MMEDIUM-SCALE RICCI CURVATURE FOR HYPERBOLICGROUPS
ANDREW KEISLING
Abstract.
We study the relationship between a notion of medium-scale Riccicurvature for finitely generated groups and that of hyperbolicity in the sense ofGromov. We give an example of a generating set that gives zero curvature withpositive density for the free group of rank 2. We prove that, by making theradius used in computing the curvature sufficiently large, we can always havenegative curvature outside of a ball in non-elementary hyperbolic groups. Onthe other hand, we give an example of a group which has negative curvaturefor all non-identity points but is not hyperbolic.
Contents
Acknowledgments 11. Introduction 12. Background 32.1. Conjugation Curvature 32.2. Hyperbolic Groups 52.3. Computing GenCon 63. Curvature of Free Group 73.1. Negative Curvature Examples in Free Group 73.2. Zero Curvature with Positive Density in Free Group 94. Relationship with δ -Hyperbolicity 164.1. Negative Curvature at Larger Radii for Hyperbolic Groups 164.2. Negatively Curved Non-Hyperbolic Groups 21References 22 Acknowledgments
The author would like to thank Thang Nguyen for providing helpful mentorshipand feedback throughout this project. They would also like to thank the Universityof Michigan Math REU for providing the working conditions that made this projectpossible. The author was partially supported by NSF grant DMS-2003712.1.
Introduction
In Riemannian geometry, Ricci curvature measures the difference of the averagedistance between points on infinitesimal spheres and the distance between theircenters. In [Oll09], Ollivier generalized this intuition to more general metric spaces
Date : January 2021. a r X i v : . [ m a t h . G R ] J a n ANDREW KEISLING equipped with Markov chains with his transportation curvature . Here we focus onthe conjugation curvature for finitely generated groups introduced by Bar-Natan,Duchin, and Kropholler in [BDK20]. They called this curvature medium-scale be-cause, rather than being based on infinitesimal or global properties, it depends onsome fixed comparison radius.In this paper, following the notation introduced by Bar-Natan et al., we use κ S k ( g ) to denote the k -spherical conjugation curvature at the group element g . Asa special case, we write κ ( g ) to denote the 1-spherical conjugation curvature. Itfollows from the definition that this type of curvature depends on the choice ofgenerating set of the group.The first question that we seek to answer is whether the free group of rank 2must have negative curvature for all but finitely many points with respect to anygenerating set. Such a result would answer a question asked by Nguyen and Wangin [NW20, Question 5.3] and show whether torsion-free, non-elementary hyperbolicgroups can have infinitely many points with non-negative curvature. We show thatthey indeed can by proving the following theorem: Theorem 1.
There exists a generating set of the free group F = (cid:104) a, b (cid:105) such that F has zero curvature with positive density. F also has negative curvature withpositive density with respect to this generating set. By positive density, we mean that there exist (cid:15) > N ≥ n is bounded belowby (cid:15) for all n ≥ N .In addition to this result for F , we also explore the relationship between con-jugation curvature and non-elementary hyperbolic groups in general. By [BDK20,Theorem 20], there exist non-elementary hyperbolic groups with sequences of ele-ments ( g n ) ∞ n =1 such that κ ( g n ) > n . We prove the following theorem toshow that, by using a sufficiently large comparison radius k to compute the curva-ture, we can have at most finitely many points with non-negative curvature in anynon-elementary hyperbolic group: Theorem 2.
Let G be a non-elementary hyperbolic groups with some fixed gener-ating set S . Then there exists K ≥ such that, for every k ≥ K and g ∈ G outsidethe ball of radius k , we have κ S k ( g ) < . For this result for hyperbolic groups to be meaningful, it is significant that thereexist infinite groups that have positive k -conjugation curvature for infinitely manypoints for arbitrarily large k . Although the existence of such groups is non-obvious,Kropholler and Mallery proved in [KM20, Theorem 4.1] that the discrete Heisenberggroup H ( Z ) = (cid:104) a, b | [ a, b ] is central (cid:105) satisfies this property.On the other hand, we show that negative curvature for all non-identity pointsdoes not imply that a group is hyperbolic by proving the following counterexample: Theorem 3.
The group Z ∗ Z = (cid:104) t, a, b | ab = ba (cid:105) is not hyperbolic, but there existsa generating set such that, for all g ∈ Z ∗ Z − { } , we have κ ( g ) < . Due to the result of Theorem 1, it is natural to ask whether every non-elementaryhyperbolic group can have non-negative curvature for a positive density of pointswith respect to some generating set. A similar interesting question for future re-search is whether F can have strictly positive curvature for infinitely many elementswith respect to some generating set. Even if this is not the case, it seems possible EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 3 that a similar proof to that of Theorem 1 can be applied to the virtually free group Z ∗ Z = (cid:104) a, b | a = b = 1 (cid:105) to give positive curvature with positive density. Ifthis turns out to be true, then another interesting question to study is whether theexistence of order 2 elements gives a necessary and sufficient condition for strictlypositive curvature for infinitely elements in non-elementary hyperbolic groups. Onereason that this might be the case is that a generating set of odd size must containan element of order 2 (in the definition of conjugation curvature, we use generatingsets that are closed under inversion and never contain the group identity). Notethat the generating set used to give positive curvature in a hyperbolic group in[BDK20, Theorem 20] has odd size, while F has no elements of order 2 because itis torsion-free.This paper is organized as follows: we begin by reviewing the definitions of con-jugation curvature and hyperbolicity in Section 2, going over some simple examplesand proving lemmas that will help with the calculation of curvature. In Section 3,we give examples of generating sets giving negative curvature for F before provingTheorem 1. We conclude in Section 4 by proving Theorems 2 and 3.2. Background
In this section we review the definitions of conjugation curvature and Gromov’s δ -hyperbolicity, looking at some specfic examples. We then prove some preliminaryresults that will help us compute the curvature.2.1. Conjugation Curvature.
Let G be a group that is generated by some finiteset S , which we assume to be closed under inversion and not contain the groupidentity. Define the word length | g | S of an element g ∈ G as the minimum numberof generators g , . . . , g n ∈ S such that g = g . . . g n . We call g . . . g n a word ofgenerators and a spelling of g . When n = | g | S , we further call g . . . g n a geodesicspelling of g (such spellings need not be unique). In general, a geodesic word ofgenerators is a word of generators such that no other word spelling the same groupelement has less generators.The norm | · | S : G → Z ≥ induces a left-invariant metric d S : G × G → Z ≥ defined by d S ( g, h ) = | h − g | S , so that d S ( g,
1) = | g | S where 1 denotes the groupidentity. Although the metric space ( G, d S ) depends on the choice of generating set S , we will henceforth not include the subscript S because the generating set willbe clear from context.Let ( X, d ) be a metric space. Then we call (
X, d ) geodesic if any two points in X can be connected by a continuous curve c : [ a, b ] ⊂ R → X such that d ( c ( t ) , c ( t )) = | t − t | for all t , t ∈ [ a, b ].Given a finitely generated group G and finite generating set S , define the Cayleygraph of (
G, S ) as the graph whose vertices are the elements of G and whose edgesconnect the points g, h ∈ G if d ( g, h ) = 1. This graph can be equipped withthe natural graph metric such that the length of each edge is 1, and the distancebetween two points is the length of the shortest path connecting them. With thismetric, the Cayley graph is a geodesic metric space such that the distance betweentwo vertices is exactly the word distance between the corresponding group elements.An example of a Cayley graph is shown in Figure 1 on page 4.By combining this geometric structure for a finitely generated group G with thealgebraic structure of G , we can define a notion of curvature which shares many ANDREW KEISLING
Figure 1.
