Lecture Notes

Reflection Groups

Dr Dmitriy Rumynin

Anna Lena Winstel

Autumn Term 2010

Contents

1 Finite Reflection Groups 3

2 Root systems 6

3 Generators and Relations 14

4 Coxeter group 16

5 Geometric representation of W (mij) 21

6 Fundamental chamber 28

7 Classification 34

8 Crystallographic Coxeter groups 43

9 Polynomial invariants 46

10 Fundamental degrees 54

11 Coxeter elements 57

1 Finite Reflection Groups

V = (V, 〈 , 〉) - Euclidean Vector space where V is a finite dimensional vector space over R and〈 , 〉 : V × V −→ R is bilinear, symmetric and positiv definit.

Example: (Rn, ·) : 〈(αi), (βi)〉 =n∑i=1

αiβi

Gram-Schmidt theory tells that for all Euclidean vector spaces, there exists an isometry (linearbijective and ∀ x, y ∈ V : T (x) · T (y) = 〈x, y〉) T : V −→ Rn.In (V, 〈 , 〉) you can

• measure length: ||x|| =√〈x, x〉

• measure angles: xy = arccos(〈x,y〉||x||·||y||

)• talk about orthogonal transformations

O(V ) = T ∈ GL(V ) : ∀ x, y ∈ V : 〈Tx, Ty〉 = 〈x, y〉 ≤ GL(V )

T ∈ GL(V ). Let V T = v ∈ V : Tv = v the fixed points of T or the 1-eigenspace.Definition. T ∈ GL(V ) is a reflection if T ∈ O(V ) and dim V T = dim V − 1.

Lemma 1.1. Let T be a reflection, x ∈ (V T )⊥ = v : ∀ w ∈ V T : 〈v, w〉 = 0, x 6= 0. Then

1. T (x) = −x

2. ∀ z ∈ V : T (z) = z − 2 〈x,z〉〈x,x〉 · x

Proof.

1. Pick v ∈ V T =⇒ Tv = v =⇒ 〈Tx, v〉 = 〈Tx, Tv〉 = 〈x, v〉 = 0. Hence Tx ∈ (V T )⊥.Since dim (V T )⊥ = dim V − dim V T = 1 : Tx = α · x for some α ∈ R. Then

α2 〈x, x〉 = 〈αx, αx〉 = 〈Tx, Tx〉 = 〈x, x〉

Since 〈x, x〉 6= 0, α2 = 1 =⇒ α ∈ −1, 1.If α = 1 =⇒ Tx = x =⇒ x ∈ V T ∩ (V T )⊥ =⇒ x = 0 which is a contradiction.So α = −1.

2.⟨z − 〈x,z〉〈x,x〉 · x, x

⟩= 〈z, x〉 − 〈x,z〉〈x,x〉 · 〈x, x〉 = 0.

So z − 〈x,z〉〈x,x〉 · x ∈ x⊥ = V T and T (z − 〈x,z〉〈x,x〉x) = z − 〈x,z〉〈x,x〉x. Hence

T (z) = T ((z − 〈x, z〉〈x, x〉

x) +〈x, z〉〈x, x〉

x) = T (z − 〈x, z〉〈x, x〉

x) +〈x, z〉〈x, x〉

T (x)

= z − 〈x, z〉〈x, x〉

x− 〈x, z〉〈x, x〉

x = z − 2〈x, z〉〈x, x〉

x

3

2

For each x ∈ V, x 6= 0 define Sx(z) = z − 2 〈x,z〉〈x,x〉x.

Lemma 1.1 implies that

1. any reflection T is equal to Sx for x determined up to a scalar. Any such x ∈ V is calleda root of T .

2. ∀ x ∈ V, x 6= 0 : Sx is a reflection.

3. any reflection T satisfies T 2 = I.

Definition. A finite reflection group is a pair (G,V ) where V is Euclidean space, G is a finitesubgroup of O(V ) and G = 〈Sx : Sx ∈ G〉, generated by all reflections in G.

Generation means: if G ⊇ X, then X generates G if G = 〈X〉 where X is defined as one of thefollowing equivalent definitions:

1. (semantic) 〈X〉 =⋂

G≥H⊃XH

2. (syntactic) 〈X〉 = 1 ∪ a±11 a±1

2 · · · a±1n : ai ∈ X

Equivalence:(G1, V1) ∼ (G2, V2) if there is an isometry ϕ : V1 −→ V2 s.t. ϕG1ϕ

−1︸ ︷︷ ︸ϕTϕ−1: T∈G1

= G2.

Problem: Classify all finite reflection groups (G,V ) up to ∼.Example: of finite reflection groups

1.

y

x

α

We want to know 〈Sx, Sy〉det(Sx, Sy) = 1 =⇒ SxSy ∈ SO2(R) =⇒ SxSy = Rotβ .Since Sx(Sy(y)) = Sx(−y) =⇒ SxSy = 2α.

• απ /∈ Q =⇒ |SxSy| =∞ =⇒ 〈Sx, Sy〉 is not finite

• απ = m

n ∈ Q (for m,n rel. prime) =⇒ |SxSy| = |Rot 2πmn | = n =⇒ 〈Sx, Sy〉 = D2n

Dihedral

I2(n) = (D2n,R2), |I2(n) = 2n|

2. Sn-symmetric group acting on 1, . . . , n extend to action of Sn on Rn.If ei ∈ basis of Rn, if σ ∈ Sn, then

Tσ : ei 7−→ eσ(i), Tσ ∈ On(R)

4

x =

11...1

has the property that forall σ : Tσ(x) = x =⇒ x ∈ (Rn)Sn ,

x⊥ =

α1...αn

:∑αi = 0

An−1 = (Sn, x⊥)

Reflection since Sn = 〈(i, j)〉 and T(i,j) = Sei−ej because Tij(ei − ej) = −(ei − ej) and

y =

α1...αn

∈ (ei − ej)⊥ ⇐⇒ αi = αj ⇐⇒ T(i,j)(y) = y

In particular, |Am| = (m+ 1)! and A2 ∼ I2(3)

3. F = Z/2Z = 0, 1 - field of two elements. Consider action of Sn on Fn, ε1, . . . , εn basisof Fn.

∀ σ ∈ Sn : tσ(εi) = εσ(i)

Consider semidirect product Sn n Fn. As a set: Sn n Fn = Sn × Fn, the product is

(σ, a) · (τ, b) = (στ, tτ−1(a) + b)

Sn n Fn acts on Rn:T(σ,α) : ei 7−→ (−1)ai · eσi

Let us check that this is the action of Sn n Fn:

T(1,α)(T(τ,0)(ei)) = T(1,α)(eτ(i))

= (−1)aτ(i)eτ(i)

= (−1)[tτ−1 (a)]ieτ(i)

= T(τ,tτ−1 (a))(ei)

Bn = (Sn n Fn,Rn)

It is reflection since Sn n Fn = 〈((i, j), 0), (1, εi)〉 and T((i,j),0) = Sei−ej , T(1,εi) = Sei .|Bn| = n!2n, B1 ∼ A1, B2 ∼ I2(4)

4. In Example 3, let Fn0 =

α1...αn

∈ Fn :∑αi = 0 codim-1 subspace in Fn or index 2

subgroup.σ ∈ Sn, a ∈ Fn0 =⇒ tσ(a) ∈ Fn0 , so Sn acts on Fn0 and

Dn = (Sn n Fn0 ,Rn)

It is a reflection since Sn n Fn0 = 〈((i, j), 0), (1, εi + εj)〉,T((i,j),0) = Sei−ej , T(1,εi+εj) = Sei+ej .

|Dn| = n! · 2n−1, D1 is trivial (1,R), D2 ∼ I2(2), D3 ∼ A3

5

2 Root systems

Let (G,V ) be a finite reflection group. The root system of (G,V ) is

Φ(G,V ) := x ∈ V : ‖x‖ = 1, Sx ∈ G

Φ has the following properties:

1. x ∈ Φ =⇒ Rx ∩ Φ = x,−x

2. |Φ| = 2·( number of reflections in G)

3. T ∈ G, x ∈ Φ =⇒ T (x) ∈ Φ

Property 3 follows from

Lemma 2.1. T ∈ O(V ), x ∈ V \ 0 =⇒ ST (x) = TSxT−1.

Proof. RHS: T (x) 7−→ TSxT−1Tx = TSxx = T (−x) = −T (x)

RHS : T (x)⊥ 3 y 7−→ TSx(T−1y) = T (T−1y − 2

⟨x, T−1y

⟩〈x, x〉

· x)

= T (T−1y − 2〈Tx, y〉〈x, x〉

x)

= T (T−1y)

= y

Hence RHS = ST (x). 2

Example: I2(3) ∼ A2 -dihedral group of order 6, symmetry of regular triangle.

6 roots sitting at the vertices of a regular hexagon.

Definition. A root system is a finite subset Φ ⊂ V s.t.

1. 0V /∈ Φ

2. x ∈ Φ =⇒ Rx ∩ Φ = x,−x

3. x, y ∈ Φ =⇒ Sx(y) ∈ Φ

6

Example:

1. Φ(G,V ) where (G,V ) is a finite reflection group

2. ∅ = Φ(1,V )

3. is a root system, not Φ(G,V ) because there are vectors of 2 different length. Chop-ping long vectors by

√2 gives ΦB2 .

Let Φ ⊂ V be a root system. Definition. A simple subsystem is Π ⊂ Φ s.t.

1. Π is linearly independent

2. ∀x ∈ Φ : x =∑y∈Π

αy · y where either all αy ≥ 0 or all αy ≤ 0

Example: Inα

β

α, β is a simple system but

γβ

is not simple since α−β ∈Φ.

Lemma 2.2. Π ⊂ Φ simple system in a root system. Then ∀ x, y ∈ Π, x 6= y =⇒ 〈x, y〉 ≤ 0(angles in a simple system are obtuse)

Proof. Suppose x 6= y, 〈x, y〉 > 0.

Φ 3 Sx(y) = y−2 〈x,y〉〈x,x〉 ·x = y+αx, α < 0. Since Π is lin. independent and x 6= y, Sx(y) = y+αx

is the only way to write Sx(y) as a linear combination of elements of Π and both positive and

negative coefficients are present. This is a contradiction. 2

Definition. A total order on R-vector space V is a linear order on V s.t. (≥-order, x > y ifx ≥ y and x 6= y)

1. x ≥ y =⇒ x+ z ≥ y + z

2. x ≥ y, α > 0 =⇒ αx ≥ αy

3. x ≥ y, α < 0 =⇒ αx ≤ αy

Example: Phonebook-order on Rn:α1

· · ·αn

>

β1...βn

⇐⇒ ∃k ∈ 1, . . . , n s.t. αi = βi ∀ i < k and αk > βk

Given an ordered basis e1, . . . , en of V we get phonebook order on V by writing∑αiei =

α1

· · ·αn

.

If ≥ is a total order on V then

x ∈ V \ 0 =⇒ either x > 0 > −x or x < 0 < −x

and V = V+∪0∪V− where

7

• V+ = x ∈ V : x > 0

• V− = x ∈ V : x < 0

Definition. A positive system in a root system Φ is a subset Θ ⊂ Φ s.t. ∃ total order on Vs.t. Θ = Φ ∩ V+.

Example:

1. Boxα

β

. Let ≥ be the total order associated to the ordered basis α, β. Then

2α+ β > α+ β > α > β > 0 > −β > −α > −α− β > −2α− β

So α, β, α+ β, 2α+ β is a positive system.

2. I2(n)-symmetry of n-gon, n reflections, so 2n roots, at vertices of regular 2n gon.

(α, β) is simple ⇐⇒ Φ ⊂ the cone ⇐⇒ −β, α =π

n⇐⇒ α, β =

(n− 1

nπ

• Every root lies in 2 simple systems

• number of simple systems = n

• positive systems = vectors between some α and β where α, β = (n−1n π

3. An : χT(a1,...,ak)(z) = (zk − 1)(z − 1)m (minimal polynomial of T(a1,...,ak) is (zk − 1).

Tσ reflection ⇐⇒ χTσ = (1 + z)(1− z)m ⇐⇒ σ = (i, j)

Reflections are T(i,j), ∃n(n−1)

n of them and Φ = ei − ej : i 6= j ∈ 1, . . . , n+ 1.A typical simple system is

e1 − e2, e2 − e3, . . . , en−1 − en

For i < j we have: ei − ej = (ei − ei+1) + (ei+1 − ei+2) + . . .+ (ej−1 − ej).

Notation: Φ - root system, Φ+ ⊂ Φ - positive system.Definition. Ω ⊂ Φ+ is a quasisimple system if it satisfies:

1. ∀ x ∈ Φ+ ∃ αt ≥ 0 s.t. x =∑t∈Ω

αtt

2. No proper subset of Ω satisfies (1).

Hint: simplicity ⇐⇒ quasisimplicity

Lemma 2.3. Let Ω ⊂ Φ+ be a quasisimple system. Then ∀ x, y ∈ Ω, x 6= y =⇒ 〈x, y〉 ≤ 0.

Proof. Suppose x 6= y ∈ Ω and 〈x, y〉 > 0. Then:Φ 3 Sx(y) = y − 2 〈x,y〉〈x,x〉x = y − λx, λ > 0. Consider two cases:

8

Case 1 Sx(y) ∈ Φ+. By (1): y − λx =∑t∈Ω

λtt, λt ≥ 0. What is λy?

If λy < 1 then (1−λy)y = λx+∑t6=y

λtt. So y is a positive linear combination of elements

of Ω \ y. Hence Ω \ y satisfies (1), contradiction to (2).Hence λy ≥ 1, then 0 = λx+ (λy−1)y+

∑t6=y

λtt, coefficients in the RHS ≥ 0. Hence RHS

≥ 0. Since RHS = 0 =⇒ λ = 0 which is a contradiction.

Case 2 Sx(y) ∈ Φ− =⇒ y − λx =∑t∈Ω

(−µt)t, µt ≥ 0.

