Asymptotic Dimension of Graphs of Groups and One Relator Groups.

We prove a new inequality for the asymptotic dimension of HNN-extensions. We deduce that the asymptotic dimension of every ﬁnitely generated one relator group is at most two, conﬁrming a conjecture of A.Dranishnikov. Asfurthercorollaries we calculate the exact asymptotic dimension of Right-angled Artin groups and we give a new upper bound for the asymptotic dimension of fundamental groups of graphs of groups.


Introduction
In 1993, M. Gromov introduced the notion of the asymptotic dimension of metric spaces (see [15]) as an invariant of finitely generated groups.It can be shown that if two metric spaces are quasi isometric then they have the same asymptotic dimension.
The asymptotic dimension asdimX of a metric space X is defined as follows: asdimX ≤ n if and only if for every R > 0 there exists a uniformly bounded covering U of X such that the R-multiplicity of U is smaller than or equal to n + 1 (i.e.every R-ball in X intersects at most n + 1 elements of U ).There are many equivalent ways to define the asymptotic dimension of a metric space.It turns out that the asymptotic dimension of an infinite tree is 1 and the asymptotic dimension of E n is n.
In 1998, the asymptotic dimension achieved particular prominence in geometric group theory after a paper of Guoliang Yu, (see [27]) which proved the Novikov higher signature conjecture for manifolds whose fundamental group has finite asymptotic dimension.Unfortunately, not all finitely presented groups have finite asymptotic dimension.For example, Thompson's group F has infinite asymptotic dimension since it contains Z n for all n.However, we know for many classes of groups that they have finite asymptotic dimension, for instance, hyperbolic, relative hyperbolic, Mapping Class Groups of surfaces and one relator groups have finite asymptotic dimension (see [4], [21], [6], [16]).The exact computation of the asymptotic dimension of groups or finding the optimal upper bound is more delicate.Another remarkable result is that of Buyalo and Lebedeva (see [7]) where in 2006 they established the following equality for hyperbolic groups: The inequalities of G.Bell and A.Dranishnikov (see [2] and [10]) play a key role on finding an upper bound for the asymptotic dimension of groups.However, in some cases the upper bounds that the inequalities of G.Bell and A.Dranishnikov provide us are quite far from being optimal.An example is the asymptotic dimension of one relator groups.
In this paper we prove some new inequalities that can be a useful tool for the computation of the asymptotic dimension of groups.As an application we give the optimal upper bound for the asymptotic dimension of one relator groups which was conjectured by A.Dranishnikov.As a further corollary we calculate the exact asymptotic dimension of any Right-angled Artin groupthis has been proven earlier by N.Wright [26] by different methods.
The first inequality and one of the main results we prove is the following: Theorem 1.1.Let G * N be an HNN-extension of the finitely generated group G over N .We have the following inequality asdim G * N ≤ max{asdimG, asdimN + 1}.
Next, we calculate the asymptotic dimension of the Right-angled Artin groups.To be more precise, let Γ be a finite simplicial graph, we denote by A(Γ) the Right-angled Artin group (RAAG) associated to the graph Γ.We set Sim(Γ) = max{n | Γ contains the 1-skeleton of the standard (n − 1)-simplex ∆ n−1 }.
In 2005, G.Bell, and A.Dranishnikov (see [3]) gave a proof that the asymptotic dimension of one relator groups is finite and also they gave an upper bound, namely the length of the relator plus one.Let G = S | r be a finitely generated one relator group such that | r |= n.Then To prove this upper bound G. Bell, and A. Dranishnikov used an inequality for the asymptotic dimension of HNN-extensions (see [2]).In particular, let G be a finitely generated group and let N be a subgroup of G.Then, In 2006, D. Matsnev (see [16]) proved a sharper upper bound for the asymptotic dimension of one relator groups.D. Matsnev proved the following: let G = S | r be a one relator group then Here by ⌈a⌉ (a ∈ R) we denote the minimal integer greater than or equal to a.
Applying Theorem 1.1 we answer a conjecture of A.Dranishnikov (see [8]) giving the optimal upper bound for the asymptotic dimension of one relator groups.
