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.

Invariant lattices

As we have seen in Section 2.3, one of the ingredients of the description of infinite complex crystallographic r-groups W is the lattice TranW. We shall show in this chapter how one can find all the invariant lattices of full rank for an arbitrary fixed finite r-group.

We use the following notation: KGL(V) a finite essential r-group, n=dimkV,
ΓV a K-invariant lattice,
Rj=Rej,μj, 1js, a fixed generating system of reflections of K,
the set of all root lines of K,
Γ1=Γ, ,
Γj=Γj, j=1,,s.

Root lattices

Definition. Γ0=Γ is called the root lattice associated with Γ.

If Γ=Γ0 then Γ is called a root lattice.

Theorem. Γ0 is a K-invariant lattice and rkΓ=rkΓ0.


It is clear that Γ0 is K-invariant, so let us prove the assertion about ranks.

We can assume that V=k=1nk because K is essential. Put S=(1-Re1,μ1) ++(1-Ren,μn). If vKerS, then Sv=0=(1-μ1)v|e1e1++(1-μn)v|enen therefore (1-μj)v|ejej=0, because ejj, ej|ej=1, 1js. Hence v|ej=0, 1js, i.e. v=0. So, S is non-singular. But SΓΓ1ΓnΓ0Γ and rkSΓ=rkΓ. Therefore rkΓ0=rkΓ.

Corollary. If rkΓ=2n (hence k=) then rkΓ=2 for every L. If rkΓ=n then rkΓ=1 for every .


We may take 1= in the previous proof. As this proof shows that rk(Γ1Γn)=rkΓ1++rkΓn=2n, the first assertion follows from the evident inequality rkΓ2. The second assertion can be proved similarly by means of reduction to the real form.

It appears that one can reconstruct Γ0 from the Γj, 1js.

Theorem. Γ0=Γ1++Γs.


Let and uΓ; then there exists gK such that guΓj for certain j (because every reflection in K is conjugate to the a power of some Rj, 1js, see Section 3.2). Let Γ=Γ1++Γs. It is easy to see that Γ is invariant, hence uΓ and Γ=Γ.

The problem of finding of all K-invariant lattices can be solved in two steps: 1) description of all K-invariant root lattices; 2) description of all K-invariant lattices with a fixed associated root lattice. We shall first show how to solve the second problem.

The lattices with a fixed root lattice.

Definition. Γ*={vV|(1-P)vΓfor everyPK}. Clearly, Γ* is a subgroup of V. It is more convenient to use another description of Γ*.

Let π:VV/Γ be a natural map and (V/Γ)K be the set of points fixed by K. Then Γ*=π-1((V/Γ)K). Therefore, Γ*= {vV|(1-Rj)vΓj,1js}. It is clear that Γ* is K-invariant and ΓΓ*.

Theorem. Γ is a lattice and rkΓ*=rkΓ.


Let us first show that Γ* is a lattice. If it is not a lattice, then there exists a vector vΓ* such that αvΓ* for every α (because Γ* a closed subgroup of V). Then (1-P)αvΓ for every α and PK. Therefore (1-P)v=0, as Γ is a lattice and, hence, v=0, because K is an essential group. Therefore Γ* is a lattice.

V is a euclidean space with respect to Re|. Let Γ be the orthogonal complement of Γ in this space. Let vΓ* and v=u+w, uΓ, wΓ. Then (1-P)v=(1-P)u +(1-P)wΓ Γfor everyPK. Therefore (1-P)w=0, hence w=0 (again because K is essential). Thus Γ*Γ. This completes the proof.

Cohomological meaning of Γ*.

We have an exact sequence of groups 0ΓVV/Γ 0. It gives the exact cohomological sequence H0(K,V) H0(K,V/Γ) H1(K,Γ) H1(K,V) But H0(K,V)=0 because K is essential, H0(K,V/Γ)=(V/Γ)K=Γ*/Γ and H1(K,V)=0 because V is divisible. Therefore Γ*/Γ H1(K,Γ).

Now we can explain how to find all K-invariant lattices with a fixed K-invariant root lattice.

