Lie algebras and the Killing form. The Killing form is defined by $K(x,y) = \text{tr}(\text{ad} x, \text{ad} y)$, right? In this lecture, we assume that $\{x_1, ... , x_n\}$ is a basis for $g$ and $\{y_1, ... ,y_n\}$ is a dual basis with respect to the Killing form...which is equivalent to saying $K(x_i, y_j) = \delta_{ij}$ for $1 \leq i, j \leq n$. 
This is kind of confusing...we know that the dual of $g$ is the set of linear functionals $f: g \rightarrow k$, right? So a basis of that is a set of functions from $g$ to $k$. However, we know that the domain of the Killing form is $g \times g$. How does that make sense?
 A: If we assume as in the notes that $\mathfrak{g}$ is semisimple, then the Killing form is nondegenerate (in fact this is an equivalent condition). So, it determines an isomorphism $\Phi: \mathfrak{g} \to \mathfrak{g}^*$ given by
$$\Phi(x)(y) := K(x, y),$$
and in particular its inverse $\Phi^{-1}$ is characterized by
$$\eta = K(\Phi^{-1}(\eta), \,\cdot\,) \qquad \text{(for all $\eta \in \mathfrak{g}^*$)}.$$
Now, given any (vector space) basis $(x_i)$ of $\mathfrak{g}$, there is a unique dual basis $(\eta_j)$ of $\mathfrak{g}^*$ such that $\eta_j(x_i) = \delta_{ij}$. Then, the isomorphism $\Phi^{-1}: \mathfrak{g}^* \to \mathfrak{g}$ sends each element $\eta_j$ of the dual basis to an element $y_j := \Phi^{-1}(\eta_j)$, and since $\Phi^{-1}$ is an isomorphism, $(y_j)$ is a basis of $\mathfrak{g}$. Unwinding gives that $(y_j)$ is related to $(x_i)$ by (and by uniqueness implicit in the above construction is characterized by)
$$K(x_i, y_j) = K(x_i, \Phi^{-1}(\eta_j)) = \eta_j(x_i) = \delta_{ij}$$
as desired.
