September 7, 2010 - Tufts Universitymath.tufts.edu/documents/ruane/GGTNotes.pdfGEOMETRIC GROUP THEORY NOTES 5 Example. Consider F 2 = ha,bi. (Picture of inﬁnite 4-valent tree.) Consider

GEOMETRIC GROUP THEORY NOTES

ANDY EISENBERG

1. September 7, 2010

Definition. Let Γ be a group. Take A ⊂ Γ to be any subset. The subgroup gener-ated by A, denoted 〈A〉, is the smallest subgroup of Γ containing A. Equivalently,〈A〉 is the intersection of all subgroups of Γ containing A. Equivalently,

〈A〉 = {aǫ11 · · · aǫn

n | ai ∈ A, ǫi = ±1},the set of finite products of elements of A ∪A−1.

Notation. We will frequently use Γ to denote a group. If A is a set, A−1 = {a−1 |a ∈ A}.Remark. 〈A〉 is characterized by the following properties:

(1) A ⊂ 〈A〉(2) 〈A〉 ≤ Γ(3) If G ≤ Γ and A ⊂ G, then 〈A〉 ⊂ G.

Definition. If A ⊂ Γ and 〈A〉 = Γ, we say that A generates Γ. If A is a finite setwhich generates Γ, then we say that Γ is finitely generated.

Notation. If A = {a1, . . . , an} and A geneates Γ, then we will write Γ = 〈a1, . . . , an〉.Definition. If ∃a ∈ Γ such that Γ = 〈a〉, then we say that Γ is cyclic.

Remark. If Γ is cyclic, then Γ is either Γ ∼= Z or Γ ∼= Zn depending on the order ofthe generator.

Example (Generating sets are not unique). Let Γ = Z = 〈a〉 = {an | n ∈ Z}.Then Γ = 〈a2, a3〉 = 〈a−2, a5〉, etc.

Example. (1) Finite groups are finitely generated.(2) The free Abelian group of rank n (that is, Zn = 〈e1, . . . , en〉) is finitely

generated.(3) Finitely generated free groups are finitely generated.(4) Gromov hyperbolic groups are finitely generated.(5) CAT (0) groups(6) π1(M) for a compact manifold M(7) SLn(Z),GLn(Z).(8) π1(K), where K is a finite simplicial complex(9) Braid groups, mapping class groups

(10) GLn(R) is NOT finitely generated. Other non-examples: Q.

Date: Fall 2010.

1

2 ANDY EISENBERG

Fact. If Γ is a finitely generated Abelian group, then any subgroup of Γ is alsofinitely generated. However, there exist finitely generated groups with subgroupswhich are not finitely generated.

Definition. Let S be a set, which we will call the alphabet. A finite sequence ofelements from S is called a word in S. We use the notation S∗ to denote the set ofall words in S. A general w ∈ S∗ looks like w = s1s2 · · · sn, si ∈ S.

Notation. Let S be a set in bijective correspondence with S. We think of theelements of this set as S = {s | s ∈ S}. We will use the convention that if a ∈ S,then a = s for some s ∈ S. We let a be an alternate notation for s.

Notation. If w ∈ {S ∪ S}∗, then, for example, w = s1s2s3 · · · sk. We use thenotation w to mean:

w = sk · · · s3s2s1= sk · · · s3s2s1

Definition. Let S be a subset of the group Γ. Then w ∈ {S ∪ S}∗ is called freelyreduced if it does not contain a subword of the form xx for x ∈ S ∪ S.

Example. The word xyxy is freely reduced, but xyxxyyy is not.

We have a map, called evaluation, ǫ : {S ∪ S}∗ → Γ. For example:

ǫ(xyxy) = xyx−1y−1.

Example. If Γ ∼= Z ⊕ Z = 〈x, y〉, S = {x, y}, then

ǫ(xyxy) = xyx−1y−1 = eΓ.

Definition. We say that Γ is free with basis S if 〈S〉 = Γ, and no freely reducedword in {S ∪ S}∗ is trivial in Γ (after evaluating with ǫ).

Exercise. Free groups are torsion free.

Definition. A group F is freely generated by a subset S ⊂ F if for any group Γand any map ϕ : S → Γ, there exists a unique homomorphism ϕ : F → Γ such thatthe following diagram commutes:

Sϕ //

� _

i

��

Γ

Fϕ

??��

Theorem. Free groups exist.

Proof. We will construct F (S), the free group with basis S (this is just terminologyfor now; it remains to show this forms a group), as the set of freely reduced wordsin {S ∪ S}∗ with concatenation as the operation. Our identity is the empty word,denoted e. (NB: the concatenation of two freely reduced words might not be freelyreduced.)

We define an equivalence relation on {S ∪ S}∗ as follows. Two words w,w′ areequivalent if they differ by a finite sequence of reductions (or expansion). Let [w]be the equivalence class of w with respect to this equivalence relation. Now, takeF (S) = {[w] | w ∈ {S ∪ S}∗}, with the operation given by [w][w′] = [ww′]. (Checkthat this operation is well-defined.) The inverse is given by [w]−1 = [w]. �

GEOMETRIC GROUP THEORY NOTES 3

Notation. When S is finite with n elements, we write F (S) = Fn for the free groupof rank n.

Exercise. If Fm∼= Fn, then m = n.

Definition. A graph K is a set of vertices, denoted V (K), and a set of edges,denoted E(K). We will sometimes take an edge e to be an unordered pair ofvertices e = {v, w}. Other times, we will want them to be ordered (directed),e = (v, w). The distinction will be stated or clear from context. We do allow loopsand multi-edges.

Definition. Recall some terminology from graph theory:

(1) Valence or degree is the number of edges coming out of a vertex.(2) A graph is locally finite if all vertices have finite valence.(3) A graph is regular if all vertices have the same valence.(4) An edge path is an alternating sequence of vertices and edges, {v0, e1, . . . , vn−1, en, vn},

where consecutive vertices and edges are adjacent.(5) K is connected if for all v, w ∈ V (K), there is an edge path with v0 = v

and vn = w.(6) A backtrack is a path of the form {v, e, w, e, v}.(7) We call a path reduced if it has no backtracks.(8) A nontrivial path is a circuit if it starts an ends at the same vertex. A

reduced circuit is called a cycle.

