February 16

Definition 1 A finite Coxeter group W is a finite group with presentation

W = ⟨s_{1}, s_{2}, …, s_{n} : (s_{i}s_{j})^{m_{i,j}} = 1⟩

with m_{i,i} = 1 and 1 < m_{i,j} = m_{j,i} < ∞ for i ≠ j.

Write S = {s_{1}, s_{2}, … s_{n}}.

Observation 1 Define the map ϵ : S → {1, −1} by s_{j} ↦ −1. Then ϵ extends to a homomorphism ϵ : F (S) → {1, −1} by the Universal Property of Free Groups. Dividing by the subgroup generated by the elements (s_{i}s_{j})^{m_{i,j}} for i ≠ j, we have a homomorphism ϵ : W → {1, −1}. Thus,

|s_{j}| ≥ |ϵ (s_{j})| = 2,

but since m_{j,j} = 1, we have s_{j}^{2} = 1 so that |s_{j}| ≤ 2. This shows that |s_{j}| = 2 so that s_{j} is an involution.

Let W be a coxeter group. The goal is to show that W is isomorphic to a GGR. This is accomplished through the following series of steps.

Let V be a vector space over ℝ of dimension n = |S| with basis {v_{1}, v_{2}, … v_{n}}. Define the bilinear form B : V × V → ℝ by

B (e_{i}, e_{j}) = − cos (\frac{π}{m_{i,j}} )

for 1 ≤ i, j ≤ n and define the linear transformation T (s_{j}) ∈ GL (V) by

T (s_{j}) (v) = v − 2B (v, e_{j}) e_{j}

for 1 ≤ j ≤ n.

Step 1 T (s_{i}) is an involution.

Since m_{i,i} = 1, we have B (e_{i}, e_{i}) = − cos (π) = 1 so that

V = ⟨e_{i}⟩ ⊕ ⟨e_{i}⟩^{⊥} .

Here ⊥ denotes the orthogonal compliment with respect to B. Also, T (s_{i}) e_{i} = −e_{i} and T (s_{i}) β = β for β ∈ ⟨e_{i}⟩^{⊥}. This shows that T (s_{i}) leaves ⟨e_{i}⟩ invariant and is trivial on ⟨e_{i}⟩^{⊥}. Then if v = αe_{i} + w for α ∈ ℝ and w ∈ ⟨e_{i}⟩^{⊥}, we have

T (s_{i})^{2} (αe_{i} + w) = αe_{i} + w

so that T (s_{i}) is an involution.

Step 2 V = ⟨e_{i}, e_{j}⟩ ⊕ ⟨e_{i}, e_{j}⟩^{⊥} for i ≠ j We begin by observing that B restricted to ⟨e_{i}, e_{j}⟩ is non-degenerate. This is because if

B (ae_{i} + be_{j}, ce_{i} + de_{j}) = ac + (ad + bc) B (e_{i}, e_{j}) + bd = 0

for all c, d, then we would have a = b = 0. You show this by substituting c = −a, d = b if B (e_{i}, e_{j}) ≠ 0 and c = a, d = b if B (e_{i}, e_{j}) = 0. Now if β ∈ ⟨e_{i}, e_{j}⟩ ∩ ⟨e_{i}, e_{j}⟩^{⊥}, then B (α, β) = 0 for all α ∈ ⟨e_{i}, e_{j}⟩ so that β = 0. Therefore, we have the decomposition

V = ⟨e_{i}, e_{j}⟩ ⊕ ⟨e_{i}, e_{j}⟩^{⊥} .

Moreover, T (s_{i}) T (s_{j}) fixes ⟨e_{i}, e_{j}⟩ and acts trivially on ⟨e_{i}, e_{j}⟩^{⊥}. This follows directly from the definition of B.

Step 3 (T (s_{i}) T (s_{j}))^{m_{i,j}} = 1.

