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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 $<n$. We
will prove that it holds for $\g$ of dimension $n\ge2$. 

Take a proper subalgebra $\h\subset\g$ of the maximal possible dimension.
Since $\F a$ is a subalgebra of $\g$. $\dim\h\ge1$. Consider a
representation of $\h$ given by $\pi:h\rightarrow\ad h,h\in\h$,
$\pi$ acting on $\g$. The fact that $\h$ is a subalgebra means
that $\h$ is an invariant subspace with respect to $\pi$.

Take $\pi'=\ad_{\g/\h}$ to be the representation of $\h$ in $\g/\h$.
In Lecture 2, we proved a lemma: if $g\in\g$ is nilpotent, then so
is $\ad g$. Therefore, $\ad h$, considered as a Lie algebra over
$\g/\h$, consists of nilpotent operators. We know that $\dim\h<\dim\g$,
so by induction, there is a non-zero vector $\bar{a}\in V$ that's
annihilated by all $\pi(h),h\in\h$. Let $a$ be some preimage of
$\bar{a}$ in $\g$ under the map $\g\rightarrow\g/\h$. Then, we
have $[\h,a]\subset\h$; in other words, $\h\oplus\F a$ is a subalgebra
of $\g$. It isn't equal to $\h$, and so, by maximality of $\h$
it must be $\g$. Hence, $\h$ must have codimension 1.

Applying the inductive assumption to $\h$, we get a nonzero $v\in V$
such that $\h v=0$. Thus, the subspace $U\subset V$ of all vectors
annihilated by $\h$ is non-empty. 

$U$ is $a$-invariant. Take $u\in U$, then we have this equivalence,
$a(u)\in U\Leftrightarrow h(a(u))=0\forall h\in\h$. Expanding the
second formulation, we have $ha(u)=([h,a]+ah)u$. $[h,a]\in\h$, so
the first term is zero; similarly $h(u)=0$, so $a(h(u))=0$. Hence,
$U$ is indeed $a$-invariant. Thus, we can consider the restriction
of $a$ to $U$. $a$ is nilpotent, so $\exists u\in U$ such that
$au=0$, but $u\ne0$. This $u$ is the desired vector.
\end{proof}
It's natural to ask: What if $\g$ is a subspace, but not a subalgebra?
Does the theorem still hold? It turns out that the answer is no. The
following exercise provides a counterexample.

\begin{xca}
Find a 2-dimensional subspace $S$ of $3\times3$ matrices that consists
of nilpotent matrices, so that no $0\ne v\in\F^{3}$ is annihilated
by every $s\in S$.\\

\end{xca}
We want two matrices $A,B$ such that any linear combination has characteristic
polynomial $x^{3}$, but with non-overlapping eigenspaces. If we take
\[
A=\left(\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0\end{array}\right),\]
 it has a 0th eigenspace, which is spanned by $e_{1}=(1,0,0)$. Then,
$B$ mustn't annihilate $e_{1}$, but needs to kill some linear combination
of $e_{2}$ and $e_{3}$. Suppose $Be_{3}=0$. Then, \[
B=\left(\begin{array}{ccc}
a & b & 0\\
c & d & 0\\
e & f & 0\end{array}\right),(B-\lambda I)=\left(\begin{array}{ccc}
a-\lambda & b+g & 0\\
c & d-\lambda & g\\
e & f & -\lambda\end{array}\right)\]
We want \begin{multline*}
-\lambda^{3}=(a-\lambda)(-\lambda(d-\lambda)-fg)-(b+g)(-\lambda c-eg)=\\
=g(be-af+eg)+(g(f+c)+bc-ad)\lambda+(a+d)\lambda^{2}-\lambda^{3}\end{multline*}
 no matter what $g$ is. Thus, $e=0$, $f=-c$,$af=0$, $a=-d$, and
$ad=bc$. Assume $f\ne0$; then, $a=d=0$, and $b=0$. We can thus
take \[
B=\left(\begin{array}{ccc}
0 & 0 & 0\\
1 & 0 & 0\\
0 & -1 & 0\end{array}\right).\]


We see that its 0th eigenspace is spanned by $(0,0,1)$, and so no
$v\in\F$is killed by both $A$ and $B$, as we wanted. By construction,
the subspace spanned by $A$ and $B$ consists of nilpotent matrices,
so we have the desired result.


