\classheader{2012-12-03}


Today, we'll start discussing Lie theory. 

First, I want to raise some philosophical questions: in what sense can $\C^\times$ be thought of as a complexification of $\S^1$? How can we complexify the symmetric group $S_n$?

We have the map $\exp:\C\to\C^\times$, defined by $a\mapsto e^a$, and the line $i\R$ maps onto $\S^1$. We can think of $\C$ as a complexification of the line $i\R$, and so the exponential map tells us that $\C^\times$ should be thought of as a complexification of $\S^1$.

More generally, if we have a $\ast$-algebra and the exponential map $\exp:A\to A^\times$, then the skew-hermitian operators $iH(A)$ are mapped to the unitary operators $U(A)$, and because $A=iH(A)\oplus H(A)$ is a complexification of $iH(A)$, we ought to think of $A^\times$ as a complexification of $U(A)$.

For example, if $A=\M_n(\C)$ with $\ast$ being the hermitian adjoint, then $A^\times=\GL_n(\C)$, $U(A)=U_n$, and $\GL_n(\C)$ is to be thought of as a complexification of $U_n$.

\begin{definition}
A finite-dimensional representation $\rho:\GL_n(\C)\to \GL_N(\C)$ is called holomorphic if $g\mapsto \rho_{ij}(g)$ is a holomorphic function on $\GL_n(\C)\subset\M_n(\C)$ for all $1\leq i,j\leq N$.
\end{definition}
\begin{proposition}
Let $\rho:\GL_n(\C)\to\GL(V)$ be a holomorphic representation, and suppose we have a subspace $W\subset V$ that is $U_n$-stable. Then in fact $W$ is $\GL_n(\C)$-stable.
\end{proposition}
\begin{corollary}[The ``unitary trick'']
Any finite-dimensional holomorphic representation of $\GL_n(\C)$ is completely reducible.
\end{corollary}
\begin{proof}[Proof of corollary]
Let $V$ be a holomorphic representation of $\GL_n(\C)$. Because $U_n$ is compact,  we know that any unitary representation is completely reducible, so considering $V$ now as a $U_n$-representation, we have a decomposition $V=V_1\oplus\cdots\oplus V_k$ into $U_n$-irreps, and the proposition implies these are also $\GL_n(\C)$-irreps (enlarging the group can't make reducible something irreducible).
\end{proof}
\begin{lemma}
Let $f:\C^r\to\C$ be a holomorphic function such that $f|_{\R^n}= 0$. Then $f=0$.
\end{lemma}
\begin{proof}[Proof of lemma]
This is easy -  use the Cauchy-Riemann equations and induction on $r$.
\end{proof}
\begin{proof}[Proof of proposition]
Suppose that $W\subset V$ is a $U_n$-stable subspace that is not $\GL_n(\C)$-stable. Thus, we can find $g\in\GL_n(\C)$ and $v\in W$ such that $gv\notin W$. Choose $\phi\in V^\ast$ such that $\phi|_{W}=0$ and $\phi(gv)\neq 0$.

Let $f:\M_n(\C)\to\C$ be the map defined by $a\mapsto \phi(\rho(\exp(ia))v)$. For any $a\in H(\M_n(\C))$, we have $\exp(ia)\in U_n$, hence $\rho(\exp(ia))v\in W$, and therefore $f(a)=0$. By the lemma, we must have that $f=0$, but this is a contradiction; for example, there is an $x\in \M_n(\C)$ such that $g=e^{ix}$, and then we must have $f(x)\neq 0$.
\end{proof}

This was a demonstration of a Lie theory argument. Now let's move to a more general setting.

\begin{definition}
Let $G\subset\GL_n(\R)$ be a closed subgroup, and denote the unit element by $e$. The Lie algebra of $G$ is defined to be $\Lie(G)=\{a\in\M_n(\R)\mid e^{ta}\in G\text{ for all }t\in\R\}$.
\end{definition}
\begin{lemma}
$\text{}$

\begin{enumerate}
\item $\Lie(G)$ is a vector subspace of $\M_n(\R)$.
\item $G$ is a submanifold in $\M_n(\R)$ with tangent space $\Lie(G)$.
\end{enumerate}
\end{lemma}