A portion of the Cayley graph of the additive group Z with respect to the generating set S := { , − , i, − i } .properties with Ricci curvature for Riemannian manifolds. For k ≥ g ∈ G ,define the ball and sphere of radius k centered at g to respectively be B k ( g ) := { h ∈ G | d ( g, h ) ≤ k } ,S k ( g ) := { h ∈ G | d ( g, h ) = k } . Usually we care about balls and spheres centered at the group identity 1, which wewrite as B k and S k respectively. In particular, note that we have S = S . Givenany distinct g, h ∈ G and k ≥
1, define the k -spherical comparison distance between g and h to be S k ( g, h ) := 1 | S k | (cid:88) s ∈ S k d ( gs, hs ) . The quantity S k ( g, h ) measures the average distance between corresponding pointson the spheres of radius k about g and h , where the correspondence is by left-translation. Comparing this quantity to the distance between g and h , we canwrite the definition of the type of curvature that we are concerned with: Definition 4. [BDK20] The k -spherical conjugation curvature between g, h ∈ G is κ S k ( g, h ) := d ( g, h ) − S k ( g, h ) d ( g, h ) = d ( g, h ) − | S k | (cid:80) s ∈ S k d ( gs, hs ) d ( g, h ) . Notice that we get positive curvature when corresponding points on S k ( g ) and S k ( h ) are closer, on average, than g and h , a property that agrees with classical Riccicurvature. By left-invariance of our metric, it suffices to consider the conjugationcurvature between a point g ∈ G − { } and the group identity 1, which we write as κ S k ( g ) := κ S k ( g, k = 1, we use the notation κ ( g ) := κ S ( g ). Notethat S ( g,
1) = 1 | S | (cid:88) s ∈ S d ( gs, s ) = 1 | S | (cid:88) s ∈ S | s − gs | =: GenCon( g ) , EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 5 where we use the notation GenCon( g ) to emphasize the fact that this quantity isthe average word length of the conjugation of g by all the generators in S (hencethe name conjugation curvature). Therefore, κ ( g ) = | g | − GenCon( g ) | g | , and in order to find the sign of the curvature it suffices to evaluate whether conju-gation by generators on average increases, decreases, or preserves the word lengthof g . Example 5.
Suppose that G is an abelian group. Then for all g ∈ G − { } wehave κ ( g ) = | g | − GenCon( g ) | g | = | g | − | S | (cid:80) s ∈ S | s − gs || g | = | g | − | S | (cid:80) s ∈ S | gs − s || g | = | g | − | S | (cid:80) s ∈ S | g || g | = | g | − | g || g | = 0 . Therefore, abelian groups are everywhere “flat” with respect to the conjugationcurvature.2.2.
Hyperbolic Groups.
The notion of a hyperbolic group was introduced byGromov in [Gro87]. This is a group that, when equipped with a word metric,satisfies the following:
Definition 6. [Gro87] A metric space (
X, d ) is called δ -hyperbolic for δ ≥ x, y, z, w ∈ X , the four-point condition ( x, y ) w ≥ min { ( x, z ) w , ( y, z ) w } − δ is satisfied. Here the quantity ( p, q ) r denotes the Gromov product of p, q, r ∈ X defined by ( p, q ) r := 12 ( d ( p, r ) + d ( r, q ) − d ( p, q )) . Gromov proved in [Gro87] that, if a group is δ -hyperbolic with respect to somefinite generating set S , then with respect to another finite generating set S (cid:48) it is δ (cid:48) -hyperbolic for some δ (cid:48) ≥
0. Therefore, we can speak of groups being hyperbolic,but in order to speak of a group being δ -hyperbolic we must first fix a generatingset.Another useful formulation of hyperbolicity for geodesic metric spaces (such as aCayley graph with a graph metric) is based on δ -thin triangles , which are defined asfollows: Given a geodesic triangle ∆ consisting of the vertices x , y , and z connectedby geodesic segments, there exists a map τ from ∆ to a tripod whose endpointsare τ ( x ), τ ( y ), and τ ( z ) such that d ( p , p ) = d ( τ ( p ) , τ ( p )) whenever p and p lie on the same edge of ∆. The length of the branch of this tripod that ends in τ ( z ) is exactly the product ( x, y ) z , and an analogous result holds for the other twobranches. We call τ a triangle-tripod transformation , and its image is unique upto isometry. We call ∆ a δ -thin triangle if, whenever τ ( p ) = τ ( q ) for two points p, q ∈ ∆, we have d ( p, q ) ≤ δ (see Figure 2 on page 6). ANDREW KEISLING
Figure 2. A δ -thin triangle ∆ in a geodesic metric space, alongwith its image under a triangle-tripod transformation τ . Anypoints that map to the same point under τ must lie within a dis-tance of δ in ∆.According to [Gro87, Proposition 6.3.C], all geodesic triangles in a δ -hyperbolicspace are 2 δ -thin, and if all geodesic triangles in a geodesic metric space are δ -thinthen the space is 2 δ -hyperbolic. In this paper, when we write that a metric spaceis δ -hyperbolic, the constant δ is determined by the four-point condition. Example 7.
Consider a metric space (
X, d ) which is a tree. Any geodesic trianglein X will either be a tripod or a single geodesic segment containing all three of itsvertices. Therefore, all geodesic triangles in X are 0-thin and X is a 0-hyperbolicmetric space.2.3. Computing GenCon.
The following basic results will be used throughoutthis paper to transform the problem of computing the quantity GenCon( g ) thatappears in the formula for κ ( g ) to that of finding spellings of g that begin and endwith certain generators: Lemma 8.
Let S be a generating set for the group G . Let g ∈ G and s ∈ S .Then | sg | = | g | − if and only if there exists a geodesic spelling g . . . g n of g withrespect to S such that g = s − . Similarly, | gs | = | g | − if and only if there existsa geodesic spelling g . . . g n of g with respect to S such that g n = s − .Proof. Assume that a geodesic spelling s − g . . . g n = g exists. Then | sg | = | ss − g . . . g n | = | g . . . g n | ≤ | g | −
1, and from the triangle inequality we have | sg | ≥ | g | −
1, so | sg | = | g | − | sg | = | g | −
1. Then there exists a geodesic spelling h . . . h n − of sg , where | g | = n . Consider the word s − h . . . h n − , which is equalto s − sg = g . Because this word consists of n generators, it is a geodesic spellingof g , and it begins with s − as desired.Note that we have | g | = | g − | for all g ∈ G because any geodesic spelling g . . . g n of g induces a spelling g − n . . . g − of g − with the same number of generators. The EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 7 proof for | gs | = | g | − | gs | = | ( gs ) − | = | s − g − | and applying the result for when | s − g − | = | g − | − | g | − (cid:3) Lemma 9.
Let S , G , s , and g be as in the statement of Lemma 8 with | g | = n .Then | sg | = | g | if and only if there exists a spelling g . . . g n +1 of g with respect to S such that g = s − and g . . . g n +1 is a geodesic word. Similarly, | gs | = | g | if andonly if there exists a spelling g . . . g n +1 of g with respect to S such that g n +1 = s − and g . . . g n is a geodesic word.Proof. Assume that a spelling s − g . . . g n +1 = g exists with g . . . g n +1 a geodesicword. Then | sg | = | ss − g . . . g n +1 | = | g . . . g n +1 | = n = | g | .Conversely, assume that | sg | = | g | . Then there exists a geodesic spelling h . . . h n of sg , so g = s − sg = s − h . . . h n . This is indeed a spelling of g that is one letterlonger than a geodesic spelling and that begins with s − as desired.As in the proof of Lemma 8, the proof for | gs | = | g | follows from applying theresult for | sg | = | g | to the inverse of gs . (cid:3) Remark . By the triangle inequality for the word metric, | g |−| s | ≤ | sg | ≤ | g | + | s | ,so | sg | = | g | − | g | , or | g | + 1 because | s | = 1. Lemmas 8 and 9 give necessary andsufficient conditions for the first two of these three possibilities, so if the hypothesesof neither of these lemmas hold then | sg | = | g | + 1. Similarly, if the necessaryand sufficient conditions for | gs | = | g | − | gs | = | g | both fail to hold, then | gs | = | g | + 1. 3. Curvature of Free Group
Throughout this section, let F denote the free group of rank 2 with the presen-tation (cid:104) a, b (cid:105) . We begin by giving examples of generating sets of F for which allnon-identity points have negative curvature. We then prove Theorem 1.3.1. Negative Curvature Examples in Free Group.Example 11.
