A short note about diffuse Bieberbach groups

TL;DR

This paper classifies low-dimensional diffuse Bieberbach groups up to dimension 6 and addresses a question about the minimal dimension of non-diffuse Bieberbach groups lacking a three-dimensional Hantzsche-Wendt subgroup.

Contribution

It provides a complete classification of diffuse Bieberbach groups in dimensions up to 6 and answers an open question about their minimal non-diffuse counterparts.

Findings

Classified diffuse Bieberbach groups up to dimension 6.

Answered an open question on minimal dimension of certain non-diffuse Bieberbach groups.

Identified properties related to the presence of Hantzsche-Wendt groups.

Abstract

We consider low dimensional diffuse Bieberbach groups. In particular we classify diffuse Bieberbach groups up to dimension 6. We also answer a question from [S. Kionke, J. Raimbault, On geometric aspects of diffuse groups, Doc. Math. 21 (2016), page 887] about minimal dimension of a non-diffuse Bieberbach group which does not contain three-dimensional Hantzsche-Wendt group.

Figures

Equations

$${\Delta }_P=⟨x,y\mid x^{-1}{y}^{2}x=y^{-2},y^{-1}{x}^{2}y=x^{-2}⟩$$

$$f:\Gamma \to {\mathbb{Z}}^k,\text{ where }k={\beta }_1(\Gamma ).$$

$$\left[\begin{array}{cc}A& a\\ 0& 1\end{array}\right],$$

$$p\left(\left[\begin{array}{cc}A& a\\ 0& 1\end{array}\right]\right)=A\text{ for every }\left[\begin{array}{cc}A& a\\ 0& 1\end{array}\right]\in \Gamma ,$$

$$\left[\begin{array}{ccc}A& B& a\\ 0& I& b\\ 0& 0& 1\end{array}\right]$$

$$f\left(\left[\begin{array}{ccc}A& B& a\\ 0& I& b\\ 0& 0& 1\end{array}\right]\right)=b$$

$$F\left(\left[\begin{array}{ccc}A& B& a\\ 0& I& 0\\ 0& 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}A& a\\ 0& 1\end{array}\right]$$

$$A=\left[\begin{array}{ccccc}0& -1& 1& 0& 1/2\\ -1& 0& 1& 0& 0\\ 0& 0& 1& 0& 1/2\\ 0& 0& 0& -1& 1/2\\ 0& 0& 0& 0& 1\end{array}\right]\text{ and }B=\left[\begin{array}{ccccc}0& 1& -1& 0& 1/2\\ 0& 1& 0& 0& 1/2\\ -1& 1& 0& 0& 0\\ 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 1\end{array}\right].$$

$$Q=\left[\begin{array}{ccccc}1& 1& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1\end{array}\right]\in \mathrm{GL}(5,\mathbb{Z})$$

$$A^Q=\left[\begin{array}{ccccc}1& 0& 0& 0& 1/2\\ 0& -1& 0& 1& 0\\ 0& 0& -1& 0& 1/2\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\text{ and }{B}^Q=\left[\begin{array}{ccccc}-1& 0& 0& -1& 0\\ 0& 1& 0& 0& 1/2\\ 0& 0& -1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right].$$

$$A^{\prime }=\left[\begin{array}{cccc}1& 0& 0& 1/2\\ 0& -1& 0& 0\\ 0& 0& -1& 1/2\\ 0& 0& 0& 1\end{array}\right]\text{ and }{B}^{\prime }=\left[\begin{array}{cccc}-1& 0& 0& 0\\ 0& 1& 0& 1/2\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}\right]$$

$$\begin{array}[]{llll}[a,b]=1&&a^{\beta}=a^{-1}&a^{\gamma}=a^{-1}\cr[a,c]=1&b^{\alpha}=b^{-1}&&b^{\gamma}=b^{-1}\cr[b,c]=1&c^{\alpha}=c^{-1}&c^{\beta}=c^{-1}&\cr\end{array}$$

