Let $L$ be a semisimple lie algebra, Then $L$ can be decomposed as $L=m_{1}L_{1} \oplus m_{2}L_{2}\oplus...\oplus m_{r}L_r$
I want to show that :
If $\varphi : L \longrightarrow L $ is an homomorphism Then
$\varphi=\varphi_{1} \oplus \varphi_{2} \oplus... \oplus \varphi_{r} $
Any help.
Where $\varphi_{i} =\varphi |_{m_{i}L} $
Write $m L_i=L_i^{j_1}\oplus ..\oplus L_i^{j_m}$.
The restriction $\varphi_{j_l}:L_i^{j_j}\rightarrow L$ is a morphism of Lie algebras. Since $L_i^{j_l}$ is simple, $Ker(\varphi_{j_l})=0$ or $L_i^{j_l}$. If $Ker(\varphi_{j_l})=0$, then the image of $\varphi_{j_l}$ is a sub Lie algebra of $L$ isomorphic to $L_i^{j_l}$ this implies that $Im(\varphi_{j_l})$ is $L_i^{j_p}$, we deduce that the image of $\varphi_i$ is contained in $mL_i$ and $\varphi=\varphi_1\oplus...\oplus\varphi_m$.