Consider the generating set S α := { a, a − , b, b − } of F . Becausethe only way to shorten a word in these generators is by applying free reductions,any two words can only represent the same group element g ∈ F − { } if theydiffer in length by an even number of generators. Therefore, by Lemma 9, thereexist no generators s ∈ S α or group elements g ∈ F − { } such that | sg | = | g | or | gs | = | g | . Moreover, because this group is free, every group element has only onegeodesic spelling with respect to S α . For each g ∈ F − { } , we therefore have byLemma 8 that there exists only one s ∈ S α such that | sg | = | g | − s ∈ S α such that | gs | = | g | −
1. For all other s ∈ S α we have | sg | = | g | + 1 and | gs | = | g | + 1, so we can conclude that κ ( g ) = | g | − GenCon( g ) | g | = | g | − (cid:80) s ∈ S α | s − gs || g | = | g | − (4 | g | + 4) | g | = − | g | . The Cayley graph of ( F , S α ) is shown in Figure 3 on page 8. Note that this graphis a tree, proving that F is a hyperbolic group.Now consider the generating set S β := { a, a − , b, b − , aba, a − b − a − } . We havethe following result: Proposition 12.
With respect to the generating set S β , we have κ ( g ) < for all g ∈ F − { } . ANDREW KEISLING
Figure 3.
The Cayley graph of the group F = (cid:104) a, b (cid:105) with respectto the generating set S α := { a, a − , b, b − } . Proof.
Let w = aba so that w − = a − b − a − . With respect to the genera-tors in S β , F has the presentation (cid:104) a, b, w | abaw − = 1 (cid:105) . Suppose that we have g . . . g n = h . . . h n +1 for some generators g , . . . , g n , h , . . . , h n +1 ∈ S β . Then h − n +1 . . . h − g . . . g n = 1, so this word with 2 n + 1 letters can be reduced to theidentity by applying free reductions and removing all occurrences of any inverse orcyclic permutation of abaw − . Because all of these operations change the numberof letters by an even number and 2 n + 1 is odd, we have a contradiction. Therefore,given any word of generators in S β , there does not exist a word with exactly onemore generator that represents the same group element in F . By Lemma 9, thisimplies that there do not exist generators s ∈ S β or group elements g ∈ F − { } such that | sg | = | g | or | gs | = | g | .For s ∈ S α = { a, a − , b, b − } , let A s ⊂ F be the set of all g ∈ F such that g begins with s in its geodesic spelling with respect to S α . Every non-identityelement of F belongs to exactly one of the four sets A a , A a − , A b , or A b − , whicheach correspond with one of the four primary branches stemming from the point1 in Figure 3. We claim that, for all g ∈ A a , every geodesic spelling of g withrespect to S β must begin with either a or w . Note that a and w are the only twogenerators in S β that begin with a when spelled with respect to S α and thus lie in A a . Suppose that there exists g ∈ A a with a geodesic spelling g = g . . . g n suchthat g (cid:54) = a and g (cid:54) = w . Then for some 1 ≤ k < n we have g . . . g k / ∈ A a ∪ { } and ( g . . . g k ) g k +1 ∈ A a . The only non-identity point that is not in A a for whichright multiplication by a generator gives a point in A a is b − a − , for which we have( b − a − ) w = a . Therefore, we must have ( g . . . g k ) g k +1 = a , contradicting ourclaim that the spelling g . . . g n was geodesic. Hence, the first letter of a geodesicspelling of any element of A a must be either a or w as desired. An identicalargument shows that a geodesic spelling of an element of A a − must begin witheither a − or w − .We now claim that every geodesic spelling of an element of A b must begin witheither b or a − . Suppose that g ∈ A b has a geodesic spelling g = g . . . g n such EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 9 that g (cid:54) = b . Then again there exists 1 ≤ k < n such that g . . . g k / ∈ A b ∪ { } and ( g . . . g k ) g k +1 ∈ A b . The only non-identity point that is not in A b for whichright multiplication by a generator gives a point in A b is a − , for which we have a − w = ba . Therefore, g . . . g k = a − . Because the spelling g . . . g n is geodesic,we must have k = 1 and g = a − . Hence, the first letter of a geodesic spellingof an element of A b must be either b or a − . An identical argument shows that ageodesic spelling of an element of A b − must begin with either b − or a .We thus conclude by Lemma 8 that, for all g ∈ F − { } , there exist at most twodistinct s ∈ S β such that | sg | = | g | −
1. By Remark 10 we must have | sg | = | g | + 1for all other s . By applying the results of the previous paragraphs to the inversesof group elements, we also get that there exist at most two distinct s ∈ S β suchthat | gs | = | g | − κ ( g ) < g = ba and g = b − a − , andwhenever g is a generator in S β . For all other g ∈ F − { } , note that g is in A s for some s ∈ S α = { a, a − , b, b − } if and only if gs (cid:48) is also in A s for allgenerators s (cid:48) ∈ S β . Hence, the two distinct s ∈ S β for which it is possible to have | sg | = | g | − s ∈ S β for which it is possible tohave | sgs (cid:48) | = | gs (cid:48) | −
1. Therefore, κ ( g ) = | g | − GenCon( g ) | g | = | g | − (cid:80) s ∈ S β | s − gs || g | = (cid:80) s ∈ S β ( | g | − | gs | ) + (cid:80) s ∈ S β ( | gs | − | s − gs | ) | g |≤ ( −
2) + ( − | g | = − | g | < . (cid:3) Zero Curvature with Positive Density in Free Group.
The generat-ing set S β from Proposition 12 did not allow for points of non-negative curva-ture because different geodesic words for the same group element could have nomore than two different starting letters. In order to prove Theorem 1, we there-fore seek a six-element generating set which allows for different geodesic wordsfor the same group element to begin with three different generators. Let S γ := { a, a − , b, b − , ababa, a − b − a − b − a − } . We define the word w = ababa so that w − = a − b − a − b − a − , and we split S γ into the two subsets S γ := { a, b − , w } and S γ := { a − , b, w − } . Then we have the following useful results: Lemma 13.
