rumynin - reflection groups

7/28/2019 Rumynin - Reflection Groups

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Lecture Notes

Reflection Groups

Dr Dmitriy Rumynin

Anna Lena Winstel

Autumn Term 2010


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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


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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, ) : (i), (i) =

ni=1

ii

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 : T x , T y = x, y} GL(V)

T GL(V). Let VT = {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 VT = dim V 1.

Lemma 1.1. Let T be a reflection, x (VT) = {v : w VT : v, w = 0}, x = 0. Then

1. T(x) = x2. z V : T(z) = z 2 x,zx,x x

Proof.

1. Pick v VT = T v = v = T x , v = T x , T v = x, v = 0. Hence T x (VT).Since dim (VT) = dim V dim VT = 1 : T x = x for some R. Then

2 x, x = x,x = T x , T x = x, xSince x, x = 0, 2 = 1 = {1, 1}.If = 1 = T x = x = x VT (VT) = x = 0 which is a contradiction.So =

1.

2.

z x,zx,x x, x

= z, x x,zx,x x, x = 0.So z x,zx,x x x = VT and T(z x,zx,xx) = z x,zx,xx. Hence

T(z) = T((z x, zx, xx) +x, zx, xx) = T(z

x, zx, xx) +

x, zx, xT(x)

= z x, zx, xx x, zx, xx = z 2

x, zx, xx

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2

For each x V, x = 0 define Sx(z) = z 2 x,zx,xx.

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 = 0 : Sx is a reflection.3. any reflection T satisfies T2 = 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 = GHX

H

2. (syntactic) X = {1} {a11 a12 a1n : ai X}Equivalence:

(G1, V1) (G2, V2) if there is an isometry : V1 V2 s.t. G11

{T 1: TG1}

= G2.

Problem: Classify all finite reflection groups (G, V) up to .Example: of finite reflection groups1.

y

x

We want to know Sx, Sydet(Sx, Sy) = 1 = SxSy SO2(R) = SxSy = Rot.Since Sx(Sy(y)) = Sx(y) = SxSy = 2.

/

Q =

|SxSy

|=

=

Sx, Sy

is not finite

= mn Q (for m, n rel. prime) = |SxSy| = |Rot 2mn | = n = Sx, Sy = D2nDihedral

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 ofRn, if Sn, then

T : ei e(i), T On(R)

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x =

11...1

has the property that forall : T(x) = x = x (Rn)Sn,

x = {1...n

: i = 0}An1 = (Sn, x)

Reflection since Sn = (i, j) and T(i,j) = Seiej because Tij(ei ej) = (ei ej) and

y =

1...n

(ei ej) i = j T(i,j)(y) = yIn particular, |Am| = (m + 1)! and A2 I2(3)

3. F = Z/2Z = {0, 1} - field of two elements. Consider action of Sn on Fn, 1, . . . , n basisofFn.

Sn : t(i) = (i)Consider semidirect product Sn Fn. As a set: Sn Fn = Sn Fn, the product is

(, a) (, b) = (,t1(a) + b)Sn Fn acts on Rn:

T(,) : ei (1)ai eiLet us check that this is the action of Sn Fn:

T(1,)(T(,0)(ei)) = T(1,)(e(i))

= (1)a(i)e(i)= (1)[t1(a)]ie(i)= T(,t

1 (a))(ei)

Bn = (Sn Fn,Rn)

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

4. In Example 3, let Fn0 = {

1...

n

Fn :

i = 0} codim-1 subspace in Fn or index 2

subgroup. Sn, a Fn0 = t(a) Fn0 , so Sn acts on Fn0 and

Dn = (Sn Fn0 ,R

n)

It is a reflection since Sn Fn0 = ((i, j), 0), (1, i + j),T((i,j),0) = Seiej , T(1,i+j) = Sei+ej .

|Dn| = n! 2n1, D1 is trivial ({1},R), D2 I2(2), D3 A3

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2 Root systems

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

(G,V) := {x V : x = 1, Sx G}

has the following properties:

1. x = Rx = {x, x}2. || = 2( number of reflections in G)3. T

G, x

=

T(x)

Property 3 follows from

Lemma 2.1. T O(V), x V \ {0} = ST(x) = T SxT1.Proof. RHS: T(x) T SxT1T x = T Sxx = T(x) = T(x)

RHS : T(x) y T Sx(T1y) = T(T1y 2

x, T1y

x, x x)

= T(T1y 2T x , yx, x x)

= T(T1y)

= 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 V s.t.1. 0V / 2. x = Rx = {x, x}3. x, y = Sx(y)

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Example:

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

2. = ({1},V)

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

ping long vectors by

2 gives B2 .

Let V be a root system. Definition. A simple subsystem is s.t.1. is linearly independent

2. x : x = y

y y where either all y 0 or all y 0

Example: In

{, } is a simple system but

is not simple since .

Lemma 2.2. simple system in a root system. Then x, y , x = y = x, y 0(angles in a simple system are obtuse)

Proof. Suppose x = y, x, y > 0. Sx(y) = y2 x,yx,x x = y+x, < 0. Since is lin. independent and x = y, Sx(y) = y+xis the only way to write Sx(y) as a linear combination of elements of 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 = y)

1. x y = x + z y + z2. x y, > 0 = x y3. x y, < 0 = x y

Example: Phonebook-order on Rn:

1 n

>

1...

n

k {1, . . . , n} s.t. i = i i < k and k > k

Given an ordered basis e1, . . . , en ofV we get phonebook order on V by writing

iei =

1 n

.If is a total order on V then

x V \ {0} = either x > 0 > x or x < 0 < xand V = V+{0}V where

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V+ = {x V : x > 0} V = {x V : x < 0}

Definition. A positive system in a root system is a subset s.t. total order on Vs.t. =

V+.

Example:

1. Box

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

2 + > + > > > 0 > > > > 2

So {,, + , 2 + } is a positive system.

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

(, ) is simple the cone , = n

, = (n 1n

Every root lies in 2 simple systems number of simple systems = n positive systems = { vectors between some and where , = (n1n }

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

T

reflection

T

= (1 + z)(1

z)m

= (i, j)

Reflections are T(i,j), n(n1)n of them and = {ei ej : i = j {1, . . . , n + 1}}.A typical simple system is

e1 e2, e2 e3, . . . , en1 enFor i < j we have: ei ej = (ei ei+1) + (ei+1 ei+2) + . . . + (ej1 ej).

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

1.

x

+

t

0 s.t. x = t tt

2. No proper subset of satisfies (1).

Hint: simplicity quasisimplicityLemma 2.3. Let + be a quasisimple system. Then x, y , x = y = x, y 0.Proof. Suppose x = y and x, y > 0. Then: Sx(y) = y 2 x,yx,xx = y x, > 0. Consider two cases:

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Case 1 Sx(y) +. By (1): y x =

ttt, t 0. What is y?

If y < 1 then (1 y)y = x +t=y

tt. So y is a positive linear combination of elements

of \ {y}. Hence \ {y} satisfies (1), contradiction to (2).Hence y

1, then 0 = x + (y

1)y + t=y tt, coefficients in the RHS 0. Hence RHS 0. Since RHS = 0 = = 0 which is a contradiction.

