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\newtheorem{theorem}{Theorem}
\newtheorem{proposition}{Proposition}
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\begin{center}
\textbf{18.745 Introduction to Lie algebra\\
Victor Kac , Fall 2004 \\
Lecture 6 , noted by Liu Ruochuan}
\end{center}

\noindent Throughout this lecture $\mathbb{F}$ will be assumed to
be an algebraically closed field.
\subsection*{Jordan Decomposition}
 Now suppose $V$ is a finite dimensional vector
space over $\mathbb{F}$ and $A$ is a linear operator on $V$. Let
$\{\lambda_{1},\cdots,\lambda_{s}\}$ denote the set of distinct
eigenvalues of $A$. By the theorem of Jordan normal form, in some
basic of $V$, the matrix of $A$ is a direct sum of the Jordan
blocks $J_{\lambda_{i}}$ assigned to some eigenvalue
$\lambda_{i}$. If we denote by $V_{\lambda_{i}}$ the span of the
vectors in the basis which correspond to all Jordan blocks
assigned to $\lambda_{i}$, we obtain the following generalized
eigenspace decomposition of $V$.
\begin{center}
 $V=\bigoplus_{i=1}^{s}V_{\lambda_{i}}$,
\end{center}
where the generalized eigenspace $V_{\lambda_{i}}$ can also be
defined as
\begin{center}
$V_{\lambda_{i}}=\{v\in V|(A-\lambda_{i}I)^{N}v=0$ for some $N$ \}
\end{center}
We take $A_{s}$ as the diagonal part of the Jordan normal form of
$A$, $A_{n}=A-A_{s}$. Then $A_{s}$ is semisimple, i.e
diagonalizable operator and $A_{n}$ is
a nilpotent operator. Moreover, $A_{s}A_{n}=A_{n}A_{s}$. This decomposition is called a Jordan decomposition of $A$. \\

\noindent \textbf{Ex6.1} Show that there exist polynomials $P(x)$
and $Q(x)$ such that $A_{s}=P(A)$ and $A_{n}=Q(A)$\\
\textbf{Solution}. By Chinese remainder theorem, there exists a
polynomial $P(x)$ such that
$P(x)\equiv\lambda_{i}(mod(x-\lambda_{i})^{n})$ for every
$\lambda_{i}$, where $n=$dim$V$. Then for $v\in V_{\lambda_{i}}$,
we have $P(A)v=\lambda_{i}v$. That means $P(A)=A_{s}$ and
$(1-P)(A)=A_{n}$.\\

\noindent \textbf{Ex6.2} If linear operators $A$ and $B$ commute,
then any eigenspace and generalized eigenspace of $A$ is
$B$-invariant. Conclude that two commuting semi-simple operators
can be diagonalized in the same basic.\\
\textbf{Solution}. Let $V_{\lambda_{i}}$ be the generalized
eigenspace of A with eigenvalue $\lambda_{i}$. For every $v\in
V_{\lambda_{i}}$, we have $(A-\lambda_{i})^{N}v=0$ for some $N$.
Since $A$ commutes with $B$, we get
$(A-\lambda_{i})^{N}Bv=B(A-\lambda_{i})^{N}v=0$. Thus $Bv$ is in
$V_{\lambda_{i}}$ by definition. That just means the generalized
eigenspace of $A$ is $B$-invariant. The method for eigenspace case
is the same. Now suppose $A$ and $B$ are both semi-simple. Let
$V=\bigoplus_{i=1}^{s}V_{\lambda_{i}}$ be the eigenspace
decomposition of $A$. Since $B$ is semi-simple and
$V_{\lambda_{i}}$ is an invariant subspace w.r.t $B$, $B$ is
semi-simple on $V_{\lambda_{i}}$. Then $B$ can be diagonalized in
$V_{\lambda_{i}}$ under some basis. Now under the basis, which is
the union of these basis, $A$ and $B$ are all diagonal.
   \\

