The Weyl Group of a Root System

Let’s take a root system $\Phi$ in the inner product space $V$ . Each vector $\alpha$ in $\Phi$ gives rise to a reflection in $\sigma_\alpha\in\mathrm{O}(V)$ , the group of transformations preserving the inner product on $V$ . So what sorts of transformations can we build up from these reflections? The subgroup of $\mathrm{O}(V)$ generated by the reflections $\sigma_\alpha$ for all $\alpha\in\Phi$ is called the Weyl group $\mathcal{W}$ of the root system. It’s pronounced “vile”, but we don’t mean that as any sort of value judgement.

Anyway, we can also realize $\mathcal{W}$ as a subgroup of the group of permutations on the vectors in $\Phi$ . Indeed, by definition each $\sigma_\alpha$ sends each vector in $\Phi$ back to another vector in $\Phi$ , and so shuffles them around. So if $\Phi$ has $k$ vectors, the Weyl group can be realized as a subgroup of $S_k$ .

In particular, $\mathcal{W}$ is a finite group, as a subgroup of another finite group. In fact, we even know that the number of transformations in $\mathcal{W}$ divides $k!$ . It may well (and usually does) have elements which are not of the form $\sigma_\alpha$ , but there are still only a finite number of them.

The first thing we want to take note of is how certain transformations in $\mathrm{GL}(V)$ act on $\mathcal{W}$ by conjugation. Specifically, if $\tau$ leaves $\Phi$ invariant, then it induces an automorphism on $\mathcal{W}$ that sends the generator $\sigma_\alpha$ to $\tau\sigma_\alpha\tau^{-1}$ — which (it turns out) is the generator $\sigma_{\tau(\alpha)}$ — for all $\alpha\in\Phi$ . Further, it turns out that $\beta\rtimes\alpha=\tau(\beta)\rtimes\tau(\alpha)$ for all $\alpha,\beta\in\Phi$ .

Indeed, we can calculate

$\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\tau(\sigma_\alpha(\beta))=\tau(\beta-(\beta\rtimes\alpha)\alpha)=\tau(\beta)-(\beta\rtimes\alpha)\tau(\alpha)$

Now, every vector in $\Phi$ is of the form $\tau(\beta)$ for some $\beta$ , and so $\tau\sigma_\alpha\tau^{-1}$ sends it to the vector $\tau(\sigma_\alpha(\beta))$ , which is again in $\Phi$ , so it leaves $\Phi$ invariant. The transformation $\tau\sigma_\alpha\tau^{-1}$ also fixes every vector in the hyperplane $\tau(P_\alpha)$ , for if $\beta$ is orthogonal to $\alpha$ , then the above formula shows that $\tau(\beta)$ is left unchanged by the transformation. Finally, $\tau\sigma_\alpha\tau^{-1}$ sends $\tau(\alpha)$ to $-\tau(\alpha)$ .

This is all the data we need to invoke our lemma, and conclude that $\tau\sigma_\alpha\tau^{-1}$ is actually equal to $\sigma_{\tau(\alpha)}$ . Specifying the action on the generators of $\mathcal{W}$ is enough to determine the whole automorphism. Of course, we can also just let $\sigma$ act on each element of $\mathcal{W}$ by conjugation, but it’s useful to know that the generating reflections are sent to each other exactly as their corresponding vectors are.

Now we can calculate from the definition of a reflection

$\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\sigma_{\tau(\alpha)}(\tau(\beta))=\tau(\beta)-(\tau(\beta)\rtimes\tau(\alpha))\tau(\alpha)$

Comparing this with the equation above, we find that $\tau(\beta)\rtimes\tau(\alpha)=\beta\rtimes\alpha$ , as asserted.

January 21, 2010 - Posted by John Armstrong | Algebra, Geometry, Group Actions, Group theory, Root Systems

16 Comments »

[…] from this, we find that the Weyl group of not only acts on itself, but on . Indeed, induces a homomorphism that sends the generator […]

Pingback by The Category of Root Systems « The Unapologetic Mathematician | January 22, 2010 | Reply
You write that:

…by definition each \sigma_\alpha sends each vector in \Phi back to another vector in \Phi, and so shuffles them around. So if \Phi has k vectors, the Weyl group can be realized as a subgroup of S_k.

It’s clear that this defines a homomorphism from the Weyl group into $S_k$ . But it doesn’t seem trivially obvious that this is injective — a priori, couldn’t there be two funny products of $\sigma_\alpha$ ‘s which aren’t equal but agree on elements of $\Phi$ ?

Comment by Peter LeFanu Lumsdaine | January 24, 2010 | Reply
That’s a good question, Peter. But there can’t be, because $\Phi$ spans the whole vector space. So if any two linear transformations agree on $\Phi$ , they must agree on the whole vector space, and thus must be the same transformation.

On the other hand, you might be thinking that some linear transformation in $\mathcal{W}$ can be written as a product of the $\sigma_\alpha$ in two different ways. This is possible, but it leads to a relation in the Weyl group. Notice that I never said that $\mathcal{W}$ was freely-generated (beyond the obvious fact that each generator is an involution).

Does that make sense?

Comment by John Armstrong | January 25, 2010 | Reply
- Ah! I’d forgotten that $\Phi$ spans the space. OK, it’s obvious 🙂 (It was indeed your first interpretation that I meant, not your second.)
  
  In fact, I guess even if we don’t assume that $\Phi$ spans the space, they agree on the span of $\Phi$ as you say, and they also act trivially on the orthogonal complement of $\Phi$ (since each $\sigma_\alpha$ does), so still agree on the whole space.
  
  Comment by Peter LeFanu Lumsdaine | January 25, 2010 | Reply
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“pronounced “vile”, but we don’t mean that as any sort of value judgement”, lovely really enjoyed reading this blog.

Comment by James Pearson | April 2, 2013 | Reply

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