The Unapologetic Mathematician

Mathematics for the interested outsider

Weyl Chambers

A very useful concept in our study of root systems will be that of a Weyl chamber. As we showed at the beginning of last time, the hyperplanes P_\alpha for \alpha\in\Phi cannot fill up all of V. What’s left over they chop into a bunch of connected components, which we call Weyl chambers. Thus every regular vector \gamma belongs to exactly one of these Weyl chambers, denoted \mathfrak{C}(\gamma).

Saying that two vectors share a Weyl chamber — that \mathfrak{C}(\gamma)=\mathfrak{C}(\gamma') — tells us that \gamma and \gamma' lie on the same side of each and every hyperplane P_\alpha for \alpha\in\Phi. That is, \langle\gamma,\alpha\rangle and \langle\gamma',\alpha\rangle are either both positive or both negative. So this means that \Phi^+(\gamma)=\Phi^+(\gamma'), and thus the induced bases are equal: \Delta(\gamma)=\Delta(\gamma'). We see, then, that we have a natural bijection between the Weyl chambers of a root system \Phi and the bases for \Phi.

We write \mathfrak{C}(\Delta)=\mathfrak{C}(\gamma) for \Delta=\Delta(\gamma) and call this the fundamental Weyl chamber relative to \Delta. Geometrically, \mathfrak{C}(\Delta) is the open convex set consisting of the intersection of all the half-spaces \{\gamma\vert\langle\gamma,\alpha\rangle>0\} for \alpha\in\Delta.

The Weyl group \mathcal{W} of \Phi shuffles Weyl chambers around. Specifically, if \sigma\in\mathcal{W} and \gamma is regular, then \sigma(\mathfrak{C}(\gamma))=\mathfrak{C}(\sigma(\gamma)).

On the other hand, the Weyl group also sends bases of \Phi to each other. If \Delta\subseteq\Phi is a base, then \sigma(\Delta) is another base. Indeed, since \sigma is invertible \sigma(\Delta) will still be a basis for V. Further, for any \beta\in\Phi we can write \beta=\sigma(\beta'), and then use the base property of \Delta to write \beta' as a nonnegative or nonpositive integral combination of \Delta. Hitting everything with \sigma makes \beta a nonnegative or nonpositive integral combination of \sigma(\Delta), and so this is indeed a base.

And, just as we’d hope, these two actions of the Weyl group are equivalent by the bijection above. We have \sigma(\Delta(\gamma))=\Delta(\sigma(\gamma)) because \sigma preserves the inner product, and so \langle\sigma(\gamma),\sigma(\alpha)\rangle=\langle\gamma,\alpha\rangle. Thus we write \Delta=\Delta(\gamma) for some regular \gamma and find that

\displaystyle\begin{aligned}\sigma(\mathfrak{C}(\Delta))&=\sigma(\mathfrak{C}(\Delta(\gamma)))\\&=\sigma(\mathfrak{C}(\gamma))\\&=\mathfrak{C}(\sigma(\gamma))\\&=\mathfrak{C}(\Delta(\sigma(\gamma)))\\&=\mathfrak{C}(\sigma(\Delta(\gamma)))\\&=\mathfrak{C}(\sigma(\Delta))\end{aligned}

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February 3, 2010 - Posted by | Geometry, Root Systems

3 Comments »

  1. Neat!

    Comment by Jonathan Vos Post | February 4, 2010 | Reply

  2. […] we’re ready to describe the action of the Weyl group of a root system on the set of its Weyl chambers. Specifically, the action is “simply transitive”, and the group itself is generated by […]

    Pingback by The Action of the Weyl Group on Weyl Chambers « The Unapologetic Mathematician | February 5, 2010 | Reply

  3. […] Fundamental Weyl Chamber When we first discussed Weyl chambers, we defined the fundamental Weyl chamber associated to a base as the collection of all the […]

    Pingback by The Fundamental Weyl Chamber « The Unapologetic Mathematician | February 9, 2010 | Reply


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