Small non-Leighton two-complexes

Abstract

How many 2-cells must two finite CW-complexes have to admit a common, but not finite common, covering? Leighton’s theorem says that both complexes must have 2-cells. We construct an almost (?) minimal example with two 2-cells in each complex.

0. Introduction

Leighton Theorem [10] If two finite graphs have a common covering, then they have a common finite covering.

Alternative proofs and various generalisations of this result can be found, e.g., in [2], [4], [14], [15], [19], and references therein.

Does a similar result hold for any CW-complexes, i.e.

is it true that, if, for finite CW-complexes $K_1$ and $K_2$ , there exist a CW-complex K and cellular coverings $K_1\leftarrow K\to K_2$ , then there exists a finite CW-complex K with this property?

This natural question was posed (in other terms) in [1] and [16]. Notice the cellularity requirement. Surely, we would obtain an equivalent question if we replace this condition with a formally stronger combinatorialness one: the image of each cell is a cell. However, without the cellularity condition, the answer would be negative: indeed, the torus and the genus-two surface have no finite common coverings (as the fundamental group the genus-two orientable surface $\left\langle{x,y,z,t\;|\;[x,y][z,t]=1}\right\rangle$ contains no abelian subgroups of finite index), while the universal coverings of these surfaces are homeomorphic, because they are the plane. The cellularity condition rules out such examples: if we take, e.g., the standard one-vertex cell structures on the torus and genus-two surface, then, on the covering plane, we obtain:

1. (i) the usual square lattice on the (Euclidean) plane (in the torus-case);

2. (ii) and an octagonal lattice on the (Lobachevskii) plane (in the genus-two case);

(i.e. though the universal coverings are homeomorphic, the cell structure on them are principally different). This example cannot be saved by a complication of the cell structures on the torus and genus-two surface (as was noted in [1] and [16]; the authors of [1] even conjectured that the answer to the (cellular version of) the question is positive).

Nevertheless, the answer turned out to be negative as was shown in [18] (and actually, much earlier in [17]); the complexes $K_1,\;K_2$ forming such a non-Leighton pair from [18] contain as few as six 2-cells each. In [8], this number was reduced to four:¹

there exist two two-complexes containing four 2-cells each that have a common covering but have not finite common coverings.

(Henceforth, we omit the prefix “CW-” and word “cellular”: a complex means a CW-complex, and all mapping between complexes are assumed to be cellular in this paper.) The non-Leighton complexes $K_1$ and $K_2$ from [8] are the standard complexes of the following group presentations $\Gamma_i$ , i.e. one-vertex complexes with edges corresponding to the generators and 2-cells attached by the relators:

\begin{align*}\Gamma_1=F_2\times F_2=\left\langle{a,b,x,y\;|\;[a,x],\;[a,y],\;[b,x],\;[b,y]}\right\rangle\ \hbox{ and }\\[4pt]\Gamma_2=\left\langle{a,b,x,y\;\bigg|\; axay,\;ax^{-1}by^{-1},\;ay^{-1}b^{-1}x^{-1},\;bxb^{-1}y^{-1}}\right\rangle .\end{align*}

Both of these complex are covered by the Cartesian product of two trees (Cayley graphs of the free group $F_2$ ); and no finite common cover exists, because the fundamental group of such hypothetical covering complex would embed in both groups $\Gamma_i$ as finite-index subgroups, but, in $\Gamma_1$ , any finite-index subgroup contains a finite-index subgroup which is the direct product of free groups, while $\Gamma_2$ has no such finite-index subgroups [8] ( $\Gamma_2$ is not even residually finite [3], [5]). The results of [8] imply also a minimality of this example in the sense that:

if we restrict ourselves to complexes $K_i$ covered by products of trees, then four two-dimensional cells is the minimum among all non-Leighton pairs.

If we do not restrict ourselves, then smaller non-Leighton pairs arise.

Main Theorem (a simplified version). There exist two finite two-complexes containing two 2-cells each that have a common covering, but have not finite common coverings.

