The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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


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.


February 5, 2010 Posted by | Geometry, Root Systems | 7 Comments