Theorem (Cayley’s better theorem). Every finitely generated group can be faith-fully represented as a symmetry group of a connected, directed, locally finite, regulargraph.


Definition. Let Γ = 〈S〉 be a finitely generated group with generating set S. Wedefine a labeled, directed graph ∆(Γ, S) with vertices V (∆) = Γ. Elements g, h areconnected by an edge if g−1h ∈ S ∪ S−1. This graph is called the Cayley graph.

That is, for each g ∈ Γ and s ∈ S, we have an edge

g s // gs

Right multiplication by elements of s give the edges.

Example. Consider Z = 〈1〉. The graph ∆(Z, {1}) looks like:

· · · // −1 // 0 // 1 // · · ·

Example. Now consider {an | n ∈ Z} = Z = 〈a, a2〉. Then the graph looks like:

· · · // a //

��???

????

? a3 // · · ·

· · · // e //

@@��a2 //

>>~~~~~~~· · ·

4 ANDY EISENBERG

Example. Consider the symmetric group S3 = 〈(1 2), (1 2 3)〉 = 〈a, b〉. The Cayleygraph is:

(1 2)

e

(1 3 2) (1 2 3)

(2 3) (1 3)

b

b

bb b

b

a

a a

Example. Consider Z ⊕ Z = 〈(1, 0), (0, 1)〉, then the Cayley graph is the integerlattice with horizontal and vertical edges.

Given w ∈ {S ∪ S}∗, we can draw a path in the Cayley graph that begins at e.

Example. From the previous example, take a = (1, 0) and b = (0, 1). Let w =

abaaba. Then the path looks like:

ab // aba // abaa

��e // a

OO

abaabb abaaboo

Conversely, given an edge path beginning at e in the Cayley graph, we can rad offa word in {S ∪S}∗. This correspondence gives a bijection between words {S ∪S}∗and ∆(Γ, S). The collection of paths with no backtracking corresponds to reducedwords.

Remark. Observe that if we can draw a nontrivial edge path that begins and endsat e in the Cayley graph, then there is a corresponding word in {S ∪ S}∗ thatevaluates to the identity.

Corollary. If Γ = Fn and S is any basis, then ∆(Fn, S) is a regular 2n-valent tree.

Proof. In Fn, there are no nontrivial reduced words that evaluate to e. Thus thereare no cycles in ∆(Fn, S). �

Proof of Cayley’s better theorem. Let Γ = 〈S〉 be a finitely generated group. Γ actson the Cayley graph ∆(Γ, S) by left multiplication. That is, if vg is the vertexlabeled by g and x ∈ Γ, then x · vg = vxg. (“x takes the vertex labeled g to thevertex labeled xg.”) Consider an arbitrary edge (g, gs). We have:

gs x // xgs

g

s

OO

x// xg

s

OO

Hence this action takes edges to edges, and is therefore a group of symmetries.The action is clearly faithful. Since ∆(Γ, S) is connected (since S generates), thiscompletes the proof. �


Example. Consider F2 = 〈a, b〉. (Picture of infinite 4-valent tree.) Consider theword w = abab. The corresponding path in the Cayley graph is:

aba

��

aboo

abab

e // a

OO

But if the element aba−1b−1 acts on e, b−1 acts first, then a−1, and so on. Soreading off the word gives you the endpoint, but it will not correctly show theaction of that element on the vertex e.

Definition. Let A be a subset of the group Γ. Then the normal closure of A,denoted 〈〈A〉〉, is the smallest normal subgroup of Γ that contains A.

Remark. 〈〈A〉〉 is characterized by the following properties:

(1) A ⊂ 〈〈A〉〉(2) 〈〈A〉〉 ⊳ Γ(3) If N ⊳ Γ and A ≤ N , then 〈〈A〉〉 ⊂ N .

We can also think of the normal closure as 〈〈A〉〉 = 〈{gag−1 | g ∈ G, a ∈ A}〉.

Remark. If A is finite, then 〈A〉 is finitely generated, but 〈〈A〉〉 might not be.

Theorem. Every group is the quotient of a free group.

Proof. Let Γ be any group and let S be some generating set for Γ (possibly infinite).Let F (S) be the free group on S. By the universal mapping property for free groups,we have

S� _

ι

��

� � ι // Γ

F (S)ϕ

=={{

{{

ϕ is onto precisely because S generated Γ. By the first isomorphism theorem, weknow Γ ∼= F (S)/ ker(ϕ). �

Definition. A presentation of a group Γ is an isomorphism with a group of theform

F/〈〈R〉〉where F is free and R ⊂ F . If Γ is finitely generated, F = Fn for some n, and R isfinite, then we say that Γ is finitely presented.

Definition. Equivalently, suppose Γ = 〈S〉 with S finite and Γ ∼= F (S)/〈〈R〉〉. IfR is finite, then we write 〈S | R〉 for the presentation of Γ using S and R.

Fact. Fn = 〈a1, . . . , an | ∅〉 is a presentation for the free group.

6 ANDY EISENBERG

Example. Let Γ = Z ⊕ Z = 〈c, d〉. Then we claim Γ = 〈a, b | aba−1b−1〉. We haveF2 = 〈a, b〉. By the universal mapping property:

{a, b} ϕ //

��

Γ

F2

ϕ

=={{

{{

{

where ϕ(a) = c and ϕ(b) = d. Then Γ ∼= F2/ ker ϕ. We want to show thatK = 〈〈aba−1b−1〉〉, which we will denoteN . We have thatN ⊂ K because cdc−1d−1

is trivial in Γ. Conversely, we have:

F2

π

��

ϕ // F2/K

F2/N

α

;;vvvvvvvvv

If x ∈ K, then ϕ(x) is trivial. Then α(π(x)) is also trivial. F2/N is Abelian(since it’s a quotient by the commutator subgroup). So π(x) = Nambn, henceα(π(x)) = Ncmdn = e implies m = n = 0.