Let h . . . h n and h (cid:48) . . . h (cid:48) n be geodesic words spelling the same elementof F . Then h ∈ S γ if and only if h (cid:48) ∈ S γ . Similarly, h n ∈ S γ if and only if h (cid:48) n ∈ S γ .Proof. Let the sets A a , A a − , A b , and A b − be as in the proof of Proposition 12, andrecall that every non-identity element of F must lie in exactly one of these four sets.Let g ∈ A a , and we claim that every geodesic spelling of g begins with an elementof S γ . Let g . . . g n be a geodesic spelling of g such that g (cid:54) = a and g (cid:54) = w . Thenthere must exist 1 ≤ k < n such that g . . . g k / ∈ A a ∪ { } and ( g . . . g k ) g k +1 ∈ A a .The only non-identity elements that are not in A a for which right multiplication bya generator gives elements of A a are b − a − and b − a − b − a − . For b − a − , wehave ( b − a − ) w = aba , while for b − a − b − a − , we have ( b − a − b − a − ) w = a . In this second case, though, note that if g . . . g k = b − a − b − a − = aw − and g k +1 = w , then the spelling g . . . g n would not have been geodesic and we wouldhave a contradiction. Therefore, the only way we could have g . . . g n ∈ A a and g / ∈ A a is if g . . . g k = b − a − and g k +1 = w . The word b − a − is the onlygeodesic spelling of itself, so we must in particular have g = b − . We can concludethat either g ∈ A a , in which case it is either a or w , or g = b − . Therefore, g ∈ S γ as desired. An identical argument shows that, if g ∈ A a − , then everygeodesic spelling of g must begin with an element of S γ .We now claim that, if g ∈ A b , then every geodesic spelling of g begins with anelement S γ . Let g . . . g n be a geodesic spelling of g such that g (cid:54) = b . Then theremust exist 1 ≤ k < n such that g . . . g k / ∈ A b ∪ { } and ( g . . . g k ) g k +1 ∈ A b . Theonly non-identity elements not in A b for which right multiplication by a generatorgives elements of A b are a − and a − b − a − . For a − , we have a − w = baba , whilefor a − b − a − , we have ( a − b − a − ) w = ba . In this second case, we would have g . . . g k = a − b − a − = baw − is a geodesic spelling, and g k +1 = w , so the spelling g . . . g n is not geodesic and we would have a contradiction. This leaves the casewhere g . . . g k = a − , and because this spelling is geodesic we must have k = 1 and g = a − . Therefore, a geodesic spelling of g ∈ A b must begin with either a − or b ,both of which are elements of S γ . An identical argument shows that, if g ∈ A b − ,then every geodesic spelling of g must begin with an element of S γ , either b − or a .Now suppose we have two geodesic words h . . . h n and h (cid:48) . . . h (cid:48) n such that h . . . h n = h (cid:48) . . . h (cid:48) n := h . If h ∈ S γ , then the results of the previous paragraphstell us that h ∈ A a or A b − . Therefore, we must also have h (cid:48) ∈ S γ . If h / ∈ S γ ,then h ∈ S γ , so h ∈ A a − or A b and h (cid:48) ∈ S γ . Finally, applying these results to h − gives the desired result for h n and h (cid:48) n . (cid:3) Lemma 14.
With respect to the generating set S γ , there do not exist generators s ∈ S γ or group elements g ∈ F − { } such that | sg | = | g | or | gs | = | g | .Proof. With respect to S γ , F has the presentation (cid:104) a, b, w | ababaw − = 1 (cid:105) . ByLemma 9, it suffices to show that no word of these generators has exactly onemore letter than a geodesic word. Suppose, on the other hand, that we have g . . . g n = h . . . h n +1 for some generators g , . . . , g n , h , . . . , h n +1 ∈ S γ , where theword g . . . g n is geodesic. Then h − n +1 . . . h − g . . . g n = 1, so this word with 2 n + 1letters can be reduced to the identity by applying free reductions and removing alloccurrences of any inverse or cyclic permutation of ababaw − . Because all of theseoperations change the number of letters by an even number and 2 n + 1 is odd, wehave a contradiction. (cid:3) Lemma 14 immediately implies the following when considered with Lemma 8and Remark 10.
Corollary 15.
With respect to the generating set S γ , we must have | sg | = | g | − or | sg | = | g | + 1 for all g ∈ F − { } and s ∈ S γ . Moreover, we have | sg | = | g | − if and only if there exists a geodesic spelling of g beginning with s − . Similarly, wemust have | gs | = | g | − or | gs | = | g | + 1 , and | gs | = | g | − if and only if thereexists a geodesic spelling of g ending with s − . (cid:3) We have the following observations that use similar ideas to the proof of Lemma13:
EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 11
Lemma 16.
Suppose that, for some g ∈ F − { } , there exists a geodesic spellingof g of the form aag . . . g n with g , . . . , g n ∈ S γ . Then there exists no geodesicspelling of g of the form b − h . . . h n +1 with h , . . . , h n +1 ∈ S γ . Similarly, if ageodesic spelling of some g (cid:48) ∈ F − { } is of the form a − a − g (cid:48) . . . g (cid:48) n , then thereexists no geodesic spelling of g (cid:48) of the form bh (cid:48) . . . h (cid:48) n +1 .Proof. In the proof of Lemma 13, we saw that a geodesic word can only beginwith the letter a if it spells a group element that lies in either A a or A b − . Wecannot have g ∈ A b − in this case because this would imply that the product of thefirst two letters of a geodesic spelling of g must lie in A b − , but we have aa ∈ A a .Thus, g ∈ A a . If a geodesic spelling b − h . . . h n +1 of g exists, then we must have h = a − and h = w because we saw in the proof of Lemma 13 that all geodesicwords in A a that begin with b − must begin with the prefix b − a − w . Therefore,because b − a − w = aba , we have aag . . . g n = g = b − a − wh . . . h n +1 = abah . . . h n +1 = ⇒ ag . . . g n = bah . . . h n +1 , which contradicts Lemma 13 because we have equal geodesic words beginning with a ∈ S γ and b / ∈ S γ . The argument for g (cid:48) is identical. (cid:3) Lemma 17.
Let g ∈ F − { } such that there exists a geodesic spelling of g of theform baag . . . g n with g , . . . , g n ∈ S γ . Then there exists no geodesic spelling of g of the form a − h . . . h n +2 with h , . . . , h n +2 ∈ S γ . Similarly, if g (cid:48) ∈ F − { } has a geodesic spelling of the form b − a − a − g (cid:48) . . . g (cid:48) n , then there exists no geodesicspelling of g (cid:48) of the form ah (cid:48) . . . h (cid:48) n +2 .Proof. Because aag . . . g n = b − g is in A a , we must have that baag . . . g n = g is in A b . Therefore, if a geodesic spelling a − h . . . h n +2 of g exists, then we must have h = w because we saw in the proof of Lemma 13 that all geodesic words in A b that begin with a − must begin with the prefix a − w . Because aba = wa − b − , wehave baag . . . g n = g = a − wh . . . h n +2 = ⇒ abaag . . . g n = wh . . . h n +2 = ⇒ wa − b − ag . . . g n = wh . . . h n +2 = ⇒ b − ag . . . g n = ah . . . h n +2 . The word ag . . . g n is geodesic because it is a subword of a geodesic word for g .Because this word begins with a , there exists no other geodesic spelling representingthe same group element that begins with b by Lemma 13. Thus, by Corollary 15, b − ag . . . g n is a geodesic word, so ah . . . h n +2 is also geodesic because it consistsof the same number of generators. Note that b − ag . . . g n − must either lie in A a or A b − because it begins with b − . If a geodesic word beginning with b − lies in A a , then it must begin with the prefix b − a − w , so b − ag . . . g n − instead lies in A b − . Therefore, ah . . . h n +2 must also lie in A b − , which implies that h = w − because all geodesic words in A b − that begin with a must begin with the prefix aw − . We then have b − ag . . . g n = aw − h . . . h n +2 = ⇒ wa − b − ag . . . g n = h . . . h n +2 = ⇒ abaag . . . g n = h . . . h n +2 = ⇒ aag . . . g n = b − a − h . . . h n +2 , which is a contradiction by Lemma 16. The argument for g (cid:48) is identical. (cid:3) The following proposition gives a sufficient condition for a point in F to havezero curvature with respect to S γ . In our proof of Theorem 1, we show that pointssatisfying this condition not only exist, but exist with positive density. Proposition 18.
