Actions and Invariants of Algebraic groups

Recordar

car 0

\section{Algebraic Lie algebras}

%\begin{defi}
%\label{defi:basicalglie}
%\index{Lie algebra!algebraic}
%Let $k$ be a field of arbitrary characteristic, $G$ be an affine algebraic group and
%$\mathfrak h \subset {\operatorname {\mathcal L}(G)}$ be a $k$--Lie subalgebra. Then
%$\mathfrak h$ is said to be {\em algebraic}\/ if there exists an
%affine algebraic subgroup $H \subset G$ with $\operatorname{\mathcal L}(H) =
%\mathfrak h$. 
%\end{defi}

begin{defi}
\label{defi:basicalglie}
\index{Lie algebra!algebraic}
Let $k$ be a field of arbitrary characteristic,
$\mathfrak h \subset {\operatorname {\mathcal L}(G)}$ be a $k$--Lie algebra. Then
$\mathfrak h$ is said to be {\em algebraic}\/ if there exists an
affine algebraic  $H$ with $\operatorname{\mathcal L}(H) =
\cong h$. 
\end{defi}


%See Exercises \ref{exer:car01c}, \ref{exer:car01d} and
%\ref{exer:car01e}, for examples of non-algebraic Lie algebras and related topics.

See Exercises \ref{exer:car01c}, \ref{exer:car01d} and
\ref{exer:car01e}, for examples of non-algebraic Lie algebras.

\begin{defi}\label{defi:alg-hull-lie}
\index{Lie algebra!algebraic hull}
\index[sim]{algebraic hull@$\mathfrak h^+$, algebraic hull}
 Let $G$ be an affine algebraic group and $\operatorname{\mathcal L}(G)$  its associated
Lie algebra. If $\mathfrak h \subset \operatorname{\mathcal L}(G)$
is  a Lie subalgebra we say that $\mathfrak h$ admits an {\em
algebraic hull}\/ if there exists an algebraic Lie subalgebra ${\mathfrak h}^{+}
\subset \operatorname{\mathcal L}(G)$ such that: (a) $\mathfrak h \subset {\mathfrak
h}^{+}$; (b) if $\mathfrak k$ is an algebraic Lie subalgebra of
$\operatorname{\mathcal L}(G)$ that contains $ \mathfrak h$, then ${\mathfrak h}^{+}
\subset \mathfrak k$.
\end{defi}

\begin{obse} 
It is clear that the algebraic hull has to be unique {\em if
it exists}. Next theorem guarantees the existence of the algebraic
hull
in the case of characteristic zero. Not too much seems
to be known about this concept in the general case.
See \cite{kn:Selalgliealg} for a survey about this subject with
special emphasis in the case of non-zero characteristic.
\end{obse}

\begin{theo} 
Suppose that $\operatorname{char} k=0$. Let $G$ be an affine algebraic group 
 and $\mathfrak h \subset \operatorname{\mathcal L}(G)$
be an arbitrary Lie subalgebra. Then $\mathfrak h$ admits an
algebraic hull. 
\end{theo}

\proof 
Consider the  closed subgroup $G\supset G_{\mathfrak h}=\bigcap\bigl\{K 
\subset G \tq K  \text{ is a closed
subgroup of  }G \text{ and } \mathfrak h \subset \operatorname{\mathcal
L}(K)\bigr\}$. As 
the Zariski topology is noetherian, there are a finite
number of subgroups $K_{i}\subset G$ , 1 \leq i \leq n$ such that 
$G_{\mathfrak h}=K_{1} \cap \dots \cap K_{n}$ and
$\mathfrak h \subset \operatorname{\mathcal L}(K_{i})$ for $i=1,\dots, n$.Using 
Theorem \ref{theo:crucial2}, we deduce that $\operatorname{\mathcal
L}(G_{\mathfrak h})=\operatorname{\mathcal L}(K_{1})\cap\dots \cap
\operatorname{\mathcal L}(K_{n}) \supset \mathfrak h$. Next we prove that 
 $G_{\mathfrak h}$ is
connected. Indeed, if we call ${G_{\mathfrak h}}_{1}$ the
identity component of $G_{\mathfrak h}$, then
if $\operatorname{\mathcal L}({G_{\mathfrak h}}_{1})= \operatorname{\mathcal L}({G_{\mathfrak h}}) 
\supset \mathfrak h$ and we can conclude that
the algebraic subgroup ${G_{\mathfrak h}}_{1}$ is one of the subgroups
we have to intersect in order to obtain $G_{\mathfrak h}$. Then 
${G_{\mathfrak h}}_{1} \supset G_{\mathfrak h}$.