Remark. Recall from last time: if Γ is a finitely generated group wwith generatingset S, then we can construct a locally finite graph ∆(Γ, S) on which Γ acts by graphsymmetries. If eΓ /∈ S, then ∆ has no loops.

For each edge e = gs−→ gs, identify the edge isometrically with [0, 1]. That is,

fe : [0, 1] → ∆ such that fe(0) = g and fe(1) = gs. We can choose these isometriesequivariantly. Define a metric on ∆ as follows:

d(p, q) = inf{ℓ(c) | c is an piecewise linear path from p to q in ∆}.Definition. A piecewise linear path is a map c : I → ∆ for which there exists apartition 0 = t0 ≤ t1 ≤ · · · ≤ tn = 1 such that

c[ti,ti+1] = fei◦ ci

where ci : [ti, ti+1] → [0, 1] is an affine map. The length of the path c is given by

ℓ(c) =n−1∑

i=0

|ci(ti) − ci(ti+1)|

Remark. The function d is a metric on ∆ which turns ∆ into a proper, geodesicmetric space. We had Γ ≤ Sym(∆). We now have Γ ≤ Isom(∆).

We can define a distance function on the group Γ with respect to the generatingset S by taking dS(g, h) to be the length of the shortest word w ∈ {S ∪ S}∗ suchthat ǫ(w) = g−1h. This is called the word metric associated to S.

Example. Consider Γ = Z⊕Z with the standard generating set S = {(1, 0), (0, 1)}.Then ∆(Γ, S) is given isometrically by the integer lattice grid in E2 with the taxicab metric.


Definition. Let (X, d) be a metric space. A (unit speed) geodesic in X is a pathγ : I = [a, b] → M such that d(γ(t), γ(u)) = |t− u| for all u, t ∈ I. In particular, ifγ(a) = x and γ(b) = y, then d(x, y) = b − a.

A geodesic ray is γ : [0,∞) → X such that d(γ(t), γ(u)) = |t − u| for all u, t ∈[0,∞).

(X, d) is called a geodesic metric space if there exists a geodesic between anypair of points. It is called uniquely geodesic if there is only one.

Definition. A metric space (X, d) is proper if it’s (complete and?) closed metricballs are compact.

Remark. Does the metric ball condition imply completeness?

Definition. A map f : (X, d) → (X ′, d′) between metric spaces is called an iso-metric embedding if d′(f(x), f(y)) = d(x, y) for all x, y ∈ X . If f is also surjective,it is called an isometry.

Definition. Let (X, d) and (X ′, d′) be proper, geodesic metric spaces. A mapϕ : X → X ′ is a λ quasi-isometric embedding if λ ≥ 1 and

1

λd(x, y) − λ ≤ d′(ϕ(x), ϕ(y)) ≤ λd(x, y) + λ.

If d′(x′, ϕ(X)) ≤ λ for all x′ ∈ X ′ (equivalently, if for each x′ ∈ X ′ there existsx ∈ X such that d′(x′, ϕ(x)) ≤ λ), then we say ϕ is almost surjective. If ϕ is analmost surjective quasi-isometric embedding, then ϕ is called a quasi-isometry. Wewrite X ∼QI X

′ to mean that X is quasi-isometric with X ′.

Example. (0) Any bounded set is quasi-isometric to a point.(1) The projection map π : R× [0, 1] → R is a quasi-isometry.(2) ∆(Z, {a, a2}) is quasi-isometric with ∆(Z, {a}). In fact, any Cayley graph

of Z is quasi-isometric to the real line.(3) ∆(Z ⊕ Z, {(1, 0), (0, 1)}) ∼QI E2. The inclusion map ι : ∆ → E2 is almost

surjective. We have

1√2d∆(x, y) ≤ dE2(x, y) ≤ d∆(x, y)

So we can take λ = 2, for example, and we have a quasi-isometry.(4) If Γ = 〈S〉 = 〈S′〉 for finite generating sets S, S′, then ∆(Γ, S) ∼QI

∆(Γ, S′) := ∆′. We take f : ∆ → ∆′ where fV (∆) is the identity. Takeλ1 = max{d′(e, a) | a ∈ S} and λ2 = max{d(e, a′) | a′ ∈ S′}. Takeλ = max{λ1, λ2}. Then

1

λ2d(g, h) ≤ d′(g, h) ≤ λ1d(g, h)


Remark. The definition of quasi-isometry makes sense for any metric space. Wedid not need to assume proper or geodesic.

Let Γ = 〈S〉 be a finitely generated group, let ∆ = ∆(Γ, S) be the Cayley graph,and let ΓS be the group with the word metric. The inclusion map i : ΓS → ∆is an isometric embedding that is almost onto with λ = 1, hence i is a quasi-isometry. If S, S′ both generate Γ, then ΓS ∼QI ΓS′ . We have ∆ ∼QI ΓS and

8 ANDY EISENBERG

ΓS′ ∼QI ∆′ = ∆(Γ, S′), so ∆ ∼QI ∆′. (We will prove as an exercise that ∼QI istransitive, and we will show in the following proposition that it is reflexive.)

Proposition. Suppose ϕ : X → Y is a quasi-isometry with constant λ. Then thereexists ψ : Y → X which is a quasi-isometry (with constant Λ) such that

d(ψ ◦ ϕ, 1X) ≤ Λ

d(ϕ ◦ ψ, 1Y ) ≤ Λ

(Without loss of generality, we will take Λ ≥ λ at least.)

