Reflection Groups 3

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, h , i) - Euclidean Vector space where V is a ﬁnite dimensional vector space over R and h , i : V × V −→ R is bilinear, symmetric and positiv deﬁnit. n n P Example: (R , ·): h(αi), (βi)i = αiβi i=1 Gram-Schmidt theory tells that for all Euclidean vector spaces, there exists an isometry (linear n bijective and ∀ x, y ∈ V : T (x) · T (y) = hx, yi) T : V −→ R . In (V, h , i) you can • measure length: ||x|| = phx, xi

hx,yi • measure angles: xyc = arccos ||x||·||y|| • talk about orthogonal transformations O(V ) = {T ∈ GL(V ): ∀ x, y ∈ V : hT x, T yi = hx, yi} ≤ GL(V )

T ∈ GL(V ). Let V T = {v ∈ V : T v = 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 : hv, wi = 0}, x 6= 0. Then 1. T (x) = −x

hx,zi 2. ∀ z ∈ V : T (z) = z − 2 hx,xi · x Proof. 1. Pick v ∈ V T =⇒ T v = v =⇒ hT x, vi = hT x, T vi = hx, vi = 0. Hence T x ∈ (V T )⊥. T ⊥ T Since dim (V ) = dim V − dim V = 1 : T x = α · x for some α ∈ R. Then α2 hx, xi = hαx, αxi = hT x, T xi = hx, xi Since hx, xi= 6 0, α2 = 1 =⇒ α ∈ {−1, 1}. If α = 1 =⇒ T x = x =⇒ x ∈ V T ∩ (V T )⊥ =⇒ x = 0 which is a contradiction. So α = −1. D hx,zi E hx,zi 2. z − hx,xi · x, x = hz, xi − hx,xi · hx, xi = 0. hx,zi ⊥ T hx,zi hx,zi So z − hx,xi · x ∈ x = V and T (z − hx,xi x) = z − hx,xi x. Hence hx, zi hx, zi hx, zi hx, zi T (z) = T ((z − x) + x) = T (z − x) + T (x) hx, xi hx, xi hx, xi hx, xi hx, zi hx, zi hx, zi = z − x − x = z − 2 x hx, xi hx, xi hx, xi

3 2

hx,zi For each x ∈ V, x 6= 0 deﬁne Sx(z) = z − 2 hx,xi 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 called a 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 finite subgroup of O(V ) and G = h{Sx : Sx ∈ G}i, generated by all reflections in G.

Generation means: if G ⊇ X, then X generates G if G = hXi where X is deﬁned as one of the following equivalent deﬁnitions:

1. (semantic) hXi = T H G≥H⊃X

±1 ±1 ±1 2. (syntactic) hXi = {1} ∪ {a1 a2 ··· an : ai ∈ X} Equivalence: −1 (G1,V1) ∼ (G2,V2) if there is an isometry ϕ : V1 −→ V2 s.t. ϕG1ϕ = G2.

| −{z1 } {ϕT ϕ : T ∈G1} Problem: Classify all finite reflection groups (G, V ) up to ∼. Example: of finite reflection groups

1. x

α y

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

Tσ : ei 7−→ eσ(i),Tσ ∈ On(R)

4 1 1 x =  has the property that forall σ : T (x) = x =⇒ x ∈ ( n)Sn , . σ R . 1   α1 ⊥  .  P x = { .  : αi = 0} αn ⊥ An−1 = (Sn, x )

Reﬂection since Sn = h(i, j)i and T(i,j) = Sei−ej because Tij(ei − ej) = −(ei − ej) and   α1  .  ⊥ y =  .  ∈ (ei − ej) ⇐⇒ αi = αj ⇐⇒ T(i,j)(y) = y αn

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

n 3. F = Z/2 = {0, 1} - ﬁeld of two elements. Consider action of Sn on F , ε1, . . . , εn basis n Z of F . ∀ σ ∈ Sn : tσ(εi) = εσ(i) n n n Consider semidirect product Sn n F . As a set: Sn n F = Sn × F , the product is

(σ, a) · (τ, b) = (στ, tτ −1 (a) + b) n n Sn n F acts on R : ai T(σ,α) : ei 7−→ (−1) · eσi n Let us check that this is the action of Sn n F :

T(1,α)(T(τ,0)(ei)) = T(1,α)(eτ(i))

aτ(i) = (−1) eτ(i)

[t −1 (a)]i = (−1) τ eτ(i)

= T (ei) (τ,tτ−1 (a)) n n Bn = (Sn n F , R ) n It is reﬂection since Sn n F = h((i, j), 0), (1, εi)i and T((i,j),0) = Sei−ej ,T(1,εi) = Sei . n |Bn| = n!2 ,B1 ∼ A1,B2 ∼ I2(4)   α1 n  .  n P n 4. In Example 3, let F0 = { .  ∈ F : αi = 0} codim-1 subspace in F or index 2 αn subgroup. n n n σ ∈ Sn, a ∈ F0 =⇒ tσ(a) ∈ F0 , so Sn acts on F0 and n n Dn = (Sn n F0 , R ) n It is a reﬂection since Sn n F0 = h((i, j), 0), (1, εi + εj)i,

T((i,j),0) = Sei−ej ,T(1,εi+εj ) = Sei+ej . n−1 |Dn| = n! · 2 ,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 : kxk = 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

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

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

x, T −1y RHS : T (x)⊥ 3 y 7−→ TS (T −1y) = T (T −1y − 2 · x) x hx, xi hT x, yi = T (T −1y − 2 x) hx, xi = 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 because there are vectors of 2 diﬀerent length. Chop- √ (G,V ) 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 P 2. ∀x ∈ Φ: x = αy · y where either all αy ≥ 0 or all αy ≤ 0 y∈Π β γ α β Example: In {α, β} is a simple system but 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 =⇒ hx, yi ≤ 0 (angles in a simple system are obtuse)

Proof. Suppose x 6= y, hx, yi > 0. hx,yi Φ 3 Sx(y) = y−2 hx,xi ·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 if x ≥ 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 n Example: Phonebook-order on R :     α1 β1  .  ··· >  .  ⇐⇒ ∃k ∈ {1, . . . , n} s.t. αi = βi ∀ i < k and αk > βk αn βn   α1 P Given an ordered basis e1, . . . , en of V we get phonebook order on V by writing αiei = ···. 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 V s.t. Θ = Φ ∩ V+.

Example:

β α 1. Box . 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. π (n − 1 (α, β) is simple ⇐⇒ Φ ⊂ the cone ⇐⇒ −\β, α = ⇐⇒ α,d β = π n n • Every root lies in 2 simple systems • number of simple systems = n (n−1 • positive systems = { vectors between some α and β where α,d β = n π} k m k 3. An : χT (z) = (z − 1)(z − 1) (minimal polynomial of T is (z − 1). (a1,...,ak) (a1,...,ak) m Tσ reﬂection ⇐⇒ χTσ = (1 + z)(1 − z) ⇐⇒ σ = (i, j)

n(n−1) Reﬂections are T(i,j), ∃ 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: P 1. ∀ x ∈ Φ+ ∃ αt ≥ 0 s.t. x = αtt t∈Ω 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 =⇒ hx, yi ≤ 0.

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

8 P Case 1 Sx(y) ∈ Φ+. By (1): y − λx = λtt, λt ≥ 0. What is λy? t∈Ω P If λy < 1 then (1 − λy)y = λx + λtt. So y is a positive linear combination of elements t6=y of Ω \{y}. Hence Ω \{y} satisfies (1), contradiction to (2). P Hence λy ≥ 1, then 0 = λx + (λy − 1)y + λtt, coefficients in the RHS ≥ 0. Hence RHS t6=y ≥ 0. Since RHS = 0 =⇒ λ = 0 which is a contradiction. P Case 2 Sx(y) ∈ Φ− =⇒ y − λx = (−µt)t, µt ≥ 0. t∈Ω P =⇒ λx − y = µtt. Similarly to case 1: t∈Ω P • µx < λ =⇒ (λ − µx)x = y + µtt contradicts (2) t6=x P • µx ≥ λ =⇒ 0 = y + (µx − λ)x + µtt. RHS has a positive coefficient, hence > 0 t6=x 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 suffices to prove that Ω is linearly independent. P Suppose αtt = 0. Sorting positive and negative coefficients to two different sides, t∈Ω X X v = βtt = γtt, βt, γt ≥ 0 t t Now we consider * + X X X hv, vi = βtt, γt = βtγs ht, si ≤ 0 t t s,t |{z} |{z} ≥0 ≤0

Hence v = 0 =⇒ βi = 0 = γt =⇒ αt = 0. Hence Ω is linearly independent and simple. Let Π, Π0 ⊂ Φ be two dinstinct simple systems. Wlog ∃ x ∈ Π0 \ Π. P Since x ∈ Φ+, x = αyy, αy ≥ 0. We know y∈Π