Let us call ${\mathfrak h}^{\diamond}= \operatorname{\mathcal L}(G_{\mathfrak
h})$. It is clear that ${\mathfrak h}^{\diamond} \supset
\mathfrak h$, and if we take $\mathfrak k$ an algebraic Lie
subalgebra of $\mathfrak g$ such that $\mathfrak k \supset {\mathfrak
h}^{\diamond}$, being $\mathfrak k = \operatorname{\mathcal L}(K)$ for some
connected subgroup $K$ of $G$, 
% we deduce that $K$ is one of the subgroups that appear in the definition of $G_{\mathfrak h}$
% and then, $G_{\mathfrak h} \subset K$ and $\mathfrak h^{\diamond} \subset \mathfrak k$.
% Then, in accordance with Definition \ref{defi:alg-hull-lie}, we deduce that $\mathfrak h^{\diamond}= 
% \mathfrak h^{+}$.
from the inclusion relation
between the corresponding Lie algebras we conclude 
that $K \supset G_{\mathfrak h}$. Then
$\mathfrak k \supset  {\mathfrak h}^{\diamond}$ and hence  
${\mathfrak h}^{\diamond}$ is the algebraic hull of $\mathfrak h$.
\qed

\begin{theo} 
\label{theo:invhh+} 
Suppose that $\operatorname{char} k=0$. Let $G$ be an affine algebraic, $W$ 
a finite dimensional rational $G$--module and
$U \subset V \subset W$ a pair of subspaces. If $\mathfrak h \subset
\operatorname{\mathcal L}(G)$ is a Lie subalgebra with the property that
$\mathfrak h \cdot V \subset U$, then $\mathfrak h^{+} \cdot V \subset U$. 
\end{theo}

\proof 
Defining $H$ as in Theorem \ref{theo:transfer} and using
the corresponding characterization of $\operatorname{\mathcal L}(H)$, we 
conclude that
$\mathfrak h \subset \operatorname{\mathcal L}(H)$. Then 
${\mathfrak h}^{+} \subset \operatorname{\mathcal L}(H)$ or equivalently
${\mathfrak h}^{+}\cdot V \subset U$.
\qed



\begin{theo} 
\label{theo:conm+}
Assume $\operatoraname{char}k=0$. Let $G$ be an affine algebraic group and 
$\mathfrak h \subset \operatorname{\mathcal L}(G)$ a Lie subalgebra. Then
$[\mathfrak h, \mathfrak h]= [{\mathfrak h}^{+}, {\mathfrak h}^{+}]$.
\end{theo}

\proof
 Consider ${\operatorname {Ad}}: G \times \operatorname{\mathcal L}(G)
\rightarrow \operatorname{\mathcal L}(G)$ and apply Theorem \ref{theo:invhh+} to
the case that $W=\operatorname{\mathcal L}(G)$, $U=[\mathfrak h, \mathfrak h]$ and
$V= \mathfrak h$. We first conclude that 
$[{\mathfrak h}, {\mathfrak h}^{+}]=[{\mathfrak h}^{+}, {\mathfrak h}]
\subset [{\mathfrak h}, {\mathfrak h}]$. Applying again the mentioned
theorem this time with 
$W=\operatorname{\mathcal L}(G)$, $U=[\mathfrak h, \mathfrak h]$ and $V= {\mathfrak
h}^{+}$, we conclude that $[{\mathfrak h}^{+}, {\mathfrak h}^{+}]
\subset [{\mathfrak h}, {\mathfrak h}]$. The other inclusion is obvious. 
\qed


\begin{obse}
\label{obse:generalization2}
Theorem \ref{theo:conm+} can be generalized as follows. Let
$\mathfrak h$ be a
Lie subalgebra of $\operatorname{\mathcal L}(G)$. 
For $i \geq 1$, we have that $D^{i}(\mathfrak h)=D^{i}({\mathfrak
h}^{+})$ and $D^{[i]}(\mathfrak h)=D^{[i]}({\mathfrak
h}^{+})$ --see Definition \ref{chapter:liealg}.\ref{defi:derivados} and Exercise \ref{exer:car01a}--.
\end{obse}


\begin{coro} 
\label{coro:radalg} 
Assume that $\operatorname{char}k=0$ and let $G$ be an affine algebraic group
The radical $\operatorname{\mathcal L}(G)_{r}$
of $\operatorname{\mathcal L}(G)$ is an algebraic Lie subalgebra.
\end{coro}