\noindent The Jordan decomposition of a linear operator $A$ is
unique in sense of \\
\begin{theorem} Let $A=A_{s}^{'}+A_{n}^{'}$ where
$A_{s}^{'}$ and
$A_{n}^{'}$ are also linear operators which satisfy\\
(1)$A_{s}^{'}$ is diagonalizable\\
(2)$A_{n}^{'}$ is nilpotent\\
(3)$A_{s}^{'}A_{n}^{'}=A_{n}^{'}A_{s}^{'}$ \\
Then $A_{s}^{'}=A_{s}$, $A_{n}^{'}=A_{n}$.
\end{theorem}
\begin{proof} We first have
$A_{s}^{'}-A_{s}=A_{n}-A_{n}^{'}$. Note that $A_{s}^{'}$ and
$A_{n}^{'}$ commute with $A$. So by Ex 6.1, $A_{s}^{'}$ and
$A_{n}^{'}$ commute with both $A_{s}$ and $A_{n}$. Then by Ex 6.2,
$A_{s}^{'}-A_{s}$ is semisimple. But $A_{n}-A_{n}^{'}$ is
nilpotent by the binomial formula. Since the only nilpotent
semisimple operator is $0$, we conclude that $A_{s}^{'}=A_{s}$ and
$A_{n}^{'}=A_{n}$ from the equation given above.
\end{proof}

\subsection*{Generalized Weight Space Decomposition}
Let $\mathfrak{g}$ be a finite dimensional Lie algebra over
$\mathbb{F}$ and $\pi$ a representation of $\mathfrak{g}$ in a
finite dimensional vector space $V$ over $\mathbb{F}$. Consider
the generalized eigenspace decomposition of $\mathfrak{g}$ w.r.t
ad$a$ and of $V$ w.r.t $\pi(a)$.
\begin{center}
 $\mathfrak{g}=\bigoplus_{\alpha}\mathfrak{g}_{\alpha}^{a}$ where
 $\mathfrak{g}_{\alpha}^{a}=\{g\in\mathfrak{g}|($ad$a-\alpha)^{N}g=0$
 for some $N\}$
\end{center}
\begin{center}
 $V=\bigoplus_{\lambda}V_{\lambda}^{a}$
 where
 $V_{\lambda}^{a}=\{v \in V|(\pi(a)-\lambda)^{N}v=0$
 for some $N\}$
\end{center}
These two decompositions are related by \\

\begin{proposition}
$\pi(\mathfrak{g}_{\alpha}^{a})V_{\lambda}^{a} \subseteq
V_{\lambda+\alpha}^{a}$
\end{proposition}

\noindent We need the following lemma to finish the proof.\\

\begin{lemma} Let $A$ be a unital associate algebra over $\mathbb{F}$. Let
$a$, $b \in A$ and $\alpha$, $\lambda\in\mathbb{F}$. Then we have
the following identity
\begin{center}
$(a-\alpha-\lambda)^{N}b=\sum_{j=0}^{N}(^{N}_{j})($ad$a-\alpha)^{j}b(a-\lambda)^{N-j}$
\end{center}
\end{lemma}
\begin{proof} Let $L_{x}$ and $R_{x}$ denote the operators of
left and right multiplication in $A$ by $x$. Then associativity
means that $L_{x}R_{y}=R_{y}L_{x}$. We have
$L_{a-\alpha-\lambda}=($ad$a-\alpha)+R_{a-\lambda}$. Note that
ad$a-\alpha$ commute with $R_{a-\lambda}$, hence
$L_{a-\alpha-\lambda}^{N}=\sum_{j=0}^{N}(^{N}_{j})
($ad$a-\alpha)^{j}R_{\alpha-\lambda}^{N-j}$ by the binomial
formula. Now apply both sides to $b$ to obtain the identity.
\end{proof}

\noindent $Proof$ $of$ $the$ $proposition$. Suppose
$g\in\mathfrak{g}_{\alpha}^{a}$, $v\in V_{\lambda}^{a}$. Apply the
lemma to $A=$End(V), $a$ is $\pi(a)$, $b=\pi(g)$, hence
\begin{center}
$(\pi(a)-\alpha-\lambda)^{N}\pi(g)v=\sum_{j=0}^{N}(^{N}_{j})($ad$\pi(a)-\alpha)^{j}\pi(g)(\pi(a)-\lambda)^{N-j}v$
\end{center}
Take $N>$dim$\mathfrak{g}_{\alpha}^{a}+$dim$V_{\lambda}^{a}$, then
each summand of the right hand side of the above equation is zero.
That means the desired result.
\\

