Recall that the reflections ${s_{\alpha} \in Aut(E)}$ for ${\alpha \in \Phi}$ preserve ${\Phi}$. They generate a group ${W}$ called the Weyl group. Moreover, since ${\Phi}$ spans ${E}$, the map ${W \rightarrow S_{\Phi}}$, the symmetric group on ${\Phi}$, is injective. So ${W}$ is a finite group of orthogonal isomorphisms of ${E}$, i.e. leaving invariant the bilinear form ${(\cdot, \cdot)}$.

Everything here actually makes sense for root systems in general, but we are restricting ourselves to the case of a root system associated to semisimple Lie algebra and a Cartan subalgebra.  The only difference is that one has to prove the result on maximal strings (which was proved in the case of Lie algebras here), though it can be done for root systems in general.

Now choose a base ${\Delta}$ for ${\Phi}$ and a corresponding partition ${\Phi = \Phi^+ \cup \Phi^-}$; the ${s_{\delta}}$ for ${\delta \in \Delta}$ are called simple reflections.

The goal we are aiming for is the following theorem, which gives a large amount of information about the Weyl group.

Theorem 1 ${W}$ acts simply transitively on Weyl chambers and on bases. Any root can be moved by an element of the Weyl group into a given base. ${W}$ is generated by the simple reflections.

Action of simple reflections on a base

So, first we’re going to look at simple reflections and show

Lemma 2 The simple reflection ${s_{\delta}}$ permutes ${\Phi^+ - \{ \delta \}}$ and sends ${\delta}$ into its inverse.

Indeed, suppose ${\alpha \in \Phi^+}$. If ${\alpha \neq \delta}$, then ${\alpha}$ is not a multiple of ${\delta}$ and contains some other element of the base with positive coefficient, say ${\delta'}$. Then applying ${s_{\delta}}$ to ${\alpha}$ only changes the coefficient of ${\delta}$, so ${s_{\delta}(\alpha)}$ still has a positive coefficient of ${\delta'}$ and is thus a positive root. It is clear that ${s_{\delta}(\delta) = -\delta}$; this is just a fact about reflections in general.

Transitivity on Weyl chambers

We are now going to prove that the subgroup ${W'}$ generated by the simple reflections (associated to a base ${\Delta}$) acts transitively on the Weyl chambers. In particular, given a regular element ${v}$ (one that is not orthogonal to any element of ${\Phi}$), we will show that there is ${s \in W}$ such that

$\displaystyle (s v, \alpha) > 0 \ \forall \alpha \in \Phi^+.$

The idea is to start by considering the element ${\rho = \frac{1}{2} \sum_{\alpha \in \Phi^+} \alpha}$, which will show up frequently in the sequel. A quantitative consequence of Lemma 1 is

$\displaystyle s_{\delta} \rho = \rho - \delta, \ \forall \delta \in \Delta.$

So, given ${v}$, choose ${s \in W'}$ maximizing ${(sv, \rho)}$. Then I claim ${(sv, \delta) \geq 0}$ for all simple roots ${\delta}$. Indeed,

$\displaystyle (sv, \rho) \geq (s_{\delta} s v, \rho) = (sv, \rho - \delta) = (sv, \rho) - (sv, \delta).$

Now the claim follows. If we prove that we have strict inequality (${(sv, \delta)>0 \ \forall \delta}$) then we will have proved transitivity of ${W'}$ on the Weyl chambers. But if ${(sv, \delta)=0}$ for some ${\delta}$, then ${v}$ would be orthogonal to ${s \delta \in \Phi}$, a contradiction by regularity.

Let ${\Delta(v), \Delta(v')}$ be two bases associated to ${v,v'}$. We can use an element of ${W'}$ to move ${v}$ into the same Weyl chamber as ${v'}$, which means that ${\Phi^+(v)}$ gets transformed into ${\Phi^+(v')}$—as a result, ${\Delta(v)}$ gets transformed into ${\Delta(v')}$. So ${W'}$ acts transitively on the set of bases.

${W}$ is generated by the simple reflections

The next step towards the proof of this theorem will be to show:

Lemma 3 Any root ${\alpha}$ is contained in a base ${\Delta(v)}$.

We will choose a regular vector ${v}$ such that ${\alpha}$ brackets positively with ${v}$ but to a very small number, so ${\alpha}$ is indecomposable in the corresponding choice of positive roots ${\Phi^+}$.