=⇒ λx− y =∑t∈Ω

µtt. Similarly to case 1:

• µx < λ =⇒ (λ− µx)x = y +∑t 6=x

µtt contradicts (2)

• µx ≥ λ =⇒ 0 = y + (µx − λ)x+∑t6=x

µtt. RHS has a positive coefficient, hence > 0

which is a contradiction.

2

Theorem 2.4. 1. Every positive system contains a unique simple system.

2. Every simple system is contained in an unique positive system.Hence there is a natural bijection between

Π ⊂ Φ : Π is simple and Φ+ ⊂ Φ : Φ is positive

Proof.

1. Let Φ+ ⊂ Φ be a positive system. Consider all subsets of Φ+ that satisfy condition (1)of the definition of a quasisimple system. A minimal subset Ω among them must satisfy(2), so Ω is a quasisimple system in Φ+. To prove that Ω is a simple system, it sufficesto prove that Ω is linearly independent.Suppose

∑t∈Ω

αtt = 0. Sorting positive and negative coefficients to two different sides,

v =∑t

βtt =∑t

γtt, βt, γt ≥ 0

Now we consider

〈v, v〉 =

⟨∑t

βtt,∑t

γt

⟩=∑s,t

βtγs︸︷︷︸≥0

〈t, s〉︸︷︷︸≤0

≤ 0

Hence v = 0 =⇒ βi = 0 = γt =⇒ αt = 0. Hence Ω is linearly independent and simple.Let Π, Π′ ⊂ Φ be two dinstinct simple systems. Wlog ∃ x ∈ Π′ \Π.Since x ∈ Φ+, x =

∑y∈Π

αyy, αy ≥ 0. We know

Π 3 y =∑z∈Π′

βzyz, βzy ≥ 0

9

So

x =∑z∈Π′

∑y∈Π

αyβzy︸ ︷︷ ︸

≥0

z

=⇒ ∃ y0 ∈ Π \ Π′ : αy0 6= 0 =⇒ ∃ at least two z1, z2 s.t. βz1y0 6= 0, βz2y0 6= 0 which is acontradiction with the lineary independece of Π′.

2. Let Π ⊂ Φ be a simple system. Π is linearly independent =⇒ we can extend Π to a basisB of V . Choose an order on B and consider the phonebook total order on V . Clearly,Π ⊂ V+ =⇒ Π ⊂ Φ+ = V+ ∩ Φ. Uniqueness follows from the definition of a simplesystem: Every x ∈ Φ is a nonnegative or nonpositive linear combination of elementsof Π =⇒ nonnegative linear combinations ∈ V+, nonpositive linear combinations ∈ V−.Hence Φ+ =

∑t∈Π

αtt ∈ Φ : αt ≥ 0.

2

Proposition 2.5. Let Φ ⊃ Φ+ ⊃ Π be a root system, positive system, simple system. Then∀ x ∈ Π, ∀ y ∈ Φ+, x 6= y:

Sx(y) ∈ Φ+

Note: x 6= y is essential since Sx(x) = −x ∈ Φ−.

Proof. Let y =∑t∈Π

αtt, αt ≥ 0. y 6= x =⇒ ∃t0 ∈ Π s.t. t0 6= x, αt0 > 0.

Hence Sx(y) =∑t∈Π

αtt− λx with αt0 > 0. Hence Sx(y) ∈ Φ+. 2

Example: Box: simple: , simple:

∀ g ∈ G, let L(g) = x ∈ Φ+ : g(x) ∈ Φ− = Φ+ ∩ g−1(Φ−). We know that L(1) =

∅, L(Sx)x∈Π= x.

Define function l : G −→ Z≥0 called length by l(g) = |L(g)|. In particular l(1) = 0, l(Sx) = 1.

Proposition 2.6. The following statements about l( ) hold:

1. l(g) = l(g−1)

2. l(Sxg) =

l(g) + 1 if g−1(x) ∈ Φ+

l(g)− 1 if g−1(x) ∈ Φ−

3. l(gSx) =

l(g) + 1 if g(x) ∈ Φ+

l(g)− 1 if g(x) ∈ Φ−

Proof.

1. x 7−→ −g(x) is a bijection between L(g) and L(g−1)

x ∈ Φ+ ⇐⇒ −x = g−1(−g(x)) ∈ Φ−

g(x) ∈ Φ− ⇐⇒ −g(x) ∈ Φ+

10

2. By 2.5, ±x is the only root that changes the sigh under Sx.

g−1(x)g7−→ x

Sx7−→ −x

if g−1(x) ∈ Φ+ =⇒ g−1(x) /∈ L(g) but g−1(x) ∈ L(Sxg). So L(Sxg) = L(g) ∪ g−1(x).if g−1(x) ∈ Φ− =⇒ −g−1(x) ∈ L(g) but not in L(Sxg) and L(Sxg) = L(g) \ −g−1(x).

2

G acts on V,Φ, Π ⊂ Φ : Π is simple .

Theorem 2.7. Let Π,Π′ ⊂ Φ be two simple systems. Then ∃ g ∈ G s.t.

Π′ = g(Π) = gx : x ∈ Π

Note: such g is unique but we need some more tools to prove it.

Proof. Let Φ+,Φ′+ be the corresponding positive systems, Φ−,Φ

′− the corresponding negative

systems. Procced by induction on |Φ+ ∩ Φ′−| (distance between Π and Π′).If |Φ+ ∩ Φ′−| = 0 =⇒ Φ+ = Φ′+ =⇒ Π = Π′.|Φ+ ∩ Φ′−| = k suppose proved.|Φ+ ∩ Φ′−| = k + 1. Since k + 1 ≥ 1 =⇒ Π 6= Π′ =⇒ Π 6⊂ Φ′+ =⇒ ∃ x ∈ Π ∩ Φ′−. ConsiderΠ = Sx(Π) corresponding positive system is

Φ+ = (Φ+ \ x) ∪ −x

Hence Φ+ ∩Φ′− = (Φ+ ∩Φ′−) \ x. In particular |Φ+ ∩Φ′−| = k and by induction assumptionthere is a g s.t. Π′ = g(Π), hence

Π′ = g(Sx(Π)) = gSx(Π)

2

Define a height function h : Φ −→ R by

h

(∑t∈Π

αtt

)=∑t∈Π

αt

Theorem 2.8. G,Φ ⊃ Π as before.

1. ∀ x ∈ Φ ∃y ∈ Π ∃g ∈ G s.t. x = gy.

2. G = 〈Sx〉x∈Π

Proof. Let H = 〈Sx〉x∈Π ≤ G. Pick any t ∈ Φ let

Σt = Ht ∩ Φ+

11

If Σt 6= ∅ (obvious if t ∈ Φ+) then ∃z ∈ Σt with minimal height h(z) in Σt.Let z =

∑s∈Π

αss, αs ≥ 0. Then

0 < 〈z, z〉 =∑s∈Π

αs︸︷︷︸≥0

〈s, z〉

Hence ∃x ∈ Π s.t. 〈x, z〉 > 0. Sx(z) =∑s∈Π

αss − 2〈x, z〉〈x, x〉︸ ︷︷ ︸>0

x hence height goes down. So

h(Sx(z)) < h(z). Moreover Sx(z) ∈ Ht. By minimality of heigt of z:

Sx(z) /∈ Σt =⇒ Sx(z) ∈ Φ−

By 2.4: z = x = gt︸ ︷︷ ︸∈Ht

for some g ∈ H. Hence Sx = gStg−1 =⇒ St = g−1Sxg ∈ H

t = g−1x ∈ Hx ⊂ HΠ ⊂ GΠ

(1) follows because if x is positive this proof shows that x ∈ GΠ, if x is negative we use t = −xto show that Sx ∈ H, t = −x ∈ GΠ. Hence x = Sx(−x) ∈ GΠ.

Finally we have shown that all reflections Sx are in H, so H = G. 2

l : G −→ Z≥0. Now since G = 〈Sx〉x∈Π, is l related to group theoretic length in thesegenerators?

Theorem 2.9. YES

∀ g ∈ G : l(g) = mink : ∃ x1, . . . , xk ∈ Π : g = Sx1Sx2 · · ·Sxk

Proof. 2.6: l(Sxh) ≤ l(h) + 1 =⇒ if g = Sx1 · · ·Sxk then l(g) ≤ k. Hence l(g) ≤ RHS.Note: the opposite direction is based on deletion principle.Assume that g = Sx1 · · ·Sxk , xi ∈ Π but l(g) < k. Hence ∃ j s.t.

l(Sx1 · · ·Sxj = j but l(Sx1 · · ·Sxj+1 = j − 1

(we pick the position where the length goes down for the first time)=⇒ Sx1 · · ·Sxj (xj+1) ∈ Φ− and xj+1 ∈ Π ⊂ Φ+. Find largest i s.t. i ≤ j and

Sxi(y) = Sxi · · ·Sxj (xj+1) ∈ Φ−, y = Sxi+1 · · ·Sxj (xj+1) ∈ Φ+

=⇒ y = xi = Sxi+1 · · ·Sxj︸ ︷︷ ︸T

(xj+1) = T (xj+1). Hence Sxi = TSxj+1T−1. Then

g = Sx1 · · ·Sxi · · ·Sxj · · ·Sxk = Sx1 · · ·Sxi−1SxiTSxj+1 · · ·Sxk= Sx1 · · ·Sxi−1(TSxj+1T

−1TSxj+1)Sxj+2 · · ·Sxk = Sx1 · · ·Sxi−1TSxj+2 · · ·Sxk

2

12

Theorem 2.10. If Π,Π′ ⊂ Φ are two simple systems then ∃! g ∈ G s.t. Π′ = g(Π).

Proof. Existence is Theorem 2.7. Suppose g, h ∈ G satisfy Π′ = g(Π) = h(Π). Then

Π = h−1gΠ =⇒ Π+ = h−1g(Φ+) hence l(h−1g) = 0 =⇒ h−1g = 1. 2

Corollary 2.11. Consider action of G and X = Π ⊂ Φ : Π is simple . For all Π ∈ X, theorbit map

γΠ : G −→ X : g 7−→ g(Π)

is a bijection.

Proof. See orbit-stabiliser Theorem. 2

13

3 Generators and Relations

• 〈X〉-free group on a set X

• 〈∅〉 = C1

• 〈x〉 ∼= Z-infinite cyclic group

• 〈X〉 = all finite words (including ∅) in alphabet x+1, x−1 as x ∈ X without subwortsx+1x−1 or x−1x+1.

• v · w = Reduction of concenated word vw

• It is a Theorem that · is well-defined.

• Universal property:

Xf //

_

G

〈X〉

∃!ϕ==

∀ groups G ∀ functions f : X −→ G ∃! group homomorphism

ϕ : 〈X〉 −→ G s.t. ∀ a ∈ X : ϕ(a) = f(a)

• 〈X | R〉: R is a set of relations i.e. either words in alphabet x±1 : x ∈ X or u = v whereu, v are two words. Relation between words and equations:

u −→ u = 1, uv−1 −→ uv−1 = 1 ⇐⇒ u = v

E.g. dihedral group I2(n) =

⟨a︸︷︷︸rot

, b︸︷︷︸refl.

: b2, an, bab−1 = a−1

⟩.

Definition. 〈X | R〉 = 〈X〉/R. If the relations are words u1, u2, . . . thenR =

⋂〈X〉.H, all ui∈H

H.

Universal property of 〈X | R〉:〈X | R〉

∃!ϕ

##X

q

OO

f // G

∀ functions f : X −→ G s.t. relations in R hold in G for elements f(x), x ∈ X then ∃! grouphomomorphism ϕ : 〈X | R〉 −→ G s.t. ∀ x ∈ X: ϕ(xR) = f(x).

14

Example: B(2, 5) =⟨x, y | w5

⟩where w is any word in the alphabet x±1, y±1.

Universal property: ∀ groups G s.t. ∀ x ∈ G : x5 = 1, ∀a, b ∈ G ∃! ϕ : B(2, 5) −→ G s.t.ϕ(x) = a, ϕ(y) = b.Known: B(2, 5) is either infinite or of order 534.

15

4 Coxeter group

A Coxeter graph is a non-oriented graph without loops or double edges s.t. each edge is markedby a symbol m ∈ Z≥3 ∪ ∞.

Convention: Draw instead of 3 We will study only graphs with finitely many vertices.

Definition. A Coxeter matrix (on set X) is an X ×X matrix (mij)i,j∈X s.t.

• mij ∈ Z≥1 ∪ ∞

• mij = mji

• mij = 1 ⇐⇒ i = j

There is a 1:1 correspondance between Coxeter graphs and Coxeter matrices

Graph ←→ Matrix (cij)

X = vertices of the graph

imij

j −→ cij = mij

i j −→ cij = 2

Example:

1∞

>>>>>>> 2

77

>>>>>>>

3 4

−→

1 3 ∞ 23 1 3 77∞ 3 1 22 77 1 1

Definition 4.1. The Coxeter group of a Coxeter graph (or corresponding X×X-matrix (mij))is

W = W (graph) = 〈X | (ab)mab = 1〉

Remark 4.2. 1. mab =∞ =⇒ no relation

2. a = b,maa = 1 =⇒ (aa)1 = 1 =⇒ a2 = 1

3. no edge ⇐⇒ mab = 2 =⇒ (ab)2 = 1 ⇐⇒ ab = b−1a−1 = ba Therefore:no edge ⇐⇒ a, b commute

Lemma 4.3. Let G be a group generated by a1, . . . , an, all of order 2. Then G is a quotient ofW (mij = |aiaj |).

16

Proof. f : 1, . . . , n −→ G where f(i) = ai is a function. Then f(i)2 = a2i = 1 and

(f(i)f(j))mij = (aiaj)|aiaj | = 1. So f(i) satisfy relations of the Coxeter group and ∃! group

homomorphism ϕ : W (mij) −→ G s.t. ϕ(i) = ai (by the universal property). Im ϕ 3 ai =⇒ ϕ

is surjective. 2

Big monster M is the largest sporadic simple group,

M ∼ 1053, M ∼=W

/H

Theorem 4.4. G is a finite reflection group Π ⊂ Φ-simple system in the root system. Thenthe natural surjection for Π × Π-matrix mx,y = |SxSy|, W ((mx,y)) G : x 7−→ Sx is anisomorphism.