$$⟨x=\left[\begin{array}{cccc}1& 0& 0& 1/2\\ 0& -1& 0& 1/2\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}\right],y=\left[\begin{array}{cccc}-1& 0& 0& 0\\ 0& 1& 0& 1/2\\ 0& 0& -1& 1/2\\ 0& 0& 0& 1\end{array}\right]⟩\subset \mathrm{GL}(4,\mathbb{Q}).$$

$${\gamma }_1=\left[\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 1/2\\ 0& 0& 0& -1& 0& 1/4\\ 0& 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\text{ and }{\gamma }_2=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 1/2\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 1& 1/2\\ 0& 0& 0& 0& 0& 1\end{array}\right]$$

$$l_i:=\left[\begin{array}{cc}{I}_5& e_i\\ 0& 1\end{array}\right]$$

$$1⟶L⟶\Gamma \stackrel{\pi }{⟶}{D}_8⟶1$$

$$\left[\begin{array}{cc}A& a\\ 0& 1\end{array}\right]\mapsto A$$

$$1⟶T⟶{\Gamma }^{\prime }⟶C_2^2⟶1,$$

$$1$$

$$T\cap L\subset {\Gamma }^{\prime }\cap L\subset T\cap L,$$

$$1$$

$$G=⟨\mathrm{diag}(1,-1,-1,-1,1),\mathrm{diag}(-1,-1,1,1,1).⟩$$

$$\alpha =\left[\begin{array}{cccccc}1& 0& 0& 0& 0& x_1-\frac{1}{2}\\ 0& -1& 0& 0& 0& x_2\\ 0& 0& -1& 0& 0& x_3\\ 0& 0& 0& -1& 0& x_4\\ 0& 0& 0& 0& 1& x_5-\frac{1}{2}\\ 0& 0& 0& 0& 0& 1\end{array}\right]\text{ and }\beta =\left[\begin{array}{cccccc}-1& 0& 0& 0& 0& y_1+\frac{1}{2}\\ 0& -1& 0& 0& 0& y_2-\frac{1}{2}\\ 0& 0& 1& 0& 0& y_3\\ 0& 0& 0& 1& 0& y_4+\frac{1}{2}\\ 0& 0& 0& 0& 1& y_5\\ 0& 0& 0& 0& 0& 1\end{array}\right],$$

$$c^{\alpha }-c^{-1}=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 4y_5+4x_5-2\\ 0& 0& 0& 0& 0& 0\end{array}\right].$$

$$\alpha =\left[\begin{array}{cccccc}0& -1& 0& 0& 0& x_1\\ 1& 0& 0& 0& 0& x_2-\frac{1}{2}\\ 0& 0& -1& 0& 0& x_3-\frac{1}{2}\\ 0& 0& 0& 1& 0& x_4+\frac{1}{4}\\ 0& 0& 0& 0& -1& x_5+\frac{1}{2}\\ 0& 0& 0& 0& 0& 1\end{array}\right]\text{ and }\beta =\left[\begin{array}{cccccc}1& 0& 0& 0& 0& y_1-\frac{1}{2}\\ 0& -1& 0& 0& 0& y_2\\ 0& 0& -1& 0& 0& y_3\\ 0& 0& 0& -1& 0& y_4\\ 0& 0& 0& 0& 1& y_5-\frac{1}{2}\\ 0& 0& 0& 0& 0& 1\end{array}\right]$$

$$ab-ba=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 2-4y_1\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right]$$

Full text

††footnotetext: E-mail addresses: rafal.lutowski@mat.ug.edu.pl, aszczepa@mat.ug.edu.pl, anna.gasior@poczta.umcs.lublin.pl ††footnotetext: 2010 Mathematics Subject Classification: Primary 20E40; Secondary 20H15, 20F65.††footnotetext: Key words and phrases: unique product property; diffuse groups, Bieberbach groups.