\section{Equivalent Formulations}

In this section, we present some corollaries of Engel's Theorem. 

\begin{enumerate}
\item \label{repcor}Let $\pi:\rightarrow\gl V$ be a representation of
$\g$ in a finite dimensional vector space $V$, such that all $\pi(g),g\in\g$
are nilpotent operators. Then, there exists nonzero $v\in V$ that
gets annihilated by all $\pi(g)$. Moreover, there exists a basis
of $V$, in which all $\pi(g)$ are strictly upper-triangular. 

\begin{proof}
$\pi(\g)$ is a subalgebra of $\gl V$; applying Engel's theorem,
we find a nonzero $v\in V$ such that $\pi(g)v=0\forall g\in\g$.

To construct the basis, take a $v_{1}=v$ as above. Then, $\F v_{1}$
is $\pi(g)$-invariant, since it gets annihilated. (Hence, every $\pi(g)$
has a column of zero coefficients corresponding to $v_{1}$ if it's
used as a basis vector). By invariance of $\F v_{1}$, we may consider
$V_{2}=V/\F v_{1}$, and the corresponding representation $\pi_{1}$
of $\g$ induced by $\pi$. The operators $\pi_{1}(g)$ are also nilpotent;
we apply Engel's Theorem again to get $v_{2}'$ that's annihilated
by all $\pi_{1}(g)$. Then, a pre-image $v_{2}\in V$ of $v_{2}'$
is sent to $\F v_{1}$by every $\pi(g)$. We proceed to make $V_{3}$
and get $v_{3}$, etc. In the end, we have the desired basis. 
\end{proof}
\item A subalgebra $\g\subset\gl V,\dim V<\infty$ consists of nilpotent
matrices iff in some basis of $V$, all matrices of $g\in\g$ are
strictly upper-triangular.

\begin{proof}
If all matrices of $\g$ are strictly upper-triangular, they are all
nilpotent. To see this, take $e_{1},\dots e_{n}$ to be the standard
basis vectors. Then (with $c$ standing for arbitrary constants),
$Ae_{1}=0,Ae_{2}=ce_{1}\Rightarrow A^{2}e_{2}=0,Ae_{3}=ce_{1}+ce_{2}\Rightarrow A^{3}e_{3}=0,\dots$.
Hence, $A^{n}e_{i}=0$ for all $i$.

The converse follows entirely from \ref{repcor}, by applying it with
$\pi$ chosen to be the defining representation.
\end{proof}
\end{enumerate}
Here is another exercise to conclude the lecture.

