Let $L = L_1 \oplus···\oplus L_t$ be the decomposition of a semisimple Lie algebra $L$ into its simple ideals. Show that the semisimple and nilpotent parts of $x \in L$ are the sums of the semisimple and nilpotent parts in the various $L_i$ of the components of $x$.
Comments: Let $x \in L$, $x=x_1+···+x_t$ with $x_i \in L_i$, and $x_i=u_i+v_i$ is the Jordan decomposition of $x_i \in L_i$, $u_i$ is
semisimple and $v_i$ is nilpotent. Define $u=u_1+···+u_t$ and $v=v_1+···+v_t$.
I'm thinking of showing that $u$ is semisimples in $L$ and $v$ is nilpotent in $L$ and for this to use abstract Jordan decomposition, that is, show that, $ad_L(x)=ad_L(u)+ad_L(v)$. My idea is correct?
Best Answer
This is a homework question for Humphrey's book on Lie algebras. It has solutions online here, on page $20$. There is a typo in the solution, it should be $[u_i,u_j]=0$ for all $i\neq j$.