X z z Π 3 y = βy z, βy ≥ 0 z∈Π0

9 So   X X z x =  αyβy  z z∈Π0 y∈Π | {z } ≥0

0 z1 z2 =⇒ ∃ y0 ∈ Π \ Π : αy0 6= 0 =⇒ ∃ at least two z1, z2 s.t. βy0 6= 0, βy0 6= 0 which is a contradiction with the lineary independece of Π0. 2. Let Π ⊂ Φ be a simple system. Π is linearly independent =⇒ we can extend Π to a basis B 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 simple system: Every x ∈ Φ is a nonnegative or nonpositive linear combination of elements of Π =⇒ nonnegative linear combinations ∈ V+, nonpositive linear combinations ∈ V−. P Hence Φ+ = { αtt ∈ Φ: αt ≥ 0}. t∈Π 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−. P Proof. Let y = αtt, αt ≥ 0. y 6= x =⇒ ∃t0 ∈ Π s.t. t0 6= x, αt0 > 0. t∈Π P Hence Sx(y) = αtt − λx with αt0 > 0. Hence Sx(y) ∈ Φ+. 2 t∈Π

Example: Box: simple: , simple:

−1 ∀ g ∈ G, let L(g) = {x ∈ Φ+ : g(x) ∈ Φ−} = Φ+ ∩ g (Φ−). We know that L(1) = x∈Π ∅,L(Sx) = {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) ( l(g) + 1 if g−1(x) ∈ Φ 2. l(S g) = + x −1 l(g) − 1 if g (x) ∈ Φ− ( l(g) + 1 if g(x) ∈ Φ+ 3. l(gSx) = l(g) − 1 if g(x) ∈ Φ− Proof. 1. x 7−→ −g(x) is a bijection between L(g) and L(g−1)

−1 x ∈ Φ+ ⇐⇒ −x = g (−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 g−1(x) 7−→ x 7−→Sx −x

−1 −1 −1 −1 if g (x) ∈ Φ+ =⇒ g (x) ∈/ L(g) but g (x) ∈ L(Sxg). So L(Sxg) = L(g) ∪ {g (x)}. −1 −1 −1 if g (x) ∈ Φ− =⇒ −g (x) ∈ L(g) but not in L(Sxg) and L(Sxg) = L(g) \ {−g (x)}. 2

G acts on V, Φ, {Π ⊂ Φ:Π is simple }.

Theorem 2.7. Let Π, Π0 ⊂ Φ be two simple systems. Then ∃ g ∈ G s.t.

Π0 = g(Π) = {gx : x ∈ Π}

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

0 0 Proof. Let Φ+, Φ+ be the corresponding positive systems, Φ−, Φ− the corresponding negative 0 0 systems. Procced by induction on |Φ+ ∩ Φ−| (distance between Π and Π ). 0 0 0 If |Φ+ ∩ Φ−| = 0 =⇒ Φ+ = Φ+ =⇒ Π = Π . 0 |Φ+ ∩ Φ−| = k suppose proved. 0 0 0 0 |Φ+ ∩ Φ−| = k + 1. Since k + 1 ≥ 1 =⇒ Π 6= Π =⇒ Π 6⊂ Φ+ =⇒ ∃ x ∈ Π ∩ Φ−. Consider Πe = Sx(Π) corresponding positive system is

Φe+ = (Φ+ \{x}) ∪ {−x}

0 0 0 Hence Φe+ ∩ Φ− = (Φ+ ∩ Φ−) \{x}. In particular |Φe+ ∩ Φ−| = k and by induction assumption there is a g s.t. Π0 = g(Π)e , hence

0 Π = g(Sx(Π)) = gSx(Π) 2

Deﬁne a height function h :Φ −→ R by ! X X h αtt = αt t∈Π t∈Π Theorem 2.8. G, Φ ⊃ Π as before.

1. ∀ x ∈ Φ ∃y ∈ Π ∃g ∈ G s.t. x = gy.

2. G = hSxix∈Π

Proof. Let H = hSxix∈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. P Let z = αss, αs ≥ 0. Then s∈Π X 0 < hz, zi = αs hs, zi |{z} s∈Π ≥0

P hx, zi Hence ∃x ∈ Π s.t. hx, zi > 0. Sx(z) = αss − 2 x hence height goes down. So s∈Π hx, xi | {z } >0 h(Sx(z)) < h(z). Moreover Sx(z) ∈ Ht. By minimality of heigt of z:

Sx(z) ∈/ Σt =⇒ Sx(z) ∈ Φ−

−1 −1 By 2.4: z = x = gt for some g ∈ H. Hence Sx = gStg =⇒ St = g Sxg ∈ H | {z } ∈Ht

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 = −x to 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 = hSxix∈Π, is l related to group theoretic length in these generators?

Theorem 2.9. YES

∀ g ∈ G : l(g) = min{k : ∃ x1, . . . , xk ∈ Π: g = Sx1 Sx2 ··· 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+

−1 =⇒ y = xi = Sxi+1 ··· Sxj (xj+1) = T (xj+1). Hence Sxi = TSxj+1 T . Then | {z } T

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

12 Theorem 2.10. If Π, Π0 ⊂ Φ are two simple systems then ∃! g ∈ G s.t. Π0 = g(Π).

Proof. Existence is Theorem 2.7. Suppose g, h ∈ G satisfy Π0 = g(Π) = h(Π). Then −1 −1 −1 −1 Π = h gΠ =⇒ Π+ = h g(Φ+) hence l(h g) = 0 =⇒ h g = 1. 2

Corollary 2.11. Consider action of G and X = {Π ⊂ Φ:Π is simple }. For all Π ∈ X, the orbit map γΠ : G −→ X : g 7−→ g(Π) is a bijection.

Proof. See orbit-stabiliser Theorem. 2

13 3 Generators and Relations

•h Xi-free group on a set X

• h∅i = C1 ∼ •h xi = Z-inﬁnite cyclic group •h Xi = { all ﬁnite words (including ∅) in alphabet x+1, x−1 as x ∈ X without subworts x+1x−1 or x−1x+1}.

• v · w = Reduction of concenated word vw

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

• Universal property: f X / G _ = ∃!ϕ hXi ∀ groups G ∀ functions f : X −→ G ∃! group homomorphism

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

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

u −→ u = 1, uv−1 −→ uv−1 = 1 ⇐⇒ u = v

* + 2 n −1 −1 E.g. dihedral group I2(n) = a , b : b , a , bab = a . |{z} |{z} rot refl.

Deﬁnition. hX | Ri = hXi/ . If the relations are words u , u ,... then R = T H. R 1 2 hXi.H, all ui∈H Universal property of hX | Ri: hX | Ri O ∃!ϕ q f # X / G ∀ functions f : X −→ G s.t. relations in R hold in G for elements f(x), x ∈ X then ∃! group homomorphism ϕ : hX | Ri −→ 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 marked by 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

mij i j −→ cij = mij

i j −→ cij = 2

Example:  1 3 ∞ 2   3 1 3 77 1 2 −→   >> Ð >> ∞ 3 1 2  >>∞ ÐÐ >>77 >> ÐÐ >> 2 77 1 1 > ÐÐ > 3 4

Deﬁnition 4.1. The Coxeter group of a Coxeter graph (or corresponding X ×X-matrix (mij)) is W = W (graph) = hX | (ab)mab = 1i

Remark 4.2. 1. mab = ∞ =⇒ no relation

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

2 −1 −1 3. no edge ⇐⇒ mab = 2 =⇒ (ab) = 1 ⇐⇒ ab = b a = 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 of W (mij = |aiaj|).

16 2 2 Proof. f : {1, . . . , n} −→ G where f(i) = ai is a function. Then f(i) = ai = 1 and m |a a | (f(i)f(j)) ij = (aiaj) i j = 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,   W   53 ∼ M ∼ 10 , M = /H Theorem 4.4. G is a ﬁnite reﬂection group Π ⊂ Φ-simple system in the root system. Then the natural surjection for Π × Π-matrix mx,y = |SxSy|, W ((mx,y)) G : x 7−→ Sx is an isomorphism. * + α Example: O2(R) ≥ I2(∞) = Sa , Rotα , π ∈/ Q |{z} | {z } order 2 inf. order | ∼ {z } =C2nC∞ abb W ( ∞ ) hS ,S i where ∈/ , =∼ hS , Rot i a b π Q a α ∞ 2 action of W ( ) = I2(∞) or R is a mess. 1 But I2(∞) has action on R with geometric meaning.

