Let $H,X,Y$ be the standard triple for $\mathfrak{sl}(2,\mathbb C)$. Let $\alpha$ be the root of $i\mathbb R H$ determined by $\alpha(H)=2$. Let $(\pi, V)$ be a finite dimensional continous representation of $SU(2)$. This gives $(\pi_*,V)$ a representation of $\mathfrak{sl}(2,\mathbb C)$. Let $V_\lambda$ denote the weight space of the weight $\lambda$ ie $V_\lambda=\{v\in V \text{ such that } Hv=\lambda v\}$. Therefore $$V_0=\{v\in V \text{ such that } Hv=0 v\}$$ and $$V_{\alpha/2}=\{v\in V \text{ such that } Hv= \alpha v/2 \}$$.
Show that $\pi_*$ is a direct sum of $\dim V_0+\dim V_{\alpha/2}$ irreducibles.
I know for compact groups we can decompose a representaiton by $\pi \sim \bigoplus_{\rho\in \hat G}\langle\pi,\rho \rangle \rho $. But $\mathfrak{sl}(2,\mathbb C)$ is not compact so we cannot use this.
My guess is there should be a way to decompose the representation into the different weights but I don't know how one would do this
Update:
I now that if $\rho$ is irreducible then $0$ and $\alpha/2$ cannot both be weights. If $\pi$ decomposes into a sum of $\rho_j$ does $\dim_{\pi}(V_k)$ in $\pi$ equal the $\sum_j \dim_{\rho_j}(V_k)$? I am also not sure on what I mean by $\dim_{\pi}(V_k)$ versus $\dim_{\rho_j}(V_k)$
