/S$ is a finite-index
subgroup of $N_G(S)/S$, $KS/S$ is $HS/S$-cr if and only if $KS/S$ is
$N_G(S)/S$-cr \cite[Prop.\ 2.12]{BMR2}.
This completes the proof.
\end{proof}

\begin{cor}
\label{cor:KS<=>K}
 Let $G$, $H$, $S$ and $K$ be as in Proposition \ref{prop:KN<=>K}.
 Suppose that $(H\cap S)^0$ is a torus.  Then $H$ is reductive and the
 following are equivalent:
 \begin{itemize}
  \item[(i)] $K$ is $H$-completely reducible;
  \item[(ii)] $KS$ is $G$-completely reducible;
  \item[(iii)] $KS$ is $HS$-completely reducible;
  \item[(iv)] $KS$ is $N_G(S)$-completely reducible;
  \item[(v)] $KS/S$ is $N_G(S)/S$-completely reducible.
 \end{itemize}
\end{cor}

\begin{proof}
 Since $N_G(S)^0 = C_G(S)^0S^0$, and $C_G(S)^0 \subseteq H \subseteq N_G(S)$,
 we have $H^0= C_G(S)^0(H\cap S)^0$, so $H$ is
 reductive.  Moreover, $H^0/C_G(S)^0$ is a quotient of $(H\cap S)^0$,
 which is a torus by hypothesis.  Hence $H/C_G(S)^0$ is a finite
 extension of a torus, which implies that any subgroup of $H/C_G(S)^0$
 is $H/C_G(S)^0$-cr.  Thus $K$ is $H$-cr if and only if $KS/S$ is $N_G(S)/S$-cr, by Proposition \ref{prop:KN<=>K}(c).
 The result now follows from Proposition \ref{prop:KN<=>K}(b).
\end{proof}

\begin{rems}
\label{rem:KvsKN}
(i).
\begin{comment}
If $H$ is assumed to be reductive, then it follows from Proposition
\ref{prop:KN<=>K} that one set of implications in Corollary
\ref{cor:KS<=>K} above holds without the restriction on $(H\cap S)^0$:
\end{comment}
Suppose $H$ is reductive.  If $K$ is $H$-cr, then Proposition
\ref{prop:KN<=>K}(c) says that $KS/S$ is $N_G(S)/S$-cr, whence $KS$ is
$G$-cr, by Proposition \ref{prop:KN<=>K}(b).  The converse,
however, is not true in this generality; just take $S = H = G$ and $K$
a non-$G$-cr subgroup of $G$.

(ii). Corollary \ref{cor:KS<=>K} holds in particular if $S$ is
linearly reductive, since then the condition that $(H\cap S)^0$ is a
torus is automatic.  However, Example \ref{exmp:KnotKS} shows that
even when $S$ is linearly reductive, the situation for subgroups of $N_G(S)$ is not as
straightforward as for subgroups of $C_G(S)$ (cf.\ Corollary \ref{cor:SnormalizesK}).
\end{rems}

\begin{comment}
\begin{proof}
Note that $N_G(N)$ is $G$-cr and thus reductive (\cite[Cor.\
3.16]{BMR}).  By \cite[Cor.\ 3.3]{BMR2}, $KN$ is $G$-cr if and only if
$KN$ is $N_G(N)$-cr, which gives the second equivalence.

Since $N_G(N)^0 = C_G(N)^0N^0$ (\cite[Lem.\ 6.8]{martin1}) and
$C_G(N)^0\subseteq H^0 \subseteq N_G(N)$, it follows that $N^0H^0 =
N_G(N)^0$. Thus $N^0H^0$ is $G$-cr, hence reductive.  Because $(H\cap
N)^0$ is a torus, $H$ itself is reductive.  Since $H^0 \subseteq
N_G(N)$, $N_H(N) := H\cap N_G(N)$ has finite index in $H$, and so
$N_H(N)$ is also reductive.  Again, since $N_G(N)^0 = C_G(N)^0N^0$ and
$C_G(N)^0\subseteq H^0$, we see that $N_H(N)N$ has finite index in
$N_G(N)$.  Consequently, by \cite[Prop.\ 2.12]{BMR2}, $K$ is $H$-cr if
and only if $K$ is $N_H(N)$-cr, and $KN$ is $N_G(N)$-cr if and only if
$KN$ is $N_H(N)N$-cr.  So we can replace $H$ by $N_H(N)$, and assume
that $H$ normalizes $N$.  Now, by \cite[Thm.\ 3.4]{BMR2}, we have that
$KN$ is $G$-cr if and only if $KN$ is $HN$-cr if and only if $KN/N$ is
$HN/N$-cr.

Let $f:H \to HN/N$ be the morphism given by composing the inclusion of
$H$ in $HN$ with the canonical projection of $HN$ onto $HN/N$.  The
connected kernel of $f$ is $(H\cap N)^0$, which is a torus by
assumption.
Moreover, $f(K) = KN/N$.  By \cite[Lem.\ 2.7(ii)]{BMR2}, $KN/N$ is
$HN/N$-cr if and only if $K$ is $H$-cr, which gives the result.
\end{proof}

\begin{cor}\label{cor:KS<=>K}
Suppose $S$ is a linearly reductive subgroup of $G$ and $K \subseteq
N_G(S)$.  Then $K$ is $N_G(S)$-completely reducible if and only if
$KS$ is $G$-completely reducible if and only if $KS$ is
$N_G(S)$-completely reducible.
\end{cor}

\begin{proof}
This follows directly from Corollary \ref{lem:KN<=>K}, setting $H =
N_G(S)$; note that $(N_G(S) \cap S)^0 = S^0$ is a torus.
\end{proof}
\end{comment}

\begin{comment}

[{\tt Pf of old [BMR2, Prop. 3.15] begins.}]  MB:- This is now
incorporated in Cor KS<=>K and the subsequent remark.

The reverse direction of Proposition \ref{prop:AB<=>CGA} actually
holds under a weaker condition on $M$, as is shown in the next result
(note that $C_G(N)$ is normal in $N_G(N)$, so if $M$ is $C_G(N)$-cr,
then $M$ is $N_G(N)$-cr, by Proposition \ref{prop:10}).

\begin{cor}
\label{cor:normcr=>gcr}
Suppose $S$ is a $G$-completely reducible subgroup of $G$ and $K$ is a
subgroup of $N_G(S)$.  If $K$ is $N_G(S)$-completely reducible, then
$KS$ is $G$-completely reducible.
\end{cor}

\begin{proof}
By Proposition \ref{prop:KN<=>K}(c), if $H$ is $N_G(S)$-cr, then
$HS/S$ is $N_G(S)/S$-cr.  But then Proposition \ref{prop:KN<=>K}(b)
says that $KS$ is $G$-cr, as required.
\end{proof}

The converse of Proposition \ref{prop:normcr=>gcr} is false: we can
just take $N$ to be $G$ and $M$ to be a non-$G$-cr subgroup of $G$.
Thus the converse of Proposition \ref{prop:AB<=>CGA} fails if we
replace $C_G(N)$ with $N_G(N)$.

[{\tt Proof ends.}]

\end{comment}

\begin{comment}
So far in this subsection, we have considered the case where $K
\subseteq N_G(S)$ for a $G$-cr subgroup $S$ of $G$.  We finish by
briefly considering a related case, where $S$ is a linearly reductive
subgroup of $G$ which normalizes $K$, rather than $K$ normalizing $S$.

\begin{lem}\label{lem:KS<=>K}
Let $S$ be a linearly reductive subgroup of $G$ and suppose $K$ is a
subgroup of $G$ such that $S \subseteq N_G(K)$.  Then $K$ is
$G$-completely reducible if and only if $KS$ is $G$-completely
reducible.
\end{lem}

\begin{proof}
Since $KS/K$ is linearly reductive, this follows immediately from
\cite[Cor.\ 3.7(ii)]{BMR2}.
\end{proof}
\end{comment}

\begin{comment}
\subsection{Regular Subgroups}
\label{subsec:reg}

We move on to some considerations how complete reducibility and
separability behave when we pass from $G$ to regular reductive
subgroups of $G$.  Our prototype is the following result, \cite[Thm.\
3.26]{BMR}, which shows that for connected groups in good
characteristic, we can pass between $G$ and regular subgroups without
affecting complete reducibility.

\begin{prop}
\label{prop:regGcr}
Let $G$, $H$ be connected reductive groups with $H$ a regular subgroup
of $G$, and let $K$ be a subgroup of $H$.  If $p$ is good for $G$,
then $K$ is $G$-completely reducible if and only if it is
$H$-completely reducible.
\end{prop}

