February 3

Let W ≤ O (V) be the group generated by the reflections {s_{α} : α ∈ Δ}. Then W acts on Δ.

Remark 1 W is finite since the map Φ : W → S_{Δ} has kernel 1 so that

W ≅ W/ ker Φ ≅ Im Φ ≤ S_{Δ}.

Remark 2 Given W a group generated by reflections in some vector space V, we can construct a root system Δ (possibly a root system for a subspace of V) whose associated g.g.r. is W. Take Δ to be the collection of unit vectors orthogonal to the hyperplanes fixed by the reflections generating W. We want to confirm that Δ is a root system in ⟨Δ⟩ ≤ V.

We need to show that if α, β ∈ Δ, then s_{α}β ∈ Δ. We must have that s_{α}, s_{β} ∈ W so that s_{s_{α}β} = s_{α}s_{β}s_{α} ∈ W. But this means that s_{α}β ∈ Δ. This shows that s_{α}Δ ⊂ Δ for α ∈ Δ.

Theorem 1 W is actually generated by {s_{α} : α ∈ Π}.

Lemma 1 (Main Lemma) Let α ∈ Π and β ∈ Δ_{+} with α ≠ β. Then s_{α} (β) ∈ Δ_{+}.

Proof. Suppose α = α_{1}. Write

β = ∑_{j} c_{j}α_{j}

and assume that c_{j} > 0 for some c_{j}, say c_{2} > 0, since α ≠ β. Then

s_{α_{1}} (β) = β − \frac{2 ⟨α_{1}, β⟩}{⟨α_{1}, α_{1}⟩}α_{1} = ∑_{j≠1} c_{j}α_{j} + (c_{1} − \frac{2 ⟨α_{1}, β⟩}{⟨α_{1}, α_{1}⟩}) α_{1}.

Either s_{α_{1}}β is in Δ_{+} or in Δ_{−}, so the coefficients of s_{α_{1}}β are either all ≥ 0 or all ≤ 0. But the coefficient of α_{2} is c_{2} > 0, so s_{α_{1}}β ∈ Δ_{+}.

Definition 1 Define the height of α ∈ Δ_{+} by

ht (α) = ∑_{j} a_{j} if α = ∑_{j} a_{j}α_{j} for α_{j} ∈ Π.

Proposition 1 Any positive root α ∈ Δ_{+} of minimal height is in Π.

Proof. Let α ∈ Δ_{+} be of minimal height. If α ∉ Π, take α_{i} ∈ Π with ⟨α, α_{i}⟩ > 0. If this were not possible, we would have ⟨α, α_{j}⟩ ≤ 0 for all α_{j} ∈ Π. But we showed last time that this means that Π ∪ α is a linearly independent set by, a contradiction. Then

s_{α_{i}}α = α − \frac{2 ⟨α, α_{i}⟩}{⟨α_{i}, α_{i}⟩}α_{i} ∈ Δ_{+}

by the Main Lemma. Then ht (s_{α_{i}}α) < ht (α), a contradiction.

Let W_{0} be the group generated by {s_{α} : α ∈ Π}. We will show that W = W_{0}.

Proposition 2 Every root α is in the W_{0}-orbit of some fundamental root.

Proof. Assume first that α ∈ Δ_{+}. We will prove this by induction on ht (α).

We have by Proposition 1 that the positive roots of minimal height are in Π and so such roots are certainly in the W_{0}-orbit of Π.

Let α be any positive root and assume now by induction that any root of smaller height is in the W_{0}-orbit of Π. By the proof of Proposition 1, we have an α_{i} ∈ Δ_{+} with s_{α_{i}}α ∈ Π and ht (s_{α_{i}}α) < ht (α). We then have that

s_{α_{i}}α = w · α_{j} for some α_{j} ∈ Π and w ∈ W_{0}.

Then

α = s_{α_{i}}s_{α_{i}}α = s_{α_{i}}w · α_{j}

so that α is in the W_{0}-orbit of Π.

Now if α is a negative root, then −α is a positive root, so by the above, we have that −α = w · α_{i}. Then α = w · s_{α_{i}}α_{i}.

Proof of Theorem 1. It suffices to show that s_{α} ∈ W_{0} for any root α. By Proposition 2, we have α = w · α_{i} for w ∈ W_{0} and α_{i} ∈ Π. Then

s_{α} = s_{wα_{i}} = ws_{α_{i}}w^{−1} ∈ W_{0}.

For w ∈ W, write

Δ_{w}^{+} = {α ∈ Δ_{+} : wα ∈ Δ_{+}}
Δ_{w}^{−} = {α ∈ Δ_{+} : wα ∈ Δ_{−}}

We abbreviate s_{i} = s_{α_{i}} and write S = {s_{1}, s_{2} … s_{m}} for the set of fundamental reflections.

Theorem 2 For w ∈ W and s_{i} ∈ S, we have

|Δ_{ws_{i}}^{−}| = |Δ_{w}^{−}| + 1

if wα_{i} ∈ Δ_{+} and

|Δ_{ws_{i}}^{−}| = |Δ_{w}^{−}| − 1

if wα_{i} ∈ Δ_{−}.

Proof. Assume first wα_{i} ∈ Δ_{+}. We will show that

Δ_{ws_{i}}^{−} = s_{i} (Δ_{w}^{−}) ∪^{·} α_{i}.  (1)

Since wα_{i} ∈ Δ_{+}, we have

ws_{i}α_{i} = w (−α_{i}) = −wα_{i} ∈ Δ_{−}

by assumption so that α_{i} ∈ Δ_{ws_{i}}^{−}. Next, observe that s_{i} (Δ_{w}^{−}) ⊂ Δ_{ws_{i}}^{−} since if s_{i} (α ∈ s_{i}Δ_{w}^{−}), we have

Δ_{−} ∋ wα = ws_{i}s_{i}α

so that s_{i}α ∈ (Δ_{ws_{i}}). Also, note that the sets in the RHS of (1) are disjoint since if α_{i} ∈ s_{i} (Δ_{w}^{−}), then s_{i}α_{i} = −α_{i} ∈ Δ_{w}^{−}, a contradiction since Δ_{w}^{−} contains only positive roots.

