## Discrete complex reflection groups

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 $r\text{-groups}$ $W$ is the lattice $\text{Tran} W\text{.}$ We shall show in this chapter how one can find all the invariant lattices of full rank for an arbitrary fixed finite $r\text{-group.}$

We use the following notation: $K\subset GL\left(V\right)$ a finite essential $r\text{-group,}$ $n={\text{dim}}_{k}V,$
$\mathrm{\Gamma }\subset V$ a $K\text{-invariant}$ lattice,
${R}_{j}={R}_{{e}_{j},{\mu }_{j}},$ $1\le j\le s,$ a fixed generating system of reflections of $K,$
$ℒ$ the set of all root lines of $K,$
${\mathrm{\Gamma }}_{1}=\ell \cap \mathrm{\Gamma },$ $\ell \in ℒ,$
${\mathrm{\Gamma }}_{j}={\mathrm{\Gamma }}_{{\ell }_{j}},$ $j=1,\dots ,s\text{.}$

### Root lattices

Definition. ${\mathrm{\Gamma }}^{0}=\sum _{\ell \in ℒ}{\mathrm{\Gamma }}_{\ell }$ is called the root lattice associated with $\mathrm{\Gamma }\text{.}$

If $\mathrm{\Gamma }={\mathrm{\Gamma }}^{0}$ then $\mathrm{\Gamma }$ is called a root lattice.

Theorem. ${\mathrm{\Gamma }}^{0}$ is a $K\text{-invariant}$ lattice and $\text{rk} \mathrm{\Gamma }=\text{rk} {\mathrm{\Gamma }}^{0}\text{.}$ Proof. It is clear that ${\mathrm{\Gamma }}^{0}$ is $K\text{-invariant,}$ 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 $v\in \text{Ker} S,$ then $Sv=0=\left(1-{\mu }_{1}\right)⟨v | {e}_{1}⟩{e}_{1}+\dots +\left(1-{\mu }_{n}\right)⟨v | {e}_{n}⟩{e}_{n}$ therefore $\left(1-{\mu }_{j}\right)⟨v | {e}_{j}⟩{e}_{j}=0,$ because ${e}_{j}\in {\ell }_{j},$ $⟨{e}_{j} | {e}_{j}⟩=1,$ $1\le j\le s\text{.}$ Hence $⟨v | {e}_{j}⟩=0,$ $1\le j\le s,$ i.e. $v=0\text{.}$ So, $S$ is non-singular. But $S\mathrm{\Gamma }\subset {\mathrm{\Gamma }}_{{\ell }_{1}}\oplus \dots \oplus {\mathrm{\Gamma }}_{{\ell }_{n}}\subset {\mathrm{\Gamma }}^{0}\subset \mathrm{\Gamma }$ and $\text{rk} S\mathrm{\Gamma }=\text{rk} \mathrm{\Gamma }\text{.}$ Therefore $\text{rk} {\mathrm{\Gamma }}^{0}=\text{rk} \mathrm{\Gamma }\text{.}$ $\square$

Corollary. If $\text{rk} \mathrm{\Gamma }=2n$ (hence $k=ℂ\text{)}$ then $\text{rk} {\mathrm{\Gamma }}_{\ell }=2$ for every $\ell \in L\text{.}$ If $\text{rk} \mathrm{\Gamma }=n$ then $\text{rk} {\mathrm{\Gamma }}_{\ell }=1$ for every $\ell \in ℒ\text{.}$ Proof. We may take ${\ell }_{1}=\ell$ in the previous proof. As this proof shows that $\text{rk}\left({\mathrm{\Gamma }}_{{\ell }_{1}}\oplus \dots \oplus {\mathrm{\Gamma }}_{{\ell }_{n}}\right)=\text{rk} {\mathrm{\Gamma }}_{{\ell }_{1}}+\dots +\text{rk} {\mathrm{\Gamma }}_{{\ell }_{n}}=2n,$ the first assertion follows from the evident inequality $\text{rk} {\mathrm{\Gamma }}_{\ell }\le 2\text{.}$ The second assertion can be proved similarly by means of reduction to the real form. $\square$

It appears that one can reconstruct ${\mathrm{\Gamma }}^{0}$ from the ${\mathrm{\Gamma }}_{j},$ $1\le j\le s\text{.}$

Theorem. ${\mathrm{\Gamma }}^{0}={\mathrm{\Gamma }}_{1}+\dots +{\mathrm{\Gamma }}_{s}\text{.}$ Proof. Let $\ell \in ℒ$ and $u\in {\mathrm{\Gamma }}_{\ell }\text{;}$ then there exists $g\in K$ such that $gu\in {\mathrm{\Gamma }}_{j}$ for certain $j$ (because every reflection in $K$ is conjugate to the a power of some ${R}_{j},$ $1\le j\le s,$ see Section 3.2). Let $\mathrm{\Gamma }\prime ={\mathrm{\Gamma }}_{1}+\dots +{\mathrm{\Gamma }}_{s}\text{.}$ It is easy to see that $\mathrm{\Gamma }\prime$ is invariant, hence $u\in \mathrm{\Gamma }\prime$ and $\mathrm{\Gamma }=\mathrm{\Gamma }\prime \text{.}$ $\square$

The problem of finding of all $K\text{-invariant}$ lattices can be solved in two steps: 1) description of all $K\text{-invariant}$ 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. ${\mathrm{\Gamma }}^{*}=\left\{v\in V | \left(1-P\right)v\in \mathrm{\Gamma } \text{for every} P\in K\right\}\text{.}$ Clearly, ${\mathrm{\Gamma }}^{*}$ is a subgroup of $V\text{.}$ It is more convenient to use another description of ${\mathrm{\Gamma }}^{*}\text{.}$

Let $π:V→V/Γ$ be a natural map and ${\left(V/\mathrm{\Gamma }\right)}^{K}$ be the set of points fixed by $K\text{.}$ Then ${\mathrm{\Gamma }}^{*}={\pi }^{-1}\left({\left(V/\mathrm{\Gamma }\right)}^{K}\right)\text{.}$ Therefore, $Γ*= {v∈V | (1-Rj)v∈Γj,1≤j≤s}.$ It is clear that ${\mathrm{\Gamma }}^{*}$ is $K\text{-invariant}$ and $\mathrm{\Gamma }\subset {\mathrm{\Gamma }}^{*}\text{.}$

