The Weyl group of $\Phi$ permutes the set $\Phi$

Question

i'm reading Humphreys's book on Lie algebra and I can't understand a passage about the Weyl group (page 43) https://www.math.uci.edu/~brusso/humphreys.pdf:

Let $\Phi$ be a root system in $E$. Denote by $W$ the subgroup of $GL(E)$ generated by the reflections $\sigma_{\alpha}(\alpha\in \Phi).$ By axiom R3 (If $\alpha\in \Phi$, the reflection $\sigma_{\alpha}$ leaves $\Phi$ invariant), $W$ permutes the set $\Phi$

What does he mean by $W$ permutes the set $\Phi$? What is the exactly definition of this? I can't find this "terminology" anywhere!

Any clarification would be appreciated

Answer 1

Consider the elements of the Weyl group just as simple reflections on an euclidian space. These reflections form a group. When these reflections acts on specific vectors called "roots" that belongs to a "root system" they leave the "root system" unchanged.

I'll give you an example. Start from an Euclidan space $E$ with base $\left\{ e_{1},\,\,e_{2},\,\,e_{3}\right\}$ and consider the vector space $V=\left\{ x\in E\,\,:\,\,\left\langle x,\,e_{1}+e_{2}+e_{3}\right\rangle =0\right\}$. Then consider the root system $A2$ as follows: $$\Delta=\left\{ \pm\left(e_{1}-e_{2}\right),\,\pm\left(e_{2}-e_{3}\right),\,\pm\left(e_{1}-e_{3}\right)\right\}. $$ A base of the system is given by $\Phi=\left\{ e_{1}-e_{2},\,\,e_{2}-e_{3}\right\}$ and let's call the simple roots $\alpha_{1}=e_{1}-e_{2}$ and $\alpha_{2}=e_{2}-e_{3}$. Standard way of writing a reflection belonging to the Weyl group is. $$s_{\alpha}\left(x\right)=x-\frac{2\left\langle x,\,\alpha\right\rangle }{\left\langle \alpha,\,\alpha\right\rangle }\alpha.$$ These reflections as you may notice leave the hiperplane $U_{\alpha}=\left\{ x\in V\,\,:\,\,\left\langle x,\,\alpha\right\rangle =0\right\}$ unchanged.

Then try to apply a specific reflection to the root system. For example let's try $s_{\alpha_{2}}$. Then we obtain
$$ s_{\alpha_{2}}(\alpha_{1})=\alpha_{1}+\alpha_{2},$$ $$s_{\alpha_{2}}(\alpha_{2})=-\alpha_{2},$$ $$s_{\alpha_{2}}(\alpha_{1}+\alpha_{2})= \alpha_{1} ,$$ $$...$$ and so on. So the element of the Weyl group (namely the reflection $s_{\alpha_{2}}$) acts as a permutation on the roots of the root system.

Answer 2

By permutation he means that the reflection is a bijection from the root system to itself. Since the reflection leaves the root system invariant, you need only check it's injective and surjective. Both of these follow easily from the definition of reflection (and the fact that it's order 2).