1 Introduction

The class of diffuse groups was introduced by B. Bowditch in [2]. By definition a group $\Gamma$ is diffuse, if every finite non-empty subset $A\subset\Gamma$ has an extremal point, i.e. an element $a\in A$ such that for any $g\in\Gamma\setminus\{1\}$ either $ga$ or $g^{-1}a$ is not in $A.$ Equivalently (see [7]) a group $\Gamma$ is diffuse if it does not contain a non-empty finite set without extremal points.

The interest in diffuse groups follows from Bowditch’s observation that they have the unique product property 111The group $\Gamma$ is said to have the unique product property if for every two finite non-empty subsets $A,B\subset\Gamma$ there is an element in the product $x\in A\dot{B}$ which can be written uniquely in the form $x=ab$ with $a\in A$ and $b\in B.$. Originally unique products were introduced in the study of group rings of discrete, torsion-free groups. More precisely, it is easily seen that if a group $\Gamma$ has the unique product property, then it satisfies Kaplansky’s unit conjecture. In simple terms this means that the units in the group ring ${\mathbb{C}}[\Gamma]$ are all trivial, i.e. of the form $\lambda g$ with $\lambda\in{\mathbb{C}}^{\ast}$ and $g\in\Gamma.$ For more information about these objects we refer the reader to [1], [9, Chapter 10] and [7]. In part 3 of [7] the authors prove that any torsion-free crystallographic group (Bieberbach group) with trivial center is not diffuse. By definition a crystallographic group is a discrete and cocompact subgroup of the group $O(n)\ltimes{\mathbb{R}}^{n}$ of isometries of the Euclidean space ${\mathbb{R}}^{n}.$ From Bieberbach’s theorem (see [12]) the normal subgroup $T$ of all translations of any crystallographic group $\Gamma$ is a free abelian group of finite rank and the quotient group (holonomy group) $\Gamma/T=G$ is finite.

In [7, Theorem 3.5] it is proved that for a finite group $G$:

If $G$ is not solvable then any Bieberbach group with holonomy group isomorphic to $G$ is not diffuse. 2. 2.

If every Sylow subgroup of $G$ is cyclic then any Bieberbach group with holonomy group isomorphic to $G$ is diffuse. 3. 3.

If $G$ is solvable and has a non-cyclic Sylow subgroup then there are examples of Bieberbach groups with holonomy group isomorphic to $G$ which are and examples which are not diffuse.

Using the above the authors of [7] classify non-diffuse Bieberbach groups in dimensions $\leq 4.$ One of the most important non-diffuse groups is the 3-dimensional Hantzsche-Wendt group, denoted in [11] by $\Delta_{P}$. For the following presentation

[TABLE]

the maximal abelian normal subgroup is generated by $x^{2},y^{2}$ and $(xy)^{2}$ (see [6, page 154]). At the end of part 3.4 of [7] the authors ask the following question.

Question 1**.**

What is the smallest dimension $d_{0}$ of a non-diffuse Bieberbach group which does not contain $\Delta_{P}$ ?

The answer for the above question was the main motivation for us. In fact we prove, in the next section, that $d_{0}=5.$ Moreover, we extend the results of part 3.4 of [7] and with support of computer, we present the classification of all Bieberbach groups in dimension $d\leq 6$ which are (non)diffuse.

2 (Non)diffuse Bieberbach groups in dimension $\leq 6.$

We use the computer system CARAT [10] to list all Bieberbach groups of dimension $\leq 6.$

Our main tools are the following observations:

The property of being diffuse is inherited by subgroups (see [2, page 815]). 2. 2.

If $\Gamma$ is a torsion-free group, $N\lhd\Gamma$ such that $N$ and $\Gamma/N$ are both diffuse then $\Gamma$ is diffuse (see [2, Theorem 1.2 (1)]).

