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\begin{document}
\title{18.745: Lecture 3}
\author{Professor: Victor Ka\v{c}\\
Scribe: Alexey Spiridonov}
\maketitle
\begin{xca}
Show that the complete list of $n$-dimensional $\g$ with $\dim\Z(\g)=n-2$
is
\end{xca}
\begin{enumerate}
\item $H_{1}\oplus Ab_{n-3}$, with $H_{1}$ as in Homework 2.
\item $b\oplus Ab_{n-2}$, with $b$ and $Ab$ as in Homework 2.
\end{enumerate}
%
\begin{proof}
Let $B=\{ x_{1},\dots,x_{n-2}\}$ be a basis for $\Z(\g)$. Let $a,b\in\g$
be some vectors such that $\s(B\cup\{ a,b\})=\g$. $a,b$ are linearly
independent since they span a 2-dimensional subspace. The commutator
of $b$ is 0 on an $n-1$ dimensional subspace of $\g$: $[b,x_{i}]=0,[b,b]=0$,
so we must have $[a,b]\neq0$, since $b\notin\Z(\g)$. Let $c=[a,b]$;
we then have two cases:
\begin{enumerate}
\item If $c\in\Z(\g)$, we call $a=p_{1}$ and $b=q_{1}$, and we see that
$\{ a,b,c\}$ is just $H_{1}$.
\item Otherwise, write $c$ in the basis $B\cup\{ a,b\}$: $c=\sum z_{i}x_{i}+z_{a}a+z_{b}b$.
We know that at least one of $z_{a},z_{b}$is non-zero. Without loss
of generality, let this be $z_{a}$. Then, $[c,b]=z_{a}[a,b]=z_{a}c$
(since $\ad b$ kills the other basis vectors). Renormalizing $b:=z_{a}^{-1}b$,
we have $[c,b]=c$, so $b,c$ span the non-commutative subalgebra
of dimension 2. Observe that (since $z_{a}\ne0$), $B\cup\{ b,c\}$
is a basis for $\g$, so we have the desired result.
\end{enumerate}
\end{proof}
\section{Homework policy}
The exercises from the previous two lectures are due every Tuesday.
Lecture write-ups should be handed in within two weeks of the date
of the lecture.
\section{Representations}
Representations are very useful tools for analysing the structure
of Lie algebras. There are some theorems that only have proofs using
representation theory, although they can be stated without it.
\begin{defn}
A \emph{representation} of a Lie algebra $\g$ in a vector space $V$
over a field $\F$ is a homomorphism $\pi:\g\rightarrow\gl V$. In
particular, this means that $\pi([a,b])=[\pi(a),\pi(b)]$.
\end{defn}
\section{Examples}
\begin{example}
$\pi(a)=0$ for all $a\in\g$. This is called the \emph{trivial representation}.
\end{example}
%
\begin{example}
Let $\g$ be a subalgebra of $\gl V$, and let $\pi(a)=a$. This is
called the defining (tautological, identity) representation.
\end{example}
%
\begin{example}
The adjoint representation, $\ad:\g\rightarrow\gl\g$ is defined by
$\ad(a)\rightarrow(\ad a)$.
To check that $\ad[a,b]=[\ad a,\ad b]$, we apply it to $c$, getting
\[
[[a,b],c]\overset{?}{=}(\ad a\ad b-\ad b\ad a)c=[a,[b,c]]-[b,[a,c]],\]
which is just the Jacobi identity. So, this linear map is indeed
a representation.
\end{example}
\begin{note*}
In these lecture notes, I will denote the center of $\g$ as $\Z(\g)$.
\end{note*}
\begin{cor}
$\ad$ defines an embedding of $\g/\Z(\g)$ into $\gl\g$, since $\ker\ad=\Z(\g)$.
\end{cor}
In particular, if $\Z(g)=0$, we embed $\g$ as a subalgebra of $\gl\g$.
It's thus natural to ask: can always embed $\g$ in some $\gl V$?
The answer is (stated without proof):
\begin{thm}
(Ado's Theorem) In fact, any finite-dimensional Lie algebra can be
embedded in some $\gl V$, $V$ finite-dimensional.
\end{thm}
There are some general ways of constructing representations.
\begin{enumerate}
\item Let $\pi:\g\rightarrow\gl V$ be a representation, and there $U\subset V$
is an invariant subspace, i.e.$\pi(\g)U\subset U$. Then, the restriction
$\pi|_{U}:U\rightarrow\gl U$ is a representation.
\item Taking $U$ as above, we can look at the quotient space $V/U$ as
well. Then, $\pi|_{V/U}:\g\rightarrow\gl(V/U)$. This warrants some
explanation. $\pi$ takes $g\in\g$ to $A\in\End V$; $U$ is invariant,
so $A\subset AU$. Thus, it can be viewed as an operator on $V/U$:
$A(v+U)=Av+U$.
\item Take 2 representation $\pi_{1},\pi_{2}$ of $\g$ in vector spaces
$V_{1}$ and $V_{2}$, respectively. Then, we can consider the direct
sum: for $g\in\g$, $(\pi_{1}\oplus\pi_{2})(g)=\pi_{1}(g)\oplus\pi_{2}(g)\in V_{1}\oplus V_{2}$.
\end{enumerate}
\section{Engel's Theorem}
Now we prove the first non-trivial theorem about Lie algebras. First,
a reminder:
\begin{defn*}
An operator $A$ on a vector space is nilpotent if there exists $n>0$
such that $A^{n}=0$.
\end{defn*}
\begin{thm*}
(Engel's Theorem) Let $\g\subset\gl V$ be a subalgebra consisting
of nilpotent operators. Assume $V$ is finite-dimensional. Then, there
exists a $v\ne0$ in $V$ such that every $g\in\g$ annihilates $v$.
\end{thm*}
\begin{proof}
We will proceed by induction on the dimension of $\g$ (which is a
subspace of the finite-dimensional $\End V$).
The base case is $\dim V=1$. Then, $\g=\F a$, with $a$ is a nilpotent
operator on $V$. Take $N$ such that $a^{N}=0$, but $a^{N-1}\ne0$.
Then, $\exists v$ such that $a^{N-1}v\ne0$. However, $a(a^{N-1}v)=0$,
so $v'=a^{N-1}v$ is the desired vector.
Assume that the theorem holds for all $\g$ of dimension $