We showed in Step 2 that B is non-degenerate on ⟨e_{i}, e_{j}⟩. It is clearly symmetric. Then by the Sylvester Law of Inertia, we can regard (⟨e_{i}, e_{j}⟩ , B) as ℝ^{2} with the standard inner product. Then T (s_{i}) and T (s_{j}) act as reflections over the lines ⟨e_{i}⟩ and ⟨e_{j}⟩. Since B (e_{i}, e_{j}) = − cos (\frac{π}{m_{i,j}} ) = cos (π − \frac{π}{m_{i,j}} ), we have that the angle between e_{i} and e_{j} is π − \frac{π}{m_{i,j}}. This means that the (acute) angle between the reflecting lines ⟨e_{i}⟩ and ⟨e_{j}⟩ is \frac{π}{m_{i,j}}. We conclude, knowing all about how dihedral groups work, that T (s_{i}) T (s_{j}) is a rotation through \frac{2π}{m_{i,j}} so that |T (s_{i}) T (s_{j})| = m_{i,j}.

Step 4 The map T : S → GL (V) can be extended to a representation T : W → GL (V).

Since the elements T (s_{i}) ∈ T (W) ⊂ GL (V) satisfy the relations defining W by the above steps, we have a homomorphism, also called T : W → GL (V) which agrees with the previous definition of T on S.

Step 5 B is invariant with respect to T (w) for w ∈ W.

B(T (s_{j}) u, T (s_{j}) v) = B(u − 2B (u, e_{j}) e_{j}, v − 2B (v, e_{j}) e_{j}) = B (u, v) .

Relabel the vertices of the Coxeter graph {s_{i,j} : 1 ≤ i ≤ k, 1 ≤ j ≤ n_{i}} so that all the s_{i,j} are in the ith connected component of the Coxeter graph with n_{1}, n_{2}, … n_{k} ≥ 1 the sizes of the k connected components. Let V_{i} = ⟨e_{i,1}, e_{i,2} … e_{i,n_{i}}⟩. Then since s_{i_{1},j_{2}} is not connected to s_{i_{2},j_{2}}, for i_{1} ≠ i_{2}, we must have that B (e_{i_{1},j_{1}}, e_{i_{2},j_{2}}) = 0 so that V_{i_{1}} ⊂ V_{i_{2}}^{⊥} Therefore, we have

V = V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{k}

and moreover, T (s_{j}) V_{i} ⊂ V_{i} for all 1 ≤ j ≤ n and all 1 ≤ i ≤ k. so that the V_{i} are ℝT (W)-submodules

Step 6 Each V_{i} is an absolutely simple ℝT (W)-module.

This will follow when we show that End_{ℝW} (V_{i}) = ℝ by the following results.

Exercise 1 (Converse of Schur’s Lemma, CR1 Exercise 27.1) Let M be a completely reducible A-module where A is an algebra over K. Show that End_{A} (M) = K implies that M is irreducible.

Theorem 1 (CR2 Theorem 3.43) A simple left A-module M is absolutely simple if and only if End_{A} (M) = K.

Let X ∈ End_{ℝT (W)}, that is, suppose that XT (s_{j}) = T (s_{j}) X. We observed in Step 1 that V = ⟨e_{j}⟩ ⊕ ⟨e_{j})^{⊥}. Suppose Xe_{j} = αe_{j} + β for α ∈ ℝ and β ∈ ⟨e_{j}⟩^{⊥}. Then

XT (s_{j}) e_{j} = −Xe_{j} = −αe_{j} − β

but on the other hand

T (s_{j}) Xe_{j} = T (s_{j}) (αe_{j} + β) = −αe_{j} + β

so that β = 0. Now let β ∈ ⟨e_{j}⟩^{⊥} and suppose Xβ = αe_{j} + γ for α ∈ ℝ and γ ∈ ⟨e_{j})^{⊥}. Then

XT (s_{j}) β = Xβ = αe_{j} + γ

but on the other hand

T (s_{j}) Xβ = T (s_{j}) (αe_{j} + γ) = −αe_{j} + γ

so that α = 0. This shows that X acts as scalar multiplication on ⟨e_{j}⟩ and fixes ⟨e_{j}⟩^{⊥} pointwise. Write Xe_{j} = α_{j}e_{j} for each j.

Let s_{j} be connected to s_{k}, that is, suppose B (e_{j}, e_{k}) ≠ 0. Write T (s_{j}) e_{k} = e_{k} − γe_{j} and note that γ ≠ 0. Then

XT (s_{j}) e_{k} = X (e_{k} − γe_{j}) = α_{k}e_{k} − γα_{j}e_{j}

but on the other hand

T (s_{j}) Xe_{k} = α_{k}T (s_{j}) e_{k} = α_{k}e_{k} − γα_{k}e_{j}

which shows that α_{j} = α_{k}. Thus X acts as scalar multiplication on each subspace V_{i}

Step 7 B is positive definite and hence an inner product on V.