We note that R. C. Lyndon (see [19]) has shown that the cohomological dimension of a torsion-free one-relator group is smaller than or equal to 2. Our result can be seen as a large scale analog of this.We note that the large scale geometry of one relator groups can be quite complicated, for example one relator groups can have very large isoperimetric functions (see e.g.[22]).
It is worth noting that L.Sledd showed that the Assouad-Nagata dimension of any finitely generated C ′ (1/6) group is at most two (see [24]).Theorem 1.3 combined with the results of M.Kapovich and B.Kleiner (see [17]) leads us to a description of the boundary of hyperbolic one relator groups.
We determine also the one relator groups that have asymptotic dimension exactly two.We prove that every infinite finitely generated one relator group G that is not a free group or a free product of a free group and a finite cyclic group has asymptotic dimension equal to 2 (Proposition 3.5).We obtain the following: Corollary.Let G be finitely generated freely indecomposable one relator group which is not cyclic.Then asdim G = 2.Moreover, we describe the finitely generated one relator groups in the following corollary: Corollary.Let G be a finitely generated one relator group.Then one of the following is true: (i) G is finite cyclic, and asdim G = 0 (ii) G is a nontrivial free group or a free product of a nontrivial free group and a finite cyclic group, and asdim G = 1 (iii) G is an infinite freely indecomposable not cyclic group or a free product of a nontrivial free group and an infinite freely indecomposable not cyclic group, and asdim G = 2.
Using Theorem 1.1 and an inequality of A.Dranishnikov about the asymptotic dimension of amalgamated products (see [10]) we obtain a more general theorem for the asymptotic dimension of fundamental groups of graphs of groups.
Theorem 1.4.Let (G, Y ) be a finite graph of groups with vertex groups Then the following inequality holds: Acknowledgments: I would like to thank Panos Papasoglu for his valuable advices during the development of this research work.I would also like to offer my special thanks to Mark Hagen and Richard Wade for their very useful comments.
Let X be a metric space and U a covering of X, we say that the covering U is d-bounded or d-uniformly bounded if sup U ∈U {diam U } ≤ d.The Lebesgue number L(U ) of the covering U is defined as follows: We recall that the order ord(U ) of the cover U is the smallest number n (if it exists) such that each point of the space belongs to at most n sets in the cover.
For a metric space X, we say that (r, d) − dimX ≤ n if for r > 0 there exists a d-bounded cover U of X with ord(U ) ≤ n + 1 and with Lebesgue number L(U ) > r.We refer to such a cover as an (r, d)-cover of X.
The following proposition is due to G.Bell and A.Dranishnikov (see [2]).
Proposition 2.1.For a metric space X, asdimX ≤ n if and only if there exists a function d(r) such that (r, d(r)) − dimX ≤ n for all r > 0.
We recall that the family X i of subsets of X satisfies the inequality asdimX i ≤ n uniformly if for every R > 0 there exists a D-bounded covering U i of X i with R−mult(U i ) ≤ n+1, for every i.For the proofs of the following theorems 2.2 and 2.3 see [1].
Theorem 2.2.(Infinite Union Theorem) Let X = ∪ a X a be a metric space where the family {X a } satisfies the inequality asdimX a ≤ n uniformly.Suppose further that for every r > 0 there is a subset Theorem 2.3.(Finite Union Theorem) For every metric space presented as a finite union X = ∪ i X i we have A partition of a metric space X is a presentation as a union X = ∪ i W i such that Int(W i ) ∩ Int(W j ) = ∅ whenever i = j.We denote by ∂W i the topological boundary of W i and by Int(W i ) the topological interior.We have that ∂W ∩ Int(W ) = ∅.The boundary can be written as For the proof of the following theorem see [10].
Theorem 2.4.(Partition Theorem) Let X be a geodesic metric space.Suppose that for every R > 0 there is d > 0 and a partition X = ∪ i W i with asdimW i ≤ n uniformly in i, and such that (R, where ∂W i is taken with the metric restricted from X. Then asdimX ≤ n. Let G be a finitely generated group, N a subgroup of G and φ : N → G a monomorphism.We set G = G * N the HNN-extension of G over the subgroup N with respect to the monomorphism φ.We fix a finite generating set S for the group G. Then the set S = S ∪ {t, t −1 } is a finite generating set for the group G and we set C(G) = Cay(G, S) its Cayley graph.