Proof. Suppose ϕ : X → Y is a λ-QI. Let y ∈ Y . There exists x ∈ X such thatd(y, ϕ(x)) ≤ λ. Define ψ(y) = x. Now d(ϕ ◦ ψ(y), y) ≤ λ by definition of ψ. Also,let x′ = ψ ◦ ϕ(x). Then

1

λd(x, x′) − λ ≤ d(ϕ(x), ϕ(x′)) ≤ λ

hence d(x, x′) ≤ 2λ2.Now let x ∈ X . We will find y ∈ Y such that d(x, ψ(y)) is bounded independent

of x as follows. Pick y = ϕ(x).It remains to show the quasi-isometry inequality:

1

Λd(y, y′) − Λ ≤ d(ψ(y), ψ(y′)) ≤ Λd(y, y′) + Λ.

Let x = ψ(y) and x′ = ψ(y′). From the QI inequality for ϕ, we have

1

λd(x, x′) − λ ≤ d(ϕ(x), ϕ(x′))

≤ λ[d(ϕ(x), ϕ(x′)) + λ]

≤ λ[d(y, y′) + 2λ] + λ2

≤ λd(y, y′) + 3λ2.

Now we need1

Λd(y, y′) − Λ ≤ d(x, x′).

Use d(ϕ(x), ϕ(x′)) ≤ λd(x, x′) + λ. By the reverse triangle inequality:

d(ϕ(x), ϕ(x′)) ≥ d(y, y′) − 2λ.

So we can take Λ ≥ 3λ2, and the result follows. �

Fact. QI is an equivalence relation on metric spaces, and we use the notationX ∼QI Y .

Exercise. Only Charlie: The empty set is only QI to itself.

Fact. (1) E1 ≁QI [0, 1]. In fact, being bounded is a QI invariant.(2) E1 ≁QI [0,∞).(3) E2 ≁QI E1.(4) Let Tn be the regular tree of valence n. T3 ≁QI E1 and T3 ≁QI E2.(5) Tn ∼QI Tm for all n,m ≥ 3. (This is nontrivial.)(6) En ∼QI Em implies n = m.

Remark. For T3, T4, if you collapse edges carefully in T3, you will get T4. Thenshow that this map is a quasi-isometry.


Big theorem coming up:

Theorem. If X,Y are proper, geodesic metric spaces, then any QI f : X → Y willinduce a homeomorphism fǫ : Ends(X) → Ends(Y ).

Let Γ,Γ′ be finitely generated groups. When can we say that Γ ∼QI Γ′? (Wecan write Γ ∼QI Γ′ without reference to a generating set, since QI is independentof the choice of generating set.) Guess: maybe they contain isomorphic finite indexsubgroups.

Definition. Two groups Γ,Γ′ are commensurable if they contain isomorphic finiteindex subgroups.

Fact. Being commensurable is an equivalence relation. A big part of geomet-ric group theory looks at the relationship between commensurability and quasi-isometry.

If Γ ∼C Γ′, then Γ ∼QI Γ′. The converse is not always true. A major questionis: under what conditions does QI imply commensurability. This is called the QIrigidity question.

Theorem. There exist M,N closed, hyperbolic 3-manifolds with:

(1) no common finite cover(2) π1(M) ∼QI π1(N) (and both are QI to H3).

Example. Consider Z2 ⋊ϕ Z, ϕ ∈ GL(2,Z). The QI class is determined by thematrix for ϕ. (Bridson)

Theorem (Svarc-Milner). Suppose Γ acts cocompactly and properly discontinuouslyby isometries on a proper, geodesic metric space X. Then Γ is finitely generated,and the map Γ → X given by γ 7→ γ ·x0 is a QI for any choice of base point x0 ∈ X.

Definition. Let (X, d) be a metric space. Isom(X) is a group. Let Γ be a groupand Φ: Γ → Isom(X) a homomorphism. Then we say Γ acts on X by isometries.We write Φ(γ)(x) =: γ ·x. The action is faithful if Φ is injective. The action is freeif for all x ∈ X and for all γ 6= e, then γ · x 6= x. The action is called cocompactif there exists a compact set K ⊂ X such that Γ · K = X . The action is calledproperly discontinuous if for all compact K ⊂ X , the set {γ ∈ Γ | γ ·K ∩K 6= ∅}is finite.

If G ≤ Γ is a finite index subgroup of the finitely generated group Γ, and ifX = ∆(Γ, S) for any finite generating set S, then G y X properly discontinuouslyand cocompactly by isometries.


Theorem (Svarc–Milnor). Suppose Γ is a group acting “geometrically” (i.e. prop-erly discontinuously and cocompactly by isometries) on a proper geodesic metricspace X. Then Γ is finitely generated, and for any choice of basepoint x0 ∈ X, themap Γ → X given by γ 7→ γ · x0 is a QI.

Theorem. Let X be a topolocial space, and let Γ act on X by homeomorphisms.Suppose there exists an open set U ⊂ X such that Γ · U = X. Then if X isconnected, then S = {γ ∈ Γ | γ · U ∩ U 6= ∅} generates Γ.

10 ANDY EISENBERG

Moreover, if U,X are both path-connected and X is simply connected, then

R = {s1s2s−13 | si ∈ S,U ∩ s1U ∩ s3U 6= ∅, s1s2 = s3}

forms a set of relators for Γ.

Proof. We will only prove the first claim. Let H = 〈S〉 ≤ Γ. Let V = H · U andV ′ = (Γ\H)·U . If V ∩V ′ 6= ∅, then we have hu = h′u′. It follows that u = h−1h′u′,hence h−1h′U ∩ U 6= ∅ and h−1h′ ∈ S. This means h′ ∈ HS ⊂ H . But h′ /∈ H byassumption, a contradiction. So V ∩ V ′ = ∅. Since X is connected, V ′ = ∅, henceΓ \H = ∅ and Γ = H . �

Theorem. Suppose Γ is a group. Γ is finitely presented if and only if Γ acts prop-erly discontinuously and cocompactly by homeomorphisms on a simply connectedgeodesic space.

This was all a topological aside. For the Svarc–Milnor theorem, we’ll need tothink more metrically.