Case 2 Sx(y) = y x =t

(t)t, t 0.= x y =

ttt. Similarly to case 1:

x < = ( x)x = y +t=x

tt contradicts (2)

x = 0 = y + (x )x +t=x

tt. 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

{ : is simple } and {+ : is positive }

Proof.

1. Let +

be a positive system. Consider all subsets of + that satisfy condition (1)

of the definition of a quasisimple system. A minimal subset among them must satisfy(2), so is a quasisimple system in +. To prove that is a simple system, it sufficesto prove that is linearly independent.Suppose

t

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

v =

t

tt =

t

tt, t, t 0

Now we consider

v, v

= t tt,t t = s,t ts0 t, s0 0

Hence v = 0 = i = 0 = t = t = 0. Hence is linearly independent and simple.Let , be two dinstinct simple systems. Wlog x \ .Since x +, x =

y

yy, y 0. We know

y =z

zy z, zy 0

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So

x =z

y

yzy

0

z=

y0

\

: y0= 0 =

at least two z1, z2 s.t.

z1y0

= 0, z2y0

= 0 which is acontradiction with the lineary independece of .

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

t

tt : t 0}.

2

Proposition 2.5. Let + be a root system, positive system, simple system. Then x , y +, x = y:Sx(y) +

Note: x = y is essential since Sx(x) = x .Proof. Let y =

t

tt, t 0. y = x = t0 s.t. t0 = x, t0 > 0.Hence Sx(y) =

t

tt x with t0 > 0. Hence Sx(y) +. 2

Example: Box: simple: , simple:

g G, let L(g) = {x + : g(x) } = + g1(). We know that L(1) =, L(Sx) x= {x}.Define function l : G Z0 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(g1)

2. l(Sxg) =

l(g) + 1 if g1(x) +l(g) 1 if g1(x)

3. l(gSx) =l(g) + 1 if g(x) +

l(g) 1 if g(x) Proof.

1. x g(x) is a bijection between L(g) and L(g1)x + x = g1(g(x))

g(x) g(x) +

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

g1(x) g x Sx x

if g1(x) + = g1(x) / L(g) but g1(x) L(Sxg). So L(Sxg) = L(g) {g1(x)}.if g

1

(x) = g1

(x) L(g) but not in L(Sxg) and L(Sxg) = L(g) \ {g1

(x)}.2

G acts on V, , { : is simple }.Theorem 2.7. Let , be two simple systems. Then g G s.t.

= g() = {gx : x }

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

Proof. Let +, + be the corresponding positive systems, , the corresponding negativesystems. Procced by induction on |+ | (distance between and ).If |+ | = 0 = + = + = = .|+ | = k suppose proved.|+ | = k + 1. Since k + 1 1 = = = + = x . Consider = Sx() corresponding positive system is+ = (+ \ {x}) {x}Hence

+ = (+ ) \ {x}. In particular |

+ | = k and by induction assumption

there is a g s.t. = g(

), hence

= g(Sx()) = gSx()

2

Define a height function h : R by

h

t

tt

=t

t

Theorem 2.8. G, as before.1. x y g G s.t. x = gy.2. G = Sxx

Proof. Let H = Sxx G. Pick any t let

t = Ht +

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If t = (obvious if t +) then z t with minimal height h(z) in t.Let z =

s

ss, s 0. Then

0 < z, z =

ss

0s, z

Hence x s.t. x, z > 0. Sx(z) =

sss 2 x, zx, 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) / t = Sx(z) By 2.4: z = x = gt

Ht

for some g H. Hence Sx = gStg1 = St = g1Sxg H

t = g1

x Hx H G(1) follows because ifx is positive this proof shows that x G, if x is negative we use t = xto show that Sx H, t = x G. Hence x = Sx(x) G.Finally we have shown that all reflections Sx are in H, so H = G. 2

l : G Z0. Now since G = Sxx, is l related to group theoretic length in thesegenerators?

Theorem 2.9. YES

g G : l(g) = min{k : x1, . . . , xk : 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 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) and xj+1 +. Find largest i s.t. i j and

Sxi

(y) = Sxi

Sxj

(xj+1)

, y = Sxi+1

Sxj

(xj+1)

+

= y = xi = Sxi+1 Sxj T

(xj+1) = T(xj+1). Hence Sxi = T Sxj+1T1. Then

g = Sx1 Sxi Sxj Sxk = Sx1 Sxi1SxiT Sxj+1 Sxk= Sx1 Sxi1(T Sxj+1T1T Sxj+1)Sxj+2 Sxk = Sx1 Sxi1T Sxj+2 Sxk

2

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Theorem 2.10. If , are two simple systems then ! g G s.t. = g().

Proof. Existence is Theorem 2.7. Suppose g, h G satisfy = g() = h(). Then = h1g = + = h1g(+) hence l(h1g) = 0 = h1g = 1. 2

Corollary 2.11. Consider action of G and X = { : is simple }. For all X, theorbit map

: G X : g g()is a bijection.

Proof. See orbit-stabiliser Theorem. 2

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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, x1 as x X without subworts

x+1x1 or x1x+1}. v w = Reduction of concenated word vw

It is a Theorem that is well-defined. Universal property:

Xf //

_

G

X

!==

groups G functions f : X G ! group homomorphism

: X G s.t. a X : (a) = f(a)

X | R: R is a set of relations i.e. either words in alphabet x1 : x X or u = v whereu, v are two words. Relation between words and equations:

u u = 1, uv1 uv1 = 1 u = v

E.g. dihedral group I2(n) =

a

rot

, brefl.

: b2, an,bab1 = a1

.

Definition. X | R = X/R

. If the relations are words u1, u2, . . . then R =

XH, all uiHH.

Universal property of X | R:X | R

!

##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 : X | R G s.t. x X: (xR) = f(x).

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Example: B(2, 5) =

x, y | w5 where w is any word in the alphabet x1, y1.Universal property: groups G s.t. x G : x5 = 1, a, b G ! : B(2, 5) G s.t.(x) = a, (y) = b.Known: B(2, 5) is either infinite or of order 534.

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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 Z3 {}.

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,jX s.t. mij Z1 {}

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 = miji j cij = 2

Example:

1bb

bbbb

b 2

77

bbbb

bbb

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 XX-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 = 13. no edge mab = 2 = (ab)2 = 1 ab = b1a1 = ba Therefore:

no edge a, b commuteLemma 4.3. LetG be a group generated by a1, . . . , an, all of order 2. Then G is a quotient ofW(mij = |aiaj|).

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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 ! grouphomomorphism : W(mij) G s.t. (i) = ai (by the universal property). Im ai = 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 -simple system in the root system. Thenthe natural surjection for -matrix mx,y = |SxSy|, W((mx,y)) G : x Sx is anisomorphism.

Example: O2(R) I2() = Saorder 2

, Rot

inf. order =C2C

, / Q

W( ) Sa, Sb whereab

/ Q, = Sa,Rot

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) = 2x, S1S0(x) = 2+ x translation by 2. So S1, S0 is W( ) insidethe group of motions on R. C is a subgroup of translations (x 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

bca

= sidereflections across the sides of a triangle with angles a , b , c .

where = a , =b , = c.

