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 :)

  • 1
    $\begingroup$ See this duplicate. $\endgroup$ – Dietrich Burde Mar 10 at 19:20
  • $\begingroup$ 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 $\endgroup$ – UsernameInvalid Mar 10 at 23:21

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