Theorem. $\mathrm{\Gamma }$ is a lattice and $\text{rk} {\mathrm{\Gamma }}^{*}=\text{rk} \mathrm{\Gamma }\text{.}$ Proof. Let us first show that ${\mathrm{\Gamma }}^{*}$ is a lattice. If it is not a lattice, then there exists a vector $v\in {\mathrm{\Gamma }}^{*}$ such that $\alpha v\in {\mathrm{\Gamma }}^{*}$ for every $\alpha \in ℝ$ (because ${\mathrm{\Gamma }}^{*}$ a closed subgroup of $V\text{).}$ Then $\left(1-P\right)\alpha v\in \mathrm{\Gamma }$ for every $\alpha \in ℝ$ and $P\in K\text{.}$ Therefore $\left(1-P\right)v=0,$ as $\mathrm{\Gamma }$ is a lattice and, hence, $v=0,$ because $K$ is an essential group. Therefore ${\mathrm{\Gamma }}^{*}$ is a lattice. $V$ is a euclidean space with respect to $\text{Re} ⟨ | ⟩\text{.}$ Let $ℝ{\mathrm{\Gamma }}^{\perp }$ be the orthogonal complement of $ℝ\mathrm{\Gamma }$ in this space. Let $v\in {\mathrm{\Gamma }}^{*}$ and $v=u+w,$ $u\in ℝ\mathrm{\Gamma },$ $w\in ℝ{\mathrm{\Gamma }}^{\perp }\text{.}$ Then $(1-P)v=(1-P)u +(1-P)w⊂Γ⊂ ℝΓ for every P∈K.$ Therefore $\left(1-P\right)w=0,$ hence $w=0$ (again because $K$ is essential). Thus ${\mathrm{\Gamma }}^{*}\subset ℝ\mathrm{\Gamma }\text{.}$ This completes the proof. $\square$

Cohomological meaning of ${\mathrm{\Gamma }}^{*}\text{.}$

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 ${H}^{0}\left(K,V\right)=0$ because $K$ is essential, ${H}^{0}\left(K,V/\mathrm{\Gamma }\right)={\left(V/\mathrm{\Gamma }\right)}^{K}={\mathrm{\Gamma }}^{*}/\mathrm{\Gamma }$ and ${H}^{1}\left(K,V\right)=0$ because $V$ is divisible. Therefore $Γ*/Γ≃ H1(K,Γ).$

Now we can explain how to find all $K\text{-invariant}$ lattices with a fixed $K\text{-invariant}$ root lattice.

Theorem. Let $\mathrm{\Lambda }$ be a fixed $K\text{-invariant}$ root lattice in $V\text{.}$ Then for every lattice $\mathrm{\Gamma }$ in $V$ the following properties are equivalent:

 a) $\mathrm{\Gamma }$ is a $K\text{-invariant}$ lattice and ${\mathrm{\Gamma }}^{0}=\mathrm{\Lambda }\text{.}$ b) $\mathrm{\Lambda }\subset \mathrm{\Gamma }\subset {\mathrm{\Lambda }}^{*}$ and ${\mathrm{\Gamma }}_{j}={\mathrm{\Lambda }}_{j},$ $1\le j\le s\text{.}$
There exist only a finite number of lattices $\mathrm{\Gamma }$ with the properties a) and b). Proof. a) $⇒$ b). Let $v\in \mathrm{\Gamma }\text{.}$ Then $\left(1-{R}_{j}\right)v=\left(1-{\mu }_{j}\right)⟨v | {e}_{j}⟩{e}_{j}\in \mathrm{\Gamma }\cap {\ell }_{j}={\mathrm{\Gamma }}_{j}\subset {\mathrm{\Gamma }}^{0}=\mathrm{\Lambda }\text{.}$ Therefore $v\in {\mathrm{\Lambda }}^{*}$ and $\mathrm{\Gamma }\subset {\mathrm{\Lambda }}^{*}\text{.}$ b) $⇒$ a). Let $\mathrm{\Lambda }\subset \mathrm{\Gamma }\subset {\mathrm{\Lambda }}^{*}\text{.}$ Then $\mathrm{\Gamma }={\pi }^{-1}\pi \left(\mathrm{\Gamma }\right),$ where $\pi :V\to V/\mathrm{\Lambda }$ is the natural map, and $\pi \left(\mathrm{\Gamma }\right)\subset \pi \left({\mathrm{\Lambda }}^{*}\right)={\left(V/\mathrm{\Lambda }\right)}^{K}\text{.}$ Hence $\pi \left(\mathrm{\Gamma }\right)$ is $K\text{-invariant}$ and therefore $\mathrm{\Gamma }$ is also $K\text{-invariant.}$ If ${\mathrm{\Gamma }}_{j}={\mathrm{\Lambda }}_{j}$ then ${\mathrm{\Gamma }}^{0}=\mathrm{\Lambda }$ (because $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}={\mathrm{\Lambda }}_{1}+\dots +{\mathrm{\Lambda }}_{s}\text{).}$ $\square$

We have already seen that for irreducible $K$ one can take $s=n,$ if $k=ℝ,$ and $s=n$ or $n+1,$ if $k=ℂ\text{.}$ It appears that if $s=n$ then there exists a good constructive way to find ${\mathrm{\Lambda }}^{*}$ by means of $\mathrm{\Lambda }\text{.}$

Theorem. Let $s=n$ and $\mathrm{\Lambda }$ be a $K\text{-invariant}$ lattice in $V\text{.}$ Let also $S=(1-R1)+…+ (1-Rn).$ (Recall from Section 4.1 that $S$ is nonsingular). Then:

 a) ${\mathrm{\Lambda }}^{*}\subset {S}^{-1}\mathrm{\Lambda }\text{.}$ b) If ${\mathrm{\Lambda }}^{0}=\mathrm{\Lambda }$ then ${\mathrm{\Lambda }}^{*}={S}^{-1}\mathrm{\Lambda }\text{.}$ Proof. a) Let $a\in {\mathrm{\Lambda }}^{*},$ then, by definition, $\left(1-{R}_{j}\right)a\in \mathrm{\Lambda }$ for all $1\le j\le n,$ hence $Sa\in \mathrm{\Lambda }$ and $a\in {S}^{-1}\mathrm{\Lambda }\text{.}$ b) Let $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}$ and $u\in {S}^{-1}\mathrm{\Lambda }\text{.}$ Then $Su=\left(1-{R}_{1}\right)u+\dots +\left(1-{R}_{n}\right)u\in \mathrm{\Lambda }={\mathrm{\Lambda }}_{1}\oplus \dots \oplus {\mathrm{\Lambda }}_{n}\text{.}$ But $\left(1-{R}_{j}\right)u\in ℂ{e}_{j},$ and $\left(1-{R}_{j}\right)u\in {\mathrm{\Lambda }}_{j}\subset \mathrm{\Lambda },$ because of linear independence of ${e}_{1},\dots ,{e}_{n}\text{.}$ Therefore, by definition of ${\mathrm{\Lambda }}^{*},$ we have $u\in {\mathrm{\Lambda }}^{*}\text{.}$ $\square$