Normal forms for HNN-extensions.
We note that there exist two types of normal forms for HNN-extensions, the right normal form and the left normal form.We are going to use both of them in this paper.Right normal form: Let S N and S φ(N ) be sets of representatives of right cosets of G/N and of G/φ(N ) respectively.Then every w ∈ G has a unique normal form w = gt Left normal form: Let N S and φ(N ) S be sets of representatives of left cosets of G/N and of G/φ(N ) respectively.Then every w ∈ G has a unique normal form w = s Convention: When we write a normal form we mean the right normal form, unless otherwise stated.
The group G = G * N acts on its Bass-Serre tree T .There is a natural projection π : G * N → T defined by the action: π(g) = gG.Proof.Let g ∈ G and s ∈ S. Then the vertex g is mapped to the vertex π(g) = π(gs) = gG.
If s ∈ S, then the edge [g, gs] is mapped to the vertex π(g) = π(gs) = gG.
If s ∈ {t, t −1 }, without loss of generality we may assume that s = t, then the edge [g, gs] is mapped to the edge [π(g), π(gs)] = [gG, gtG] of T .
We observe that the simplicial map π : The base vertex G separates T into two parts T − \ G and T + \ G, where We note that both T + \ G and T − \ G are unions of connected components of T and π −1 (T + ) and π −1 (T − ) are unions of connected components of C(G).
See figure 2 for an illustration of T − and T + .
We consider the Bass-Serre tree T as a metric space with the simplicial metric d.If Y is a graph, we denote by Y 0 or V (Y ) the vertices of Y .For u ∈ T 0 we denote by | u | the distance to the vertex with label G.We note that the distance of the vertex wG from G in the Bass-Serre tree equals to the length l(w) of the normal form of w, | wG |= l(w).We denote by l(w) the length of the normal form of w, we note that the length of both the right and the left normal form of w is the same.
We recall that a full subgraph of a graph Γ is a subgraph formed from a subset of vertices V and from all of the edges that have both endpoints in the subset V .If A is a subgraph of Γ we define the edge closure E(A) of A to be the full subgraph of Γ formed from V (A).Obviously, V (E(A)) = V (A).
We fix some notation on the Bass-Serre tree T and on the Cayley graph.
In the tree T .We denote by B T r the r−ball in T centered at G (r ∈ N).There is a partial order on vertices of T defined as follows: v ≤ u if and only if v lies in the geodesic segment [G, u] joining the base vertex G with u.For u ∈ T 0 of nonzero level (i.e.u = G) and r > 0 we set For every vertex u ∈ T 0 represented by a coset g u G we have the equality  In the Cayley graph.For R ∈ N, let ) and Convention: We associate every u ∈ T 0 to an element g u ∈ G such that the following two conditions hold: We see that in this way we may define a bijective map from T 0 to the set G T which consists of the elements of G such that conditions (i) and (ii) hold.
Proposition 2.6.If 4 < 4R ≤ r, and the distinct vertices u, u ′ ∈ T 0 , satisfy Figure 4 Proof.We distinguish two cases.See figure 5(a) and figure 5(b) for the case 1 and case 2 respectively.Case 1: We recall that every path γ in C(G) projects to a path π(γ) in the tree T .Then since and π is 1-Lipschitz we have that R and let γ be a geodesic from x to y.Then the path π(γ) passes through the vertices u, u ′ and ζ 0 .So the geodesic γ intersects both g u (N ∪ φ(N )) and For w ∈ G * N , we denote by w the distance from w to 1 G in the Cayley graph Cay(G, S).
Proof.Without loss of generality we assume that ǫ k = 1.Let w = ( be a shortest presentation of w in the alphabet S (we note that s i j / ∈ {t, t −1 }).We set The first step when we rewrite w in normal form starting from the previous presentation is to write g k = ns k (where n ∈ N ).Then We note that there exists an amalgamated product analogue of the following proposition proved by A.Dranishnikov in [10].
Then asdimQ m ≤ n, for every m ∈ N.
Proof.We set P λ = {w ∈ G | l(w) = λ}.To prove the statement of the proposition it is enough to show that asdimP λ ≤ n, for every λ ∈ N. Indeed, since by the Finite Union Theorem we obtain that asdimQ m ≤ n.
We consider the family xtG where By lemma 2.7 we obtain that g Finally, by observing that xtG and G are isometric we deduce that asdim(xtG) ≤ n uniformly.Since all the conditions of the Infinite Union Theorem hold we have that For w ∈ G, we set T w = T π(w) , where π(w) = wG.
Theorem 2.9.Let G * N be an HNN-extension of the finitely generated group G over N .We have the following inequality asdim G * N ≤ max{asdimG, asdimN + 1}.
We recall that we denote by l(g) the length of the normal form of g.

