\classheader{2012-12-07}


Let's finish up the computation of $\frac{d(\text{left side})}{dt}$ from last time, where
\[\text{left side}=\rho(e^{t(a+b)+\frac{t^2}{2}[a,b]+o(t^2)})=\rho(e^{t\cdot f(t)})=e^{t\cdot d\rho(f(t))},\]
letting $f(t)=a+b+\frac{t}{2}[a,b]+\frac{o(t^2)}{t}$. Note that $f(t)\to a+b$ as $t\to 0$.

We have
\[\frac{e^{t\,d\rho(f(t))}-1}{t} = \frac{e^{t\,d\rho(f(t))}-1}{t\cdot d\rho(f(t))}\cdot\frac{t d\rho(f(t))}{t}.\]
Because $\rho$ is continuous, we also have that $d\rho$ is continuous, hence the first factor $\frac{e^{t\,d\rho(f(t))}-1}{t\cdot d\rho(f(t))}$ goes to 1 as $t\to 0$, and hence the limit of the entire expression as $t\to 0$ is $d\rho(a+b)$.

Now we will consider the action of $\SL_2(\C)\rightcirclearrow \C^m[u,v]$, and the action of $\SU_2$ obtained by the inclusion $\SU_2\hookrightarrow\SL_2(\C)$.

\begin{theorem}
For each $m\geq 0$, restricting the $\SL_2(\C)$-action on $\C^m[u,v]$ to $\SU_2$ yields an irrep of $\SU_2$, and these are all irreps of $\SU_2$ up to isomorphism.
\end{theorem}
\begin{proof}
Because $\SU_2=\SL_2(\C)\cap \U_2$, we have
\[\Lie(\SU_2)=\{A\in\M_2(\C)\mid A^*=-A,\tr(A)=0\}=\left\{\begin{pmatrix}
ia & b+ic\\ -b+ic & -ia
\end{pmatrix}\;\middle\vert\;a,b,c\in\R\right\}.\]
We can decompose $\M_2(\C)$ as a direct sum of the hermitian and skew-hermitian matrices:
\[\M_2(\C)=H(\M_2(\C))\oplus iH(\M_2(\C)),\]
and therefore
\[\underbrace{\{A\in \M_2(\C)\mid \tr(A)=0\}}_{\Lie(\SL_2(\C))} = \{A\in H(\M_2(\C))\mid \tr(A)=0\} \oplus \underbrace{i\{A\in H(\M_2(\C))\mid \tr(A)=0\}}_{\Lie(\SU_2)}.\]
We can express this as 
\[\Lie(\SL_2(\C))=i\Lie(\SU_2)\oplus \Lie(\SU_2) =\C\otimes_\R \Lie(\SU_2),\]
so that $\Lie(\SU_2)$ is like the ``real part'' of the vector space $\Lie(\SL_2(\C))$, and
$\Lie(\SL_2(\C))$ is like the complexification of $\Lie(\SU_2)$.

Let $\rho:\SU_2\to \GL(V)$ be a irrep. The differential map is $d\rho:\Lie(\SU_2)\to\End_\C(V)$. Because $\rho$ is a irrep, there is no proper subspace $W\subset V$ that is stable under $\im(d\rho)$. Because $\Lie(\SU_2)$ is a real vector space and $\End_\C(V)$ is a complex vector space, we can always extend an $\R$-linear map from $\Lie(\SU_2)$ to $\End_\C(V)$ to a $\C$-linear map from the complexification of $\Lie(\SU_2)$ to the same vector space $\End_\C(V)$,
\[(\C\otimes_\R d\rho):\underbrace{\C\otimes_\R\Lie(\SU_2)}_{\Lie(\SL_2(\C))}\to \End_\C(V).\]
Because $\Lie(\SL_2(\C))=\langle e,h,f\rangle$, we see that $V$ acquires the structure of a simple $\cal{U}$-module.
\end{proof}


Now let's discuss $\SO(\R^3)$. There is a natural action $\SO(\R^3)\rightcirclearrow\C^n[x,y,z]$.
\begin{theorem}
For each positive odd integer $2n+1$, $n=0,1,\ldots$, there is a unique irrep of $\SO(\R^3)$ of dimension $2n+1$ (up to isomorphism). Specifically, this irrep is $\mathrm{Harm}^m(\C^3,\SO(\R^3))$. These are all of the irreps of $\SO(\R^3)$.
\end{theorem}
\begin{proof}
Recall from the first homework assignment that there is a double cover map $\pi:\SU_2\to\SO(\R^3)$, which you obtained by thinking of $\SU_2$ as $\U(\H)$. The kernel of $\pi$ is just $\{\pm 1\}$.