1

a+

1

b+

1

c> 1 spherical

= 1 euclidean

< 1 hyperbolic

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tesselation of the space.

In particular, W

bca triangles in tesselation

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

x = Sa1 Sai Saj SanProof. already proved. 2

Proof. of Theorem 4.3

W(mab) G

a awe will work in here

Sa

It suffices to show that is injective.Let w = a1a2 an Ker (a12 = 1). Then (w) = Sa1 San = 1. Since det Sai =1, (1)n = 1 n is even, write n = 2k.Proceed by induction on k = n2

k = 1 n = 2 w = a1a21 = (w) = Sa1Sa2 Sa1 = Sa2 a1 = a2 ai= a1 = a2 w = a12 = 1

k m 1: done (induction assumption) k = m : n = 2m, w = a1 a2m, 1 = (w) = Sa1 Sa2m. Consider

w1 = a11wa1 = 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 , w = (ab)m.1 = (w) = (SaSb)

m = |SaSb| = mab|m = w = (ab)m = 1 W

Case 2 there is a v W s.t. w2 = v1wv = b1b2b3 b2m with all bi and b1 = b3.Let us play the trick: Observe that

x = Sb1 Sbm+1 = Sb2m Sbm+2 = yIndeed y1 = Sbm+2 Sb2m and xy1 = Sb1 Sb2m = (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+1Consider two cases:

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Case 2.1 (i, j) = (1, m + 1)Sbi Sbj = Sbi+1 Sbj1

Sbi SbjSbj1 Sbi+1 = 1

Hence w3 = bi bj bj1 bi+1 ker . It has < 2m terms hence by inductionassumption w3 = 1. Hencebi bj = bi+1 bj1

So w2 = b1 bi bj b2m. By induction assumption w2 = 1 and thereforew = vw2v

1 = vv1 = 1

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

= b3).

Let w4 = b11w2b1 = b2 bnb1 ker . Let us play the trick onx1 = Sb2 Sbm+1Sbm+2 = Sb1 Sbm+3

Deletion principle x1 = Sb2 Sbi Sbj Sbm+2 . Get 2 cases:Case 2.2.1 (i, j) = (2, m + 2)

Sbi Sbj = Sbi+1 Sbj1Sb

i Sb

jSbj1 Sbi+1 = 1

w5 = bi

bjbj

1

bi+1

ker . By induction assumption w5 = 1

bi bj = bi+1 bj1

and w2 = b1 bi bj 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

but of length n. Let w6 = b3w5b31

= b3b2b3 bm+2bm+1 b4 ker and repeat the trick on (w6) = 1.

x2 = Sb3Sb2Sb3 Sbm+1 = Sb4 Sbm+2can delete at i, j. Get 2 cases:

Case 2.2.2.1 (i, j) = (1st incidence of 3, m + 1)w7 = bi bjbj1 bi+1 ker

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

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

w7 = b3b2b3 bm+1bm b2 ker

Remember w3 = b1b2b3 bm+1bm b2. Then

w7 = b3b1w3 b3b1 = w7w13 ker Sb3Sb1 = 1 = Sb3 = Sb1 = b3 = b1 = b3 = b1

But this is a contradiction!

2

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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 = R2.

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

V = Rx x

and

Sx : V V, Sx(a) = a 2 a, xx, xx

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) = {aX

aea : a R, finitely many a = 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

mab

a = b = mab = 1 = ea, ea = 1 mab = 2 = ea, eb = 0 mab = = ea, eb = cos

= cos 0 = 1

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Lemma 5.1. V( m ) is euclidean m = .Proof. e1, e2 = cos m . If x = e1 + e2 then

x, x = 2 e1, e1 + 2e1, e2 + 2 e2, e2

= 2

+ 2

2cos m= ( + )2 2(1 + cos

m

)

0

andx, x = 0 = = 0 OR m = , =

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 = 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 equivalentto det ei, ej = 0. 2

Coxeter graph (matrix) = (mab)a,bX Coxeter group W() = X : (ab)mab = 1 vector space with bilinear form

V() = aXRe

a, ea, eb = cos mabOperators

a = Sea : V() V() : x x 2 ea, x eaSince : a(ea) = ea, a|ea = I, |a| = 2.Proposition 5.3.

|ab| = mab

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Proof.

Case 1 mab = 5.1= U = Rea Reb is euclidean vector space w.r.t , . In particularU U = 0 = V() = U U. Since U ea , a|U = I. Similarly, b|U = ISince b(U

) U, ab|U = a|U b|U = I I = I. On U, we can do explicitcalculation.

e

e mab

Hence a|U b|U = Rot by 2mab and |ab| = |ab|U| = mabCase 2 Exercise: show that in ea, eb basis

a|U b|U =

3 22 1

1 10 1

so

|ab

|U

|=

|ab

|=

.

2

Hence we have a group homomorphism

: W() GL(V()) : a (a X) a is called geometric representation of W().

Suppose = 12 (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() = W(1) W(2)Proof. Construct inverse group homomorphism : W() W(1) W(2), : W(1) W(2) W().If a X, let

(a) =

(a, 1) if a X1(1, a) if a X2

Since all relations of W() hold in W(1) W(2) such uniquely extends to a group homo-morphism.

If a Xi X, i(a) = a extends to a group homomorphism i : W(i) W() (allrelations of W(i) hold in W() )

Notice that generators of the image of 1 commute with generators ofIm 2. Hence subgroups

Im 1, Im 2 commute. Hence (x, y) = 1(x)2(y) is a well-defined group homomorphism.

Since a W(), ((a)) = a, = IW(). Similarly = IW(1)W(2). 2

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is connected if = 12 = 1 = or 2 = .Each has connected components 1, 2, . . . s.t.

= 123 . . .Corollary 5.5. In this case W()

= W(1)

W(2)

. . . and V() = V(1)

V(2)

. . .

with V(i)V(j) for i = j.

Proof. Exercise (use transfinite induction). 2

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

Example: Take a Reflection group (G, V), then (V, i : G GL(V)) is a representation.If is a Coxeter graph, then (V(), ) is a representation of W().A subrepresentation of (V, ) is a subspace U V) s.t. g G : (g)(U) U. So (U, |GL(U))is a representation.V is irreducible if V

= 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 is connected, then any proper W()-subrepresentation of V() lies in(V()).

Exercise: (V()) is a subrepresentation of V().Proof. Let U V = V() be a subrepresentation. Then a X ( = (mab)a,bX), a(U) U. Observe that ea / U. Otherwise: b X, mab = 2 holds:

U b(ea) = ea + 2 cos mab

=0

eb = eb U

Since 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 Rea . Pick any z U and write z = ea + z where z Rea . Then

U a(z) = z 2 ea, z ea= ea + z

2 ea, ea ea 2

ea, z ea

= ea + z

Hence z = 12(z + a(z)) U and 2ea U = = 0. Hence z = z Rea and thereforeU Rea and U

Rea = V. 2

Corollary 5.7. If is connected and V() nonsingular (V() = 0), thenV() is irreducible.