Theorem. Let Λ be a fixed K-invariant root lattice in V. Then for every lattice Γ in V the following properties are equivalent:

a) Γ is a K-invariant lattice and Γ0=Λ.
b) ΛΓΛ* and Γj=Λj, 1js.
There exist only a finite number of lattices Γ with the properties a) and b).


a) b). Let vΓ. Then (1-Rj)v=(1-μj)v|ejejΓj=ΓjΓ0=Λ. Therefore vΛ* and ΓΛ*.

b) a). Let ΛΓΛ*. Then Γ=π-1π(Γ), where π:VV/Λ is the natural map, and π(Γ)π(Λ*)=(V/Λ)K. Hence π(Γ) is K-invariant and therefore Γ is also K-invariant. If Γj=Λj then Γ0=Λ (because Λ=Λ0=Λ1++Λs).

We have already seen that for irreducible K one can take s=n, if k=, and s=n or n+1, if k=. It appears that if s=n then there exists a good constructive way to find Λ* by means of Λ.

Theorem. Let s=n and Λ be a K-invariant lattice in V. Let also S=(1-R1)++ (1-Rn). (Recall from Section 4.1 that S is nonsingular). Then:

a) Λ*S-1Λ.
b) If Λ0=Λ then Λ*=S-1Λ.


a) Let aΛ*, then, by definition, (1-Rj)aΛ for all 1jn, hence SaΛ and aS-1Λ.

b) Let Λ=Λ0 and uS-1Λ. Then Su=(1-R1)u++(1-Rn)uΛ=Λ1Λn. But (1-Rj)uej, and (1-Rj)uΛjΛ, because of linear independence of e1,,en. Therefore, by definition of Λ*, we have uΛ*.

This theorem is useful in practice because one can explicitly describe the operator S: its matrix with respect to the basis e1,,en is ( (1-μ1)e1|e1 (1-μ1)en|e1 (1-μn)e1|en (1-μn)en|en )

What to do if k= and s=n+1?

It is not difficult to see that one can take R1,,Rn+1 in such a way that R1,,Rn generate a subgroup K of K, which is also irreducible (the numbering in Table 1 has this property).

Example: 5 2 3 2 4 2 i 2 1 -1+i K=K31 1 2 3 4 i K=K29 Therefore the problem may be solved as follows: find all K-invariant lattices and select those lattices among them which are invariant under Rn+1.

We shall show how this can be done in Section 4.8, after we have explained (in Section 4.7) how to describe invariant root lattices.

By now we want to discuss several general properties of invariant lattices and explain the significance of root lattices in the theory of infinite r-groups.

Further remarks on lattices

Theorem. Let KGL(V) be a finite linear irreducible group and Γ be a nonzero K-invariant lattice in V. Then rkΓ=n=dimkV if k=, and rkΓ=n or 2n, if k=. Moreover, if K is an r-group, k= and rkΓ=n then K is the complexification of the Weyl group of a certain irreducible root system.


as in Section 3.1 and 3.4.

Corollary. If k= and K has an invariant lattice of rank n then K has an invariant lattice of rank 2n.


K is the complexicification of a Weyl group, let Γ be a lattice of rank n in the corresponding real form of V, which is invariant under this Weyl group. Then, for every z- a lattice Γ+zΓ is K-invariant and has rank 2n.

It should be noted that if k= and K is a Weyl group then a root lattice of rank n is (up to similarity) a lattice Λ of radical weights and Λ* is a lattice of weights, see [Bou1968]. The matrix of S with respect to a basis of simple roots, is, in this case, the Cartan matrix.

Properties of the operators S

Theorem. Let K be irreducible, s=n, Λ be a nonzero K-invariant lattice and S=(1-R1)++(1-Rn). Then:

a) |detS| and TrS if rkΛ=n,
|detS|2 and 2ReTrS if rkΛ=2n.
b) |detS| depends only on K but not on the choice of the generating system of reflections.
c) |detS| is divisible by [Λ:Λ0] if rkΛ=n,
|detS|2 is divisible by [Λ:Λ0] if rkΛ=2n.
d) Λ=Λ0[Λ*:Λ]=|detS| if rkΛ=n and =|detS|2 if rkΛ=2n.
e) If Λ=Λ0 and d1,,dr are the invariant factors of a matrix of the endomorphism of Λ, defined by S, dj>0, d1||dr, then, H1(K,Λ) Λ*/Λ /d1 /dr. Specifically, d1,,dr do not depend on the choice of the generating system of reflections.
f) Λ=Λ0|H1(K,Λ)|=|detS| if rkΛ=n and =|detS|2 if rkΛ=2n.


