# The Unapologetic Mathematician

## 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}

February 3, 2010 - Posted by | Geometry, Root Systems Comment by Jonathan Vos Post | February 4, 2010 | Reply