(Explicit forms of these two two-2-cell two-complexes can be found at the very end of this paper.) Thus, the only question remaining open concerns complexes with a single 2-cell. This question seems to be difficult (although it is closely related to the well-developed theory one-relator groups). The point is that a classification of one-relator groups up to commensurability is not an easy task even for the Baumslag–Solitar groups ${\rm BS}(n,m)\:\stackrel{\rm def}{=}\left\langle{c,d\;\big|\;c^{nd}=c^m}\right\rangle$ (though, in this special case, it was recently obtained [6]). Henceforth, $x^{ky}\:\stackrel{\rm def}{=}y^{-1}x^ky$ , where x and y are elements of a group and $k\in{\mathbb Z}$ .

In conclusion, note that results on coverings of two-complexes can imply nontrivial facts about graphs, because one can “model” 2-cells in graphs by means of additional vertices and edges, see [4]. Higher dimensional complexes are of little sense here: if complexes $K_1$ and $K_2$ form a non-Leighton pair, then their two-skeleta also form such a pair, as is easy to verify. A detailed exposition of the general theory of coverings and CW-complexes can be found, e.g., in [7].

1. Algebraic lemmata

The following fact is well known [13], we give a short proof for the reader’s convenience.

Commutator Lemma. In the group $H={\rm BS}(3,5)=\left\langle{c,d\;\big|\;c^{3d}=c^5}\right\rangle,$ the commutator $h=[c^d,c]$ belongs to any finite-index subgroup.

Proof. Each finite-index subgroup contains a normal finite-index subgroup (see, e.g., [9]) Therefore, it suffices to show that h lies in the kernel of any homomorphism $\varphi\,{:}\,H\to K$ to any finite group K.

The elements $\varphi(c^3)$ and $\varphi(c^5)$ have the same order (because they are conjugate); hence, the order of $\varphi(c)$ is not divisible by three. Therefore, $\varphi(c)\in\left\langle{\varphi(c^3)}\right\rangle$ . Thus, $\varphi(c)^{\varphi(d)}\in\left\langle{\varphi(c)}\right\rangle$ and $h=[c^d,c]$ belongs to the kernel of $\varphi$ . This completes the proof.

Bottle Lemma. If a group G has a subgroup $\left\langle{a,b}\right\rangle=\left\langle{a,b\;\big|\;a^b=a^{-1}}\right\rangle\simeq{\rm BS}(1,-1),$ and the element b lies in all finite-index subgroups of G, then any finite-index subgroup of G contains a subgroup isomorphic to the Klein-bottle group ${\rm BS}(1,-1)$ .

Proof. Any finite-index subgroup contains all elements conjugate to b, because the intersection R of all finite-index subgroups is normal. Therefore, $a^2=b^{-1}b^a\in R$ and $\left\langle{a^2,b}\right\rangle\subseteq R$ It remains to note that $a^{2b}=a^{-2},$ and the groups $\left\langle{a^2}\right\rangle$ and $\left\langle{b}\right\rangle$ are infinite; hence, the subgroup $\left\langle{a^2,b}\right\rangle$ is isomorphic to ${\rm BS}(1,-1)$ , because,

in any group, infinite-order elements x and y such that $x^y=x^{-1}$ generate a subgroup isomorphic to the Klein-bottle group. (1)

Indeed, there is obvious epimorphism

\begin{equation*}\varphi\,{:}\,{\rm BS}(1,-1)=\left\langle{a,b}\right\rangle \longrightarrow \left\langle{x,y}\right\rangle.\end{equation*}

Any element $g\in{\rm BS}(1,-1)$ can be written in the form $g=a^kb^l$ . If $g=a^kb^l\in\ker\varphi$ , then $\ker\varphi\ni[b,g]=b^{-1}b^{-l}a^{-k}ba^kb^l=a^{\pm2k}.$ Therefore, $k=0$ (because $|\!\left\langle{x}\right\rangle\!|=\infty$ ). But then $l=0$ too, because $1=\varphi(g)=\varphi(b^l)=y^l$ , and $|\!\left\langle{y}\right\rangle\!|=\infty$ . Thus, $\ker\varphi=\{1\}$ and $\varphi$ is an isomorphism. This completes the proof.