With respect to S γ , κ ( g ) = 0 for all g ∈ F such that g hasa geodesic spelling of the form g = ( aba ) ± ( h )( aba ) ± or g = ( aba ) ± ( h )( aba ) ∓ ,where h is a geodesic subword such that | h | = | g | − .Proof. Suppose first that g = ( aba )( h )( aba ). Then ga − = ( aba )( h )( aba ) a − = ( aba )( h )( ab ) , which must be a geodesic spelling of | ga − | because it consists of | g | − ga − by the triangle inequality. By Lemma 13,every geodesic spelling of ga − must begin with an element of S γ so no geodesicspelling begins with a − . Combining this fact with Corollary 15, we have | aga − | = | a ( aba )( h )( ab ) | = | ( aba )( h )( ab ) | + 1 = ( | g | −
1) + 1 = | g | . Likewise, we have | a − ga | = | a − ( aba )( h )( aba ) a | = | ( ba )( h )( aba ) a | = | ( ba )( h )( aba ) | + 1 = ( | g | −
1) + 1 = | g | . To compute the length of conjugations by the other generators, we use the relation aba = wa − b − = b − a − w to show that | bgb − | = | b ( b − a − w )( h )( aba ) b − | = | ( a − w )( h )( aba ) b − | = | ( a − w )( h )( aba ) | + 1 = ( | g | −
1) + 1 = | g | , | b − gb | = | b − ( aba )( h )( wa − b − ) b | = | b − ( aba )( h )( wa − ) | = | ( aba )( h )( wa − ) | + 1 = ( | g | −
1) + 1 = | g | , | wgw − | = | w ( aba )( h )( b − a − w ) w − | = | w ( aba )( h )( b − a − ) | = | ( aba )( h )( b − a − ) | + 1 = ( | g | −
1) + 1 = | g | , | w − gw | = | w − ( wa − b − )( h )( aba ) w | = | ( a − b − )( h )( aba ) w | = | ( a − b − )( h )( aba ) | + 1 = ( | g | −
1) + 1 = | g | . Therefore, GenCon( g ) = | S γ | (cid:80) s ∈ S γ | s − gs | = | g | , so κ ( g ) = 0.To prove the case of g = ( aba ) − ( h )( aba ) − , note that g − is a group element ofthe type considered in the previous case. Therefore, because the word length of agroup element is equal to the length of its inverse, we have | s − gs | = | s − g − s | = | g − | = | g | for all s ∈ S γ . Therefore, GenCon( g ) = | S γ | (cid:80) s ∈ S γ | s − gs | = | g | , so κ ( g ) = 0. EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 13
Now suppose that we have g = ( aba )( h )( aba ) − . Then, using the relations aba = b − a − w = wa − b − and a − b − a − = w − ab = baw − , we have | a − ga | = | a − ( aba )( h )( a − b − a − ) a | = | ( ba )( h )( a − b − ) | ≤ | g | − , | bgb − | = | b ( b − a − w )( h )( w − ab ) b − | = | ( a − w )( h )( w − a ) | ≤ | g | − , | w − gw | = | w − ( wa − b − )( h )( baw − ) w | = | ( a − b − )( h )( ba ) | ≤ | g | − . By the triangle inequality, | s − gs | ≥ | g | − s ∈ S γ . Therefore,each of the above conjugations has a length of exactly | g | − g beginning with a , no geodesic spellingof g can begin with a − by Lemma 13. Combining this fact with Corollary 15, wehave that the spelling ag = a ( aba )( h )( aba ) − is geodesic. This spelling ends with a − , so we can once again apply Lemma 13 toshow that no geodesic spelling of ag ends with a . Thus, the spelling aga − = a ( aba )( h )( aba ) − a − is geodesic, so | aga − | = | g | + 2. Likewise, no geodsic spelling of g starts with b or w − or ends with b − or w . Thus, | b − gb | = | b − ( aba )( h )( aba ) − b | = | g | + 2 , | wgw − | = | w ( aba )( h )( aba ) − w − | = | g | + 2 . Therefore, GenCon( g ) = | S γ | (cid:80) s ∈ S γ | s − gs | = (3( | g | + 2) + 3( | g | − | g | ,so κ ( g ) = 0. The proof for the case of g = ( aba ) − ( h )( aba ) is identical, with | s − gs | = | g | − s = a − , b , or w − , and | s − gs | = | g | + 2 for s = a , b − , or w . (cid:3) The following proposition gives a sufficient condition for points in F to havenegative curvature with respect to S γ , which we also show to be true with positivedensity in our proof of Theorem 1. Proposition 19.
With respect to S γ , we have κ ( g ) < for all g ∈ F such that ageodesic spelling of g has the form g = ( aa ) ± ( h )( aa ) ± or g = ( aa ) ± ( h )( aa ) ∓ ,where h is a geodesic subword such that | h | = | g | − .Proof. Suppose first that g = ( aa )( h )( a − a − ). Then no geodesic spelling of g begins with a − , b , or w − by Lemma 13, and no geodesic spelling begins with b − by Lemma 16. If s ∈ { a, b − , w, b } , then the word s ( aa )( h )( a − a − ) = sg isgeodesic by Corollary 15. Applying Lemmas 13 and 16 again, we have that nogeodesic spelling of sg ends with a , b − , w , or b . Therefore, if s ∈ { a, b − , w, b } ,then the spelling s ( aa )( h )( a − a − ) s − = sgs − is geodesic by Corollary 15, so | sgs − | = | g | + 2. Because this holds for four out of the six choices for generator s ∈ S γ , and the minimum possible value for | sgs − | is | g | − g ) = | S γ | (cid:80) s ∈ S γ | s − gs | > | g | so κ ( g ) < g . The proof for g = ( a − a − )( h )( aa ) is identical, with | sgs − | = | g | + 2for s ∈ { a − , b − , w − , b } .Now suppose that g = ( aa )( h )( aa ). Then again by Lemmas 13 and 16, nogeodesic spellings of g start or end with b or b − , so | b − gb | = | bgb − | = | g | +2. No geodesic spelling of g begins or ends with a − or w − by Lemma 13, so | s − gs | ≥ | g | for s ∈ { a, a − , w, w − } . Therefore, we again have GenCon( g ) = | S γ | (cid:80) s ∈ S γ | s − gs | > | g | so κ ( g ) < g = ( a − a − )( h )( a − a − ) follows by noting that | sgs − | = | sg − s − | = | g − | + 2 = | g | + 2 for s = b or b − because g − is a group element ofthe type considered in the previous case. Likewise, we have | sgs − | = | sg − s − | ≥| g − | = | g | for s ∈ { a, a − , w, w − } . Therefore, GenCon( g ) = | S γ | (cid:80) s ∈ S γ | s − gs | > | g | so κ ( g ) < (cid:3) In order to prove Theorem 1, we recall the following useful inequality:
Lemma 20.
Let a , . . . , a k , b , . . . , b k be positive integers such that a b ≤ . . . ≤ a k b k .Then a b ≤ (cid:80) ki =1 a i (cid:80) kj =1 b j .Proof. This follows from induction on k . For k = 2, we have a b ≤ a b = ⇒ a b ≤ a b = ⇒ a b + a b ≤ a b + a b = ⇒ a b ≤ a + a b + b . Suppose that the desired inequality holds for k ≤ l −
1. Then we have a b ≤ (cid:80) li =2 a i (cid:80) lj =2 b j .By assumption, a b ≤ a b . Therefore, we can apply the result for k = 2 to concludethat a b ≤ (cid:80) li =2 a i (cid:80) lj =2 b j = ⇒ a b ≤ a + (cid:80) li =2 a i b + (cid:80) lj =2 b j = (cid:80) li =1 a i (cid:80) lj =1 b j . (cid:3) Proof of Theorem 1.
