About Schur's lemma in lie algebra

[HDB1]

Please, I have a question about Schur's Lemma ;

Let $\phi: L \rightarrow g I((V)$ be irreducible. Then the only endomorphisms of $V$ commuting with all $\phi(x)(x \in L)$ are the scalars.

Could you explain it, and please, how we can apply this lemma on lie algebra $L=\mathfrak{s l}(2)$thanks in advance,

 

[HDB1]

Dear @fresh_42 , If you could help, I would appreciate it, thanks in advance,

 

[fresh_42]

Please, I have a question about Schur's Lemma ;

Let $\phi: L \rightarrow g I((V)$ be irreducible. Then the only endomorphisms of $V$ commuting with all $\phi(x)(x \in L)$ are the scalars.

Could you explain it, and please, how we can apply this lemma on lie algebra $L=\mathfrak{s l}(2)$thanks in advance,

It says if $\varphi \, : \,V\longrightarrow V$ is a linear transformation of the vector space $V$ and $\phi\, : \, L \rightarrow \mathfrak{gl}(V)$ an irreducible representation of the Lie algebra $L$ then
\begin{align*}
[\phi(X),\varphi ](v)&=(\phi(X)\cdot \varphi -\varphi \cdot \phi(X))(v)=\phi(X).\varphi (v)-\varphi (\phi(X).v)=0 \text{ for all }X\in L\\ &\Longrightarrow \\
\varphi(v)&=\lambda \cdot v\text{ for some }\lambda \in \mathbb{K}
\end{align*}

Note:
a) $\varphi \in \operatorname{End}(V)=\mathfrak{gl}(V)$
b) $\{\alpha \in \mathfrak{gl}(V)\,|\,[\alpha,\beta]=0\text{ for all }\beta\in \mathfrak{gl}(V)\}=Z(\mathfrak{gl}(V)).$
c) Schur's lemma can therefore be phrased as follows:

A linear transformation $\varphi$ of a finite-dimensional representation space $V$ of an irreducible representation $\phi$ of a Lie algebra $L,$ i.e. $V$ is an irreducible $L$ module, that lies in the center of $\mathfrak{gl}(V)$ is a scalar multiple of the identity matrix.

Consider the case $L=\mathfrak{sl}(2)\, , \,V_2=\mathbb{K}^2$ with $x.v=[x,v]$ being the Lie multiplication of $\mathfrak{sl}(2)\ltimes V_2$ we have seen before, will say: $x.v$ is the multiplication of a matrix $x\in \mathfrak{sl}(2)$ and a vector $v\in V_2.$ This is an irreducible representation, since $V_2$ has no one-dimensional submodule $U=\mathbb{K}u$ such that $x.u \in U$ for every $x\in \mathfrak{sl}(2).$ (Prove it!)

So all conditions of Schur's lemma are fulfilled. Now, if we have a matrix $\varphi = A= \begin{pmatrix}a&b\\c&d\end{pmatrix}\in \mathfrak{gl}(V_2)$ such that
$$
0=[\phi(X),A]=[X,A]=X\cdot A- A\cdot X\text{ for all } X\in \mathfrak{sl}(2) \;\;\Longleftrightarrow \;\;AX=XA
$$
then $A=\lambda \cdot \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for some $\lambda \in \mathbb{K}.$

You can check this yourself. Prove:
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot \begin{pmatrix}h&x\\y&-h\end{pmatrix}=\begin{pmatrix}h&x\\y&-h\end{pmatrix}\cdot \begin{pmatrix}a&b\\c&d\end{pmatrix} \text{ for all }x,h,y\;\;\Longrightarrow \;\;b=c=0 \text{ and }a=d
$$