Discrete complex reflection groups

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

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 Tran W. 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: K⊂GL(V) a finite essential r-group, n=dimkV,
Γ⊂V a K-invariant lattice,
Rj=Rej,μj, 1≤j≤s, a fixed generating system of reflections of K,
ℒ the set of all root lines of K,
Γ1=ℓ∩Γ, ℓ∈ℒ,
Γj=Γℓj, j=1,…,s.

Root lattices

Definition. Γ0=∑ℓ∈ℒΓℓ 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 rk Γ=rk Γ0.

Proof.

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

We can assume that V=⨁k=1nℓk because K is essential. Put S=(1-Re1,μ1) +…+(1-Ren,μn). If v∈Ker S, then Sv=0=(1-μ1)⟨v | e1⟩e1+…+(1-μn)⟨v | en⟩en therefore (1-μj)⟨v | ej⟩ej=0, because ej∈ℓj, ⟨ej | ej⟩=1, 1≤j≤s. Hence ⟨v | ej⟩=0, 1≤j≤s, i.e. v=0. So, S is non-singular. But SΓ⊂Γℓ1⊕…⊕Γℓn⊂Γ0⊂Γ and rk SΓ=rk Γ. Therefore rk Γ0=rk Γ.

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Corollary. If rk Γ=2n (hence k=ℂ) then rk Γℓ=2 for every ℓ∈L. If rk Γ=n then rk Γℓ=1 for every ℓ∈ℒ.

Proof.

We may take ℓ1=ℓ in the previous proof. As this proof shows that rk(Γℓ1⊕…⊕Γℓn)=rk Γℓ1+…+rk Γℓn=2n, the first assertion follows from the evident inequality rk Γℓ≤2. The second assertion can be proved similarly by means of reduction to the real form.

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It appears that one can reconstruct Γ0 from the Γj, 1≤j≤s.

Theorem. Γ0=Γ1+…+Γs.

Proof.

Let ℓ∈ℒ and u∈Γℓ; then there exists g∈K such that gu∈Γj for certain j (because every reflection in K is conjugate to the a power of some Rj, 1≤j≤s, see Section 3.2). Let Γ′=Γ1+…+Γs. It is easy to see that Γ′ is invariant, hence u∈Γ′ and Γ=Γ′.

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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*={v∈V | (1-P)v∈Γ for every P∈K}. Clearly, Γ* is a subgroup of V. It is more convenient to use another description of Γ*.

Let π:V→V/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*= {v∈V | (1-Rj)v∈Γj,1≤j≤s}. It is clear that Γ* is K-invariant and Γ⊂Γ*.

Theorem. Γ is a lattice and rk Γ*=rk Γ.

Proof.

Let us first show that Γ* is a lattice. If it is not a lattice, then there exists a vector v∈Γ* such that αv∈Γ* for every α∈ℝ (because Γ* a closed subgroup of V). Then (1-P)αv∈Γ for every α∈ℝ and P∈K. 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 ℝΓ⊥ be the orthogonal complement of ℝΓ in this space. Let v∈Γ* and v=u+w, u∈ℝΓ, w∈ℝΓ⊥. Then (1-P)v=(1-P)u +(1-P)w⊂Γ⊂ ℝΓ for every P∈K. Therefore (1-P)w=0, hence w=0 (again because K is essential). Thus Γ*⊂ℝΓ. This completes the proof.

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Cohomological meaning of Γ*.

We have an exact sequence of groups 0→Γ→V→V/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) Λ⊂Γ⊂Λ* and Γj=Λj, 1≤j≤s.
There exist only a finite number of lattices Γ with the properties a) and b).

Proof.

a) ⇒ b). Let v∈Γ. Then (1-Rj)v=(1-μj)⟨v | ej⟩ej∈Γ∩ℓj=Γj⊂Γ0=Λ. Therefore v∈Λ* and Γ⊂Λ*.

b) ⇒ a). Let Λ⊂Γ⊂Λ*. Then Γ=π-1π(Γ), where π:V→V/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).