This theorem is useful in practice because one can explicitly describe the operator $S:$ its matrix with respect to the basis ${e}_{1},\dots ,{e}_{n}$ is $( (1-μ1)⟨e1 | e1⟩ … (1-μ1)⟨en | e1⟩ … (1-μn)⟨e1 | en⟩ … (1-μn)⟨en | en⟩ )$

What to do if $k=ℂ$ and $s=n+1\text{?}$

It is not difficult to see that one can take ${R}_{1},\dots ,{R}_{n+1}$ in such a way that ${R}_{1},\dots ,{R}_{n}$ generate a subgroup $K\prime$ of $K,$ which is also irreducible (the numbering in Table 1 has this property).

Example: $5 2 3 2 4 2 i 2 1 -1+i K={K}_{31} 1 2 3 4 i K\prime ={K}_{29}$ Therefore the problem may be solved as follows: find all $K\prime \text{-invariant}$ lattices and select those lattices among them which are invariant under ${R}_{n+1}\text{.}$

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 $r\text{-groups.}$

### Further remarks on lattices

Theorem. Let $K\subset GL\left(V\right)$ be a finite linear irreducible group and $\mathrm{\Gamma }$ be a nonzero $K\text{-invariant}$ lattice in $V\text{.}$ Then $\text{rk} \mathrm{\Gamma }=n={\text{dim}}_{k} V$ if $k=ℝ,$ and $\text{rk} \mathrm{\Gamma }=n$ or $2n,$ if $k=ℂ\text{.}$ Moreover, if $K$ is an $r\text{-group,}$ $k=ℂ$ and $\text{rk} \mathrm{\Gamma }=n$ then $K$ is the complexification of the Weyl group of a certain irreducible root system. Proof. as in Section 3.1 and 3.4. $\square$

Corollary. If $k=ℂ$ and $K$ has an invariant lattice of rank $n$ then $K$ has an invariant lattice of rank $2n\text{.}$ Proof. $K$ is the complexicification of a Weyl group, let $\mathrm{\Gamma }$ be a lattice of rank $n$ in the corresponding real form of $V,$ which is invariant under this Weyl group. Then, for every $z\in ℂ-ℝ$ a lattice $\mathrm{\Gamma }+z\mathrm{\Gamma }$ is $K\text{-invariant}$ and has rank $2n\text{.}$ $\square$

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 $\mathrm{\Lambda }$ of radical weights and ${\mathrm{\Lambda }}^{*}$ 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, $s=n,$ $\mathrm{\Lambda }$ be a nonzero $K\text{-invariant}$ lattice and $S=\left(1-{R}_{1}\right)+\dots +\left(1-{R}_{n}\right)\text{.}$ Then:

 a) $|\text{det} S|$ and $\text{Tr} S\in ℤ$ if $\text{rk} \mathrm{\Lambda }=n,$ ${|\text{det} S|}^{2}$ and $2 \text{Re} \text{Tr} S\in ℤ$ if $\text{rk} \mathrm{\Lambda }=2n\text{.}$ b) $|\text{det} S|$ depends only on $K$ but not on the choice of the generating system of reflections. c) $|\text{det} S|$ is divisible by $\left[\mathrm{\Lambda }:{\mathrm{\Lambda }}^{0}\right]$ if $\text{rk} \mathrm{\Lambda }=n,$ ${|\text{det} S|}^{2}$ is divisible by $\left[\mathrm{\Lambda }:{\mathrm{\Lambda }}^{0}\right]$ if $\text{rk} \mathrm{\Lambda }=2n\text{.}$ d) $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}⇒\left[{\mathrm{\Lambda }}^{*}:\mathrm{\Lambda }\right]=|\text{det} S|$ if $\text{rk} \mathrm{\Lambda }=n$ and $={|\text{det} S|}^{2}$ if $\text{rk} \mathrm{\Lambda }=2n\text{.}$ e) If $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}$ and ${d}_{1},\dots ,{d}_{r}$ are the invariant factors of a matrix of the endomorphism of $\mathrm{\Lambda },$ defined by $S,$ ${d}_{j}>0,$ ${d}_{1}|\dots |{d}_{r},$ then, $H1(K,Λ)≃ Λ*/Λ≃ ℤ/d1ℤ⊕…⊕ ℤ/drℤ.$ Specifically, ${d}_{1},\dots ,{d}_{r}$ do not depend on the choice of the generating system of reflections. f) $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}⇒|{H}^{1}\left(K,\mathrm{\Lambda }\right)|=|\text{det} S|$ if $\text{rk} \mathrm{\Lambda }=n$ and $={|\text{det} S|}^{2}$ if $\text{rk} \mathrm{\Lambda }=2n\text{.}$ Proof. If $\text{rk} \mathrm{\Lambda }=n,$ then, by considering the corresponding real form, we reduce the problem to the case $k=ℝ\text{.}$ If $\text{rk} \mathrm{\Lambda }=2n$ then $k=ℂ\text{.}$ It is known that in this case for every $P\in GL\left(V\right)$ one has $\text{det} {P}_{ℂ|ℝ}={|\text{det} P|}^{2}$ and $\text{Tr} {P}_{ℂ|ℝ}=2 \text{Re} \text{Tr} P$ (here ${P}_{ℂ|ℝ}$ is $P,$ considered as a linear operator of a $2n\text{-dimensional}$ real vector space $V\text{).}$ Now, a) follows from the fact that a basis of $\mathrm{\Lambda }$ is an $ℝ\text{-basis}$ of $V\text{.}$ We have also: $Λ0⊂Λ⊂ (Λ0)*= S-1Λ0, so [Λ:Λ0] |[S-1Λ0:Λ0].$ But ${S}^{-1}{\mathrm{\Lambda }}^{0}/{\mathrm{\Lambda }}^{0}\simeq {\mathrm{\Lambda }}^{0}/S{\mathrm{\Lambda }}^{0},$ whence $\left[{S}^{-1}{\mathrm{\Lambda }}^{0}:{\mathrm{\Lambda }}^{0}\right]=\left[{\left({\mathrm{\Lambda }}^{0}\right)}^{*}:{\mathrm{\Lambda }}^{0}\right]=|\text{det} S|$ if $k=ℝ$ or $=|\text{det} {S}_{ℂ|ℝ}|$ if $k=ℂ\text{.}$ The assertions b), c), d), e), f) follow from these equations. $\square$

