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 '15 at 0:28
  • $\begingroup$ @TsemoAristide I've edited the question. $\endgroup$ – nihan Dec 20 '15 at 1:08

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.