G
Figure 7: An illustration of V r .
We will use the Partition Theorem (Thm 2.4).Let R, r ∈ N be such that R > 1 and r > 4R.
We set, where Let A R be the collection of the edges between the elements of M R ⊆ U r .We have that A R ⊆ U r .We define V r to be the set obtained by removing the interior of the edges of A R from U r .Formally we have that See figure 7 for an illustration of the set V r .We observe that the sets U r and V r are subgraphs of We also have To be more precise, Next, we will show that asdimV u r ≤ n and asdimN R (N ∪ φ(N )) ≤ n uniformly.This will complete our proof that all the conditions of the Partition Theorem are satisfied.It suffices to show that asdimV u r ≤ n uniformly and We observe that V r ⊆ π −1 (B T r+R ) ⊆ N 1 (Q r+R ), so by the proposition 2.8 we have that asdimV u r ≤ n.Since the sets V u r of our partition are isometric to each other we conclude that asdimV u r ≤ n uniformly.

Right-angled Artin groups.
We use the following theorem of G.Bell, A.Dranishnikov, and J.Keesling (see [5]).Let Γ be a finite simplicial graph with n vertices, the Right-angled Artin group (RAAG) A(Γ) associated to the graph Γ has the following presentation: By valency(u) of a vertex u we denote the number of edges incident to the vertex u.
If Γ is a simplicial graph, we denote by 1 − skel(Γ) the 1-skeleton of Γ. Recall that a full subgraph of a graph Γ is a subgraph formed from a subset of vertices V and from all of the edges that have both endpoints in the subset V .
Lemma 2.11.Let Γ be a finite simplicial graph.Then Proof.Since theorem 2.10 holds we observe that it suffices to show the statement of the lemma 2.11 for connected simplicial graphs.We assume that Γ is a connected simplicial graph.We use induction on the rank(A(Γ)).For rank(A(Γ)) = 1 we have that A(Γ) is the integers Z, so the statement holds.We assume that the statement holds for every k ≤ n and we show that it holds for n + 1 (n + 1 ≥ 2).
We observe that the RAAG A(Γ) is an HNN-extension of the RAAG A(Γ ′ ).To be more precise, we have that By Theorem 2.9 we obtain that Since rankA(Γ ′ ) = n, by the induction we deduce that Combining the three previous inequalities we obtain: Using the previous lemma we can compute the exact asymptotic dimension of A(Γ).We note that this has already been computed by N.Wright [26] using different methods.

We set
Theorem 2.12.Let Γ be a finite simplicial graph.Then, Proof.Since theorem 2.10 holds we observe that it suffices to show the statement of Theorem 2.12 for connected simplicial graphs.We assume that Γ is a connected simplicial graph.

