1
$\begingroup$

Let $\mathfrak{g}$ be a simple Lie algebra and $\mathfrak{h}$ a Cartan subalgebra. There is a non-degenerate symmetric bilinear form on $\mathfrak{h}$ which is a rescaling of the Killing form.

How does the non-degenerate symmetric bilinear form on $\mathfrak{h}$ induce a non-degenerate symmetric bilinear form on $\mathfrak{h}^*$?

Thank you very much.

$\endgroup$
  • $\begingroup$ The form itself induces an isomorphism $\mathfrak h\to\mathfrak h^*$ given by mapping $x$ to the functional $\langle x,-\rangle$. $\endgroup$ – Cheerful Parsnip Feb 20 '18 at 16:20
4
$\begingroup$

In fact, this has nothing to do with Lie algebras. If $B(\cdot,\cdot)$ is a symmetric non-degenerate bilinear form on a finite-dimensional space $V$, it induces an isomorphism $\psi$ from $V$ onto its dual: $v\mapsto B(v,\cdot)$. So, you can define a symmetric non-degenerate bilinear form on $V*$:$$\langle\alpha,\beta\rangle=B\bigl(\psi^{-1}(\alpha),\psi^{-1}(\beta)\bigr).$$

| cite | improve this answer | |
$\endgroup$

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.