Now let $\Gamma$ be a Bieberbach group of dimension less than or equal to $6$. By the first Betti number $\beta_{1}(\Gamma)$ we mean the rank of the abelianization $\Gamma/[\Gamma,\Gamma]$. Note that we are only interested in the case when $\beta_{1}(\Gamma)>0$ (see [7, Lemma 3.4]). Using a method of E. Calabi [12, Propostions 3.1 and 4.1], we get an epimorphism

[TABLE]

From the assumptions $\ker f$ is a Bieberbach group of dimension $<6.$ Since ${\mathbb{Z}}^{k}$ is a diffuse group our problem is reduced to the question about the group $\ker f.$

Remark 1*.*

Up to conjugation in $\operatorname{GL}(n+1,{\mathbb{R}})$, $\Gamma$ is a subgroup $\operatorname{GL}(n,{\mathbb{Z}})\ltimes{\mathbb{Z}}^{n}\subset\operatorname{GL}(n+1,{\mathbb{Q}})$, i.e. it is a group of matrices of the form

[TABLE]

where $A\in\operatorname{GL}(n,{\mathbb{Z}}),a\in{\mathbb{Q}}^{n}$. If $p\colon\Gamma\to\operatorname{GL}(n,{\mathbb{Z}})$ is a homomorphism which takes the linear part of every element of $\Gamma$

[TABLE]

then there is an isomorphism $\rho\colon G\to p(\Gamma)\subset\operatorname{GL}(n,{\mathbb{Z}})$. It is known that that the rank of the center of a Bieberbach group equals the first Betti number (see [5, Proposition 1.4]). By [12, Lemma 5.2], the number of trivial constituents of the representation $\rho$ is equal to $k$. Hence without lose of generality we can assume that the matrices in $\Gamma$ are of the form

[TABLE]

where $A\in\operatorname{GL}(n-k,{\mathbb{Z}})$, $I$ is the identity matrix of degree $k$, $B$ is an integral matrix of dimension $n-k\times k$, $a\in{\mathbb{Q}}^{n-k}$ and $b\in{\mathbb{Q}}^{k}$. Then $f$ may be defined by

[TABLE]

and one can easily see that the map $F\colon\ker f\to\operatorname{GL}(n-k+1,{\mathbb{Q}})$ given by

[TABLE]

is a monomorphism and hence its image is a Bieberbach group of rank $n-k$.

Now if $\Gamma$ has rank $4$ we know that the only non-diffuse Bieberbach group of dimension less than or equal to $3$ is $\Delta_{P}.$ Using the above facts we obtain 17 non-diffuse groups. Note that the list from [7, section 3.4] consists of 16 groups. The following example presents the one which is not in [7] and illustrates computations given in the above remark.

Example 1**.**

Let $\Gamma$ be a crystallographic group denoted by "05/01/06/006" in [3] as a subgroup of $\operatorname{GL}(5,{\mathbb{R}})$. Its non-lattice generators are as follows

[TABLE]

Conjugating the above matrices by

[TABLE]

one gets

[TABLE]

Now its easy to see that the rank of the center of $\Gamma$ equals $1$ and the kernel of the epimorphism $\Gamma\to\mathbb{Z}$ is isomorphic to a 3-dimensional Bieberbach group $\Gamma^{\prime}$ with the following non-lattice generators:

[TABLE]

Clearly the center of $\Gamma^{\prime}$ is trivial, hence it is isomorphic to the group $\Delta_{P}.$

Now we formulate our main result.

Theorem 1**.**

The following table summarizes the number of diffuse and non-diffuse Bieberbach groups of dimension $\leq 6$.

[TABLE]

Proof.

If a group has a trivial center then it is not diffuse. In other case we use the Calabi (1) method and induction. A complete list of groups was obtained using computer algebra system GAP [4] and is available here [8]. ∎

Before we answer Question 1 from the introduction, let us formulate the following lemma:

Lemma 1**.**

Let $\alpha,\beta$ be any generators of the group $\Delta_{P}$. Let $\gamma=\alpha\beta,a=\alpha^{2},b=\beta^{2},c=\gamma^{2}$. Then the following relations hold:

[TABLE]

where $x^{y}:=y^{-1}xy$ denotes the conjugation of $x$ by $y$.