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

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Proof. Pick a subrepresentation U V.Observe that direct sum representations V = U W are in bijection with linear operators

p 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) pPick any vector space decomposition and let p be the corresponding operator, : G GL(X).Define p = 1|G|

xG(x)p(x1)

if t U

p(t) = 1|G| xG

(x)(p((x1)(t) U

))p|U=I

=1

|G|xG

(x)(x1)(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 (x), x G):

(x)p = 1|G| yG

(x)(y)p(y1)

=1

|G|yG

(xy)p((xy)1)(x)

=1

|G| zG

(z)p(z1

)(x)

= p(x)Hence ker p is a G-subrepresentation and V = U ker p.

2

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

Proof. Irreducibility follows from nonsingularity by Corollary 5.?. Suppose V = 0 (V =V()). 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 is non-empty. 2

Proposition 5.10. (Schurs Lemma)If is connected and |W()| < then EndW()V() = R.

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Proof. Pick EndW()V(). Then a X : a = a. Therefore: z V() :(a(z)) = a((z)).

(z) 2 ea, z (ea) = (z 2 ea, z ea) = (z) 2 ea, (z) eaHence

ea, z

(ea) =

ea, (z)

ea. Let z = ea, then (ea) =

ea, (ea)

ea. Therefore, hasan -eigenvector, I EndW()V().0 = ker I is a subrepresentation of V(). By 5.9,

ker I = V() = I = 0 = = I

2

Theorem 5.11. If |W()| < then (V(), , ) is euclidean.Proof. |W()| < = X is finite = dim V() = |X| < .If = Ti, Ti connected, then V() = i V(i) with V(i)V(j), i = j. Hence it sufficesto consider a case of connected .Pick any Euclidean form , 1 over V(). Let us integrate , 1 over W() so let

u, v2 =

xW()x(u), x(v)1

, 2 is symmetric since , 1 is u = 0, u, u2 =

x

x(u), x(u)1 >0

hence , 2 is euclidean.

y

W():

y(u), y(v)2 =

x

xy(u), xy(v)1

=

x

xy(u), xy(v)1

=

zW()z(u), z(v)1

= u, v2Hence all y O(V(), , 2). Remember that all u O(V(), , ) and consider : V

V defined by

(u), v2 = u, v v EndW()V() since for all x W():

x((u)), v2 =

(u), 1x (v)2

=

u, 1x (v)2

= x(u), v= (x(u)), v2

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This implies x = x. By 5.10, = I for some R, so for all u, v:

u, v2 = u, v

Since ea, ea = 1 = ea, ea2 = = 1ea,ea2 > 0. 2

Section 4 = Finite Reflection group = Coxeter groupSection 5 = Finite Coxeter group Reflection group (V(), (W()))Problem: 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 areisomorphic as groups.

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6 Fundamental chamber

Aim: : W() Im() is isomorphic (at least for finite Coxeter groups)Define l : W() Z0 : l(x) = min{n : x = a1 an, ai X()}.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 acontradiction. 2

For each a X() consider half spaces

H+a = {u V() : u, ea > 0} Ha = {u V() : u, ea < 0}

the dividing hyperplane Ha = {u V() : u, ea = 0}and the cone C =

aX()

H+a which we call the fundamental chamber.

Idea: Let x W :xi(Ha)

x(C)

C

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

Let = (mab)a,bX and denote W = W(), V = V()

Lemma 6.2.

|W

| 0,

g1(z), eb

> 0

On the other hand

g1(z), g1(ea)

= z, ea = 0.

As g a, b: g1(ea) = ea + eb , R

We denote by () a root system of (W) V() and since ea, eb form a simple system of() span(ea, eb), either

0, 0 s.t. not both are 0 and theng1(z), g1(ea)

=

g1(z), ea + eb

=

g1(z), ea

+

g1(z), eb

> 0

which is a contradiction

OR 0, s.t. not both are 0 and therefore

g1(z), g1(ea) < 0which is a contradiction too.

It remains to prove the length statement if g(H+a H+b ) Ha . 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

bLba

L

L

H+aHa

gL Ha U g = a or g = ab or . . . or g = ab ab . . .b mab

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

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Remark 6.3. There are 2 finiteness assumptions:

1. W is finite

2. is finite, i.e. |X| < and a, b X : mab < .

Exercise: W is finite = is finite.Hint: a = b X = a = b GL(V()) = a = b W(). Then one needs to show thatmab < . This follows from ab|span(ea,eb) =

1 10 1

.

Note: Lemma 6.2 holds under assumption that is finite.

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

1. For allw W and for all u X: either w(C) H+a OR

w(C) Haand

l(aw) = l(w) 12. for allw W and for all a = b X there is a g a, b s.t.

w(C) g(H+a H+b ) and l(w) = l(g) + l(g1w)

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 provedStep 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) Ha . w(C) Ha = w(C) = a(w(C)) a(Ha ) = H+a w(C) H+a = w(C) = a(w(C)) a(H+a ) = Ha

Case 2: a = 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(g1w). 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 ) Ha and l(adg) = l(dg) 1.Then w(C) dg(H+a H+d ) Ha and

l(aw) = l(adgg1w) l(adg) + l(g1w) l(g) + l(g1w) = l(w) = k 1

= l(aw) = l(w) 1

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2. Now assume (1) and (2) for l(w) = k1 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 H+a H+b .

Ha

Hb

H+a H+b

H+a HbHa Hb

Ha H+b

Observe that wC Ha = . wC w(H+a H+b ) is in either H+a or Ha by 6.2.Wlog wC Ha .By part (1): l(aw) = l(w) 1.Use part (2) on w = aw (by induction assumption) then there is g a, bs.t. wC g(H+a H+b ) and l(w) = l(g) + l((g)1w).Claim: g = ag satisfies (2) for w.

wC = awC ag(H+a H+b ) = g(H+a H+b )

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

w)= 1 + l(g) + l(g1w) l(ag) + l(g1w)= l(g) + l(g1w) l(gg1w) = l(w)

Hence l(w) = l(g) + l(g1w).

2

Theorem 6.5. 1. W is finite = V = gW

gC

2. is finite, then:

g, x

W : gC

xC

=

=

g = x

Note: If W is infinite

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

Proof.

1. C = { W : a X : , a 0}.Recall the height function h : V R : v =

aXaea

aX

a.

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

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Claim: z C and this implies that v = g1(z) g1(C) and V = gC.Indeed, otherwise there is an a X s.t. z, ea < 0. Then:

a(z) = h(z) 2 z, ea eahas hight

h(a(z)) = h(z) 2 z, ea > h(z)which is a contradiction to the maximality.

2. Assume gC xC = . Then C g1xC = . If g1x = 1 then g = x and we are done.If g1x = 1 then there is an a X s.t. l(ag1x) = l(g1x) 1.Set w = ag1x, then l(aw) = l(w) + 1. By 6.3 ag1xC = wC H+a = g1xC Ha .Then C g1xC H+a Ha = which is a contradiction.