S0 S1

-1 0 1 ∞ S0(x) = x, S1(x) = −2−x, S1S0(x) = 2+x −→ translation by 2. So hS1,S0i is W ( ) inside ∞ the 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(∞) 1 tesselates R by interval [0, 1].  a   b  Example: W  c  ∼ sidereﬂections across the sides of a triangle with angles π , π , π .   = a b c  

α γ

a b where α = π , β = π , γ = cπ. 1 1 1 + + > 1 spherical a b c = 1 euclidean < 1 hyperbolic

17 −→ tesselation of the space.  a  b  c  In particular, W   ←→ 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 ··· Scai · Scaj · San

Proof. already proved. 2

Proof. of Theorem 4.3

ϕ ψ hΠi W (mab) G a 7−→ a 7−→ Sa |{z} we will work in here It suﬃces to show that ψ is injective. 2 Let w = a1a2 ··· an ∈ Ker ψ (a1 = 1). Then ψ(w) = Sa1 · San = 1. Since det Sai = −1, (−1)n = 1 −→ n is even, write n = 2k. n Proceed by induction on k = 2

• k = 1 −→ n = 2 −→ w = a1a2

ai∈Π 2 1 = ψ(w) = Sa1 Sa2 −→ Sa1 = Sa2 −→ a1 = ±a2 =⇒ a1 = a2 −→ w = a1 = 1

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

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

−1 w1 = a1 wa1 = 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.

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

−1 Case 2 there is a v ∈ W s.t. w2 = v wv = 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

−1 −1 Indeed y = Sbm+2 ··· Sb2m and xy = 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 ··· Scbi ··· Scbj ··· Sbm+1 Consider two cases:

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

Sbi ··· Sbj = Sbi+1 ··· Sbj−1

Sbi ··· Sbj Sbj−1 ··· Sbi+1 = 1

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

bi ··· bj = bi+1 ··· bj−1

So w2 = b1 ··· bbi ··· bbj ··· b2m. By induction assumption w2 = 1 and therefore

−1 −1 w = vw2v = vv = 1

Case 2.2 (i, j) = (1, m + 1) Then w3 = b1b2 ··· bm+1bm ··· b2 ∈ Ker ψ has length 2m and we seem to be stuck. (Hint: eventual contradiction is with b1 6= b3). −1 Let w4 = b1 w2b1 = b2 ··· bnb1 ∈ ker ψ. Let us play the trick on

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

Deletion principle −→ x = S ··· S 0 ··· S 0 ··· S . Get 2 cases: 1 b2 cbi cbj bm+2 Case 2.2.1 (i0, j0) 6= (2, m + 2)

S 0 ··· S 0 = S ··· S bi bj bi0+1 bj0−1

S 0 ··· S 0 S ··· S = 1 bi bj bj0−1 bi0+1

w5 = bi0 ··· bj0 bj0−1 ··· bi0+1 ∈ ker ψ. By induction assumption w5 = 1

bi0 ··· bj0 = bi0+1 ··· bj0−1

0 0 and w2 = b1 ··· bbi ··· bbj ··· b2m. By induction assumption w2 = 1 =⇒ w = −1 vw2v = 1. Case 2.2.2 (i0, j0) = (2, m + 2)

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

−1 but of length n. Let w6 = b3w5b3 = b3b2b3 ··· bm+2bm+1 ··· b4 ∈ ker ψ and repeat the trick on ψ(w6) = 1.

x2 = Sb3 Sb2 Sb3 ··· Sbm+1 = Sb4 ··· Sbm+2 can delete at i00, j00. Get 2 cases: Case 2.2.2.1 (i00, j00) 6= (1st incidence of 3, m + 1)

w7 = bi00 ··· bj00 bj00−1 ··· bi00+1 ∈ ker ψ

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

19 Case 2.2.2.2 (i00, j00) = (first 3, m + 1)

w7 = b3b2b3 ··· bm+1bm ··· b2 ∈ ker ψ

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

−1 w7 = b3b1w3 −→ b3b1 = w7w3 ∈ ker ψ

Sb3 Sb1 = 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 h , i. (the ﬁeld is R, dim V - any, h , i - any) Still ∀ W ≤ V we have

• W ⊥ = {x ∈ V : ∀ a ∈ W : ha, wi = 0}

• (W ⊥)⊥ ⊃ W

x a Example: 2, , = xa − yb R y b 1 W = · =⇒ W ⊥ = W =⇒ W + W ⊥ = W 6= 2. R 1 R Vectors v with hx, xi = 0 are called isotropic. If hx, xi= 6 0 (x is not isotropic) then

⊥ V = Rx ⊕ x and ha, xi S : V −→ V,S (a) = a − 2 x x x hx, xi is well-deﬁned and has usual properties

• Sx(x) = −x

• Sx|xbot = id 2 • Sx = id

• Sx ∈ O(V, h , i)

Start with X × X Coxeter matrix (mab) or its Coxeter graph. Let X V (mab) = { αaea : αa ∈ R, finitely many αa 6= 0} a∈X ea, a ∈ X form a basis of V (mab). Symmetric bilinear form h , i : V (mab) × V (mab) −→ R is deﬁned on the basis by π hea, ebi = − cos mab

• a = b =⇒ mab = 1 =⇒ hea, eai = 1

• mab = 2 =⇒ hea, ebi = 0

π • mab = ∞ =⇒ hea, ebi = − cos ∞ = − cos 0 = −1

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

π Proof. he1, e2i = − cos m . If x = αe1 + βe2 then

2 2 hx, xi = α he1, e1i + 2αβ he1, e2i + β he2, e2i π = α2 + β2 − 2αβ cos m π = (α + β)2 − 2αβ(1 + cos ) m ≥ 0 and hx, xi = 0 ⇐⇒ α = β = 0 OR m = ∞, α = β 2

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

V = U ⊕ U ⊥

Note: it breaks down if dim U = ∞. Example: (V, h , i) 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.

X ⊥ D X E X v − xiei ∈ U ⇐⇒ ∀ j : v − xiei, ej = 0 ⇐⇒ hv, eji − xi hei, eji = 0 i P The system of n equations xi hei, eji = hv, eii j = 1, . . . , n has a solution. It holds i ⊥ true because U ∩ U = 0 which implies that h , i |U is non-degenerate which is equivalent to det hei, eji= 6 0. 2

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

M π V (Γ) = Rea, hea, ebi = − cos mab a∈X Operators

ρa = Sea : V (Γ) −→ V (Γ) : x 7−→ x − 2 hea, xi ea

Since ρ : a(e ) = −e , ρ | ⊥ = I, |ρ | = 2. a a a ea a Proposition 5.3. |ρaρb| = mab

22 Proof. 5.1 Case 1 mab 6= ∞ =⇒ U = Rea ⊕ Reb is euclidean vector space w.r.t h , i. In particular ⊥ ⊥ ⊥ ⊥ U ∩ U = 0 =⇒ V (Γ) = U ⊕ U . Since U ⊂ ea , ρa|U ⊥ = I. Similarly, ρb|U ⊥ = I ⊥ ⊥ Since ρb(U ) ⊂ U , ρaρb|U ⊥ = ρa|U ⊥ · ρb|U ⊥ = I · I = I. On U, we can do explicit calculation. e π − π α mab

eβ

2π Hence ρa|U · ρb|U = Rot by and |ρaρb| = |ρaρ | = mab mab b|U

Case 2 Exercise: show that in ea, eb basis 3 −2 1 1 ρ | · ρ | = −→ a U b U 2 −1 0 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 ( (a, 1) if a ∈ X ψ(a) = 1 (1, a) if a ∈ X2

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

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 completely irreducible 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: π U 3 ρb(ea) = ea + 2 cos eb =⇒ eb ∈ U mab | {z } 6=0

Since Γ is connected, this would imply that eb ∈ U for all b ∈ X. So U = V and not a proper subrepresentation which is a contradiction. ⊥ 0 0 ⊥ By 5.?, V = Rea ⊕ Rea . Pick any z ∈ U and write z = αea + z where z ∈ Rea . Then

U 3 ρa(z) = z − 2 hea, zi ea 0 0 = αea + z − 2α hea, eai ea − 2 ea, z ea 0 = −αea + z

0 1 0 ⊥ Hence z = 2 (z + ρa(z)) ∈ U and 2αea ∈ U =⇒ α = 0. Hence z = z ∈ Rea and therefore ⊥ T ⊥ ⊥ U ⊂ Rea and U ⊂ Rea = 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 operators 2 p ∈ EndF(V ) s.t. p = p and Im(p) = U: V = U ⊕ W −→ p(v) = u where v = u + w is the unique decomposition with u ∈ U, w ∈ W V = Im(p) ⊕ ker(p) ←− p

Pick any vector space decomposition and let p be the corresponding operator, ρ : G −→ GL(X). 1 P −1 Deﬁne pe = |G| ρ(x)pρ(x ) x∈G • if t ∈ U

1 X −1 p|U =I 1 X −1 pe(t) = ρ(x)(p(ρ(x )(t))) = ρ(x)ρ(x )(t) = t |G| | {z } |G| x∈G ∈U x∈G

So pe|U = I.