No-Bottle Lemma. The amalgamated free product

\begin{equation*}G=\left\langle{a,c,d\;\bigg|\;[a,[c^d,c]]=1,\; c^{3d}=c^5}\right\rangle=\left\langle{a,b\;\big|\;[a,b]=1}\right\rangle\mathop*_{b=[c^d,c]}\left\langle{c,d\;\bigg|\;c^{3d}=c^5}\right\rangle\end{equation*}

of the free abelian group and the Baumslag–Solitar group ${\rm BS}(3,5)$ contains no subgroups isomorphic to the Klein-bottle group $K={\rm BS}(1,-1)$ .

Proof. The group ${\rm BS}(3,5)$ does not contain subgroups isomorphic to K [11] and is torsion-free. Therefore, applying once again (1), we obtain that the quotient

\begin{equation*}G/\left\langle\langle{{[a,G]}}\right\rangle\rangle=\left\langle{a}\right\rangle_\infty\times{\rm BS}(3,5)\end{equation*}

by the normal closure $\left\langle\langle{{[a,G]}}\right\rangle\rangle$ of the set [a, G] of commutators of a and all elements of G has no nonidentity elements conjugate their inverse. Therefore, any element of G conjugate to its inverse lies in $N=\left\langle\langle{{[a,G]}}\right\rangle\rangle$ . This subgroup intersects trivially the free factors (and their conjugates). Thus, all elements of N have length at least two, and the following conjugation criterion (see, e.g., [12]) applies:

two cyclically reduced words of length $\ge2$ in an amalgamated free product $U\mathop*\limits_WV$ are conjugate if and only if one of them can be obtained from the other by a cyclic permutation and subsequent conjugation by an element of W.

In $U\mathop*\limits_WV$ , an equality of reduced words $u_1v_1\cdots=u_1^{\prime}v_1^{\prime}\cdots$ implies the equalities of the double cosets $Wu_1W=Wu_1^{\prime}W$ , $Wv_1W=Wv_1^{\prime}W, \dots$ $Wv_1W=Wv_1^{\prime}W$ Therefore, if a cyclically reduced word $x\in N\triangleleft\,\left\langle{a,b\;|\;[a,b]=1}\right\rangle\mathop*\limits_{b=[c^d,c]}\left\langle{c,d\;\big|\;c^{3d}=c^5}\right\rangle$ is conjugate to its inverse, then, for a letter $x_1$ of x, we obtain the equality $x_1=b^kx_1^{-1}b^l$ (because the map $f_{k,n}\,{:}\,i\mapsto k-i\pmod n$ from the set (of subscripts) $\{1,\dots,n\}$ to itself has either a fixed point, or an almost fixed point: $f_{k,n}(i)=i+1\pmod n$ for some i; the latter case would imply that $x_1=b^kx_2^{-1}b^l$ for some adjacent letters $x_1$ and $x_2$ of the reduced word x, which is impossible). Substituting $x_1=b^k\widehat{x}_1$ , we obtain $\widehat{x}_1^2=b^{l-k}$ ; thus:

1. (i) either $\widehat{x}_1^2\in\left\langle{b}\right\rangle$ for some $\widehat{x}_1\in(\!\left\langle{a}\right\rangle_\infty\times\left\langle{b}\right\rangle_\infty\!)\setminus\left\langle{b}\right\rangle$ ;

2. (ii) or $\widehat{x}_1^2\in\left\langle{[c^d,c]}\right\rangle$ for some $\widehat{x}_1\in\left\langle{c,d\;\big|\;c^{3d}=c^5}\right\rangle\setminus\left\langle{[c^d,c]}\right\rangle$ .

