Verifying that the angle between $\alpha$ and $\beta$ is $\frac {2 \pi} {3}.$

Question

Consider the Lie algebra $sl_{3} (\mathbb C)$ of dimension $8.$ Let $$\begin{align*} H : & = \text {span} \left \{H_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}, H_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \right \}. \end{align*}$$ Then $H$ is the Cartan subalgebra of $sl_3 (\mathbb C).$ The roots of it are given by $\left \{\pm \alpha, \pm \beta, \pm (\alpha + \beta) \right \}.$ In my book it is claimed that the angle between $\alpha$ and $\beta$ are $\frac {2 \pi} {3},$ where the angle $\theta$ between $\alpha$ and $\beta$ is given by the following equation $:$ $$\cos \theta = \frac {K^{\ast} (\alpha, \beta)} {\sqrt {K^{\ast} (\alpha, \alpha)} \sqrt {K^{\ast} (\beta, \beta)}}$$ where $K^{\ast}$ is the dual Killing form defined on $H^{\ast} \times H^{\ast}$ by $K^{\ast} (\gamma, \delta) = K \left (i^{-1} (\gamma), i^{-1} (\delta) \right )$ where $i : H \longrightarrow H^{\ast}$ is the vector space isomorphism given by $h \mapsto K(h, \cdot)$ induced by the Killing form $K.$ With these information in mind how do I find $\theta\ $?

Any help in this regard would be warmly appreciated. Thanks for your time.

Answer 1

Let us try to understand $K$ in terms of a matrix (which we will also denote by $K$) whose entries are given as follows $:$

Choose a basis $\beta = \{e_1, \cdots, e_n \}$ of $H.$ Then the $ij$-th entry of $K$ is given by $K_{ij} = K(e_i, e_j).$

Let $v, w \in H$ with $v = \sum\limits_{i=1}^{n} v_i e_i$ and $w = \sum\limits_{i=1}^{n} w_i e_i.$ Then we have

$$\tag{1}\begin{align*} K(v,w) & = \sum\limits_{i,j = 1}^{n} v_i w_j K_{ij} \end{align*}$$

Let $\alpha \in H^{\ast}.$ We need to find out $\alpha^{\sharp} \in H$ such that for all $v \in H$ we have

$$\tag{2}K(\alpha^{\sharp}, v) = \alpha (v)$$

Let $\beta^{\ast} : = \{e_1^{\ast}, \cdots, e_n^{\ast} \}$ be the dual basis of $H^{\ast}.$ Let $\alpha = \sum\limits_{i = 1}^{n} \alpha_i e_i^{\ast}.$ Then we have $$\begin{align*} \tag{3} \alpha (v) & = \sum\limits_{i=1}^{n} \alpha_i v_i \\ & = \begin{pmatrix} \alpha_1 & \alpha_2 & \cdots & \alpha_n \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}. \end{align*}$$

This observation would lead us to think of the elements of $H$ as column vectors and the elements of $H^{\ast}$ as row vectors.

So our purpose boils down to find a column vector $\alpha^{\sharp} = \begin{pmatrix} \alpha_1^{\sharp} \\ \alpha_2^{\sharp} \\ \vdots \\ \alpha_n^{\sharp} \end{pmatrix} \in H$ such that $K (\alpha^{\sharp}, v) = \alpha (v)$ for all column vectors $v \in H.$

If $v = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}$ then by $(1)$ we have

$$\tag{4}K(\alpha^{\sharp}, v) = \sum\limits_{i,j = 1}^{n} \alpha_i^{\sharp} v_j K_{ij}$$

Combining $(2), (3)$ and $(4)$ we have $$\begin{align*}\sum\limits_{j=1}^{n} \alpha_j v_j & = \sum\limits_{i,j = 1}^{n} \alpha_i^{\sharp} v_j K_{ij} \\ & = \sum\limits_{j = 1}^{n} \left (\sum\limits_{i=1}^{n} K_{ij} \alpha_i^{\sharp} \right ) v_j \end{align*}$$ Since the above equation holds for all column vectors $v \in H$ it follows that $$\begin{align*} \alpha_j & = \sum\limits_{i=1}^{n} K_{ij} \alpha_i^{\sharp} \\ & = \sum\limits_{i=1}^{n} K_{ji} \alpha_i^{\sharp}\ \ (\because K\ \text {is symmetric}) \\ & = \left (K \alpha^{\sharp} \right )_j \end{align*}$$

Since $\alpha^{\sharp}$ is a column vector so is $K \alpha^{\sharp}$ and $\alpha$ is a row vector. So the above equation implies that $$\tag{5} \alpha^{\top} = K \alpha^{\sharp}$$

Now let $Q : H \longrightarrow H^{\ast}$ be the vector space isomorphism induced by $K.$ Recall that $Q$ is defined as $Q(x) = K(x, \cdot)$ for all $x \in H.$ Also notice that any element $\phi \in H^{\ast}$ can be written as $\phi = \sum\limits_{i=1}^{n} \phi(e_i) e_i^{\ast}.$ Thus we have $$\begin{align*} Q(e_j) & = \sum\limits_{i=1}^{n} Q(e_j) (e_i) e_i^{\ast} \\ & = \sum\limits_{i=1}^{n} K(e_j,e_i) e_i^{\ast} \\ & = \sum\limits_{i=1}^{n} K(e_i, e_j) e_{i}^{\ast}\ \ (\because K\ \text {is symmetric}) \\ & = \sum\limits_{i=1}^{n} K_{ij} e_{i}^{\ast} \end{align*}$$ This shows that the matrix representation $[Q]_{\beta \beta^{\ast}}$ of $Q$ with respect to the bases $\beta$ of $H$ and $\beta^{\ast}$ of $H^{\ast}$ is $K.$ Since $Q$ is an isomorphism it follows that $K$ is invertible.