Corollary.

 a) If $|\text{det} S|=1$ then $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0},$ i.e. any invariant lattice is a root lattice. b) Let $k=ℂ$ and suppose ${|\text{det} S|}^{2}$ is a prime number. Then $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}$ or $\mathrm{\Lambda }={\left({\mathrm{\Lambda }}^{0}\right)}^{*}\text{.}$

Remark. One can calculate $\text{det} S$ directly from the graph of $K,$ i.e. $det S=∑σsgn σ·aσ$ where $\sigma$ runs through the permutations of degree $n$ and ${a}_{\sigma }={c}_{\alpha }{c}_{\beta }\dots {c}_{\gamma }$ for $\sigma =\alpha \beta \cdots \gamma$ a decomposition of $\sigma$ as a product of cycles.

Examples.

a) $K={K}_{1},$ type ${A}_{n}\text{.}$ Write $S=S\left({A}_{n}\right)$ for the graph $1 n$ Simple cyclic products $\ne 0$ are ${c}_{j,j+1}={1}_{m}$ $1\le j\le n-1,$ and ${c}_{j}=2,$ $1\le j\le n\text{.}$

We have $\text{det} S\left({A}_{1}\right)=2\text{:}$ Assume, by induction, that $\text{det} S\left({A}_{k}\right)=k+1$ if $k Then, $\text{det} S\left({A}_{n}\right)={c}_{1}\text{det} S\left({A}_{n-1}\right)-{c}_{12}·\text{det} S\left({A}_{n-2}\right)=2n-\left(n-1\right)=n+1\text{.}$

b) $K={K}_{2},$ type $G\left(m,m,n\right)$ Let us consider the generating system of reflections given by the graph $1 2 3 n 4{\text{cos}}^{2}\frac{\pi }{m} 2{\zeta }_{2m}\text{cos}\frac{\pi }{m} {\zeta }_{2m}={e}^{\pi i/m}$

The list of all nonzero simple cyclic products is ${c}_{j}=2,$ $1\le j\le n\text{;}$
${c}_{13}={c}_{23}={c}_{34}={c}_{45}=\dots ={c}_{n-1,n}=1$
${c}_{12}=4{\text{cos}}^{2}\frac{\pi }{m}$
${c}_{123}=2\text{cos} \frac{\pi }{m} {e}^{i/m},$ ${c}_{132}=2\text{cos} \frac{\pi }{m} {e}^{-\pi i/m}\text{.}$

It is easy to see that $\text{det} S={c}_{1} \text{det} S\left({A}_{n-1}\right)-{c}_{12}\text{det} S\left({A}_{n-2}\right)-{c}_{13}·{c}_{2}\text{det} S\left({A}_{n-3}\right)+{c}_{123}\text{det} S\left({A}_{n-3}\right)+{c}_{132}\text{det} S\left({A}_{n-3}\right)=2n-4{\text{cos}}^{2} \frac{\pi }{m} \left(n-1\right)-1·2·\left(n-2\right)+2\text{cos} \frac{\pi }{m} {e}^{\pi i/m}\left(n-2\right)+2\text{cos}\frac{\pi }{m} {e}^{-\pi i/m}\left(n-2\right)=4-4{\text{cos}}^{2} \frac{\pi }{m} \left(n-1\right)+4{\text{cos}}^{2} \frac{\pi }{m} \left(n-2\right)=4{\text{sin}}^{2}\frac{\pi }{m}\text{.}$

Remark. If $k=ℝ$ and $\mathrm{\Lambda }$ is a lattice of radical weights with Weyl group $K$ then ${d}_{1},\dots ,{d}_{r}$ are invariant factors of a Cartan matrix and ${H}^{1}\left(K,\mathrm{\Lambda }\right)$ is isomorphic to the centre of a simple simply connected Lie group corresponding to $K\text{.}$

### Root lattices and infinite $r\text{-groups}$

We shall now show that if there exists a nonzero $K\text{-invariant}$ lattice $\mathrm{\Lambda }$ then there also exists an infinite $r\text{-group}$ $W$ with $\text{Lin} W=K$ and $\text{Tran} W=\mathrm{\Lambda }\text{.}$

Theorem. Let $K\subset GL\left(V\right)$ be a finite linear $r\text{-group}$ and $\mathrm{\Lambda }$ be a nonzero $K\text{-invariant}$ lattice in $V\text{.}$ Then the semidirect product of $K$ and $\mathrm{\Lambda }$ is an $r\text{-group}$ iff $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}\text{.}$ Proof. Let $W$ be the semidirect product of $K$ and $\mathrm{\Lambda }\text{.}$ Let $a$ be a special point. We identify $W$ and ${\kappa }_{a}\left(W\right)=K·\mathrm{\Lambda }\subset GL\left(V\right)V\text{.}$ Then the set of all reflections in $W$ is $\left\{\left(R,V\right) |$ $R$ is a reflection of $K$ and $v\in {l}_{R}\cap \mathrm{\Lambda }\right\}\text{.}$ Let ${W}^{0}$ be a subgroup of $W$ generated by this set. Then $K\subset {W}^{0}\text{.}$ Also ${\mathrm{\Lambda }}^{0}\subset {W}^{0},$ because ${\mathrm{\Lambda }}^{0}=\sum _{l\in ℒ}{\mathrm{\Lambda }}_{l}\text{.}$ Moreover, if $v\in {\mathrm{\Lambda }}_{l}$ and $l$ is the root line of a reflection $R,$ then $\left({R}^{-1},0\right)\left(R,v\right)=\left(0,v\right)\text{.}$ Therefore we have $K·{\mathrm{\Lambda }}^{0}\subset {W}^{0}\text{.}$ But every reflection in $W$ lies in $K·{\mathrm{\Lambda }}^{0}$ (because it has the form $\left(R,v\right),$ where $R$ is a reflection and $v\in {l}_{R}\cap \mathrm{\Lambda }={\mathrm{\Lambda }}^{0},$ see Section 1.2). Therefore, ${W}^{0}=K{\mathrm{\Lambda }}^{0}$ is generated by reflections, and we have ${W}^{0}=K\text{.}$ Hence, $K·\mathrm{\Lambda }=W={W}^{0}=K·{\mathrm{\Lambda }}^{0}$ iff $\mathrm{\Lambda }={\mathrm{\Lambda }}^{0}\text{.}$ $\square$

