Let $\mathfrak{g}$ be a finite dimensional Lie algebra and $B:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{K}$ be a non-degenerate invariant symmetric bilinear form. Then there exists a unique $C_{B} \in \mathfrak{U(g)}$, called the Casimir element of $\mathfrak{g}$ corresponding to $B$, such that $$\sum_{i=1}^{n} e_{i}(B^{\sharp}e_{i}^{*}), $$ for any basis $e_{1},...,e_{n}$ of $\mathfrak{g}$, with corresponding dual basis $e_{1}^{*},...,e_{n}^{*}$, where $B^{\sharp}:\mathfrak{g^{*}} \rightarrow \mathfrak{g}$ denotes the inverse of the linear isomorphism $B^{\flat}: \mathfrak{g} \rightarrow \mathfrak{g^{*}}, X \mapsto B(X,-)$. Additionally, $C_{B}$ is a quadratic central element of $\mathfrak{U(g)}$, i.e. $$C_{B} \in \mathfrak{U_{2}(g)} \setminus \mathfrak{U_{1}(g)}$$ and $$C_{B} \in Z(\mathfrak{U(g)}). $$
I'm studying this proposition on Casimir element and i've a question: in its proof it says that if we fix a basis $e_1,...,e_n \in \mathfrak{g}$, since $B$ is non-degenerate, also $B^{\sharp}e_{1}^{*},...,B^{\sharp}e_{n}^{*}$ is a basis of $\mathfrak{g}$, but i can't understand why.
Is there anyone who can help me?
Since $B$ is non-degenerate, we can easily compute the kernel of $B^\flat$, indeed $$\begin{align*} X \in \ker B^\flat &\iff B^\flat (X) = 0\\ &\iff \forall Y \in \mathfrak{g}, \, B(X,Y) = 0\\ &\iff X = 0 \end{align*}$$
So $B^\flat$ is injective and since $\dim \mathfrak{g} = \dim \mathfrak{g}^* < \infty$, it is bijective, thus it has an inverse linear map $B^\sharp:\mathfrak{g}^* \to \mathfrak{g}$ that sends any basis of $\mathfrak{g}^*$ on a basis of $\mathfrak{g}$.
In particular $(B^\sharp e_1^*,\dots,B^\sharp e_n^*)$ is a basis of $\mathfrak{g}$.
This can be summarized as "the composition $B^\sharp \circ (-)^*_{(e_1,\dots,e_n)}$ is an automorphism of $\mathfrak{g}$ (as a vector space).