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We have already seen that for irreducible K one can take s=n, if k=ℝ, and s=n or n+1, if k=ℂ. 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.

Proof.

a) Let a∈Λ*, then, by definition, (1-Rj)a∈Λ for all 1≤j≤n, hence Sa∈Λ and a∈S-1Λ.

b) Let Λ=Λ0 and u∈S-1Λ. Then Su=(1-R1)u+…+(1-Rn)u∈Λ=Λ1⊕…⊕Λn. But (1-Rj)u∈ℂej, and (1-Rj)u∈Λj⊂Λ, because of linear independence of e1,…,en. Therefore, by definition of Λ*, we have u∈Λ*.

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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=ℂ 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 K⊂GL(V) be a finite linear irreducible group and Γ be a nonzero K-invariant lattice in V. Then rk Γ=n=dimk V if k=ℝ, and rk Γ=n or 2n, if k=ℂ. Moreover, if K is an r-group, k=ℂ and rk Γ=n then K is the complexification of the Weyl group of a certain irreducible root system.

Proof.

as in Section 3.1 and 3.4.

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

Proof.

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∈ℂ-ℝ a lattice Γ+zΓ is K-invariant and has rank 2n.

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It should be noted that if k=ℝ 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) |det S| and Tr S∈ℤ if rk Λ=n,
|det S|2 and 2 Re Tr S∈ℤ if rk Λ=2n.
b) |det S| depends only on K but not on the choice of the generating system of reflections.
c) |det S| is divisible by [Λ:Λ0] if rk Λ=n,
|det S|2 is divisible by [Λ:Λ0] if rk Λ=2n.
d) Λ=Λ0⇒[Λ*:Λ]=|det S| if rk Λ=n and =|det S|2 if rk Λ=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≃ ℤ/d1ℤ⊕…⊕ ℤ/drℤ. Specifically, d1,…,dr do not depend on the choice of the generating system of reflections.
f) Λ=Λ0⇒|H1(K,Λ)|=|det S| if rk Λ=n and =|det S|2 if rk Λ=2n.

Proof.

If rk Λ=n, then, by considering the corresponding real form, we reduce the problem to the case k=ℝ. If rk Λ=2n then k=ℂ. It is known that in this case for every P∈GL(V) one has det Pℂ|ℝ=|det P|2 and Tr Pℂ|ℝ=2 Re Tr P (here Pℂ|ℝ 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 ℝ-basis of V. We have also: Λ0⊂Λ⊂ (Λ0)*= S-1Λ0, so [Λ:Λ0] |[S-1Λ0:Λ0]. But S-1Λ0/Λ0≃Λ0/SΛ0, whence [S-1Λ0:Λ0]=[(Λ0)*:Λ0]=|det S| if k=ℝ or =|det Sℂ|ℝ| if k=ℂ. The assertions b), c), d), e), f) follow from these equations.

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Corollary.

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

Remark. One can calculate det S directly from the graph of K, i.e. det S=∑σsgn  σ·aσ where σ runs through the permutations of degree n and aσ=cαcβ…cγ for σ=αβ⋯γ a decomposition of σ as a product of cycles.

Examples.

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

We have det S(A1)=2: Assume, by induction, that det S(Ak)=k+1 if k<n. Then, det S(An)=c1det S(An-1)-c12·det S(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, 1≤j≤n;
c13=c23=c34=c45=…=cn-1,n=1
c12=4cos2πm
c123=2cos πm ei/m, c132=2cos πm e-πi/m.

It is easy to see that det S= c1 det S(An-1) -c12det S(An-2)- c13·c2det S(An-3)+ c123det S(An-3)+ c132det S(An-3)= 2n-4cos2 πm (n-1)- 1·2·(n-2)+ 2cos πm eπi/m(n-2)+ 2cosπm e-πi/m(n-2)= 4-4cos2 πm (n-1) +4cos2 πm (n-2)= 4sin2πm.

Remark. If k=ℝ 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 Lin W=K and Tran W=Λ.

Theorem. Let K⊂GL(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.

Proof.