Example: O2(R) ≥ I2(∞) =

⟨Sa︸︷︷︸

order 2

, Rotα︸ ︷︷ ︸inf. order

⟩︸ ︷︷ ︸

∼=C2nC∞

, απ /∈ Q

W ( ∞ ) 〈Sa, Sb〉 whereab

π/∈ Q, ∼= 〈Sa, Rotα〉

action of W ( ∞ ) = I2(∞) or R2 is a mess.But I2(∞) has action on R1 with geometric meaning.

-1 0 1

S0 S1

S0(x) = x, S1(x) = −2−x, S1S0(x) = 2+x −→ translation by 2. So 〈S1, S0〉 isW ( ∞ ) insidethe group of motions on R. C∞ is a subgroup of translations (x 7−→ x+ 2n). As C2 ≤W ( ∞ )you can choose any 1, Sn where Sn(x) = 2n − x. The fundamental domain is [0, 1], I2(∞)tesselates R1 by interval [0, 1].

Example: W

bc

a ∼= ⟨sidereflections across the sides of a triangle with angles πa ,

πb ,

πc

⟩.

α γ

β

where α = aπ , β = b

π , γ = cπ.1

a+

1

b+

1

c> 1 spherical

= 1 euclidean< 1 hyperbolic

17

−→ tesselation of the space.

In particular, W

bca

←→ triangles in tesselation

Lemma 4.5. (Deletion Condition)G - finite reflection group, Φ - root system, Φ ⊃ Π - simple system.If x = Sa1 · San , ai ∈ Π and l(x) < n then ∃i < j s.t.

x = Sa1 · · · Sai · Saj · San

Proof. already proved. 2

Proof. of Theorem 4.3

〈Π〉ϕW (mab)

ψ G

a 7−→ a︸︷︷︸we will work in here

7−→ Sa

It suffices to show that ψ is injective.Let w = a1a2 · · · an ∈ Ker ψ (a1

2 = 1). Then ψ(w) = Sa1 · San = 1. Since det Sai =−1, (−1)n = 1 −→ n is even, write n = 2k.Proceed by induction on k = n

2

• k = 1 −→ n = 2 −→ w = a1a2

1 = ψ(w) = Sa1Sa2 −→ Sa1 = Sa2 −→ a1 = ±a2ai∈Π=⇒ a1 = a2 −→ w = a1

2 = 1

• k ≤ m− 1: done (induction assumption)

• k = m :−→ n = 2m, w = a1 · · · a2m, 1 = ψ(w) = Sa1 · · ·Sa2m . Consider

w1 = a1−1wa1 = a1wa1 = a2a3 · · · a2ma1

Repeating conjugation w.r.t. the first symbol, we can consider two cases:

Case 1 w = abab · · · ab for some a, b ∈ Π, w = (ab)m.

1 = ψ(w) = (SaSb)m =⇒ |SaSb| = mab|m =⇒ w = (ab)m = 1 ∈W

Case 2 there is a v ∈W s.t. w2 = v−1wv = b1b2b3 · · · b2m with all bi ∈ Π and b1 6= b3.Let us play the trick: Observe that

x = Sb1 · · ·Sbm+1 = Sb2m · · ·Sbm+2 = y

Indeed y−1 = Sbm+2 · · ·Sb2m and xy−1 = Sb1 · · ·Sb2m = ψ(w) = 1. Hence x = y.By deletion principle (note that l(x) ≤ m− 1) there are 1 ≤ i < l ≤ m+ 1 s.t.

x = Sb1 · · · Sbi · · · Sbj · · ·Sbm+1

Consider two cases:

18

Case 2.1 (i, j) 6= (1,m+ 1)

Sbi · · ·Sbj = Sbi+1· · ·Sbj−1

Sbi · · ·SbjSbj−1· · ·Sbi+1

= 1

Hence w3 = bi · · · bjbj−1 · · · bi+1 ∈ ker ψ. It has < 2m terms hence by inductionassumption w3 = 1. Hence

bi · · · bj = bi+1 · · · bj−1

So w2 = b1 · · · bi · · · bj · · · b2m. By induction assumption w2 = 1 and therefore

w = vw2v−1 = vv−1 = 1

Case 2.2 (i, j) = (1,m+ 1)Then w3 = b1b2 · · · bm+1bm · · · b2 ∈ Ker ψ has length 2m and we seem to bestuck. (Hint: eventual contradiction is with b1 6= b3).Let w4 = b1

−1w2b1 = b2 · · · bnb1 ∈ ker ψ. Let us play the trick on

x1 = Sb2 · · ·Sbm+1Sbm+2 = Sb1 · · ·Sbm+3

Deletion principle −→ x1 = Sb2 · · · Sb′i · · · Sb′j · · ·Sbm+2 . Get 2 cases:

Case 2.2.1 (i′, j′) 6= (2,m+ 2)

Sb′i · · ·Sb′j = Sbi′+1· · ·Sbj′−1

Sb′i · · ·Sb′jSbj′−1· · ·Sbi′+1

= 1

w5 = bi′ · · · bj′bj′−1 · · · bi′+1 ∈ ker ψ. By induction assumption w5 = 1

bi′ · · · bj′ = bi′+1 · · · bj′−1

and w2 = b1 · · · b′i · · · b′j · · · b2m. By induction assumption w2 = 1 =⇒ w =

vw2v−1 = 1.

Case 2.2.2 (i′, j′) = (2,m+ 2)

w5 = b2 · · · bm+1bm+2bm+1bm · · · b3 ∈ ker ψ

but of length n. Let w6 = b3w5b3−1

= b3b2b3 · · · bm+2bm+1 · · · b4 ∈ ker ψand repeat the trick on ψ(w6) = 1.

x2 = Sb3Sb2Sb3 · · ·Sbm+1 = Sb4 · · ·Sbm+2

can delete at i′′, j′′. Get 2 cases:

Case 2.2.2.1 (i′′, j′′) 6= (1st incidence of 3,m+ 1)

w7 = bi′′ · · · bj′′bj′′−1 · · · bi′′+1 ∈ ker ψ

Induction assumption =⇒ w7 = 1 =⇒ w6 = 1 =⇒ w5 = 1 =⇒ w2 =1 =⇒ w = 1.

19

Case 2.2.2.2 (i′′, j′′) = (first 3,m+ 1)

w7 = b3b2b3 · · · bm+1bm · · · b2 ∈ ker ψ

Remember w3 = b1b2b3 · bm+1bm · · · b2. Then

w7 = b3b1w3 −→ b3b1 = w7w−13 ∈ ker ψ

Sb3Sb1 = 1 =⇒ Sb3 = Sb1 =⇒ b3 = ±b1 =⇒ b3 = b1

But this is a contradiction!

2

20

5 Geometric representation of W (mij)

Recall some geometry of a vector space V with symmetric bilinear form 〈 , 〉. (the field is R,dim V - any, 〈 , 〉 - any)Still ∀ W ≤ V we have

• W⊥ = x ∈ V : ∀ a ∈W : 〈a,w〉 = 0

• (W⊥)⊥ ⊃W

Example: R2,

⟨(xy

),

(ab

)⟩= xa− yb

W = R ·(

11

)=⇒W⊥ = W =⇒W +W⊥ = W 6= R2.

Vectors v with 〈x, x〉 = 0 are called isotropic. If 〈x, x〉 6= 0 (x is not isotropic) then

V = Rx⊕ x⊥

andSx : V −→ V, Sx(a) = a− 2

〈a, x〉〈x, x〉

x

is well-defined and has usual properties

• Sx(x) = −x

• Sx|xbot = id

• S2x = id

• Sx ∈ O(V, 〈 , 〉)

Start with X ×X Coxeter matrix (mab) or its Coxeter graph. Let

V (mab) = ∑a∈X

αaea : αa ∈ R, finitely many αa 6= 0

ea, a ∈ X form a basis of V (mab). Symmetric bilinear form 〈 , 〉 : V (mab)× V (mab) −→ R isdefined on the basis by

〈ea, eb〉 = − cos

(π

mab

)• a = b =⇒ mab = 1 =⇒ 〈ea, ea〉 = 1

• mab = 2 =⇒ 〈ea, eb〉 = 0

• mab =∞ =⇒ 〈ea, eb〉 = − cos(π∞)

= − cos 0 = −1

21

Lemma 5.1. V ( m ) is euclidean ⇐⇒ m 6=∞.

Proof. 〈e1, e2〉 = − cos πm . If x = αe1 + βe2 then

〈x, x〉 = α2 〈e1, e1〉+ 2αβ 〈e1, e2〉+ β2 〈e2, e2〉

= α2 + β2 − 2αβ cos( πm

)= (α+ β)2 − 2αβ(1 + cos

( πm

))

≥ 0

and〈x, x〉 = 0 ⇐⇒ α = β = 0 OR m =∞, α = β

2

Lemma 5.2. (V, 〈 , 〉) − R-vector space with symmetric bilinear form. Let U ⊂ V be a finitedimensional subspace s.t. U ∩ U⊥ = 0. Then

V = U ⊕ U⊥

Note: it breaks down if dim U =∞.Example: (V, 〈 , 〉) Hilbert space with Hilbert base e1, . . . , U = span(ei) ( V dense.Then U⊥ = 0, so U ∩ U⊥ = 0 but U ⊕ U⊥ = U 6= V .

Proof. Pick v ∈ V , basis e1, . . . , en of U .It suffices to find x1, . . . , xn ∈ R s.t.

v −∑

xiei ∈ U⊥ ⇐⇒ ∀ j :⟨v −

∑xiei, ej

⟩= 0 ⇐⇒ 〈v, ej〉 −

∑i

xi 〈ei, ej〉 = 0

The system of n equations∑ixi 〈ei, ej〉 = 〈v, ei〉 j = 1, . . . , n has a solution. It holds

true because U ∩ U⊥ = 0 which implies that 〈 , 〉 |U is non-degenerate which is equivalent

to det 〈ei, ej〉 6= 0. 2

Coxeter graph (matrix) Γ = (mab)a,b∈X −→ Coxeter group W (Γ) = 〈X : (ab)mab = 1〉−→ vector space with bilinear form

V (Γ) =⊕a∈X

Rea, 〈ea, eb〉 = − cos

(π

mab

)Operators

ρa = Sea : V (Γ) −→ V (Γ) : x 7−→ x− 2 〈ea, x〉 eaSince ρ : a(ea) = −ea, ρa|e⊥a = I, |ρa| = 2.

Proposition 5.3.|ρaρb| = mab

22

Proof.

Case 1 mab 6= ∞5.1

=⇒ U = Rea ⊕ Reb is euclidean vector space w.r.t 〈 , 〉. In particularU ∩ U⊥ = 0 =⇒ V (Γ) = U ⊕ U⊥. Since U⊥ ⊂ e⊥a , ρa|U⊥ = I. Similarly, ρb|U⊥ = ISince ρb(U⊥) ⊂ U⊥, ρaρb|U⊥ = ρa|U⊥ · ρb|U⊥ = I · I = I. On U , we can do explicitcalculation.

eβ

eα π − πmab

Hence ρa|U · ρb|U = Rot by 2πmab

and |ρaρb| = |ρaρb|U | = mab

Case 2 Exercise: show that in ea, eb basis

ρa|U · ρb|U =

(3 −22 −1

)−→

(1 10 1

)so |ρaρb|U | =∞ −→ |ρaρb| =∞.

2

Hence we have a group homomorphism

ρ : W (Γ) −→ GL(V (Γ)) : a (a ∈ X) 7−→ ρa

ρ is called geometric representation of W (Γ).

Suppose Γ = Γ1∪Γ2 (disjoint union of 2 graphs). On the level of Coxeter matrices

X = X1 ∪X2, X1 ∩X2 = ∅

and for all a ∈ X1, b ∈ X2 : mab = 2.

Proposition 5.4. W (Γ) ∼= W (Γ1)×W (Γ2)

Proof. Construct inverse group homomorphism ψ : W (Γ) −→W (Γ1)×W (Γ2), ϕ : W (Γ1)×W (Γ2) −→W (Γ).If a ∈ X, let

ψ(a) =

(a, 1) if a ∈ X1

(1, a) if a ∈ X2

Since all relations of W (Γ) hold in W (Γ1)×W (Γ2) such ψ uniquely extends to a group homo-

morphism.

If a ∈ Xi ⊂ X, ϕi(a) = a extends to a group homomorphism ϕi : W (Γi) −→ W (Γ) (all

relations of W (Γi) hold in W (Γ) )

Notice that generators of the image of ϕ1 commute with generators of Im ϕ2. Hence subgroups

Im ϕ1, Im ϕ2 commute. Hence ϕ(x, y) = ϕ1(x)ϕ2(y) is a well-defined group homomorphism.

Since ∀ a ∈W (Γ), ϕ(ψ(a)) = a, ϕψ = IW (Γ). Similarly ψϕ = IW (Γ1)×W (Γ2). 2

23

Γ is connected if Γ = Γ1∪Γ2 =⇒ Γ1 = ∅ or Γ2 = ∅.Each Γ has connected components Γ1,Γ2, . . . s.t.

Γ = Γ1∪Γ2∪Γ3∪ . . .

Corollary 5.5. In this case W (Γ) ∼= W (Γ1)×W (Γ2)× . . . and V (Γ) = V (Γ1)⊕ V (Γ2)⊕ . . .with V (Γi)⊥V (Γj) for i 6= j.

Proof. Exercise (use transfinite induction). 2

A representation of a group G is the pair (V, ρ), where V is a vector space and ρ : G −→ GL(V ).

Example: Take a Reflection group (G,V ), then (V, i : G −→ GL(V )) is a representation.If Γ is a Coxeter graph, then (V (Γ), ρ) is a representation of W (Γ).A subrepresentation of (V, ρ) is a subspace U ⊂ V ) s.t. ∀g ∈ G : ρ(g)(U) ⊂ U . So (U, ρ|GL(U))is a representation.V is irreducible if V 6= 0 and 0, V are the only subrepresentations. V is called completelyirreducible if for all subrepresentations U ⊂ V there exists a subrepresentation W ⊂ V s.t.V = U ⊕W .

