\classheader{2012-12-05}


Let's review what we did last time.

Given a closed subgroup $G\subset \GL_n(\R)$, we consider a continuous representation $\rho:G\to\GL(V)$. For any $a\in\Lie(G)$, we define $d\rho(a)$ by
\[\rho(e^{ta})=e^{td\rho(a)}.\]
We have a diagram
\begin{center}
\begin{tikzcd}
\Lie(G) \ar{r}{d\rho} \ar{d}[swap]{\exp} & \End(V)\ar{d}{\exp} \\ G \ar{r}[swap]{\rho} & \GL(V)
\end{tikzcd}
\end{center}
which commutes essentially by the definition of $d\rho$.

Recall that the commutator of two elements of a ring $a,b\in R$ is $[a,b]:=ab-ba$.

\begin{proposition}
Let $\rho:G\to \GL(V)$ be a continuous representation.
\begin{enumerate}
\item $\rho$ is a $C^\infty$ map.
\item $d\rho$ is a linear map.
\item If $a,b\in\Lie(G)$, then $[a,b]\in\Lie(G)$ where the commutator is defined just by consdiering $a$ and $b$ as matrices, and moreover, $d\rho$ respects the commutator, i.e. 
\[d\rho([a,b])=[d\rho(a),d\rho(b)].\]
\item Suppose $G$ is connected. If $W\subseteq V$ is a vector subspace stable under $d\rho(a)$ for all $a\in\Lie(G)$, then $W$ is also stable under $\rho(g)$ for all $g\in G$.
\item If $\rho':G\to\GL(V)$ is another continuous representation on the same vector space $V$, if $d\rho=d\rho'$, then $\rho=\rho'$.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof of 1]
We know that $\exp$ is smooth, so that a neighborhood of 0 in $\Lie(G)$ is mapped diffeomorphically to a neighborhood of $e\in G$. Assuming $d\rho$ is linear, it is smooth, so $\rho$ is smooth near identity. 
For any $g_0\in G$, the map $g\mapsto \rho(g_0\cdot g)=\rho(g_0)\rho(g)$ is smooth, so we can
translate to get $\rho$ smooth everywhere.
\end{proof}
\begin{proof}[Proof of 2]
First, we will need the following lemma:
\begin{lemma}
Let $f:\R\to G$ is a smooth path with $f(0)=e$, then
\begin{enumerate}
\item $\frac{df}{dt}\big|_{t=0}\in\Lie(G)$
\item If $\frac{df}{dt}\big|_{t=0}=0$, then $\frac{d^2f}{dt^2}\big|_{t=0}\in\Lie(G)$.
\end{enumerate}
\end{lemma}
For any $A,B\in\M_n(\R)$ and $t\in\R$, we have
\[e^{tA}e^{tB}=e^{t(A+B)+\frac{t^2}{2}[A,B]+o(t^2)}.\]
Let $a,b\in\Lie(G)$. Then 
\[\rho(e^{ta})\rho(e^{tb})=\rho(e^{ta}e^{tb})=\rho(e^{t(a+b)+\frac{t^2}{2}[a,b]+o(t^2)})\]
because $\rho$ is a representation, but we also have that
\[\rho(e^{ta})\rho(e^{tb})=e^{td\rho(a)}e^{td\rho(b)}=e^{t(d\rho(a)+d\rho(b))+\frac{t^2}{2}[d\rho(a),d\rho(b)]+o(t^2)}.\]
Both sides must be the identity when $t=0$, so we can apply the lemma to get
\[\frac{d(\text{right side})}{dt}\bigg|_{t=0}\;\underset{{\substack{\text{chain}\\\text{rule}}}}{=}\;d\rho(a+b).\]
and
\[\frac{d(\text{left side})}{dt}\bigg|_{t=0}=d\rho(a)+d\rho(b)\]
so $d\rho$ is linear. To avoid circular reasoning, we need to justify use of chain rule (since we
don't want to assume part 1); see next lecture.
\end{proof}
\begin{proof}[Proof of 3]
Consider the map $f:\R\to G$ defined by $t\mapsto e^{ta}e^{tb}e^{-t(a+b)}$. As we showed last time, for any $a,b\in\Lie(G)$, we also have $a+b\in\Lie(G)$. By part 2 (?), we have that $\frac{df}{dt}\big|_{t=0}=0$, so the second part of the lemma implies that $\frac{d^2f}{dt^2}\big|_{t=0}=[a,b]\in\Lie(G)$.
\end{proof}
\begin{proof}[Proof of 4]
We will need the following lemma:
\begin{lemma}
Let $G$ be a connected topological group, and let $U$ be an open neighborhood of $e\in G$. Then the subgroup generated by $U$ is $G$.
\end{lemma}
\begin{proof}[Proof of lemma]
We improve our open neighborhood a bit by defining $\cal{U}=U\cap U^{-1}$, where $U^{-1}=\{g^{-1}\mid g\in U\}$. Because $U^{-1}$ is open and contains $e$, $\cal{U}$ is an open neighborhood of $e$. Thus, it suffices to show that the subgroup generated by $\cal{U}$, which we will write $G'=\bigcup_{k\geq 1}\cal{U}^k$, is equal to $G$.