In \cite{BMR}, the proof of this result involves reducing to the case
that $H$ is the connected centralizer of a linearly reductive group.
This is possible because of the assumptions that $G$ and $H$ are
connected and $p$ is good.  In general, not all regular subgroups of a
reductive algebraic group arise in this way; however, the class of
regular subgroups arising as centralizers or normalizers of linearly
reductive groups is an important one.  We begin this section by
exploring to what extent one can mimic Proposition \ref{prop:regGcr}
with relaxed hypotheses.  Our first result shows that it is enough to
consider regular subgroups which contain a maximal torus.
\begin{lem}
\label{lem:H<=>HT}
Suppose $H$ is a reductive subgroup of $G$ normalized by the maximal
torus $T$ of $G$.  If $K$ is a subgroup of $H$, then $K$ is
$H$-completely reducible if and only if $K$ is $HT$-completely
reducible.
\end{lem}

\begin{proof}
Since $H$ is normal in $HT$, the result
follows from \cite[Prop.\ 2.8]{BMR2}.
\end{proof}

We recall an example due to Liebeck \cite[Ex.\ 3.45]{BMR} which shows
that if we do not require $p$ to be good for $G$, then the reverse
implication of Proposition \ref{prop:regGcr} can fail: if we take $p =
2$, $m\geq 4$ to be an even integer, $H = \SP_m\times \SP_m\subseteq
\SP_{2m}: = G$ and $K \cong \SP_m$ embedded diagonally in $H$, then
$K$ is $H$-cr but not $G$-cr.  Proposition \ref{prop:GcrnotMcr} below
shows that the forward implication of Proposition \ref{prop:regGcr} can
also fail in bad characteristic.  Our next example shows that the
analogue of Proposition \ref{prop:regGcr} fails in general for
disconnected $H$, even in good characteristic.
\begin{exmp}
\label{exmp:regI}
Let $G = \SL_2(k)$, where $k$ has characteristic $2$.  Let $T$ denote
the standard maximal torus of $G$ and set $H = N_G(T)$; then $H$ is
certainly a regular subgroup of $G$.  Take $K = \langle s \rangle
\subseteq N_G(T) = H$, where $s$ is the unipotent element given by the
matrix $\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$.
Then $K$ is $H$-completely reducible, since $K$ is $H$-ir; however,
$K$ is unipotent, so is not $G$-completely reducible.
\end{exmp}

\begin{rem}
Despite the existence of examples such as this, if $p$ is good for $G$
and $H$ is a regular reductive subgroup of $G$, then we can use some
of our earlier results to provide a method for extending Proposition
\ref{prop:regGcr} to a wider class of regular subgroups, as follows:
Let $H$ be a regular reductive subgroup of $G$ and suppose $K
\subseteq H$.  By Lemma \ref{lem:H<=>HT}, $K$ is $H$-cr if and only if
$K$ is $HT$-cr, where $T$ is a maximal torus of $G$ normalizing $H$,
so we may replace $H$ with $HT$.  Thus we may assume that $H^0$ is a
regular connected reductive subgroup of $G^0$ containing a maximal
torus of $G^0$.  If $p$ is good for $G^0$, then using the arguments in
the proof of \cite[Thm.\ 3.26]{BMR},
we may assume that $H^0 = C_G(S)^0$ for some finite subgroup $S$ of
$T$.  Finally, if this subgroup $S$ we have found normalizes $K$, then
$K$ is $G$-cr if and only if $K$ is $H$-cr, by Corollary
\ref{prop:SnormalizesK}.
\end{rem}

We finish this section by considering how the property of separability
behaves when we pass from $G$ to a regular subgroup.  Our motivation
for this is as follows: Suppose we are interested in the subgroup
$K\subseteq G$, but $K$ is not separable in $G$. One might hope that
by passing to a regular subgroup of $G$, one can remove this problem.
The final result of this section is of this form.
We let $\pi_G : G\ra G_{\rm ad}$ denote the canonical projection.

\begin{prop}
\label{prop:sepovergp}
Let $K$ be a $G$-completely reducible subgroup of $G$.  Then there
exists a subgroup $M$ of $G$ containing a maximal torus of $G$
such that $K \subseteq M$, $K$ is $M$-irreducible and $\pi_M(K)$ is
separable in $M_{\rm ad}$.
\end{prop}

\begin{proof}
Since a regular reductive subgroup of an R-Levi subgroup of $G$ is a
regular reductive subgroup of $G$ and $K$ is $G$-cr, we may assume
that $K$ is $G$-ir after replacing $G$ by an R-Levi subgroup of $G$
that is minimal with respect to containing $K$.  By
\cite[Cor.~2.7(i)]{BMR}, $K$ is $M$-ir in any reductive subgroup $M$
of $G$ containing $K$.  If $\pi_G(K)$ is separable in $G_{\rm ad}$,
then we can take $M=G$, so suppose not.
By \cite[Prop.\ 3.39]{BMR}, there exists a
subgroup $M'$ of $G_{\rm ad}$ containing a maximal torus of $G_{\rm
ad}$ such that $\pi_G(K)\subseteq M'$ and $M'$ is not separable in
$G_{\rm ad}$.  As $G_{\rm ad}^0$ is of adjoint type, its
Lie algebra has trivial centre,
so $G_{\rm ad}$ is separable in $G_{\rm ad}$.
This forces $M'$ to be of strictly smaller dimension than $G_{\rm
ad}$.  Let $M= \pi_G^{-1}(M')$, a subgroup of $G$ that is of strictly
smaller dimension than $G$ and contains a maximal torus of $G$.  Since
$M$ is an overgroup of the $G$-ir group $K$, $M$ is reductive.  The
result now follows by induction on $\dim G$.
\end{proof}

Proposition \ref{prop:sepovergp} is false if we do not assume that $K$
is $G$-completely reducible: see Example \ref{exmp:nonGcr} below.
\end{comment}

\begin{comment}
In this section we explore how the properties of complete reducibility
pass from $G$ to regular subgroups of $G$ and vice versa.  The
motivation for this comes in part from the constructions in the
previous section; we try to prove results by passing to regular
subgroups and using an inductive argument (see Theorems
\ref{thm:allregss} and \ref{thm:adjsc}).

It would be useful to have an analogue of Proposition
\ref{prop:regGcr} which held for disconnected $G$ and $H$.  We
formulate this as a question:

\begin{question}
\label{q:regI}
Let $H \subseteq G$ be regular and let $K$ be a subgroup of $H$.  What
conditions ensure that $K$ is $H$-completely reducible if and only if
$K$ is $G$-completely reducible?
\end{question}

Question \ref{q:regI} is closely related to Question \ref{q:regII}
below; indeed, Example \ref{exmp:regI} also gives a counterexample to
the forward implication of Question \ref{q:regII} (just take $S = T$
above).  A counterexample to the reverse implication of both questions
in bad characteristic is provided by Example \ref{exmp:KnotKS}.  A
sufficient condition for the equivalence in Question \ref{q:regII} to
hold is that $K$ and $S$ centralize each other; see
Lemma~\ref{lem:KS<=>K} and \cite[Cor.~3.14]{BMR2}.

\begin{question}
\label{q:regII}
Let $S$ be linearly reductive, $K \subseteq N_G(S)$.  What conditions
ensure that $KS$ is $G$-cr if and only if $K$ is $G$-cr?
\end{question}

Questions \ref{q:regI} and \ref{q:regII} are related because many
regular subgroups of reductive groups can be realized as the
centralizers or normalizers of linearly reductive subgroups.  The
following result provides a concrete link between the two questions,
in the case that $N$ below is linearly reductive
(see Corollary \ref{cor:KS<=>K}).
\end{comment}

\section{An Important Example}
\label{sec:ex}

We consider a collection of important examples, which serve to illustrate
many of the points raised in the previous sections.
Throughout this section, we suppose that $p=2$ and
let $G$ be a simple group of type $G_2$.
We fix a maximal torus $T$ and a Borel subgroup $B$ of $G$ with $T\subseteq B$.
Let $\Psi$ be the set of roots of $G$ with respect to $T$.
We fix a base $\Sigma = \{\alpha,\beta\}$ for the set $\Psi^+$ of
positive roots with respect to $B$, where $\alpha$ is short and
$\beta$ is long.  The positive roots are $\alpha$, $\beta$,
$\alpha+\beta$, $2\alpha+\beta$, $3\alpha+\beta$ and $3\alpha+2\beta$.
For each root $\gamma$,
we choose an isomorphism $\kappa_\gamma\colon k\ra U_\gamma$ and set
$s_\gamma=\kappa_\gamma(1)\kappa_{-\gamma}(-1)\kappa_\gamma(1)$ (cf.\
\cite[32.3]{Hum}).  Then $s_\gamma$ represents the reflection
corresponding to $\gamma$ in the Weyl group $N_G(T)/T$ of $G$.  Since
$p=2$, the order of $s_\gamma$ is 2 for every $\gamma\in \Psi$.