If rkΛ=n, then, by considering the corresponding real form, we reduce the problem to the case k=. If rkΛ=2n then k=. It is known that in this case for every PGL(V) one has detP|=|detP|2 and TrP|=2ReTrP (here P| is P, considered as a linear operator of a 2n-dimensional real vector space V).

Now, a) follows from the fact that a basis of Λ is an -basis of V. We have also: Λ0Λ (Λ0)*= S-1Λ0, so[Λ:Λ0] |[S-1Λ0:Λ0]. But S-1Λ0/Λ0Λ0/SΛ0, whence [S-1Λ0:Λ0]=[(Λ0)*:Λ0]=|detS| if k= or =|detS|| if k=. The assertions b), c), d), e), f) follow from these equations.


a) If |detS|=1 then Λ=Λ0, i.e. any invariant lattice is a root lattice.
b) Let k= and suppose |detS|2 is a prime number. Then Λ=Λ0 or Λ=(Λ0)*.

Remark. One can calculate detS directly from the graph of K, i.e. detS=σsgn σ·aσ where σ runs through the permutations of degree n and aσ=cαcβcγ for σ=αβγ a decomposition of σ as a product of cycles.


a) K=K1, type An. Write S=S(An) for the graph 1 n Simple cyclic products 0 are cj,j+1=1m 1jn-1, and cj=2, 1jn.

We have detS(A1)=2: Assume, by induction, that detS(Ak)=k+1 if k<n. Then, detS(An)=c1detS(An-1)-c12·detS(An-2)=2n-(n-1)=n+1.

b) K=K2, type G(m,m,n) Let us consider the generating system of reflections given by the graph 1 2 3 n 4cos2πm 2ζ2mcosπm ζ2m=eπi/m

The list of all nonzero simple cyclic products is cj=2, 1jn;
c123=2cosπmei/m, c132=2cosπme-πi/m.

It is easy to see that detS= c1detS(An-1) -c12detS(An-2)- c13·c2detS(An-3)+ c123detS(An-3)+ c132detS(An-3)= 2n-4cos2πm(n-1)- 1·2·(n-2)+ 2cosπmeπi/m(n-2)+ 2cosπme-πi/m(n-2)= 4-4cos2πm(n-1) +4cos2πm(n-2)= 4sin2πm.

Remark. If k= and Λ is a lattice of radical weights with Weyl group K then d1,,dr are invariant factors of a Cartan matrix and H1(K,Λ) is isomorphic to the centre of a simple simply connected Lie group corresponding to K.

Root lattices and infinite r-groups

We shall now show that if there exists a nonzero K-invariant lattice Λ then there also exists an infinite r-group W with LinW=K and TranW=Λ.

Theorem. Let KGL(V) be a finite linear r-group and Λ be a nonzero K-invariant lattice in V. Then the semidirect product of K and Λ is an r-group iff Λ=Λ0.


Let W be the semidirect product of K and Λ. Let a be a special point. We identify W and κa(W)=K·ΛGL(V)V. Then the set of all reflections in W is {(R,V)| R is a reflection of K and vlRΛ}. Let W0 be a subgroup of W generated by this set. Then KW0. Also Λ0W0, because Λ0=lΛl. Moreover, if vΛl and l is the root line of a reflection R, then (R-1,0)(R,v)=(0,v). Therefore we have K·Λ0W0. But every reflection in W lies in K·Λ0 (because it has the form (R,v), where R is a reflection and vlRΛ=Λ0, see Section 1.2). Therefore, W0=KΛ0 is generated by reflections, and we have W0=K. Hence, K·Λ=W=W0=K·Λ0 iff Λ=Λ0.

Corollary. The infinite r-groups W for which LinW is generated by n reflections are exactly the groups LinW·Λ, where Λ is a nonzero LinW-invariant root lattice.