Let B_{i} be the restriction of B to V_{i} and starting with any positive definite form φ on V_{i}, define a new form B′_{i} : V_{i} × V_{i} → ℝ on V_{i} by

B′_{i} (e_{j}, e_{k}) = ∑_{w∈W} φ(T (w) e_{j}, T (w) e_{k}).

Then B′_{i} is both positive definite and T (W)-invariant. Identify V_{i} with its dual V_{i}^{∗} by sending

v ↦ B′_{i} (v, ·) .

Then the B′_{i} (e_{i,j}, ·) are a basis of V^{∗}. Write

B_{i} (e_{i,j}, ·) = ∑_{k} α_{j,k}B′_{i} (e_{i,k}, ·) .

Then if u = ∑_{j} β_{j}e_{i,j}, we have

B_{i} (u, ·) = ∑_{j} β_{j}B_{i} (e_{i,j}, ·) = ∑_{j,k} β_{j}α_{j,k}B′_{i} (e_{i,k}, ·) = β′_{i} (Xu, ·)

where X ∈ End (V_{i}) is given by X = (α_{j,k}). This shows that B_{i} (u, v) = B′_{i} (Xu, v). Now since B_{i} is T (W)-invariant, we have

B_{i} (u, v) = B_{i} (T (w) u, T (w) v) = B′_{i} (XT (w) u, T (w) v) .

But on the other hand, we know that B′_{i} is also T (W)- invariant so that

B_{i} (u, v) = B′_{i} (Xu, v) = B′_{i} (T (w) Xu, T (w) v)

This shows that X commutes with T (w). That is, X ∈ End_{ℝT (W)} (V) so that by Step 6, we have that X = λ1 for some λ ∈ ℝ. Moreover, since

1 = B_{i} (u, u) = λ \underbrace{B′_{i} (u, u)}_{≥0}

so that λ > 0. Now since B_{i} = λB′_{i}, we have that B_{i} is positive definite. Step 8 The image T (W) is a GGR acting on the inner product space (V, B) with root system Δ = {T (s_{i}) e_{j} : 1 ≤ i, j ≤ n}.

Step 9 ℓ (ws_{i}) = ℓ (w) ± 1

Observe that

ℓ (ws_{i}) ≤ ℓ (w) + ℓ (s_{i}) = ℓ (w) + 1

and similarly

ℓ (w) = ℓ (ws_{i}s_{i}) ≤ ℓ (ws_{i}) + ℓ (s_{i}) = ℓ (ws_{i}) + 1

so that

ℓ (w) − 1 ≤ ℓ (ws_{i}) ≤ ℓ (w) + 1.  (1)

By Observation 1, we have a homomorphism ϵ : W → {1, −1} with ϵ (s_{i}) = −1. Then ϵ (w) = (−1)^{ℓ(w)}. Then ℓ (ws_{i}) = ℓ (w) ℓ (s_{i}) = −ℓ (w) so that ℓ (ws_{i}) ≠ ℓ (w). Then by (1) we have the result.

Recall the following definition. Definition 2 A subset Π ⊂ Δ satisfying

1. Π is a basis for V

2. Every root is a linear combination of Π having either all non-negative or all nonpositive coeffecients

3. B (e_{i}, e_{j}) ≤ 0 for all e_{i}, e_{j} ∈ Π is called a fundamental system.

Step 10 Π = {e_{1}, e_{2}, … e_{n}} is a fundamental system in Δ.

We need to verify (2) from Definition 2 and for this, it will be more than sufficient to show that

if ℓ (ws_{i}) > ℓ (w) then T (w) e_{i} has non-negative coeffecients.

Note that this will mean on the other hand that if ℓ (ws_{i}) < ℓ (w), then T (w) e_{i} will have non-positive coeffecients and (2) will follow.

Step 11 T : W → GL (V) is faithful so that W is isomorphic to a GGR.

Recall that ℓ (w) = N (w) = |Δ_{w}^{−}| where

Δ_{w}^{−} = {α ∈ Δ_{+} : T (w) α ∈ Δ_{−}} .

Then if w ∈ ker T, ℓ (w) = 0 so that w = 1.