Define the sets P n := { g ∈ S n | there exists a geodesic spelling g = g . . . g n with g , g n ∈ { a, a − }} . We claim that the ratio | P n || S n | is bounded below by some positive number for all n >
0. For n > g ∈ S n − , let g . . . g n − be a geodesic spelling of g . If g ∈ S γ , then by Lemma 13 no geodesic spelling of g begins with a − . Hence, byCorollary 15 the word ag . . . g n − ∈ S n − is geodesic. Similarly, if g ∈ S γ , thenthe word a − g . . . g n − ∈ S n − is geodesic. Let h g . . . g n − be a geodesic wordwhere h ∈ { a, a − } . If g n − ∈ S γ , then no geodesic spelling of h g ends with a − by Lemma 13, so by Corollary 15 the word h g . . . g n − a is geodesic. Similarly,if g n − ∈ S γ , then h g . . . g n − a − is geodesic. Therefore, for all g ∈ S n − , thereexists at least one element of P n of the form h gh where h , h ∈ { a, a − } .Note that every element of S n for n > h gh for some h , h ∈ S γ and g ∈ S n − . Because there are six generators in S γ , there are nomore than 6 = 36 elements of S n of the form h gh for each g ∈ S n − . Therefore,for all n > | P n || S n | ≥ . EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 15 P contains the element a , and P contains the element aa , so both are nonempty.Therefore, letting η = min (cid:110) | B | , (cid:111) , we have | P n || S n | ≥ η for all n > g ∈ P n be such that a geodesic spelling of g is ag . . . g n − for some generators g , . . . , g n − ∈ S γ . Then no geodesic spelling of g begins with a − by Lemma 13,so the word a ( ag . . . g n − ) = ag is geodesic by Corollary 15. By Lemma 16,no geodesic spelling of ag begins with b − , so the word b ( aag . . . g n − ) = bag is geodesic by Corollary 15. Finally, no geodesic spelling of bag begins with a − by Lemma 17, so the word a ( baag . . . g n − ) = abag is geodesic by Corollary 15.Therefore, we have | ( aba ) g | = n + 3.An identical argument shows that, if g ∈ P n is such that a geodesic spelling of g is a − g . . . g n − for generators g , . . . , g n − ∈ S γ , then the word ( aba ) − a − g . . . g n − is geodesic with length n + 3. Applying these claims to the inverses of elementsof P n , we get that we either have | g ( aba ) | = n + 3 or | g ( aba ) − | = n + 3 forall g ∈ P n . We can thus conclude that, if a geodesic spelling of some g ∈ P n is a ± g . . . g n − a ± , then the word ( aba ) ± ( a ± g . . . g n − a ± )( aba ) ± is geodesicwith length n + 6, and if a geodesic spelling of g ∈ P n is a ± g . . . g n − a ∓ , then theword ( aba ) ± ( a ± g . . . g n − a ∓ )( aba ) ∓ is geodesic with length n + 6. This factimplies a natural bijection between the sets P n − and Q n := { g ∈ S n | g = ( aba ) ± ( h )( aba ) ± or g = ( aba ) ± ( h )( aba ) ∓ is a geodesic spelling for some h ∈ P n − } . for all n >
6. Therefore, | P n − | = | Q n | . Recall that elements of Q n have zerocurvature by Proposition 18.Every element of the sphere S n can be written as an element of the sphere S n − that has been left-multiplied by one of the six generators in S . Therefore, | S n | ≤ | S n − | . For C > and n >
6, this implies that | S n | < C | S n − | . Wetherefore have for all n > | Q n || S n | > | Q n | C | S n − | = | P n − | C | S n − | ≥ ηC . (3.1)Let (cid:15) = min (cid:110) | B | , ηC (cid:111) and N = 7. Applying Lemma 20, equation 3.1, and thefact that | Q | > aba )( a )( aba ) and ( aba ) − ( a − )( aba ) − ),we have for all n ≥ N that (cid:80) ni =7 | Q i || B n | > (cid:80) ni =8 | Q i || B | + (cid:80) nj =8 | S j |≥ min (cid:26) | B | , | Q || S | , . . . , | Q n || S n | (cid:27) ≥ min (cid:26) | B | , ηC (cid:27) = (cid:15) . Therefore, the set of points in the union of all the Q n has positive density in F .Because κ ( g ) = 0 for all such points g , this proves that F has zero curvature withpositive density with respect to S γ . We now show that points with negative curvature also have positive density. For n ≥
5, define the sets R n := { g ∈ S n | g has a geodesic spelling of the form g ( h ) g with g , g ∈ { a, a − } and h ∈ P n − } . Note that a ± ( h ) a ± is geodesic if and only if h ∈ P n − is such that a geodesicspelling of h is a ± h . . . h n − a ± , and a ± ( h ) a ∓ is geodesic if and only if a geodesicspelling of h is a ± h . . . h n − a ∓ . By definition, every element of P n − satisfiesexactly one of these conditions, so there is a bijection between R n and P n − . There-fore, | R n | = | P n − | . Recall that elements of R n have negative curvature for n ≥ D > = 36 so that | S n | < D | S n − | for all n >
2. Then we have | R n || S n | > | R n | D | S n − | = | P n − | D | S n − | ≥ ηD . (3.2)Let (cid:15) = min (cid:110) | B | , ηD (cid:111) and N = 5. Then by Lemma 20, equation 3.2, and thefact that a , a − ∈ R so | R | >
1, we have for all n ≥ N that (cid:80) ni =5 | R i || B n | > (cid:80) ni =6 | R i || B | + (cid:80) nj =6 | S j |≥ min (cid:26) | B | , | R || S | , . . . , | R n || S n | (cid:27) ≥ min (cid:26) | B | , ηD (cid:27) = (cid:15) . Therefore, the set of points in the union of all the R n has positive density in F .Because κ ( g ) < g , this proves that F has negative curvaturewith positive density with respect to S γ . (cid:3) Relationship with δ -Hyperbolicity The group F from Section 3 is an example of a non-elementary hyperbolic group,which is a hyperbolic group that is not virtually cyclic. We have thus proven thatnon-elementary hyperbolic groups can have a positive density of points with zerocurvature. In this section we prove Theorem 2 to conclude that we get negativecurvature for all but finitely many points in non-elementary hyperbolic groups if wecompute the k -spherical conjugation curvature for sufficiently large k . On the otherhand, we show that there exist non-hyperbolic groups that have negative curvaturefor all non-identity points.4.1. Negative Curvature at Larger Radii for Hyperbolic Groups.
For therest of this section, let G be a non-elementary δ -hyperbolic group that has a wordmetric d ( · , · ) and Gromov product ( · , · ) · determined by some finite generating set S . We begin by proving a series of lemmas that give constants to determine thedesired value of K in Theorem 2. This proof is analogous to and motivated by[Oll09, Example 15] for Ollivier’s transportation curvature. Lemma 21.
There exists a constant C ≥ such that, for all integers k ≥ , g ∈ G − B k , and s ∈ S k we have | (1 , gs ) g − ( s, gs ) g | ≤ C . EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 17
Proof.
