Let $\mathfrak {g}$ be a semisimple Lie algebra with a non-degenerate invariant bilinear form $B$ (e.g. Killing form). Then what is meant by saying that $C$ is a Casimir tensor with respect to the bilinear form $B\ $?
The wikipedia article on Casimir element says the following $:$
Given a basis $\{X_i \}$ of $\mathfrak {g}$ and the dual basis $\{X^{i}\}$ relative to $B$ the Casimir tensor is defined by $C = \sum\limits_{i} X_{i} X^{i}.$ But how can we guarantee that such a dual basis exists?
Any help in this regard would be warmly appreciated. Thanks for your time.
Once you have a non-degenerate bilinear form $B$ on $\mathfrak {g}$ you have an isomorphism $\mathfrak {g} \simeq \mathfrak {g}^{\ast},$ namely $x \mapsto B(x, \cdot).$ Now given a basis $\{X_i \}$ of $\mathfrak {g}$ let $\{X_i^{\ast}\}$ be the dual basis of $\mathfrak {g}^{\ast}.$ Then the element $X^i$ is just the inverse image of the dual basis element $X_i^{\ast}$ under the above isomorphism which means that $B(X^i, X_j) = \delta_{ij}.$