Discrete complex reflection groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 12 May 2014

Notes and References

This is an excerpt of the lecture notes Discrete complex reflection groups by V.L. Popov. Lectures delivered at the Mathematical Institute, Rijksuniversiteit Utrecht, October 1980.

The structure of r-groups in the case s=n+1

When s=n+1 it is no longer true in general that an infinite irreducible complex crystallograph r-group W is the semidirect product of LinW and TranW. We shall explain here how one can find the corresponding extensions of TranW by LinW in this case.

We assume that k=.

The Cocycle c

We recall that the structure of the extension is given by a 1-cocycle c of K, where K=LinW, with values in V; see Section 2.4.

Let Γ=TranW and aE. Taking a as an origin, we can identify A(E) and GL(V)V; then W consists of the elements (P,c(P)+Γ), PK.

First of all, one can assume, upon replacing c by a suitable cocycle homologuous to c, that c(r1)== c(rn)=0. We assume that the order of Rj is equal to m(HRj), 1js, see Section 3.2.


Let γ1,,γnW be the reflections with Linγj=Rj, 1jn. We assume, as usual, that the group K generated by Rj, 1jn, is irreducible. We have j=1nHRj=0, hence j=1nHγj=bE. Then we have κb(γj)= (Rj,0),j =1,,n, and we are done.

We shall assume now that c(r1)== c(rn)=0.

Therefore c is defined by only one vector c(rn+1). Moreover, one can take c(rn+1)=λ en+1,λ because there exists a reflection (Rn+1,v) in W.

Therefore, given an invariant lattice Γ of full rank, one has to solve the following problems

Find those λ for which

a) the cocycle c of K with values in V, given by condition c(rj)=0, 1jn;c (rn+1)=λ en+1,λ, satisfies c(F)Γ for any relation F of K (i.e. for every element F of Kerϕ, see Section 2.4).
b) Condition a) holds and the corresponding group W is an r-group.


1) If Γ=Γ0 then a) b).
2) If c(F)Γ0 for every relation F then b) Γ=Γ0.


1) Let Γ=Γ0. We know that Γ0=Γ1++ Γn+1. But Γ=Γ1++Γn is a root lattice for K and the condition c(r1)==c(rn)=0 shows that the semidirect product of K and Γ lies in W. This semidirect product is an r-group, see Section 4.5. We also have that W contains the reflections (Rn+1,λen+1+t), tΓn+1.

Let γW be an arbitrary element, γ=(P,v). Then P is a product of reflections Rj, 1jn+1, in some order. Therefore, multiplying γ by (Rj,0), 1jn, and by (Rn+1,λen+1) in a suitable order, one can obtain an element of the form (1,t). But the elements (1,t), tΓ, are also products of reflections. Therefore, multiplying γ by reflections, one can obtain (1,r), rΓn+1. But (Rn+1,λen+1) (Rn+1,λen+1+r) (1,r)=(1,0). Therefore γ is a product of reflections.

2) We have c(F)Γ0. Therefore c, in fact, defines a 1-cocycle of K with values in V/Γ0. Let W be the group defined by c, with LinW=K and TranW=Γ0. It is an r-group because of 1). Also we have a group W defined by c, with LinW=K and TranW=Γ.

Let γW be a reflection. By section 3.2, we know, there exists δW with δγδ-1=(Rjl,t) for certain l and j. This is also a reflection, hence tHRj. But t=c(rjl)+v, vΓ. We have c(rjl)= (1+rj+rj2++rjl-1) c(rj). By definition of c, we have c(rj)HRj. Hence, also c(rjl)HRj. Therefore, vHRj, or, in other words, vΓ0. This means that δγδ-1W. It follows now from δW that γW; hence W=W and Γ=Γ0.

The following simple observation is very useful in practice because it gives strong restrictions on the choice of λ:

Theorem. Let a) be satisfied and let PK be such that Rn+1P Rn+1-1 K. Then λ(1-Rn+1PRn+1-1) en+1Γ.


It follows from the definition of c that (P,0) and (Rn+1,λen+1)W. We have now: (Rn+1,λen+1) (P,0) (Rn+1,λen+1)-1= ( Rn+1P Rn+1-1, -Rn+1P Rn+1-1 (λen+1) +λen+1 ) and we are done.


K=K31. The graph is 1 2 2 2 2 3 4 5 i i-1 The vectors e1,e2,,e5 are given in Table 2. Note that e5=ie1+e2+e3.

There exists only one (up to similarity) K-invariant lattice Γ of full rank Γ=[1,i]e1+ +[1,i]e4.

The presentation (i.e. the sets of generators and relations) of K is known, see [Cox1974].

The relations are: R12=R22= R32=(R2R3)3 =(R3R1)3= (R1R2)3=1, (R2R1R3R1)4=1, (R4R5)3=1. R52=(R5R2)2= (R5R1R3R1)2= (R5R3)4=R1 (R5R3R2R3)R1 (R5R3R2R3)-1=1 R42=(R4R1)2 =(R4R3)2= (R4R2)3=1 It follows from R1(R5R3R2R3)R1(R5R3R2R3)-1=R52=1 that R5R1R5K.

Therefore Γλ (R1R5R1R5) e5=λ ((1+i)e1+2e2+2e3) and hence λ=a+bi2, a,b, and ab (mod 2). But [1,i]e5Γ. Therefore, we may assume that λ=1+i2.

It only remains to check whether c(F)Γ or not for this particular λ. This has to be done by straightforward computations: c(r52) = c(r5)+r5c (r5)=1+i2 (1+r5)e=0Γ. c((r4r5)3) = (1+r4r5+(r4r5)2) (c(r4)+r4c(r5)) = 1+i2 (1+r4r5+(r4r5)2) (r4e5)=0Γ, and so on (one only needs to consider the relations which involve R5). This check shows that λ=1+i2 gives a cocycle of K, indeed, and, hence, defines an r-group W with LinW=K, TranW=Γ. It can be proved in a straightforward manner that this cocycle c is not a coboundary, i.e. W is not a semidirect product.

The same considerations can be given for each K with s=n+1, and, as a result, one obtains Table 2. It appears a posteriori that in all cases either Γ is a root lattice or c(F)Γ0 for every relation F of K (therefore, for an r-group W, TranW is always a root lattice). By performing similar straightforward computations in the remaining cases, a proof of the theorems in Section 2.8. is obtained.

As for the proof of the theorem from Section 2.9, the first part of it is given in Section 4.6; the second part (about minimality of [TrK]) follows from [BSh1963]. The third part follows from Table 1.

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