# Show $H \cap L_i$ is a Cartan subalgebra of $L_i$

Suppose we have a semisimple complex Lie algebra $$L$$, with a Cartan subalgebra $$H$$.

Suppose that $$L= L_1 \oplus\cdots\oplus L_k$$ with each $$L_i$$ a simple ideal of $$L$$.

I want to show that $$H_i=H \cap L_i$$ is a Cartan subalgebra of $$L_i$$.

For this I would need to show $$3$$ things from the definition of a Cartan subalgebra:

(i) $$H_i$$ is abelian

(ii) every non-trivial $$h_i \in H_i$$ is semisimple

(iii) $$H_i$$ is maximal with respect to (i) and (ii)

I feel like the first property follows immediately from the fact that $$H$$ is abelian being a Cartan subalgebra of $$L$$ itself, and so as $$H_i \subset H$$ we have that $$H_i$$ is also abelian.

I am struggling to figure out what to do with the other two cases, for (ii) a direct approach seems like it wouldn't work since there is no specific Lie algebra we are working with so I have no basis to calculate ad$$(h_i)$$ for a general $$h_i \in H_i$$, but I cannot figure out what approach to take.

Any help would be appreciated thanks :)

• See this duplicate. – Dietrich Burde Mar 10 at 19:20
• Thanks for the reply :) although I can see from the theorem referenced that the result follows trivially I am interested in a more direct approach to the problem, as I feel it should be possible – UsernameInvalid Mar 10 at 23:21