# The Unapologetic Mathematician

## The Action of the Weyl Group on Weyl Chambers

With our latest lemmas in hand, we’re ready to describe the action of the Weyl group $\mathcal{W}$ of a root system $\Phi$ on the set of its Weyl chambers. Specifically, the action is “simply transitive”, and the group itself is generated by the reflections corresponding to the simple roots in any given base $\Delta$.

To be a bit more explicit, let $\Delta$ be any fixed base of $\Phi$. Then a number of things happen:

• If $\gamma$ is any regular vector, then there is some $\sigma\in\mathcal{W}$ so that $\langle\sigma(\gamma),\alpha\rangle>0$ for all $\alpha\in\Delta$. That is, $\sigma$ sends the Weyl chamber $\mathfrak{C}(\gamma)$ to the fundamental Weyl chamber $\mathfrak{C}(\Delta)$.
• If $\Delta'$ is another base, then there is some $\sigma\in\mathcal{W}$ so that $\sigma(\Delta')=\Delta$. That is, $\sigma$ sends $\mathfrak{C}(\Delta')$ to $\mathfrak{C}(\Delta)$. We say that the action of the Weyl group is “transitive” on bases and their corresponding Weyl chambers.
• If $\alpha\in\Phi$ is any root, then there is some $\sigma\in\mathcal{W}$ so that $\sigma(\alpha)\in\Delta$.
• The Weyl group $\mathcal{W}$ is generated by the $\sigma_\alpha$ for $\alpha\in\Delta$.
• If $\sigma(\Delta)=\Delta$ for some $\sigma\in\mathcal{W}$, then $\sigma$ is the identity transformation. That is, the only transformation in the Weyl group that sends a base back to itself is the trivial one. We say that the action of the Weyl group is “simple” on bases and their corresponding Weyl chambers.

What we’ll do is let $\mathcal{W}'$ be the group generated by the $\sigma_\alpha$ for $\alpha\in\Delta$, as in the fourth assertion. We’ll show that this group satisfies the first three assertions, and then show that $\mathcal{W}'=\mathcal{W}$.

Let $\gamma$ be a regular vector and write $\delta$ for the half-sum of the positive roots

$\displaystyle\delta=\frac{1}{2}\sum\limits_{\beta\in\Phi^+}\beta$

Choose some $\sigma\in\mathcal{W}'$ so that $\langle\sigma(\gamma),\delta\rangle$ is as large as possible. If $\sigma_\alpha$ is simple, then $\sigma_\alpha\sigma$ is in $\mathcal{W}'$ too, so we find

\displaystyle\begin{aligned}\langle\sigma(\gamma),\delta\rangle&\geq\langle\sigma_\alpha\sigma(\gamma),\delta\rangle\\&=\langle\sigma(\gamma),\sigma_\alpha(\delta)\rangle\\&=\langle\sigma(\gamma),\delta-\alpha\rangle\\&=\langle\sigma(\gamma),\delta\rangle-\langle\sigma(\gamma),\alpha\rangle\end{aligned}

which forces $\langle\sigma(\gamma),\alpha\rangle\geq0$ for all $\alpha\in\Delta$. None of these inner products can actually equal zero, because if one was then we would have $\gamma\in P_\alpha$ and $\gamma$ wouldn’t be regular. Therefore $\sigma(\gamma)$ lies in the fundamental Weyl chamber, as desired.

For the second assertion, we know that there must be some regular $\gamma$ in the positive half-space for each root $\alpha'\in\Delta'$, and the first assertion then applies to send $\Delta'$ to $\Delta$.

For the third assertion, we can invoke the second assertion as long as we know that every root $\alpha\in\Phi$ lies in some base $\Delta'$. We can find some $\gamma\in P_\alpha$ that’s in no other hyperplane perpendicular to another root (other than $-\alpha$). Then pick some close enough $\gamma'$ so that $\langle\gamma',\alpha\rangle=\epsilon>0$, but also $\lvert\langle\gamma',\beta\rangle\rvert>0$ for all $\beta\neq\pm\alpha$. The root $\alpha$ must then belong to the base $\Delta(\gamma')$.

Okay, now let’s show that $\mathcal{W}'=\mathcal{W}$. We just need to show that each reflection $\sigma_\alpha$ for $\alpha\in\Phi$ (all of which together generate $\mathcal{W}$) is an element of $\mathcal{W}'$. But using our third assertion we can find some $\tau\in\mathcal{W}'$ so that $\beta=\tau(\alpha)\in\Delta$. Then

$\displaystyle\sigma_\beta=\sigma_{\tau(\alpha)}=\tau\sigma_\alpha\tau^{-1}$

and so $\sigma_\alpha=\tau^{-1}\sigma_\beta\tau\in\mathcal{W}'$.

Finally, suppose that $\sigma$ is some non-identity element of $\mathcal{W}$ so that $\sigma(\Delta)=\Delta$. Thanks to our fourth assertion we can write $\sigma$ as a string $\sigma_1\dots\sigma_t$ of basic reflections, and we can assume that $t$ is as small as possible. Then we must have $\sigma(\alpha_t)\prec0$ by our final lemma from last time, but we also must have $\sigma(\alpha_t)\in\Delta\subseteq\Phi^+$, which gives us a contradiction.