\begin{center}
\begin{tikzpicture}
\draw[thick] (-2,0) to [bend left=20] node[midway,fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt,label=above:$e$] {} (2,0) node[label=right:$G$] {};
\draw (-1.3,0) --++ (2,0) --++ (0.7,0.9) node[label={[shift={(0,-0.2)}]$\Lie(G)$}] {} --++ (-2,0) --++ (-0.7,-0.9);
\node at (-0.8,1.6) {$\M_n(\R)$};
\end{tikzpicture}
\end{center}
\begin{proof}[Proof of 1]
For any $a\in\Lie(G)$, we know that $za\in\Lie(G)$ for any $z\in\R$ (this is clear from the
		definition). For any $a,b\in\Lie(G)$, we can consider the map $f:\R\to G$ defined by
$t\mapsto e^{ta}e^{tb}$ (we have $e^{ta}e^{tb}\in G$ because $a,b\in \Lie(G)$, so their product is as well). Clearly, 
\[\frac{df}{dt}\bigg|_{t=0}=a+b.\]
We claim that $\frac{df}{dt}|_{t=0}\in\Lie(G)$. We can deduce this from the fact that
\[\lim_{n\to\infty}(e^{\frac{1}{n}a}e^{\frac{1}{n}b})^n=e^{a+b}.\]
Indeed, it is more generally true that for any $a,b\in\M_n(\R)$ with $|a|,|b|\ll 1$, we have
\[\log(e^ae^b)=a+b+\frac{1}{2}(ab-ba)+o(|a|^2+|b|^2).\]
{\small [This is similar to the calculation of $\frac{\partial^2}{\partial t\partial s}
(e^{ta}e^{sb}e^{-ta}e^{-sb})$ on a previous homework.]}
We then conclude that
\[e^{ta+tb+\frac{t^2}{2}(ab-ba)+o(t^2(|a|^2+|b|^2))}=e^{ta}e^{tb}\in G.\]
{\small This doesn't seem to be enough to prove?}
\end{proof}

Since this is not really a course on Lie theory, proof of 2 is omitted. 

\begin{examples}
$\text{}$

\begin{itemize}
\item $\Lie(\GL_n(\R))=\M_n(\R)$
\item $\Lie(\S^1)=i\R$
\item $\Lie(U_n)=iH(\M_n(\C))$
\item $\Lie(\SL_n(\C))=\{a\in\M_n(\C)\mid \tr(a)=0\}$ (you showed on a homework that
		$e^{t\tr(a)}=\det(e^{ta})$)
\item Let $G=\mathrm{O}_n(\R)=\{g\in\M_n(\R)\mid (g^T)^{-1}=g\}$. Then
\begin{align*}
\Lie(\mathrm{O}_n) & = \{a\mid ((e^{ta})^T)^{-1}=e^{ta}\text{ for all }t\}\\
&=\{a\mid e^{-ta^T}=e^{ta}\text{ for all }t\}\\
&=\{a\mid -a^T=a\}\\
&=\{\text{skew-symmetric matrices}\}.
\end{align*}
\end{itemize}
\end{examples}
\begin{remark}
Let $N$ be the orthogonal complement of $\Lie(G)$ in $\M_n(\R)$.
\[\exp:\underbrace{\M_n(\R)}_{\mathclap{\Lie(G)\,\oplus\, N}}\longrightarrow\GL_n(\R).\]
Then $\exp$ restricts to a homeomorphism of a neighborhood of the origin in $\Lie(G)$ to a neighborhood of the identity in $G$.
\end{remark}

Let $\rho:G\to \GL(V)$ be a continuous representation, where $G\subset \GL_n(\C)$ is a closed subgroup. For any $a\in\Lie(G)$, we can form the composition
\begin{center}
\begin{tikzcd}[row sep=0.1cm]
\R\ar{r}{\exp} & G\ar{r}{\rho} & \GL(V)\\
t\ar[mapsto]{r} & e^{ta}\ar[mapsto]{r} & \rho(e^{ta})
\end{tikzcd}
\end{center}
which is a continuous map from $\R$ to $\GL(V)$, and by a previous homework assignment, we know that this implies there is a unique $d\rho(a)\in\End(V)$ such that $\rho(e^{ta})=e^{t\cdot d\rho(a)}$. We get a map $d\rho:\Lie(G)\to \End(V)$.
\begin{proposition}
$\text{}$

\begin{enumerate}
\item Any continuous representation $G\to \GL(V)$ is a $C^\infty$ map.
\item $d\rho(ab-ba)=d\rho(a)d\rho(b)-d\rho(b)d\rho(a)$ for all $a,b\in \Lie(G)$.
\item If $G$ is connected and there is no subspace $W\subset V$, $W\neq 0,V$ such that $d\rho(a)(W)\subseteq W$ for all $a\in\Lie(G)$, then $\rho$ is an irrep of $G$.
\end{enumerate}
\end{proposition}