\begin{xca}
Compute the centers of the Lie algebras $\gln,\sln,\son,\spn$.\\

\end{xca}
\begin{enumerate}
\item $\gln$ is the entire space of matrices. If $n=1$, the entire space
is commutative (and hence $\text{gl}_{1}(\F)$ is its own center).
Otherwise, denote by $M_{i,j}$ the matrix of zeros with a single
1 at position $i,j$. Then, if $x\in\Z(\gln)$, $M_{i,j}x=xM_{ij}$.
The first term is the $j$th row of $x$ placed as the $i$th row
in the product. The second term is the $i$th column of $x$ placed
as the $j$th column in the product. The other entries in both matrices
are 0. Therefore, everything in the $j$th row of $x$ (except for
$j,j$) must be equal to 0. Hence, $x$ must be diagonal. Moreover,
notice that $(M_{i,j}x)_{i,j}=x_{j,j}$, while $(xM_{i,j})_{i,j}=x_{i,i}$.
So, $x_{i,i}=x_{j,j}$. This forces $x=aI$ for some $a\in\F$.
\item $\sln$ is the algebra of matrices with trace 0. If $n=1$, the algebra
and its center are trivial. When $n\geq2$, the matrices $M_{i,j},i\ne j$
used before are in the algebra. Hence, the same argument shows that
$\Z(\sln)\in\{ aI|a\in\F\}$. But, since these are trace 0 matrices,
$a=0$, and $\Z(\sln)=0$.
\item \label{son}We have a symmetric $B$. Suppose $x\in\Z(\son)$; then,
we have $x^{T}B+Bx=0$ and $xz=zx$ for any $z\in\g$. Let $y=(M_{ij}-M_{ji})$,
and $z=B^{-1}y$, then, it is in $\g$: \[
z^{T}B+Bz=y^{T}(B^{-1})^{T}B^{T}+BB^{-1}y=y^{T}+y=0\]
Multiplying the commutation relation by $B$, we have $0=Bxz-Bzx=-x^{T}Bz-Bzx=x^{%T
}y+yx$. Since $y$ is antisymmetric, this is equivalent to $(yx)^{T}=(yx)$.
Observe that $yx$ is all zeros, except for a copy of $j$th row of
$x$ in the $i$th row place, and a negated copy of the $i$th row
of $x$ in the $j$th row place. The fact that $yx$ is symmetric
means that all but four positions of $x$ are forced to be 0. We get
no constraint on $x_{i,j}$ and $x_{j,i}$, since these land on the
diagonal in $xy$. However, we do get that $x_{i,i}=-x_{j,j}$. Supposing
that $n\geq3$, this means we can zero out all the off-diagonal elements.
Additionally, we would get $x_{i_{1},i_{1}}=-x_{i_{2},i_{2}},x_{i_{2},i_{2}}=-x_{i_{3},i_{3}},x_{i_{1},i_{1}}=-x_{i_{3},i_{3}}$,
which implies that the diagonal is zero as well. So, for $n\geq3$,
we have $\Z(\g)=0$. The same holds for $n=1$, trivially. When $n=2$,
we change the basis to make $B=y\Bigl(\begin{array}{cc}
1 & 0\\
0 & x\end{array}\Bigr)$, with $y\ne0,x\ne0$. Then, for $A=\Bigl(\begin{array}{cc}
a & b\\
c & d\end{array}\Bigr)$, we have \[
0=A^{T}B+BA=y\Bigl(\begin{array}{cc}
a & cx\\
b & dx\end{array}\Bigr)+y\Bigl(\begin{array}{cc}
a & b\\
cx & dx\end{array}\Bigr),\]
which implies that (assume that $\text{char }\F\ne2$): $a=d=0$ and
$b=-cx$. Given \[
C=\Bigl(\begin{array}{cc}
0 & -cx\\
c & 0\end{array}\Bigr),D=\Bigl(\begin{array}{cc}
0 & -dx\\
d & 0\end{array}\Bigr),\]
 they clearly always commute. So, for $n=2$, $\Z(\g)=\g$.
\item The computation is very similar to \ref{son}. This time, our $B$
is antisymmetric. Suppose $x\in\Z(\son)$; then, we have $x^{T}B+Bx=0$
and $xz=zx$ for any $z\in\g$. Let $y=(M_{ij}+M_{ji})$, and $z=B^{-1}y$,
then, it is in $\g$: \[
z^{T}B+Bz=y^{T}(B^{-1})^{T}(-B)^{T}+BB^{-1}y=y-y^{T}=0\]
Multiplying the commutation relation by $B$, we have $0=Bxz-Bzx=-x^{T}Bz-Bzx=x^{%T
}y+yx$. Since $y$ is symmetric, this means that $yx=-(yx)^{T}$. Observe
that $yx$ is all zeros, except for a copy of $j$th row of $x$ in
the $i$th row place, and a copy of the $i$th row of $x$ in the
$j$th row place. Since $yx$ is antisymmetric, all but four positions
of $x$ are forced to be 0. We get no constraint on $x_{i,j}$ and
$x_{j,i}$, since these land on the diagonal in $yx$. However, we
do get that $x_{i,i}=x_{j,j}$. Supposing that $n\geq3$, this means
we can zero out all the off-diagonal elements. Additionally, we have
equality relations between any element of the diagonal, meaning that
$x=aI$ (and we can pick any $a$). However, $aI$ can only be in
$\g$ if $2aB=0$. This is impossible, except in characteristic 2.


When $n=2$, $B=\Bigl(\begin{array}{cc}
0 & x\\
-x & 0\end{array}\Bigr)$, with $x\ne0$. Then, for $A=\Bigl(\begin{array}{cc}
a & b\\
c & d\end{array}\Bigr)$, to be in $\g$. we must have \[
0=A^{T}B+BA=x\Bigl(\begin{array}{cc}
-c & a\\
-d & b\end{array}\Bigr)+x\Bigl(\begin{array}{cc}
c & d\\
-a & -b\end{array}\Bigr)\]
 so we see that $a=-d$. If $C\in\Z(\g)$, then we also know $a=d$
(from above). So, if $C$ commutes with everything, it has the form
$\Bigl(\begin{array}{cc}
0 & u\\
v & 0\end{array}\Bigr)$. Then, the following must hold for all $A\in\g$:

\[
0=AC-CA=\Bigl(\begin{array}{cc}
bv & au\\
-av & cu\end{array}\Bigr)-\Bigl(\begin{array}{cc}
cu & -au\\
av & bv\end{array}\Bigr)\Rightarrow C=0.\]
 The only antisymmetric matrix of dimension 1 is 0, so that case is
nonsensical. Thus, the center of $\spn$ is always 0.\end{enumerate}

\end{document}