By [Gro87, Section 6.2], there exists a tree T in the Cayley graph of ( G, S )consisting of at most three geodesic segments T , T , and T such that the extremalpoints (vertices of degree 1) of T are elements of { , s, g, gs } , and which approx-imates distances between elements of { , s, g, gs } within some error for which wecan compute an upper bound. We use d T ( · , · ) and T ( · , · ) · to denote respectivelythe length metric and Gromov product induced on T by the graph metric of theCayley graph of ( G, S ). Note that, because T is a subset of this Cayley graph, wehave d T ( x, y ) ≥ d ( x, y ) for all x, y ∈ T . By the triangle inequality,( s, gs ) + ( s, gs ) g = 12 ( d ( s,
1) + d (1 , gs ) − d ( s, gs ) + d ( s, g ) + d ( g, gs ) − d ( s, gs )) ≤
12 ( k + ( | g | + k ) − ( | g | − k ) + ( | g | + k ) + k − ( | g | − k | )=4 k < | g | . Therefore, we have the following inequality:( s, gs ) + ( s, gs ) g < | g | + 12 δ = d (1 , g ) + 4(log (4) + 1) δ. By Step 2 in [Gro87, Section 6.2], this inequality implies that we can let T bea geodesic segment connecting 1 and g , and let T and T be two of the shortestgeodesic segments connecting s and gs respectively to points on T if s and gs donot already lie on T (if they do then T will consist of only one or two geodesicsegments). By Property 3 in [Gro87, Section 6.2], we have for all x, y ∈ { , s, g, gs } that d T ( x, y ) ≤ d ( x, y ) + Cδ (log (4)) = d ( x, y ) + 4 Cδ for some C ≤ d T ( x, y ) ≥ d ( x, y ), we can rewrite this statement as | d T ( x, y ) − d ( x, y ) | ≤ Cδ. (4.1)Because d (1 , s ) = d ( g, gs ) = k and d (1 , g ) > k , T and T do not intersect andthe intersection of T and T is closer to g than that of T and T . Therefore, T (1 , gs ) g = T ( s, gs ) g (if gs ∈ T , then both of these products are equal to thedistance from g to gs ). The tree T is shown in Figure 4. Figure 4.
A geodesic tree (bold segments) connecting 1, s , g , and gs such that the induced Gromov products T (1 , gs ) g and T ( s, gs ) g are equal. By the equation T (1 , gs ) g = T ( s, gs ) g , inequality 4.1, and the triangle inequality, | (1 , gs ) g − ( s, gs ) g | = | [ T (1 , gs ) g − T ( s, gs ) g ] + [(1 , gs ) g − T (1 , gs ) g ] + [ T ( s, gs ) g − ( s, gs ) g ] |≤| (1 , gs ) g − T (1 , gs ) g | + | T ( s, gs ) g − ( s, gs ) g |≤
12 ( | d (1 , g ) − d T (1 , g ) | + | d ( g, gs ) − d T ( g, gs ) | + | d T (1 , gs ) − d (1 , gs ) | + | d ( s, g ) − d T ( s, g ) | + | d ( g, gs ) − d T ( g, gs ) | + | d T ( s, gs ) − d ( s, gs ) | ) ≤
12 6(4 Cδ ) ≤ δ. Therefore, letting C = 1200 δ gives the desired result. (cid:3) The following is a well-known fact for non-elementary hyperbolic groups. See,for example, [Cha12, Corollary IV.1.3.] for a proof.
Lemma 22.
There exists a constant C > such that, for all k ≥ , we have | B k | ≥ C | B k − | . (cid:3) We have the following lemma that shows that a majority of geodesic curvesconnecting points in a ball to points far away will pass close to the center of theball. The proof is analogous to [Oll04, Proposition 21].
Lemma 23.
There exists a constant C > such that, for all g ∈ G , k ≥ , andintegers L with (cid:100) δ (cid:101) ≤ L ≤ k we have | A k ( g, L ) || B k | ≤ C C − L , where A k ( g, L ) := { s ∈ B k | ( g, s ) ≥ L } , and C is the constant from Lemma 22.Proof. Fix any such g , k , and L , and let s ∈ B k . Suppose that ( g, s ) ≥ L . Then bythe triangle inequality, ( g, s ) = ( | g | + | s | − d ( g, s )) ≤ ( | g | + | s | − ( | g | − | s | )) = | s | ,so L ≤ | s | , and we similarly have L ≤ | g | . Let h, h (cid:48) ∈ G respectively be points ona geodesic segment connecting 1 and g and a geodesic segment connecting 1 and s such that d (1 , h ) = d (1 , h (cid:48) ) = L . Applying the 2 δ -thin condition to a geodesictriangle with vertices 1, g , and s containing h and h (cid:48) , we have that d ( h, s ) ≤ d ( h (cid:48) , s ) + d ( h, h (cid:48) ) ≤ ( k − L ) + 2 δ (see Figure 5 on page 19). Therefore, A k ( g, L ) ⊂ B k − L +2 (cid:100) δ (cid:101) ( h ).By left-invariance of the word metric, it suffices to show that | B k − L +2 (cid:100) δ (cid:101) || B k | isbounded above by the desired function of L . Note that k − L + 2 (cid:100) δ (cid:101) is an in-teger bounded above by k . Therefore, letting C be the constant from Lemma 22,we have C | B k − | ≤ | B k | = ⇒ C L − (cid:100) δ (cid:101) | B k − L +2 (cid:100) δ (cid:101) | ≤ | B k | = ⇒| B k − L +2 (cid:100) δ (cid:101) || B k | ≤ C − L +2 (cid:100) δ (cid:101) = ⇒ | B k − L +2 (cid:100) δ (cid:101) || B k | ≤ C C − L . where C := C (cid:100) δ (cid:101) . The fact that C does not depend on g , k , or L implies thedesired result. (cid:3) EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 19
Figure 5.
Using the 2 δ -thin condition and triangle inequality toshow that, because ( g, s ) ≥ L , we must have d ( h, s ) ≤ k − L + 2 δ .The following lemma uses the previous result to deduce an upper bound on theaverage value of the Gromov product between all the points on a sphere and anyarbitrary point: Lemma 24.
There exists a constant C > such that for all g ∈ G and k ≥ wehave | S k | (cid:88) s ∈ S k ( g, s ) ≤ C . Proof.
Define the set D k ( g, L ) := { s ∈ S k | ( g, s ) = L } . Then we have 1 | S k | (cid:88) s ∈ S k ( g, s ) = k (cid:88) L =0 L | D k ( g, L ) || S k | because each ( g, s ) ≤ | s | = k and all the ( g, s ) are non-negative integers. Let C be the constant from Lemma 22. Then | S k || B k | = | B k | − | B k − || B k | = 1 − | B k − || B k | ≥ − C =: b > . Also, define a := (cid:80) (cid:100) δ (cid:101)− L =0 L , which is an upper bound for (cid:80) (cid:100) δ (cid:101)− L =0 L | D k ( g,L ) || S k | for all k because | D k ( g,L ) || S k | ≤
1. Then because D k ( g, L ) ⊂ A k ( g, L ), we can apply Lemma
23 to conclude that k (cid:88) L =0 L | D k ( g, L ) || S k | ≤ a + k (cid:88) L =2 (cid:100) δ (cid:101) L | D k ( g, L ) || S k | ≤ a + k (cid:88) L =2 (cid:100) δ (cid:101) L | A k ( g, L ) || S k |≤ a + k (cid:88) L =2 (cid:100) δ (cid:101) L | A k ( g, L ) | b | B k | ≤ a + C b k (cid:88) L =0 LC − L , where C is the constant from Lemma 23. For L ≥ C ) , the sequence ( LC − L ) ∞ L =0 is monotonically decreasing. Letting f := (cid:22) C ) (cid:23) ,c := (cid:24) C ) (cid:25) ,I := (cid:90) ∞ c xC − x dx < ∞ , we have by the integral test that a + C b k (cid:88) L =0 LC − L < a + C b ∞ (cid:88) L =0 LC − L ≤ a + C b (cid:32) f (cid:88) L =0 LC − L + cC − c + I (cid:33) =: C , where C does not depend on g or k . (cid:3) The constants C and C can now be applied to prove our main result: Proof of Theorem 2.
