dual basis of $\mathfrak{sl}(n,F)$ with respect to the Killing form

Question

Exercise 8.6 of Section 8 of Humphreys book

Compute the basis of $\mathfrak{sl}(n,F)$ which is dual (via the Killing form) to the standard basis.

Let $L=H\oplus (\bigoplus_{\alpha \in \Phi} L_\alpha)$ be the Cartan decomposition of $L$ for a fixed maximal toral subalgebra $H$. The standard basis that he refers is the set $\{x_\alpha, y_\alpha, h_\alpha: \alpha \in \Phi\},$ where $x_\alpha \in L_\alpha, y_\alpha \in L_{-\alpha}$ and $h_\alpha$ satisfies $h_\alpha = [x_\alpha,y_\alpha]$. Denoting by $\kappa(x,y)$ the Killing form of $L$, it turns out that $h_\alpha = \frac{2}{\kappa(t_\alpha,t_\alpha)} t_\alpha,$ where $t_\alpha\in H$ is the unique element satisfying $\alpha(h) = \kappa(t_\alpha,h), h \in H$.

My attempt: For each $\alpha \in \Phi$, define $e_\alpha = \frac{\kappa(t_\alpha,t_\alpha)}{2} y_\alpha$, so we see that

$ [x_\alpha,e_\alpha] = \frac{\kappa(t_\alpha, t_\alpha)}{2}h_\alpha = t_\alpha.$

On the other hand, since $e_\alpha \in L_{-\alpha}$ and $x_\alpha \in L_\alpha$, by a result in the book we have $$[x_\alpha, e_\alpha] = \kappa(x_\alpha, e_\alpha)t_\alpha. $$

Since $t_\alpha \neq 0$, it follows that $\kappa(x_\alpha, e_\alpha) =1$. Also, by a result in the book we have $\kappa (L_\beta, L_{-\alpha}) = 0$ for every $\beta \neq \alpha$ and $H$ being orthogonal (wrt to $\kappa$) to $L_{-\alpha}$, shows that $e_\alpha$ is the dual of $x_\alpha$.

Similar argument shows that $f_\alpha = \frac{\kappa(t_\alpha,t_\alpha)}{2}x_\alpha$ is the dual of $y_\alpha$.

But I cant find the dual of each $h_\alpha$. Initially, I tought that $g_\alpha = t_\alpha/2$ would do the trick, but I cant show that this is orthogonal to any other $h_\beta, \beta\neq \alpha$.

Any help? Thank you.

Answer 1

Let $\ell=n-1$ and $\alpha_1,...,\alpha_\ell$ a system of simple roots in the usual ordering. Abbreviate $h_i := h_{\alpha_i}$. You already noticed that the CSA $\mathfrak h = \mathrm{span}(h_1,..., h_\ell)$ is orthogonal wrt the Killing form to all the root spaces, so it suffices to look at the restriction of the Killing form to $\mathfrak h$ and find the dual basis to $(h_1, ..., h_\ell)$ here. Now on this, the Killing form is given by [EDIT: corrected scaling thanks to Hetong Xu's comment]

$$(*) \qquad \qquad \kappa(h_i, h_j)= \begin{cases} 2(\ell+1) \text{ if } i=j \\ -1(\ell+1) \text{ if } j =i\pm1 \\0 \text{ else} \end{cases}.$$

So setting $h_i^* := \sum_{j=1}^\ell a_i^{j} h_j$ you see that for each $i$, the condition $\kappa(h_i^*, h_j)= \delta_{ij}$ translates to $\ell$ equations in $\ell$ variables which are easily solved through backwards substitution. E.g. for $i=1$, we get [EDIT: corrected some typos as well]

$$\begin{align} (\ell+1)(2a_1^1-a_1^2)&=1\\ (\ell+1)(-a_1^1+2a_1^2-a_1^3)&=0\\ (\ell+1)(-a_1^2+2a_1^3-a_1^4)&=0\\ ...\\ (\ell+1)(-a_1^{\ell-2}+2a_1^{\ell-1}-a_1^\ell)&=0\\ (\ell+1)(-a_1^{\ell-1}+2a_1^{\ell})&=0 \end{align}$$

(where one can scrap the factor $\ell+1$ in all but the first line) which is solved by $a_1^j=\dfrac{\ell-j+1}{(\ell+1)^2}$ i.e.

$$h_1^* = \dfrac{1}{(\ell+1)^2} \left(\ell h_1 + (\ell-1) h_2 +(\ell-2) h_3 + ... + 2h_{\ell-1} + h_\ell \right).$$

I leave the others to you. Note that the list of end results must be somewhat symmetric under $i \leftrightarrow \ell-i$.

For example for $\ell=2$ we easily get $h_1^*= \frac29 h_1+ \frac19 h_2$ and $h_2^*= \frac19 h_1+ \frac29 h_2$ (see what I mean by "somewhat symmetric").

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Note that the scaling of the Killing form is somewhat annoying, cf. e.g. How are the roots of a Lie algebra related to its Killing form?. Indeed, one would often replace the Killing form $\kappa( \cdot , \cdot)$ above by a scaled version $\langle \cdot, \cdot \rangle$ which instead satisfies

$$(**) \qquad \qquad \langle h_i, h_j \rangle = \begin{cases} 2 \text{ if } i=j \\ -1 \text{ if } j =i\pm1 \\0 \text{ else} \end{cases}.$$

Then the equations become

$$\begin{align} 2a_1^1-a_1^2&=1\\ -a_1^1+2a_1^2-a_1^3&=0\\ -a_1^2+2a_1^3-a_1^4&=0\\ ...\\ -a_1^{\ell-2}+2a_1^{\ell-1}-a_1^\ell&=0\\ -a_1^{\ell-1}+2a_1^{\ell}&=0 \end{align}$$

etc. which are solved by $a_1^j=\dfrac{\ell-j+1}{\ell+1}$ i.e.

$$h_1^* = \dfrac{1}{\ell+1} \left(\ell h_1 + (\ell-1) h_2 +(\ell-2) h_3 + ... + 2h_{\ell-1} + h_\ell \right),$$

for example for $\ell=2$ we now get $h_1^*=\frac23 h_1+ \frac13 h_2$ and $h_2^*=\frac13h_1+\frac23 h_2$ etc. If you compare to the answers in How do I find a dual basis given the following basis?, another way of viewing what we're doing then is finding the (rows/columns of) the inverse of the Cartan matrix

$$\pmatrix{2 &-1 &0 &...&0\\ -1&2&-1&...&0\\ 0&-1&2&...&0\\ \vdots&\vdots &&\vdots \\ 0 &0&0& &2}$$

which is equivalent to the equations $(**)$.

These calculations give the fundamental dominant weights (often called $\varpi_i$ or $\omega_i$, see e.g. here) in terms of the chosen set of simple roots $\alpha_i$. E.g. for $\ell=2$ we have $\varpi_1 = \frac23 \alpha_1 + \frac13 \alpha_2$ and $\varpi_2=\frac13 \alpha_1+\frac23 \alpha_2$.