I am trying to understand what is happening on pg. 27 of Humphrey's Lie algebras and Representation theory.
We are given a semi-simple Lie algebra $L$ and a symmetric, invariant, nondegenerate bilinear form $\beta:L\times L \to \mathbb{C}$. The author claims that if $(x_1,\ldots,x_n)$ is a basis for $L$, then there is a uniquely determined dual basis $(y_1,\ldots,y_n)$ satisfying $\beta(x_i,y_j) = \delta_{ij}$.
Now, I thought a dual basis would mean a basis for the dual space $L^* = Hom_{\mathbb{C}}(L,\mathbb{C})$. So I am not understanding what the condition $\beta(x_i,y_j) = \delta_{ij}$ means. How are we putting a map $y_j : L \to \mathbb{C}$ into our bilinear form?
Best Answer
If $L$ is semisimple, then the Killing form $\beta$ is non-degenerate and thus induces an isomorphism $L \to L^*$ via $y \mapsto \beta(\cdot, y)$. The thing is that calling $\{y_1,\ldots,y_n\}$ a "dual basis" is an abuse of language, since $y_j \in L$ for all $j$. But the word "dual" is justified by the isomorphism $L\to L^*$, since given the basis $\{x_1,\ldots,x_n\}\subseteq L$, there is the dual basis $\{x_1^*,\ldots,x_n^*\}\subseteq L^*$ satisfying $x_j^*(x_i)=\delta_{ij}$. And the map $L\to L^*$ induced by $\beta$ is bijective, so for each $j$ there is a unique $y_j\in L$ such that $\beta(\cdot,y_j)=x_j^*$. Evaluate both sides at $x_i$ to get $\beta(x_i,y_j)=\delta_{ij}$, as wanted.