The proof of the above lemma is omitted. Just note that the relations are easily checked if consider the following representation of $\Delta_{P}$ as a matrix group

[TABLE]

Proposition 1**.**

There exists an example of a five dimensional non-diffuse Bieberbach group which does not contain any subgroup isomorphic to $\Delta_{P}$.

Proof.

Let $\Gamma$ be the Bieberbach group enumerated in CARAT as "min.88.1.1.15". It generated by the elements $\gamma_{1},\gamma_{2},l_{1},\ldots,l_{5}$ where

[TABLE]

and $l_{1},\ldots,l_{5}$ generate the lattice $L$ of $\Gamma$:

[TABLE]

where $e_{i}$ is the $i$-th column of the identity matrix $I_{5}$. $\Gamma$ fits into the following short exact sequence

[TABLE]

where $\pi$ takes the linear part of every element of $\Gamma$:

[TABLE]

and the image $D_{8}$ of $\pi$ is the dihedral group of order 8.

Now assume that $\Gamma^{\prime}$ is a subgroup of $\Gamma$ isomorphic to $\Delta_{P}$. Let $T$ be its maximal normal abelian subgroup. Then $T$ is free abelian group of rank $3$ and $\Gamma^{\prime}$ fits into the following short exact sequence

[TABLE]

where $C_{m}$ is a cyclic group of order $m$. Consider the following commutative diagram

[TABLE]

We get that $H$ must be an abelian subgroup of $D_{8}=\pi(\Gamma)$ and $T\cap L$ is a free abelian group of rank $3$ which lies in the center of $\pi^{-1}(H)\subset\Gamma$. Now if $H$ is isomorphic to either to $C_{4}$ or $C_{2}^{2}$ then the center of $\pi^{-1}(H)$ is of rank at most $2$. Hence $H$ must be the trivial group or the cyclic group of order $2$. Note that as $\Gamma^{\prime}\cap L$ is a normal abelian subgroup of $\Gamma^{\prime}$ it must be a subgroup of $T$:

[TABLE]

hence $T\cap L=\Gamma^{\prime}\cap L$. We get the following commutative diagram with exact rows and columns

[TABLE]

where $G=\pi(\Gamma^{\prime})$. Consider two cases:

$H$ is trivial. In this case $G$ is one of the two subgroups of $D_{8}$ isomorphic to $C_{2}^{2}$. Since the arguments for both subgroups are similar, we present only one of them. Namely, let

[TABLE]

In this case $\Gamma^{\prime}$ is generated by the matrices of the form

[TABLE]

where $x_{i},y_{i}\in{\mathbb{Z}}$ for $i=1,\ldots,5$. If $c=(\alpha\beta)^{2}$ then by Lemma 1 $c^{\alpha}=c^{-1}$, but

[TABLE]

Obviously solutions of the equation $4{{y}_{5}}+4{{x}_{5}}-2=0$ are never integral and we get a contradiction. 2. 2.

$H$ is of order $2$. Then $G=D_{8}$ and $H$ is the center of $G$. The generators $\alpha,\beta$ of $\Gamma^{\prime}$ lie in the cosets $\gamma_{1}\gamma_{2}L$ and $\gamma_{2}L$, hence

[TABLE]

where $x_{i},y_{i}\in{\mathbb{Z}}$ for $i=1,\ldots,5$, as before. Setting $a=\alpha^{2},b=\beta^{2}$ we get

[TABLE]

and again the equation $2-4{{y}_{1}}=0$ does not have an integral solution.

The above considerations show that $\Gamma$ does not have a subgroup which is isomorphic to $\Delta_{P}$. ∎

Acknowledgements

This work was supported by the Polish National Science Center grant 2013/09/B/ST1/04125.