Proof of claim 2:
We use induction on the rank(A(Γ)), for rank(A(Γ)) = 1 we have that A(Γ) is the integers Z, so the statement holds.We assume that the statement holds for every r ≤ m, we will show that holds for m + 1 as well.Let Γ be a connected simplicial graph with m + 1 vertices.
3 Asymptotic dimension of one-relator groups.
Theorem 3.1.Let G be a finitely generated one relator group.Then Proof.Let G = S | r be a presentation of G where S is finite and r is a cyclically reduced word in S S −1 .To omit trivial cases, we assume that S contains at least two elements and | r |> 0, (we denote by | r | the length of the relator r in the free group F (S)).
We may assume that every letter of S appears in r.Otherwise our group G is isomorphic to a free product H * F of a finitely generated one relator group H with relator r and generating set S H ⊆ S consisting of all letters which appear in r and a free group F with generating set the remaining letters of S. We recall that the asymptotic dmension of any finitely generated non-abelian free group is equal to one.Then asdimG = max{asdimH, asdimF } = max{asdimH, 1} (see [2]).
We denote by ǫ r (s) the exponent sum of a letter s ∈ S in a word r and by oc r (s) the minimum number of the positions of appearance of the elements of the set {s k for some 0 = k ∈ Z} in a cyclically reduced word r.For example, if r = abcab 10 a −2 c −1 , then oc r (a) = 3, oc r (b) = 2, oc r (c) = 2 and ǫ r (c) = 0.
We observe that if there exists b ∈ S such that oc r (b) = 1, then the group G is free (see [20], thm 5.1, page 198 ), so asdimG = 1.From now on we assume that for every s ∈ S we have that oc r (s) ≥ 2 (so | r |≥ 4).
The proof is by induction on the length of r.We observe that if | r |= 4 then the statement of the theorem holds since by the result of D.Matsnev [16] we have that asdimG ≤ ⌊|r|⌋ 2 = 4 2 = 2 (where ⌊ * ⌋ is the floor function).We assume that the statement of the theorem holds for all one relator groups with relator length smaller than or equal to | r | −1.
We follow the arguments of McCool, Schupp and Magnus (see [20], thm 5.1, page 198), which we shall describe in what follows.We distinguish two cases.We note that the argument we use in case 1 is slightly different from the argument of McCool and Schupp as we will consider HNN-extensions over finitely generated groups.To be more precise, the classical arguments prove that if the relation of G has exponent sum zero, then G is an HNN-extension of another one relator group over an infinitely generated non-abelian free subgroup.Our contribution here is that we showed that G is an HNNextension of another one relator group over a finitely generated non-abelian free subgroup.In case 2 with non-zero exponent sum we use the original argument of Magnus to show that G can be embedded in an one relator group Γ whose defining relator has exponent sum zero.
Case 1: There exists a letter a ∈ S such that ǫ r (a) = 0.
We shall exhibit G as an HNN-extension of a one relator group G 1 whose defining relator has shorter length than r, over a finitely generated free subgroup F .Let S = {a = s 1 , s 2 , s 3 , s 4 , ...s k }.Set s (j) i = a j s i a −j for j ∈ Z and for k ≥ i ≥ 2. Rewrite r scanning it from left to right and changing any occurrence of a j s i to s (j) i a j , collecting the powers of adjacent a−letters together and continuing with the leftmost occurrence of a or its inverse in the modified word.We denote by r ′ the modified word in terms of s (j) i .We note that by doing this we make at least one cancellation of a and its inverse.The resulting word r ′ which represents r in terms of s (j) i and their inverses has length smaller than or equal to | r | −2.For example, if r = as 2 s 3 as 4 2 a −2 s 3 then r ′ = s Let m and M be the minimal and the maximal superscript of all s (j) i (i ≥ 2) occurring in r ′ respectively.To be more precise, i occurs in r ′ }.Continuing our example, we have m = 0 and M = 2.

Claim 1.1:
In case 1 we have M − m > 0 and m ≤ 0 ≤ M .We may assume, replacing r with a suitable permutation if necessary, that r begins with a k for some k = 0. Then we can write r = a k swa n tz, where k, n = 0, a / ∈ {s, t} ⊆ S and both a and a −1 do not appear in the word z (oc z (a) = 0).Then we observe that the letter s has as superscript k in the word r ′ while t has as superscript 0 in the word r ′ .Since k = 0 we have that M − m > 0. This completes the proof of the claim 1.1.