Description of the group of linear parts; proof

We shall prove now the theorem from Section 2.5 and part of the theorem from Section 2.9.


a) b) is already proved in Section 4.3.

b) a) is trivial.

c) b) is also trivial: such a lattice is TranW.

b) c). Let Γ be an invariant lattice of rank 2n. Then the semidirect product of K and Γ0 is a crystallographic r-group because of the previous theorem.

b) d). Let us first prove that [TrK] coincides with the ring with unity generated over by all cyclic products.

We have [TrK]=[TrK]. Indeed, clearly TrKTrK, hence [TrK][TrK]. The reverse inclusion follows from the fact that Tr is an additive function. But 1-Rj, 1js, generate the ring K (here Rj, 1js, is a generating system of reflections of K). Therefore, the monomials (1-Rj1)(1-Rjr) generate as a -module. We have: (*) (1-Rj1) (1-Rjr) (1-Rjr-1) (1-Rj2)ej1= cj1jrej1, and it easily follows from this equality that Tr(1-Rj1) (1-Rjr) (1-Rjr-1) (1-Rj2) cj1jr Therefore, [TrK]=[,cj1jr,] and we are done. By now, let Γ be a K-invariant lattice of rank 2n. It follows from the equality (*) that cj1jr Γj1Γj1 for every j1,,jr. But rkΓj=2, see Section 4.1. Therefore, cj1jr is an integral element of a certain purely imaginary quadratic extension of ; denote this extension by Lj1.

Let L=L1; we shall show that cj1jrL for every j1,,jr.

K being irreducible, there exists α= c 1l1l2 ltj1lt lt-1l1 0 1 l1 l2 lt j1 j2 j3 jr for certain l1,,lt. Let β = c 1l1l2lt j1j2jr j1ltlt-1 l1 , γ = cj1j2jr. Then α,βL and β=αγ. But α0, hence γ=β/αL. Therefore the ring [TrK] lies in the maximal order of L.

d) e). This is proved in [Vin1971-2], Lemma 1.2.

d) a). Let [TrK]D, where D is the maximal order of a certain purely imaginary quadratic extension L of . K being irreducible and D being integrally closed, one can again apply Lemma 1.2 from [Vin1971-2] and obtain that K is defined over D. Therefore there exists a K-invariant D-submodule Γ of V such that ΓDV is an isomorphism. But D is a Dedekind ring and Γ is a D-module of rank n without torsion. Therefore, Γ is isomorphic to a direct sum of n fractional ideals of the field L, see [CRe1961]. Let J1,,Jn be these ideals. Then there exists a -basis e1,,en of V such that Γ=J1v1++ Jnvn. But D is a lattice of rank 2 in and for every fractional ideal J of L there exists dD, d0, with d·JD. Hence J is also a lattice of rank 2 in .

e) d). Let K be defined over a purely imaginary quadratic extension L of . Then TrPL for every PK, hence [TrK]L. But, TrP is an integral algebraic number. Therefore [TrK] lies in a maximal order of L.

d) f). Using Table 1, we can easily find those K for which [TrK] lies in the maximal order of a certain purely imaginary quadratic extension of . It is convenient to use the following necessary condition on each generator of this ring [TrK]:

z is an integral element of a certain purely imaginary quadratic extention of iff |z|2 and 2Rez.

It appears aposteriori that this condition is also sufficient. After a certain amount of calculations we obtain the list of f).

Example. Let K=K2, type G(m,m,s), s3. The graph of K is 1 2 3 s 4cos2πm 2ζ2mcosπm ζ2m=eπi/m We have [TrK]=[eπi/m]. The previous condition for the 2eπi/m gives: 2cos2πm.

Therefore m=2,3,4,6, and in these cases [TrK] lies in a maximal order of a certain purely imaginary extension of .

Description of root lattices

We assume now that k=. We shall first describe (up to similarity) all invariant root lattices of rank 2n when s=n (it follows from Section 4.5 that only these lattices are of interest to us).

Theorem. Let K be an irreducible finite linear r-group and let Λjej, 1js, be a set of lattices of rank 2 and Γ=Λ1++Λs. In order for Γ to be a K-invariant root lattice with Γj=Λj, 1js, it is necessary, and, if s=n, also sufficient, that:

a) [TrK]ΛjΛj, 1js.
b) (1-Rk)ΛjΛk and (1-Rj)ΛkΛj for every kj such that ckj0.
Moreover, b) is equivalent to
c) For every k and j, such that j<k and ckj0, one has (1-Rk)Λj Λkckj-1 (1-Rk)Λj.