• since pe(t) = t for t ∈ U, Im(pe) ⊃ U 2 • by deﬁnition of p,e Im(pe) ⊂ Im(p) ⊂ U. Hence Im(pe) = U =⇒ pe = pe

• pe is a homomorphism of representations (commutes with any ρ(x), x ∈ G): 1 X ρ(x)p = ρ(x)ρ(y)pρ(y−1) e |G| y∈G 1 X = ρ(xy)pρ((xy)−1)ρ(x) |G| y∈G 1 X = ρ(z)pρ(z−1)ρ(x) |G| z∈G = pρe (x)

Hence ker pe is a G-subrepresentation and V = U ⊕ ker pe. 2

Corollary 5.9. If Γ is connected and |W (Γ)| < ∞ then V (Γ) is nonsingular (w.r.t h , i) and irreducible (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 =⇒ h , i = 0. Nonsence since hea, eai = 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 hea, zi Θ(ea) = Θ(z − 2 hea, zi ea) = Θ(z) − 2 hea, Θ(z)i ea

Hence hea, zi Θ(ea) = hea, Θ(z)i ea. Let z = ea, then Θ(ea) = hea, Θ(ea)i ea. Therefore, Θ has | {z } λ 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), h , i) is euclidean. Proof. |W (Γ)| < ∞ =⇒ X is ﬁnite =⇒ dim V (Γ) = |X| < ∞. S˙ L If Γ = Ti,Ti connected, then V (Γ) = V (Γi) with V (Γi)⊥V (Γj), i 6= j. Hence it suﬃces i to consider a case of connected Γ.

Pick any Euclidean form h , i1 over V (Γ). Let us integrate h , i1 over W (Γ) so let X hu, vi2 = hρx(u), ρx(v)i1 x∈W (Γ)

•h , i2 is symmetric since h , i1 is P • u 6= 0, hu, ui2 = hρx(u), ρx(u)i1 hence h , i2 is euclidean. x | {z } >0 •∀ y ∈ W (Γ): X hρy(u), ρy(v)i2 = hρxρy(u), ρxρy(v)i1 x X = hρxy(u), ρxy(v)i1 x X = hρz(u), ρz(v)i1 z∈W (Γ)

= hu, vi2

Hence all ρy ∈ O(V (Γ), h , i2). Remember that all ρu ∈ O(V (Γ), h , i) and consider Θ: V −→ V deﬁned by

hΘ(u), vi2 = hu, vi ∀ v

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

−1 hρx(Θ(u)), vi2 = Θ(u), ρx (v) 2 −1 = u, ρx (v) 2 = hρx(u), vi

= hΘ(ρx(u)), vi2

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

λ hu, vi2 = hu, vi

1 Since hea, eai = 1 = λ hea, eai =⇒ λ = > 0. 2 2 hea,eai2

Section 4 =⇒ Finite Reﬂection group =∼ Coxeter group Section 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.   n 2n Hence distinct reﬂection groups I2(2n) = W and A1 × I2(n) = W   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) = min{n : 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)

m m Since relations of W hold in GL1(R) ((ab) ab = 1 −→ ((−1)(−1)) ab = 1 ok) sgn is well- deﬁned. n l(x) If l(x) = n then x = a1 ··· an and sgn(x) = sgn(a1) ··· sgn(an) = (−1) = (−1) . 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

+ • Ha = {u ∈ V (Γ) : hu, eai > 0} − • Ha = {u ∈ V (Γ) : hu, eai < 0} the dividing hyperplane Ha = {u ∈ V (Γ) : hu, eai = 0} T + and the cone C = Ha which we call the fundamental chamber. a∈X(Γ)

Idea: Let x ∈ W : xi(Ha)

ρx(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(Ha ∩ Hb ) ⊂ Ha OR + + − • g(Ha ∩ Hb ) ⊂ Ha and l(ag) = l(g) − 1 + + + + + − + + Proof. If g(Ha ∩ Hb ) 6⊂ Ha and g(Ha ∩ Hb ) 6⊂ Ha then by convexity of g(Ha ∩ Hb ): + + g(Ha ∩ Hb ) ∩ Ha 6= ∅ + + Pick z ∈ g(Ha ∩ Hb ) ∩ Ha. Then z ∈ Ha =⇒ hz, eai = 0 + + −1 + + −1 −1 z ∈ g(Ha ∩ Hb ) =⇒ g (z) ∈ Ha ∩ Hb =⇒ g (z), ea > 0, g (z), eb > 0 −1 −1 On the other hand g (z), g (ea) = hz, eai = 0. As g ∈ a, b : −1 g (ea) = αea + βeb α, β ∈ R

We denote by Φ(Γ) a root system of ρ(W ) ∈ V (Γ) and since ea, eb form a simple system of Φ(Γ) ∩ span(ea, eb), either • α ≥ 0, β ≥ 0 s.t. not both are 0 and then −1 −1 −1 −1 −1 g (z), g (ea) = g (z), αea + βeb = α g (z), ea + β g (z), eb > 0 which is a contradiction • OR α ≤ 0, β ≤ s.t. not both are 0 and therefore −1 −1 g (z), g (ea) < 0 which is a contradiction too. + + − It remains to prove the length statement if g(Ha ∩ Hb ) ⊂ Ha . Inspect U = span(ea, eb) and + + L = U ∩ Ha ∩ Hb . This is a dihedral group of order 2mab.

Hb ∩ U

bL eb baL − + Ha ea Ha abab...L

gL ⊂ H− ∩ U ⇐⇒ g = a or g = ab or ... or g = abab . . . b a | {z } 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 1 1 m < ∞. This follows from ρ ρ | = . ab a b span(ea,eb) 0 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) ⊂ Ha OR − • w(C) ⊂ Ha 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.

+ + −1 w(C) ⊂ g(Ha ∩ Hb ) and l(w) = l(g) + l(g w)

Proof. Simultaneous induction on l(w):

+ Base l(w) = 0 =⇒ w = 1 =⇒ w(C) = C ⊂ Ha 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 = dw0, d ∈ X, l(w0) = k − 1. Consider two cases: Case 1: a = d. 0 0 + 0 − Statement (1) for w gives either w (C) ⊂ Ha or w (C) ⊂ Ha . 0 − 0 − + • w (C) ⊂ Ha =⇒ w(C) = a(w (C)) ⊂ a(Ha ) = Ha 0 + 0 + − • w (C) ⊂ Ha =⇒ w(C) = a(w (C)) ⊂ a(Ha ) = Ha Case 2: a 6= d. 0 0 + + 0 By (2) for w there is a g ∈ a, b s.t. w (C) ⊂ g(Ha ∩ Hb ) and l(w ) = l(g) + l(g−1w). Then

0 + + w(C) = d(w (C)) ⊂ dg(Ha ∩ Hd )

+ + + + By Lemma 6.2 either dg(Ha ∩ Hd ) ⊂ Ha (=⇒ w(C) ⊂ Ha ) or + + − dg(Ha ∩ Hd ) ⊂ Ha and l(adg) = l(dg) − 1. + + − Then w(C) ⊂ dg(Ha ∩ Hd ) ⊂ Ha and l(aw) = l(adgg−1w0) ≤ l(adg) + l(g−1w0) ≤ l(g) + l(g−1w0) = l(w0) = 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 ⊂ Ha ∩ Hb . Then g = 1 satisﬁes (2). + + Case 2: wC 6⊂ Ha ∩ Hb .

− + + + Ha ∩ Hb Ha ∩ Hb

− − + − Ha ∩ Hb Ha ∩ Hb

+ + + − Observe that wC ∩ Ha = ∅. wC ⊂ w(Ha ∩ Hb ) is in either Ha or Ha by 6.2. − Wlog wC ⊂ Ha . By part (1): l(aw) = l(w) − 1. Use part (2) on w0 = aw (by induction assumption) then there is g0 ∈ a, b 0 0 + + 0 0 0 −1 s.t. w C ⊂ g (Ha ∩ Hb ) and l(w ) = l(g ) + l((g ) w). Claim: g = ag0 satisﬁes (2) for w.

0 0 + + + + wC = aw C ⊂ ag (Ha ∩ Hb ) = g(Ha ∩ Hb )

l(w) = 1 + l(aw) = 1 + l(g0) + l((g0)−1w0) = 1 + l(g0) + l(g−1w) ≥ l(ag0) + 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 = S gC g∈W 2. Γ is ﬁnite, then: ∀ g, x ∈ W : gC ∩ xC 6= ∅ =⇒ g = x

Note: If W is inﬁnite S gC is called Tits cone which is a proper subset of V . g∈W Proof.