\proof
 By Theorem \ref{theo:conm+} the ideal ${\operatorname {\mathcal L}(G)}_{r}^{+}$
of $\operatorname{\mathcal L}(G)$ satisfies that $[{\operatorname{\mathcal L}(G)}_{r}^{+}, {\operatorname{\mathcal L}(G)_{r}^{+}] = [{\operatorname{\mathcal
 L}(G)}_{r}, {\operatorname{\mathcal L}(G)}_{r}]$, i.e.~ 
$D^{1}({\operatorname{\mathcal L}(G)}_{r}^{+})=D^{1}({\operatorname{\mathcal 
L}(G)}_{r})$. Then, by
induction we conclude that for all $i \geq 1$, 
$D^{i}({\operatorname{\mathcal L}(G)}_{r}^{+})=D^{i}({\operatorname{\mathcal L}(G)}_{r})$, and  from
the solvability of ${\operatorname{\mathcal L}(G)}_{r}$ we deduce the solvability
 of ${\operatorname{\mathcal L}(G)}_{r}^{+}$. As ${\operatorname {\mathcal L}(G)}_r$ is the {\em maximal}  solvable ideal of ${\operatorname {\mathcal L}(G)}$, and ${\operatorname{\mathcal 
L}(G)}_{r}^{+}$ is  solvable we conclude that ${\operatorname {\mathcal L}(G)}_{r}^+ ={\operatorname{\mathcal L}(G)}_{r}$.
\qed

\begin{defi}
\label{defi:G-nil}
\index{Lie subalgebra!$G$-nilpotent}
Let $G$ be an affine algebraic group
and $\mathfrak h$ a subalgebra of $\operatorname{\mathcal L}(G)$. We
say that $\mathfrak h$ is {\em $G$--nilpotent}\/ if the action of $\mathfrak
h$ on $k[G]$ is locally nilpotent (i.e.~nilpotent on every finite
dimensional $G$--stable subspace of $k[G]$).
\end{defi}

\begin{theo}
\label{theo:nilp-unip} 
Let $G$ be an affine algebraic
group and  $H\subset G$ a closed subgroup. If $H$ is a unipotent subgroup,
then $\operatorname{\mathcal L}(H)$ is a $G$--nilpotent subalgebra. Conversely, 
if $H$ is connected, the characteristic of the base field is zero and 
$\operatorname{\mathcal L}(H)$ is $G$--nilpotent, then $H$ is unipotent.
\end{theo}

\proof
 Let $V$ be a finite dimensional $G$--stable submodule of
$k[G]$ and $\{0\}=V_{n} \subset V_{n-1} \subset \dots
\subset V_{0}=V$ a 
composition series of $V$ as an $H$--module. As $H$  is unipotent,and the quotients
$V_i/V_{i+1}$ are simple, 
Corollary \ref{chapter:alg-grp2}.\ref{coro:unip-fixed} guarantees that
the action of $H$  on $V_{i}/V_{i+1}$ is trivial for
$i=0,\dots, n-1$. This implies that  $\operatorname{\mathcal L}(H)$ acts trivially 
on the composition factors and hence the action of $\operatorname{\mathcal L}(H)$ on $V$ is by 
nilpotent linear transformations.  Then $\operatorname{\mathcal L}(H)$ is
$G$--nilpotent. 

Conversely, take $V \subset k[G]$ a finite dimensional $G$--submodule
of $k[G]$ that contains a set of algebra generators of $k[G]$. As
$\operatorname{\mathcal L}(H)$ is $G$--nilpotent, there exists a finite number of
subspaces of $V$, $\{0\}=V_{n} \subset V_{n-1} \subset \dots
\subset V_{0}=V$, with the property that for $i=0,\dots,n-1$,
${\operatorname{\mathcal L}(H)}\cdot V_{i} \subset V_{i+1}$. Consider $K=\bigl\{x \in
G \tq x\cdot V_{i}=V_{i}, \text{ and }
x\cdot(V_{i}/V_{i+1})=id_{V_{i}/V_{i+1}},\, i=1,\dots ,n-1\bigr\}$. An
easy generalization of Theorem \ref{theo:transfer} --see Exercise
\ref{exer:car03}-- guarantees that $\operatorname{\mathcal L}(K)=\bigl\{\tau \in
\operatorname{\mathcal L}(G)\tau \cdot V_{i} \subset V_{i+1}, i=0,\dots, n-1\bigr\}$. Then
$\operatorname{\mathcal L}(H)\subset \operatorname{\mathcal L}(K)$ and if 
$K_{1}$ is the connected component of the identity in $K$, we deduce that 
$\operatorname{\mathcal L}(H) \subset \operatorname{\mathcal L}(K_{1}) = 
\operatorname{\mathcal L}(K)$. We
conclude from Corollary \ref{coro:inclusion} that $H \subset K_{1}
\subset K$. As $K$ acts unipotently on $V$ and $V$ generates $k[G]$,
we conclude that $K$ --and hence $H$-- is unipotent.
\qed