Assertion a) follows from the formula cj1jrΓj1Γj1 and the fact that [TrK] is generated over by cyclic products.

The "necessary part" of b) follows from the invariance of the lattice. Let us prove the "sufficient part". If s=n then Γ is in fact a direct sum of Λj, 1jn, hence Γ is a lattice. This lattice is invariant under 1-Rk for every k: if kj then it follows from b); if k=j then it follows from a). Hence, Γ is K-invariant.

Now let us prove that b) c).

b) c). One obtains the proof by applying 1-Rk to both sides of (1-Rj)ΛkΛj.

c) b). One obtains the proof by applying 1-Rj to both sides of Λkckj-1(1-Rk)Λj.

Corollary. Let Γ be a nonzero K-invariant lattice. If |ckj|=1 then Γj and Γk determine each other uniquely by the formulas Γk= (1-Rk) Γjand Γj= (1-Rj)Γk.


We have by c): (1-Rk)Λj Λkckj-1 (1-Rk)Λj (we assume that j<k). The index of the left lattice in the right lattice is |ckj|2=1. Therefore the inclusions are in fact equalities. Applying 1-Rj, we get cjkΛj (1-Rj)Λk Λj. Again, the inclusions are in fact equalities.

If s=n+1, we also need to know Γ* for Γ a K-invariant lattice, and to select those ΓΛΓ* for which

a) Λ is Rn+1-invariant,
b) Λj=Γj, 1js.
Therefore, He also need to know (Γ*)0. We have the following description of this lattice:

Theorem. Γj*= ksuch thatcjk0 cjk-1 (1-Rj)Γk, 1js. In particular, Γj*=Γj if there is a number k such that |cjk|=1.


We have λej*Γj* iff(1-Rk)λej Γk,1ks.

Assume that cjk0, i.e. ej|ek0. If λ(1-Rk)ejΓk then, applying 1-Rj to the both sides of this relation, we obtain λcjkej(1-Rj)Γk, i.e. λejcjk-1(1-Rj)Γk.

Vice versa, if λejcjk-1(1-Rj)Γk for some λ then applying 1-Rk, we obtain: (1-Rk)λejcjk-1cjkΓk=Γk, i.e. λejΓ*.

In order to prove the second assertion, let us apply 1-Rj to Γk(1-Rk)Γj. We obtain Γj(1-Rj)ΓkcjkΓj and, if |cjk|=1 then cjkΓj=Γj. Thus the inclusion is in fact an equality; hence cjk-1(1-Rj)Γk=Γj.

Let us assume now that s=n.

An algorithm for constructing K-invariant root lattices of full rank: case 1.

We shall only consider here groups K from the theorem in Section 1.6 with the property: every two nodes of the graph of K (see Table 1) can be connected by a path of edges such that the absolute value of the weight of each edge equals 1.

These groups are: K1; K2, type G(6,1,s), s2, type G(m,m,s), m=2,4,6, s3 and m=3, s2; K3, m=2,3,4,6; K4; K8, K24, K25, K29, K32, K33, K34, K35, K36; K37.


Let Δ be an arbitrary lattice in of rank 2, which is invariant under [TrK].

Let: Λ1=Δe1.

For l2 let us consider an arbitrary path of edges from 1 to l, for which the absolute value of the weight of each edge equals 1. 1=l1 l2 l3 l=lr Let Λ1= (1-Rlr) (1-Rl2) Λ1=Δ ( j=2r (1-μ1j) elj-1|elj ) elr. We claim that:

a) Γ=Λ1++Λn is a K-invariant lattice of full rank.
b) Γ does not depend on the construction (i.e. on the choice of the paths).
c) each invariant lattice is obtained in this way.