Corollary. The infinite $r\text{-groups}$ $W$ for which $\text{Lin} W$ is generated by $n$ reflections are exactly the groups $\text{Lin} W·\mathrm{\Lambda },$ where $\mathrm{\Lambda }$ is a nonzero $\text{Lin} W\text{-invariant}$ 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 $\text{Tran} W\text{.}$ b) $⇒$ c). Let $\mathrm{\Gamma }$ be an invariant lattice of rank $2n\text{.}$ Then the semidirect product of $K$ and ${\mathrm{\Gamma }}^{0}$ is a crystallographic $r\text{-group}$ because of the previous theorem. b) $⇒$ d). Let us first prove that $ℤ\left[\text{Tr}K\right]$ coincides with the ring with unity generated over $ℤ$ by all cyclic products. We have $ℤ\left[\text{Tr}K\right]=ℤ\left[\text{Tr}ℤK\right]\text{.}$ Indeed, clearly $\text{Tr}K\subset \text{Tr}ℤK,$ hence $ℤ\left[\text{Tr}K\right]\subset ℤ\left[\text{Tr}ℤK\right]\text{.}$ The reverse inclusion follows from the fact that Tr is an additive function. But $1-{R}_{j},$ $1\le j\le s,$ generate the ring $ℤK$ (here ${R}_{j},$ $1\le j\le s,$ is a generating system of reflections of $K\text{).}$ Therefore, the monomials $\left(1-{R}_{{j}_{1}}\right)\dots \left(1-{R}_{{j}_{r}}\right)$ generate $ℤ$ as a $ℤ\text{-module.}$ 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, $ℤ\left[\text{Tr}ℤK\right]=ℤ\left[\dots ,{c}_{{j}_{1}\cdots {j}_{r}},\dots \right]$ and we are done. By now, let $\mathrm{\Gamma }$ be a $K\text{-invariant}$ lattice of rank $2n\text{.}$ It follows from the equality $\left(*\right)$ that $cj1⋯jr Γj1⊂Γj1$ for every ${j}_{1},\dots ,{j}_{r}\text{.}$ But $\text{rk} {\mathrm{\Gamma }}_{j}=2,$ see Section 4.1. Therefore, ${c}_{{j}_{1}\cdots {j}_{r}}$ is an integral element of a certain purely imaginary quadratic extension of $ℚ\text{;}$ denote this extension by ${L}_{{j}_{1}}\text{.}$ Let $L={L}_{1}\text{;}$ we shall show that ${c}_{{j}_{1}\cdots {j}_{r}}\in L$ for every ${j}_{1},\dots ,{j}_{r}\text{.}$ $K$ being irreducible, there exists $α= c 1l1l2⋯ ltj1lt lt-1⋯l1 ≠0$ $1 {l}_{1} {l}_{2} {l}_{t} {j}_{1} {j}_{2} {j}_{3} {j}_{r}$ for certain ${l}_{1},\dots ,{l}_{t}\text{.}$ Let $β = c 1l1l2⋯lt j1j2⋯jr j1ltlt-1 ⋯l1 , γ = cj1j2⋯jr.$ Then $\alpha ,\beta \in L$ and $\beta =\alpha \gamma \text{.}$ But $\alpha \ne 0,$ hence $\gamma =\beta /\alpha \in L\text{.}$ Therefore the ring $ℤ\left[\text{Tr}K\right]$ lies in the maximal order of $L\text{.}$ d) $⇒$ e). This is proved in [Vin1971-2], Lemma 1.2. d) $⇒$ a). Let $ℤ\left[\text{Tr}K\right]\subset D,$ where $D$ is the maximal order of a certain purely imaginary quadratic extension $L$ of $ℚ\text{.}$ $K$ being irreducible and $D$ being integrally closed, one can again apply Lemma 1.2 from [Vin1971-2] and obtain that $K$ is defined over $D\text{.}$ Therefore there exists a $K\text{-invariant}$ $D\text{-submodule}$ $\mathrm{\Gamma }$ of $V$ such that $\mathrm{\Gamma }{\otimes }_{D}ℂ\to V$ is an isomorphism. But $D$ is a Dedekind ring and $\mathrm{\Gamma }$ is a $D\text{-module}$ of rank $n$ without torsion. Therefore, $\mathrm{\Gamma }$ is isomorphic to a direct sum of $n$ fractional ideals of the field $L,$ see [CRe1961]. Let ${J}_{1},\dots ,{J}_{n}$ be these ideals. Then there exists a $ℂ\text{-basis}$ ${e}_{1},\dots ,{e}_{n}$ 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 $d\in D,$ $d\ne 0,$ with $d·J\subset D\text{.}$ Hence $J$ is also a lattice of rank $2$ in $ℂ\text{.}$ e) $⇒$ d). Let $K$ be defined over a purely imaginary quadratic extension $L$ of $ℚ\text{.}$ Then $\text{Tr}P\in L$ for every $P\in K,$ hence $ℤ\left[\text{Tr}K\right]\subset L\text{.}$ But, $\text{Tr}P$ is an integral algebraic number. Therefore $ℤ\left[\text{Tr}K\right]$ lies in a maximal order of $L\text{.}$ d) $⇒$ f). Using Table 1, we can easily find those $K$ for which $ℤ\left[\text{Tr}K\right]$ lies in the maximal order of a certain purely imaginary quadratic extension of $ℚ\text{.}$ It is convenient to use the following necessary condition on each generator of this ring $ℤ\left[\text{Tr}K\right]\text{:}$ $z\in ℂ$ is an integral element of a certain purely imaginary quadratic extention of $ℚ$ iff ${|z|}^{2}\in ℤ$ and $2 \text{Re} z\in ℤ\text{.}$ It appears aposteriori that this condition is also sufficient. After a certain amount of calculations we obtain the list of f). Example. Let $K={K}_{2},$ type $G\left(m,m,s\right),$ $s\ge 3\text{.}$ The graph of $K$ is $1 2 3 s 4{\text{cos}}^{2}\frac{\pi }{m} 2{\zeta }_{2m}\text{cos}\frac{\pi }{m} {\zeta }_{2m}={e}^{\pi i/m}$ We have $ℤ\left[\text{Tr}K\right]=ℤ\left[{e}^{\pi i/m}\right]\text{.}$ The previous condition for the $2{e}^{\pi i/m}$ gives: $2\text{cos} \frac{2\pi }{m}\in ℤ\text{.}$ Therefore $m=2,3,4,6,$ and in these cases $ℤ\left[\text{Tr}K\right]$ lies in a maximal order of a certain purely imaginary extension of $ℚ\text{.}$ $\square$

