Asymptotic Dimension of Finitely Presented Groups

It is proven that if a finitely presented group is one ended it has asymptotic dimension bigger than one. It follows that finitely presented groups with asdim 1 are virtually free. A counterexample is given for the finitely generated case.


Introduction
The notion of asymptotic dimension of a metric space was introduced by Gromov in [6]. It is a large scale analog of topological dimension and it is invariant by quasiisometries. This notion has proved relevant in the context of Novikov's higher signature conjecture and it was investigated further by other people (see [13], [1], [10]).
In this paper we show the following: If G is a one-ended finitely presented group then G has asymptotic dimension greater or equal to 2.
Also we deduce as corollary: Theorem 2. If G is a finitely presented group with asdim G = 1 then G is virtually free.
For finitely generated groups the statement above doesn't hold. We give a counter-example at the end.
After we completed this work T.Januszkiewicz brought to our attention his joint paper with J.Swiatkowski ( [8]) where they proved the same results independently.
Finally I would like to thank pr. P.Papasoglu for his help and guidance during the writing of this paper.

Preliminaries
Metric Spaces. Let (X, d) be a metric space. If A, B are subsets of X we set d(A, B) = inf{d(a, b) : a ∈ A, b ∈ B}. A path in X is a map γ : I → X where I is an interval in R. A path γ joins two points x and y in X if I = [a, b], γ(a) = x and γ(b) = y. The path γ is called an infinite ray starting from x 0 if I = [0, ∞) and γ(0) = x 0 . A geodesic, a geodesic ray, or a geodesic segment in X, is an isometry γ : I → X where I is R or [0, ∞) or a closed interval in R. We use the terms geodesic, geodesic ray etc. for the images of γ without discrimination. On a path-connected space X given two points x, y we define the path metric to be ρ(x, y) = inf {length(p)} where the infimum is taken over all paths p that join x and y. A space is called a geodesic metric space if for every x, y in X there exists a geodesic segment which joins them. In a geodesic space the path metric is indeed a metric. A geodesic metric space X is said to be one-ended if for every bounded K, X − K has exactly one unbounded connected component. We say that X is uniformly one-ended, if for every n ∈ R + there is an m ∈ R + such that for every K ⊂ X with diam(K) < n, X − K has exactly one connected component of diameter bigger than m.
Groups. The Cayley graph of G with respect to a generating set S is the 1dimensional complex having a vertex for each element of G and an edge joining vertex x to vertex xs for every vertex x and every s ∈ S . The Cayley graph has a natural metric which makes it a geodesic metric space, where each edge has length 1 (see [10]). In fact any connected graph can be made geodesic metric space in the same way.
We will use the same letter G for both the group and its Cayley graph as a metric space. We also use van -Kampen Diagrams (see [9] chapter V pg 236 -240 and [5]). A van Kampen diagram D for a word w in S representing the identity element of G, is a finite, planar, contractible, combinatorial 2-complex; its 1-cells are directed and labeled by generators and the boundary labels of each of its 2-cells are cyclic conjugates of relators or inverse relators. Further the boundary label for D is w when read (by convention anticlockwise) from a base point in ∂D. We recall here that a word w represents the identity element of G if and only if the path in the Cayley graph labeled by w is closed.
There is a natural map f from the 1-skeleton D (1) of D to the Cayley graph of G. f sends the base point to a vertex v of the Cayley graph and edges of D (1) to edges of the Cayley graph with the same label. Obviously f is determined by the image of the base point, v. f is not necessarily injective. If we consider D (1) as a geodesic metric space giving each edge length 1 then d(x, y) ≥ d(f (x), f (y)) for every x, y in D (1) .
We say that a group G is virtually free if there exists a finite index subgroup H of G which is a free group.
A group G is called free-by-finite if it contains a normal subgroup N of finite index such that N is free.
Generally if X is a property of groups then we say that G is virtually Obviously if G is X-by-finite then G is virtually X. The converse also holds if the property X is inherited to subgroups. Now since if G is free every subgroup of G is free we have that G is virtually free if and only if G is free by-finite.
Definition. (Asymptotic Dimension 1) We say that a space X has asymptotic dimension n if n is the minimal number such that for every d > 0 we have: X = X k for k = 0, 1, 2, ..., n and all X k are d-disconnected. We then write asdim X = n.
We say that a covering {B i } of X has d -multiplicity ≤ k, if and only if every d-ball B(x, d) in X meets no more than k sets B i of the covering. The covering has Definition. (Asymptotic Dimension 2) We say that a space X has asdimX = n, if n is the minimal number such that ∀ d > 0 there exists a D-bounded covering of X with d -multiplicity ≤ n + 1.
The two definitions used here are the first two definitions Gromov gave in his paper [6]. It is not difficult to see that the two definitions are equivalent.