To show the reverse containment, let β ∈ Δ_{ws_{i}}^{−} and assume β ≠ α_{1}. Then ws_{i}β ∈ Δ_{−} and s_{i}β ∈ Δ_{+} because β ≠ α_{1} so that s_{i}β ∈ Δ_{w}^{−}. Then

β = s_{i} (s_{i}β) ∈ s_{i}Δ_{w}^{−}.

Now if wα_{i} ∈ Δ_{−}, then (ws_{i}) α_{i} ∈ Δ_{+} so that

Δ_{w}^{−} = Δ_{ws_{i}s_{i}}^{−} = s_{i}Δ_{ws_{i}}^{−} ∪^{·} α_{i}

so that

|Δ_{ws_{i}}^{−}| = |Δ_{w}^{−}| − 1.

Write N (w) = |Δ_{w}^{−}| for w ∈ W. Using this notation, Theorem (2) says that N (ws_{i}) = N (w) ± 1 depending on whether wα_{i} is in Δ_{+} or Δ_{−}.

Theorem 3 (Cancellation Law) If

w = s_{k_{1}}s_{k_{2}} … s_{k_{m}}  (2)

with s_{j_{k}} ∈ S and N (w) < m, then there exist 1 ≤ i ≤ j ≤ m − 1 with

w = s_{k_{1}} … ŝ_{k_{i}} … ŝ_{k_{j+1}} … s_{k_{m}}

Proof. We have by Theorem 2 all the following.

N (w) = N (s_{k_{1}}s_{k_{2}} … s_{k_{m}})
= N (s_{k_{1}}s_{k_{2}} … s_{k_{m−1}}) ± 1
= N (s_{k_{1}}s_{k_{2}} … s_{k_{m−2}}) ± 1 ± 1
= ⋮
= ±1 ± 1 ⋯ ± 1 ≤ m  (3)

Now since N (w) < m, one of the ±’s above must be a −, say

N (s_{k_{1}}s_{k_{2}} … s_{k_{j}}) ± 1 ± 1 ⋯ − 1.

But this means by Theorem 2 that

s_{k_{1}}s_{k_{2}} … s_{k_{j}}α_{k_{j+1}} ∈ Δ_{−}

Let i be such that

α_{k_{j+1}} ∈ Δ_{+}
s_{k_{j}}α_{k_{j+1}} ∈ Δ_{+}
s_{k_{j−1}}s_{k_{j}}α_{k_{j+1}} ∈ Δ_{+}
s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}α_{k_{j+1}} ∈ Δ_{+}

and

s_{k_{i}}s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}α_{k_{j+1}} ∈ Δ_{−}.

Note that we could have i = j or i = 1. Since s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}α_{k_{j+1}} ∈ Δ_{+}, we have by the Main Lemma that

α_{k_{i}} = s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}α_{k_{j+1}}.

Recall from last time that s_{gα} = gs_{α}g^{−1} for g ∈ O (V). Taking g = s_{k_{i+1}} … s_{k_{j}}, we have that

s_{k_{i}} = (s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}) s_{k_{j+1}} (s_{k_{j}}s_{k_{j−1}} … s_{k_{i+1}}) .

Unwinding, we have

s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}s_{k_{j+1}} = s_{k_{i}}s_{k_{i+1}} … s_{k_{j−1}}s_{k_{j}}.

Substituting this into (2), we have

w = s_{k_{1}} … s_{k_{i}}s_{k_{i}} … s_{k_{j+1}}s_{k_{j+1}} … s_{k_{m}} = s_{k_{1}} … ŝ_{k_{i}} … { \atop s_{k_{j+1}}} … s_{k_{m}}.

Definition 2 For w ∈ W, define the length of w, denoted ℓ (w), to be the smallest number of factors needed to express w as a product of fundamental reflections. An expression

w = s_{j_{1}}s_{j_{2}} … s_{j_{k}}

is called reduced if k = ℓ (w).

Proposition 3 ℓ (w) = N (w).

Proof. Let w = s_{k_{1}}s_{k_{2}} … s_{k_{m}} be a reduced expression for w. If we had N (w) < m, then by Theorem (3), we would be able to reduce w further, a contradiction. So N (w) ≥ m. But we have N (w) ≤ m from (3) so that ℓ (w) = m.

Proposition 4 (Exchange Condition) If

ℓ (s_{k_{1}}s_{k_{2}} … s_{k_{m}}) = m
and ℓ (s_{k_{0}}s_{k_{1}}s_{k_{2}} … s_{k_{m}}) < m + 1

then for some 0 ≤ j ≤ m − 1 we have

s_{k_{0}}s_{k_{1}} … s_{k_{j}} = s_{k_{1}}s_{k_{2}} … s_{k_{j+1}}.

Proposition 5 If w (Δ_{+}) = Δ_{+}, then w = 1.

Proposition 6 There exists an element w_{0} ∈ W of maximal length for which

ℓ (w_{0}) = |Δ_{+}| ,  w_{0} (Δ_{+}) = Δ_{+},  and  w_{0}^{2} = 1.

Proposition 7 An element w ∈ W has a reduced expression ending in a given s_{i} ∈ S if and only if wα_{i} ∈ Δ_{−}.