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.

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February 5, 2010 - Posted by John Armstrong | Geometry, Root Systems

7 Comments »

Mathworld has a cute animation,

http://mathworld.wolfram.com/WeylGroup.html

and begins their text with:

Let L be a finite-dimensional split semisimple Lie algebra over a field of field characteristic 0, H a splitting Cartan subalgebra, and Lambda a weight of H in a representation of L. Then

Lambda’ = LambdaS_alpha = lambda-(2(Lambda,alpha))/((alpha,alpha))(alpha)

is also a weight. Furthermore, the reflections S_alpha with alpha a root, generate a group of linear transformations in H_0^* called the Weyl group W of L relative to H, where H^* is the conjugate space of H and H_0^* is the Q-space spanned by the roots (Jacobson 1979, pp. 112, 117, and 119).

The Weyl group acts on the roots of a semisimple Lie algebra, and it is a finite group. The animations above illustrate this action for Weyl Group acting on the roots of a homotopy from one Weyl matrix to the next one (i.e., it slides the arrows from g to h) in the first two figures, while the third figure shows the Weyl Group acting on the roots of the Cartan matrix of the infinite family of semisimple lie algebras A_3 (cf. Dynkin diagram), which is the special linear Lie algebra, sl_4.

SEE ALSO: Cartan Matrix, Dynkin Diagram, Lie Algebra, Lie Algebra Root, Lie Group, Macdonald’s Constant-Term Conjecture, Root System, Root Lattice, Semisimple Lie Algebra, Weight Lattice, Weyl Chamber

Portions of this entry contributed by Todd Rowland

REFERENCES:

Jacobson, N. Lie Algebras. New York: Dover, 1979.

Comment by Jonathan Vos Post | February 6, 2010 | Reply
It’s cute, but it goes so fast that if you don’t already know what it’s supposed to be, you’ll never understand it.

If you want more visuals, there are some coming.

Comment by John Armstrong | February 6, 2010 | Reply
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