Complex Reflection Groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 10 January 2014

Section III

Let Γ𝒞 and put R={r1,,r} the set of generators for W=W(Γ). We define an equivalence relation "" on R as follows. We say riri and if ij we say rirj if and only if there is a sequence of vertices of Γ, ai=ai1,ai2,,aim=aj, such that qikik+1 is odd for all 1km-1. We then write R=R1R2Rn as the disjoint union of equivalence classes. Since ri is conjugate to rj if qij is odd all the elements of an equivalence class Rk are conjugate and we denote by pik the integer which is the common label on all those vertices in Γ corresponding to the generators in Rk.

Lemma 3. Let Γ𝒞 and let W=W(Γ). Let W denote the commutator subgroup of W. Then using the notation just introduced above we have |W/W|= k=1npik.

Proof.

We denote the natural map WW/W by xx. Then since W is generated by r1,,r, W is generated by r1,,r. But in the map WW/W all the elements of an equivalence class Rk are identified. Thus W/W is generated by n elements (perhaps not all different) of orders at most pi1,,pin. Hence |W|k=1npik.

Now let A=εi1××εin the direct product of n cyclic groups of orders pi1,,pin. We define γ:RA by γ(rj)= ( 1,,1, εik kthposition ,1,,1 ) if rjRk. We claim that γ can be extended to a homomorphism of W onto A. For clearly we have (γ(rj))pj=1. Further if the jth and jth vertices of Γ are not joined by an edge, then γ(rj) and γ(rj) commute as A is abelian. Now if qj,j is odd, then rj and rj are in the same equivalence class and hence γ(rj)=γ(rj) so the relation γ(rj) γ(rj) γ(rj) = γ(rj) γ(rj) γ(rj) with qjj factors on each side is obviously satisfied. Finally if qjj is even the relation γ(rj) γ(rj) γ(rj) = γ(rj) γ(rj) γ(rj) with qjj factors on each side becomes ( γ(rj) γ(rj) ) qjj2 = ( γ(rj) γ(rj) ) qjj2 which is obvious as A is abelian. So we have our claim and thus |W||A| =k=1npik.

Corollary 6. Let Γ𝒞 and let W=W(Γ). If ri and rj are in distinct equivalence classes under the equivalence relation "", then no non-identity element of ri is conjugate in W to any element of rj.

Let V be a finite dimensional complex vector space and let G be a finite subgroup of GL(V). We say RG is a reflection if R1 and if there is a hyperplane UV fixed by R. U is called the reflecting hyperplane for the reflection R. We let 𝒰=𝒰(G) be the collection of hyperplanes U in V such that U is the reflecting hyperplane for some reflection RG. Note that if U𝒰 is the reflecting hyperplane for RG and if TG, then T(U)𝒰 as it is the reflecting hyperplane for the reflection TRT-1. Thus G acts on 𝒰. For U𝒰 we let C(U) be the subgroup of G consisting of all elements of G which are the identity on U. Since G is finite each C(U), for U𝒰, is isomorphic to a finite subgroup of \{0} and hence is a cyclic group. We denote its order by e(U). Then if Δ is an orbit of G on 𝒰 and if U,UΔ we have C(U) and C(U) are conjugate, so e(U)=e(U). We thus denote by e(Δ) the common value of e(U) for UΔ.

The following result appears in [Spr1974].

Proposition 9. Let Γ𝒞+ and let G=θ(W(Γ)). Let G denote the commutator subgroup of G. Then G/G is the direct product of cyclic groups of orders e(Δ) as Δ runs through the orbits of G on 𝒰(G). In particular |G/G|=Δe(Δ).

Since for Γ𝒞+, θ is a faithful representation of W(Γ), Proposition 9 gives another way of computing the order of W/W. By using both these methods and comparing results we can give a case by case verification of the following theorem which is a known result for Coxeter groups [Bou1968, p.74].

Theorem 5. Let Γ𝒞+ and let G=θ(W(Γ)). Then every reflection in G is conjugate in G to a power of one of the generating reflections, S1,,S.

Proof.

We may assume Γ is connected. Thus Γ occurs in List 1. Now if Γ has just one vertex there is nothing to prove. So we consider the case where Γ has two vertices. Say Γ is p1[q]p2. For i=1,2, we put Ui= the reflecting hyperplane for Si, Δi= the orbit containing Ui, and ei=e(Δi). Since SiC(Ui) we have pi|ei so in particular piei.

Now if q is odd we apply Lemma 3 to obtain |G/G|=p1. Now clearly Proposition 9 implies |G/G|e1. We thus obtain the sequence |G/G| e1p1= |G/G|. Hence e1=p1 and Δ1 is the only orbit of G on 𝒰(G). Thus every reflection in G is conjugate in G to a power of S1.