Proof of Svarc–Milnor. Let C ⊂ X be a compact set such that Γ ·C = X . Choosex0 ∈ X and D > 0 such that C ⊂ B(x0, D/3) and let

A = {γ ∈ Γ | γ ·B(x0, D) ∩B(x0, D) 6= ∅}.Since the action is properly discontinuous, A is finite.

We have a map f : ΓA → X given by γ 7→ γ · x0. We need to show f is a QI.We know that every point of X is within D/3 of an orbit point. So f is almostsurjective. (The more common phrase is: the image is quasi-dense.) We must showthat

1

λdA(e, γ) − λ ≤ dX(f(e), f(γ)) ≤ λdA(e, γ) + λ.

(Since both metrics are equivariant, we may assume we’re at the origin.) Takeλ = max{dX(x0, a · x0) | a ∈ A}. Suppose dA(e, γ) = n. Then γ = a1a2 · · ·an,ai ∈ A. Let gi = a1 · · · ai. Then

dX(gix0, gi+1x0) = dX(x0, g−1i gi+1x0) ≤ λ.

Hence dX(x0, γx0) ≤ λn = λdA(e, γ).For the other inequality, we can consider a geodesic from x0 to γx0 as an isometry

c : [0, d(x0, γx0)] → X . Consider the points c(iD/3), and for each one choose anorbit point γix0 within distance D/3. Then d(c(iD/3), c((i + 1)D/3)) ≤ D/3,d(γix0, γi+1x0) ≤ D, and d(x0, γx0) ≥ (n − 1)D/3 ≥ d(e, γ) − 1. We have γ =γ0(γ

−10 γ1) · · · (γ−1

n−1γn), where γ0 = e and γ−1i γi+1 ∈ A. So for this inequality,

choose λ = 1, and for the overall proof, take the maximum λ. �

As a useful application of this theorem, if we want to show that two groups areQI, we can make them both act geometrically on the same space.

Example. Let M,N be compact hyperbolic manifolds. Their fundamental groupsare QI, because they both act on Hn geometrically.

Corollary. If Γ is a finitely generated group and G a subgroup of finite index, thenG is finitely generated and G ∼QI Γ.

Proof. G acts geometrically on any Cayley graph for Γ. (The generating set doesn’tmatter, since all the Cayley graphs are QI.) �


Definition. For any property P , we say Γ is virtually P if there is some finiteindex subgroup G ≤ Γ which has property P .

Question: when does Γ ∼QI Γ′ imply Γ ∼C Γ′? (We already know the converseis always true by the previous theorem.)

Fact. In the following cases, Γ ∼QI Γ′ implies Γ ∼C Γ′:

(0) Γ or Γ′ is finite.(1) Γ or Γ′ is virtually Z.(2) If both Γ and Γ′ are finitely generated and virtually Abelian, then they are

both virtually Zn for some n: let Γ be virtually Abelian. Then there existsa finite index group G ≤ Γ, which is finitely generated. By the structuretheorem, G ∼= H × T , where T is the torsion part and H ∼= Zn for some n.For QI purposes, we do not lose generality by assuming G is torsion free.Similarly, we’ll have a G′ ≤ Γ′ which is Zm for some m. So we really needto show that Zn ∼QI Zm implies n = m.

It’s easier to prove that En ∼QI Em implies n = m. (The claim follows

from Svarc–Milnor.) To show this, we would start by showing R2 ∼QI R.(3) If Γ is a finitely generated group and Γ ∼QI Zn, then Γ is virtually Zn.

Proposition. If Γ is a finitely generated group and Γ ∼QI Z, then Γ is virtuallyZ.


Recall from last time:

Corollary (Corollary to Gromov’s theorem). If Γ is a finitely generated group andΓ ∼QI Zn then Γ is virtually Zn.

Corollary. If Γ is a group of polynomial growth, then Γ is virtually nilpotent.

Continuing our examples from last time:

Fact. (4) Any finitely generated group Γ that is QI to Hn can be realized as afinite extension of a proper cocompact subgroup of Isom(Hn). That is:

1 → F → Γ → Γ′ → 1

where Γ′ is a proper cocompact subgroup of Isom(Hn) and F is finite.(Result of Tukia/Gabai, Casson–Jungreis.)

Remark. (E. Rieffel) If Γ ∼QI H2 × R for a finitely generated group Γ, then

1 → F → Γ → Γ′ → 1

where Γ′ is a proper cocompact subgroup of either Isom(H2×R) or Isom(PSL(2,R)).

Proposition (Special case of Borsak–Ulam theorem). Suppose g : S1 → E1 isa continuous map. Then there exists an antipodal pair x = (cos t0, sin t0) and−x = (− cos t0,− sin t0) so that g(x) = g(−x).Proof. Let f : [0, π] → E1 be given by

f(t) = g(cos t, sin t) − g(− cos t,− sin(t)).

Now we have f(0) = g(1, 0) − g(−1, 0) and f(π) = g(−1, 0) − g(1, 0). If f(0) =0 = f(π), then the antipodal pair is (1, 0), (−1, 0). Otherwise, f(0) and f(π) haveopposite sign. By the intermediate value theorem, we’re done. �

12 ANDY EISENBERG

Proposition. E2 ≁QI E1.

Proof. Suppose ϕ : E2 → E1 is a QI with constant λ. Choose n large enough,and take 2n equally spaced points x0, x1, . . . , x2n = x0 on the circle of radiusn, with xi and xi+n antipodal, and having dE2(xi, xi+1) ≤ π. Since ϕ is a QI,|ϕ(xi) − ϕ(xi+1)| ≤ λπ + λ. Define a map f : S1 → E1 by f(xi) = ϕ(xi), and thearc between xi and xi+1 onto the interval between ϕ(xi) and ϕ(xi+1). there existsan antipodal pair x,−x so that f(x) = f(−x).

Choose an xi that is closest to x. Then d(x, xi) ≤ π and d(−x,−xi) ≤ π. Now|f(x) − f(xi)| ≤ λπ + λ, and |f(−x) − f(−xi)| ≤ λπ + λ. Now we have

|ϕ(xi) − ϕ(−xi)| = |f(xi) − f(−xi)|≤ |f(xi) − f(x)| + |f(x) − f(−xi)|≤ 2(λπ + λ).

