Consider the following standard triple for $sl(2,\mathbb C)$, $$H=\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right) \qquad X=\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right) \qquad Y=\left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}\right) $$ Let $\rho:\mathfrak{su}(2)\rightarrow \text{End}(V)$ be a lie algebra homomorphism such that for any $A\in \mathfrak{su(2)}$ we have $\rho(A)^*=-\rho(A)$. We extend this to a complex linear map $\rho:\mathfrak{sl}(2,\mathbb V)\rightarrow \text{End}_{\mathbb C}(V)$. Consider the angular momentum operators defined by $L_j=\frac 12 \rho(\sigma_j)$, where $\sigma_j$ are the pauli matrices.
I want to show the following
Let $v$ be an eigenvector of both $L^2$ and $L_3$. Show that there is a unique linear space $W\subset V$ that is invariant under $\rho$, such that $v\in W$ and $\rho |_W$ is a finite dimensional irreducible representation of $\mathfrak{su}(2)$.
I think I need to show that the space of vectors which are eigenvectors of both $L^2$ and $L_3$ is finite. I believe this should be the set $L^n_+v$ and $L^m_-v$ for $0\leq n, m \leq k$ for some $k$. I am not sure how to show this $k$ must be finite. If we had this then this should give that $dim(W)=2k$ or perhaphs $dim(W)=2k+1$ which would determine an irreducible representation of $\mathfrak{su}(2)$.
Note here $L_+=L_1+iL_2$ and $L_-=L_1-iL_2$.
update: I have shown that if $v$ is an eigen vector for both $L^2$ and $L_3$ with eigen values $j$ and $m$ respectively, then $L_+ v$ is also eigenvector with eigen values $j$ and $m+1$ respectively.
I used this to show that $\|L_+v\|^2=j-m(m+1)\|v\|^2$. This shows that $L^k_+v=0$ for some $k$. We can do something similar for $L_-$.
Since $L_+=\rho(X)$, $L_-=\rho(Y)$ and $L_3=2\rho(H)$, it follows that $L_+,L_-,L_3$ is a also basis?. Therefore to show that a space $W$ is invariant it is enough to show it is invariant under $L_+,L_-,L_3$.