Convention: The Coxeter graph ∅ is not connected.

Proposition 5.6. If Γ is connected, then any proper W (Γ)-subrepresentation of V (Γ) lies in(V (Γ))⊥.

Exercise: (V (Γ))⊥ is a subrepresentation of V (Γ).Proof. Let U ⊂ V = V (Γ) be a subrepresentation. Then ∀ a ∈ X (Γ = (mab)a,b∈X), ρa(U) ⊂U . Observe that ea /∈ U . Otherwise: ∀b ∈ X, mab 6= 2 holds:

U 3 ρb(ea) = ea + 2 cosπ

mab︸ ︷︷ ︸6=0

eb =⇒ eb ∈ U

Since Γ is connected, this would imply that eb ∈ U for all b ∈ X. So U = V and not a propersubrepresentation which is a contradiction.By 5.?, V = Rea ⊕ Re⊥a . Pick any z ∈ U and write z = αea + z′ where z′ ∈ Re⊥a . Then

U 3 ρa(z) = z − 2 〈ea, z〉 ea= αea + z′ − 2α 〈ea, ea〉 ea − 2

⟨ea, z

′⟩ ea= −αea + z′

Hence z′ = 12(z + ρa(z)) ∈ U and 2αea ∈ U =⇒ α = 0. Hence z = z′ ∈ Re⊥a and therefore

U ⊂ Re⊥a and U ⊂⋂Re⊥a = V ⊥. 2

Corollary 5.7. If Γ is connected and V (Γ) nonsingular ( V (Γ)⊥ = 0), then V (Γ) is irreducible.

Theorem 5.8. (Maschke)If G is a finite group and F a field s.t. char F 6 | |G| and V is a representation of G over F,then V is completely irreducible.

24

Proof. Pick a subrepresentation U ⊂ V .Observe that direct sum representations V = U ⊕ W are in bijection with linear operatorsp ∈ EndF(V ) s.t. p2 = p and Im(p) = U :

V = U ⊕W −→ p(v) = u where v = u+ w is the unique decomposition with u ∈ U,w ∈WV = Im(p)⊕ ker(p)←− p

Pick any vector space decomposition and let p be the corresponding operator, ρ : G −→ GL(X).Define p = 1

|G|∑x∈G

ρ(x)pρ(x−1)

• if t ∈ U

p(t) =1

|G|∑x∈G

ρ(x)(p(ρ(x−1)(t)︸ ︷︷ ︸∈U

))p|U=I

=1

|G|∑x∈G

ρ(x)ρ(x−1)(t) = t

So p|U = I.

• since p(t) = t for t ∈ U, Im(p) ⊃ U

• by definition of p, Im(p) ⊂ Im(p) ⊂ U . Hence Im(p) = U =⇒ p2 = p

• p is a homomorphism of representations (commutes with any ρ(x), x ∈ G):

ρ(x)p =1

|G|∑y∈G

ρ(x)ρ(y)pρ(y−1)

=1

|G|∑y∈G

ρ(xy)pρ((xy)−1)ρ(x)

=1

|G|∑z∈G

ρ(z)pρ(z−1)ρ(x)

= pρ(x)

Hence ker p is a G-subrepresentation and V = U ⊕ ker p.

2

Corollary 5.9. If Γ is connected and |W (Γ)| <∞ then V (Γ) is nonsingular (w.r.t 〈 , 〉) andirreducible (as W (Γ)-representation).

Proof. Irreducibility follows from nonsingularity by Corollary 5.?. Suppose V ⊥ 6= 0 (V =

V (Γ)). By 5.8, V = V ⊥ ⊕ U for some G-subrepresentation U . By 5.6, U ⊂ V ⊥ =⇒ U = 0 =⇒V = V ⊥ =⇒ 〈 , 〉 = 0. Nonsence since 〈ea, ea〉 = 1 and Γ is non-empty. 2

Proposition 5.10. (Schur’s Lemma)If Γ is connected and |W (Γ)| <∞ then EndW (Γ)V (Γ) = R.

25

Proof. Pick Θ ∈ EndW (Γ)V (Γ). Then ∀ a ∈ X : Θρa = ρaΘ. Therefore: ∀ z ∈ V (Γ) :Θ(ρa(z)) = ρa(Θ(z)).

Θ(z)− 2 〈ea, z〉Θ(ea) = Θ(z − 2 〈ea, z〉 ea) = Θ(z)− 2 〈ea,Θ(z)〉 ea

Hence 〈ea, z〉Θ(ea) = 〈ea,Θ(z)〉 ea. Let z = ea, then Θ(ea) = 〈ea,Θ(ea)〉︸ ︷︷ ︸λ

ea. Therefore, Θ has

an λ-eigenvector, Θ− λI ∈ EndW (Γ)V (Γ).0 6= ker Θ− λI is a subrepresentation of V (Γ). By 5.9,

ker Θ− λI = V (Γ) =⇒ Θ− λI = 0 =⇒ Θ = λI

2

Theorem 5.11. If |W (Γ)| <∞ then (V (Γ), 〈 , 〉) is euclidean.

Proof. |W (Γ)| <∞ =⇒ X is finite =⇒ dim V (Γ) = |X| <∞.If Γ =

⋃Ti, Ti connected, then V (Γ) =

⊕iV (Γi) with V (Γi)⊥V (Γj), i 6= j. Hence it suffices

to consider a case of connected Γ.Pick any Euclidean form 〈 , 〉1 over V (Γ). Let us integrate 〈 , 〉1 over W (Γ) so let

〈u, v〉2 =∑

x∈W (Γ)

〈ρx(u), ρx(v)〉1

• 〈 , 〉2 is symmetric since 〈 , 〉1 is

• u 6= 0, 〈u, u〉2 =∑x〈ρx(u), ρx(u)〉1︸ ︷︷ ︸

>0

hence 〈 , 〉2 is euclidean.

• ∀ y ∈W (Γ):

〈ρy(u), ρy(v)〉2 =∑x

〈ρxρy(u), ρxρy(v)〉1

=∑x

〈ρxy(u), ρxy(v)〉1

=∑

z∈W (Γ)

〈ρz(u), ρz(v)〉1

= 〈u, v〉2

Hence all ρy ∈ O(V (Γ), 〈 , 〉2). Remember that all ρu ∈ O(V (Γ), 〈 , 〉) and considerΘ : V −→ V defined by

〈Θ(u), v〉2 = 〈u, v〉 ∀ v

Θ ∈ EndW (Γ)V (Γ) since for all x ∈W (Γ):

〈ρx(Θ(u)), v〉2 =⟨Θ(u), ρ−1

x (v)⟩

2

=⟨u, ρ−1

x (v)⟩

2

= 〈ρx(u), v〉= 〈Θ(ρx(u)), v〉2

26

This implies ρxΘ = Θρx. By 5.10, Θ = λI for some λ ∈ R, so for all u, v:

λ 〈u, v〉2 = 〈u, v〉

Since 〈ea, ea〉 = 1 = λ 〈ea, ea〉2 =⇒ λ = 1〈ea,ea〉2

> 0. 2

Section 4 =⇒ Finite Reflection group ∼= Coxeter groupSection 5 =⇒ Finite Coxeter group −→ Reflection group (V (Γ), ρ(W (Γ)))Problem: ρ could have a kernel. Section 6 rules this out.

Example: Exercise: D4n∼= C2 ×D2n when n is odd.

Hence distinct reflection groups I2(2n) = W

(2n

)and A1 × I2(n) = W

n are

isomorphic as groups.

27

6 Fundamental chamber

Aim: ρ : W (Γ) −→ Im(ρ) is isomorphic (at least for finite Coxeter groups)Define l : W (Γ) −→ Z≥0 : l(x) = minn : x = a1 · · · an, ai ∈ X(Γ).In particular: l(1) = 0, l(ai) = 1.

Lemma 6.1. For all a ∈ X, g ∈W holds: l(ag) is either l(g) + 1 or l(g)− 1.

Proof. g = a1 · · · an with l(g) = n, then ag = aa1 · · · an so l(ag) ≤ l(g) + 1. Since g =a(ag), l(g) ≤ l(ag) + 1 so l(ag) ≥ l(g)− 1.Need to rule out l(ag) = l(g). Consider sign representation:

sgn : W −→ GL1(R) : sgn(a) = (−1)

Since relations of W hold in GL1(R) ((ab)mab = 1 −→ ((−1)(−1))mab = 1 ok) sgn is well-

defined.

If l(x) = n then x = a1 · · · an and sgn(x) = sgn(a1) · · · sgn(an) = (−1)n = (−1)l(x).

If l(ag) = l(g) then sgn(ag) = sgn(g), sgn(a)sgn(g) = sgn(g) =⇒ sgn(a) = 1 which is a

contradiction. 2

For each a ∈ X(Γ) consider half spaces

• H+a = u ∈ V (Γ) : 〈u, ea〉 > 0

• H−a = u ∈ V (Γ) : 〈u, ea〉 < 0

the dividing hyperplane Ha = u ∈ V (Γ) : 〈u, ea〉 = 0and the cone C =

⋂a∈X(Γ)

H+a which we call the fundamental chamber.

Idea: Let x ∈W :xi(Ha)

ρx(C)

C

28

l(x) is the minimum of hyperplanes one has to cross to walk from C to ρx(C).

Let Γ = (mab)a,b∈X and denote W = W (Γ), V = V (Γ)

Lemma 6.2. |W | <∞, a, b ∈ X. Then ∀ g ∈⟨a, b⟩≤W : either

• g(H+a ∩H+

b ) ⊂ H+a OR

• g(H+a ∩H+

b ) ⊂ H−a and l(ag) = l(g)− 1

Proof. If g(H+a ∩H+

b ) 6⊂ H+a and g(H+

a ∩H+b ) 6⊂ H−a then by convexity of g(H+

a ∩H+b ):

g(H+a ∩H+

b ) ∩Ha 6= ∅

Pick z ∈ g(H+a ∩H+

b ) ∩Ha. Then z ∈ Ha =⇒ 〈z, ea〉 = 0

z ∈ g(H+a ∩H+

b ) =⇒ g−1(z) ∈ H+a ∩H+

b =⇒⟨g−1(z), ea

⟩> 0,

⟨g−1(z), eb

⟩> 0

On the other hand⟨g−1(z), g−1(ea)

⟩= 〈z, ea〉 = 0.

As g ∈⟨a, b⟩:

g−1(ea) = αea + βeb α, β ∈ RWe denote by Φ(Γ) a root system of ρ(W ) ∈ V (Γ) and since ea, eb form a simple system ofΦ(Γ) ∩ span(ea, eb), either

• α ≥ 0, β ≥ 0 s.t. not both are 0 and then⟨g−1(z), g−1(ea)

⟩=⟨g−1(z), αea + βeb

⟩= α

⟨g−1(z), ea

⟩+ β

⟨g−1(z), eb

⟩> 0

which is a contradiction

• OR α ≤ 0, β ≤ s.t. not both are 0 and therefore⟨g−1(z), g−1(ea)

⟩< 0

which is a contradiction too.

It remains to prove the length statement if g(H+a ∩H+

b ) ⊂ H−a . Inspect U = span(ea, eb) andL = U ∩H+

a ∩H+b . This is a dihedral group of order 2mab.

ea

Hb ∩ U

eb

abab...L

bLbaL

L

H+aH−a

gL ⊂ H−a ∩ U ⇐⇒ g = a or g = ab or . . . or g = abab . . . b︸ ︷︷ ︸mab

All such g admit a reduced expression starting with a. Hence l(ag) = l(g)− 1. 2

29

Remark 6.3. There are 2 finiteness assumptions:

1. W is finite

2. Γ is finite, i.e. |X| <∞ and ∀ a, b ∈ X : mab <∞.

Exercise: W is finite =⇒ Γ is finite.Hint: a 6= b ∈ X =⇒ ρa 6= ρb ∈ GL(V (Γ)) =⇒ a 6= b ∈ W (Γ). Then one needs to show that

mab <∞. This follows from ρaρb|span(ea,eb) =

(1 10 1

).

Note: Lemma 6.2 holds under assumption that Γ is finite.

Proposition 6.4. Let Γ be finite and let C be the fundamental chamber. Then:

1. For all w ∈W and for all u ∈ X: either

• w(C) ⊂ H+a OR

• w(C) ⊂ H−a and l(aw) = l(w)− 1

2. for all w ∈W and for all a 6= b ∈ X there is a g ∈⟨a, b⟩s.t.

w(C) ⊂ g(H+a ∩H+

b ) and l(w) = l(g) + l(g−1w)

Proof. Simultaneous induction on l(w):

Base l(w) = 0 =⇒ w = 1 =⇒ w(C) = C ⊂ H+a and g = 1 works for (2).

Assumption if l(w) = k − 1 then (1) and (2) are assumed to have been proved

Step Let l(w) = k.

1. Write w = dw′, d ∈ X, l(w′) = k − 1. Consider two cases:

Case 1: a = d.Statement (1) for w′ gives either w′(C) ⊂ H+

a or w′(C) ⊂ H−a .• w′(C) ⊂ H−a =⇒ w(C) = a(w′(C)) ⊂ a(H−a ) = H+

a

• w′(C) ⊂ H+a =⇒ w(C) = a(w′(C)) ⊂ a(H+

a ) = H−a

Case 2: a 6= d.By (2) for w′ there is a g ∈

⟨a, b⟩s.t. w′(C) ⊂ g(H+

a ∩ H+b ) and l(w′) =

l(g) + l(g−1w). Then

w(C) = d(w′(C)) ⊂ dg(H+a ∩H+

d )

By Lemma 6.2 either dg(H+a ∩H+

d ) ⊂ H+a (=⇒ w(C) ⊂ H+

a ) ordg(H+

a ∩H+d ) ⊂ H−a and l(adg) = l(dg)− 1.