We use various equations from \cite[33.5]{Hum}.  Some are reproduced
below.  For brevity, we do not give the commutation relations between
the root subgroups: these are the equations of the form
$$ \kappa_\gamma(a)\kappa_{\gamma'}(b) =
\kappa_{\gamma'}(b)\kappa_\gamma(a)g,
$$ where $g$ is a product of elements of the form
$\kappa_{\gamma''}(p_{\gamma''}(a,b))$ over certain roots $\gamma''$,
each $p_{\gamma''}$ being a monomial in $a$ and $b$.  (Recall,
however, that $U_\gamma$ and $U_{\gamma'}$ commute if no positive
integral combination of $\gamma$ and $\gamma'$ is a root.)  We refer
to these equations collectively below as ``the CRs''.

Since $G$ is simply connected, we have $G_\gamma\cong \SL_2(k)$ for
every $\gamma\in \Psi$, by Lemma~\ref{lem:sc}.

We have
\begin{equation}\label{eqn:fundalphaveepairings}
 \langle \alpha,\alpha^\vee\rangle=2,\ \langle
 \beta,\alpha^\vee\rangle=-3,\ \langle \alpha,\beta^\vee\rangle=-1,\
 \langle \beta,\beta^\vee\rangle=2
\end{equation}
(see \cite[32.3]{Hum}; note that $\inprod{\beta,\alpha}$ in
Humphreys's notation coincides with $\inprod{\beta,\alpha^\vee}$).
This yields
\begin{equation}
\label{eqn:alphaveepairings}
 \langle \alpha+\beta,\alpha^\vee\rangle=-1,\ \langle
 2\alpha+\beta,\alpha^\vee\rangle=1,\ \langle
 3\alpha+\beta,\alpha^\vee\rangle=3,\ \langle
 3\alpha+2\beta,\alpha^\vee\rangle=0.
\end{equation}
We have
$$ s_\alpha\cdot \alpha= -\alpha ,\ s_\alpha\cdot \beta=
3\alpha+\beta,
$$ and it follows that

\begin{equation}
\label{eqn:corootaction}
 s_\alpha\cdot \alpha^\vee= -\alpha^\vee,\ s_\alpha\cdot \beta^\vee =
\alpha^\vee+\beta^\vee.
\end{equation}

We need to know how $s_\alpha$ acts on the $U_\gamma$.  We can choose
the homomorphisms $\kappa_\gamma$ so that $s_\alpha$ maps each
$U_\gamma$ to $U_{s_\alpha\cdot \gamma}$ by conjugation in the
following way (see \cite[33.1 and 33.5]{Hum}):
\begin{equation}
\label{eqn:srootgps}
 s_\alpha\kappa_\beta(a)s_\alpha = \kappa_{3\alpha+\beta}(a),\
 s_\alpha\kappa_{\alpha+\beta}(a)s_\alpha =
 \kappa_{2\alpha+\beta}(a),\ s_\alpha\kappa_{2\alpha+\beta}(a)s_\alpha
 = \kappa_{\alpha+\beta}(a),
\end{equation}
$$ \ s_\alpha\kappa_{3\alpha+\beta}(a)s_\alpha=\kappa_\beta(a), \
s_\alpha\kappa_{3\alpha+2\beta}(a)s_\alpha=\kappa_{3\alpha+2\beta}(a),
$$ and we can choose $0\neq e_\gamma\in \uu_\gamma$ for each positive
root $\gamma$ such that the adjoint action of $s_\alpha$ on the
$e_\gamma$ is given by
\begin{equation}
\label{eqn:sonLie}
 \Ad {s_\alpha}(e_\beta) = e_{3\alpha+\beta},\ \Ad
 {s_\alpha}(e_{\alpha+\beta}) = e_{2\alpha+\beta},\ \Ad
 {s_\alpha}(e_{2\alpha+\beta}) = e_{\alpha+\beta},
\end{equation}
$$ \ \Ad {s_\alpha}(e_{3\alpha+\beta}) = e_\beta,\ \Ad
 {s_\alpha}(e_{3\alpha+2\beta}) = e_{3\alpha+2\beta}.
$$ Using Eqn.\ \eqref{eqn:srootgps} and the CRs, we get
\begin{eqnarray}
\label{eqn:scommrelns}
 & & s_\alpha \kappa_\beta(a) \kappa_{\alpha+\beta}(a')
              \kappa_{2\alpha+\beta}(b) \kappa_{3\alpha+\beta}(b')
              \kappa_{3\alpha+2\beta}(c) s_\alpha \\ & = &
              \kappa_\beta(b') \kappa_{\alpha+\beta}(b)
              \kappa_{2\alpha+\beta}(a') \kappa_{3\alpha+\beta}(a)
              \kappa_{3\alpha+2\beta}(ab'+a'b+c) \nonumber.
\end{eqnarray}

Let
\[
L:=\langle G_\alpha\cup T\rangle.
\]
Let $P$ be the parabolic subgroup of $G$ that contains $B$ and has $L$
as a Levi subgroup.  The roots of $R_u(P)$ with respect to $T$ are
$\beta$, $\alpha+\beta$, $2\alpha+\beta$, $3\alpha+\beta$,
$3\alpha+2\beta$.

Let $S$ be the torus $\alpha^\vee(k^*)$; we have $S=G_\alpha\cap T$.
Fix $t\in S$ such that $t$ has order 3.  By Eqns.\
\eqref{eqn:fundalphaveepairings} and \eqref{eqn:alphaveepairings},
\begin{equation}
\label{eqn:taction}
 \mbox{$t$ acts trivially on $U_\beta$, $U_{3\alpha+\beta}$,
                             $U_{3\alpha+2\beta}$, $\uu_\beta$,
                             $\uu_{3\alpha+\beta}$,
                             $\uu_{3\alpha+2\beta}$, and}
\end{equation}
$$ \mbox{$t$ acts non-trivially on $U_\alpha$, $U_{\alpha+\beta}$,
$U_{2\alpha+\beta}$, $\uu_\alpha$, $\uu_{\alpha+\beta}$,
$\uu_{2\alpha+\beta}$.}
$$

Set
\[
H := \langle \{s_\alpha,t\} \rangle\subseteq G_\alpha.
\]
Note that $H \cong S_3$.  Since $G_\alpha\cong \SL_2(k)$ (Lemma
\ref{lem:sc}), $\alpha^\vee$ is an isomorphism from $k^*$ onto
$G_\alpha\cap T$.  Set
$$ z := d\alpha^\vee(1)\in \gg_\alpha = \Lie G_\alpha
$$ (where we regard $1$ as an element of the tangent space
$T_1(k^*)\cong k$); then $z \neq 0$ and $k\cdot z$ is the centre of
$\gg_\alpha$.  In particular, $H$ centralizes $z$.  If $\gamma\in
\Psi$, then
\begin{equation}
\label{eqn:hcent}
 \mbox{$U_\gamma$ centralizes $z$}\iff \mbox{2 divides $\langle
\gamma,\alpha^\vee\rangle$}.
\end{equation}

By the CRs, $G_\alpha$ commutes with $G_{3\alpha+ 2\beta}$.  The
multiplication map $G_\alpha \times G_{3\alpha+2\beta}\ra G_\alpha
G_{3\alpha+2\beta}$ is bijective because $G_\alpha \cap
G_{3\alpha+2\beta}$, being a proper normal subgroup of $G_\alpha$,
must be trivial, but this map is not an isomorphism.  For
$\zz(\gg_{3\alpha+2\beta})\subseteq \Lie(G_{3\alpha+ 2\beta}\cap T)$,
so we have $\zz(\gg_{3\alpha+2\beta})\subseteq \cc_\frakt(G_\alpha)
\subseteq \cc_\frakt(H)$, which equals $k\cdot z$, so $\gg_\alpha\cap
\gg_{3\alpha+2\beta}$ is non-empty.  In fact $\gg_\alpha\cap
\gg_{3\alpha+2\beta}= k\cdot z$, because $k\cdot z$ is the unique
proper non-zero normal subalgebra of $\Lie G_\alpha= \Sl_2(k)$.