\begin{defi}
Assuma that $\operatorname{char} k=0$  and let $G$ be an affine algebraic group
with unipotent radical $R_u(G)$. The ideal
$\operatorname{\mathcal L}(R_u(G))$ is called the {\em $G$--nilpotent radical}\/ of
$\operatorname{\mathcal L}(R_u(G))$.
\end{defi}

\begin{theo}
\label{theo:largest-unip}
Assume that $\operatorname{char}k=0$ and let $G$ be an affine algebraic
group. Then $\operatorname{\mathcal L}(R_u(G))$ is the largest $G$--nilpotent ideal
of $\operatorname{\mathcal L}(G)$.
\end{theo}

\proof
 If $\mathfrak a$ is a $G$--nilpotent ideal of $\operatorname{\mathcal
L}(G)$, then using Exercise \ref{exer:car01}  we
deduce  that $\mathfrak a^{+}$ is also a $G$--nilpotent ideal of
 $\operatorname{\mathcal L}(G)$. Let $U$ be the connected normal
 unipotent subgroup of $G$ that has 
$\mathfrak a^{+}$ for Lie algebra. By definition of the unipotent
radical, we conclude that $U \subset R_u(G)$ and this implies that
$\mathfrak a \subset \operatorname{\mathcal L}(R_u(G))$.
\qed

\begin{defi}
\index{Lie subalgebra!$G$-linearly reductive} 
\label{defi:reductive-lie}
 Let $G$ be an affine algebraic
group. A Lie subalgebra $\mathfrak
r$ of $\operatorname{\mathcal L}(G)$ is said to be {\em $G$-linearly reductive}\/
 if every 
rational $G$--module is semisimple as an $\mathfrak r$--module.
\end{defi}

\begin{theo}
\label{theo:mostowlinear}
\index{Mostow's Theorem}
\index[nam]{Mostow, G.}
 Assume that $\operatorname{char}k=0$ and let $G$ be an affine algebraic
group. There exists a $G$--linearly reductive algebraic Lie subalgebra
$\mathfrak r \subset \operatorname{\mathcal L}(G)$ such that
$\operatorname{\mathcal 
L}(G)=\operatorname{\mathcal L}(R_u(G)) \oplus \mathfrak r$.
\end{theo}

\proof
 It is an immediate consequence of Theorem
\ref{theo:nilp-unip} that $\operatorname{\mathcal L}(R_u(G))$ is a nilpotent (and
hence solvable) ideal of $\operatorname{\mathcal L}(G)$. Then it 
follows that $\operatorname{\mathcal L}(R_u(G)) \subset  \operatorname{\mathcal L}(G)_{r}$, where
as usual $\operatorname{\mathcal L}(G)_{r}$ denotes the radical of $\operatorname{\mathcal L}(G)$. 
As $[\operatorname{\mathcal L}(G), \operatorname{\mathcal L}(G)_{r}]$ 
 is an ideal --see Lemma
\ref{chapter:liealg}.\ref{lema:derivados}-- that acts nilpotently on any finite dimensional rational
$G$--module,   
it follows from Theorem \ref{theo:largest-unip} that 
$[\operatorname{\mathcal L}(G), \operatorname{\mathcal L}(G)_{r}] \subset \operatorname{\mathcal L}(R_u(G))$.


