Cartan subalgebras of a semisimple Lie algebra

Question

Let $L$ be a semisimple complex Lie algebra and $H$ a Cartan subalgebra. Suppose that $L=L_{1}\oplus ...\oplus L_{k}$ with each $L_{i}$ a simple ideal of $L$. Show that for $1\leq i \leq k$, $H\cap L_{i}$ is a Cartan subalgebra of $L_{i}$.

I have had no thoughts on this problem and no idea how to solve it hence I cannot make an attempt. I have seriously severe difficulties.

Answer 1

Hint: Let $C_i$ be the Cartan subalgebra of $L_i$, $\oplus_iC_i$ is a Cartan subalgebra of $L$ since it is Abelian and equal to its normalizer. $(\oplus_iC_i)\cap L_i=C_i$.

If $D$ is another Cartan subalgebra, there exists an automorphism $f$ of $L$ such that $f(\oplus_iC_i)=D$ (see property 3) in the reference. Such an automorphism sends a simple component to a simple component, We can write $f(C_i)=D_{f(i)}$ where $f(C_i)\subset L_{f(i)}$. We can write $D=\oplus_iD_i$.

http://mathworld.wolfram.com/CartanSubalgebra.html

Answer 2

Let $\phi_i\colon L\rightarrow L\cap L_i$ be the canonical projection to the $i$-th factor. We can use now the following lemma:

Lemma: Let $\phi:L\rightarrow L'$ be a surjective Lie algebra homomorphism, and $H$ be a Cartan subalgebra of $L$. Then $\phi(H)$ is a Cartan subalgebra of $L'$.

Proof: See Lemma $3.6.2$ here. The proof works also for not necessarily semisimple Lie algebras.