1. C = {α ∈ W : ∀ a ∈ X : hα, ai ≥ 0}. P P Recall the height function h : V −→ R : v = βaea 7−→ βa. a∈X a∈X Pick any v ∈ V . Since W is ﬁnite, the height achieves a maximum on the set W v = {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 = S gC. Indeed, otherwise there is an a ∈ X s.t. hz, eai < 0. Then:

ρa(z) = h(z) − 2 hz, eai ea

has hight h(ρa(z)) = h(z) − 2 hz, eai > 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. −1 −1 + −1 − Set w = ag x, then l(aw) = l(w) + 1. By 6.3 ag xC = wC ⊂ Ha =⇒ g xC ⊂ Ha . −1 + − Then C ∩ g xC ⊂ Ha ∩ Ha = ∅ 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 equivalence classes 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−→ αxx x∈X x∈X W (Γ) −→ G

a1a2 ··· an 7−→ Sa1 ··· San

This is an isomorphism by 4.3 and 6.5. 2

Some fun bits:

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

32 n n−1 • C \ C : is connected with fundamantal group Z a a 0 S a Try CC = C ⊗R V ⊃ CH = C ⊗R H . Then CV = CV \ g(CH ) is connected. g∈W, a∈X 0 0 W acts freely on CV and the universal cover of CV is simply connected (if W is ﬁnite). V 0 One can draw the fundamental group of C /W :

ρa(x) Ta x 0 C ⊂C V

V 0 [x] = [ρa(x)] ∈ C /W . V 0 Ta is a loop in C /W . In fact: * + V 0 B(Γ) = π1 C /W = Ta, a ∈ X : TaTbTa ··· = TbTaTb ··· | {z } | {z } | {z } Braid group mab mab

0 V 0 V −→ /W gives 1 −→ PB(Γ) ←→ B(Γ) W −→ 1 | {z } pure Braid group

33 7 Classification

Let (V, h , i) be an Euclidean vector space. Therefore it has a norm kvk = phv, vi

It induces a norm on LinR(V,V ): kT vk kT k = sup /kvk v∈V \{0} 1. kT xk ≤ kT k · kxk 2. kT k ≥ 0 and [kT k = 0 ⇐⇒ T = 0] 3. kT + Sk ≤ kT k + kSk 4. T ∈ O(V ) =⇒ kT vk = kvk =⇒ kT k = 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 mus be finite. Long version: pick v ∈ C s.t. kvk = 1. Pick ε > 0 s.t. Bε(v) ⊂ C.

g 6= x ∈ W =⇒ gC ∩ xC = ∅ =⇒ Bε(gv) ∩ Bε(xv) = ∅ =⇒ kgv − xvk > ε =⇒ kρ(g) − ρ(x)k > ε By fact 5, O(V ) is compact. Suppose W is inﬁnite, W ⊂ O(V ) compact. Then there is a 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 : kTi − Tjk < 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 what we proved, we may conclude that any finite reflection group has a form k (W (Γ1) × ... × W (Γn),V (Γ1) ⊕ ... ⊕ V (Γn) ⊕ R ) where the Γi are connected positive definite Coxeter graphs and each W (Γi) acting trivially on k V (Γj) with i 6= j and R .

Problem: What are the connected positive deﬁnite Coxeter graphs?

34   a11 ··· a1n  . .  Proposition 7.2. A =  . .  symmetric matrix over R. an1 ··· ann   a11 ··· a1j  . .  A is positive deﬁnite ⇐⇒ ∀ j ∈ {1, . . . , n} : det  . .  > 0. aj1 ··· ajj Proof. Induction on n: n=1 obvious n-1 we assume to be done   a1n  .   B .  n" =⇒" A =   where B is (n − 1) × (n − 1) matrix.  .   .  an1 ······ ann   ∗    α1 .   .  .  B represents the same bilinear form restricted to U =   =  .  : αn = 0    α  0 n in the standard basis of U. In particular: for all x ∈ U \{0} : xT Bx > 0. By induction assimption:   a11 ··· a1j  . .   . .  > 0 aj1 ··· ajj

for j < n. Since there exists an orthogonal Q s.t.   λ1 −1  ..  QAQ =  .  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, hx, yi = xT By) is Eu- clidean. By Lemma ??: n ⊥ R = U ⊕ U |{z} under xT Ay n ⊥ Any x ∈ R can be written as x = y + z, y ∈ U, z ∈ U and

hx, xi = hy + z, y + zi = hy, yi + hz, zi

So it suﬃces to check that for z ∈ U ⊥, z 6= 0 and hz, zi > 0. ⊥ T Let f1, . . . , fn−1 be an orthonormal basis of U, fn ∈ U , fn 6= 0. x Ay in the basis

35 f1, . . . , fn will look like  0   .   In .  C =    0  0 ··· 0 hfn, fni

Hence A is positive deﬁnite ⇐⇒ C is positive deﬁnite ⇐⇒ hfn, fni > 0. 2

0 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 Γ0 is a Coxeter subgraph of Γ and Γ is positive deﬁnite then Γ0 is positive deﬁnite. Proof. π π 2 ≤ nab ≤ mab =⇒ − cos ≥ − cos nab mab P 0 P . Pick any v = αaea ∈ V (Γ ) and suppose hv, vi ≤ 0. Let w = |αa|ea. Then: a∈Y a∈Y X π hw, wiV (Γ) = |αa| · |αb|(− cos ) mab a,b∈Y X π ≤ |αa| · |αb|(− cos ) nab a,b∈Y X π ≤ αaαb(− cos ) nab a,b∈Y ≤ 0

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

π Let d(Γ) = det(2 hea, eni)X×X = det(−2 cos m )a,b∈X . | {z } ab C(Γ) 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

Γ n n-1 2

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 −1 0 ··· 0 −1 2 Expanding the last row:  0   .  n+(n−1)  C(Γ2) .  d(Γ) = 2d(Γ1) + (−1) (−1)det   = 2d(Γ1) − d(Γ2)  0  ∗ · · · ∗ −1 2

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

An ◦ ◦ ··· ◦ ◦ 4 Bn ◦ ◦ ··· ◦ ◦ ◦ GG GGG Dn ◦ ··· ◦ ◦ www ◦ ww

E6 ◦ ◦ ◦ ◦ ◦ ◦

E7 ◦ ◦ ◦ ◦ ◦ ◦ ◦

E8 ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 4 F4 ◦ ◦ ◦ ◦ 5 H2 ◦ ◦ 5 H3 ◦ ◦ ◦ 5 H4 ◦ ◦ ◦ ◦

m I2(m) ◦ ◦

Note: A2 = I2(3),B2 = I2(4),H2 = I2(5). Sometimes G2 = I2(6) is used. Proof. Let xn denote d(Xn). π 2 −2 cos m π 2 • i2(m) = det π = 4 − 4(cos m ) > 0 −2 cos m 2

37 • a1 = det(2) = 2 π 2 a2 = i2(3) = 4 − 4(cos 3 ) = 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 · · · · 4 B3 ◦ ◦ ◦

A1 ·

B2 ·

b3 = 2b2 − a1 = 2 · 2 − 2 = 2 And for Bn: · · · · 4 Bn ◦ ◦ · · · ◦ ◦ ◦

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: ◦ 0 · 00 · 00 ◦ 0 · 00 · 00 ◦ ◦ ◦ · · ·

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. √ 2 π 5− 5 • h2 = i2(5) = 4 − 4 cos 5 = 2 √ √ 5− 5 • h3 = 2h2 − a1 = 2 2 = 3 − 5 > 0 √ √ √ 5− 5 7−3 5 • h4 = 3h3 − h2 = 2(3 − 5) − 2 = 2 > 0 √ • Notice that h5 = 3h4 − h3 = 5 − 2 5 < 0. 2

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

A ◦ ◦ en DD zzz DD ◦ z ◦ ◦ ◦ zzz ◦ ◦ z

◦ DD DD 4 Ben ◦ ··· ◦ ◦ ◦ zzz ◦ z 4 4 Cen ◦ ◦ ··· ◦ ◦ ◦ ◦ ◦ DDD z D zzz B ◦ ◦ ··· ◦ ◦ en DD zzz DD ◦ z ◦ ◦ ◦ D DDD Ee6 ◦ ◦ ◦ zzz ◦ ◦ z ◦ ◦ ◦ D DDD Ee7 ◦ ◦ ◦ zzz ◦ z ◦ ◦ D DDD Ee8 ◦ ◦ ◦ ◦ ◦ ◦ zzz ◦ z

39 4 Fe4 ◦ ◦ ◦ ◦ ◦

6 Ge2 ◦ ◦ ◦

5 He3 ◦ ◦ ◦ ◦

5 He4 = H5 ◦ ◦ ◦ ◦ ◦ Proof.

• We have computed e8 = e9 = 9 − 9 = 0. √ • eh4 = h5 = 4 − 2 5 < 0  2 −1 −1 −1 2 −1     ......  • an = det  . . .  = 0 because the sum of the rows is zero e    .. ..   . . −1 −1 −1 2

• ebn = 2bn − a1bn−2 = 4 − 4 = 0

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

• e6 = 2e6 − a5 = 0

• e7 = 2e7 − d6 = 0

• fe4 = 2f4 − b3 = 0

• ge2 = 2i2(6) − a1 = 0 √ • eh3 = 2h3 − a2 = 3 − 2 5 < 0  ∗ 0   c(Bn−1) ∗ 0    • c(Cen) looks like  ∗ 0 .    ∗ 2 −2 0 −2 2 Similarly to Lemma 7.4, expanding the last row gives

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

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

Proof. The graphs on 7.5 are closed under taking connected Coxeter subgraphs. Therefore 7.2 tells us that all c(Γ) are positive definite in 7.5 and 7.1 tells us that all W (Γ) are finite in 7.5. Now let Γ be a connected graph with finite W (Γ). Then c(Γ) is positive definite by 7.1. In particular, Γ contains no subgraphs in 7.6

• No Aen =⇒ Γ has no cycles

• No De4 =⇒ Γ has no vertex with ≥ 4 edges

• No Den, n ≥ 5 =⇒ Γ contains at most one vertex with 3 edges. Case 1 No vertex has 3 edges, hence Γ looks like this:

x x x ◦ 1 ◦ 2 ··· ◦ m ◦

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 Ge3 =⇒ edges may have multiplicity at most 5

– No Cen =⇒ 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 Fe4 =⇒ Γ = F4 Case 1.3 There is one edge of multiplicity 5: No He3 =⇒ the multiple edge is on a side and no He4 =⇒ 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 going right. Then we have a + b + c + 1 vertices in total and wlog a ≥ b ≥ c.