Now set
\[
M := G_\alpha G_{3\alpha+ 2\beta}.
\]
Note that $T\subseteq M$ and $M$ is a semisimple maximal rank subgroup of $G$
of type $\widetilde A_1 A_1$.
Consequently, $M$ is $G$-ir.
Moreover, we have
\[
C_G(M)= \ker(\alpha)\cap \ker(3\alpha+2\beta)
      = \ker(\alpha)\cap \ker(2\beta)
      = \ker(\alpha)\cap \ker(\beta)
      = Z(G)
      = \{1\}.
\]
It is easy to see that $\zz(\mm)=k\cdot z$.

\begin{lem}
\label{lem:M}
With the notation as above, we have
 \begin{itemize}
  \item[(a)] $(G,M)$ is a reductive pair;
  \item[(b)] $N_G(M)=M$;
  \item[(c)] $M= C_G(z)$.
 \end{itemize}
\begin{comment}
  (a) $(G,M)$ is a reductive pair.\smallskip\\ (b)
  $N_G(M)=M$.\smallskip\\ (c) $M= C_G(z)$.
\end{comment}
\end{lem}

\begin{proof}
\begin{comment}
 \begin{itemize}
  \item[(a)] Let $V$ be the subspace of $\gg$ spanned by $\uu_{\pm
  \beta}$, $\uu_{\pm (\alpha+\beta)}$, $\uu_{\pm (2\alpha+\beta)}$ and
  $\uu_{\pm (3\alpha+\beta)}$.  Suppose $\gamma\in \Psi(M)$,
  $\gamma'\in \Psi$ and $b$ is a positive integer such that
  $\gamma'+b\gamma\in \Psi(M)$.  It is easy to check that $\gamma'\in
  \Psi(M)$.  It follows from \cite[Prop.\ 27.2]{Hum} that $V$ is an
  $\Ad(M)$-stable complement to $\mm$ in $\gg$, so $(G,M)$ is a
  reductive pair.
  \item[(b)] Let $g\in N_G(M)$.  Without loss of generality, we can
assume that $g$ normalizes $T$, so $g$ permutes $\Psi(M)=\{\pm \alpha,
\pm (3\alpha+2\beta)\}$.  As $\alpha$ is short and $3\alpha+2\beta$ is
long, $g$ must normalize $G_\alpha$ and $G_{3\alpha+2\beta}$.  But
$G_\alpha$ and $G_{3\alpha+2\beta}$ are rank one groups, so they have
no outer automorphisms.  Since $C_G(M)=\{1\}$, we therefore have that
$g\in M$, as required.
  \item[(c)] Since $C_G(z)^0\supseteq M$ and $M$ is $G$-ir, $C_G(z)^0$
is $G$-ir, so $C_G(z)^0$ is reductive.  This implies that $C_G(z)^0$
is generated by the root subgroups that it contains together with $T$.
It follows from Eqns.\ \eqref{eqn:fundalphaveepairings},
\eqref{eqn:alphaveepairings} and \eqref{eqn:hcent} that $C_G(z)^0= M$.
Hence $C_G(z)\subseteq N_G(C_G(z)^0)= N_G(M)= M$.  Thus $C_G(z) = M$.
 \end{itemize}
\end{comment}
  (a). Since $\Psi(M)$ is a closed subsystem of $\Psi$, this follows immediately from Lemma~\ref{lem:redpair}.
  \begin{comment}
  Let $V$ be the subspace of $\gg$ spanned by $\uu_{\pm \beta}$,
  $\uu_{\pm (\alpha+\beta)}$, $\uu_{\pm (2\alpha+\beta)}$ and
  $\uu_{\pm (3\alpha+\beta)}$.  Suppose $\gamma\in \Psi(M)$,
  $\gamma'\in \Psi$ and $b$ is a positive integer such that
  $\gamma'+b\gamma\in \Psi(M)$.  It is easy to check that $\gamma'\in
  \Psi(M)$.
Since $M$ is generated by $U_{\pm \alpha}$ and $U_{\pm (3\alpha+ 2\beta)}$, it follows from \cite[Prop.\ 27.2]{Hum} that $V$ is an
  $\Ad(M)$-stable complement to $\mm$ in $\gg$, so $(G,M)$ is a
  reductive pair.
  \end{comment}

  (b). Let $g\in N_G(M)$.
Without loss of generality, we can assume that $g$ normalizes $T$,
so $g$ permutes $\Psi(M)=\{\pm \alpha, \pm (3\alpha+2\beta)\}$.
As $\alpha$ is short and $3\alpha+2\beta$ is long,
$g$ must normalize $G_\alpha$ and $G_{3\alpha+2\beta}$.
But $G_\alpha$ and $G_{3\alpha+2\beta}$ are rank one groups,
so they have no outer automorphisms.
Since $C_G(M)=\{1\}$, we therefore have that $g\in M$, as required.

  (c). Since $C_G(z)^0\supseteq M$ and $M$ is $G$-ir, $C_G(z)^0$ is $G$-ir,
so $C_G(z)^0$ is reductive.
This implies that $C_G(z)^0$ is generated by the root
subgroups that it contains together with $T$.
It follows from Eqns.\ \eqref{eqn:fundalphaveepairings},
\eqref{eqn:alphaveepairings} and
\eqref{eqn:hcent} that $C_G(z)^0= M$.
Hence $C_G(z)\subseteq N_G(C_G(z)^0)= N_G(M)= M$, by part (b).
Thus $C_G(z) = M$.
\end{proof}

\begin{lem}
\label{lem:HGcr}
With the notation as above, we have
\begin{itemize}
\item[(a)] $H$ is $G$-completely reducible and $M$-completely reducible;
\item[(b)] $C_G(H)= G_{3\alpha+ 2\beta}$ and $N_G(H)=HG_{3\alpha+ 2\beta}$.
\end{itemize}
\begin{comment}
 (a) $H$ is $G$-completely reducible and $M$-completely reducible.\smallskip\\
 (b) $C_G(H)= G_{3\alpha+ 2\beta}$ and $N_G(H)=HG_{3\alpha+ 2\beta}$.
\end{comment}
\end{lem}

\begin{proof}
\begin{comment}
 \begin{itemize}
  \item[(a)] It is easily checked that $H$ is not contained in any
  Borel subgroup of $L$, so $H$ is $L$-ir.  Now $L$ is a Levi subgroup
  both of $G$ and of $M$, so $H$ is both $G$-cr and $M$-cr by
  \cite[Cor.\ 3.22]{BMR}.
 \item[(b)] Since $H$ is $G$-cr, $C_G(H)$ is $G$-cr by \cite[Cor.\
3.17]{BMR}, so $C_G(H)^0$ is reductive.  Now $C_G(H)^0$ cannot have
rank $2$, because $H$ is not centralized by any maximal torus of $G$.
Hence $C_G(H)^0$ must be equal to its rank $1$ subgroup $G_{3\alpha+
2\beta}$.  By a similar argument, $C_G(G_{3\alpha+ 2\beta})^0=
G_\alpha$.

By \cite[Thm. 3.10]{BMR}, $G_{3\alpha+ 2\beta}=C_G(H)^0$ is $G$-cr.
Let $g\in N_G(H)$.  Then $g$ normalizes $C_G(H)^0= G_{3\alpha+
2\beta}$, so $g$ normalizes $C_G(G_{3\alpha+ 2\beta})^0= G_\alpha$, so
$g$ normalizes $M$.  Lemma \ref{lem:M}(a) now implies that $N_G(H)=
N_M(H)= HG_{3\alpha+ 2\beta}$ and $C_G(H)= C_M(H)= G_{3\alpha+
2\beta}$.
 \end{itemize}
\end{comment}
  (a). It is easily checked that $H$ is not contained in any Borel
 subgroup of $L$, so $H$ is $L$-ir.  Now $L$ is a Levi subgroup both
 of $G$ and of $M$, so $H$ is both $G$-cr and $M$-cr by \cite[Cor.\
 3.22]{BMR}

  (b). Since $H$ is $G$-cr, $C_G(H)$ is $G$-cr by \cite[Cor.\ 3.17]{BMR},
  so $C_G(H)^0$ is reductive.  Now $C_G(H)^0$ cannot have
 rank $2$, because $H$ is not centralized by any maximal torus of $G$.
 Hence $C_G(H)^0$ must be equal to its rank $1$ subgroup $G_{3\alpha+
 2\beta}$.  Now $G_{3\alpha+ 2\beta}=C_G(H)^0$ is $G$-cr by \cite[Thm.\ 3.10]{BMR},
 so $C_G(G_{3\alpha+ 2\beta})^0= G_\alpha$ by a similar argument to that for $C_G(H)^0$.

Let $g\in N_G(H)$.  Then $g$ normalizes $C_G(H)^0= G_{3\alpha+
2\beta}$, so $g$ normalizes $C_G(G_{3\alpha+ 2\beta})^0= G_\alpha$, so
$g$ normalizes $M$.  Lemma \ref{lem:M}(b) now implies that $N_G(H)=
N_M(H)= HG_{3\alpha+ 2\beta}$ and $C_G(H)= C_M(H)= G_{3\alpha+
2\beta}$.
\end{proof}

The following example shows that the converse of Corollary
\ref{cor:LGsep} can fail, even when $H$ is $L$-ir.  It also gives a
counterexample to the converse of the first assertion of Theorem
\ref{thm:redpairs} (note that $(G,L)$ is a reductive pair by Lemma \ref{lem:redpair}).

\begin{prop}
\label{prop:LsepGnonsep}
 The subgroup $H$ is separable in $L$, but not in $G$.
\end{prop}

\begin{proof}
Since $DL = G_\alpha\cong \SL_2(k)$,
$L$ is isomorphic either to $\GL_2(k)$ or to $\SL_2(k)\times k^*$.
To rule out the latter case, it's enough to show
that $s_\alpha$ acts non-trivially on $\frakt$;
this follows from Eqn.\ (\ref{eqn:corootaction}).
It now follows easily that $\cc_\frakl(H)= k\cdot z$.
Now $z$ is tangent to $\Lie Z(L)$ as $L\cong \GL_2(k)$.
We deduce that $H$ is separable in $L$.