Consider the hyperplane ${\alpha^{\perp}}$. We can choose ${w \in \alpha^{\perp}}$ that is not perpendicular to any other root in ${\Phi}$ (except ${-\alpha}$). Modifying ${w}$ slightly, we can get a vector ${v}$ such that ${(v, \alpha)>0}$ is very small, smaller than ${|(v, \beta)|}$ for any root ${\beta \neq \pm \alpha}$. Then ${\alpha}$ is a simple root in ${\Delta(v)}$.

It is now possible for us to show that ${W' = W}$; we must show that ${s_{\alpha}}$ for ${\alpha \in \Phi}$ belongs to ${W'}$. Now we can find ${w' \in W}$ with ${w \alpha = \delta \in \Delta}$ by the above lemma. Then

$\displaystyle s_{\alpha} = s_{w'^{-1} \delta} = w'^{-1} s_{\delta} w' \in W'.$

${W}$ acts simply transitively on the bases

We first prove a lemma on shrinking expressions of products of simple reflections.

Lemma 4 Given simple roots ${\delta_1, \dots, \delta_n}$ and the corresponding reflections ${s_i := s_{\delta_i}}$. If ${s_1 \dots s_{t-1}(\delta_t) \in \Phi^-}$, then we can write ${s_1 \dots s_t}$ as a product of smaller length using only ${s_1, \dots, s_{t-1}}$.

There has to be a first index ${j}$ where ${s_j s_{j+1} \dots s_{t-1}(\delta_t) \in \Phi^-}$. By the above lemma, since these are simple reflections, it must be the case that

$\displaystyle s_{j+1} \dots s_{t-1} (\delta_t) = \delta_j.$

Now if ${T}$ is an orthogonal transformation, then ${T s_{v} T^{-1} = s_{Tv}}$ as is easily checked. Therefore,

$\displaystyle s_{j+1} \dots s_{t-1} s_t = s_j s_{j+1} \dots s_{t-1}$

and it follows that

$\displaystyle s_1 \dots s_t = s_1 \dots s_{j-1} s_{j+1} \dots s_{t-1}$

since ${s_j^2 = 1}$.

Suppose ${w \in W}$ satisfies ${w(\Delta) = \Delta}$; I claim ${w=1}$.

Write ${w = s_1 \dots s_t}$ as a minimal product of simple reflections associated to ${\delta_1, \dots, \delta_t \in \Delta}$. By assumption ${w \delta_t \in \Delta \subset \Phi^+}$, so

$\displaystyle s_1 \dots s_{t-1} \delta_t \in \Phi^- .$

This means we can shrink the expression for ${w}$ by the second lemma, contradiction.

As a result, it follows that ${w}$ acts simply transitively on Weyl chambers too, since they are in bijection with bases.

The size of the minimal expression

I discussed in the last section how one could exploit properties of the minimal expression for ${w \in W}$ as a product ${s_1 \dots s_t}$ of simple reflections. A minimal expression is called reduced and ${t}$ is called the length ${l(w)}$.

If ${w}$ is a simple reflection, then ${w}$ sends precisely one positive root into a negative root, and ${t=1}$.

More generally:

Proposition 5 The length ${l(w)}$ equals the number ${n(w)}$ of positive roots sent into negative roots.

This is proved by induction on the length; we have just stated it for ${l(w)=1}$ above. We can write ${w = s_1 \dots s_{l(w)-1} s_{l(w)}}$ where the expression ${s_1 \dots s_{l(w)-1}}$ is itself reduced. Suppose these reflections are associated to ${\delta_1, \dots, \delta_{l(w)}}$. Then ${s_1 \dots s_{l(w)-1} \delta_{l(w)} \in \Phi^+}$ (if it were in ${\Phi^-}$, we could use the previous lemma to shrink the expression). In particular, ${w \delta_{l(w)} \in \Phi^-}$—in this respect, ${w}$ behaves differently from the subproduct ${s_1 \dots s_{l(w)-1}}$. Now ${s_{l(w)}}$ permutes the positive roots in ${\Phi^+ - \{ \delta_{l(w)}}$. So in particular,

$\displaystyle l(w) = l(s_1 \dots s_{l(w)-1}) + 1 .$

But we have ${l(s_1 \dots s_{l(w)-1}) = n(s_1 \dots s_{l(w)-1})}$, so the inductive argument completes the proof.