2

Corollary 6.6. is finite. Then : W

GL(V) is injective.

Proof. Suppose (g) = (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 (G)).Say (G, V) is essential if V1 = V.

Corollary 6.7.

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

Proof.

(W(), V())X = is a simple system in (G), mab = |SaSb| (G, V)

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

V() V

xXxex xXxxW() G

a1a2 an Sa1 San

This is an isomorphism by 4.3 and 6.5. 2

Some fun bits:

Rn \Rn1: disjoint union of 2 contractible half-spaces

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Cn \Cn1: is connected with fundamantal group ZTry CC = CR V CHa = CR Ha. Then CV0 = CV \

gW, aX

g(CHa) is connected.

W acts freely on CV0 and the universal cover ofCV

0 is simply connected (if W is finite).

One can draw the fundamental group of CV0

/W

:

Ha

x

a(x)Ta

C C V0

[x] = [a(x)] CV0

/W.

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

B() Braid group

= 1 CV0/W = Ta, a X : TaTbTa mab

= TbTaTb mab

V0 V0/W gives

1 P B() pure Braid group

B()W 1

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7 Classification

Let (V, , ) be an Euclidean vector space. Therefore it has a normv =

v, v

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

T = supvV\{0}

T v/v

1. T x T x2.

T

0 and [

T

= 0

T = 0]

3. T + S T + S4. T O(V) = T v = v = T = 15. O(V) is a closed bounded subset of LinR(V, V).

Theorem 7.1. Let be finite. Then W() is finite V() is Euclidean.Proof.

"=" Theorem 5.9"=" short version: By 6.4 W() acts discretely on the unit sphere in V(). Hence it mus

be finite.Long version: pick v C s.t. v = 1. Pick > 0 s.t. B(v) C.g = x W = gC xC = = B(gv) B(xv) =

= gv xv > = (g) (x) > By fact 5, O(V) is compact. Suppose W is infinite, W O(V) compact. Then there isa convergent sequence T1, T2, . . . , T n in W s.t. Ti = Tj for i = j.Hence Ti T O(V), so there is a N s.t i, j > N : Ti Tj < .But Ti, Tj (W) with Ti = Tj, so this is a contradiction.

2

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

(W(1) . . . W(n), V(1) . . . V(n) Rk)where the i are connected positive definite Coxeter graphs and each W(i) acting trivially onV(j) with i = j and Rk.

Problem: What are the connected positive definite Coxeter graphs?

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Proposition 7.2. A =

a11 a1n... ...an1 ann

symmetric matrix overR.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

=1...

n

: n = 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.

QAQ1 =

1 . . .n

and A is positive definite all i > 0, we conclude that det(A) = 1 n > 0.

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

Rn = U

Uunder xTAyAny 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 = 0 and z, z > 0.Let f1, . . . , f n1 be an orthonormal basis ofU, fn U, fn = 0. xTAy in the basis

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f1, . . . , f n 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 = (X, mab). We say that = (Y, nab) is a subgraph of if

Y X a, b X : nab mab

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

definite.Proof.

2 nab mab = cos nab

cos mab

. Pick any v =

aYaea V() and suppose v, v 0. Let w =

aY

|a|ea. Then:

w, wV() =

a,bY|a| |b|( cos

mab)

a,bY|a| |b|( cos

nab)

a,bY

ab( cos nab

)

0

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

Let d() = det(2 ea, en C()

)XX = det(2cos mab )a,bX.

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

Lemma 7.4. Let be of the following form:

1

n n-12

Then d() = 2d(1) d(2).

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

0C(2)

......

0 2 10 0 1 2

Expanding the last row:

d() = 2d(1) + (1)n+(n1)(1)det

0

C(2)...0

1

= 2d(1) d(2)2

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

An Bn 4

qq

qqq

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 2cos m2cos m 2

= 4 4(cos m )2 > 0

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a1 = det(2) = 2a2 = i2(3) = 4 4(cos 3 )2 = 3

Lemma 7.4 allows us to compute an by

An

An2 An1

Then an = 2(an1) an2 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 Bn2

Bn1 = bn = 2bn1 bn2 = bn = 2

Before computing Dn notice that:d(12) = d(1) d(2)

The decomposition for Dn is:

HH

HHH

HHHHH

A1An3

An1 Therefore dn = 2an1 a1an3 = 2n 2(n 3 + 1) = 4.

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Similarly: en = 2an1 a2an4 = 9 n.Hence en is positive for n 8.

h2 = i2(5) = 4 4cos2 5 = 55

2

h3 = 2h2

a1 = 2

552

= 3

5 > 0

h4 = 3h3 h2 = 2(3

5) 55

2 =735

2 > 0

Notice that h5 = 3h4 h3 = 5 2

5 < 0.

2

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

An

zzzz

hhhh

zzzz

hhhhBn 4

zzzz

Cn 4 4

hh

hh

zzzzBn

zzzz

hhhh

hh

hhE6

zzzz

hh

hh

E7

zzzz

hh

hhE8

zzzz

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F4 4

G2 6

H3 5 H4 = H5 5

Proof.

We have computed

e8 = e9 = 9 9 = 0.

h4 = h5 = 4 25 < 0 an = det

2 1 1

1 2 1. . .

. . .. . .

. . .. . . 1

1 1 2

= 0 because the sum of the rows is zero

bn = 2bn a1bn2 = 4 4 = 0dn = 2dn a1dn2 = 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 25 < 0

c( Cn) looks like

0c(Bn1) 0

0 2 20 2 2

.

Similarly to Lemma 7.4, expanding the last row gives

d( Cn) = 2d(Bn) 2d(Bn1) = 2 2 2 2 = 02

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Theorem 7.7. The connected Coxeter graphs which give a finite W() 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() are positive definite in 7.5 and 7.1 tells us that all W() are finite in

7.5.Now let be a connected graph with finite W(). Then c() is positive definite by 7.1. Inparticular, contains no subgraphs in 7.6

No An = has no cycles No D4 = has no vertex with 4 edges No Dn, n 5 = contains at most one vertex with 3 edges.

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

x1 x2 xm

If has 2 vertices it is either A1 or I2(m), both in Lemma 7.5.Let 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 = = AnCase 1.2 There is one edge of multiplicity 4:

if it is on a side, = Bn

if it is in the middle then no

F4 = = F4

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

Case 2 there is a vertex with 3 edges. Then 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 = has no multiple edges. No E6 = c = 1 No E7 = b 2 if b = c = 1 = = Da+3 if b = 2, c = 1, then no E8 = a < 5. As a b = 2 there are 3 cases:

1. a = 2 = = E6

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2. a = 3 = = E73. a = 4 = = E8

2

crystallographic NON crystallographic

An, Bn, Dn, F4, En, G2 = I2(6) I2(m) for m = 5, m 7, H3, H4d() Z d() / Z

the character table of is integral non-integralall but one representation of W()

can be realised by integral matrices

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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() is crystallographic if there is a lattice L V() s.t.

g W() : g(L) L

Informally this means that there is a basis e1, . . . , en of V() s.t. (g) is an integer matrix foreach g W().Lemma 8.1. Let |X| < . If W() is crystallographic then a, b X :

mab {2, 3, 4, 6, }

Proof. Assume W() is crystallographic. Then g W : (g) can be written as an integermatrix, in particular T r(()) Z.If mab = then we are done.Suppose mab = = U = span(ea, eb) is Euclidean = W() = U U.In the basis ea, eb, any basis of U:

(SaSb) =

cos 2mab sin 2mab 0 sin 2mab cos 2mab 00 0 I

T r((SaSb)) = |X| 2 + 2 cos 2mab hence

2cos2

mab Z cos 2

mab 1

2Z mab {2, 3, 4, 6}

2

Note: = is true but the proof is harder unless W() is finite.Theorem 8.2. LetW() be a finite Coxeter group. ThenW() is crystallographic a, b X : mab {2, 3, 4, 6}.Proof.