– No Ben =⇒ Γ has no multiple edges.

– No Ee6 =⇒ c = 1

– No Ee7 =⇒ b ≤ 2

– if b = c = 1 =⇒ Γ = Da+3

– if b = 2, c = 1, then no Ee8 =⇒ 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 crystallographic An,Bn,Dn,F4,En,G2 = I2(6) I2(m) for m = 5, m ≥ 7,H3,H4 d(Γ) ∈ Z d(Γ) ∈/ Z the character table of Γ is integral non-integral all 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 = he1, . . . , eni where e1, . . . , en is a basis of V . n n Informally, L is Z inside R .

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 for each 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 integer matrix, in particular T r(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 :

 cos 2π ± sin 2π 0 mab mab 2π 2π ρ(SaS ) = ∓ sin cos 0 b  mab mab  0 0 I

2π T r(ρ(SaSb)) = |X| − 2 + 2 cos hence mab 2π 2π 1 2 cos ∈ Z ⇐⇒ cos ∈ Z ⇐⇒ mab ∈ {2, 3, 4, 6} mab mab 2 2

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

Theorem 8.2. Let W (Γ) be a ﬁnite Coxeter group. Then W (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

hαa, αbi ρa(αb) = αb − 2 hαb, eai ea = αb − 2 αa hαa, αai

Since ρa = Sea = Sαa and ρ(a1 ··· an) = ρa1 ··· ρan it suﬃces to ensure that for all a, b ∈ X : hαa, αbi 2 ∈ Z hαa, αai Wlog Γ is connected. • m = 2 =⇒ hα , α i = 0 =⇒ 2 hαa,αbi = 0 ab a b hαa,αai π cacb • mab = 3 =⇒ hαa, αbi = cacb hea, ebi = caca − cos = − 2 mab

| {z1 } =− 2

hα , α i −c c c =⇒ 2 a b = a b = − b hαa, αai caca ca

Need in this case ca = ±cb • m = 4 =⇒ hα , α i = c c (− cos π ) = − c√acb ab a b a b 4 2 hα , α i √ c =⇒ 2 a b = − 2 b hαa, αai ca √ √ Need ca = 2cb or cb = 2ca √ 3 • mab = 6 =⇒ hαa, αbi = − 2 cacb √ 3 hα , α i −cac √ c =⇒ 2 a b = 2 b 2 = − 3 b hαa, αai caca ca √ √ 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 4 Γ2 |{z} |{z} √ choose ca=1 choose ca= 2

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

4 1 2 3 ··· n √ √ • α1 = e1, α2 = 2e2, . . . , αk = 2ek which is called the crystallographic reﬂection group Bn √ • α1 = 2e1, α2 = e2, . . . , αk = ek which is called the crystallographic reﬂection group Cn

W eyl group and Root lattice - semisimple Lie algebra crystallographic reflection group n b i Serre T heorem

Kac−Moody algebra 1 2 Nakajima construction

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 polynomial functions on V . Let e1, . . . , en be a basis of V and Xi : V −→ R the linear maps which form the 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. G ∼ We will prove that S = R[z1, . . . , zk] (when G is a reﬂection group). In fact G ∼ S = R[z1, . . . , zk] ⇐⇒ G is a reﬂection group (no proof)

∞ L S is a graded algebra, i.e. S = Sk where Sk consists of polynomials where each monomial k=0 has degree k. Example:

• 2010 ∈ S0

2 2 • x1 + x2 + x1x3 ∈ S2

• 1 + x1 ∈ S0 ⊕ S1 but not in any Si

If F ∈ Si we say deg F = i. m G P G Lemma 9.1. Let F ∈ S ,F = Fj,Fj ∈ Sj. Then ∀ j: Fj ∈ S . j=0 ∗ ∗ 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)

−j In particular, f ∈ Sj : (1, α) · f = α · f. ∗ ∗ Since actions of G and R commutes, ∀α ∈ R : m X −j G (1, α)F = α Fj ∈ S j=0 1, . . . , m + 1 : m + 1 diﬀerent real numbers

46 G • 1 · F = F0 + ... + Fm+1 ∈ S

1 1 G • 2 · F = F1 + 2 F1 + ... + 2m+1 Fm+1 ∈ S • ...

1 1 G • (m + 1) · F = F0 + m+1 F1 + ... + (m+1)m+1 Fm+1 ∈ S

 1 ··· 1  1 1 m+1  2 ··· 2    (m+1)×(m+1) If M =  . .  ∈ R is a VanDerMonde matrix then  . .   m+1 1 1 m+1 ··· m+1   F0  .  Gm+1 M ·  .  ∈ S Fm | {z } ∈Sm+1     F0 H0  .  −1  .  G Since M is invertible,  .  = M  .  where Hi ∈ S . Fm HM G Hence all Fi ∈ S . 2

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

∞ G M G G G S = (S ∩ Sk) and Sk = S ∩ Sk k=0

∞ G L G G Let S+ = Sk be the invariants of positive degree. Let I = S+ be the ideal of S generated k=1 G by S+ . By Hilbert’s Basis Theorem, any ideal in S is generated by ﬁnitely many elements. G P Say I = (f1, . . . , fk)S, fi ∈ S . Write fi = fij where fij ∈ Sj. With Lemma 9.1: j G fij ∈ Sj , j < 0. Hence I = (I ,...,I ) where I ∈ SG . If for any k: I ∈ (I ,...,I ,I ,...,I ) we throw 1 r S j dj k 1 k−1 k+1 r S Ik away. Wlog, I1,...,Ir is a minimal system of such generators.

Deﬁnition. I1,...,Ir are called fundamental invariants if they are a minimal system of G homogeneous invariant generators of (S+ )S. S • Coinvariants: SG = /I • Integral: R : S −→ SG G Z 1 X F = gF |G| G g∈G

47 Note that F ∈ SG =⇒ gF = F =⇒ R F = F and ∀ H ∈ S : G

Z 1 X 1 X Z (FH) = (gF )(gH) = F gH = F H |G| |G| g g G G

So R is an SG-linear map S −→ SG. G

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

G ∀F ∈ S ∃ 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 r G G P d = N > 0: F ∈ S =⇒ F ∈ (S )S =⇒ F = HjIj,Hj ∈ S. Since d = deg(F ), dj = j=1 deg(Ij): r X F = (Hj)d−dj Ij j=1 and therefore r Z X Z F = F = Ij (Hj)d−dj G j=1 G R By induction assumption ∀ j ∃ pj ∈ R[z1, . . . , zr] s.t. (Hj)d−dj = pj(I1,...,Ir). Hence G

r X p = zjpj j=1

does the job. 2

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

G R[z1, . . . , zr] −→ S , 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

G 3. S0 = R · 1 constant functions

48 G G 4. f ∈ S1 , f 6= 0 =⇒ V = Ker(f) ⊕ R. Hence S1 6= 0 ⇐⇒ GV has a trivial constituent 5. if h , i is a Euclidean form on V then [x, y] = R hx, yi makes G orthogonal. Hence G G F (x) = [x, x] deﬁned F ∈ S2 .