Given an irrep $\phi:\SO(\R^3)\to\GL(V)$, the composition $\phi\circ\pi$ is an irrep of $\SU_2$ since $\phi$ is surjective, and the map $\phi\mapsto \phi\circ \pi$ clearly gives a bijection 
\[\widehat{\SO(\R^3)}\overset{\pi^*}{\cong}\{\rho\in\widehat{\SU_2}\mid \rho(-1)=1\}\]
between irreps $\phi$ of $\SO(\R^3)$ and irreps $\rho$ of $\SU_2$ that annihilate $\ker(\pi)$.

Any irrep of $\SU_2$ is some $\C^m[u,v]$. Note that $-\id$ maps $u^av^b$ to $(-1)^{a+b} u^av^b$, so a representation descends to $\SO(\R^3)$ if and only if $m=a+b$ is even, i.e. $m=2n$ for some $m$. Note that $\dim(\C^m[u,v])=\dim(\C^{2n}[u,v])=2n+1$.

Now we need to identify this representation with the representation $\mathrm{Harm}^m(\C^3,\SO(\R^3))$. Recall that by a homework problem,
\[\C^n[x,y,z]\cong \eucal{H}^n\oplus r^2\eucal{H}^{n-2}\oplus\cdots\]
where $r^2=x^2+y^2+z^2$, and also that $\dim(\eucal{H}^k)=2k+1$. Note that $\eucal{H}^k$ is an $\SO(\R^3)$-stable subspace of $\C^k[x,y,z]$.

There are two possibilities:
\begin{enumerate}
\item $\eucal{H}^n$ is an irrep of dimension $2n+1$ (which is what we want), or
\item it isn't.
\end{enumerate}
If it isn't, then each irreducible direct summand of $\eucal{H}^n$ has dimension $<2n+1$, and observing the decomposition of $\C^n[x,y,z]$ we wrote above, we see that statement 2 is equivalent to
\begin{itemize}
\item[$2'$.] Each irreducible direct summand of $\C^n[x,y,z]$ has dimension $<2n+1$.
\end{itemize}
Now, it will suffice to show that $2'$ is impossible. Indeed, we claim that the $(2n+1)$-dimensional irrep of $\SO(\R^3)$ does occur in $\C^n[x,y,z]$.

Consider the map $\R\to\SU_2$ defined by $t\mapsto \big(\begin{smallmatrix}
e^{it} & 0 \\ 0 & e^{-it}.
\end{smallmatrix}\big)$. Let's check what it does to monomials:
\[\begin{pmatrix}
e^{it} & 0 \\ 0 & e^{-it}
\end{pmatrix}:u^av^b\mapsto e^{iat}e^{-ibt}u^av^b=e^{i(a-b)t}u^av^b.\]
Thus, the eigenvalues of this matrix on $\C^{2n}[u,v]$ are $\{e^{ikt}\mid -2n\leq k\leq 2n\}$, and in particular, the maximum possible $k$ is $2n$. 

Therefore, if we can show that $\pi\big(\begin{smallmatrix}
e^{it} & 0 \\ 0 & e^{-it}.
\end{smallmatrix}\big)$ does have $e^{2nit}$ as an eigenvalue when acting on $\C^n[x,y,z]$, we will be done, because no lower dimensional irrep could have produced it.

On the homework, you classified continuous homomorphisms $\R\to\SO_n$. Then we know that the composition $\rho:\R\to\SU_2\xrightarrow{\;\pi\;}\SO(\R^3)$ must map $t$ to $e^{At}$ for some $A$, and moreover, we now know that we must have $A\in\Lie(\SO(\R^3))$, or in other words, $A\in\M_3(\R)$ satisfies $A^t=-A$. Therefore, there exists an orthogonal basis $x,y,z$ in which
\[A=\begin{pmatrix}
0 & a & 0\\ -a & 0 & 0 \\ 0 & 0 & 0
\end{pmatrix}.\]
Note that
\begin{align*}
\{t\in\R\mid \rho(t)=1\} & = \left\{t\in\R\middle\vert \begin{pmatrix}
e^{it} & 0 \\ 0 & e^{it}
\end{pmatrix}=\pm 1\right\}\\
&=\{t\in\pi\Z\}
\end{align*}
But on the other hand,
\begin{align*}
\{t\in\R\mid \rho(t)=1\} & = \{t\mid e^{At}=1\}\\
&=\{t\mid e^{\pm iat}=1\}\\
&=\{at\in 2\pi\Z\}
\end{align*}
Therefore, $a=2$, and the matrix $e^{At}$ has eigenvalues $e^{2it}$, $e^{-2it}$, and 1. Acting on $\C^n[x,y,z]$, we can see that it has as $e^{2nit}$ as an eigenvalue, because (for example) $x^n$ is mapped to $(e^{2it}x)^n=e^{2nit}x^n$. Thus, we are done.
\end{proof}