Using Levi's decomposition --Theorem
\ref{chapter:liealg}.\ref{theo:levidecomposition}--  we can 
guarantee the existence of a 
semisimple Lie subalgebra $\mathfrak s$ of $\operatorname{\mathcal L}(G)$ such
that $\operatorname{\mathcal L}(G)_{r} \oplus \mathfrak s = \operatorname{\mathcal L}(G)$. By
Weyl's theorem --Theorem
\ref{chapter:liealg}.\ref{theo:weylliesemisimple}-- all the 
representations of  
the Lie algebra $\mathfrak s$ are completely reducible. This implies
in particular that $\mathfrak s$ is $G$--linearly reductive. 
Among the $G$--linearly reductive Lie subalgebras of $\operatorname{\mathcal
L}(G)$ that contain $\mathfrak s$, take a maximal one and denote it as
$\mathfrak r$. It is easy to prove that $\mathfrak r$ is an algebraic Lie
algebra. In fact, the algebraic hull ${\mathfrak r}^{+}$ of $\mathfrak r$ is
also $G$--linearly reductive --see Exercise \ref{exer:car02}-- and contains
$\mathfrak r$. Hence ${\mathfrak r}^{+}=\mathfrak r$. Our
goal is  to prove that $\operatorname{\mathcal L}(G)=\operatorname{\mathcal L}(R_u(G)) +
\mathfrak r$. In order to do this 
we use the $G$--linear reductivity of $\mathfrak r$
to conclude that there exists an $\mathfrak r$--submodule of
$\operatorname{\mathcal L}(G)_{r}$, that we call $\mathfrak p$, such that 
$\operatorname{\mathcal L}(G)_{r}= \operatorname{\mathcal L}(R_u(G)) \oplus \mathfrak p$. In this
situation $[\mathfrak r,\mathfrak p]=\{0\}$.  
Indeed, from $[\operatorname{\mathcal L}(G), \operatorname{\mathcal
L}(G)_{r}] \subset \operatorname{\mathcal L}(R_u(G))$ we deduce that 
$[\operatorname{\mathcal L}(G), \operatorname{\mathcal L}(R_u(G)) + \mathfrak p] \subset 
\operatorname{\mathcal L}(R_u(G))$ and then that $[\operatorname{\mathcal L}(G), \mathfrak p] \subset 
\operatorname{\mathcal L}(R_u(G))$. As $\operatorname{\mathcal L}(G)$ is the sum of 
$\operatorname{\mathcal L}(G)_{r}$ and $\mathfrak r$, we conclude that
$[\mathfrak r, \mathfrak p] \subset \operatorname{\mathcal L}(R_u(G))$, but as we
also have that $[\mathfrak r, \mathfrak p] \subset \mathfrak p$ 
 the conclusion that $[\mathfrak r, \mathfrak p]=0$ follows.

Next we prove the equality $\operatorname{\mathcal L}(G)_{r}= 
\operatorname{\mathcal L}(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$.
Assume that the inclusion $\operatorname{\mathcal L}(G)_{r} \supset 
\operatorname{\mathcal L}(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$ is
strict. In this case we can find $\rho \in  \operatorname{\mathcal L}(R_u(G))$ and
$\nu \in \mathfrak p$ such that $\rho + \nu \not \in \operatorname{\mathcal L}
(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$. It follows that $\nu
\not \in \operatorname{\mathcal L}(R_u(G)) + \mathfrak r$. Indeed, if $\nu=\alpha + \beta$
with $\alpha \in \operatorname{\mathcal L}(R_u(G)), \beta \in \mathfrak r$ as $\nu
\in \mathfrak p \subset \operatorname{\mathcal L}(G)_{r}$ we conclude that $\beta
\in \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$ and then $\rho + \nu =
\rho + \alpha + \beta \in \operatorname{\mathcal L}(R_u(G)) +
\operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$. 

Consider now ${\nu}^{s}, {\nu}^{n}$ the
semisimple and nilpotent parts of $\nu$ in $\operatorname{\mathcal L}(G)$. 
Corollary \ref{coro:radalg} and Exercise \ref{exer:car04} guarantee that
${\nu}^{s}, {\nu}^{n} \in \operatorname{\mathcal L}(G)_{r}$.
The subspace $\operatorname{\mathcal L}(R_u(G)) + k {\nu}^{n} \subset
\operatorname{\mathcal L}(G)$ is a $G$--nilpotent  ideal. In fact if
we compute $[\operatorname{\mathcal L}(G), \operatorname{\mathcal
  L}(R_u(G)) + k {\nu}^{n}] \subset \operatorname{\mathcal L}(R_u(G)) +
[\operatorname{\mathcal L}(G), {\nu}^{n}] \subset
\operatorname{\mathcal L}(R_u(G)) + [\operatorname{\mathcal L}(G), \operatorname{\mathcal L}(G)_{r}] \subset \operatorname{\mathcal
L}(R_u(G))$, we deduce that ${\mathcal
L}(R_u(G)) + k {\nu}^{n} \subset \operatorname{\mathcal L}(G)$
is an ideal and also that $[\operatorname{\mathcal L}(R_u(G)),{\nu}^{n}] \subset
\operatorname{\mathcal L}(R_u(G))$. 
Using  Exercise
\ref{chapter:liealg}.\ref{exer:nilp4} we deduce that $\operatorname{\mathcal L}(R_u(G)) + k
{\nu}^{n}$ is 
a $G$--nilpotent ideal and then that $\operatorname{\mathcal L}(R_u(G)) + k
{\nu}^{n} \subset \operatorname{\mathcal L}(R_u(G))$. Hence, ${\nu}^{n} \in \operatorname{\mathcal
L}(R_u(G))$. As $\nu = {\nu}^{s} + {\nu}^{n} \not \in \operatorname{\mathcal
L}(R_u(G)) + \mathfrak r$, it follows that ${\nu}^{s} \not \in
\mathfrak r$. Consider now the subspace 
$\mathfrak r + k {\nu}^{s}$ that is strictly larger than $\mathfrak
r$. 
As $\mathfrak r$ and $\mathfrak p$ commute, $\mathfrak r$ commutes with
$\nu$ and then we deduce from general results about Lie algebras 
--see Section \ref{chapter:alg-grp2}.\ref{section:jordan-coalg}-- that
$\mathfrak r$ 
%
%esto falta poner mejor
%
commutes with $\nu^{s}$. 
But then $\mathfrak r + k {\nu}^{s}$ is a $G$--linearly reductive sub-Lie
algebra of $\operatorname{\mathcal L}(G)$ that is larger than $\mathfrak r$. 
This contradicts the maximality of $\mathfrak r$ and then, we conclude
that $\operatorname{\mathcal L}(G)_{r}= 
\operatorname{\mathcal L}(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$.