"=" Lemma 8.1

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"=" Let us choose a new basis a = caea, a X, ca R \ {0}. Then

a(b) = b 2 b, ea ea = b 2 a, ba, aa

Since a = Sea = Sa and (a1

an) = a1

an it suffices to ensure that for all

a, b X :2a, ba, a Z

Wlog is connected.

mab = 2 = a, b = 0 = 2 a,ba,a = 0 mab = 3 = a, b = cacb ea, eb = caca cos

mab = 12

= cacb2

= 2 a, b

a, a = cacb

caca = cbca

Need in this case ca = cb mab = 4 = a, b = cacb( cos 4 ) = cacb2

= 2 a, ba, a =

2cbca

Need ca =

2cb or cb =

2ca

mab = 6 = a, b = 32 cacb

= 2 a, ba, a = 2cacb32

caca=

3

cbca

Need ca =

3cb or cb =

3ca

In each component of we choose

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

Case 2 Bn, F41

choose ca=1

4 2

choose ca=2

Case 3 In G2 = I2(6):

a 6 b

a =

3ea, b = eb

2

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

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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

1 = e1, 2 =

2e2, . . . , k =

2ek which is called the crystallographic reflection groupBn

1 = 2e1, 2 = e2, . . . , k = ek which is called the crystallographic reflection group Cn

semisimple Lie algebra

Weyl group and Root lattice--

crystallographic reflection groupSerre Theorem

nn

KacMoody algebra

ii

Nakajima construction

12

bb

We have seen An, Bn, Dn, I2(m) explicitely.

Can we compute |W()|, |()| for all ?

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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) = ij. In fact, S = R[X1, . . . , X n].G acts on the algebra S: g G, F S, v V : g F(v) = F(g1v).Exercise: check that G acts on S by algebra automorphisms.Note: for F S the following are equivalent:

1. g G : gF = F2. 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 + x22 + 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=0Fj, Fj Sj . Then j: Fj SG.

Proof. G R acts on V: (g, )v = g(v) = g(v). Hence G R also acts on S:(g, ) F : v F(g11v)

In particular, f Sj : (1, ) f = j f.Since actions of G and R commutes, R :

(1, )F =m

j=0

jFj SG

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

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1 F = F0 + . . . + Fm+1 SG

2 F = F1 + 12F1 + . . . + 12m+1 Fm+1 SG

. . .

(m + 1) F = F0 +1

m+1F1 + . . . +1

(m+1)m+1 Fm+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 SGm+1

Since M is invertible,

F0...Fm

= M1H0...

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 S

Gk = S

G

Sk

Let SG+ =

k=1

SGk be the invariants of positive degree. Let I =

SG+

be the ideal ofS generated

by SG+ . By Hilberts Basis Theorem, any ideal in S is generated by finitely many elements.Say I = (f1, . . . , f k)S, fi SG. Write fi =

j

fij where fij Sj. With Lemma 9.1:fij SGj , j < 0.Hence I = (I1, . . . , I r)S where Ij SGdj . If for any k: Ik (I1, . . . , I k1, Ik+1, . . . , I r)S we throwIk away. Wlog, I1, . . . , I r is a minimal system of such generators.

Definition. I1

, . . . , I r

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|gG

gF

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Note that F SG = gF = F = G

F = F and H S :G

(F H) =1

|G|

g

(gF)(gH) = F1

|G|

g

gH = F

G

H

So G

is an SG-linear map S SG.

Proposition 9.2. LetG be any finite group acting on V, I1, . . . , I r SG = R[V]G fundamentalinvariants. Then:

F SG p(z1, . . . , zr) R[z1, . . . , zr] s.t. F = p(I1, . . . , I r)

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=1HjIj, Hj S. Since d = deg(F), dj =

deg(Ij ):

F =r

j=1

(Hj)ddjIj

and therefore

F =

G

F =r

j=1

Ij

G

(Hj)ddj

By induction assumption j pj R[z1, . . . , zr] s.t. G

(Hj)ddj = pj (I1, . . . , I r). 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 Iiis surjective.

1. Injectivity holds only for reflection groups

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

3. SG0 = R 1 constant functions

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4. f SG1 , f = 0 = V = Ker(f) R. Hence SG1 = 0 GV has a trivial constituent5. if , is a Euclidean form on V then [x, y] =

G

x, y makes G orthogonal. HenceF(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 = xn + b(x1, . . . , xn1) with = 0. Let us divide P byL with remainder by replacing every xn with 1 b(x1, . . . , xn1). xn = 1 L 1 b(x1, . . . , xn1).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, . . . , xn1]. If B = 0 we are done. IfB

= 0 pick

a1, . . . , an1 R s.t. B(a1, . . . , an1) = 0.Let an = 1 b(a1, . . . , an1) = L(a1, . . . , an) = 0 = P(a1, . . . , an) = 0.But P(a1, . . . , an) = B(a1, . . . , an1) = 0 which is a contradiction. 2

Proposition 9.4. Let G = W be a finite reflection group. Let J1, . . . , J k SW s.t.J1 / (J2, . . . , J k)SW. If Pi Sbi s.t.

ki=1

PiJi = 0 then P1 (SW+ )S.

Proof. Observe that J1 / (J2, . . . , J k)S: if J1 =k

t=2HtJt with Ht S then

J1 =

J1 =

t

Jt

Ht (J2, . . . , J k)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, . . . , J k)S =

P1 = 0 (SW+ )Sdeg(P1) < N: done

deg(P1

) = N: Let

W be the root system of G.

V let L

S1

be L

(v) =, v

.

L = 0 = v = S(v) = v

= j : (SPj Pj)(v) = Pj(S1 v) Pj(v) = 0

With Lemma 9.3: SPj Pj = LPj for some Pj S.Note that deg(Pj) < deg(Pj)

P1J1 + . . . + PkJk = 0 (SP1)J1 + . . . + (SPk)Jk = 0

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(P1 SP1)J1 + . . . + (Pk SPk)Jk = 0Hence L(P1J1+. . .+PkJk) = 0. Since S is a domain P1J1+. . . P kJk = 0. By inductionassumption, P1 (SW+ )S. Hence SP1 P1 = LP1 (SW+ )S.For each polynomial F, denote

F = F + (SW+ )W SW. Hence : S

P1 =

P1.