### Description of root lattices

We assume now that $k=ℂ\text{.}$ 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 $r\text{-group}$ and let ${\mathrm{\Lambda }}_{j}\subset ℂ{e}_{j},$ $1\le j\le s,$ be a set of lattices of rank $2$ and $\mathrm{\Gamma }={\mathrm{\Lambda }}_{1}+\dots +{\mathrm{\Lambda }}_{s}\text{.}$ In order for $\mathrm{\Gamma }$ to be a $K\text{-invariant}$ root lattice with ${\mathrm{\Gamma }}_{j}={\mathrm{\Lambda }}_{j},$ $1\le j\le s,$ it is necessary, and, if $s=n,$ also sufficient, that:

 a) $ℤ\left[\text{Tr}K\right]{\mathrm{\Lambda }}_{j}\subset {\mathrm{\Lambda }}_{j},$ $1\le j\le s\text{.}$ b) $\left(1-{R}_{k}\right){\mathrm{\Lambda }}_{j}\subset {\mathrm{\Lambda }}_{k}$ and $\left(1-{R}_{j}\right){\mathrm{\Lambda }}_{k}\subset {\mathrm{\Lambda }}_{j}$ for every $k\ne j$ such that ${c}_{kj}\ne 0\text{.}$
Moreover, b) is equivalent to
 c) For every $k$ and $j,$ such that $j and ${c}_{kj}\ne 0,$ one has $(1-Rk)Λj⊂ Λk⊂ckj-1 (1-Rk)Λj.$ Proof. Assertion a) follows from the formula ${c}_{{j}_{1}\cdots {j}_{r}}{\mathrm{\Gamma }}_{{j}_{1}}\subset {\mathrm{\Gamma }}_{{j}_{1}}$ and the fact that $ℤ\left[\text{Tr}K\right]$ 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 $\mathrm{\Gamma }$ is in fact a direct sum of ${\mathrm{\Lambda }}_{j},$ $1\le j\le n,$ hence $\mathrm{\Gamma }$ is a lattice. This lattice is invariant under $1-{R}_{k}$ for every $k\text{:}$ if $k\ne j$ then it follows from b); if $k=j$ then it follows from a). Hence, $\mathrm{\Gamma }$ is $K\text{-invariant.}$ Now let us prove that b) $⇔$ c). b) $⇒$ c). One obtains the proof by applying $1-{R}_{k}$ to both sides of $\left(1-{R}_{j}\right){\mathrm{\Lambda }}_{k}\subset {\mathrm{\Lambda }}_{j}\text{.}$ c) $⇒$ b). One obtains the proof by applying $1-{R}_{j}$ to both sides of ${\mathrm{\Lambda }}_{k}\subset {c}_{kj}^{-1}\left(1-{R}_{k}\right){\mathrm{\Lambda }}_{j}\text{.}$ $\square$

Corollary. Let $\mathrm{\Gamma }$ be a nonzero $K\text{-invariant}$ lattice. If $|{c}_{kj}|=1$ then ${\mathrm{\Gamma }}_{j}$ and ${\mathrm{\Gamma }}_{k}$ 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 The index of the left lattice in the right lattice is ${|{c}_{kj}|}^{2}=1\text{.}$ Therefore the inclusions are in fact equalities. Applying $1-{R}_{j},$ we get $cjkΛj⊂ (1-Rj)Λk ⊂Λj.$ Again, the inclusions are in fact equalities. $\square$

If $s=n+1,$ we also need to know ${\mathrm{\Gamma }}^{*}$ for $\mathrm{\Gamma }$ a $K\prime \text{-invariant}$ lattice, and to select those $\mathrm{\Gamma }\subset \mathrm{\Lambda }\subset {\mathrm{\Gamma }}^{*}$ for which

 a) $\mathrm{\Lambda }$ is ${R}_{n+1}\text{-invariant,}$ b) ${\mathrm{\Lambda }}_{j}={\mathrm{\Gamma }}_{j},$ $1\le j\le s\text{.}$
Therefore, He also need to know ${\left({\mathrm{\Gamma }}^{*}\right)}^{0}\text{.}$ We have the following description of this lattice:

Theorem. $Γj*= ⋂k such thatcjk≠0 cjk-1 (1-Rj)Γk, 1≤j≤s.$ In particular, ${\mathrm{\Gamma }}_{j}^{*}={\mathrm{\Gamma }}_{j}$ if there is a number $k$ such that $|{c}_{jk}|=1\text{.}$ Proof. We have $λej*∈Γj* iff(1-Rk)λej ∈Γk,1≤k≤s.$ Assume that ${c}_{jk}\ne 0,$ i.e. $⟨{e}_{j} | {e}_{k}⟩\ne 0\text{.}$ If $\lambda \left(1-{R}_{k}\right){e}_{j}\in {\mathrm{\Gamma }}_{k}$ then, applying $1-{R}_{j}$ to the both sides of this relation, we obtain $\lambda {c}_{jk}{e}_{j}\in \left(1-{R}_{j}\right){\mathrm{\Gamma }}_{k},$ i.e. $\lambda {e}_{j}\in {c}_{jk}^{-1}\left(1-{R}_{j}\right){\mathrm{\Gamma }}_{k}\text{.}$ Vice versa, if $\lambda {e}_{j}\in {c}_{jk}^{-1}\left(1-{R}_{j}\right){\mathrm{\Gamma }}_{k}$ for some $\lambda$ then applying $1-{R}_{k},$ we obtain: $\left(1-{R}_{k}\right)\lambda {e}_{j}\in {c}_{jk}^{-1}{c}_{jk}{\mathrm{\Gamma }}_{k}={\mathrm{\Gamma }}_{k},$ i.e. $\lambda {e}_{j}\in {\mathrm{\Gamma }}^{*}\text{.}$ In order to prove the second assertion, let us apply $1-{R}_{j}$ to ${\mathrm{\Gamma }}_{k}\supset \left(1-{R}_{k}\right){\mathrm{\Gamma }}_{j}\text{.}$ We obtain ${\mathrm{\Gamma }}_{j}\supset \left(1-{R}_{j}\right){\mathrm{\Gamma }}_{k}\supset {c}_{jk}{\mathrm{\Gamma }}_{j}$ and, if $|{c}_{jk}|=1$ then ${c}_{jk}{\mathrm{\Gamma }}_{j}={\mathrm{\Gamma }}_{j}\text{.}$ Thus the inclusion is in fact an equality; hence ${c}_{jk}^{-1}\left(1-{R}_{j}\right){\mathrm{\Gamma }}_{k}={\mathrm{\Gamma }}_{j}\text{.}$ $\square$