This last equality 
guarantees that $\operatorname{\mathcal L}(G)=\operatorname{\mathcal L}(R_u(G)) +
\mathfrak r$ because if $\operatorname{\mathcal L}(G)_{r}= 
\operatorname{\mathcal L}(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r$,
we deduce that $\operatorname{\mathcal L}(G)=\operatorname{\mathcal L}(G)_{r} + \mathfrak r = 
\operatorname{\mathcal L}(R_u(G)) + \operatorname{\mathcal L}(G)_{r} \cap \mathfrak r + \mathfrak
r = \operatorname{\mathcal L}(R_u(G)) + \mathfrak r $.

Next, we observe that the sum is direct. Call $\mathfrak k= \operatorname{\mathcal
L}(R_u(G)) \cap \mathfrak r$. It is clear that $\mathfrak k$ is at the same time
$G$--nilpotent as well as $G$--linearly reductive. Then, it has to be
trivial (see Exercise \ref{exer:car05}). 
\qed


\section{Exercises}

\begin{exer}
\label{exer:car0-1}
Assume that  $\operatorname{char}k=p>0$ and consider the following closed
 subgroups of $G_a^{2}$: $H=\{(a,a^{p})\tq a \in k\}$ and $K=\{(a,0)\tq a \in k\}$.
 Show that $\operatorname{\mathcal L}(H)=\operatorname{\mathcal L}(K)$.
\end{exer}

\begin{exer}
\label{exer:car00} 
In the situation of Definition
\ref{defi:infin} prove that $I_{\tau,\infty} \subset k[G]$ is a Hopf ideal.
\end{exer}

\begin{exer}
\label{exer:car00a} 
Let  $G$ be an affine algebraic
group. If $f \in k[G]$ and $\sigma
\in \operatorname{\mathcal L}(G)$, then $\sigma\cdot f=\sum \sigma(f_{1})f_{2}$.
In this situation prove that if $\tau_{i},\, 1 \leq n$ is a
finite set of elements of $\operatorname{\mathcal L}(G)$ and $f,g \in k[G]$, then
$(\tau_{1}\dots \tau_{n})(fg)=(\tau_{2}\dots
\tau_{n})(f\tau_{1}\cdot g+ (\tau_{1}\cdot f)g)$ and $(\tau_{1}\dots
\tau_{n})(f)=(\tau_{2}\dots \tau_{n})(\tau_{1}\cdot f)$. Prove that 
$(\tau_{1}\dots \tau_{n})(Sf)=(-1)^{n}(\tau_{n}\dots \tau_{1})(f)$.
\end{exer}

\begin{exer}
\label{exer:car00b} 
Let $G$ be an affine algebraic group. Define the action of $G$
on $\operatorname{\mathcal L}(G)$ by the formula: if $x
\in G$ and $\tau \in \operatorname{\mathcal L}(G)$, then $(x\cdot \tau\cdot
x^{-1})(f)=
\tau(x^{-1}\cdot f\cdot x)$. Prove that for $\tau_{1},\dots,\tau_{n}
\in \operatorname{\mathcal L}(G)$ we have that $x\cdot (\tau_{1}\dots 
\tau_{n})\cdot
x^{-1}=(x\cdot \tau_{1}\cdot x^{-1})\dots (x\cdot\tau_{n}\cdot x^{-1})$. 
\end{exer}