It follows from Eqns.\ \eqref{eqn:sonLie} and \eqref{eqn:taction}
that $H$ centralizes
$e_\beta+ e_{3\alpha+ \beta}$.
But $C_G(H)= G_{3\alpha+ 2\beta}$ by Lemma~\ref{lem:HGcr}(b),
so $e_\beta+ e_{3\alpha+ \beta}$
is not tangent to $\Lie C_G(H)= \gg_{3\alpha+ 2\beta}$.
Hence $H$ is not separable in $G$.
\end{proof}

\begin{rem}
 It is easily checked that for every semisimple $x\in \cc_\gg(H)$, we have
 $x\in \Lie C_G(H)$.  The same result cannot hold if we replace $H$ with a $G$-irreducible and
 non-separable subgroup of $G$; cf.\ the proof of
 \cite[Thm.~3.39]{BMR}.
\end{rem}

For $a\in k$, set
$$ u(a)= \kappa_\beta(a)\kappa_{3\alpha+\beta}(a).
$$ The CRs yield
\begin{equation}
\label{eqn:umult}
 u(a)u(b)= u(a+b)\kappa_{3\alpha+2\beta}(ab)
\end{equation}
and
\begin{equation}
\label{eqn:uinv}
 u(a)^{-1}= u(a)\kappa_{3\alpha+2\beta}(a^2).
\end{equation}
Define
\[
H_a= u(a)Hu(a)^{-1}.
\]
By Eqn.\ \eqref{eqn:taction}, $u(a)$ centralizes $t$; by Eqn.\
\eqref{eqn:scommrelns}, we have $u(a)s_\alpha u(a)^{-1}= s_\alpha
\kappa_{3\alpha+2\beta}(a^2)$.  Hence
$H_a = \langle \{ t, s_\alpha \kappa_{3\alpha+ 2\beta}(a^2) \}\rangle
\subseteq M$ for all $a\in k$.

The following example shows that Theorem \ref{thm:finorbit}(b) can
fail if we do not require $K$ to be separable in $G$ (recall that
$(G,M)$ is a reductive pair by Lemma \ref{lem:M}(a)).
This example also proves Theorem \ref{thm:countereg}.

\begin{exmp}
\label{exmp:orbitcountereg}
Let $(m_1,m_2)= (s_\alpha, t \kappa_{3\alpha+2\beta}(1))$.  For
$a\in k$, set
\[
(m_1(a),m_2(a))= u(a)\cdot (m_1,m_2).
\]
By construction,
$(m_1(a),m_2(a))$ is $G$-conjugate to $(m_1(b),m_2(b))$
for all $a,b\in k$.  We now show that if $a\neq b$, then
$(m_1(a),m_2(a))$ is not $M$-conjugate to
$(m_1(b),m_2(b))$.  This shows that $G\cdot (m_1,m_2)$ is
an infinite union of $M$-conjugacy classes.

For $a \in k$ let
\[
\widehat{H}_a := \langle \{m_1(a),m_2(a)\} \rangle.
\]
Note that $t$ commutes with $\kappa_{3\alpha+2\beta}(1)$ by Eqn.\ (\ref{eqn:taction}), so
\[
\widehat{H}_0 =  \langle \{ m_1, m_2 \} \rangle= \langle \{ s_\alpha, t, \kappa_{3\alpha+2\beta}(1) \} \rangle = \langle H\cup \{\kappa_{3\alpha+2\beta}(1)\}\rangle
\]
and $\widehat{H}_a = \langle H_a\cup \{\kappa_{3\alpha+2\beta}(1)\}\rangle$ similarly.  Hence
\[
C_G(\widehat{H}_0) = C_G(H)\cap C_G(\kappa_{3\alpha+2\beta}(1)) =
           G_{3\alpha+2\beta}\cap C_G(\kappa_{3\alpha+2\beta}(1)) =
           U_{3\alpha+2\beta},
\]
by Lemma \ref{lem:HGcr}(b), so
\[
C_G(\widehat{H}_a)= C_G(u(a)\widehat{H}_0u(a)^{-1})= u(a)C_G(\widehat{H}_0)u(a)^{-1}= u(a)U_{3\alpha+2\beta}u(a)^{-1}=
U_{3\alpha+2\beta}.
\]
Now let $a, b \in k$, and suppose that $(m_1(a),m_2(a))$ and
$(m_1(b),m_2(b))$ are $M$-conjugate.
Then there exists $m\in M$
such that $(mu(a))\cdot (m_1,m_2)=u(b)\cdot (m_1,m_2)$.  We
have $mu(a)u(b)^{-1}\in C_G(\langle \{m_1(b),m_2(b)\}\rangle)=
C_G(\widehat{H}_b)= U_{3\alpha+2\beta}\subseteq M$, so
$u(a)u(b)^{-1}\in M$.  But $u(a)u(b)^{-1}= u(a+b)
\kappa_{3\alpha+2\beta}(ab+b^2)$ by Eqns.\ (\ref{eqn:umult}) and
(\ref{eqn:uinv}), so $u(a+b)\in M$.  So we must have $u(a+b)=1$,
whence $a=b$.
Thus if $a \neq b$, then $(m_1(a),m_2(a))$ and
$(m_1(b),m_2(b))$ are not $M$-conjugate.

These calculations show that even though $(G,M)$ is a reductive pair,
$G\cdot(m_1,m_2) \cap M^2$ consists of an infinite union of $M$-conjugacy classes.
Observe that this is consistent with Theorem \ref{thm:finorbit}(b); a
similar calculation to the one in the proof of Proposition
\ref{prop:LsepGnonsep} shows that $\widehat{H}_0$
is not separable in $G$.
\end{exmp}

\begin{rem}
 The $n$-tuple $(m_1,m_2,\ldots, m_n)$ yields a similar example for any $n\geq 2$, where $m_1$ and $m_2$ are as above and $m_3= \cdots = m_n=1$.
\end{rem}

\begin{comment}
K\"ulshammer [ref] asked the following question.

\begin{question}
 Let $G$ be an algebraic group and let $F$ be a finite group.  Fix a
Sylow $p$-subgroup $F_p$ of $F$.  Given a homomorphism $\rho_0\colon
F\ra G$, is it true that there are only finitely many $G$-conjugacy
classes of homomorphisms $\rho\colon F\ra G$ such that the
restrictions $\rho|_{F_p}$ and $\rho_0|_{F_p}$ are $G$-conjugate?
\end{question}

The answer is yes in many cases: if the characteristic is zero or if
$G=\GL_n(k)$, for example [refs].  The answer is no for general
non-reductive groups [ref].
\end{comment}

Our next example shows that the second assertion
of Theorem~\ref{thm:redpairs} is false with the separability
assumption on $K$ removed, even though $(G,M)$ is a reductive pair by
Lemma \ref{lem:M}(a).
Since $M$ is a regular subgroup of $G$, this is also a new example of the failure of
\cite[Thm.\ 3.26]{BMR} in bad characteristic (compare
\cite[Ex.\ 3.45]{BMR}, which gives subgroups $H' \subseteq M' \subseteq G'$,
with $G'$ connected reductive, $M'$ regular in $G'$ such that
$H'$ is $M'$-cr, but not $G'$-cr).

\begin{prop}
\label{prop:GcrnotMcr}
Let $a\in k^*$.  Then $H_a$ is $G$-completely reducible but not
$M$-completely reducible.
\end{prop}

\begin{proof}
 Since $H_a$ is $G$-conjugate to $H$ and $H$ is $G$-cr (Lemma
\ref{lem:HGcr}(a)), $H_a$ is $G$-cr.
Let $\lambda= \alpha^\vee+ 2\beta^\vee\in Y(T)$: then $\langle
\alpha,\lambda\rangle= 0$ and $\langle \beta,\lambda\rangle= 1$.  It
is clear that $P=P_\lambda$ and $L = L_\lambda$.  We have a
homomorphism $c_\lambda\colon P_\lambda\ra L_\lambda$ as defined in
Subsection \ref{subsec:noncon}.  If $h\in H$ then $u(a)hu(a)^{-1}= hu$ for
some $u\in U_{3\alpha+ 2\beta}\subseteq \ker(c_\lambda)$, so
$c_\lambda(H_a)= H$.  To prove that $H_a$ is not $M$-cr, it suffices
by \cite[Lem.\ 2.17 and Thm.\ 3.1]{BMR} to show that $H_a$ and $H$ are
not $M$-conjugate.  But this is the case since $u(a)\not\in M$ and
$N_G(H)\subseteq M$ (Lemma \ref{lem:HGcr}(b)), so we are done.
\end{proof}

\begin{rem}
\label{rem:thm1.3(c)fails}
Proposition \ref{prop:GcrnotMcr} shows that
part (c) of Theorem \ref{thm:finorbit} fails
without the separability hypothesis on $K$:
for, by Theorem \ref{thm:gcrcrit},
the $G$-conjugacy class $G\cdot (m_1(a),m_2(a))$ is closed
in $G \times G$ but $M\cdot (m_1(a),m_2(a))$ is not.
\end{rem}

\begin{comment}
\begin{lem}
\label{lem:Hmaximality}
Let $K$ be a proper connected regular reductive subgroup of $G$ such
that $H\subseteq G$.  Then some conjugate of $K$ is contained in
$G_\alpha G_{3\alpha+2\beta}$.  In particular, every connected
unipotent subgroup of $K$ is abelian.
\end{lem}

\begin{proof}
 The semisimple element $t$ lies in a maximal torus of $K$.  Since $K$
is regular, this torus is also a maximal torus of $G$, so we can
assume without loss of generality that this torus is $T$.

