0
$\begingroup$

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).

1 Answers 1

1

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.

  • 0
    Very good, thanks. I really need to review the linear algebra.2011-10-09