Let us assume now that $s=n\text{.}$

An algorithm for constructing $K\text{-invariant}$ 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 $1\text{.}$

These groups are: ${K}_{1}\text{;}$ ${K}_{2},$ type $G\left(6,1,s\right),$ $s\ge 2,$ type $G\left(m,m,s\right),$ $m=2,4,6,$ $s\ge 3$ and $m=3,$ $s\ge 2\text{;}$ ${K}_{3},$ $m=2,3,4,6\text{;}$ ${K}_{4}\text{;}$ ${K}_{8},$ ${K}_{24},$ ${K}_{25},$ ${K}_{29},$ ${K}_{32},$ ${K}_{33},$ ${K}_{34},$ ${K}_{35},$ ${K}_{36}\text{;}$ ${K}_{37}\text{.}$

Algorithm:

Let $\mathrm{\Delta }$ be an arbitrary lattice in $ℂ$ of rank $2,$ which is invariant under $ℤ\left[\text{Tr}K\right]\text{.}$

Let: ${\mathrm{\Lambda }}_{1}=\mathrm{\Delta }{e}_{1}\text{.}$

For $l\ge 2$ let us consider an arbitrary path of edges from $1$ to $l,$ for which the absolute value of the weight of each edge equals $1\text{.}$ $1={l}_{1} {l}_{2} {l}_{3} l={l}_{r}$ Let $Λ1= (1-Rlr)… (1-Rl2) Λ1=Δ ( ∏j=2r (1-μ1j) ⟨elj-1 | elj⟩ ) elr.$ We claim that:

 a) $\mathrm{\Gamma }={\mathrm{\Lambda }}_{1}+\dots +{\mathrm{\Lambda }}_{n}$ is a $K\text{-invariant}$ lattice of full rank. b) $\mathrm{\Gamma }$ 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 ${k}_{1}=1={j}_{1} {j}_{2} {j}_{3} {j}_{q-1} j={j}_{q} {k}_{2} {k}_{3} {k}_{p-1} k={k}_{p}$ ${\mathrm{\Lambda }}_{k}=\left(1-{R}_{{k}_{p}}\right)\dots \left(1-{R}_{{k}_{2}}\right){\mathrm{\Lambda }}_{1}$ and ${\mathrm{\Lambda }}_{j}=\left(1-{R}_{{j}_{q}}\right)\dots \left(1-{R}_{{j}_{2}}\right){\mathrm{\Lambda }}_{1}\text{.}$ Let $P=\left(1-{R}_{{k}_{1}}\right)\dots \left(1-{R}_{{k}_{p-1}}\right)\text{.}$ Then $P{\mathrm{\Lambda }}_{k}={c}_{{k}_{1}{k}_{2}\dots {k}_{p-1}{k}_{p}{k}_{p-1}\dots {k}_{2}}{\mathrm{\Lambda }}_{1}\subset {\mathrm{\Lambda }}_{1}\text{.}$ (thanks to the construction of ${\mathrm{\Lambda }}_{1}\text{).}$ But ${c}_{{k}_{1}{k}_{2}\dots {k}_{p-1}{k}_{p}{k}_{p-1}\dots {k}_{2}}={c}_{{k}_{1}{k}_{2}}{c}_{{k}_{2}{k}_{3}}\dots {c}_{{k}_{p-1}{k}_{p}}\text{.}$ Hence $|{c}_{{k}_{1}{k}_{2}\dots {k}_{p-1}{k}_{p}{k}_{p-1}\dots {k}_{2}}|=1,$ so that $PΛk=Λ1.$ Let us consider $\left(1-{R}_{k}\right){\mathrm{\Lambda }}_{j}\text{.}$ 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 $ℂ{e}_{k}$ has a trivial kernel, because $P{\mathrm{\Lambda }}_{k}={\mathrm{\Lambda }}_{1}\text{.}$ Therefore, $\left(1-{R}_{k}\right){\mathrm{\Lambda }}_{j}\subset {\mathrm{\Lambda }}_{k}\text{.}$ Using the same arguments we obtain also $\left(1-{R}_{j}\right){\mathrm{\Lambda }}_{k}\subset {\mathrm{\Lambda }}_{j}\text{.}$ $\square$

Therefore we only need to find all $\mathrm{\Delta }\subset ℂ$ such that $ℤ\left[\text{Tr}K\right]$ 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 $ℤ\left[\text{Tr}K\right]\ne ℤ$ then $ℤ\left[\text{Tr}K\right]$ (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.

$K={K}_{2},$ type $G\left(4,4,s\right),$ $s\ge 3\text{.}$ The graph is $1 2 3 s=n 1+i$

The path we choose connecting $1$ and $l\ge 3$ will be $1,3,4,\dots ,l\text{;}$ and the path, connecting $1$ and $2$ will be $1,3,2\text{.}$

Here $ℤ\left[\text{Tr}K\right]=ℤ\left[i\right]$ and $⟨{e}_{1} | {e}_{3}⟩=⟨{e}_{2} | {e}_{3}⟩=⟨{e}_{3} | {e}_{4}⟩=\dots =⟨{e}_{n-1} | {e}_{n}⟩=-\frac{1}{2}\text{.}$

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 $\text{Tran} W$ with $\text{Lin} W=K$ of Table 2.

An algorithm for constructing $K\text{-invariant}$ root lattices of full rank: case 2.

We shall consider now the remaining irreducible finite $r\text{-groups}$ $K,$ 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 ${c}_{12},{c}_{23},\dots ,{c}_{n-1,n}$ are the only nonzero ${c}_{jk}$ (the numbering in Table 1 has this property).