Let W be the semidirect product of K and Λ. Let a be a special point. We identify W and κa(W)=K·Λ⊂GL(V)V. Then the set of all reflections in W is {(R,V) | R is a reflection of K and v∈lR∩Λ}. Let W0 be a subgroup of W generated by this set. Then K⊂W0. Also Λ0⊂W0, because Λ0=∑l∈ℒΛl. Moreover, if v∈Λl and l is the root line of a reflection R, then (R-1,0)(R,v)=(0,v). Therefore we have K·Λ0⊂W0. But every reflection in W lies in K·Λ0 (because it has the form (R,v), where R is a reflection and v∈lR∩Λ=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.

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

Proof.

a) ⇒ b) is already proved in Section 4.3.

b) ⇒ a) is trivial.

c) ⇒ b) is also trivial: such a lattice is Tran W.

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 ℤ[TrK] coincides with the ring with unity generated over ℤ by all cyclic products.

We have ℤ[TrK]=ℤ[TrℤK]. Indeed, clearly TrK⊂TrℤK, hence ℤ[TrK]⊂ℤ[TrℤK]. The reverse inclusion follows from the fact that Tr is an additive function. But 1-Rj, 1≤j≤s, generate the ring ℤK (here Rj, 1≤j≤s, is a generating system of reflections of K). Therefore, the monomials (1-Rj1)…(1-Rjr) generate ℤ as a ℤ-module. We have: (*) (1-Rj1) (1-Rjr) (1-Rjr-1)… (1-Rj2)ej1= cj1⋯jrej1, and it easily follows from this equality that Tr(1-Rj1) (1-Rjr) (1-Rjr-1)… (1-Rj2) cj1⋯jr Therefore, ℤ[TrℤK]=ℤ[…,cj1⋯jr,…] and we are done. By now, let Γ be a K-invariant lattice of rank 2n. It follows from the equality (*) that cj1⋯jr Γj1⊂Γj1 for every j1,…,jr. But rk Γj=2, see Section 4.1. Therefore, cj1⋯jr is an integral element of a certain purely imaginary quadratic extension of ℚ; denote this extension by Lj1.

Let L=L1; we shall show that cj1⋯jr∈L for every j1,…,jr.

K being irreducible, there exists α= c 1l1l2⋯ ltj1lt lt-1⋯l1 ≠0 1 l1 l2 lt j1 j2 j3 jr for certain l1,…,lt. Let β = c 1l1l2⋯lt j1j2⋯jr j1ltlt-1 ⋯l1 , γ = cj1j2⋯jr. Then α,β∈L and β=αγ. But α≠0, hence γ=β/α∈L. Therefore the ring ℤ[TrK] lies in the maximal order of L.

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

d) ⇒ a). Let ℤ[TrK]⊂D, where D is the maximal order of a certain purely imaginary quadratic extension L of ℚ. 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 Γ⊗Dℂ→V 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 ℂ-basis e1,…,en of V such that Γ=J1v1+…+ Jnvn. But D is a lattice of rank 2 in ℂ and for every fractional ideal J of L there exists d∈D, d≠0, with d·J⊂D. Hence J is also a lattice of rank 2 in ℂ.

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

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

z∈ℂ is an integral element of a certain purely imaginary quadratic extention of ℚ iff |z|2∈ℤ and 2 Re z∈ℤ.

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), s≥3. The graph of K is 1 2 3 s 4cos2πm 2ζ2mcosπm ζ2m=eπi/m We have ℤ[TrK]=ℤ[eπi/m]. The previous condition for the 2eπi/m gives: 2cos 2πm∈ℤ.

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

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Description of root lattices

We assume now that k=ℂ. 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 Λj⊂ℂej, 1≤j≤s, 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, 1≤j≤s, it is necessary, and, if s=n, also sufficient, that:

a) ℤ[TrK]Λj⊂Λj, 1≤j≤s.
b) (1-Rk)Λj⊂Λk and (1-Rj)Λk⊂Λj for every k≠j such that ckj≠0.
Moreover, b) is equivalent to
c) For every k and j, such that j<k and ckj≠0, one has (1-Rk)Λj⊂ Λk⊂ckj-1 (1-Rk)Λj.