Hence d(xi,−xi) is bounded independent of n by the other QI inequality. Butwe know this distance to be 2n, so we can choose n large enough that this is acontradiction. (Exercise: figure out how large n needs to be.) �

Exercise. For homework, E1 ≁QI [0,∞).

Lemma (Ping-pong lemma). Suppose Γ is a group generated by a set S = {s1, . . . , sn}.Suppose Γ acts on a set X. That is, we have a homomorphism ϕ : Γ → Sym(X).Suppose we can choose 2n subsets of X, X+

1 , X+2 , . . . , X

+n , X

−1 , . . . , X

−n , one for

each of si and s−1i , which are nonempty, pairwise disjoint subsets of X such that

si(X −X−i ) ( X+

i and s−1i (X −X+

i ) ( X−i . Then Γ ∼= F (S).

Proof. For simplicity, suppose there exists p ∈ X \ ⋃ni=1(X

+i ∪ X−

i ). We need to

show that if w is a nontrivial reduced word in {S ∪ S}∗, then ε(w) 6= eΓ. We willshow that w(p) 6= p (abusing notation), so that ϕ(w) 6= 1X .

Claim. Suppose w = aǫ11 · · · aǫk

k , ai ∈ S, ǫ = ±1. Then w(p) ∈ X+j if ǫ1 = 1 or

w(p) ∈ X−j if ǫ1 = −1, where a1 = sj.

Proof. We will induct on the length k of the reduced word w. Suppose k = 1. Thenw = sj or w = s−1

j for some j. Without loss of generality, suppose w = sj . Since

p /∈ X−j , w(p) ∈ X+

j , proving the base case.Now suppose the claim is true for reduced words of length up to k− 1. Suppose

without loss of generality that aǫ11 = sj . We want to show w(p) ∈ X+

j . Suppose not.

Then s−1j (w(p)) ∈ X−1

j . But s−1j w = aǫ2

2 · · · aǫk

k has length k − 1, and aǫ22 6= s−1

j

since w was reduced. By induction, s−1j w(p) /∈ X−

j (because the X±i are pairwise

disjoint), thus the claim is proven. �

By the claim, then w(p) ∈ X±i , but p is not in any of them. Hence ϕ(w) 6= 1X ,

and every freely reduced word is nontrivial. �

Here’s one application. Let Γ be an interesting group (like the mapping classgroup of some surface). Take a finite set of elements f1, . . . , fk in Γ, and ask: doesthere exists an N0 so that for all N ≥ N0, 〈fN

1 , . . . , fNk 〉 ∼= Fk?

Open Question. Is the Burau representation faithful? Equivalently, do a partic-ular pair of matrices in GL3(Z[t, t−1]) generate an F2?


Example. Let Γ = F2 and X = ∆(Γ, {a, b}). The origin with the four half-edgesleaving it forms a fundamental domain. Removing this domain, we are left withfour maximal subtrees. These will be the X±

a,b.

Let H ≤ F2 = 〈a, b〉 be the subgroup consisting of even length words. We wantto show that H ∼= F3. We have a fundamental domain F (Charlie has the picture).After removing it, there are six connected components remaining. We have

{g ∈ H | g · F ∩ F 6= ∅} = {a2, a−2, ab, b−1a−1, ab−1, ba−1}.So S = {a1, ab, ab−1} is a free basis for H .


Finishing a thought from last time:

Theorem. A group Γ is free if and only if Γ acts freely on a tree.

Corollary. Every subgroup of a free group is free.

Example. Let Γ = Z3 ⋆Z4. Then Γ acts on a tree but not freely. However, Γis virtually free. Let T3,4 be the infinite tree constructed as follows. Start with avalence 4 vertex. Each of its neighbors has valence 3. Each of their neighbors hasvalence 4, and so on. Suppose Z3 = 〈a〉 and Z4 = 〈b〉. Then the action of b on T3,4

is the rotation about the center vertex, and the action of a is rotation about, say,the vertex north of the center.

Remark. The free product Z3 ⋆Z4 is given by

{ai1bj1 · · · ainbjn | 1 ≤ ik ≤ 2, 1 ≤ jk ≤ 3, 0 ≤ i1 ≤ 2, 0 ≤ jn ≤ 3}.Consider ϕ : Z3 ⋆Z4 → Z3 ⊕ Z4. Then kerϕ is free.

Fact. (1) Any finite group acting on a tree will fix a point.(2) The free product of two finite groups is virtually free.

Recall: if Γ is a group and S ⊂ Γ is any subset, we can build a Cayley graph∆ = ∆(Γ, S). We have seen that ∆ is connected if and only if S generates.

Theorem. Let Γ = 〈S〉, S finite. Thus F (S) ։ Γ. Let N be the kernel of thismap, and let R ⊂ N . Consider the 2-complex obtained by attaching 2-cells to alledge loops in ∆(Γ, S) that are labeled by r ∈ R. The resulting 2-complex is simplyconnected if and only if 〈〈R〉〉 = N .

Proof. Consider T = ∆(F (S), S). This is a tree. Then we can view ∆(Γ, S) as thequotient of T by the action of N . A word w ∈ {S ∪ S}∗ defines an element of Nif and only if it is the label of an edge loop in ∆(Γ, S) at eΓ. (That is, we have anidentification π1(∆, eΓ) ∼= N .)

Let u ∈ F (S) be any reduced word, r ∈ R fixed, and ε(u) be the evaluationsof the words in Γ. Then r traces out an edge loop at each ε(u) in ∆(Γ, S). Weattach a 2-cell along the word r at every ε(u). Then the resulting complex hasfundamental group given by N/〈〈u−1ru〉〉.

If we attach a 2-cell along every such r-labeled edge loop for r ∈ R, then theresulting 2-complex K has π1(K) ∼= N/〈〈R〉〉. The result follows. �

Proposition. Suppose (Γ1, A1) and (Γ2, A2) are finitely generated groups and Γ2 =〈A2 | R2〉 is a finite presentation. If Γ1 ∼QI Γ2 with constant λ, then Γ1 is alsofinitely presented with generating set A1.

