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

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.

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

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 $V*$:$$\langle\alpha,\beta\rangle=B\bigl(\psi^{-1}(\alpha),\psi^{-1}(\beta)\bigr).$$