# Prove identity map of a Lie algebra is unique Cartan involution when Killing form is negative definite

First some Definitions for convenience:

Let ${\mathfrak {g}}$ be a real semisimple Lie algebra and let $B(\cdot ,\cdot )$ be its Killing form. An involution on ${\mathfrak {g}}$ is a automorphism whose square is the identity. Such an involution is called a Cartan involution on ${\mathfrak {g}}$ if $B_{\theta }(X,Y):=-B(X,\theta Y)$ is a positive definite bilinear form.

The question:

Prove the identity map of ${\mathfrak {g}}$ is the unique Cartan involution if the Killing form is negative definite.

• How would $\theta$ need to ask on a basis where the quadratic form is diagonalized? – AHusain Dec 20 '15 at 0:28
• @TsemoAristide I've edited the question. – nihan Dec 20 '15 at 1:08

## 1 Answer

Since $\theta^2=Id$, its eigenvalues are 1 or -1. Suppose $\theta^2(u)=-u$, this implies that $-B(u,\theta(u))=B(u,u)<0$ since $B$ is negative definite. This is a contradiction since $B_{\theta}$ is definite positive. An involution which as only $1$ as eigenvalue is the identity.