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.

  • $\begingroup$ How would $\theta$ need to ask on a basis where the quadratic form is diagonalized? $\endgroup$
    – AHusain
    Dec 20, 2015 at 0:28
  • $\begingroup$ @TsemoAristide I've edited the question. $\endgroup$
    – nihan
    Dec 20, 2015 at 1:08

1 Answer 1


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.


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