Main Theorem
Before we get to the main theorem we will prove two lemmas that we will need below.
Lemma 1. Let G be a finitely generated, infinite group then asdim G > 0.
Proof. Let asdim G = 0 and fix d > 0. Then according to the first definition we But since G is finitely generated we have immediately that G is finite which is a contradiction.

Lemma 2.
If G is an one-ended finitely presented group then G contains a biinfinite geodesic.
Proof. Take any n ∈ N. Then define C n = {geodesic paths starting from e of length n} We note that: a) C n = ∅ for all n ∈ N. b) C n is a finite set for all n ∈ N. Consider a map π n n1 : C n → C n−1 which takes every geodesic path of length n from G n and cuts off the last edge of that path. Then obviously what is left is also a geodesic and now the length is n − 1. Thus it is contained in C n−1 . So for every n ∈ N, π n n−1 is a natural well defined map. So now consider the inverse limit sequence with bonding maps π n n−1 . By using the well known statement that the inverse limit of compact spaces is compact and hence nonempty, we obtain that Obviously r is an infinite geodesic ray starting from e. Now lets fix a geodesic segment ξ in C n and denote by Y ξ all the infinite geodesics in F starting with ξ. Then obviously Y ξ is the inverse image (π ∞ n ) −1 (ξ) where π ∞ n : F → C n is the natural projection. Thus if Y ξ is non empty it is compact.

Now define:
S n = {g ∈ G such that d(g, e) = n}, L n = {all the geodesic paths starting from S n and ending at e}, E n = {r : r are geodesic rays, r(0) ∈ K n , e ∈ r i } Obviously E n is not empty since let r ′ one of the geodesic rays we defined starting from e. Let g = r ′ (n) be the vertex of r ′ in S n . Define g −1 · r ′ . Since multiplying with an element doesn't change the respective distances g −1 · r ′ is also a geodesic starting from g −1 ·e = g −1 which is obviously in S n and passing through g −1 ·g = e. Thus, E n = ∅.
Now fix an η in L n and its corresponding h ∈ S n . Define the inverse limit The above inverse limit gives us all the geodesic rays starting from h i.e. h · F . The set Y ′ η of all geodesic rays starting with the path η being homeomorphic to Y h −1 ·η , is a compact by the above remark (if it is not empty of course!).
Then by the definition we get that: Since E n is non-empty, it is compact as a finite union of compact sets. Thus the inverse limit: with the restrictions as the bonding maps is compact and nonempty. Now let k in E. Then k is a path that passes through e and k is our bi-infinite geodesic line. This concludes the lemma.