Fix k ≥ g ∈ G − B k . For all s ∈ S k , d ( s, gs ) = d (1 , g ) + d (1 , s ) + d ( g, gs ) − [ d ( g,
1) + d (1 , s ) − d ( g, s )] − [ d ( s, g ) + d ( g, gs ) − d ( s, gs )]= | g | + 2 k − g, s ) − s, gs ) g = | g | + 2 k − g, s ) − , gs ) g + 2[(1 , gs ) g − ( s, gs ) g ] . The curvature is then given by κ S k ( g ) = | g | − | S k | (cid:80) s ∈ S k d ( s, gs ) | g | = | g | − | S k | (cid:80) s ∈ S k | g | + 2 k − g, s ) − , gs ) g + 2[(1 , gs ) g − ( s, gs ) g ] | g | = − k + (cid:16) | S k | (cid:80) s ∈ S k g, s ) + 2(1 , gs ) g − , gs ) g − ( s, gs ) g ] (cid:17) | g | , so | g | κ S k ( g ) + 2 k =2 (cid:32) (cid:88) s ∈ S k ( g, s ) | S k | + (cid:88) s ∈ S k (1 , gs ) g | S k | − | S k | (cid:88) s ∈ S k [(1 , gs ) g − ( s, gs ) g ] (cid:33) ≤ C + C + C ) . where C and C are the constants from Lemmas 21 and 24 respectively (the in-equality for the (1 , gs ) g term follows from using left-invariance to rewrite it as( g − , s ) and applying Lemma 24). We recall that the constants C and C are EDIUM-SCALE RICCI CURVATURE FOR HYPERBOLIC GROUPS 21 independent of k . Therefore, letting K = (cid:100) C + C + 1 (cid:101) , we have for all k ≥ K and g ∈ G − B k that κ S k ( g ) ≤ C + C + C ) − k | g | < . (cid:3) Negatively Curved Non-Hyperbolic Groups.
With our proof of Theo-rem 2, we have shown that hyperbolicity implies negative conjugation curvature forsufficiently large comparison radius. It is natural to ask whether negative curvatureeverywhere conversely implies that a group is hyperbolic. We conclude this paperby proving that this is not the case.
Proof of Theorem 3.
Gromov proved in [Gro87] that no hyperbolic group contains Z as a subgroup. Therefore, Z ∗ Z is not a hyperbolic group.Using the presentation Z ∗ Z = (cid:104) a, b, t | ab = ba (cid:105) , let S neg := { a, a − , b, b − , t, t − } ,and we will show that we always have negative curvature with respect to S neg .Suppose that g . . . g n = h . . . h n +1 for some generators g , . . . , g n , h , . . . , h n +1 ∈ S neg . Then h − n +1 . . . h − g . . . g n = 1, so this word with 2 n +1 letters can be reducedto the identity by applying free reductions and commuting the letters a , a − , b , and b − with each other. Because free reductions reduce the number of letters by aneven number and 2 n + 1 is odd, we have a contradiction. Therefore, given anygeodesic word of generators in S neg , there does not exist a word with exactly onemore generator that represents the same group element in Z ∗ Z . Thus, by Lemma9 we do not have | sg | = | g | or | gs | = | g | for any g ∈ Z ∗ Z or s ∈ S neg .Let g ∈ Z ∗ Z − { } . From the definition of Z ∗ Z − { } , we see that a certaingeodesic spelling of g contains the letter t ± in the i th position if and only if everygeodesic spelling of g does. Likewise, a geodesic spelling of g contains an element of { a, a − , b, b − } in the i th position if and only if every geodesic spelling of g containsan element of this set in the i th position, and no geodesic spelling of g can contain s in the i th position for s ∈ { a, a − , b, b − } if another geodesic spelling of g contains s − in the i th position.Suppose first that every geodesic spelling of g does not include the letters t or t − . Then no geodesic spelling of tg or t − g will end with t or t − , so | tgt − | = | t − gt | = | g | + 2 by applying Remark 10 twice. On the other hand, all the lettersof a geodesic spelling of g commute with the letters in { a, a − , b, b − } , so for all s ∈ { a, a − , b, b − } we have | s − gs | = | gs − s | = | g | . Adding these together, we getthat GenCon( g ) = (cid:80) s ∈ S neg | s − gs | = | g | + and κ ( g ) = − | g | < g starts and ends with elements of { a, a − , b, b − } and also contains at least one element of { t, t − } . Then we still have | tgt − | = | t − gt | = | g | + 2. There exist at most two generators s ∈ { a, a − , b, b − } such that | sg | = | g | −
1, and at most two such that | gs | = | g | −
1. Because everygeodesic spelling of g contains a letter t or t − that does not commute with theletters in { a, a − , b, b − } , we have | s − gs | = | gs | − | s − g | = | g | − Therefore, κ ( g ) = | g | − GenCon( g ) | g | = | g | − (cid:80) s ∈ S neg | s − gs || g | = (cid:80) s ∈ S neg ( | g | − | gs | ) + (cid:80) s ∈ S neg ( | gs | − | s − gs | ) | g |≤ ( −
2) + ( − | g | = − | g | < . If every geodesic spelling of g starts and ends with elements of { t, t − } , then forall s ∈ { a, a − , b, b − } we have | s − gs | = | g | + 2. Therefore, because the minimumpossible value of | s − gs | for any s ∈ S neg is | g | − g ) = (cid:80) s ∈ S neg | s − gs | ≥ | g | + and κ ( g ) ≤ − | g | < g either starts or ends with t ± and ends or starts with an element of { a, a − , b, b − } . Without loss of generality,say that every geodesic spelling of g starts with t , and that at least one geodesicspelling of g ends with a . Then we must have | tgt − | = | g | + 2 , | a − ga | = | g | + 2 , and | s − gs | ≥ | g | for all s ∈ { t, a − , b, b − } . Therefore, we have GenCon( g ) = (cid:80) s ∈ S neg | s − gs | ≥ | g | + and κ ( g ) ≤ − | g | < κ ( g ) ≤ − | g | < g ∈ Z ∗ Z − { } . Therefore, Z ∗ Z is a non-hyperbolic group that has negative curvature for all non-identitypoints. (cid:3) References [BDK20] A. Bar-Natan, M. Duchin, and R. Kropholler.
Conjugation curvature forCayley graphs . 2020. arXiv: .[Cha12] V. Chaynikov.
Properties of hyperbolic groups: Free normal subgroups,quasiconvex subgroups and actions of maximal growth . Thesis (Ph.D.)–Vanderbilt University. ProQuest LLC, Ann Arbor, MI, 2012, p. 71. isbn :978-1267-81402-9.[Gro87] M. Gromov. “Hyperbolic groups”. In:
Essays in group theory . Vol. 8.Math. Sci. Res. Inst. Publ. Springer, New York, 1987, pp. 75–263. doi : .[KM20] R. Kropholler and B. Mallery. Medium-scale curvature at larger radii infinitely generated groups . 2020. arXiv: .[NW20] T. Nguyen and S. Wang.
Cheeger-Gromoll Splitting Theorem for groups .2020. arXiv: .[Oll04] Y. Ollivier. “Sharp phase transition theorems for hyperbolicity of ran-dom groups”. In:
Geom. Funct. Anal. issn :1016-443X. doi : .[Oll09] Y. Ollivier. “Ricci curvature of Markov chains on metric spaces”. In: J.Funct. Anal. issn : 0022-1236. doi :10.1016/j.jfa.2008.11.001