Reflection Groups 3

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 8 x; y 2 V : T (x) · T (y) = hx; yi) T : V −! R . In (V; h ; i) you can • measure length: jjxjj = phx; xi hx;yi • measure angles: xyc = arccos jjx||·||yjj • talk about orthogonal transformations O(V ) = fT 2 GL(V ): 8 x; y 2 V : hT x; T yi = hx; yig ≤ GL(V ) T 2 GL(V ). Let V T = fv 2 V : T v = vg the fixed points of T or the 1-eigenspace. Definition. T 2 GL(V ) is a reflection if T 2 O(V ) and dim V T = dim V − 1. Lemma 1.1. Let T be a reflection, x 2 (V T )? = fv : 8 w 2 V T : hv; wi = 0g; x 6= 0. Then 1. T (x) = −x hx;zi 2. 8 z 2 V : T (z) = z − 2 hx;xi · x Proof. 1. Pick v 2 V T =) T v = v =) hT x; vi = hT x; T vi = hx; vi = 0. Hence T x 2 (V T )?. T ? T Since dim (V ) = dim V − dim V = 1 : T x = α · x for some α 2 R. Then α2 hx; xi = hαx; αxi = hT x; T xi = hx; xi Since hx; xi= 6 0; α2 = 1 =) α 2 {−1; 1g. If α = 1 =) T x = x =) x 2 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 2 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 2 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 2 V is called a root of T . 2. 8 x 2 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 = hfSx : Sx 2 Ggi, 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 = f1g [ fa1 a2 ··· an : ai 2 Xg Equivalence: −1 (G1;V1) ∼ (G2;V2) if there is an isometry ' : V1 −! V2 s.t. 'G1' = G2. | −{z1 } f'T ' : T 2G1g Problem: Classify all finite reflection groups (G; V ) up to ∼. Example: of finite reflection groups 1. x α y We want to know hSx;Syi det(Sx;Sy) = 1 =) SxSy 2 SO2(R) =) SxSy = Rotβ. Since Sx(Sy(y)) = Sx(−y) =) SxSy = 2α. α • π 2= Q =) jSxSyj = 1 =) hSx;Syi is not finite α m 2πm • π = n 2 Q (for m; n rel. prime) =) jSxSyj = jRot n j = n =) hSx;Syi = D2n Dihedral 2 I2(n) = (D2n; R ); jI2(n) = 2nj n 2. Sn-symmetric group acting on f1; : : : ; ng extend to action of Sn on R . n If ei 2 basis of R , if σ 2 Sn, then Tσ : ei 7−! eσ(i);Tσ 2 On(R) 4 011 B1C x = B Chas the property that forall σ : T (x) = x =) x 2 ( n)Sn ; B.C σ R @.A 1 0 1 α1 ? B . C P x = f@ . A : αi = 0g 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 0 1 α1 B . C ? y = @ . A 2 (ei − ej) () αi = αj () T(i;j)(y) = y αn In particular, jAmj = (m + 1)! and A2 ∼ I2(3) n 3. F = Z/2 = f0; 1g - field of two elements. Consider action of Sn on F ;"1;:::;"n basis n Z of F . 8 σ 2 Sn : tσ("i) = "σ(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;"i)i and T((i;j);0) = Sei−ej ;T(1;"i) = Sei . n jBnj = n!2 ;B1 ∼ A1;B2 ∼ I2(4) 0 1 α1 n B . C n P n 4. In Example 3, let F0 = f@ . A 2 F : αi = 0g codim-1 subspace in F or index 2 αn subgroup. n n n σ 2 Sn; a 2 F0 =) tσ(a) 2 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;"i + "j)i, T((i;j);0) = Sei−ej ;T(1;"i+"j ) = Sei+ej . n−1 jDnj = n! · 2 ;D1 is trivial (f1g; 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 ) := fx 2 V : kxk = 1;Sx 2 Gg Φ has the following properties: 1. x 2 Φ =) Rx \ Φ = fx; −xg 2. jΦj = 2·( number of reflections in G) 3. T 2 G; x 2 Φ =) T (x) 2 Φ Property 3 follows from −1 Lemma 2.1. T 2 O(V ); x 2 V n f0g =) 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 2= Φ 2. x 2 Φ =) Rx \ Φ = fx; −xg 3. x; y 2 Φ =) Sx(y) 2 Φ 6 Example: 1. Φ(G;V ) where (G; V ) is a finite reflection group 2. ; = Φ(f1g;V ) 3. is a root system, not Φ because there are vectors of 2 different length. Chop- p (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. 8x 2 Φ: x = αy · y where either all αy ≥ 0 or all αy ≤ 0 y2Π β γ α β Example: In fα; βg is a simple system but is not simple since α −β 2 Φ. Lemma 2.2. Π ⊂ Φ simple system in a root system. Then 8 x; y 2 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 : 0 1 0 1 α1 β1 B . C @···A > @ . A () 9k 2 f1; : : : ; ng s.t. αi = βi 8 i < k and αk > βk αn βn 0 1 α1 P Given an ordered basis e1; : : : ; en of V we get phonebook order on V by writing αiei = @···A. αn If ≥ is a total order on V then x 2 V n f0g =) either x > 0 > −x or x < 0 < −x and V = V+[f_ 0g[_ V− where 7 • V+ = fx 2 V : x > 0g • V− = fx 2 V : x < 0g Definition.

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