Then w(C) ⊂ dg(H+a ∩H+

d ) ⊂ H−a and

l(aw) = l(adgg−1w′) ≤ l(adg) + l(g−1w′) ≤ l(g) + l(g−1w′) = l(w′) = k − 1

=⇒ l(aw) = l(w)− 1

30

2. Now assume (1) and (2) for l(w) = k−1 and note that (1) for l(w) = k is established.Consider two Cases:

Case 1: wC ⊂ H+a ∩H+

b . Then g = 1 satisfies (2).

Case 2: wC 6⊂ H+a ∩H+

b .

Ha

Hb

H+a ∩H+

b

H+a ∩H−bH−a ∩H−b

H−a ∩H+b

Observe that wC ∩Ha = ∅. wC ⊂ w(H+a ∩H+

b ) is in either H+a or H−a by 6.2.

Wlog wC ⊂ H−a .By part (1): l(aw) = l(w)− 1.Use part (2) on w′ = aw (by induction assumption) then there is g′ ∈

⟨a, b⟩

s.t. w′C ⊂ g′(H+a ∩H+

b ) and l(w′) = l(g′) + l((g′)−1w).Claim: g = ag′ satisfies (2) for w.

wC = aw′C ⊂ ag′(H+a ∩H+

b ) = g(H+a ∩H+

b )

l(w) = 1 + l(aw) = 1 + l(g′) + l((g′)−1w′)

= 1 + l(g′) + l(g−1w) ≥ l(ag′) + l(g−1w)

= l(g) + l(g−1w) ≥ l(gg−1w) = l(w)

Hence l(w) = l(g) + l(g−1w).

2

Theorem 6.5. 1. W is finite =⇒ V =⋃g∈W

gC

2. Γ is finite, then: ∀ g, x ∈W : gC ∩ xC 6= ∅ =⇒ g = x

Note: If W is infinite⋃g∈W

gC is called Tits cone which is a proper subset of V .

Proof.

1. C = α ∈W : ∀ a ∈ X : 〈α, a〉 ≥ 0.Recall the height function h : V −→ R : v =

∑a∈X

βaea 7−→∑a∈X

βa.

Pick any v ∈ V . Since W is finite, the height achieves a maximum on the setWv = g(v) : g ∈W. Let z = g(v) be such a maximum.

31

Claim: z ∈ C and this implies that v = g−1(z) ∈ g−1(C) and V =⋃gC.

Indeed, otherwise there is an a ∈ X s.t. 〈z, ea〉 < 0. Then:

ρa(z) = h(z)− 2 〈z, ea〉 ea

has highth(ρa(z)) = h(z)− 2 〈z, ea〉 > h(z)

which is a contradiction to the maximality.

2. Assume gC ∩ xC 6= ∅. Then C ∩ g−1xC 6= ∅. If g−1x = 1 then g = x and we are done.If g−1x 6= 1 then there is an a ∈ X s.t. l(ag−1x) = l(g−1x)− 1.Set w = ag−1x, then l(aw) = l(w) + 1. By 6.3 ag−1xC = wC ⊂ H+

a =⇒ g−1xC ⊂ H−a .Then C ∩ g−1xC ⊂ H+

a ∩H−a = ∅ which is a contradiction.

2

Corollary 6.6. Γ is finite. Then ρ : W −→ GL(V ) is injective.

Proof. Suppose ρ(g) = ρ(w). Then gC = wC. Hence g = w by 6.4. 2

Let (G,V ) be a reflection group, G ≤ O(V ), V is Euclidean. Let V1 = span(x : x ∈ Φ(G)).Say (G,V ) is essential if V1 = V .

Corollary 6.7.Γ←→ (W (Γ), V (Γ))

is a bijection between isomorphism classes of Coxeter graphs with finite W (Γ) and equivalenceclasses of essential reflection groups.

Proof.

Γ −→ (W (Γ), V (Γ))

X = Π is a simple system in Φ(G), mab = |SaSb| ←− (G,V )

By definition, (W (Γ), V (Γ)) is essential. Going Γ 7−→ (W (Γ), V (Γ)) 7−→ (X with mab = |SaSb|)gives Γ back by 6.5 and 4.3 (?).(G,V ) 7−→ Γ = (X,mab) 7−→ (W (Γ), V (Γ)). Then

V (Γ) −→ V∑x∈X

αxex 7−→∑x∈X

αxx

W (Γ) −→ G

a1a2 · · · an 7−→ Sa1 · · ·San

This is an isomorphism by 4.3 and 6.5. 2

Some fun bits:

• Rn \ Rn−1: disjoint union of 2 contractible half-spaces

32

• Cn \ Cn−1: is connected with fundamantal group Z

Try CC = C⊗R V ⊃ CHa = C⊗R H

a. Then CV0 = CV \

⋃g∈W, a∈X

g(CHa) is connected.

W acts freely on CV0 and the universal cover of CV

0 is simply connected (if W is finite).One can draw the fundamental group of CV

0/W :

Ha

x

ρa(x)Ta

C ⊂C V0

[x] = [ρa(x)] ∈ CV0/W .

Ta is a loop in CV0/W . In fact:

B(Γ)︸ ︷︷ ︸Braid group

= π1

(CV

0/W

)=

⟨Ta, a ∈ X : TaTbTa · · ·︸ ︷︷ ︸

mab

= TbTaTb · · ·︸ ︷︷ ︸mab

⟩

V 0 −→ V 0/W gives

1 −→ PB(Γ)︸ ︷︷ ︸pure Braid group

←→ B(Γ) W −→ 1

33

7 Classification

Let (V, 〈 , 〉) be an Euclidean vector space. Therefore it has a norm

‖v‖ =√〈v, v〉

It induces a norm on LinR(V, V ):

‖T‖ = supv∈V \0

‖Tv‖/‖v‖

1. ‖Tx‖ ≤ ‖T‖ · ‖x‖

2. ‖T‖ ≥ 0 and [‖T‖ = 0 ⇐⇒ T = 0]

3. ‖T + S‖ ≤ ‖T‖+ ‖S‖

4. T ∈ O(V ) =⇒ ‖Tv‖ = ‖v‖ =⇒ ‖T‖ = 1

5. O(V ) is a closed bounded subset of LinR(V, V ).

Theorem 7.1. Let Γ be finite. Then W (Γ) is finite ⇐⇒ V (Γ) is Euclidean.

Proof.

"=⇒" Theorem 5.9

"⇐=" short version: By 6.4 W (Γ) acts discretely on the unit sphere in V (Γ). Hence it musbe finite.Long version: pick v ∈ C s.t. ‖v‖ = 1. Pick ε > 0 s.t. Bε(v) ⊂ C.

g 6= x ∈W =⇒ gC ∩ xC = ∅ =⇒ Bε(gv) ∩Bε(xv) = ∅

=⇒ ‖gv − xv‖ > ε =⇒ ‖ρ(g)− ρ(x)‖ > ε

By fact 5, O(V ) is compact. Suppose W is infinite, W ⊂ O(V ) compact. Then there isa convergent sequence T1, T2, . . . , Tn in W s.t. Ti 6= Tj for i 6= j.Hence Ti −→ T ∈ O(V ), so there is a N s.t ∀ i, j > N : ‖Ti − Tj‖ < ε.But Ti, Tj ∈ ρ(W ) with Ti 6= Tj , so this is a contradiction.

2

Γ is positive definite if V (Γ) is Euclidean. This is equivalent to W (Γ) being finite. From whatwe proved, we may conclude that any finite reflection group has a form

(W (Γ1)× . . .×W (Γn), V (Γ1)⊕ . . .⊕ V (Γn)⊕ Rk)

where the Γi are connected positive definite Coxeter graphs and each W (Γi) acting trivially onV (Γj) with i 6= j and Rk.

Problem: What are the connected positive definite Coxeter graphs?

34

Proposition 7.2. A =

a11 · · · a1n...

...an1 · · · ann

symmetric matrix over R.

A is positive definite ⇐⇒ ∀ j ∈ 1, . . . , n : det

a11 · · · a1j...

...aj1 · · · ajj

> 0.

Proof. Induction on n:

n=1 obvious

n-1 we assume to be done

n "=⇒" A =

a1n

B......

an1 · · · · · · ann

where B is (n− 1)× (n− 1) matrix.

B represents the same bilinear form restricted to U =

∗...

0

=

α1

...αn

: αn = 0

in the standard basis of U .In particular: for all x ∈ U \ 0 : xTBx > 0. By induction assimption:a11 · · · a1j

......

aj1 · · · ajj

> 0

for j < n. Since there exists an orthogonal Q s.t.

QAQ−1 =

λ1

. . .λn

and A is positive definite ⇐⇒ all λi > 0, we conclude that det(A) = λ1 · · ·λn > 0.

"⇐=" By induction assumption B is positive definite, so (U, 〈x, y〉 = xTBy) is Eu-clidean. By Lemma ??:

Rn = U ⊕ U⊥︸︷︷︸under xTAy

Any x ∈ Rn can be written as x = y + z, y ∈ U, z ∈ U⊥ and

〈x, x〉 = 〈y + z, y + z〉 = 〈y, y〉+ 〈z, z〉

So it suffices to check that for z ∈ U⊥, z 6= 0 and 〈z, z〉 > 0.Let f1, . . . , fn−1 be an orthonormal basis of U , fn ∈ U⊥, fn 6= 0. xTAy in the basis

35

f1, . . . , fn will look like

C =

0

In...0

0 · · · 0 〈fn, fn〉

Hence A is positive definite ⇐⇒ C is positive definite ⇐⇒ 〈fn, fn〉 > 0.

2

Let Γ = (X,mab). We say that Γ′ = (Y, nab) is a subgraph of Γ if

• Y ⊂ X

• ∀ a, b ∈ X : nab ≤ mab

Lemma 7.3. If Γ′ is a Coxeter subgraph of Γ and Γ is positive definite then Γ′ is positivedefinite.

Proof.2 ≤ nab ≤ mab =⇒ − cos

π

nab≥ − cos

π

mab

. Pick any v =∑a∈Y

αaea ∈ V (Γ′) and suppose 〈v, v〉 ≤ 0. Let w =∑a∈Y|αa|ea. Then:

〈w,w〉V (Γ) =∑a,b∈Y

|αa| · |αb|(− cosπ

mab)

≤∑a,b∈Y

|αa| · |αb|(− cosπ

nab)

≤∑a,b∈Y

αaαb(− cosπ

nab)

≤ 0

Since V (Γ) is Euclidean, w = 0. Then all αa = 0 and v = 0. 2

Let d(Γ) = det(2 〈ea, en〉︸ ︷︷ ︸C(Γ)

)X×X = det(−2 cos πmab

)a,b∈X .

Note: d(Γ) is independent of the order on X one chooses to write a matrix.

Lemma 7.4. Let Γ be of the following form:

Γ1

n n-1Γ2

Then d(Γ) = 2d(Γ1)− d(Γ2).

36

Proof. Let |X| = n, call the two vertices n and n− 1. Then the matrix C(Γ) looks like∗ 0

C(Γ2)...

...∗ 0

∗ · · · ∗ 2 −10 · · · 0 −1 2

Expanding the last row:

d(Γ) = 2d(Γ1) + (−1)n+(n−1)(−1)det

0

C(Γ2)...0

∗ · · · ∗ −1

= 2d(Γ1)− d(Γ2)

2

Lemma 7.5. The following Coxeter graphs Γ have d(Γ) > 0:

An · · ·

Bn 4 · · ·

GGGGG

Dn · · ·

wwwww

E6

E7

E8

F4 4

H2 5

H3 5

H4 5

I2(m) m

Note: A2 = I2(3), B2 = I2(4), H2 = I2(5). Sometimes G2 = I2(6) is used.Proof. Let xn denote d(Xn).

• i2(m) = det

(2 −2 cos π

m−2 cos π

m 2

)= 4− 4(cos π

m)2 > 0

37

• a1 = det(2) = 2a2 = i2(3) = 4− 4(cos π3 )2 = 3

• Lemma 7.4 allows us to compute an by

· ·· ·

An · · ·An−2 ·

An−1 ·

Then an = 2(an−1)− an−2 and by induction an = n+ 1.

• b2 = i2(4) = 2· ·

· ·

B3 4

A1 ·

B2 ·

b3 = 2b2 − a1 = 2 · 2− 2 = 2And for Bn:

· ·· ·

Bn 4 · · ·

Bn−2 ·

Bn−1 ·

=⇒ bn = 2bn−1 − bn−2 =⇒ bn = 2

• Before computing Dn notice that:

d(Γ1∪Γ2) = d(Γ1) · d(Γ2)

The decomposition for Dn is:

00000· ·

· ·

00000 · · ·

A1∪An−3 ·

An−1 ·

Therefore dn = 2an−1 − a1an−3 = 2n− 2(n− 3 + 1) = 4.

38

• Similarly: en = 2an−1 − a2an−4 = 9− n.Hence en is positive for n ≤ 8.

• h2 = i2(5) = 4− 4 cos2 π5 = 5−

√5

2

• h3 = 2h2 − a1 = 25−√

52 = 3−

√5 > 0

• h4 = 3h3 − h2 = 2(3−√

5)− 5−√

52 = 7−3

√5

2 > 0

• Notice that h5 = 3h4 − h3 = 5− 2√

5 < 0.

2

Lemma 7.6. The following Coxeter graphs have d(Γ) < 0:

An

zzzz

DDDD

zzzz

DDDD

Bn · · · 4

zzzz

Cn 4 · · · 4

DDDD

zzzz

Bn · · ·

zzzz

DDDD

DDDD

E6

zzzz

DDDD

E7

zzzz

DDDD

E8

zzzz

39

F4 4

G2 6

H3 5

H4 = H5 5

Proof.

• We have computed e8 = e9 = 9− 9 = 0.

• h4 = h5 = 4− 2√

5 < 0

• an = det

2 −1 −1−1 2 −1

. . . . . . . . .. . . . . . −1

−1 −1 2

= 0 because the sum of the rows is zero

• bn = 2bn − a1bn−2 = 4− 4 = 0

• dn = 2dn − a1dn−2 = 2 · 4− 2 · 4 = 0

• e6 = 2e6 − a5 = 0

• e7 = 2e7 − d6 = 0

• f4 = 2f4 − b3 = 0

• g2 = 2i2(6)− a1 = 0

• h3 = 2h3 − a2 = 3− 2√

5 < 0

• c(Cn) looks like

∗ 0

c(Bn−1) ∗ 0∗ 0

∗ 2 −20 −2 2

.

