# Longest element of Weyl Group for $G_2$

Let $$\mathfrak{g}$$ be a semisimple Lie Algebra, $$\mathfrak{t}$$ a Cartan Subalgebra, $$\Phi$$ the corresponding set of roots, $$\Delta \subset \Phi$$ a root basis and $$W$$ the Weyl Group with respect to $$\Delta$$

I am having trouble finding the longest element of the Weyl Group $$w_0$$ as a product of the simple reflections $$w_\alpha \in W$$ in the case of $$G_2$$.

Here are my thoughts:

Taking $$\Delta = \{\alpha, \beta\}$$ where in the drawing of $$G_2$$ (that takes the shape of the star of david), the upper-left corner is $$\beta$$ and the point to the right of $$0$$ is $$\alpha$$.

Then this indeed qualifies as a root basis, and using the fact that for $$G_2%$$ we have $$W \cong D_{12}$$ (a quoted result from earlier in my course), then:

Letting $$w_\alpha = s \in D_{12}$$, we see that $$w_\beta = r^2s$$ where $$r$$ is a clockwise rotation by $$\frac{\pi}{3}$$.

Further, we may note that $$w_0$$ must send $$\Delta$$ to $$-\Delta$$ and so $$w_0 = w_\alpha w_{3\alpha + 2\beta}$$

But, continuing to identify $$W$$ with $$D_{12}$$, we find that $$w_{3\alpha+2\beta} = r^3s$$ which cannot be generated by $$s, r^2s$$ which seems to imply that $$W$$ is not generated by the simple reflections.

Clearly something has gone wrong here, and I am really struggling to find what that might be.

## 2 Answers

While Travis' answer gives a nice hands-on calculation, I like to point out two answers to related questions which put things in perspective:

Anton Geraschenko's answer here states, among other things, that the longest element in most simple types (actually, all except $$A_{n \ge 2}, D_{2n+1}$$ and $$E_6$$) is just $$-id$$. So this is also the case here, $$w_0$$ must be multiplication with $$-1$$ (which, since we are in two dimensions, is the same as rotating by $$\pi$$).

Allen Knutson's answer here on MathOverflow gives a nice general method to express $$w_0$$ as product of simple reflections. In the case of type $$G_2$$, we can choose $$w=w_\alpha, b=w_\beta$$, and the Coxeter number for type $$G_2$$ is $$h=6$$, so the general formula there gives $$w_0 =(w_\alpha w_\beta)^{h/2} = (w_\alpha w_\beta)^{3}$$. Switching the roles of $$w$$ and $$b$$, which is allowed, gives alternatively $$w_0=(w_\beta w_\alpha)^{3}$$, as in the other answer.

• These are great references, cheers! – Travis Nov 14 '18 at 5:53

If we denote $$s = w_{\alpha}$$, then we can check that $$w_{\beta} = r s$$ (not $$r^2 s$$), where $$r$$ is a clockwise rotation by $$\frac{\pi}{3}$$. (Indeed, $$s$$ and $$r^2 s$$ only generate $$\langle r^2, s \rangle \cong D_6$$, not all of $$D_{12}$$.)

To verify the claim, it's enough to check that it holds for a basis; it is convenient to take the basis consisting of $$\beta$$ and a root orthogonal to $$\beta$$, say, $$2 \alpha + \beta$$: \begin{align} rs \cdot (2 \alpha + \beta) &= r \cdot (\alpha + \beta) = 2 \alpha + \beta = w_{\beta}(2 \alpha + \beta) \\ rs \cdot \beta &= r \cdot (3 \alpha + \beta) = -\beta = w_{\beta}(\beta) . \end{align} Exhausting all possibilities we find that the longest element is $$w_{\alpha} w_{\beta} w_{\alpha} w_{\beta} w_{\alpha} w_{\beta} = w_{\beta} w_{\alpha} w_{\beta} w_{\alpha} w_{\beta} w_{\alpha}$$, or under our identification, $$(w_{\beta} w_{\alpha})^3 = (rs \cdot s)^3 = r^3$$.