 Any root is $W$-conjugate either to $\alpha$ or $3\alpha+2\beta$, so
if $K$ has semisimple rank at most $1$, then $K$ is conjugate to one
of $G_\alpha$, $G_{3\alpha+2\beta}$, $\langle G_\alpha$, $T\rangle$,
$\langle G_{3\alpha+2\beta},T\rangle$ or to a subtorus of $T$.  So
assume that $K$ has semisimple rank 2 and that $T\subseteq K$.  ...
\end{proof}
\end{comment}

The next example shows that Proposition \ref{prop:sepovergp} can fail
if we allow $H$ to be non-$G$-cr.  First we need a refinement of Lemma \ref{lem:sep}.
If $G_1$ is a reductive group and $H_1$ is a subgroup of $G_1$, then we say
that $x\in \gg_1$ is a {\em witness to the non-separability of $H_1$} if
$x\in \cc_{\gg_1}(H_1)$ but $x\not\in \Lie C_{G_1}(H_1)$.

\begin{lem}
\label{lem:nilptwitness}
 Let $f\colon G_1\ra G_2$ be an epimorphism of reductive groups such
that $\ker df$ consists of semisimple elements and let $H_1$ be a
subgroup of $G_1$.  Let $x\in \cc_{\gg_1}(H_1)$ be nilpotent.
Then $x$ is a witness to the non-separability of $H_1$ if and only
if $df(x)$ is a witness to the non-separability of $f(H_1)$.
\end{lem}

\begin{proof}
 It is clear that $df(x)\in \cc_{\gg_2}(f(H_1))$ and that
$df(x)$ is tangent to $C_{G_2}(f(H_1))$ if $x$ is tangent
to $C_{G_1}(H_1)$.  Conversely, suppose that $df(x)$ is
tangent to $C_{G_2}(f(H_1))$.  By \cite[Prop.~IV.14.26]{borel},
there exists a connected unipotent subgroup $U_2$ of $C_{G_2}(f(H_1))$
such that $df(x)\in \Lie U_2$.  By \cite[V.22.1]{borel} the restriction
of $f$ is an isogeny from $D(G_1^0)$ to $D(G_2^0)$.  Let $U_1= (f^{-1}(U_2)\cap D(G_1^0))^0$.
Any semisimple element of $U_1$ must belong to the finite group $\ker f\cap D(G_1^0)$, so $U_1$
has only finitely many semisimple elements.  Hence $U_1$ is a unipotent group.  As $\ker df$
consists of semisimple elements and $U_2\subseteq D(G_2^0)$, it follows that the restriction
of $f$ is an isomorphism from $U_1$ onto $U_2$.  Hence we can choose $x'\in \Lie U_1$ such that
$df(x')=df(x)$.  Now $\ker df\subseteq \zz(\gg_1)$, by \cite[V.22.2]{borel}, and $x,x'$ are both
nilpotent, so we must have $x=x'$, whence $x\in \Lie U_1$.

To complete the proof, it suffices to show that $U_1\subseteq C_{G_1}(H_1)$.
Fix $h\in H_1$.  If $u\in U_1$, then $f(h)f(u)f(h)^{-1}= f(u)$,
so we have $huh^{-1}=cu$ for some $c\in \ker f$.
The map $u\mapsto huh^{-1}u^{-1}$
is therefore a morphism from the connected
set $U$ to the finite set $\ker f\cap D(G_1^0)$,
so $huh^{-1}u^{-1}=1$ for all $h\in H_1$ and all $u\in U_1$.
Hence $U_1\subseteq C_{G_1}(H_1)$, as required.
\end{proof}

\begin{exmp}
\label{exmp:nonGcr}
 Let $C= \{u(a) \mid a\in k\}$.  Let $H'= \langle H\cup C\rangle$ and
suppose that $K$ is any reductive subgroup of $G$ containing $H'$.
Set $y:=e_\beta+ e_{3\alpha+ \beta}$.  Then $y$ belongs to the tangent
space $T_1(C)$, which is contained in $\Lie K$.  Now $\uu_{3\alpha+
2\beta}$ is centralized by $U_\beta$ and $U_{3\alpha+ 2\beta}$ and we
have $\Ad \kappa_{3\alpha+\beta}(a)(e_\beta) = e_\beta+ ae_{3\alpha+
2\beta}$, $\Ad \kappa_\beta(a)(e_{3\alpha+ \beta}) = e_{3\alpha+
\beta}+ ae_{3\alpha+2\beta}$ by the CRs, so $\Ad u(a)(y)= y$.  Hence $y\in \cc_\kk(H')$, but $y$ is not
tangent to $C_K(H')$ since it is not tangent to $C_G(H)$ (cf.\ the
proof of Proposition \ref{prop:LsepGnonsep}).  By Lemma \ref{lem:nilptwitness}, $y$ is a witness to the non-separability of $\pi_K(H')$, so $\pi_K(H')$ is not separable in $K_{\rm ad}$.
\end{exmp}

\begin{comment}
\begin{exmp}
\label{exmp:nonGcr}
 Let $C= \{u(a) \mid a\in k\}$.  Let $H'= \langle H\cup C\rangle$ and
suppose that $K$ is any reductive subgroup of $G$ containing $H'$.
Set $y:=e_\beta+ e_{3\alpha+ \beta}$.  Then $y$ belongs to the tangent
space $T_1(C)$, which is contained in $\Lie K$.  Now $\uu_{3\alpha+
2\beta}$ is centralized by $U_\beta$ and $U_{3\alpha+ 2\beta}$ and we
have $\Ad \kappa_{3\alpha+\beta}(a)(e_\beta) = e_\beta+ ae_{3\alpha+
2\beta}$, $\Ad \kappa_\beta(a)(e_{3\alpha+ \beta}) = e_{3\alpha+
\beta}+ ae_{3\alpha+2\beta}$ by the CRs, so $\Ad u(a)(y)= y$.  Hence $y\in \cc_\kk(H')$, but $y$ is not
tangent to $C_K(H')$ since it is not tangent to $C_G(H)$ (cf.\ the
proof of Proposition \ref{prop:LsepGnonsep}).

Suppose that $d\pi_K(y)$ is tangent to $C_{K_{\rm ad}}(\pi_K(H'))$.  By \cite[Prop.~IV.14.26]{borel}, there exists a connected unipotent subgroup $V$ of $C_{K_{\rm ad}}(\pi_K(H'))$ such that $d\pi_K(y)\in \Lie V$.  Note that $V\subseteq (K_{\rm ad})^0$ and $\pi_K$ gives an isogeny from $D(K^0)$ to $\pi_K(D(K^0))= (K_{\rm ad})^0$.  Let $U= (\pi_K^{-1}(V)\cap D(K^0))^0$.  Any semisimple element of $U$ must belong to the finite group $\ker \pi_K\cap D(K^0)$, so $U$ has only finitely many semisimple elements.  Hence $U$ is a unipotent group.  Since $\pi_K$ is an isogeny from $D(K^0)$ to $(K_{\rm ad})^0$ and $\ker d\pi_K$ consists of semisimple elements, it follows that the restriction of $\pi_K$ is an isomorphism from $U$ onto $V$.  Hence there exists $y'\in \Lie U$ such that $d\pi_K(y')= d\pi_K(y)$.  As $\ker d\pi_K\subseteq \zz(\kk)$ and $y,y'$ are both nilpotent, we must have $y=y'$, so $y\in \Lie U$.