Theorem 1.
If G is an one-ended finitely presented group then asdim G ≥ 2.
Proof. By lemma 1 and since G is one-ended we have that G is infinite and thus asdim G > 0 We will show that asdim G = 1. Let's suppose that asdim G = 1. Let M = max{|r i |, i = 1, 2, ..., n: r i relation of G}, where |r| = length of the word r. We fix d > 100M + 100. Since asdim G = 1 there is a covering B = {B i } with: G = i∈I B i and diamB i < D, ∀ i ∈ I, such that every ball B(x, d) intersects at most 2 sets of the covering B. We may assume without loss of generality that if r is a path in the Cayley graph labeled by a relator r i then r is contained in some B j ∈ B.
Since G is one-ended we have that G has a bi-infinite geodesic S(Lemma 2). Let N = max{100D 100 , 300M }. Choose an x 0 ∈ S and consider the ball B(x 0 , N ) which separates the geodesic S into two geodesic rays S 1 and S 2 . Since G is oneended there is an x in S 1 , a y in S 2 and a path p with p(0) = x and p(t) = y such that p B(x 0 , N ) = ∅.
We denote by [x, y] the part of the geodesic S, that connects x and y. Obviously length([x, y]) ≥ 2N . We denote by w the path that corresponds to [x, y] p We have then that So in order to cover the path w we need at least 3 sets of the covering {B i }.
We consider now the van-Kampen diagram D that corresponds to the path w and the function f from D (1) to the Cayley graph G. So f (∂D) = w. For notational convenience we label vertices and edges of ∂D in the same way as w. So for example we denote the vertex on ∂D which is mapped to x 0 ∈ w by f also by x 0 .
Let B be a set of the covering that intersects [x, y]. We consider f −1 (B). Let C(B) be the union of all 2-cells of D which have the property that their boundary is contained in f −1 (B). Let U be the collection of all such sets B with the following property: For some connected component, d(a, b) for such a component and let d(B) be the maximal value of all d(K) for K component of Let P = D − K K. P is connected since K is connected. Each edge of P is contained in two 2-cells. One of these 2-cells lies in f −1 (B 1 B 2 ) and one does not lie in this set. It is not possible that all edges of P are contained in a 2-cell of f −1 (B 2 ). Indeed in this case we would have d(B 2 ) > d(B 1 ), which is impossible. Since r is contained in C(B 2 ) some edge of P is not contained in a 2-cell of f −1 (B 1 ). It follows that there are 2 adjacent edges e 1 , e 2 in P such that one of them is contained in a 2-cell of f −1 (B 1 ) and the other in a 2-cell of f −1 (B 2 ). If c is the 2-cell that contains e 1 and is not in The edges e 1 , e 2 and the 2-cell c have a vertex v in common. So This concludes the proof.
Remark. The result above holds also for uniformly one-ended simply connected simplicial complexes. So if X is a uniformly one-ended simply connected simplicial complex then asdim X ≥ 2.
Of course the result does not hold for one-ended simply connected simplicial complexes, a half-line gives a counterexample. We remark finally that if a Cayley graph is one ended then it is uniformly one ended.
We note that the following theorem holds: Theorem. ) If G is a finitely presented group then G is the fundamental group of a graph of groups such that all the edge groups are finite and all the vertex groups are 0 or 1 ended.
Also it is known that: Lemma 3. If all the vertex groups are 0-ended (i.e. finite) then G is virtually free (see [11], page 120, prop.11).
Furthermore it is not difficult to prove the following lemma (see [7]): Lemma 4. If H < G and H is finitely generated then asdim G ≥ asdim H.
Using the lemma above and theorem 1 we have the stronger result: If G is a finitely presented group with asdim G = 1 then G is virtually free.
Proof. Let G be a finitely presented group with asdim G = 1. Let Γ be the graph of groups of the Dunwoody-Stallings theorem. If a vertex group H is one-ended then from the theorem 1, asdim H ≥ 2. But H < G which means that asdim G ≥ 2 which is a contradiction. So all vertex groups are 0-ended. It follows that G is virtually free. Now we give an example of a finitely generated group which is not finitely presented, not virtually free and that has asymptotic dimension 1. Namely: Proposition. Let G = Z 2 ≀ r Z be the restricted wreath product of Z 2 and Z. Then asdim G = 1 and G is not virtually free.
Proof. Since G = Z 2 ≀ r Z there exists a short exact sequence: By the Hurewicz type formula (see [3]) we have: Since every finitely generated subgroup F of (⊕ Z Z 2 ) is finite, we have that all F of that type have asymptotic dimension 0. Following the definition of asymptotic dimension for arbitrary discrete groups found in [3] we get: asdim(⊕ Z Z 2 ) = sup{asdimF |F < G, finitely generated} = 0 Another way to get the same result is by using the following corollary found in [12] Corollary. Let G be a countable abelian group. Then asdim G = 0 if and only if G is torsion.
Obviously ⊕ Z Z 2 is abelian and torsion so asdim ⊕ Z Z 2 = 0. Thus we get: asdim G ≤ 1 + 0 = 1 Since G is a finitely generated infinite group (see [2] for the description of the generators), by lemma 1 we have that asdim G > 0. So asdim G = 1.
We will prove that G is not virtually free. Let G be virtually free. Then G is free-by-finite. So there exists a normal subgroup N of G such that N is free and the index |G : N | is finite. Recall the exact sequence: 0 → (⊕ Z Z 2 ) → G → Z → 0 and lets denote the image of ⊕ Z Z 2 in G to be H through the one to one mapping f . Then obviously H < G. Define q : G → G N and restrict q to H. Since H is infinite and G N is finite then H kerq = ∅. But kerq = N thus H N = ∅ which is a contradiction because if a ∈ H a is torsion. Thus a / ∈ N since N is free.