HDB1 said:
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:
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot \begin{pmatrix}h&x\\y&-h\end{pmatrix}=\begin{pmatrix}h&x\\y&-h\end{pmatrix}\cdot \begin{pmatrix}a&b\\c&d\end{pmatrix} \text{ for all }x,h,y\;\;\Longrightarrow \;\;b=c=0 \text{ and }a=d
$$