Fix $h\in H'$.  If $u\in U$, then $\pi_K(h)\pi_K(u)\pi_K(h)^{-1}= \pi_K(u)$,
so we have $huh^{-1}=cu$ for some $c\in \ker \pi_K$.
The map $u\mapsto huh^{-1}u^{-1}$
is a morphism from the connected
set $U$ to the finite set $\ker \pi_K\cap D(K^0)$,
so $huh^{-1}u^{-1}=1$ for all $h\in H'$ and all $u\in U$.
Hence $U\subseteq C_K(H')$.  This implies that $y\in \Lie C_K(H')$,
a contradiction.  We conclude that $d\pi_K(y)$ is not tangent to
$C_{K_{\rm ad}}(\pi_K(H'))$.
Since $d\pi_K(y)\in \cc_{\kk_{\rm ad}}(\pi_K(H'))$,
it follows that $\pi_K(H')$ is not separable in $K_{\rm ad}$.
\end{exmp}
\end{comment}

Here is a further example arising from this construction which relates
to the discussions in Sections \ref{sec:redpairs} and \ref{sec:cent+norm}
(see in particular
Remarks \ref{rem:C_G(S)redpair}(ii) and \ref{rem:KvsKN}(ii)).

\begin{exmp}
\label{exmp:KnotKS}
Recall that $S$ is the torus $\alpha^\vee(k^*)$
and therefore, $(G,N_G(S))$ is a reductive pair by
Proposition~\ref{prop:C_G(S)redpair}(b).
We have that $H_a \subseteq N_G(S)$, since $U_{3\alpha+2\beta}$
centralizes $S$.  Let $a\in k^*$. We will show that $H_aS$ is not
$G$-cr, although $H_a$ is (Proposition \ref{prop:GcrnotMcr}).  Let
$\lambda\in Y(T)$ and $c_\lambda$ be as in the proof of Proposition
\ref{prop:GcrnotMcr}.  Then
$$ c_\lambda(H_aS)= \langle S\cup\{s_\alpha\} \rangle.
$$
If $H_aS$ lies in a Levi subgroup of $P_\lambda$ then $uH_aSu^{-1}\subseteq L_\lambda$ for some $u\in R_u(P_\lambda)$, so $c_\lambda(H_aS)= uH_aSu^{-1}$.  Thus to show that $H_aS$ is not $G$-cr, it is enough to show that
$\langle S\cup\{s_\alpha\} \rangle$ is not $R_u(P_\lambda)$-conjugate
to $H_aS$.  To see this, note that if $u\in R_u(P_\lambda)$ with
$u\langle S\cup\{s_\alpha\} \rangle u^{-1}= H_aS$, then $uSu^{-1}= S$,
whence $u$ centralizes $S$ (since $S$ normalizes $R_u(P_\lambda)$).
Since the centralizer of $S$ in $R_u(P_\lambda)$ is $U_{3\alpha+
2\beta}$ by Eqns.\ (\ref{eqn:fundalphaveepairings}) and
(\ref{eqn:alphaveepairings}), we have $u\in U_{3\alpha+ 2\beta}$.  But
$U_{3\alpha+ 2\beta}$ centralizes $H_aS$, so $\langle
S\cup\{s_\alpha\} \rangle= H_aS$, a contradiction.

Since $H_aS$ is not $G$-cr, $H_aS$ is not $N_G(S)$-cr (Corollary \ref{cor:KS<=>K}).  Since the canonical projection $f\colon N_G(S)\ra N_G(S)/S$ is non-degenerate and $f(H_a)= f(H_aS)$, it follows from \cite[Lem.\ 2.12(ii)]{BMR} that $H_a$ is not $N_G(S)$-cr.  Thus we have an example of a subgroup
$H_a \subseteq N_G(S)$, with $S$ linearly reductive --- in fact, a
torus --- such that $H_a$ is $G$-cr, but not $N_G(S)$-cr.

\end{exmp}

Finally, we consider a rationality question.  Let $k_0$ be a subfield
of an algebraically closed field $k_1$.  Suppose that $G_1$ is a
reductive algebraic group defined over $k_0$.  If $K_1$ is a subgroup
of $G_1$ defined over $k_0$, then we say that $K_1$ is
\emph{$G_1$-completely reducible over $k_0$} if whenever $P_1$ is an
R-parabolic subgroup of $G_1$ such that $K_1\subseteq P_1$ and $P_1$
is defined over $k_0$, there exists an R-Levi subgroup $L_1$ of $P_1$
such that $K_1\subseteq L_1$ and $L_1$ is defined over $k_0$ (see
\cite[Sec.~5]{BMR} for further details).  In particular, $K_1$ is
$G_1$-cr if and only if $K_1$ is $G_1$-cr over $k_1$.  An example of McNinch
\cite[Ex.\ 5.11]{BMR} shows that if $K_1$ is $G_1$-cr over $k_0$, then $K_1$
need not be $G_1$-cr over $k_1$.  The next example shows that the converse
can also happen.

\begin{comment}
\begin{lem}
\label{lem:levirat}
 Let $P$ be a parabolic subgroup of $G$, let $L$ be a Levi subgroup of
$P$ and suppose that $P,L$ are defined over $k_0$.  Given $p\in P$,
write $p=lu$, where $l\in L$ and $u\in R_u(P)$.  Then $p\in P(k_0)$ if
and only if $l\in L(k_0)$ and $u\in R_u(P)(k_0)$.
\end{lem}

\begin{proof}
 It is immediate that $p\in P(k_0)$ if $l\in L(k_0)$ and $u\in
R_u(P)(k_0)$.  Conversely, suppose that $p\in P(k_0)$.  By
\cite[Prop.\ 20.5, Thm.\ 6.8]{borel}, $R_u(P)$ is defined over $k_0$
and the canonical projection $\pi_P\colon P\ra P/R_u(P)$ is defined
over $k_0$.  Hence the map $\psi= \pi_P\circ i$ is defined over $k_0$,
where $i$ is the inclusion of $L$ in $P$.  So $l=\psi(p)\in L(k_0)$
and $u= l^{-1}p\in R_u(P)(k_0)$, as required.
\end{proof}
\end{comment}

\begin{exmp}
\label{exmp:mcninchcvse}
Suppose that $k_0$ is a subfield of $k$ such that $G$ is defined over
$k_0$ and $k_0$-split.  We can assume that $T$ is chosen so that $T$
is defined over $k_0$ and $k_0$-split and so that for every $\gamma\in
\Psi$, the homomorphisms $\gamma\colon T\ra k^*$, $\gamma^\vee\colon k^*\ra T$ and
$\kappa_\gamma\colon k\ra U_\gamma$ are defined over $k_0$.  Now
suppose that $k/k_0$ is not separable.  Then $k_0$ is not perfect, so
there exists $a\in k\setminus k_0$ such that $a^2\in k_0$.  Consider
$H_a= \langle \{t,s_\alpha \kappa_{3\alpha+ 2\beta}(a^2) \}\rangle$.  Then
$H_a$ is defined over $k_0$, since $H_a$ is a finite subgroup of $G(k_0)$.  As $H_a$ is
$G$-cr (Proposition \ref{prop:GcrnotMcr}), $H_a$ is $G$-cr over $k$.  We
show that $H_a$ is not $G$-cr over $k_0$.

\begin{comment}
Let $P$ be a parabolic subgroup of $G$ such that $P$ is defined over
$k_0'$ and $H\subseteq P$.  We may assume that $P\neq G$.  Since $H$
is $G$-cr, there is a Levi subgroup $L$ of $P$ such that $H\subseteq
P$.  Choose indivisible $\mu\in Y(G)$ such that $P= P_\mu$ and
$L=L_\mu$.  We can regard $\mu$ as an element of $Y(G_{3\alpha+
2\beta})$, and $\mu$ is non-trivial as $P\neq G$.  Since $G_{3\alpha+
2\beta}\cong \PGL_2(k)$, there exists $z\in G_{3\alpha+ 2\beta}$ such
that $\mu= z\cdot \lambda$.  Hence $zP_\lambda z^{-1}= P_\mu= P$.  By
\cite[Thm.\ 20.9(iii)]{borel}, there exists $g\in G(k_0')$ such that
$gP_\lambda g^{-1}= P$.  We can write $g= zp$ for some $p\in
N_G(P_\lambda)= P_\lambda$.

Let $P$ be a parabolic subgroup of $G$ such that $P$ is defined over
$k_0'$ and $H\subseteq P$.  We may assume that $P\neq G$.  Since any
Levi subgroup of a Borel subgroup of $G$ is abelian and $H$ is
non-abelian, $P$ is not a Borel subgroup of $G$.  By \cite[?]{borel},
$P\cap P_\lambda$ contains a maximal torus $S$ of $G$ such that $S$ is
defined over $k_0'$.  There exists $u\in R_u(P_\lambda)$ such that
$uSu^{-1}= T$.  Hence $uPu^{-1}$ is a parabolic subgroup of $G$
containing $T$ and $uHu^{-1}$.
\end{comment}

