Second cohomology group $H^2(\mathfrak{g},\mathbb{R})$ of a semisimple Lie algebra

Question

I was reading proposition 3.13 in http://www.staff.science.uu.nl/~ban00101/lecnotes/repq.pdf in which they prove that $H^2(\mathfrak{g},\mathbb{R})=0$ if $\mathfrak{g}$ is semisimple. Let $\omega\in\wedge^2\mathfrak{g}^*$ be closed. Why does there exist a linear map $\phi:\mathfrak{g}\to\mathfrak{g}$ such that $\omega(X,Y)=\kappa(X,\phi(Y))$ where $\kappa$ is the Killing form. I don't see where it comes from. Is it due to the nondegeneracy of the Killing form?

Answer 1

Yes, the claim follows from the nondegeneracy of the Killing form:

Fix any basis of $\mathfrak{g}$; then, we can rephrase the condition as the existence of a matrix $[\phi]$ such that $$[\omega] = [\kappa] [\phi] ,$$ where $[\omega]$ and $[\kappa]$ are respectively the matrix representations of $\omega, \kappa$ with respect to the basis. Since $\kappa$ is nondegenerate, so is $[\kappa]$, and we may rearrange to find an explicit expression for $\phi$.