killing form is positive definite on cartan subalgebra  I am studying Theorem 10 on page 147 of Jaboson's Lie algebra book, and at the end of the proof I believe he uses the fact that the restriccion of the Killing form to $H$, the Cartan subalgebra is positive definite, so my question is:

If L is a semisimple Lie algebra over an algebraic closed field $F$ of characteristic 0 and $H$ is a Cartan subalgebra, is the Killing form positive definite on $H$?

If so, how can I prove that? (or where can I find the proof?) I know how to prove that it is non degenerate.

If not, why does Jacobson assume so, or under what hypothesis is that so?

I did the example $L=sl(2,F)$ and it is definite positive on $H$.

• Not everybody has the book, so I think it would be better if you write down the statement here. I don't understand the topic very well to answer your question, but it seems the answer is no. According to the wikipedia, the Killing form is bilinear. Then it cannot be positive definite over $F = \mathbb{C}$, since $i^2 = -1$. – Hwang Jan 22 '18 at 7:23

The statement in Theorem $10$ to prove is that the real Lie algebra $L_u$ is compact. So it has to be shown that its Killing form over the real numbers is negative definite. Jacobson defines just before Theorem $10$ a real Lie algebra to be compact if its Killing form is negative definite. This implies semisimple.
• yes that is true but to prove that it uses that $(h,h)>0$ and that is what I don't understand – allizdog Jan 23 '18 at 3:43