We will show that $G'$ is open, as a union of open sets, and $G'$ is closed, because for any $g\in\overline{G}$,
\end{proof}
Now, recall that we showed that the exponential map $\exp:\Lie(G)\to G$ is a local isomorphism. Then the image of $\exp$ contains an open neighborhood $U$ of $e\in G$. Let $W\subset V$ be stable under $d\rho(a)$ for all $a\in\Lie(G)$. Then $W$ is stable under $e^{d\rho(a)}=\rho(e^a)$ because $e^{d\rho(a)}$ is just a sum of powers of $d\rho(a)$, and therefore $W$ is stable under $\rho(g)$ for any $g\in \im(\exp)$. Thus, $W$ is stable under $\rho(g)$ for any $g\in\cal{U}$, hence stable under $\rho(g)$ for any $g\in\cal{U}^k$, hence stable under $\rho(g)$ for any $g\in G$.
\end{proof}
\begin{proof}[Proof of 5]
By assumption, $d\rho=d\rho'$. Exponentiate both sides. Then the same argument as in proof of 4
shows $\rho= \rho'$ everywhere.
\end{proof}

Let $G\subset\GL_2(\C)$. There is a natural action $G\rightcirclearrow \C^m[u,v]$ for each $m=0,1,\ldots$

\begin{theorem}
For each $m\geq 0$, $\SL_2(\C)$ has a unique (up to isomorphism) holomorphic irrep $L_m$ of $\dim(L_m)=m+1$ and $L_m=\C^m[u,v]$.
\end{theorem}
\begin{proof}
First, we need to compute $\Lie(\SL_2(\C))$. As we showed last time,
\[\Lie(\SL_2(\C))=\{A\in \M_2(\C)\mid \tr(A)=0\}=\left\{\begin{pmatrix}
a & b \\ c & -a
\end{pmatrix}\;\middle\vert\;a,b,c\in\C\right\}.\]
Thus, $\Lie(\SL_2(\C))$ has as a basis
\[e=\begin{pmatrix}
0 & 1\\ 0 & 0
\end{pmatrix},\quad h=\begin{pmatrix}
1 & 0\\ 0 &-1
\end{pmatrix},\quad f=\begin{pmatrix}
0 & 0\\ 1 & 0
\end{pmatrix}.\]
As you've shown on a homework,
\[[e,f]=h,\quad [h,e]=2e,\quad [h,f]=2f.\]
Let $\rho:\SL_2(\C)\to \GL(V)$ be a holomorphic representation. Consider the map $d\rho:\Lie(G)\to \End(V)$, and let $E=d\rho(e)$, $H=d\rho(h)$, $F=d\rho(f)$. Then $E$, $H$, $F$ satisfy the same commutation relations.

This gives $V$ the structure of a $\cal{U}$-module, where $\cal{U}$ is the algebra defined on the same homework assignment.

We claim that if $V$ is irreducible as an $\SL_2(\C)$-representation, then $V$ is simple as a $\cal{U}$-module. Suppose otherwise; then let $W\subset V$ be a non-trivial $\cal{U}$-submodule. Then part 4 of our proposition implies that, since $W$ is stable under $d\rho(a)$ for any $a\in \Lie(G)$, we must have that $W$ is a subrepresentation of $V$, but this is a contradiction with the assumption that $V$ is irreducible.

As you showed on your homework, this means that $V\cong \C^m[u,v]$ as a $\cal{U}$-module for some $m$.

Thus, we get some representation $\rho:\SL_2(\C)\to \GL(\C^m[u,v])$, which we can compare with the natural representation. But by construction, their differentials are equal since the $\cal{U}$-action on $\C^m[u,v]$ is the natural one. Because $\SL_2(\C)$ is connected, we are done by part 5 of the proposition.
\end{proof}

Next time, we will look at $\SU_2(\C)$, $\SO_3(\R)$.