Similarly to Lemma 7.4, expanding the last row gives

d(Cn) = 2d(Bn)− 2d(Bn−1) = 2 · 2− 2 · 2 = 0

2

40

Theorem 7.7. The connected Coxeter graphs Γ which give a finite W (Γ) are precisely thegraphs listed in Lemma 7.5.

Proof. The graphs on 7.5 are closed under taking connected Coxeter subgraphs. Therefore7.2 tells us that all c(Γ) are positive definite in 7.5 and 7.1 tells us that all W (Γ) are finite in7.5.Now let Γ be a connected graph with finite W (Γ). Then c(Γ) is positive definite by 7.1. Inparticular, Γ contains no subgraphs in 7.6

• No An =⇒ Γ has no cycles

• No D4 =⇒ Γ has no vertex with ≥ 4 edges

• No Dn, n ≥ 5 =⇒ Γ contains at most one vertex with 3 edges.

Case 1 No vertex has 3 edges, hence Γ looks like this:

x1 x2 · · · xm

If Γ has ≤ 2 vertices it is either A1 or I2(m), both in Lemma 7.5.Let Γ contain ≥ 3 vertices (m ≥ 2).

– No G3 =⇒ edges may have multiplicity at most 5

– No Cn =⇒ at most one edge has multplicity ≥ 4.

Case 1.1 No edge of multiplicity ≥ 4 =⇒ Γ = An

Case 1.2 There is one edge of multiplicity 4:

– if it is on a side, Γ = Bn

– if it is in the middle then no F4 =⇒ Γ = F4

Case 1.3 There is one edge of multiplicity 5:No H3 =⇒ the multiple edge is on a side and no H4 =⇒ Γ = H3 or H4

Case 2 there is a vertex with 3 edges. Then Γ look like this:

· ·

Let b denote the vertices going up, a the vertices going left and c the vertices goingright. Then we have a+ b+ c+ 1 vertices in total and wlog a ≥ b ≥ c.– No Bn =⇒ Γ has no multiple edges.

– No E6 =⇒ c = 1

– No E7 =⇒ b ≤ 2

– if b = c = 1 =⇒ Γ = Da+3

– if b = 2, c = 1, then no E8 =⇒ a < 5. As a ≥ b = 2 there are 3 cases:

1. a = 2 =⇒ Γ = E6

41

2. a = 3 =⇒ Γ = E7

3. a = 4 =⇒ Γ = E8

2

crystallographic NON crystallographicAn, Bn, Dn, F4, En, G2 = I2(6) I2(m) for m = 5, m ≥ 7, H3, H4

d(Γ) ∈ Z d(Γ) /∈ Zthe character table of Γ is integral non-integralall but one representation of W (Γ)can be realised by integral matrices

42

8 Crystallographic Coxeter groups

Let V be a vector space over R.A lattice in V is an abelian subgroup L of V s.t. L = 〈e1, . . . , en〉 where e1, . . . , en is a basis ofV .Informally, L is Zn inside Rn.

Definition. W (Γ) is crystallographic if there is a lattice L ⊂ V (Γ) s.t.

∀ g ∈W (Γ) : g(L) ⊂ L

Informally this means that there is a basis e1, . . . , en of V (Γ) s.t. ρ(g) is an integer matrix foreach g ∈W (Γ).

Lemma 8.1. Let |X| <∞. If W (Γ) is crystallographic then ∀ a, b ∈ X :

mab ∈ 2, 3, 4, 6,∞

Proof. Assume W (Γ) is crystallographic. Then ∀ g ∈ W : ρ(g) can be written as an integermatrix, in particular Tr(ρ(Γ)) ∈ Z.If mab =∞ then we are done.Suppose mab =∞ =⇒ U = span(ea, eb) is Euclidean =⇒W (Γ) = U ⊕ U⊥.In the basis ea, eb, any basis of U⊥:

ρ(SaSb) =

cos 2πmab

± sin 2πmab

0

∓ sin 2πmab

cos 2πmab

0

0 0 I

Tr(ρ(SaSb)) = |X| − 2 + 2 cos 2π

mabhence

2 cos2π

mab∈ Z ⇐⇒ cos

2π

mab∈ 1

2Z ⇐⇒ mab ∈ 2, 3, 4, 6

2

Note: ⇐= is true but the proof is harder unless W (Γ) is finite.

Theorem 8.2. LetW (Γ) be a finite Coxeter group. ThenW (Γ) is crystallographic ⇐⇒ ∀ a, b ∈X : mab ∈ 2, 3, 4, 6.

Proof.

"=⇒" Lemma 8.1

43

"⇐=" Let us choose a new basis αa = caea, a ∈ X, ca ∈ R \ 0. Then

ρa(αb) = αb − 2 〈αb, ea〉 ea = αb − 2〈αa, αb〉〈αa, αa〉

αa

Since ρa = Sea = Sαa and ρ(a1 · · · an) = ρa1 · · · ρan it suffices to ensure that for alla, b ∈ X :

2〈αa, αb〉〈αa, αa〉

∈ Z

Wlog Γ is connected.

• mab = 2 =⇒ 〈αa, αb〉 = 0 =⇒ 2 〈αa,αb〉〈αa,αa〉 = 0

• mab = 3 =⇒ 〈αa, αb〉 = cacb 〈ea, eb〉 = caca− cosπ

mab︸ ︷︷ ︸=− 1

2

= − cacb2

=⇒ 2〈αa, αb〉〈αa, αa〉

=−cacbcaca

= − cbca

Need in this case ca = ±cb• mab = 4 =⇒ 〈αa, αb〉 = cacb(− cos π4 ) = − cacb√

2

=⇒ 2〈αa, αb〉〈αa, αa〉

= −√

2cbca

Need ca =√

2cb or cb =√

2ca

• mab = 6 =⇒ 〈αa, αb〉 = −√

32 cacb

=⇒ 2〈αa, αb〉〈αa, αa〉

= 2−cacb

√3

2

caca= −√

3cbca

Need ca = ±√

3cb or cb = ±√

3ca

In each component of Γ we choose

Case 1 An, Dn, En: choose all ca = 1.

Case 2 Bn, F4

Γ1︸︷︷︸choose ca=1

4 Γ2︸︷︷︸choose ca=

√2

Case 3 In G2 = I2(6):a 6

b

αa =√

3ea, αb = eb

2

Corollary 8.3. The finite crystallographic reflection groups are W (Γ) where Γ is either

44

• An, n ≥ 1

• Bn, n ≥ 2

• Dn, n ≥ 4

• En, 6 ≤ n ≤ 8

• F4

• G2

Note: In Bn, n ≥ 3 there are 2 non-isomorphic choices of crystallographic structure

14

2 3 · · · n

• α1 = e1, α2 =√

2e2, . . . , αk =√

2ek which is called the crystallographic reflection groupBn

• α1 =√

2e1, α2 = e2, . . . , αk = ek which is called the crystallographic reflection group Cn

semisimple Lie algebra

Weyl group and Root lattice--

crystallographic reflection groupSerre Theorem

nn

Kac−Moody algebra

ii

Nakajima construction

12

bb

We have seen An, Bn, Dn, I2(m) explicitely.Can we compute |W (Γ)|, |Φ(Γ)| for all Γ?

45

9 Polynomial invariants

Let G be a finite group acting on VR, dim V < ∞. Let S = R[V ], algebra of all polynomialfunctions on V . Let e1, . . . , en be a basis of V and Xi : V −→ R the linear maps which formthe dual basis. Then Xi(ej) = δij . In fact, S = R[X1, . . . , Xn].G acts on the algebra S: g ∈ G,F ∈ S, v ∈ V : g · F (v) = F (g−1v).Exercise: check that G acts on S by algebra automorphisms.Note: for F ∈ S the following are equivalent:

1. ∀ g ∈ G : gF = F

2. F is constant on G-orbits in V .

We call such functions invariant and they form a subalgebra SG.We will prove that SG ∼= R[z1, . . . , zk] (when G is a reflection group). In fact

SG ∼= R[z1, . . . , zk] ⇐⇒ G is a reflection group

(no proof)

S is a graded algebra, i.e. S =∞⊕k=0

Sk where Sk consists of polynomials where each monomial

has degree k.Example:

• 2010 ∈ S0

• x21 + x2

2 + x1x3 ∈ S2

• 1 + x1 ∈ S0 ⊕ S1 but not in any Si

If F ∈ Si we say deg F = i.

Lemma 9.1. Let F ∈ SG, F =m∑j=0

Fj , Fj ∈ Sj. Then ∀ j: Fj ∈ SG.

Proof. G× R∗ acts on V : (g, α)v = g(αv) = αg(v). Hence G× R∗ also acts on S:

(g, α) · F : v 7−→ F (g−1α−1v)

In particular, f ∈ Sj : (1, α) · f = α−j · f .Since actions of G and R∗ commutes, ∀α ∈ R∗ :

(1, α)F =

m∑j=0

α−jFj ∈ SG

1, . . . ,m+ 1 : m+ 1 different real numbers

46

• 1 · F = F0 + . . .+ Fm+1 ∈ SG

• 2 · F = F1 + 12F1 + . . .+ 1

2m+1Fm+1 ∈ SG

• . . .

• (m+ 1) · F = F0 + 1m+1F1 + . . .+ 1

(m+1)m+1Fm+1 ∈ SG

If M =

1 · · · 112 · · ·

(12

)m+1

......

1m+1 · · ·

(1

m+1

)m+1

∈ R(m+1)×(m+1) is a VanDerMonde matrix then

M ·

F0...Fm

︸ ︷︷ ︸∈Sm+1

∈(SG)m+1

Since M is invertible,

F0...Fm

= M−1

H0...

HM

where Hi ∈ SG.

Hence all Fi ∈ SG. 2

Lemma 9.1 means that SG is a graded subalgebra of S, i.e.

SG =

∞⊕k=0

(SG ∩ Sk) and SGk = SG ∩ Sk

Let SG+ =∞⊕k=1

SGk be the invariants of positive degree. Let I =(SG+)be the ideal of S generated

by SG+ . By Hilbert’s Basis Theorem, any ideal in S is generated by finitely many elements.Say I = (f1, . . . , fk)S , fi ∈ SG. Write fi =

∑jfij where fij ∈ Sj . With Lemma 9.1:

fij ∈ SGj , j < 0.Hence I = (I1, . . . , Ir)S where Ij ∈ SGdj . If for any k: Ik ∈ (I1, . . . , Ik−1, Ik+1, . . . , Ir)S we throwIk away. Wlog, I1, . . . , Ir is a minimal system of such generators.

Definition. I1, . . . , Ir are called fundamental invariants if they are a minimal system ofhomogeneous invariant generators of (SG+)S .

• Coinvariants: SG = S/I

• Integral:∫G

: S −→ SG ∫G

F =1

|G|∑g∈G

gF

47

Note that F ∈ SG =⇒ gF = F =⇒∫G

F = F and ∀ H ∈ S :

∫G

(FH) =1

|G|∑g

(gF )(gH) = F1

|G|∑g

gH = F

∫G

H

So∫G

is an SG-linear map S −→ SG.

Proposition 9.2. Let G be any finite group acting on V , I1, . . . , Ir ∈ SG = R[V ]G fundamentalinvariants. Then:

∀F ∈ SG ∃ p(z1, . . . , zr) ∈ R[z1, . . . , zr] s.t. F = p(I1, . . . , Ir)

Proof. Lemma 9.1: we may restrict to homogeneous F . We do induction on d = deg(F ).

d = 0: =⇒ F = const =⇒ p = const · 1 does the job

d < N : is done

d = N > 0: F ∈ SG =⇒ F ∈ (SG)S =⇒ F =r∑j=1

HjIj , Hj ∈ S. Since d = deg(F ), dj =

deg(Ij):

F =r∑j=1

(Hj)d−djIj

and therefore

F =

∫G

F =

r∑j=1

Ij

∫G

(Hj)d−dj

By induction assumption ∀ j ∃ pj ∈ R[z1, . . . , zr] s.t.∫G

(Hj)d−dj = pj(I1, . . . , Ir). Hence

p =

r∑j=1

zjpj

does the job.

2

So for any real representation V of any finite group G the algebra map

R[z1, . . . , zr] −→ SG, zi 7−→ Ii

is surjective.

1. Injectivity holds only for reflection groups

2. Hilbert’s 14-th problem is that SG is a finitely generated algebra for any group

3. SG0 = R · 1 constant functions

48

4. f ∈ SG1 , f 6= 0 =⇒ V = Ker(f)⊕ R. Hence SG1 6= 0 ⇐⇒ GV has a trivial constituent

5. if 〈 , 〉 is a Euclidean form on V then [x, y] =∫G

〈x, y〉 makes G orthogonal. Hence

F (x) = [x, x] defined F ∈ SG2 .

Lemma 9.3. Let P ∈ S, L ∈ S1 \ 0, if ∀ x ∈ V : L(x) = 0 =⇒ P (x) = 0 then L|P .

Proof. S = R[x1, . . . , xn]. Wlog L = αxn + b(x1, . . . , xn−1) with α 6= 0. Let us divide P byL with remainder by replacing every xn with − 1

αb(x1, . . . , xn−1). xn = 1αL−

1αb(x1, . . . , xn−1).

Example: L = x1 − x2

P = x21 = x1 · x1 = x1(L+ x2)

= x1L+ x1x2 = x1L+ (L+ x2)x2

= (x1 + x2)L+ x22

P = AL + B, A ∈ R[x1, . . . , xn], B ∈ R[x1, . . . , xn−1]. If B = 0 we are done. If B 6= 0 pick

a1, . . . , an−1 ∈ R s.t. B(a1, . . . , an−1) 6= 0.

Let an = − 1αb(a1, . . . , an−1) =⇒ L(a1, . . . , an) = 0 =⇒ P (a1, . . . , an) = 0.