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 by 1 1 1 L with remainder by replacing every xn with − α b(x1, . . . , xn−1). xn = α L − α b(x1, . . . , xn−1). Example: L = x1 − x2

2 P = x1 = x1 · x1 = x1(L + x2) = x1L + x1x2 = x1L + (L + x2)x2 2 = (x1 + x2)L + x2

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. 1 Let an = − α b(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

W Proposition 9.4. Let G = W be a ﬁnite reﬂection group. Let J1,...,Jk ∈ S s.t. k P W J1 ∈/ (J2,...,Jk)SW . If Pi ∈ Sbi s.t. PiJi = 0 then P1 ∈ (S+ )S. i=1

k P Proof. Observe that J1 ∈/ (J2,...,Jk)S: if J1 = HtJt with Ht ∈ S then t=2

Z X Z J1 = J1 = Jt Ht ∈ (J2,...,Jk)SW t

Proceed by induction on deg(P1):

k P deg(P1) = 0: =⇒ P1 = c · 1 =⇒ c · J1 = (−Pt)Jt =⇒ c = 0 because J1 ∈/ (J2,...,Jk)S =⇒ t=2 W P1 = 0 ∈ (S+ )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) = hα, vi. Lα = 0 =⇒ α⊥v =⇒ Sα(v) = v

−1 =⇒ ∀ j :(SαPj − Pj)(v) = Pj(Sα 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 induction W W assumption, P 1 ∈ (S+ )S. Hence SαP1 − P1 = LαP 1 ∈ (S+ )S. W For each polynomial F , denote Fb = F + (S+ )W ∈ SW . Hence ∀ α ∈ Φ: SαPb1 = Pb1. R Since W is generated by Sα, ∀ g ∈ W : gPb1 = Pb1. Hence Pb1 = Pb1 (W acts on SW and W R R Fb = dF ) R W W So P1 ∈ P1 + (S+ )S ⊂ (S+ )S. W 2

Lemma 9.5. (Euler’s formula) F ∈ Sk = R[x1, . . . , xn]k, then n X ∂F x = kF j ∂x j=1 j Proof. Both sides are linear in F , hence it suﬃces to verify the formula on monomials. Let a1 an P F = x1 ··· xn , ai = k. Then ∂F xj = ajF ∂xj n Hence P ∂F = (P a )F = kF . 2 ∂xj i j=1

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

Note: I1,...,Ir are algebraically independent. P Proof. Consider a grading on R[z1, . . . , zr] with deg zi = di = deg Ii. Let P = Pj,Pj homogeneous of degree j. Then X P (I1,...,Ir) = Pj(I1,...,Ir) = 0 j | {z } ∈Sj

Hence Pj(I1,...,Ir) = 0. WLOG, we assume P is homogeneous. Let P = ∂P and f = P (I ,...,I ) ∈ SW is homogeneous. Let us ﬁddle around with the (j) ∂zj j (j) 1 r ideal W K = (f1, . . . , fr)SW /S

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 m W P Qij ∈ S ,Qij homogeneous of degree deg Ii − deg Ij s.t. fi = Qijfj. j=1

r ∂P (I1,...,Ir) X ∂P ∂Ii P (I ,...,I ) = 0 =⇒ 0 = = (I ,...,I ) ∀ k ∈ {1, . . . , n} 1 r ∂x ∂z 1 r ∂x k i=1 i k

50 Hence: m r m X ∂Ii X X ∂Ii ∀ k : f + Q f = 0 i ∂x ij j ∂x i=1 k i=m+1 j=1 k m  r  X ∂Ii X ∂Ij ∀ k : f + Q = 0 i ∂x ji ∂x  i=1 k j=m+1 k r ∂II X ∂Ij By 9.4, ∀ k ∀ i ∈ {1, . . . , m} : + Q ∈ (SW ) ∂x ji ∂x + S k j=m+1 k

k Hence for some Ht ∈ S : r r ∂II X ∂Ij X ∀ i, k : + Q = I Hk ∂x ji ∂x t t k j=m+1 k t=1

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

r r X X ∀ i : diIi + djQjiIj = ItHet j=m+1 t=1

P k where Het have no free terms (Het = xkHt ). Ii enters both sides of the equality. If we consider the homogeneous degree di parts of both sides, then

r X X diIi + cjIj = Itbt j=m+1 t where bi has to be zero because Hei has no free terms. Therefore X Ii = IjGj =⇒ Ii ∈ (I1,...,Ii−1,Ii+1,...,Ir) j6=i which is a contradiction. Therefore all ∂P = 0 =⇒ P = 0. 2 ∂xk

n Example: Sn acting on R by permutations.     α1  !  X  1 n = U ⊕ where U =  .  : α = 0 and = · R R  .  i R R .   .1  αn 

(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 j j is a system of fundamental invariants. The power functions πj = x1 + ... + xn, j = 1, . . . , n is another fundamental system.

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

n n W (Bn) Sn C C Bn: group has more transformations xi 7−→ ±xi, S = (S ) 2 = R[σ1, . . . , σn] 2 . 2 2 Hence the fundamental invariants are σj(x1, . . . , xn) with degrees 2 · 1, 2 · 2,..., 2 · n.

n W (Dn) Dn: in Dn, instead of the full group C2 , it is a subgroup of index 2. So σn = x1 ··· xn ∈ S . 2 2 So the fundamental invariants are σ1(xi ), . . . , σn−1(xi ), 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,...,Ir fundamental invariants. Then

G S = R[I1,...,Ir] and r = n

Proof. The natural algebra homomorphism

G ψ : R[z1, . . . , zr] −→ S ,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 that n = r. Consider the ﬁeld of rational functions f F = (x , . . . , x ) = { : f, g ∈ S, g 6= 0} R 1 n g

Transcendence degree is trdegR F = n. The smaller field A = R(I1,...,Ir) ≤ F, trdegR A = r. From Galois theory, trdegA F = trdegR F −trdegR A = 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 F G = A. Consider h(z) ∈ F [z] defined by Y h(z) = (z − g.f) g∈G

Since ∀ t ∈ G, t.h = Q (z − tg.f) = h =⇒ h ∈ A[z]. g∈G Y Y h(f) = (f − g.f) = (f − f) (f − g.f) = 0 g∈G g6=1

Hence f is algebraic over A. 2

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

0 0 0 0 0 Theorem 9.8. If I1,...,In is another system of fundamental invariants with degrees d1, . . . , dn, dj = 0 deg Ij then ∃ σ ∈ Sn s.t. 0 ∀ j : dj = dσ(j)

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

0 0 0 Ij = Fj(I1,...,In) and Ij = Gj(I1,...,In) ∀ j

Then ∀ j : Ij = Fj(G1(I1,...,In),...,Gn(I1,...,In)). Hence, since I1,...,In are algebraically independent: zj = Fj(G1(z1, . . . , zn),...,Gn(z1, . . . , zn)) and therefore n ∂zj X ∂Fj ∂Gs = (G (z , . . . , z ),...,G (z , . . . , z )) · (z , . . . , z ) ∂z ∂z 1 1 n n 1 n ∂z 1 n k s=1 s k And on matrix level: ∂Fi ∂Gs In = · ∈ Matn(S) ∂zs 0 0 ∂zk I1,...,In I1,...,In Hence det ∂Fi is invertible in S = [x , . . . , x ]. Hence ∃ σ ∈ S s.t. ∂zs R 1 n n

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

0 0 Since deg Ii = deg zi = di, deg Fi = deg Ii = di:

0 ∀ i : di = dσ(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 absoulute value 1.

n n Proof. |G| = n < ∞ =⇒ x = 1 =⇒ ρ(x) = In. n Hence the minimal poynomial of ρ(x), µx(z) divides z − 1. Therefore eigenvalues λ satisfy λn − 1 = 0, hence λn = 1 =⇒ |λ|n = 1 =⇒ |λ| = 1. n Further, since z − 1 has no multiple rootes, µx(z) has no multiple roots and all Jordan blocks of ρ(x) have size 1. 2

∞ ∞ P n L Let C[[t]] = { αnz } be the ring of formal power series. If V = Vn is a graded vector n=0 n=0 space with dim Vn < ∞, the "right" notion of dimension of V is the Poincaré polynomial ∞ X n pV (t) = dim Vn · t n=0 Example: d 2d 3d 1 • pC[z](t) = 1 + t + t + t + ... = 1−td where deg z = d.