If q is even we apply Lemma 3 to obtain |G/G|=p1p2. Now if Δ1Δ2 we can apply Proposition 9 to get |G/G|e1e2. Thus if Δ1Δ2 we have the sequence |G/G|e1e2p1p2=|G/G| which forces pi=ei implying the desired result. Thus we must show Δ1Δ2. Assume to the contrary that Δ1=Δ2. Then C(U1) is conjugate in G to C(U2). Let d=(p1,p2). If d1, Si has a subgroup Di of order d>1 (i=1,2). Now D1C(U1) and D2C(U2) and thus D1 is conjugate in G to D2 since C(U1) and C(U2) are cyclic and conjugate. Thus r1 has a non trivial subgroup which is conjugate in W(Γ) to a subgroup of r2. This is a contradiction to Corollary 6. Hence we are forced to assume d=1. Thus p1p2|e1 and we have the existence of a reflection RC(U1) of order p1p2. Now Z(G) consists of scalar matrices as G is irreducible. Thus R also has order p1p2 when viewed as an element in G/Z(G). But G/Z(G) is isomorphic to the alternating group on four letters, the symmetric group on four letters, or the alternating group on five letters [STo1954, p.279]. Now none of these groups contain an element of order higher than five. But (p1,p2)=1 forces p1p26. So we have the theorem for graphs with two vertices.

If Γ has three vertices but is not in any of the infinite families in List 1, we see it is one of

(1) 2[3]2[5]2
(2) 3[3]3[3]3
(3) 3[3]3[4]2
We number the vertices from left to right and let Ui= the reflecting hyperplane for Si, Δi= the orbit of Ui, and ei=e(Δi). Again we remark that SiC(Ui) and thus pi|ei so in particular, piei.

For (1) we apply Lemma 3 to get |G/G|=2. Proposition 9 certainly implies that |G/G|e1. We thus obtain the sequence |G/G|e1 p1=2= |G/G|. Hence, e1=p1 and Δ1 is the only orbit.

For (2) the argument is the same as for (1).

For (3) we apply Lemma 3 to obtain |G/G|=6. Thus if Δ1Δ3 we can apply Proposition 9 as we have before to obtain the result. If Δ1=Δ3, there is a reflection RG of order 6. So the eigenvalues of R are 1,1,δ where δ is a primitive sixth root of unity. We obtain a contradiction as follows:

SL(2,3) acts naturally on a two dimensional vector space P over GF(3). Viewing the vector space as an abelian group of row vectors we can form the semi-direct product K=PSL(2,3). K can be viewed as the matrix group { ( A 00 xy 1 ) |ASL(2,3) x,yGF(3) } . We note that |K|=216. We define a homomorphism GK, denoted by xx, by putting S1= ( 110 010 001 ) , S2= ( 100 -110 001 ) ,S3= ( -100 0-10 010 ) . It is easy to see that G=K. Further (S1S2S3)3=1K and in G we compute (S1S2S3)3=-ε12I3 a central element of order six. Now comparing, [3] with [2] we see that |G|=1296, |Z(G)|=6. Hence the kernel of the homomorphism defined above is Z(G). Hence, since R is a reflection, |R|=|R|. Now it is easy to see that there are exactly two conjugacy classes of elements of order six in K -- one determined by S1S3 and the other by its inverse. Thus, in G, R must be conjugate to an element of Z(G)S1S3 or of Z(G)(S1S3)-1. Now the eigenvalues of S1S3 are 1, -1, ε1 and those of (S1S3)-1 are 1, -1, ε12. In particular for either S1S3 or (S1S3)-1 the eigenvalues are distinct. Hence the same is true for any scalar multiple of either S1S3 or (S1S3)-1. Thus the element R, with eigenvalues 1,1,δ, cannot be conjugate to such a scalar multiple.

If Γ has four vertices but is not in one of the infinite families in List 1 it is either H4, F4, or 3[3]3[3]3[3]3. The first two are Coxeter groups for which the result is known. For the last graph we number the vertices from left to right and use the familiar notation Ui, Δi, ei. We apply Lemma 3 to obtain the equality |G/G|=3. Now Proposition 9 gives us the inequality |G/G|e1. Putting them together with the fact that e1p1 we have the sequence |G/G|e1 p1=3=|G/G|. Thus e1=p1 and Δ1 is the only orbit.

Now all of E6, E7, E8, A, D are graphs of Coxeter groups for which the result is known. Or, just as easily for all these groups, an application of Lemma 3 gives |G/G|=2 and thus applying Proposition 9 yields the result. The only remaining case is Bp. Applying Lemma 3 gives |G/G|=2p. So numbering from the left as usual we see that if Δ1Δ2 an application of Proposition 9 will again yield the result. Now if p is even we can argue just as we did for the graph p1[q]p2 with q even and (p1,p2)1 to show that Δ1Δ2. So we assume p is odd. Then Δ1=Δ2 forces the existence of a reflection in C(U1) of order 2p. In fact, by Proposition 9, in this situation we must have |C(U1)|=2p and Δ1 is the only orbit. But in the basis {x1,,x} defined by xi=1αv1+v2++vi (α=(2sinπ/p)12) the matrices for the generators Sk are the standard generators for the full monomial group of p! Namely S1(x1)=ε1x1, S1(xj)=xj for j1. If k1, Sk interchanges xk and xk-1 and fixes the other basis vectors. In this basis it becomes clear that C(U1)=S1 a cyclic group of order p. Thus, Δ1Δ2.

Notes and references

This is a typed version of David W. Koster's thesis Complex Reflection Groups.

This thesis was submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy (Mathematics) at the University of Wisconsin - Madison, 1975.

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