Since W is generated by S,

g

W : gP1 = P1. Hence WP1 = P1 (W acts on SW andF = F)So P1

W

P1 + (SW+ )S (SW+ )S.

2

Lemma 9.5. (Eulers formula)F Sk = R[x1, . . . , xn]k, then

nj=1

xjF

xj= kF

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

ai = k. Then

xjF

xj= ajF

Hencen

j=1

Fxj

= (

ai)F = kF. 2

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

1, . . . , I

rare algebraically independent.

Proof. Consider a grading on R[z1, . . . , zr] with deg zi = di = deg Ii. Let P = Pj, Pjhomogeneous of degree j. Then

P(I1, . . . , I r) =

j

Pj(I1, . . . , I r) Sj

= 0

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

Pzj

and fj = P(j)(I1, . . . , I r) SW is homogeneous. Let us fiddle around with theideal

K = (f1, . . . , f r)SW SW

WLOG, f1, . . . , f m 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, . . . , f m)SW so there areQij SW, Qij homogeneous of degree deg Ii deg Ij s.t. fi =

mj=1

Qijfj.

P(I1, . . . , I r) = 0 = 0 = P(I1, . . . , I r)xk

=r

i=1

P

zi(I1, . . . , I r)

Iixk

k {1, . . . , n}

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Hence:

k :m

i=1

fiIixk

+r

i=m+1

mj=1

QijfjIixk

= 0

k :m

i=1fi Iixk +

rj=m+1

Qji Ijxk = 0

By 9.4, k i {1, . . . , m} : IIxk

+r

j=m+1

QjiIjxk

(SW+ )S

Hence for some Hkt S :

i, k : IIxk

+r

j=m+1

QjiIjxk

=r

t=1

ItHkt

Multiply each equality by xk and sum over k. By Eulers formula:

i : diIi +r

j=m+1

djQjiIj =r

t=1

It Htwhere Ht have no free terms ( Ht = xkHkt ). Ii enters both sides of the equality. If we considerthe 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. ThereforeIi =

j=i

IjGj = Ii (I1, . . . , I i1, Ii+1, . . . , I r)

which is a contradiction. Therefore all Pxk = 0 = P = 0. 2

Example: Sn acting on Rn by permutations.

Rn = U R where U =

1...

n

:

i = 0

and R = R

1...1

(Sn, U) is then a type An1 essential reflection group with elementary symmetric functions:

1(x1, . . . , xn) = x1 + . . . + xn 2(x1, . . . , xn) = x1x2 + x1x3 + . . . + xn1xn . . . n(x1, . . . , xn) = x1 xn

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is a system of fundamental invariants. The power functions j = xj1 + . . . + x

jn, j = 1, . . . , n is

another fundamental system.

An1: 1|U = 02|U, . . . , n|U are fundamental invariants with fundamental degrees 2, 3, . . . , n.

Bn: group has more transformations xi xi, SW(Bn) = (SSn)Cn2 = R[1, . . . , n]Cn2 .Hence the fundamental invariants are j(x

21, . . . , x

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

Dn: in Dn, instead of the full group Cn2 , it is a subgroup of index 2. So n = x1 xn SW(Dn).

So the fundamental invariants are 1(x2i ), . . . , n1(x

2i ), n(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, . . . , I r

fundamental invariants. Then

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

Proof. The natural algebra homomorphism

: R[z1, . . . , zr] SG, F(z1, . . . , zr) F(I1, . . . , I r)

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

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

: f, g S, g = 0}

Transcendence degree is trdegR F = n. The smaller field A = R(I1, . . . , I r)

F, trdegRA = r.

From Galois theory, trdegA F = trdegR FtrdegRA = nr. 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) =gG

(z g.f)

Since t G, t.h = gG

(z tg.f) = h = h A[z].

h(f) =

gG(f g.f) = (f f)

g=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. IfI1, . . . , I n is another system of fundamental invariants with degreesd

1, . . . , d

n, d

j =

deg Ij then Sn s.t. j : dj = d(j)

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

Ij = Fj(I1, . . . , I

n) and I

j = Gj(I1, . . . , I n) j

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

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

and therefore

zjzk

=n

s=1

Fjzs

(G1(z1, . . . , zn), . . . , Gn(z1, . . . , zn)) Gszk

(z1, . . . , zn)

And on matrix level:

In =

Fizs

I1,...,I

n

Gszk

I1,...,In

M atn(S)

Hence detFizs is invertible in S = R[x1, . . . , xn]. Hence Sn s.t.F1

z(1) Fn

z(n)= c 1 + higher degree terms, c = 0

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

i : di = d(i)

2

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10 Fundamental degrees

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

Lemma 10.1. A finite group G acts in a finite dimensional vectorspace overC(by : G GLn(C)). Then x G : (x) is diagonalisable with eigenvalues of absoulutevalue 1.

Proof. |G| = n < = xn = 1 = (x)n = In.Hence the minimal poynomial of (x), x(z) divides z

n 1. Therefore eigenvalues satisfyn 1 = 0, hence n = 1 = ||n = 1 = || = 1.Further, since zn 1 has no multiple rootes, x(z) has no multiple roots and all Jordan blocksof (x) have size 1. 2

Let C[[t]] = {

n=0nz

n} be the ring of formal power series. If V =

n=0Vn is a graded vector

space with dim Vn < , the "right" notion of dimension of V is the Poincar polynomial

pV(t) =

n=0

dim Vn tn

Example:

pC[z](t) = 1 + td + t2d + t3d + . . . = 11td where deg z = d. pVW(t) = pV(t) +pW(t) pVW(t) = pV(t)pW(t) pC[z1,...,zn] =

j

pC[zj ](t) =1

1td1 11tdn where deg zi = di, C[z1, . . . , zn] = C[z1] . . . C[zn].If d1 = . . . = dn = 1 then pC[z1,...,zn](t) =

1(1t)n .

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

x(t) = det(IVt(x))1 = j

11 j t = j (1+jt+2j t2+. . .) = j=0 tj ( a1+...+an=j a11 ann )where 1, . . . , n are the eigenvalues of (x).

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

1

|G|xG

x(t) =n

j=1

1

1 tdj C[[t]]

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

x|S1 belongs to

1 0. . .0 n

in some basis e1, . . . , en of S1. Then ea11 eann witha1 + . . . + an = j form a basis of Sj and

xea11 eann = a11 ann ea11 eannhence

tr(x|Sj ) =

a1+...+an=j

a11 ann

Hence

LHS =1

|G|

xG,j=0tr(x|Sj )tj

Remember j =

G: Sj SGj , j(v) = 1|G| gG

gv. In particular LHS =

j=0tr(j)t

j.

2j = j = j

I 00 0

and tr j = rk j = dim S

Gj

Hence LHS = pSG(t). 2

Corollary 10.3. |G| = d1 dn.

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

2|(G)|

number of reflections

.