\begin{exer}
 \label{exer:car01b}
Let $G$ be a connected affine algebraic group and $\tau\in \operatorname{\mathcal L}(G)$.  
Prove that if $\operatorname{char}k>0$ then the closed
 subgroup
$G_{\tau,\infty}$ is trivial. In $\operatorname{char}k=0$, 
prove that $I_{\tau,\infty}$ is a prime ideal, and that the group
$G_{\tau,\infty}$ is contained in every algebraic subgroup of $G$
whose Lie algebra contains $\tau$. {\sc Hint: }to prove that
$I_{\tau,\infty}$ is prime take $f,g \not \in I_{\tau,\infty}$ and
choose $p$ and $q$ minimal with the property that $\tau^{p}(f) \neq 0$
and similarly for $q$ and $g$. Prove that if we write $\tau^{p+q}(fg)$
as a sum of products of $\tau^{n}(f)\tau^{m}(g)$ only one of the
summands is non-zero and corresponds to $n=p, m=q$.   
\end{exer}

\begin{exer}
\label{exer:car00f} 
Assume that $\operatorname{char}k=0$. Let $G$ be an affine algebraic group
 and  $U \subset V$ finite dimensional rational $G$--modules. Call $H=\{x
\in G\tq x\cdot v - v \in U,\, \forall v \in V\}$. Prove that
$\operatorname{\mathcal L}(H)=\{\tau \in \operatorname{\mathcal L}(G)\tq \tau\cdot V \subset
U\}$. {\sc Hint: }Observe that --if
$\{v_{1},\dots, v_{n}\}$ is a basis of $V$-- $H$ can be
viewed as $H=\bigcap_{i=1,\dots,n}G_{v_{i}+U}$. Deduce the result from
Theorem \ref{theo:lie-of-fixed}.
\end{exer}


\begin{exer}
\label{exer:car00x} 
Let $G$ be an affine algebraic
group, $V$  a finite
dimensional rational $G$--module  and $W \subset V$ a linear subspace
of $V$. Define $G_{W}=\{x \in G\tq x\cdot W = W\}$ and ${\operatorname{\mathcal 
L}(G)}_{W}=\{\tau \in \operatorname{\mathcal L}(G) \tq \tau \cdot W \subset W\}$. 
Prove that
$\operatorname{\mathcal L}(G_{W}) \subset {\operatorname{\mathcal L}(G)}_{W}$ and 
that if $\operatorname{char}k=0$, then $\operatorname{\mathcal L}(G_{W})
= {\operatorname{\mathcal L}(G)}_{W}$. {\sc Hint: }reduce the assertion to the 
case that $\dim W=1$  by taking a convenient exterior power.
\end{exer}

\begin{exer}
\label{exer:car00c}
 Let $G$ and $H$ be affine algebraic
groups and $\rho:G \rightarrow H$ a morphism of algebraic groups. Prove that in this
situation $\operatorname{\mathcal L}(\rho(H))=\rho^{\bullet}(\operatorname{\mathcal L}(H))$.
\end{exer}

\begin{exer}
\label{exer:car00d}
 Assume that $\operatorname{char}k=p>0$ and consider the action of $G_{a}$
on $k^{2}$ given by the rule: $a\cdot(x,y)=(x,y+a^{p})$. Compute the
action of the corresponding Lie algebra on $k^{2}$ and conclude that
there are examples of subspaces of $k^{2}$ stable with respect to the
action of the Lie algebra but not with respect to the action of $G_{a}$.
\end{exer}

\begin{exer}
\label{exer:car00e} 
Asusme that $\operatorname{char}k=p>0$. Consider the closed subgroup $G=\left\{
\left(\begin{smallmatrix} a & 0 & 0 \\
0 & a^{p} & b \\ 0 & 0 & 1 \end{smallmatrix}\right) \tq a,b \in k, a\neq
0\right\} \subset {\operatorname {GL}}_3$ --see Exercise
\ref{chapter:alg-grp0}.\ref{exer:otherunip2}--. Prove  
that 
\[
\operatorname{\mathcal L}(G)= 
\left\{\left(\begin{smallmatrix} a & 0 & 0 \\
0 & 0  & b \\ 0 & 0 & 0 \end{smallmatrix}\right) \tq a,b \in k \right\}\
.
\]

