Suppose that L is a complex semisimple Lie algebra containing an abelian subalgebra H consisting of semisimple elements. I am wondering how to see that L has a basis of common eigenvectors for the elements of ad(H).
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
Here's how it works:
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Commutative diagonalizable operators over algebraically closed field have simultaneous eigenspace decomposition. This is a standard fact from linear algebra. In this case, $ad(H)$ exactly consists of commutative semisimple operators.