Claim 1.2:
We claim that G has a presentation a, s i , (i ∈ {2, . . ., k}), (j ∈ {m, . . ., M }) | r ′ , as To verify the claim, let H be the group defined by the presentation given above.The map φ : G −→ H defined by a −→ a, s i −→ s (0) i is a homomorphism since φ(r) = r ′ .On the other hand, the map ψ : is also a homomorphism since all relators of H are sent to 1 G .
It is easy to verify that ψ•φ is the identity map of G.The homomorphism we combine these equations and we get s Since φ • ψ and ψ • φ are the identity maps on H and G respectively we deduce that φ is an isomorphism.This completes the proof of claim 1.2.
We set i , (i ∈ {2, . . ., k}), (j ∈ {m, . . ., M }) | r ′ .We note that there exists a letter s im ∈ S such that s This claim follows by the Freiheitssatz (see [20], thm 5.1, page 198), since X omits a generator of G 1 occurring in r ′ (this is the letter s (M ) i M ) the subgroup F is free.The same holds for Λ, since Y omits the letter s In particular, the map s from X to Y extends to an isomorphism from F to Λ. Thus H is exhibited as the HNN extension of G 1 over the finitely generated free group F using a as a stable letter.Since G ≃ H (claim 1.2) we have that By the fact that | r ′ |<| r | and the inductive assumption we have that asdimG 1 ≤ 2. To conclude we apply the inequality for HNN-extensions (Theorem 2.9): asdimG ≤ max{asdimG 1 , asdimF + 1} = max{asdimG 1 , 2} = 2.
Case 2: For every letter s ∈ S we have | ǫ r (s) |≥ 1.Let S = {a = s 1 , b = s 2 , s 3 , s 4 , ...s k } and S 1 = {t, x, s i , (3 ≤ i ≤ k)}.We consider the following homomorphism between the free group F (S) and the free group F (S 1 ) We set where we denote by r(t, x, s i , ...(i > 2)) the modified word in terms of t, x, s i , (3 ≤ i ≤ k) which is obtained from r when we replace a generator s with φ(s).Then φ induces a homomorphism φ : G → Γ.
The following claim shows that the homomorphism φ is actually a monomorphism into Γ, so we have an embedding of G into Γ via φ: Claim 2.1: The homomorphism φ : G → Γ is monomorphism.Proof of the claim: We set S 2 = {a, t, s i , (3 ≤ i ≤ k)} and S 1 = {x, t, s i , (3 ≤ i ≤ k)}.We define g : F (S) → F (S 2 ) and f : F (S 2 ) → F (S 1 ), by We set: We can easily see that f is an isomorphism.Indeed, the homomorphism ψ : Γ → G 2 given by is the inverse homomorphism of f .
It is enough to prove that g is monomorphism.This follows by the fact that the group G 2 is the amalgamated product G * Z < t >, where Z =< λ > and ψ 1 (λ) = b , ψ 2 (λ) = t ǫr(a) are the corresponding monomorphisms.We can see that the homomorphism g is the inclusion of G into the amalgamated product, so g is injective.This completes the proof of the claim 2.1.
We denote by r(t, x, s i , ...(i ≥ 3)) the modified word in terms of t, x, s i , (3 ≤ i ≤ k) which can be obtained from r when we replace a generator s with φ(s) and by p the cyclically reduced r(t, x, s i , (3 ≤ i ≤ k)).We observe that ǫ p (t) = 0 and that x occurs in p.
If the letter t occurs in the word p, from Case 1 we have that Γ is an HNN extension of some group H over a free subgroup F , namely, Γ = H * F .As in Case 1 by assuming that p starts with t or t −1 we introduce new variables s (j) i = t j s i t −j .Using these variables, we rewrite p as a word w, eliminating all occurrences of t and its inverse.Then we observe that | w |≤| r | −1.By using the inductive assumption for w we obtain that asdimG ≤ asdimΓ ≤ 2.
Then by the inductive assumption for p we have that asdimΓ ′ ≤ 2. Finally, we conclude that asdimG ≤ asdimΓ ≤ 2.