The assertions b) and c) follow from the corollary above. Let us prove a). We check the conditions a) and b) of the theorem proved at the beginning of this section. The condition a) is clearly fulfilled, so we need only check b). We have k1=1=j1 j2 j3 jq-1 j=jq k2 k3 kp-1 k=kp Λk=(1-Rkp)(1-Rk2)Λ1 and Λj=(1-Rjq)(1-Rj2)Λ1. Let P=(1-Rk1)(1-Rkp-1). Then PΛk=ck1k2kp-1kpkp-1k2Λ1Λ1. (thanks to the construction of Λ1). But ck1k2kp-1kpkp-1k2=ck1k2ck2k3ckp-1kp. Hence |ck1k2kp-1kpkp-1k2|=1, so that PΛk=Λ1. Let us consider (1-Rk)Λj. We have P(1-Rk)Λj = (1-Rk1) (1-Rkp-1) (1-Rkp) (1-Rjq) (1-Rj2)Λ1 = ck1kp-1kpjqj2 Λ1 Λ1. Therefore, P(1-Rk)Λj Λ1=PΛk. But the restriction of P to ek has a trivial kernel, because PΛk=Λ1. Therefore, (1-Rk)ΛjΛk. Using the same arguments we obtain also (1-Rj)ΛkΛj.

Therefore we only need to find all Δ such that [TrK] is in the ring of integers of the field it generates. It is well known how to do it (see [BSh1963]). We obtain, after checking, that in fact if [TrK] then [TrK] (in the cases under consideration) is the maximal order in its fraction field. Hence, up to similarity, we have Δ=[TrK], if[TrK].


K=K2, type G(4,4,s), s3. The graph is 1 2 3 s=n 1+i

The path we choose connecting 1 and l3 will be 1,3,4,,l; and the path, connecting 1 and 2 will be 1,3,2.

Here [TrK]=[i] and e1|e3=e2|e3=e3|e4==en-1|en=-12.

Hence, Δ = [1,i], Λ1 = [1,i]e1, Λl = [1,i] (1-(-1))l-2 (-12)l-2e1 =[1,i]elif l3, Λ2 = [1,i] (1-(-1))2 (-12)2e2 =[1,i]e2. Therefore Λ=[1,i]e1+ +[1,i]en. Checking all cases as done in this example, one obtains exactly those lattices that are in the column TranW with LinW=K of Table 2.

An algorithm for constructing K-invariant root lattices of full rank: case 2.

We shall consider now the remaining irreducible finite r-groups K, i.e. the groups K2,type G(m,1,s), s2,m=2, 3,4,and G(6,6,2); K5;K26; K28.

We see that the graph of K in these cases is a chain. Taking a suitable numbering, we can assume that c12,c23,,cn-1,n are the only nonzero cjk (the numbering in Table 1 has this property).


Let Δ1 be a [TrK]-invariant lattice in . We have Δ1c12-1 Δ1. Let us take an arbitrary [TrK]-invariant lattice Δ2 between these two lattices (such a lattice exists, e.g. Δ1 has this property): Δ1Δ2 c12-1Δ1. And so on: Δ1 Δ2 c12-1 Δ1, Δ2 Δ3 c23-1 Δ2, Δn-1 Δn cn-1,n-1 Δn-1. Let Λ1 = Δ1 (1-Rl) (1-Rl-1) (1-R2)e1 = Δ1 ( j=2l (1-μj) ej-1|ej ) ej. We claim that Γ=Λ1++Λn is an invariant lattice and that any invariant lattice is of this form.


Let us check the conditions a) and c) of the theorem proved in the beginning of this section. We need in fact only check c), because a) is obvious. We have Λ1 = Δ1 (1-Rl) (1-Rl-1) (1-R2)e1 Λl+1 = Δl+1 (1-Rl+1) (1-Rl) (1-R2)e1. Therefore, (1-Rl+1) Λ1 = Δ1(1-Rl+1) (1-R2)e1 Δl+1(1-Rl+1) (1-R2)e1 = Λl+1 cl,l+1-1 Δl (1-Rl+1) (1-R2)e1= cl,l+1-1 (1-Rl+1) Λ1.

It appears that, as in the case 1, [TrK]is the maximal order and Δ1is similar to [TrK], if [TrK].


K=K2, type G(3,1,s), s2. The graph is 1 2 3 n 1-ω We have [TrK]=[ω] and e1|e2=1/2, e2|e3==en-1|en=-12, c12=1-ω, c23==cn-1,n=1. Therefore, (1-R1)(1-R2)e1=(1-(-1))l-1(-12)l-212e1=(-1)l2el, l=2,,n. We have Δ1=[1,ω], Δ2==Δn, [1,ω]Δ2 (1-ω)-1 [1,ω].