Proof. Multiply 10.2 by (1

t)n:

1

|G|xG

x(t)(1 t)n =n

j=1

1 t1 tdj

1

|G|xG

(1 t)ndet(1 tg) =

nj=1

1

1 + t + t2 + . . . + tdj1 =:(t)

=1

|G|( 1

g=1

+1

2||1 t

1 + t

gSx, x+ 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 :

||2|G|

2(1 + t)2

+h(t)(1t) = nj=1

(t)(1+ t+ . . .+tdj1)

1 1 + 2t + 3t2 + . . . + (dj 1)tdj2

(1 + t + . . . + tdj1)2

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Plug in t = 1:

||4|G| =

nj=1

djd1 dn

(dj 1)dj

2d2j

As |G| = d1 dn

||4

=n

j=1

dj 12

= 12

(( dj ) n)2

Example:

I2(m) has fundamental degrees 2, m.

|G| = 2m, || = 2((m + 2) 2) = 2m

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

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

n(n + 3)

2 n

= n2+3n2n = n2+n

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11 Coxeter elements

Let (G, V) be a finite reflection group, V a root system.A Coxeter element is w = S1 Sn for some simple system = {1, . . . , n}.There are two choices involved:

simple system order on

In Sn the choices will lead to

S1 = (a1a2), S2 = (a2a3) . . . S n1 = (an1an)

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

Lemma 11.1. LetG 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

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

i : 1(Xi) Xi, i(c) = (c)

Then

g = g0f(a)g1f(b)

= g0f(a)f(b)g1

g1(g0f(a)f(b)g1)g11= g1g0f(a)f(b)

This is conjugate to g by Case 1.

Case 3 (j) = b, j < n, X0, X1, 0, 1 as above. Then

g = g0f(b)g1 g1g0f(b)done by Case 2.

2

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

Proof. Let C = { Coxeter elements }. ThenxS1 Snx1 = (xS1x1) (xSnx1) = Sx(1) Sx(n) C

Hence xCx1 = C, so C is the union of some conjugacy classes. Let w1 = S1 Sn, w2 =S1 Sn C. Then there is a g G s.t. {g(1), . . . , g(n)} = {1, . . . , n}. Hence gw1g1 =Sg(1)

Sg(n) 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 1, . . . , n be the eigenvalues of w|V(W). Since wh = 1, hk = 1 = k = e

2imkh for a

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

Finite Coxeter graphs are bipartite (X = X1 X2) Example: D6:

dd

dddd

d

~~~~~~~

dd

dddd

d

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E8:

~~~~~~~

ddddd

dd

dd

dddd

d

For a, b Xi = ab = ba. Hence j =

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

2j =

a2 = 1. = 12 is called canonical Coxeter element.G = 1, 2 is a dihedral group of order 2h.

Proposition 11.3. Let us order the exponents 0 m1 . . . mn < h. Then1. m1 12. j : h mj is an exponent

3.n

j=1mj = h

n2

Proof.

1. ifm1 = 0 then e0 = 1 is a eigenvalue of . Hence x V(W) s.t. x = 0, x = x.

2x = 112x = 1x = 1x

Let a Xj : Sa(c) = a x = x 2 ea, x ea. Hence

1x = x +

aX1aeq = 2x = x +

bX2

beb

=

aX1aea =

bX2

beb

and therefore all a, b = 0. So 1x = 2x = 0. Consider

1|V(W) ajX1= Sa1 Sak : V1 y yV1 y ywhere Vi =

aXj

Rea.

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

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2. G acts on V(W). Let f(z) be the characteristic polynomial of |V(W). Thenf(z) = f1(z) fs(z)

where fj(z) R[z] and monic irreducible in R[z].Case 1 fj(z) = z

,

R.

Since h = 1 = h = 1 = = 1. Note that 1 is not an eigenvalue of bypart (1), so = 1, h is even and = e2ih h2 , h2 is the exponent.

Case 2 fj(z) = (z )(z ), / R.Then if = e

2ih

mk then = e2ih

mk = e2ih

(hmk).

3. This also gives part (3) because the average of exponents in each fj(z) ish2 .

2

Lemma 11.4. LetW 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-linear3. i|P = Szi4. zi C, the closure of the fundamental chamber5. P C = R>0z1 +R>0z2

Proof. ai is the dual basis to eai . That means

ai, eaj

= ij.Easy claims:

C =

x :

x, e

a 0

= a R0a Vj =

a /Xj

Ra

angle between a and b = angle between ea and ebThe angles between a and b are accute and a, b 0 and a, b = 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(

a) = e

ain the basis

ai.

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

the new form [ , ] given by Q I in the basis eai .Note: [x, x] 0 and [x, x] = 0 x ker[ , ] = K. K = 0, K = eigenspace of q.Pick z = 0, z =

aaa K and let z = |a|a.

Now, the idea is to use that zj =

aXj|a|a, but we have no time to finish the proof. 2

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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 12.

|

|= hn

Proof. (Idea) H P is G-conjugate to Rz1 or Rz2.

if h is odd:

Rz1

Rz2

There are n2 hyperplanes H s.t. H P is one of these h lines if h is even:

Rz1

Rz2

There are h2 lines with one intersection pattern andh2 lines with another intersection

pattern

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

Ijzk

in two different ways. 2

Corollary 11.7. |W| = j

(mj + 1), || = 2j

mj.

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Corollary 11.8. Let W be connected.

1 =

1. . .

1

(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

(|V(W))h2

e2ih

mk

. . .

e2ih

mk

h2

=

eimk

. . .

eimk

=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

h(z) =

1mh1

gcdm,h=1

(z e 2ih m)

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

e2ih )

|f(z) in C[z]. Therefore

h(z)|f(z) in Z[z]

and therefore all e2ih

m are eigenvalues. 2

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

2408 = 30. The number of coprimes

of 30 is(30) = (2)(5)(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

|| h = ||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 32H3 30 10 1, 9 5 2 6 10 = 120H4 120 30 1, 29 11, 19 2 12 20 30 = (120)2

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Calculations:Since exponents come in pairs (m, hm) 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 = 1264321if = e

2im12 where m is a missing component, there are 4 possible cases:

2() = 0 6, 6 4() = 0 3, 9 3() = 0 4, 8 6() = 0 2, 10

One needs to write the Coxeter transformations itself to see that 3() = 0.

H4: can be realised by matrices with coefficients inQ(5), z301 = 30 . OverQ(5), 30 =(1)30

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

(0)30 and we know that

(0)30 (e

2i30 ) = 0. The roots of

(0)30 are:

e2i30 , e

2i30

29, e2i30

11, e2i30

19

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

Every reflection of the Dodecahedron fixes two opposite edges. Hence

|| = 2|reflections| = 2 |edges|2

= 30

and|W| = 20

|vertices|6 = 120

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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

: USU2(C)

SO(3)(R) = SO(Him) : q (x qxq)which is 2:1. Then H4 =

1(Rot.Symm(Dodec)) = |H4 | = 2 60 = 120.F4 =

1(Rot.Symm.(Cube)) = |F4 | = 2 24 = 48.E8, E7: There exists a 8-dimensional algebra (octonions). If G , |G| < , |G| is even

= G is a root system. E8 is the "group" of invertible octavian integers E7 are the elements of order 4 in it.

rumynin - reflection groups

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