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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?

Artus
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    We're using $K$ to identify the basis of $\mathfrak{g}^*$ dual to $(x_i)$ with elements of $\mathfrak{g}$ itself, namely, $(y_i)$. – Travis Willse Oct 12 '14 at 15:48
  • @Travis So we look for another basis for $g$, ${y_1, ... ,y_n}$ such that $K(x_i, y_j) = \delta_{ij}$? – Artus Oct 12 '14 at 15:50
  • @Artos: the Killing form need not be definite, so that in the general semisimple case the best one can hope of a single basis is that $K(x_{i}, x_{i}) = \pm 1$ and $K(x_{i}, x_{j}) = 0$ when $i \neq j$. – Dan Fox Oct 12 '14 at 15:55
  • better to capitalize the K http://en.wikipedia.org/wiki/Wilhelm_Killing – Will Jagy Oct 12 '14 at 16:20
  • @Artos Not so much look for one, but produce one, in short we form the basis of $\mathfrak{g}^*$ dual to $(x_i)$ and "raise indices" using the Killing form, see my answer for more details. – Travis Willse Oct 13 '14 at 04:53
  • @DanFox So...is the lecture wrong when it says $K(x_i, y_i) = \delta_{ij}$? – Artus Oct 16 '14 at 10:39
  • That isn't the sense in which "dual basis" is being used here. By the "dual basis" to the $x_i$ wrt $K$ they literally mean a basis $y_i$ of $g$ such that $K(x_i,y_j)=\delta_{ij}$. It's not necessary to think about the dual space at all, although as other people pointed out there is a connection. – Matthew Towers Oct 16 '14 at 11:15
  • @Artos: A symmetric bilinear form $B$ on $g$ is nondegenerate if $B(x, y) = 0$ for all $y \in g$ implies $x = 0$. Such a bilinear form gives an isomorphism between $g$ and its dual $g^{\ast}$ by sending $x$ to $B(x, \cdot)$. Pulling a basis of $g^{\ast}$ back via this isomorphism yields a basis of $g$. Given a basis ${x_{i}}$ of $g$ choose a basis ${z_{j}}$ of $g^{\ast}$ such that $z_{j}(x_{i}) = \delta_{ij}$. Let ${y_{i}}$ be the basis of $g$ such that $B(y_{i}, \cdot) = z_{i}$. Then $B(x_{i}, y_{j}) = \delta_{ij}$. If ${x_{i}}$ is $B$-orthogonal, then $y_{i} = \pm x_{i}$. – Dan Fox Oct 17 '14 at 08:51
  • https://math.stackexchange.com/questions/1441119/dual-basis-with-respect-to-bilinear-form/1442538 – user5826 Apr 26 '20 at 20:18

1 Answers1

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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.

Travis Willse
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  • The isomorphism $\Phi$ that you give takes two arguments of $\mathfrak{g}$ - are we fixing some $x$ here? – mi.f.zh Jan 11 '20 at 14:51
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    As written $\Phi$ is a map $\mathfrak{g} \to \mathfrak{g}^$, so $\color{bf0000}{\Phi(x)}$ is an element of $\mathfrak g^$, which we can view as a map $\mathfrak{g} \to \Bbb F$, where $\Bbb F$ is the field underlying $\mathfrak{g}$. Thus, for each $y \in \mathfrak{g}$, $\color{bf0000}{\Phi(x)}(y)$ is an element of $\Bbb F$, and we declare in the answer that the map $\Phi$ is characterized by $$\color{bf0000}{\Phi(x)}(y) = K(x, y) .$$ – Travis Willse Jan 13 '20 at 21:02
  • Huge brain fart there. I know comments shouldn't be used for this only, but thanks! – mi.f.zh Jan 14 '20 at 08:34