• pV ⊕W (t) = pV (t) + pW (t)

• pV ⊗W (t) = pV (t)pW (t) • p = Q p (t) = 1 ··· 1 where deg z = d , [z , . . . , z ] = [z ] ⊗ ... ⊗ C[z1,...,zn] C[zj ] 1−td1 1−tdn i i C 1 n C 1 j C[zn]. 1 If d1 = ... = dn = 1 then pC[z1,...,zn](t) = (1−t)n . Example: If G acts on V , then ∀ x ∈ G let ∞ −1 Y 1 Y 2 2 X j X a1 an χx(t) = det(IV −tρ(x)) = = (1+λjt+λj t +...) = t ( λ1 ··· λn ) 1 − λjt j j j=0 a1+...+an=j 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 n 1 X Y 1 χx(t) = ∈ C[[t]] |G| 1 − tdj x∈G j=1

54 Proof. By 9.7 and the Example before, the RHS is p G (t).   S λ1 0 x| belongs to  ..  in some basis e , . . . , e of S . Then ea1 ··· ean with S1  .  1 n 1 1 n 0 λn a1 + ... + an = j form a basis of Sj and

a1 an a1 an a1 an xe1 ··· en = λ1 ··· λn e1 ··· en hence X a1 an tr(x|Sj ) = λ1 ··· λn a1+...+an=j Hence ∞ 1 X LHS = tr(x| )tj |G| Sj x∈G,j=0 ∞ R G 1 P P j Remember πj = : Sj −→ Sj , πj(v) = |G| gv. In particular LHS = tr(πj)t . G g∈G j=0 I 0 π2 = π =⇒ π ∼ and tr π = rk π = dim SG j j j 0 0 j j j

Hence LHS = pSG (t). 2

Corollary 10.3. |G| = d1 ··· dn. 1 Corollary 10.4. d + ... + d = n + |Φ(G)| . 1 n 2 | {z } number of reflections Proof. Multiply 10.2 by (1 − t)n: n 1 X n Y 1 − t χx(t)(1 − t) = |G| 1 − tdj x∈G j=1 n 1 X (1 − t)n Y 1 = |G| det(1 − tg) 1 + t + t2 + ... + tdj −1 x∈G j=1 | {z } =:α(t) 1 1 1 − t = ( 1 + |Φ| + h(t) · (1 − t)2 ) |G| |{z} 2 1 + t | {z } g=1 | {z } others have 1 as eigenvalue of mult.≤2 g∈Sx, x∈Φ [h(1) is deﬁned ] n Plug in t = 1, then 1 = Q 1 |G| dj j=1 =⇒ |G| = d1 ··· dn ∂ Apply ∂t : n 2 dj −2 |Φ| −2 X 1 + 2t + 3t + ... + (dj − 1)t +eh(t)(1−t) = α(t)(1+t+...+tdj −1) −1 2|G| (1 + t)2 (1 + t + ... + tdj −1)2 j=1

55 Plug in t = 1: n ! |Φ| X dj (dj − 1)dj − = − 4|G| d ··· d 2d2 j=1 1 n j

As |G| = d1 ··· dn n |Φ| X dj − 1 1 X = = (( d ) − n) 4 2 2 j j=1 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 n(n + 3) |G| = (n+1)!, |Φ| = 2((2+3+...+n+1)−n) = 2 − n = n2+3n−2n = n2+n 2

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 cycles of 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}, the conjugacy 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

gψ = f(ψ(1)) ··· f(ψ(n − 2))f(a)f(b) = f 0(ψ(1)) ··· f 0(ψ(n − 1))

0 0 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 −1 ψi : ψ (Xi) −→ Xi, ψi(c) = ψ(c) Then

gψ = gψ0 f(a)gψ1 f(b)

= gψ0 f(a)f(b)gψ1 ∼ g (g f(a)f(b)g )g−1 ψ1 ψ0 ψ1 ψ1

= gψ1 gψ0 f(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 = gψ0 f(b)gψ1 ∼ gψ1 gψ0 f(b) done by Case 2. 2

Proposition 11.2. Coxeter elements form a conjugancy class in G. Proof. Let C = { Coxeter elements }. Then −1 −1 −1 xSα1 ··· Sαn x = (xSα1 x ) ··· (xSαn x ) = Sx(α1) ··· Sx(αn) ∈ C

−1 Hence xCx = C, so C is the union of some conjugacy classes. Let w1 = Sα1 ··· Sαn , w2 = −1 Sβ1 ··· Sβn ∈ C. Then there is a g ∈ G s.t. {g(α1), . . . , g(αn)} = {β1, . . . , βn}. Hence gw1g =

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 of the Coxeter graph. Then h = |w| is called the Coxeter number of W . 2πim h h k Let λ1, . . . , λn be the eigenvalues of w|V (W ). Since w = 1, λk = 1 =⇒ λk = e h 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: ◦ ◦ ~~ ~~ ~~ ◦ ~ ◦ @ @@ @@ @@ ◦ ◦ @ @@ @@ @@ ◦ ◦ Q For a, b ∈ Xi =⇒ ab = ba. Hence ωj = a does not depend on the order of Xj. Moreover, a∈Xj 2 Q 2 ωj = a = 1. ω = ω1ω2 is called canonical Coxeter element. G = hω1, ω2i 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

n P n 3. mj = h 2 j=1 Proof.

0 1. if m1 = 0 then e = 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 hea, xi ea. Hence X X ω1x = x + αaeq = ω2x = x + αbeb

a∈X1 b∈X2 X X =⇒ αaea = αbeb

a∈X1 b∈X2

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

( ⊥ aj ∈X1 V1 3 y 7−→ y ω1|V (W ) = Sa1 ··· Sak : V1 3 y 7−→ −y L where Vi = Rea. a∈Xj ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ Hence x ∈ V1 ∩ V2 and V1 ∩ V2 = 0 because V = V1 ⊕ V2 = V1 ⊕ V2 since the form is 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 by 2πi h h 2 h part (1), so λ = −1, h is even and λ = e , 2 is the exponent. ∗ Case 2 fj(z) = (z − λ)(z − λ ), λ∈ / R. 2πi m ∗ − 2πi m 2πi (h−m ) Then if λ = e h k then λ = e h k = e h k .

h 3. This also gives part (3) because the average of exponents in each fj(z) is 2 . 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 = S ⊥ zi

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: P • C = hx : hx, eai ≥ 0i = R≥0ωa a ⊥ L • Vj = Rωa a/∈Xj

• angle between ωa and ωb = π− angle between ea and eb The angles between ωa and ωb are accute and hωa, ωbi ≥ 0 and hωa, ωbi = 0 ⇐⇒ hea, ebi = 0 ⇐⇒ a and b commute.

Q = (hea, ebi)a,b is the matrix of

• symmetric positiv deﬁnite bilinear form h , i 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 of Q. 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. P P Pick z 6= 0, z = αaωa ∈ K and let ze = |αa|ωa. a P Now, the idea is to use that zj = |αa|ωa, but we have no time to ﬁnish the proof. 2 a∈Xj

60 Theorem 11.5. If W is ﬁnite connected of rank n (i.e. the number of vertices in the Coxeter graph 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:

Rz2

Rz1

n There are 2 hyperplanes Hα s.t. Hα ∩ P is one of these h lines • if h is even:

Rz2

Rz1

h h There are 2 lines with one intersection pattern and 2 lines with another intersection pattern 2

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

∀ j : dj = mj + 1

Proof. (idea) Compute Jac = det ∂Ij in two diﬀerent ways. 2 ∂zk Q P Corollary 11.7. |W | = (mj + 1), |Φ| = 2 mj. j j

61 Corollary 11.8. Let W be connected. −1   ..  −1 =  .  ∈ ρ(W ) ⇐⇒ all mi are odd −1 Proof. W d d "=⇒" (−1) ◦ Ij = Ij ∈ S . Therefore (−1) · xk = xk =⇒ (−1)Ij = (−1) j Ij =⇒ (−1) j = 1. So all dj are even and therefore all mj are odd.

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

h  2πi m  2  πim    e h k e k −1 h 2  ..   ..   ..  (ω|V (W )) ∼  .  =  .  =  .  2πi m πimk e h k e −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

Y 2πi m Φh(z) = (z − e h ) 1≤m≤h−1 gcd m,h=1 is irreducible in Z[z]. The characteristic polynomial fω(z) of ω|V (W ) is in Z[z]. 2πi For m = 1: (z − e h )|fω(z) in C[z]. Therefore

Φh(z)|fω(z) in Z[z]

2πi m and therefore all e h are eigenvalues. 2

|Φ| 240 Application: E8 has rank 8 and 240 roots. Hence h = n = 8 = 30. The number of coprimes of 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| |Φ| h = n ”easy exponents” missing exponents |W | E6 72 12 1, 5, 7, 11 4, 8 E7 126 18 1, 5, 7, 11, 13, 17 9 E8 240 30 1, 7, 11, 13, 17, 19, 23, 29 none 7 2 F4 48 12 1, 5, 7, 11 none 2 · 6 · 8 · 12 = 2 · 3 H3 30 10 1, 9 5 2 · 6 · 10 = 120 2 H4 120 30 1, 29 11, 19 2 · 12 · 20 · 30 = (120)

62 Calculations: h Since exponents come in pairs (m, h−m) the only stand alone exponent is 2 . This gives missing E7 and H3 components. 12 E6: two exponents are missing. We know that z − 1 is satisﬁed by Coxeter transformations.

12 z − 1 = φ12φ6φ4φ3φ2φ1

2πim if λ = e 12 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. √ √ 30 H4: can be realised by matrices with coeﬃcients in Q( 5), z −1 = φ30 ··· . Over Q( 5), φ30 = (1) (0) φ30 φ30 , the product of 2 irreducibles of degree 4. Exponents correspond to the factor (0) (0) 2πi (0) φ30 and we know that φ30 (e 30 ) = 0. The roots of φ30 are:

2πi 2πi 29 2πi 11 2πi 19 e 30 , e 30 , e 30 , e 30

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

Every reflection of the Dodecahedron fixes two opposite edges. Hence |edges| |Φ| = 2|reflections| = 2 = 30 2 and |W | = 20 ·6 = 120 |{z} |vertices|

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

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

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