But P (a1, . . . , an) = B(a1, . . . , an−1) 6= 0 which is a contradiction. 2

Proposition 9.4. Let G = W be a finite reflection group. Let J1, . . . , Jk ∈ SW s.t.

J1 /∈ (J2, . . . , Jk)SW . If Pi ∈ Sbi s.t.k∑i=1

PiJi = 0 then P1 ∈ (SW+ )S.

Proof. Observe that J1 /∈ (J2, . . . , Jk)S : if J1 =k∑t=2

HtJt with Ht ∈ S then

J1 =

∫J1 =

∑t

Jt

∫Ht ∈ (J2, . . . , Jk)SW

Proceed by induction on deg(P1):

deg(P1) = 0: =⇒ P1 = c · 1 =⇒ c · J1 =k∑t=2

(−Pt)Jt =⇒ c = 0 because J1 /∈ (J2, . . . , Jk)S =⇒

P1 = 0 ∈ (SW+ )S

deg(P1) < N : done

deg(P1) = N : Let Φ ⊂ W be the root system of G. ∀α ∈ V let Lα ∈ S1 be Lα(v) = 〈α, v〉.Lα = 0 =⇒ α⊥v =⇒ Sα(v) = v

=⇒ ∀ j : (SαPj − Pj)(v) = Pj(S−1α v)− Pj(v) = 0

With Lemma 9.3: SαPj − Pj = LαP j for some P j ∈ S.Note that deg(P j) < deg(Pj)

• P1J1 + . . .+ PkJk = 0

• (SαP1)J1 + . . .+ (SαPk)Jk = 0

49

• (P1 − SαP1)J1 + . . .+ (Pk − SαPk)Jk = 0

Hence Lα(P 1J1+. . .+P kJk) = 0. Since S is a domain P 1J1+. . . P kJk = 0. By inductionassumption, P 1 ∈ (SW+ )S . Hence SαP1 − P1 = LαP 1 ∈ (SW+ )S .For each polynomial F , denote F = F + (SW+ )W ∈ SW . Hence ∀ α ∈ Φ : SαP1 = P1.Since W is generated by Sα, ∀ g ∈W : gP1 = P1. Hence

∫W

P1 = P1 (W acts on SW and∫F =

∫F )

So P1 ∈∫W

P1 + (SW+ )S ⊂ (SW+ )S .

2

Lemma 9.5. (Euler’s formula)F ∈ Sk = R[x1, . . . , xn]k, then

n∑j=1

xj∂F

∂xj= kF

Proof. Both sides are linear in F , hence it suffices to verify the formula on monomials. LetF = xa11 · · ·xann ,

∑ai = k. Then

xj∂F

∂xj= ajF

Hencen∑j=1

∂F∂xj

= (∑ai)F = kF . 2

Proposition 9.6. I1, . . . , Ir fundamental invariants for a (finite) reflection group (W,V ). IfP (I1, . . . , Ir) = 0 for some P ∈ R[z1, . . . , zr] then P = 0.

Note: I1, . . . , Ir are algebraically independent.Proof. Consider a grading on R[z1, . . . , zr] with deg zi = di = deg Ii. Let P =

∑Pj , Pj

homogeneous of degree j. Then

P (I1, . . . , Ir) =∑j

Pj(I1, . . . , Ir)︸ ︷︷ ︸∈Sj

= 0

Hence Pj(I1, . . . , Ir) = 0. WLOG, we assume P is homogeneous.Let P(j) = ∂P

∂zjand fj = P(j)(I1, . . . , Ir) ∈ SW is homogeneous. Let us fiddle around with the

idealK = (f1, . . . , fr)SW / SW

WLOG, f1, . . . , fm is a minimal system of generators of K, m ≤ r. Note that deg fi =deg P − deg Ii and fi is homogeneous. If i ≥ m + 1: fi ∈ K = (f1, . . . , fm)SW so there are

Qij ∈ SW , Qij homogeneous of degree deg Ii − deg Ij s.t. fi =m∑j=1

Qijfj .

P (I1, . . . , Ir) = 0 =⇒ 0 =∂P (I1, . . . , Ir)

∂xk=

r∑i=1

∂P

∂zi(I1, . . . , Ir)

∂Ii∂xk

∀ k ∈ 1, . . . , n

50

Hence:

∀ k :m∑i=1

fi∂Ii∂xk

+r∑

i=m+1

m∑j=1

Qijfj∂Ii∂xk

= 0

∀ k :m∑i=1

fi

∂Ii∂xk

+r∑

j=m+1

Qji∂Ij∂xk

= 0

By 9.4, ∀ k ∀ i ∈ 1, . . . ,m :∂II∂xk

+r∑

j=m+1

Qji∂Ij∂xk

∈ (SW+ )S

Hence for some Hkt ∈ S :

∀ i, k :∂II∂xk

+r∑

j=m+1

Qji∂Ij∂xk

=r∑t=1

ItHkt

Multiply each equality by xk and sum over k. By Euler’s formula:

∀ i : diIi +

r∑j=m+1

djQjiIj =

r∑t=1

ItHt

where Ht have no free terms (Ht =∑xkH

kt ). Ii enters both sides of the equality. If we consider

the homogeneous degree di parts of both sides, then

diIi +

r∑j=m+1

cjIj =∑t

Itbt

where bi has to be zero because Hi has no free terms. Therefore

Ii =∑j 6=i

IjGj =⇒ Ii ∈ (I1, . . . , Ii−1, Ii+1, . . . , Ir)

which is a contradiction. Therefore all ∂P∂xk

= 0 =⇒ P = 0. 2

Example: Sn acting on Rn by permutations.

Rn = U ⊕ R where U =

α1

...αn

:∑

αi = 0

and R = R ·

(1...1

)

(Sn, U) is then a type An−1 essential reflection group with elementary symmetric functions:

• σ1(x1, . . . , xn) = x1 + . . .+ xn

• σ2(x1, . . . , xn) = x1x2 + x1x3 + . . .+ xn−1xn

• . . .

• σn(x1, . . . , xn) = x1 · · ·xn

51

is a system of fundamental invariants. The power functions πj = xj1 + . . .+ xjn, j = 1, . . . , n isanother fundamental system.

An−1: σ1|U = 0σ2|U , . . . , σn|U are fundamental invariants with fundamental degrees 2, 3, . . . , n.

Bn: group has more transformations xi 7−→ ±xi, SW (Bn) = (SSn)Cn2 = R[σ1, . . . , σn]C

n2 .

Hence the fundamental invariants are σj(x21, . . . , x

2n) with degrees 2 · 1, 2 · 2, . . . , 2 · n.

Dn: inDn, instead of the full group Cn2 , it is a subgroup of index 2. So σn = x1 · · ·xn ∈ SW (Dn).So the fundamental invariants are σ1(x2

i ), . . . , σn−1(x2i ), σn(xi) with degrees 2 · 1, 2 ·

2, . . . , 2 · (n− 1), n.

Theorem 9.7. Let (G,V ) be a finite reflection group, n = dim V, S = R[V ], I1, . . . , Irfundamental invariants. Then

SG = R[I1, . . . , Ir] and r = n

Proof. The natural algebra homomorphism

ψ : R[z1, . . . , zr] −→ SG, F (z1, . . . , zr) 7−→ F (I1, . . . , Ir)

is surjective by 9.2 and injective by 9.6. Hence ψ is an isomorphism. It remains to show thatn = r.Consider the field of rational functions

F = R(x1, . . . , xn) = fg

: f, g ∈ S, g 6= 0

Transcendence degree is trdegR F = n. The smaller field A = R(I1, . . . , Ir) ≤ F, trdegRA = r.From Galois theory, trdegA F = trdegR F−trdegRA = n−r. It suffices to show that trdegA F =0, i.e. every element of F is algebraic over A.Pick f ∈ F, G acts on F with FG = A. Consider h(z) ∈ F [z] defined by

h(z) =∏g∈G

(z − g.f)

Since ∀ t ∈ G, t.h =∏g∈G

(z − tg.f) = h =⇒ h ∈ A[z].

h(f) =∏g∈G

(f − g.f) = (f − f)∏g 6=1

(f − g.f) = 0

Hence f is algebraic over A. 2

Let d1, . . . , dn with di = deg Ii be the fundamental degrees.

Theorem 9.8. If I ′1, . . . , I′n is another system of fundamental invariants with degrees d′1, . . . , d

′n, d

′j =

deg I ′j then ∃ σ ∈ Sn s.t.∀ j : dj = d′σ(j)

52

Proof. By 9.7 there are F1, . . . , Fn, G1, . . . , Gn ∈ R[z1, . . . , zn] s.t.

Ij = Fj(I′1, . . . , I

′n) and I ′j = Gj(I1, . . . , In) ∀ j

Then ∀ j : Ij = Fj(G1(I1, . . . , In), . . . , Gn(I1, . . . , In)). Hence, since I1, . . . , In are algebraicallyindependent:

zj = Fj(G1(z1, . . . , zn), . . . , Gn(z1, . . . , zn))

and therefore

∂zj∂zk

=n∑s=1

∂Fj∂zs

(G1(z1, . . . , zn), . . . , Gn(z1, . . . , zn)) · ∂Gs∂zk

(z1, . . . , zn)

And on matrix level:

In =

(∂Fi∂zs

)I′1,...,I

′n

·(∂Gs∂zk

)I1,...,In

∈Matn(S)

Hence det(∂Fi∂zs

)is invertible in S = R[x1, . . . , xn]. Hence ∃ σ ∈ Sn s.t.

∂F1

∂zσ(1)· · · ∂Fn

∂zσ(n)= c · 1 + higher degree terms, c 6= 0

Since deg I ′i = deg zi = d′i, degFi = deg Ii = di:

∀ i : di = d′σ(i)

2

53

10 Fundamental degrees

Discuss the role of d1, . . . , dn.

Lemma 10.1. A finite group G acts in a finite dimensional vectorspace over C(by ρ : G −→ GLn(C)). Then ∀ x ∈ G : ρ(x) is diagonalisable with eigenvalues of absoulutevalue 1.

Proof. |G| = n <∞ =⇒ xn = 1 =⇒ ρ(x)n = In.

Hence the minimal poynomial of ρ(x), µx(z) divides zn − 1. Therefore eigenvalues λ satisfy

λn − 1 = 0, hence λn = 1 =⇒ |λ|n = 1 =⇒ |λ| = 1.

Further, since zn− 1 has no multiple rootes, µx(z) has no multiple roots and all Jordan blocks

of ρ(x) have size 1. 2

Let C[[t]] = ∞∑n=0

αnzn be the ring of formal power series. If V =

∞⊕n=0

Vn is a graded vector

space with dimVn <∞, the "right" notion of dimension of V is the Poincaré polynomial

pV (t) =∞∑n=0

dimVn · tn

Example:

• pC[z](t) = 1 + td + t2d + t3d + . . . = 11−td where deg z = d.

• pV⊕W (t) = pV (t) + pW (t)

• pV⊗W (t) = pV (t)pW (t)

• pC[z1,...,zn] =∏jpC[zj ](t) = 1

1−td1 · · ·1

1−tdn where deg zi = di, C[z1, . . . , zn] = C[z1]⊗ . . .⊗

C[zn].If d1 = . . . = dn = 1 then pC[z1,...,zn](t) = 1

(1−t)n .

Example: If G acts on V , then ∀ x ∈ G let

χx(t) = det(IV−tρ(x))−1 =∏j

1

1− λjt=∏j

(1+λjt+λ2j t

2+. . .) =∞∑j=0

tj(∑

a1+...+an=j

λa11 · · ·λann )

where λ1, . . . , λn are the eigenvalues of ρ(x).

Theorem 10.2. Let (G,V ) be a finite reflection group, d1, . . . , dn the fundamental degrees.Then

1

|G|∑x∈G

χx(t) =

n∏j=1

1

1− tdj∈ C[[t]]

54

Proof. By 9.7 and the Example before, the RHS is pSG(t).

x|S1 belongs to

λ1 0. . .

0 λn

in some basis e1, . . . , en of S1. Then ea11 · · · eann with

a1 + . . .+ an = j form a basis of Sj and

xea11 · · · eann = λa11 · · ·λ

ann e

a11 · · · e

ann

hencetr(x|Sj ) =

∑a1+...+an=j

λa11 · · ·λann

Hence

LHS =1

|G|

∞∑x∈G,j=0

tr(x|Sj )tj

Remember πj =∫G

: Sj −→ SGj , πj(v) = 1|G|∑g∈G

gv. In particular LHS =∞∑j=0

tr(πj)tj .

π2j = πj =⇒ πj ∼

(I 00 0

)and trπj = rkπj = dimSGj

Hence LHS = pSG(t). 2

Corollary 10.3. |G| = d1 · · · dn.

Corollary 10.4. d1 + . . .+ dn = n+1

2|Φ(G)|︸ ︷︷ ︸

number of reflections

.

Proof. Multiply 10.2 by (1− t)n:

1

|G|∑x∈G

χx(t)(1− t)n =n∏j=1

1− t1− tdj

1

|G|∑x∈G

(1− t)n

det(1− tg)=

n∏j=1

1

1 + t+ t2 + . . .+ tdj−1︸ ︷︷ ︸=:α(t)

=1

|G|( 1︸︷︷︸g=1

+1

2|Φ|1− t

1 + t︸ ︷︷ ︸g∈Sx, x∈Φ

+ h(t) · (1− t)2︸ ︷︷ ︸others have 1 as eigenvalue of mult.≤2

)

[h(1) is defined ]

Plug in t = 1, then 1|G| =

n∏j=1

1dj

=⇒ |G| = d1 · · · dnApply ∂

∂t :

|Φ|2|G|

−2

(1 + t)2+ h(t)(1−t) =

n∑j=1

α(t)(1+t+ . . .+tdj−1)

(−1

1 + 2t+ 3t2 + . . .+ (dj − 1)tdj−2

(1 + t+ . . .+ tdj−1)2

)

55

Plug in t = 1:

− |Φ|4|G|

=

n∑j=1

djd1 · · · dn

(−(dj − 1)dj

2d2j

)As |G| = d1 · · · dn

|Φ|4

=

n∑j=1

dj − 1

2=

1

2((∑

dj)− n)

2

Example:

• I2(m) has fundamental degrees 2,m.

|G| = 2m, |Φ| = 2((m+ 2)− 2) = 2m

• An has fundamental degrees 2, 3, . . . , n+ 1

|G| = (n+1)!, |Φ| = 2((2+3+. . .+n+1)−n) = 2

(n(n+ 3)

2− n

)= n2+3n−2n = n2+n

56

11 Coxeter elements

Let (G,V ) be a finite reflection group, Φ ⊂ V a root system.A Coxeter element is w = Sα1 · · ·Sαn for some simple system Π = α1, . . . , αn.There are two choices involved:

• simple system

• order on Π

In Sn the choices will lead to

Sα1 = (a1a2), Sα2 = (a2a3) . . . Sαn−1 = (an−1an)

where a1, . . . , an = 1, . . . , n. Hence w = (a1 . . . an). In Sn, Coxeter elements are cyclesof order n.