Observe that $G$ 
is non abelian and $\operatorname{\mathcal L}(G)$ is abelian. Prove that
$\operatorname{\mathcal Z}(G)=\{1\}$ and ${\operatorname {Ker}}{\operatorname {Ad}} \neq
\{1\}$. Observe that this also provides a counter--example to
Corollary \ref{coro:estab}.
\end{exer}

\begin{exer}
\label{exer:car01c} 
Let $\mathfrak g$ the 3-dimensional Lie algebra generated by $x,y,z
\in \mathfrak g$ with bracket defined by the following rules:
$[x,y]=0,\,[z,x]=x+y,\, [z,y]=y$. Prove that this Lie algebra is not associated to an 
affine algebraic group.
\end{exer}

\begin{exer}
\label{exer:car01d} 
Show that the Lie algebra of an algebraic subgroup of a
complex torus is defined over ${\mathbb Q }$. Exhibit a non algebraic
Lie subalgebra of the abelian Lie algebra of dimension $n$ over
$\mathbb C$.
\end{exer}

\begin{exer}
\label{exer:car01e} 
Let $k$ be an arbitrary field and
consider the Lie subalgebra of ${\mathfrak {gl}}_{4}$ defined as
follows: $\mathfrak g = \left\{\left( \begin{smallmatrix} a & a & c & d \\
                                                     0 & a & 0 & e \\
                                                     0 & 0 & b & b \\
                                                     0 & 0 & 0 & b
                                 \end{smallmatrix}\right): a,b,c,d,e \in
k\right\}$.
Prove that $\mathfrak g$ is not algebraic.
\end{exer}


\begin{exer}
\label{exer:car01a} 
Assume that $\operatorname{char}k=0$ and let $G$ be an affine algebraic group.
 Let $\mathfrak h$ be a Lie subalgebra of \operatorname{\mathcal L}(G)$. Prove
 that for all $i \geq 1$, we have that:  $D^{i}(\mathfrak
h)=D^{i}({\mathfrak h}^{+})$ and $D^{[i]}(\mathfrak
h)=D^{[i]}({\mathfrak h}^{+})$.
\end{exer}

\begin{exer}
\label{exer:car03}
 State and prove a generalization of
Theorem \ref{theo:transfer} for a family of $n$ subspaces. 
\end{exer}

\begin{exer}
\label{exer:car01} 
Assume that $G$ is an affine algebraic group defined
over a field of characteristic zero. Let $\mathfrak a$ be an ideal of 
$\operatorname{\mathcal L}(G)$. Then ${\mathfrak a}^{+}$ is also an ideal. If
$\mathfrak a$ is $G$--nilpotent prove that ${\mathfrak a}^{+}$ is also
$G$--nilpotent. 
\end{exer} 

\begin{exer}
\label{exer:car02} 
Assume that $\operatorname{char}k=0$. Let  $G$ be an affine algebraic group and 
let $\mathfrak r$  a $G$--linearly
reductive Lie subalgebra of $\operatorname{\mathcal L}(G)$. Prove that the algebraic
hull of $\mathfrak r$ is also $G$--linearly reductive.
\end{exer}


\begin{exer}
 \label{exer:car04}
Assume that $\operatorname{char}k=0$. Let $G$ be an affine algebraic group and  $\mathfrak r$  an  algebraic
Lie subalgebra of $\operatorname{\mathcal L}(G)$. Prove that if $\nu \in \mathfrak r$
then ${\nu}^{s} \in \mathfrak r$ and ${\nu}^{n} \in \mathfrak r$, where
${\nu}^{s}$ and ${\nu}^{n}$ are the semisimple and nilpotent parts of $\nu$. 
\end{exer}



\begin{exer}
 \label{exer:car05}
Assume that $\operatorname{char}k=0$ and let $G$ be an affine algebraic group.

\noindent (a) Let $\mathfrak r$ be a $G$--linearly reductive 
Lie subalgebra of $\operatorname{\mathcal L}(G)$ and $\mathfrak a \subset
\mathfrak r$  an ideal. Then $\mathfrak a$ is $G$--linearly reductive.

\noindent (b) Assume that $\mathfrak t$ is a $G$--nilpotent subalgebra
that is at the same time $G$-linearly reductive. Conclude that
$\mathfrak t$ is trivial. 

\noindent (c) Conclude that if $\mathfrak n$ is a $G$--nilpotent ideal and
$\mathfrak r$ a $G$--linearly reductive subalgebra, then $\mathfrak n
\cap \mathfrak r =\{0\}$.
\end{exer}

AlgebraicGroups (última edición 2004-04-28 15:10:30 efectuada por )