Cartan subalgebras of a semisimple Lie algebra [closed]

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.

closed as off-topic by Moishe Kohan, Scientifica, Dietrich Burde, Leucippus, user91500Mar 12 '17 at 6:05

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• It is a bad idea to write «its Cartan subalgebra», for the Cartan subalgebra is not unique. – Mariano Suárez-Álvarez Mar 11 '17 at 20:03
• Same question here. – Dietrich Burde Mar 11 '17 at 21:38
• @MarianoSuárez-Álvarez I have reworded this. – pilgrim Mar 11 '17 at 21:46
• Yes, the question was deleted and then asked again. But the link is still there. – Dietrich Burde Mar 12 '17 at 15:24

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.

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

• You showed that there exist Cartan subalgebras whose intersections with the summands are Cartan subalgebras there, but the question is whether every Cartan subalgebra had that property. – Mariano Suárez-Álvarez Mar 11 '17 at 20:02