\noindent Suppose $\mathfrak{h}$ is a finite dimensional Lie
algebra over $\mathbb{F}$. Let $\pi$ be a representation of
$\mathfrak{h}$ in a finite dimensional vector space $V$ over
$\mathbb{F}$ and let $\lambda\in\mathfrak{h}^{*}$. Then the
generalized weight space attached to $\lambda$ is defined as
\begin{center}
$V_{\lambda}=\{v\in V|(\pi(h)-\lambda(h))^{N}v=0$ for every
$h\in\mathfrak{h}$ and some $N\}$.
\end{center}
\begin{theorem} Notation as above and further assume
char.$\mathbb{F}=0$ and $\mathfrak{h}$ is nilpotent. Then
\begin{center}
$V=\bigoplus_{\lambda\in\mathfrak{h}^{*}}V_{\lambda}$
\end{center}
\end{theorem}
\textbf{Ex 6.3} a)Deduce the theorem from Ex 6.2 in the case
when
$\mathfrak{h}$ is abelian.\\
b)Consider the adjoint representation of the unique 2-dimensional
non-abelian Lie algebra to show that theorem $2$ fails if
$\mathfrak{h}$ is not nilpotent.\\
\textbf{Solution}. a) For any $a\in\mathfrak{h}$,
$\pi(\mathfrak{h})V_{\lambda}^{a} \subset V_{\lambda}^{a}$ by Ex
6.2 . So if there is at least one $\pi(a)$ has distinct
eigenvalues, then we can apply induction on dim$V$. Otherwise we
just need to prove that the only eigenvalue of every $\pi(a)$ is
linear functional on $\mathfrak{h}$. Let $a$ and $b$ are elements
of $\mathfrak{h}$. Suppose eigenvalues of $\pi(a)$ and $\pi(b)$
are $\lambda$ and $\mu$ respectively. Then since $\mathfrak{h}$ is
abelian, we have
\begin{center}
$(\pi(a+b)-\lambda-\mu)^{N}=\sum_{j=0}^{N}(^{N}_{j})(\pi(a)-\lambda)^{j}(\pi(b)-\mu)^{N-j}$
\end{center}
Now choose $N> 2n$ where $n=$dim$V$. Then apply the above equation
to every $v\in V$ to conclude that the eigenvalue of $\pi(a+b)$ is
$\lambda+\mu$.\\
b)Denote the unique 2-dimensional non-abelian Lie algebra by
$\mathfrak{b}=\mathbb{F}a+\mathbb{F}b$ which satisfies $[a,b]=b$.
It is easy to see that $\mathbb{F}a$ is a generalized eigenspace
of ad$a$, but it is not a generalized eigenspace of $b$. Thus the
theorem may fails if the given Lie algebra is not nilpotent.
\begin{proof}Take any $a\in\mathfrak{h}$, then $\mathfrak{h}=\mathfrak{h}^{a}_{0}$.
Hence by the proposition which we just proved
$\pi(\mathfrak{h})V_{\lambda}^{a} \subset V_{\lambda}^{a}$ for all
$a\in\mathfrak{h}$ and eigenvalue $\lambda$ of $a$. So if there is
at least one $\pi(a)$ has distinct eigenvalues, then we can apply
induction on dim$V$. Otherwise we just need to prove that the only
eigenvalue of every $\pi(a)$ is linear functional on
$\mathfrak{h}$. Apply Lie's theorem that all $\pi(a)$ are upper
triangular in some basis, then we deduce the desired result since
the eigenvalue are just the numbers on diagonal.
\end{proof}
\noindent \textbf{Remark} It is easy to see that we can still get
the decomposition even if char.$\mathbb{F}\neq 0$.
But $\lambda$ may not be linear functional on $\mathfrak{h}$ in this case.\\

\noindent \textbf{Ex 6.4} Consider the 2-dimensional
representation of $H_{1}$ for char.$\mathbb{F}=2$
\begin{center}
$V=\mathbb{F}[x]/x^{2}\mathbb{F}[x]$, $p=\frac{\partial}{\partial
x}$, $q=x$, $c=1$.
\end{center}
Then $V=V_{\lambda}$, but $\lambda$ is not a linear function.\\
\textbf{Solution}. Note that $p$ and $q$ are all nilpotent. But
$p+q$ is not nilpotent since $(p+q)^{2}(x)=x$. So in this case
$\lambda$ is not a linear function.\\

 \noindent \textbf{Ex 6.5}$^{*}$ If
$\mathfrak{h}^{p}=0$, then the theorem still holds for
char.$\mathbb{F}=p$.


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