The first is impossible of course. The impossibility of the second case can be verified, e.g., as follows:

1. (i) the quotient group $Q=\left\langle{c,d\;\big|\;c^{3d}=c^5}\right\rangle/\left\langle\left\langle{{[c^d,c]}}\right\rangle\right\rangle$ is torsion-free; indeed, Q is the HNN-extension $Q\ =\ \left\langle{c,e,d\;|\;[e,c]\ =\ 1,\; e^3\ = \ c^5,\; c^d=e}\right\rangle$ of the abelian group $A=\left\langle{c,e\;|\;[e,c]\,=\,1,\;\ e^3\,=\,c^5}\right\rangle$ , which is torsion-free (moreover, it is easy to verify that $A\simeq{\mathbb Z}$ and $Q\simeq{\rm BS}(3,5)$ );

2. (ii) therefore, $\widehat{x}_1$ lies in the normal closure $F=\left\langle\left\langle{{[c^d,c]}}\right\rangle\right\rangle$ , which is a free group, because, by the Karrass–Solitar theorem (see, e.g., [12]), any subgroup of an HNN-extension is free if it intersects conjugates of the base trivially. It remains to show that $[c^d,c]$ is not a square in F (because in a free group an inclusion $\alpha^2\in\left\langle\beta\right\rangle$ implies that $\left\langle{\alpha,\beta}\right\rangle$ is cyclic by the Nielsen–Schreier theorem and, hence, $\alpha\in\left\langle\beta\right\rangle$ if $\beta$ is not a square). The commutator $[c^d,c]$ is not a square in F, because, assuming the contrary and noting that automorphic images of squares are squares too, we obtain $F=\left\langle\left\langle{{[c^d,c]}}\right\rangle\right\rangle=\left\langle\left\langle{{\widehat{x}^2}}\right\rangle\right\rangle\subseteq\left\langle{\{f^2\;\;|\;\;f\in F\}}\right\rangle,$ which cannot hold in a nontrivial free group F. This completes the proof.

2. Proof of the main theorem

Take the fundamental groups of the torus and the Klein bottle:

\begin{equation*}G_1={\rm BS}(1,1)=\left\langle{a,b\;|\;[a,b]=1}\right\rangle\ \hbox{and}\ G_{-1}={\rm BS}(1,-1)=\left\langle{a,b\;\bigg|\;a^b=a^{-1}}\right\rangle\end{equation*}

and consider the amalgamated free products $H_\varepsilon=G_\varepsilon\mathop*\limits_{b=h}H$ of $G_\varepsilon$ and a group

\begin{equation*}H=\left\langle{X\;|\;R}\right\rangle\supseteq\left\langle{h}\right\rangle_\infty\end{equation*}

(henceforth $\varepsilon=\pm1$ ). Let $K_\varepsilon$ be the standard complex of the (standard) presentation of $H_\varepsilon$ :

\begin{equation*}H_\varepsilon=\left\langle{\{a\}\sqcup X\;\bigg|\;\{a^{\widehat{h}}a^{-\varepsilon}\}\sqcup R}\right\rangle,\end{equation*}

where $\widehat{h}$ is a word in the alphabet $X^{\pm1}$ representing the element $h\in H$ .

The Cayley graphs of $G_\varepsilon$ are isomorphic surely (as abstract undirected graphs), the same is true for the universal coverings of the standard complexes of presentations of the groups $G_\varepsilon$ (these covering complexes are planes partitioned on squares, Figure 1).

Fig. 1. Universal coverings of the standard complexes of presentations $G_1$ (left) and $G_{-1}$ (right); vertical/horizontal edge are labelled by a and b, respectively; each small square is filled with a 2-cell.

A slightly less trivial observation is that, for groups $H_\varepsilon$ , the universal coverings are isomorphic too:

for any infinite-order element h of any group H, the universal coverings of complexes $K_\varepsilon$ are isomorphic. (*)

In what follows, we explain this simple fact in details; the readers who regard this fact as obvious, can skip to Observation $(^{**})$ .