So from $(5)$ we have $$\tag{6} \alpha^{\sharp} = K^{-1} \alpha^{\top}$$

So if $\alpha^{\sharp}$ and $\beta^{\sharp}$ are the pre-images of $\alpha$ and $\beta$ under $Q$ then by $(1)$ we have $$\begin{align*} K^{\ast} (\alpha, \beta) & = K (\alpha^{\sharp}, \beta^{\sharp}) \\ & = \sum\limits_{i,j = 1}^{n} \alpha_i^{\sharp} \beta_j^{\sharp} K_{ij} \\ & = \sum\limits_{i,j,k,l = 1}^{n} \left (K^{-1} \right )_{ik} \alpha_k \left (K^{-1} \right )_{jl} \beta_l K_{ij} \\ & = \sum\limits_{i,k,l=1}^{n} \delta_{il} \left (K^{-1} \right )_{ik} \alpha_k \beta_l \ \ (\because \sum\limits_{j=1}^{n} K_{ij} \left (K^{-1} \right )_{jl} = \left (K K^{-1} \right )_{il} = I_{il} = \delta_{il}) \\ & = \sum\limits_{k,l=1}^{n} \left (K^{-1} \right )_{lk} \alpha_k \beta_l \\ & = \sum\limits_{k,l=1}^{n} \left (K^{-1} \right )_{kl} \alpha_k \beta_l \ \ (\because K^{-1}\ \text {is symmetric since so is}\ K) \\ & = K^{-1} (\alpha, \beta) \ \ (\text {similar to}\ (1)) \end{align*}$$

In particular, $$\tag{7} \left (K^{\ast} \right )_{ij} = \left (K^{-1} \right )_{ij}$$ for all $i,j = 1, 2, \cdots, n.$

The above procedure gives a nice way to compute $K^{\ast} (\gamma, \delta)$ for any $\gamma, \delta \in H^{\ast}.$

In our case one can check that $K_{11} = K (H_1, H_1) = 12,$ $K_{22} = K(H_2, H_2) = 12,$ $K_{12} = K(H_1, H_2) = 6$ and $K_{21} = K(H_2, H_1) = 6.$ So the matrix that is represented by the Killing form is given by $$K = \begin{pmatrix} 12 & 6 \\ 6 & 12 \end{pmatrix}.$$ Hence $K^{-1}$ is given by $$K^{-1} = \frac {1} {108} \begin{pmatrix} 12 & -6 \\ -6 & 12 \end{pmatrix}.$$ By $(7)$ the entries of $K^{\ast}$ are precisely the entries of $K^{-1}.$ So $K^{\ast} (H_1^{\ast}, H_1^{\ast}) = K^{\ast} (H_2^{\ast}, H_2^{\ast}) = \frac {12} {108} = \frac {1} {9}$ and $K^{\ast} (H_1^{\ast}, H_2^{\ast}) = K^{\ast} (H_2^{\ast}, H_1^{\ast}) = -\frac {6} {108} = - \frac {1} {18}.$

Also it is easy to see that $\alpha = H_1^{\ast} - H_2^{\ast}$ and $\beta = H_1^{\ast} + 2 H_2^{\ast}.$ From here it is easy to check that $K^{\ast} (\alpha, \alpha) = K^{\ast} (\beta, \beta) = \frac {36} {108} = \frac {1} {3}$ and $K^{\ast} (\alpha, \beta) = K^{\ast} (\beta, \alpha) = - \frac {18} {108} = - \frac {1} {6}.$ So it follows that $\cos \theta = - \frac {1} {2}$ i.e. $\theta = \frac {2 \pi} {3},$ as required.

Answer 2

I strongly recommend that you do the explicit matrix computations here yourself. I remember going through examples like this was how I truly understood what roots and coroots etc. do and what they are good for.

Call $H_\alpha := \pmatrix{1&0&0\\0&-1&0\\0&0&0}$, and $H_\beta$ your $H_2$, so that your $H_1$ is $H_\alpha + H_\beta$.

I recommend checking the following things, in this order:

1. $\alpha$ is the linear map $\in H^*$ which sends $\pmatrix{h_1&0&0\\0&h_2&0\\0&0&h_3}$ to $h_1-h_2$. $\beta$ is the linear map $\in H^*$ which sends $\pmatrix{h_1&0&0\\0&h_2&0\\0&0&h_3}$ to $h_2-h_3$.

2. The Killing form on $H \times H$ is given by $K(\pmatrix{h_1&0&0\\0&h_2&0\\0&0&h_3}, \pmatrix{k_1&0&0\\0&k_2&0\\0&0&k_3}) = 6 \cdot (h_1k_1 + h_2k_2 +h_3k_3)$ i.e. 6 times the trace form.

3. Consequently, the map $i : h \mapsto K(h, \cdot)$ sends $H_\alpha$ to what (hint: it is almost just $\alpha$), and $H_\beta$ to what?

With this information you can certainly compute $\cos(\theta)$ with either the formula in your question or in the comment. What do you get?