14 ANDY EISENBERG

Proof. Let ρ be the longest r ∈ R2. Let K2 be the 2-complex obtained by attaching2-cells along any loop of length at most ρ in ∆(Γ2, A2). By the previous theoremthis complex will be simply connected.

Claim. There exists an M = M(ρ, λ) such that if we attach 2-cells along all loopsof length at most M in ∆(Γ1, A1), then the resulting 2-complex is simply connected.

Proof. Let the resulting complex be K1. Consider a loop ϕ : ∂D → K1 with verticesg1, g2, . . . , gn. Let vi be the preimage of gi in D. If f is the QI from Γ1 to Γ2, thenf(gi) give some collection of vertices in K2. We will define a new map α : ∂D → K2

so that α(vi) = f(gi). Then we extend to a map α : D → K2. We sketch theremainder of the proof.

Divide the resulting disk in K2 into smaller disks of boundary length at most ρ.Use the QI inequalities to pull these back to K1 to show that ϕ(∂D) bounds a diskin K1. M will be determined by ρ and the QI inequalities. The claim will followby construction of K1. �

The claim finishes the proposition. �

We are working up to the following theorem.

Theorem (mostly Hopf, 1940s). Let Γ be a finitely generated group. Then:

(1) Γ has 0, 1, 2 or ∞ many ends.(2) Γ has 0 ends if and only if Γ is finite.(3) Γ has 2 ends if and only if Γ is virtually Z.(4) End(Γ) is compact. If it’s infinite, theen it is uncountable and each point

is an accumulation point.(5) (Stallings) Γ has infinitely many ends if and only if Γ splits as Γ ∼= A⋆C B

or Γ ∼= A⋆C for a finite group C, and |A/C| ≥ 3, |B/C| ≥ 2.

Remark. A⋆C B is the amalgamated free product of A and B. A⋆C is the HNNextension.

Definition. Let Γ be a group, and let ϕ : H → K be an isomorphism betweensubgroups of Γ. The HNN extension of Γ relative to ϕ is

Γ ⋆ϕ = 〈S, t | R, tht−1 = ϕ(h), h ∈ H〉.The new generator t is called the stable letter.

Example. (1) Zn is 1-ended for n ≥ 2.(2) Fn is infinite-ended for n ≥ 2.(3) F1 is 2-ended.

Definition. Recall that a continuous map f : X → Y between topolocial spaces iscalled proper if f−1(C) is compact for all compact C ⊂ Y .

Proposition. Let x0 ∈ X be a fixed base point in a metric space X. Considerthe function X → R given by x 7→ d(x, x0). Then X is proper if and only if thisfunction is proper.

Definition. Let X be a topological space. A ray in X is a continuous mapr : [0,∞) → X , and a proper ray is a ray which is proper.

Definition. Suppose r1, r2 : [0,∞) → X are proper rays. Then we say r1 and r2converge to the same end if for all compact C ⊂ X there exists N > 0 such thatr1([N,∞)) and r2([N,∞)) lie in the same path component of X \ C.


Notation. Converging to the same end gives an equivalence relation on the set ofproper rays. We write end(r) = [r] for the equivalence class of a ray r.

Definition. The ends of X are the classes of proper rays. We write Ends(X) ={end(r)}, where r ranges over all proper rays.

Definition. We say rn converges to r, written end(rn) → end(r), if for all compactC ⊂ X , there exist {Nn} such that rn([Nn,∞)) and r([Nn,∞)) are in the samepath component of X \ C for n sufficiently large.

We give a topology to Ends(X) as follows. A subset B ⊂ Ends(X) is closed ifend(r) ∈ B whenever {end(rn)} ⊂ B converges to r.


If X is a proper geodesic metric space, then any compact set C is contained insome open ball B(x0, R), which is contained in the closed ball B(x0, R). So we mayreplace compact sets with open balls in the definition of ends.

Definition. Let k > 0. A k-path in X connecting x to y is x = x0, x1, . . . , xn = ysuch that d(xi, xi+1) ≤ k.

Lemma. Let k > 0, and let r1, r2 be proper rays.

(1) end(r1) = end(r2) if and only if for all R > 0 there exists a T > 0 suchthat r1(t) can be joined to r2(t) by a k-path in X \B(x0, R) for all t > T .

(2) Let Gx0be the set of all geodesic rays in X emanating from x0. There is a

map Gx0→ Ends(X) given by c 7→ end(c). This map is surjective.

Proof. (1) Suppose we have a k-path from r1(t) to r2(t) in X \B(x0, R). Joinsuccessive points by geodesics to get a path in X \ B(x0, R − k). So r1(t)and r2(t) are in the same path component of the complement of a ball. Ina proper geodesic metric space, since we can replace compact sets by openballs in the definition of ends, end(r1) = end(r2). The converse is trivial.

(2) Let end(r) ∈ Ends(X), with r : [0,∞) → X a proper ray. Without lossof generality (by connecting r(0) with a geodesic to x0) we may assumer(0) = x0.

Let cn : [0, dn] → X bee a geodesic joining x0 to r(n). Extend eachcn by the constant map so that cn : [0,∞) → X and cn(t) = dn for t ≥dn. The collection {cn} is an equicontinuous family of functions. ApplyArzela–Ascoli to obtain a subsequence that converges to a geodesic (byconstruction) map c : [0,∞) → X .

�

Theorem. Let X1, X2 be proper geodesic metric spaces and f : X1 → X2 a QI withconstant λ. Then f induces a homeomorphism fε : Ends(X1) → Ends(X2).

Proof. Let end(r) ∈ Ends(X1). Without loss of generality, assume r is a geodesic.Connect the images f(r(i)) by geodesics. This creates a proper ray in X2 (whichfollows from the fact that f is a QI, but there is a little bit to show here). Denotethis ray by f∗(r). Then define fε(end(r)) = end(f∗(r)).