Algorithm.

Let ${\mathrm{\Delta }}_{1}$ be a $ℤ\left[\text{Tr}K\right]\text{-invariant}$ lattice in $ℂ\text{.}$ We have $Δ1⊂c12-1 Δ1.$ Let us take an arbitrary $ℤ\left[\text{Tr}K\right]\text{-invariant}$ lattice ${\mathrm{\Delta }}_{2}$ between these two lattices (such a lattice exists, e.g. ${\mathrm{\Delta }}_{1}$ 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 $\mathrm{\Gamma }={\mathrm{\Lambda }}_{1}+\dots +{\mathrm{\Lambda }}_{n}$ 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, $(1-Rl+1) Λ1 = Δ1(1-Rl+1) …(1-R2)e1⊂ Δl+1(1-Rl+1) …(1-R2)e1 = Λl+1⊂ cl,l+1-1 Δl (1-Rl+1)… (1-R2)e1= cl,l+1-1 (1-Rl+1) Λ1.$ $\square$

It appears that, as in the case 1, $ℤ[TrK] is the maximal order and Δ1 is similar to ℤ[TrK],$ if $ℤ\left[\text{Tr}K\right]\ne ℤ\text{.}$

Example.

$K={K}_{2},$ type $G\left(3,1,s\right),$ $s\ge 2\text{.}$ The graph is $1 2 3 n 1-\omega$ We have $ℤ\left[\text{Tr}K\right]=ℤ\left[\omega \right]$ and $⟨{e}_{1} | {e}_{2}⟩=1/\sqrt{2},$ $⟨{e}_{2} | {e}_{3}⟩=\dots =⟨{e}_{n-1} | {e}_{n}⟩=-\frac{1}{2},$ ${c}_{12}=1-\omega ,$ ${c}_{23}=\dots ={c}_{n-1,n}=1\text{.}$ Therefore, $\left(1-{R}_{1}\right)\dots \left(1-{R}_{2}\right){e}_{1}={\left(1-\left(-1\right)\right)}^{l-1}{\left(-\frac{1}{2}\right)}^{l-2}\frac{1}{\sqrt{2}}{e}_{1}={\left(-1\right)}^{l}\sqrt{2}{e}_{l},$ $l=2,\dots ,n\text{.}$ We have $Δ1=[1,ω], Δ2=…=Δn, [1,ω]⊂Δ2 ⊂(1-ω)-1 [1,ω].$

But ${|1-\omega |}^{2}=3\text{.}$ 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 $\text{Tran} W$ with $\text{Lin} W=K$ of Table 2 as well as other lattices in the cases $K={K}_{2},$ types $G\left(2,1,2\right),$ $G\left(6,6,2\right)\text{;}$ $K={K}_{5}$ and $K={K}_{28}\text{.}$ As a matter of fact, these last lattices are not included in Table 2 because they do not give new $r\text{-groups}$ (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 $s=n+1\text{.}$

Now we shall briefly consider the case $s=n+1\text{.}$

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) $K={K}_{2},$ type $G\left(6,2,n\right)$ or $G\left(6,3,n\right)\text{.}$ The graphs of these groups are $1 2 3 n n+1 2+\omega 3 G\left(6,2,n\right) 3 1 2 3 n n+1 2+\omega 3 G\left(6,3,n\right)$

We see that $K\prime ={K}_{2},$ type $G\left(6,6,n\right),$ cf. Table 1. Using the previous theory, one checks that, up to similarity, there exists only one $K\prime \text{-invariant}$ root lattice of full rank, i.e. $Γ=[1,ω]e1 +…+[1,ω]en.$ But for $K\prime$ 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, $\mathrm{\Gamma }$ is a unique $K\text{-invariant}$ 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 $\mathrm{\Gamma }$ (of course, one can check that $\mathrm{\Gamma }$ is a $K\text{-invariant}$ lattice also by means of a straightforward computation).

b) $K={K}_{2},$ type $G\left(4,2,n\right),$ $n\ge 3\text{.}$ The graph is $1 2 3 n n+1 1+i 2 G\left(6,3,n\right)$ We see that $K\prime ={K}_{2},$ type $G\left(4,4,n\right)\text{.}$ One can check that there exists only one (up to similarity) $K\prime \text{-invariant}$ root lattice of full rank: $Λ=[1,i]e1 +…+[1,i]en.$

So, to describe all $K\text{-invariant}$ lattices of full rank we need to find all $\mathrm{\Gamma }$ such that:

 a) ${\mathrm{\Gamma }}^{0}=\mathrm{\Lambda }$ (with respect to $K\prime \text{);}$ b) $\mathrm{\Lambda }\subset \mathrm{\Gamma }\subset {\mathrm{\Lambda }}^{*}={S}^{-1}\mathrm{\Lambda }$ (with respect to $K\prime \text{);}$ c) $\left(1-{R}_{n+1}\right)\mathrm{\Gamma }\subset \mathrm{\Gamma }\text{.}$
We see that on every node there is an edge, whose weight is $1\text{.}$ Hence ${\left({\mathrm{\Lambda }}^{*}\right)}^{0}=\mathrm{\Lambda },$ see the theorem above. Therefore, b) $⇒$ a).

The matrix of $S$ with respect to the basis ${e}_{1},\dots ,{e}_{n}$ has the form $( 2 1-i -1 1+i 2 -1 -1 2 -1 -1 2 ⋱ 2 -1 -1 2 )$ We have $\text{det} S=4{\text{sin}}^{2} \frac{\pi }{4}=2,$ see Section 4.4. Therefore the coefficients of ${S}^{-1}$ lie in $ℤ\left[i,\frac{1}{2}\right]$ and (cf. Section 4.4), the order of ${\mathrm{\Lambda }}^{*}/\mathrm{\Lambda }$ is equal to $4\text{.}$ It follows from these facts that ${\mathrm{\Lambda }}^{*}/\mathrm{\Lambda }=ℤ/2ℤ\oplus ℤ/2ℤ\text{.}$ It is not difficult to see that ${e}_{j}=S{f}_{j},$ $j=1,2,$ 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 ${f}_{1},{f}_{2},{f}_{3}$ are representatives of different nonzero elements of ${\mathrm{\Lambda }}^{*}/\mathrm{\Lambda }\text{.}$ Therefore every $K\text{-invariant}$ 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 $\oplus$ are $K\text{-invariant.}$

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