But |1-ω|2=3. Hence, Λ2=[1,ω] or(1-ω)-1 [1,ω]=13 [1,ω]. Thus, we have only two possibilities: either Λ=[1,ω]e1 +[1,ω]2e2+ +[1,ω]2en, or Λ=[1,ω]e1+ [1,ω]i23e2 ++[1,ω]i 23en.

One can check that, using this algorithm, one obtains, the lattices that are in the column TranW with LinW=K of Table 2 as well as other lattices in the cases K=K2, types G(2,1,2), G(6,6,2); K=K5 and K=K28. As a matter of fact, these last lattices are not included in Table 2 because they do not give new r-groups (i.e. the semidirect product of K and the corresponding lattice is equivalent to one of the groups from Table 2). We omit the check of these last assertions.

Invariant lattices in the case s=n+1.

Now we shall briefly consider the case s=n+1.

These are the groups: G(4,2,n), G(6,2,n), G(6,3,n), K12andK31. We shall explain the approach by several examples.


a) K=K2, type G(6,2,n) or G(6,3,n). The graphs of these groups are 1 2 3 n n+1 2+ω 3 G(6,2,n) 3 1 2 3 n n+1 2+ω 3 G(6,3,n)

We see that K=K2, type G(6,6,n), cf. Table 1. Using the previous theory, one checks that, up to similarity, there exists only one K-invariant root lattice of full rank, i.e. Γ=[1,ω]e1 ++[1,ω]en. But for K we have detS=4sin2 πm|m=6 =1, see the example b) in Section 4.4. It follows from the corollary in Section 4.4 that, up to similarity, Γ is a unique K-invariant lattice of full rank. But K must have an invariant lattice because of the theorem in Section 2.5. Therefore this lattice has to be Γ (of course, one can check that Γ is a K-invariant lattice also by means of a straightforward computation).

b) K=K2, type G(4,2,n), n3. The graph is 1 2 3 n n+1 1+i 2 G(6,3,n) We see that K=K2, type G(4,4,n). One can check that there exists only one (up to similarity) K-invariant root lattice of full rank: Λ=[1,i]e1 ++[1,i]en.

So, to describe all K-invariant lattices of full rank we need to find all Γ such that:

a) Γ0=Λ (with respect to K);
b) ΛΓΛ*=S-1Λ (with respect to K);
c) (1-Rn+1)ΓΓ.
We see that on every node there is an edge, whose weight is 1. Hence (Λ*)0=Λ, see the theorem above. Therefore, b) a).

The matrix of S with respect to the basis e1,,en has the form ( 2 1-i -1 1+i 2 -1 -1 2 -1 -1 2 2 -1 -1 2 ) We have detS=4sin2π4=2, see Section 4.4. Therefore the coefficients of S-1 lie in [i,12] and (cf. Section 4.4), the order of Λ*/Λ is equal to 4. It follows from these facts that Λ*/Λ=/2/2. It is not difficult to see that ej=Sfj, j=1,2, where f1 = n2e1+ -1-(n-1)i2 e2+ (1-i)(n-2)2 e3+ (1-i)(n-3)2 e4++ (1-i)12en, f2 = -(n-1)-i2e1+ ni2e2+ (i-1)(n-2)2 e3+(i-1)(n-3)2 e4++(i-1)12 en. It is readily checked that f1,f2,f3 are representatives of different nonzero elements of Λ*/Λ. Therefore every K-invariant lattice of full rank has to be similar to one of the lattices: (*) Λ;Λ (Λ+fj),j= 1,2,3;Λ j=13 (Λ+fj) Using the equalities en+1 = -1-i2e1+ -1+i2e2- 2e3+2e4- -2en, (1-Rn+1)v = { 0 ifv=e1,, en-1,f3 2en+1 ifv=en, 1-i2 en+1 ifv=f1, i-12 en+1 ifv=f2, one can straightforwardly verify that all the lattices are K-invariant.

The same considerations can be carried out for other groups K from the list above, and the description of allˆ(up to similarity) K-invariant lattices of full rank will be obtained. We leave this to the reader (the most complicated case is K=K2, type G(4,2,2)).

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