Proof.

Assertion a) follows from the formula cj1⋯jrΓj1⊂Γj1 and the fact that ℤ[TrK] is generated over ℤ 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, 1≤j≤n, hence Γ is a lattice. This lattice is invariant under 1-Rk for every k: if k≠j 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)Λk⊂Λj.

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

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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) Γj and  Γj= (1-Rj)Γk.

Proof.

We have by c): (1-Rk)Λj⊂ Λk⊂ckj-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.

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

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

Theorem. Γj*= ⋂k such thatcjk≠0 cjk-1 (1-Rj)Γk, 1≤j≤s. In particular, Γj*=Γj if there is a number k such that |cjk|=1.

Proof.

We have λej*∈Γj* iff(1-Rk)λej ∈Γk,1≤k≤s.

Assume that cjk≠0, i.e. ⟨ej | ek⟩≠0. If λ(1-Rk)ej∈Γk then, applying 1-Rj to the both sides of this relation, we obtain λcjkej∈(1-Rj)Γk, i.e. λej∈cjk-1(1-Rj)Γk.

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

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

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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), s≥2, type G(m,m,s), m=2,4,6, s≥3 and m=3, s≥2; K3, m=2,3,4,6; K4; K8, K24, K25, K29, K32, K33, K34, K35, K36; K37.

Algorithm:

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

Let: Λ1=Δe1.

For l≥2 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.

Proof.

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=ck1k2…kp-1kpkp-1…k2Λ1⊂Λ1. (thanks to the construction of Λ1). But ck1k2…kp-1kpkp-1…k2=ck1k2ck2k3…ckp-1kp. Hence |ck1k2…kp-1kpkp-1…k2|=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 = ck1…kp-1kpjq…j2 Λ1 ⊂ Λ1. Therefore, P(1-Rk)Λj ⊂Λ1=PΛk. But the restriction of P to ℂek has a trivial kernel, because PΛk=Λ1. Therefore, (1-Rk)Λj⊂Λk. Using the same arguments we obtain also (1-Rj)Λk⊂Λj.

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Therefore we only need to find all Δ⊂ℂ such that ℤ[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 ℤ[TrK]≠ℤ then ℤ[TrK] (in the cases under consideration) is the maximal order in its fraction field. Hence, up to similarity, we have Δ=ℤ[TrK],  if ℤ[TrK]≠ℤ.

Example.

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

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

Here ℤ[TrK]=ℤ[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]el if  l≥3, 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 Tran W with Lin W=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),  s≥2, 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).

Algorithm.

Let Δ1 be a ℤ[TrK]-invariant lattice in ℂ. We have Δ1⊂c12-1 Δ1. Let us take an arbitrary ℤ[TrK]-invariant lattice Δ2 between these two lattices (such a lattice exists, e.g. Δ1 has this property): Δ1⊂Δ2⊂ 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.

Proof.

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, ℤ[TrK] is the maximal order and  Δ1 is similar to  ℤ[TrK], if ℤ[TrK]≠ℤ.

Example.

K=K2, type G(3,1,s), s≥2. The graph is 1 2 3 n 1-ω We have ℤ[TrK]=ℤ[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 Tran W with Lin W=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),  K12 and K31. We shall explain the approach by several examples.

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 det S=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), n≥3. 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) Λ⊂Γ⊂Λ*=S-1Λ (with respect to K′);
c) (1-Rn+1)Γ⊂Γ.
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 det S=4sin2 π4=2, see Section 4.4. Therefore the coefficients of S-1 lie in ℤ[i,12] and (cf. Section 4.4), the order of Λ*/Λ is equal to 4. It follows from these facts that Λ*/Λ=ℤ/2ℤ⊕ℤ/2ℤ. 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 if v=e1,…, en-1,f3 2en+1 if v=en, 1-i2 en+1 if v=f1, i-12 en+1 if v=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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