It suffices to show that some coverings $\widehat{K}_\varepsilon\to K_\varepsilon$ have isomorphic $\widehat{K}_\varepsilon$ ; we prefer to take the coverings corresponding to the normal closure $\left\langle\!\langle{{a}}\right\rangle\!\rangle$ of $a\in H_\varepsilon$ . In explicit form, these complexes $\widehat{K}_\varepsilon$ are the following ones:

1. (i) the vertices are elements of H;

2. (ii) the edges with labels from X are drawn as in the Cayley graph of the group H: an edge with label $x\in X$ go from each vertex $h'\in H$ to the vertex $h'x\in H$ ;

3. (iii) in addition, to each vertex $h'\in H$ , a directed loop (edge) $a_{h'}$ labelled by a is attached;

4. (iv) to each cycle whose label is a relator from R, an oriented 2-cell is attached;

5. (v) to each cycle with label $a^{\widehat{h}}a^{-\varepsilon}$ , an oriented 2-cell (a special cell) is attached; thus, going along the boundary of a special cell in the positive direction, we meet two edges labelled by a, namely, $a_{h'}$ and $a_{h'h}^{-\varepsilon}$ , where, as usual, $a_{h'h}^{-1}$ means that the edge $a_{h'h}$ is traversed against its direction.

The isomorphism $\Phi\,{:}\,{\widehat{K}_1}\to\widehat{K}_{-1}$ is the following:

1. (i) the vertices, edges with labels from X and nonspecial 2-cells (corresponding to relators from R) are mapped identically;

2. (ii) to define the mapping $\Phi$ on edges labelled by a and special 2-cells, we choose a set T of left-coset representatives of $\left\langle{h}\right\rangle$ in H, and put $\Phi(a_{th^k})=a_{th^k}^{(-1)^k}$ for all $t\in T$ and $k\in{\mathbb Z}$ (i.e., in each coset, each second loop labelled by a is inverted); then the mapping of singular cells are defined naturally: a cell of $\widehat{K}_1$ with edges $a_{h'}$ and $a_{h'h}^{-1}$ on its boundary is mapped to the cell of $\widehat{K}_{-1}$ containing $a_{h'}$ and $a_{h'h}$ on its boundary.

The next simple observation is that:

if $h\in H$ belongs to all finite-index subgroups of H, and the complexes $K_\varepsilon$ have a finite common covering, then the group $H_1$ contains a subgroup isomorphic to the Klein-bottle group ${\rm BS}(1,-1)$ .(**)

Indeed, in $H_{-1}$ , the element $b=h$ is contained in all subgroups of finite index (because the intersection of each such subgroup with H is of finite index in H and, therefore, contains h). By the bottle lemma (applied to $G=H_{-1}$ ), we obtain that each finite-index subgroup contains a subgroup isomorphic to the Klein-bottle group. It remains to note that, if a finite complex $\widehat{K}$ covers $K_1$ and $K_{-1}$ , then its fundamental group $\pi_1(\widehat{K})$ embeds into $\pi_1(K_\varepsilon)=H_\varepsilon$ as a finite-index subgroup.

Now, we take a particular group H, namely, let H be the Baumslag–Solitar group: $H={\rm BS}(3,5)=\left\langle{c,d\;\big|\;c^{3d}=c^5}\right\rangle,$ and let $h\in H$ be the commutator: $h=[c^d,c].$ This element h is contained in any finite-index subgroup of H by the commutator lemma. According to $(^{**})$ , this means that, if complexes $K_\varepsilon$ would have a common finite covering, then $H_1=\left\langle{a,c,d\;\big|\;[a,[c^d,c]]=1,c^{3d}=c^5}\right\rangle$ would contain the Klein-bottle group as a subgroup, which contradicts the no-bottle lemma. Therefore, there are no finite common coverings for complexes $K_\varepsilon$ ; while an infinite common covering exists according to $(\!*\!)$ . Thus, the following fact is proven.

Main Theorem. The standard complexes of presentations

\begin{equation*}H_\varepsilon=\left\langle{a,c,d\;\bigg|\;a^{[c^d,c]}=a^\varepsilon,\; c^{3d}=c^5}\right\rangle,\end{equation*}

where $\varepsilon=\pm1$ , containing two 2-cells and one vertex, and three edges have a common covering, but have no finite common coverings.