# The Unapologetic Mathematician

## 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 -

1. […] 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

2. 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

3. 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 đź™‚ (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

4. […] are the same as the original reflections, and so they generate the same subgroup of . That is, the Weyl group of is the same as the Weyl group of […]

Pingback by Dual Root Systems « The Unapologetic Mathematician | January 26, 2010 | Reply

5. […] Weyl group of shuffles Weyl chambers around. Specifically, if and is regular, then […]

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6. […] our assumption, and . Thus there is some smallest index so that . Then , and we must have . But we know that . In […]

Pingback by Some Lemmas on Simple Roots « The Unapologetic Mathematician | February 4, 2010 | Reply

7. […] Weyl Chambers With our latest lemmas in hand, 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 […]

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

8. […] from earlier about simple roots and reflections, we can define a few notions that make discussing Weyl groups easier. Any Weyl group element can be written as a composition of simple […]

Pingback by Lengths of Weyl Group Elements « The Unapologetic Mathematician | February 8, 2010 | Reply

9. […] set — allowing — and show that it’s a fundamental domain for the action of the Weyl group […]

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

10. […] to each one in , then we will find a similar decomposition of . But we know from our study of the Weyl group that every root in can be sent by the Weyl group to some simple root in . So we define to be the […]

Pingback by Properties of Irreducible Root Systems I « The Unapologetic Mathematician | February 10, 2010 | Reply

11. […] is irreducible, then the Weyl group acts irreducibly on . That is, we cannot decompose the representation of on as the direct sum of […]

Pingback by Properties of Irreducible Root Systems II « The Unapologetic Mathematician | February 11, 2010 | Reply

12. […] that any combination of the serves to permute the coefficients of a given vector. That is, the Weyl group of the system is naturally isomorphic to the symmetric group . Possibly related posts: […]

Pingback by Construction of A-Series Root Systems « The Unapologetic Mathematician | March 2, 2010 | Reply

13. […] of all, right when we first talked about the category of root systems, we saw that the Weyl group of is a normal subgroup of . This will give us most of the structure we need, but there may be […]

Pingback by The Automorphism Group of a Root System « The Unapologetic Mathematician | March 11, 2010 | Reply

14. […] a collection of reflections, and these reflections generate a group of transformations called the Weyl group of the root system. It’s one of the most useful tools we have at our disposal through the […]

Pingback by Root Systems Recap « The Unapologetic Mathematician | March 12, 2010 | Reply

15. “pronounced â€śvileâ€ť, but we donâ€™t mean that as any sort of value judgement”, lovely really enjoyed reading this blog.

Comment by James Pearson | April 2, 2013 | Reply