Recall that the reflections for preserve . They generate a group called the **Weyl group**. Moreover, since spans , the map , the symmetric group on , is injective. So is a finite group of orthogonal isomorphisms of , i.e. leaving invariant the bilinear form .

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 for and a corresponding partition ; the for 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 1acts simply transitively on Weyl chambers and on bases. Any root can be moved by an element of the Weyl group into a given base. 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 2The simple reflection permutes and sends into its inverse.

Indeed, suppose . If , then is not a multiple of and contains some other element of the base with positive coefficient, say . Then applying to only changes the coefficient of , so still has a positive coefficient of and is thus a positive root. It is clear that ; this is just a fact about reflections in general.

**Transitivity on Weyl chambers **

We are now going to prove that the subgroup generated by the simple reflections (associated to a base ) acts transitively on the Weyl chambers. In particular, given a regular element (one that is not orthogonal to any element of ), we will show that there is such that

The idea is to start by considering the element , which will show up frequently in the sequel. A quantitative consequence of Lemma 1 is

So, given , choose maximizing . Then I claim for all simple roots . Indeed,

Now the claim follows. If we prove that we have strict inequality () then we will have proved transitivity of on the Weyl chambers. But if for some , then would be orthogonal to , a contradiction by regularity.

Let be two bases associated to . We can use an element of to move into the same Weyl chamber as , which means that gets transformed into —as a result, gets transformed into . So acts transitively on the set of bases.

** is generated by the simple reflections **

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

Lemma 3Any root is contained in a base .

We will choose a regular vector such that brackets positively with but to a very small number, so is indecomposable in the corresponding choice of positive roots .

Consider the hyperplane . We can choose that is not perpendicular to any other root in (except ). Modifying slightly, we can get a vector such that is very small, smaller than for any root . Then is a simple root in .

It is now possible for us to show that ; we must show that for belongs to . Now we can find with by the above lemma. Then

** acts simply transitively on the bases **

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

Lemma 4Given simple roots and the corresponding reflections . If , then we can write as a product of smaller length using only .

There has to be a first index where . By the above lemma, since these are simple reflections, it must be the case that

Now if is an orthogonal transformation, then as is easily checked. Therefore,

and it follows that

since .

Suppose satisfies ; I claim .

Write as a minimal product of simple reflections associated to . By assumption , so

This means we can shrink the expression for by the second lemma, contradiction.

As a result, it follows that 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 as a product of simple reflections. A minimal expression is called **reduced** and is called the **length** .

If is a simple reflection, then sends precisely one positive root into a negative root, and .

More generally:

Proposition 5The length equals the number of positive roots sent into negative roots.

This is proved by induction on the length; we have just stated it for above. We can write where the expression is itself reduced. Suppose these reflections are associated to . Then (if it were in , we could use the previous lemma to shrink the expression). In particular, —in this respect, behaves differently from the subproduct . Now permutes the positive roots in . So in particular,

But we have , so the inductive argument completes the proof.

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