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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.

Travis Willse
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user366818
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2 Answers2

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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, $(w_{\beta} w_{\alpha})^3 = (rs \cdot s)^3 = r^3$.

Travis Willse
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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.

Torsten Schoeneberg
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