Recall that $P = P_\lambda$ and $L = L_\lambda$, where $\lambda$ is as
in the proof of Proposition \ref{prop:GcrnotMcr}.  Since $G$ and $T$ are split, $P$
and $L$ are defined over $k_0$ (cf.\ \cite[Props.~V.20.4 and V.20.5]{borel}).  Suppose there exists a Levi subgroup
$L'$ of $P$ such that $L'$ is defined over $k_0$ and $H_a$ is
contained in $L'$.  By \cite[Prop.~V.20.5]{borel}, there exists $u\in
R_u(P)(k_0)$ such that $L'= uLu^{-1}$.  Then $u^{-1}H_au$,
$u(a)^{-1}H_au(a)\subseteq L$, so $c_\lambda(u^{-1}gu)= u^{-1}gu$ and
$c_\lambda(u(a)^{-1}gu(a))= u(a)^{-1}gu(a)$ for all $g\in H_a$.  Since
$c_\lambda(u)= c_\lambda(u(a))= 1$, we have
\[
u^{-1}gu = c_\lambda(u^{-1}gu) = c_\lambda(g)
= c_\lambda(u(a)^{-1}gu(a)) =
u(a)^{-1}gu(a)
\]
for all $g\in H_a$.  Thus $u= u(a)c$ for some $c \in C_G(H)\cap
R_u(P)= G_{3\alpha+ 2\beta}\cap R_u(P) = U_{3\alpha+ 2\beta}$ (using Lemma \ref{lem:HGcr}(b)): say $u=
\kappa_\beta(a)\kappa_{3\alpha+ \beta}(a) \kappa_{3\alpha+ 2\beta}(y)$
for some $y\in k$.  But then $u\not\in P(k_0)$ since $a\not\in k_0$, a
contradiction.  Thus no such $L'$ can exist, and $H_a$ is not
$G$-cr over $k_0$.
\begin{comment}
We may regard $u$ as an element of $P_{\alpha^\vee}$, with $z\in
L_{\alpha^\vee}$ and $u(a)\in R_u(P_{\alpha^\vee})$.  By \cite[20.5
Prop.,\ 6.8 Thm.]{borel}, $R_u(P_{\alpha^\vee})$ is defined over $k_0$
and the canonical projection $\pi\colon P_{\alpha^\vee}\ra
P_{\alpha^\vee}/R_u(P_{\alpha^\vee})$ is defined over $k_0$.  Hence
the isomorphism $\psi= \pi\circ i$ from $L_{\alpha^\vee}$ to
$P_{\alpha^\vee}/R_u(P_{\alpha^\vee})$ is defined over $k_0$, where
$i$ is the inclusion of $L_{\alpha^\vee}$ in $P_{\alpha^\vee}$.  We
have $z= \psi^{-1}(\pi(u))\in L_{\alpha^\vee}(k_0)$, so $u\in
P_{\alpha^\vee}(k_0)$.  But $u\not\in P_{\alpha^\vee}(k_0)$, because
$a\not \in k_0$, a contradiction.  Thus no such $L$ can exist, and
$H_a$ is not $G(k_0)$-cr.
\end{comment}
\end{exmp}

\begin{comment}

Serre has asked the following question.

\begin{question}
\label{qn:sep}
 Suppose that $k_0'/k_0$ is a finite separable extension of fields,
and suppose that $G$ is defined over $k_0$.  Let $H$ be a subgroup of
$G$ defined over $k_0$.  Is it true that $H$ is $G(k_0)$-completely
reducible if and only if $H$ is $G(k_0')$-completely reducible?
\end{question}

In \cite[Thm.\ 5.8]{BMR}, it was shown that the answer is yes under
the additional hypothesis that $k_0$ and $k_0'$ are perfect.  We give
a sufficient criterion for the answer to be yes.  First we need some
terminology: if $H$ is a subgroup of $G$, then we say that a parabolic
subgroup $P$ of $G$ \emph{destabilizes} $H$ is $H\subseteq P$ but $H$
is not contained in any Levi subgroup of $P$.

\begin{thm}
\label{thm:sep}
 Suppose that $G$ and $H$ have the following property: for every
subfield $\Omega$ of $k$ such that $G$ and $H$ are defined over
$\Omega$, if $H$ is not $G(\Omega)$-cr, then there exists a parabolic
subgroup $P$ of $G$ defined over $\Omega$ such that $P$ destabilizes
$H$.  Then the answer to Serre's question is yes.
\end{thm}

\begin{proof}
(Sketch): Suppose $H$ is not $G(k_0')$-cr.  Then by hypothesis, there
exists a parabolic subgroup $P$ of $G$ defined over $k_0'$ such that
$P$ destabilizes $H$.  Let $P'$ be the optimising destablizing
parabolic subgroup for $H$ in the sense of [my generalisation of
Kempf's results: one takes the optimal destabilising parabolic
subgroup not over the whole of the building $X(G)$, but over the
subcomplex consisting of all parabolic subgroups of $G$ that are
defined over $k_0'$; this makes sense because this subcomplex is
non-empty.]  Then $P'$ is defined over $k_0'$ and is ${\rm
Gal}(k_0'/k_0)$-stable (since it is ``canonical'').  Since $k_0'/k_0$
is separable, $P'$ is therefore defined over $k_0$.  By definition of
destabilising, $H$ is not contained in any Levi subgroup of $P'$.
Thus $H$ is not $G(k_0)$-cr.

 Conversely, suppose $H$ is not $G(k_0)$-cr.  By hypothesis, there
exists a parabolic subgroup $P$ of $G$ defined over $k_0$ such that
$P$ destabilizes $H$.  Then $P$ is defined over $k_0'$ and $H$ is not
contained in any Levi subgroup of $P$ defined over $k_0'$, so $H$ is
not $G(k_0')$-cr.
\end{proof}

\begin{rem}
 The hypothesis of Theorem \ref{thm:sep} fails in Example
\ref{exmp:mcninchcvse}; $H_x$ is $G(k)$-cr but not $G(k_0)$-cr.  I
suspect, however, that this failure is a low-characteristic
phenomenon.  This is interesting because it suggests that the answer
to Question \ref{qn:sep} might depend on the characteristic.  At any
rate, the method of attack via passing to the algebraic closure is
sensitive to the characteristic.  But maybe there is an alternative
approach.
\end{rem}

\begin{exmp}
 [Sketch] Let $G_1$ and $H_1$ be $G$ and $H_x$ respectively from
Example \ref{exmp:mcninchcvse}.  Let $G_2$ and $H_2$ be $G$ and $H$
from McNinch's counterexample (see above).  Then the subgroup
$H_1\times H_2$ of $G_1\times G_2$ has the following properties:
$H_1\times H_2$ is not $G(k_0)$-cr and is not $G(k)$-cr, but there
does not exist any parabolic subgroup $P$ of $G_1\times G_2$ such that
$P$ is defined over $k_0$ and $P$ destabilizes $H_1\times H_2$.  For
any parabolic $P$ of $G_1\times G_2$ is of the form $P_1\times P_2$,
where $P_i$ is a parabolic subgroup of $G_i$; there are no
destabilising parabolic subgroups $P_2$ of $G_2$ for $H_2$ such that
$P_2$ is defined over $k_0$ (see \cite[Ex.\ 5.11]{BMR}), and there are
no destabilising parabolic subgroups $P_1$ of $G_1$ for $H_1$ at all.
\end{exmp}

\begin{question}
 Are there conditions that prevent counterexamples like our $G_2$ one
from occurring?  E.g., $p$ good, the unipotent radical of certain
parabolic subgroups abelian?  Can we express these conditions as a
single condition involving, say, the cohomology of $R_u(P)$ as a
(non-abelian) $P$-module?  Can we use these ideas to get some positive
results?  E.g., show that the conjecture works (and hence Serre's
question has a positive answer) under certain conditions?
\end{question}

\begin{rem}
If $U_\alpha$ centralizes $\uu_\gamma$, then $U_\alpha$ centralizes
$U_\gamma$; true by inspection of rank $2$ simple groups.
\end{rem}

\end{comment}

\begin{comment}
\begin{rem}
We can have $H \subseteq M \subseteq G$ with $M$ regular in $G$ and
$H$ $M$-cr but not $G$-cr (Liebeck's example from \cite{BMR}).  {\tt
[BM:- Fill this in.  The counterexample is $H:=\SP_m$ embedded
diagonally in $M:=\SP_m\times \SP_m$ inside $G:=\SP_{2m}$.]}  So this
proposition is maybe not so useful!
\end{rem}
\begin{comment}
If we can find a counterexample to Question \ref{q:sepI}, then this
gives another such counterexample.  (BM: I don't understand.  What
gives a counterexample to what?)  For: then $H \subseteq G$ acts
completely reducibly on $\gg$ but $H$ is not $G$-cr.  This forces $H$
to be non-separable in $G$ (\cite{BMR}).  DOES THIS WORK????????
\end{comment}

\begin{comment}
\begin{question}\label{q:sepIV}
Does $H$ separable in $G$ imply that $\pi_G(H)$ is separable in
$G_{\rm ad}$?
\end{question}
\end{comment}

\bigskip {\bf Acknowledgements}: The authors acknowledge the financial
support of EPSRC Grant EP/C542150/1 and Marsden Grant UOC0501.  Part
of the research for this paper was carried out while the authors were
staying at the Mathematical Research Institute Oberwolfach supported
by the ``Research in Pairs'' programme.  We are grateful to Martin
Liebeck and Gary Seitz for helpful discussions.

\bigskip

redpajama_set_name: RedPajamaArXiv