2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ These are great references, cheers! $\endgroup$ – Travis Nov 14 '18 at 5:53
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.