One relator groups with asymptotic dimension two.
We recall that a nontrivial group H is freely indecomposable if H can not be expressed as a free product of two non-trivial groups.
A natural question derived from Theorem 3.1 is, which one relator groups have asymptotic dimension two.In this subsection, we will show that the asymptotic dimension of every finitely generated one relator group that is not a free group or a free product of a free group and a finite cyclic group is exactly two.
Proposition 3.2.Let G be an infinite finitely generated one relator group with torsion.If G has more than one ends, then G is a free product of a nontrivial free group and a freely indecomposable one relator group.Proposition 3.3.Let G be a torsion free infinite finitely generated group.If G is virtually free, then it is free.Lemma 3.4.Let G be an infinite finitely generated one relator group such that is not a free group or a free product of a nontrivial free group and a freely indecomposable one relator group.Then G is not virtually free.
Proof.If G has torsion, by proposition 3.2 we have that G has exactly one end, so G can not be virtually free.If G is torsion free, by proposition 3.3 we obtain that G is free and this is a contradiction by the assumption of the lemma.
We note that every finite one relator group is cyclic.To see that it is enough to observe that every one relator group with at least two generators has infinite abelianization.
The following proposition is the main result of this subsection.We assume that G is an infinite virtually free group.So if G is torsion free then by proposition 3.3 we obtain that G is free.If G has torsion then by lemma 3.4 G is a free product of a nontrivial free group and a freely indecomposable one relator group G 1 .Observe that G 1 is an infinite noncyclic group, then by the same lemma G 1 is not virtually free so G is not virtually free either, which is a contradiction.We conclude that asdim G = 2.
Corollary 1.Let G be a finitely generated freely indecomposable one relator group which is not cyclic.Then asdim G = 2.
Using the above results we sum up to following corollary which describes the finitely generated one relator groups.
Corollary 2. Let G be a finitely generated one relator group.Then one of the following is true: (i) G is finite cyclic, and asdim G = 0 (ii) G is a nontrivial free group or a free product of a nontrivial free group and a finite cyclic group, and asdim G = 1 (iii) G is an infinite freely indecomposable not cyclic group or a free product of a nontrivial free group and an infinite freely indecomposable not cyclic group, and asdim G = 2.
We could further describe the boundaries of hyperbolic one relator groups.We recall the following result of Buyalo and Lebedeva (see [7]) for hyperbolic groups: Let G be an infinite finitely generated hyperbolic one relator group that is not virtually free.By T.Gentimis ([14]) we obtain that asdimG = 1,so asdimG = 2. Using the previous equality we obtain that G has one dimensional boundary.Applying a theorem of M. Kapovich and B. Kleiner (see [17]) we can describe the boundaries of hyperbolic one relator groups.asdimπ 1 (G, Y, T) ≤ max{asdimπ 1 (G, Γ, T ′ ), asdim G u , asdimG e ′ + 1}.

Lemma 2 . 5 .
The map π : G → T extends to a simplicial map from the Cayley graph, π : C(G, S) → T which is 1-Lipschitz.

Figure 2 :
Figure 2: An illustration of T + and T − .

See figure 3
for an illustration of the sets T u and B u r Here we have an illustration of B u r , where r = 2.

Figure 3
Figure 3 An illustration of case 1 of proposition 2.6.
An illustration of case 2 of proposition 2.6.

Figure 5 :
Figure 5: In figure (a) we have u ′ = G

Proposition 4 . 2 .
Let (G, Y ) be a finite graph of groups with vertex groups {G v | v ∈ Y 0 } and edge groups {G e | e ∈ Y 1 + }.
Proof.By Theorem 3.1 we have that asdim G ≤ 2. If G is finite then it is cyclic.If G is infinite we have that 1 ≤ asdim G.By a theorem of Gentimis ([14]) we have that asdim G = 1 if and only if G is virtually free.