Lemma 11.1. Let G a group, X a finite forest (graph without cycles). Suppose f : V er(X) −→G is a function s.t. if two vertices a, b ∈ V er(X) are not connected, then

f(a)f(b) = f(b)f(a)

Then for any choice of a total order on the vertices of X, say V er(X) = a1, . . . , an, theconjugacy class ClG(g) of g = f(a1) · · · f(an) does not depend on the order chosen.

Proof. Induction on n = |V er(X)|.

n=1: g = f(a1), nothing to prove

n<k proved

n=k Easy case: X has no edges, then all f(a)f(b) = f(b)f(a), so g = f(a1) · · · f(an) isindependent of the choice of the order.If X has an edge, there is a vertex b connected with exactly one other vertex a. For eachfunction ψ : 1, . . . , n −→ V er(X), let gψ = f(ψ(1)) · · · f(ψ(n)).Pick a particular π : 1, . . . , −→ V er(X) s.t. π(n) = b, π(n− 1) = a. It suffices to finda s ∈ G s.t. gπ = sgψs

−1.

Case 1 ψ(n) = n, ψ(n− 1) = a.

Let Y = X \ b and f ′ : Y −→ G be f ′(c) =

f(a)f(b) c = a

f(c) c 6= a. Then

gψ = f(ψ(1)) · · · f(ψ(n− 2))f(a)f(b)

= f ′(ψ(1)) · · · f ′(ψ(n− 1))

By induction assumption it is conjugate to f ′(π(1)) · · · f ′(π(n− 1)) = gπ.

57

Case 2 ψ(n) = b, ψ(j) = a, j < n− 1.Let X0 = ψ(k) : k < j, X1 = ψ(k) : j < k < n. Then define

ψi : ψ−1(Xi) −→ Xi, ψi(c) = ψ(c)

Then

gψ = gψ0f(a)gψ1f(b)

= gψ0f(a)f(b)gψ1

∼ gψ1(gψ0f(a)f(b)gψ1)g−1ψ1

= gψ1gψ0f(a)f(b)

This is conjugate to gπ by Case 1.

Case 3 ψ(j) = b, j < n, X0, X1, ψ0, ψ1 as above. Then

gψ = gψ0f(b)gψ1 ∼ gψ1gψ0f(b)

done by Case 2.

2

Proposition 11.2. Coxeter elements form a conjugancy class in G.

Proof. Let C = Coxeter elements . Then

xSα1 · · ·Sαnx−1 = (xSα1x−1) · · · (xSαnx−1) = Sx(α1) · · ·Sx(αn) ∈ C

Hence xCx−1 = C, so C is the union of some conjugacy classes. Let w1 = Sα1 · · ·Sαn , w2 =

Sβ1 · · ·Sβn ∈ C. Then there is a g ∈ G s.t. g(α1), . . . , g(αn) = β1, . . . , βn. Hence gw1g−1 =

Sg(α1) · · ·Sg(αn) is conjugate to w2 by Lemma 11.1. 2

Let W be a finite Coxeter group, w = a1 · · · an a Coxeter element, a1, . . . , an the vertices ofthe Coxeter graph. Then h = |w| is called the Coxeter number of W .Let λ1, . . . , λn be the eigenvalues of w|V (W ). Since wh = 1, λhk = 1 =⇒ λk = e

2πimkh for a

unique 0 ≤ mk < h.We call m1, . . . ,mn the exponents of W .

Finite Coxeter graphs are bipartite (X = X1 ∪X2) Example:

• D6:

@@@@@@@

~~~~~~~

@@@@@@@

58

• E8:

~~~~~~~

@@@@@@@

@@@@@@@

For a, b ∈ Xi =⇒ ab = ba. Hence ωj =∏a∈Xj

a does not depend on the order of Xj . Moreover,

ω2j =

∏a2 = 1.

ω = ω1ω2 is called canonical Coxeter element.G = 〈ω1, ω2〉 is a dihedral group of order 2h.

Proposition 11.3. Let us order the exponents 0 ≤ m1 ≤ . . . ≤ mn < h. Then

1. m1 ≥ 1

2. ∀ j : h−mj is an exponent

3.n∑j=1

mj = hn2

Proof.

1. if m1 = 0 then e0 = 1 is a eigenvalue of ω. Hence ∃ x ∈ V (W ) s.t. x 6= 0, ωx = x.

ω2x = ω1ω1ω2x = ω1ωx = ω1x

Let a ∈ Xj : Sa(c) = a · x = x− 2 〈ea, x〉 ea. Hence

ω1x = x+∑a∈X1

αaeq = ω2x = x+∑b∈X2

αbeb

=⇒∑a∈X1

αaea =∑b∈X2

αbeb

and therefore all αa, αb = 0. So ω1x = ω2x = 0. Consider

ω1|V (W )aj∈X1

= Sa1 · · ·Sak :

V ⊥1 3 y 7−→ y

V1 3 y 7−→ −y

where Vi =⊕a∈Xj

Rea.

Hence x ∈ V ⊥1 ∩ V ⊥2 and V ⊥1 ∩ V ⊥2 = 0 because V = V1 ⊕ V2 = V ⊥1 ⊕ V ⊥2 since the formis non-degenerate. But this is a contradiction (x = 0) proving that m1 > 0.

59

2. G acts on V (W ). Let f(z) be the characteristic polynomial of ω|V (W ). Then

f(z) = f1(z) · · · fs(z)

where fj(z) ∈ R[z] and monic irreducible in R[z].

Case 1 fj(z) = z − λ, λ ∈ R.Since ωh = 1 =⇒ λh = 1 =⇒ λ = ±1. Note that 1 is not an eigenvalue of ω bypart (1), so λ = −1, h is even and λ = e

2πih

h2 , h

2 is the exponent.

Case 2 fj(z) = (z − λ)(z − λ∗), λ /∈ R.Then if λ = e

2πihmk then λ∗ = e−

2πihmk = e

2πih

(h−mk).

3. This also gives part (3) because the average of exponents in each fj(z) is h2 .

2

Lemma 11.4. Let W be finite and connected. Then there are z1, z2 ∈ V s.t.

1. z1, z2 are linearly independent

2. P = Rz1 ⊕ Rz2 is G = D2h-linear

3. ωi|P = Sz⊥i

4. zi ∈ C, the closure of the fundamental chamber

5. P ∩ C = R>0z1 + R>0z2

Proof. ωai is the dual basis to eai . That means⟨ωai , eaj

⟩= δij .

Easy claims:

• C = 〈x : 〈x, ea〉 ≥ 0〉 =∑aR≥0ωa

• V ⊥j =⊕a/∈Xj

Rωa

• angle between ωa and ωb = π− angle between ea and ebThe angles between ωa and ωb are accute and 〈ωa, ωb〉 ≥ 0 and 〈ωa, ωb〉 = 0 ⇐⇒ 〈ea, eb〉 =0 ⇐⇒ a and b commute.

Q = (〈ea, eb〉)a,b is the matrix of

• symmetric positiv definite bilinear form 〈 , 〉 in the basis eai

• linear map q : V −→ V, q(ωa) = ea in the basis ωai .

In particular, its eigenvalues are positive and real. Let λ be the largest eigenvalue ofQ. Consider

the new form [ , ] given by Q− λI in the basis eai .

Note: [x, x] ≤ 0 and [x, x] = 0 ⇐⇒ x ∈ ker[ , ] = K. K 6= 0, K = eigenspace of q.

Pick z 6= 0, z =∑aαaωa ∈ K and let z =

∑|αa|ωa.

Now, the idea is to use that zj =∑a∈Xj

|αa|ωa, but we have no time to finish the proof. 2

60

Theorem 11.5. If W is finite connected of rank n (i.e. the number of vertices in the Coxetergraph is n), then

1. m1 = 1, mn = h− 1

2. |Φ| = hn

Proof. (Idea)α ∈ Φ −→ Hα ∩ P is G-conjugate to Rz1 or Rz2.

• if h is odd:

Rz1

Rz2

There are n2 hyperplanes Hα s.t. Hα ∩ P is one of these h lines

• if h is even:

Rz1

Rz2

There are h2 lines with one intersection pattern and h

2 lines with another intersectionpattern

2

Theorem 11.6. Let W be finite and connected of rank n, 1 ≤ m1 ≤ . . . ≤ mn = h − 1 thefundamental exponents, 2 ≤ d1 ≤ . . . ≤ dn the fundamental degrees. Then:

∀ j : dj = mj + 1

Proof. (idea)

Compute Jac = det(∂Ij∂zk

)in two different ways. 2

Corollary 11.7. |W | =∏j

(mj + 1), |Φ| = 2∑jmj.

61

Corollary 11.8. Let W be connected.

−1 =

−1. . .

−1

∈ ρ(W ) ⇐⇒ all mi are odd

Proof.

"=⇒" (−1) Ij = Ij ∈ SW . Therefore (−1) · xk = xk =⇒ (−1)Ij = (−1)djIj =⇒ (−1)dj = 1.So all dj are even and therefore all mj are odd.

"⇐=" m1 = 1, mn = h− 1 are both odd =⇒ h is even

(ω|V (W ))h2 ∼

e2πihmk

. . .e

2πihmk

h2

=

eπimk

. . .eπimk

=

−1. . .

−1

2

Proposition 11.9. Let W be finite, connected, crystallographic. Then ∀ m s.t. 1 ≤ m ≤h− 1, gcd(m,h) = 1 then m is an exponent.

Proof. Cyclotomic polynomial

Φh(z) =∏

1≤m≤h−1

gcdm,h=1

(z − e2πihm)

is irreducible in Z[z].The characteristic polynomial fω(z) of ω|V (W ) is in Z[z].For m = 1: (z − e

2πih )|fω(z) in C[z]. Therefore

Φh(z)|fω(z) in Z[z]

and therefore all e2πihm are eigenvalues. 2

Application: E8 has rank 8 and 240 roots. Hence h = |Φ|n = 240

8 = 30. The number of coprimesof 30 is

ϕ(30) = ϕ(2)ϕ(5)ϕ(3) = 1 · 4 · 2 = 8

Hence the exponents are 1, 7, 11, 13, 17, 19, 23, 29 and so

|W | = 2 · 8 · 12 · 14 · 18 · 20 · 24 · 30

|Φ| h = |Φ|n ”easy exponents” missing exponents |W |

E6 72 12 1, 5, 7, 11 4, 8E7 126 18 1, 5, 7, 11, 13, 17 9E8 240 30 1, 7, 11, 13, 17, 19, 23, 29 noneF4 48 12 1, 5, 7, 11 none 2 · 6 · 8 · 12 = 27 · 32

H3 30 10 1, 9 5 2 · 6 · 10 = 120H4 120 30 1, 29 11, 19 2 · 12 · 20 · 30 = (120)2

62

Calculations:Since exponents come in pairs (m,h−m) the only stand alone exponent is h2 . This gives missingE7 and H3 components.

E6: two exponents are missing. We know that z12 − 1 is satisfied by Coxeter transformations.

z12 − 1 = φ12φ6φ4φ3φ2φ1

if λ = e2πim12 where m is a missing component, there are 4 possible cases:

• φ2(λ) = 0 −→ 6, 6

• φ4(λ) = 0 −→ 3, 9

• φ3(λ) = 0 −→ 4, 8

• φ6(λ) = 0 −→ 2, 10

One needs to write the Coxeter transformations itself to see that φ3(λ) = 0.

H4: can be realised by matrices with coefficients inQ(√

5), z30−1 = φ30 · · · . OverQ(√

5), φ30 =

φ(1)30 φ

(0)30 , the product of 2 irreducibles of degree 4. Exponents correspond to the factor

φ(0)30 and we know that φ(0)

30 (e2πi30 ) = 0. The roots of φ(0)

30 are:

e2πi30 , e

2πi30

29, e2πi30

11, e2πi30

19

H3: H3 = Sym(Icosahedron) = Sym(Dodecahedron).

Every reflection of the Dodecahedron fixes two opposite edges. Hence

|Φ| = 2|reflections| = 2|edges|

2= 30

and|W | = 20︸︷︷︸

|vertices|

·6 = 120

63

F4, H4: W (F4) = Sym(24− cells), W (H4) = Sym(120− cells) = Sym(600− cells). We wantto use a trick here:Let x, y ∈ H, then Sx(y) = −xyx. Hence if G ≤ H∗, |G| <∞, |G| even =⇒ G is a rootsystem. Let q : ‖q‖ = 1 = U ⊂ H∗ and consider

ψ : U”SU2(C)”−→ SO(3)(R) = SO(Him) : q 7−→ (x 7−→ qxq)

which is 2:1. Then ΦH4 = ψ−1(Rot.Symm(Dodec)) =⇒ |ΦH4 | = 2 · 60 = 120.ΦF4 = ψ−1(Rot.Symm.(Cube)) =⇒ |ΦF4 | = 2 · 24 = 48.

E8, E7: There exists a 8-dimensional algebra Φ (octonions). If G ≤ Φ∗, |G| < ∞, |G| is even=⇒ G is a root system.

• ΦE8 is the "group" of invertible octavian integers

• ΦE7 are the elements of order 4 in it.

64