# 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