This map is well-defined because k-paths in X1 turn become (λk + λ)-paths inX2. By the lemma, this map is defined on the whole domain. fε is continuous, butwe won’t do this because there’s a little more work to do. Use the QI to find aninverse map which shows fε is bijective, and therefore a homeomorphism. �

16 ANDY EISENBERG

Definition. Let Γ be a finitely generated group and ∆ = ∆(Γ, S) any Cayleygraph. Then we define Ends(Γ) = Ends(∆).

Theorem. Let Γ be a finitely generated group. Then Γ has 0, 1, 2 or infinitelymany ends.

Proof. Γ acts by homeomorphisms on Ends(Γ). (Recall that Γ acts by isometrieson ∆(Γ, S).) Thus we have a homomorphism ϕ : Γ → Homeo(Ends(Γ)).

If Ends(Γ) is finite, then Homeo(Ends(Γ)) is finite, so K = kerϕ has finite indexin Γ. By quasi-density, there exists µ such that for any γ ∈ Γ, there exists k ∈ Kwith d(γ, k) ≤ µ.

Suppose there are three distinct ends e0, e1, e2. Draw geodesic rays r1, r2 at eΓwith end(ri) = ei for i = 1, 2. Take a geodesic ray r′0 with end(r′0) = e0. Define aproper ray r0 with end(r0) = e0 such that d(r0(n), eΓ) ≥ n, r0(n) ∈ K, and r0(n)is at least µ-close to r′0. r0 is a proper ray.

Choose ρ so that r0([ρ,∞)), r1([ρ,∞)), and r2([ρ,∞)) are in different pathcomponents of X \B(eΓ, ρ). If t, t′ ≥ 2ρ, then d(r1(t), r2(t

′)) ≥ 2ρ.Choose n > 3ρ. Then γn = r0(n) is in K, and d(eΓ, γn) > eρ by construction of

r0. Since its in K, γn acts like the identity on Ends(Γ), and γn acts like an isometryon ∆. So γn(r1) is a geodesic ray based at γn. But end(γn(r1)) = e1 since γn actstrivially on Ends(Γ). Since γn lies in a different path component of X \ B(eΓ, ρ),there must be some t such that γn(r1(t)) ∈ B(eΓ, ρ). Since n > 3ρ, t > 2ρ.

Similarly, there exists t′ > 2ρ so that γn(r2(t′)) ∈ B(eΓ, ρ). so that

d(γn(r1(t)), γn(r2(t′))) ≤ 2ρ.

Since γn acts like an isometry, d(r1(t), r2(t′)) ≤ 2ρ, a contradiction. So Γ cannot

have three distinct ends. �

9. October 5, 2010

Consider Ends(X). Recall that we defined end(rn) → end(r) to mean that forall compact C ⊂ X , there exist {Nn} such that rn([Nn,∞)) and r([Nn,∞)) lie inthe same path component of X \ C for sufficiently large n.

Proof continued from last time. If f : X1 → X2 is a QI, let fε : Ends(X1) → Ends(X2)be the induced map. To show continuity, we must show that if end(rn) → end(r)in X1, then fε(end(rn)) → fε(end(r)).

Claim. If rn and r are geodesic rays with rn([N,∞)) and r([N,∞)) in the same

path component of X1 \ B(x0,M), then there exists M so that f∗(rn), f∗(r) from

N to ∞ are in the same path component of X2 \B(x0, M).

Proof. Let p be a point on rn outside a ball of radius M , and let q be a point in routside a ball of radius M . Suppose f(p) is on f∗(rn) and f(q) is on f∗(r). Thend(f(x0), f(p) ≥ 1

λM − λ. Partition a geodesic path from f(p) to f(q) with points

γi which are each distance at most 1. Then f(γi) are at most 2λ apart. Connecting

these by geodesics and using the lemma from last time, if we take M = 1λM − 3λ,

the claim follows. �

Continuity follows. Recall other parts of the proof are still left to the reader(well-defined, inverse). �


Definition. For a finitely generated group Γ = 〈S〉, we define Ends(Γ) to beEnds(∆(Γ, S)) for any finite generating set S. (We need finite to know that ∆ is aproper space.)

Example. Let Γ1 = Zn for n ≥ 2. Then Γ1 has one end. Let Γ2 = Fn for n ≥ 2.Then |Ends(Γ2)| = ∞. So Zn ≁QI Fn for n ≥ 2.

Definition. A right-angled Coxeter group is a group with a presentation

〈a1, a2, . . . , an | a2i , [ai, aj ] for some i, j}.

That is, the group is finitely generated by elements of order 2, and any additionalrelations are given by commutators. We often right RACG for right-angled Coxetergroup.

We can represent a RACG by a graph as follows. We have one vertex for eachgenerator, and we connect two generators by an edge if their commutator is arelation.

Fact. If the graph is disconnected, then the group can be written as a free product.Any time there is a separating subgraph, we can write the group as an amalgamatedfree product.

Definition. If v is a vertex in a graph, then the link of v, denoted Lk(v), is the1-neighborhood of v. That is, Lk(v) is the subgraph consisting of v and all of theneighbors of v.

Theorem. A RACG is infinite ended if and only if there exists a complete sepa-rating subgraph K such that there exists a vertex v ∈ Γ \ K such that Lk(v) doesnot contain K.

Definition. Let X be a proper geodesic metric space. A quasi-geodesic in X isa quasi-isometric embedding ϕ : I → X of an interval (or, sometimes, I is theintersection of an interval with Z). Note: that I may be a ray or all of E1. In theformer case, ϕ is called a quasi-ray, and in the latter case, ϕ is called a quasi-line.

Fact. Let c : [0,∞) → E2 be given by c(t) = (t, log(1 + t)) = (r, θ) in polar coor-dinates. (This is the logarithmic spiral going counter-clockwise outwards from theorigin.) This is a quasi-ray, which is scary because it looks nothing like a geodesic.

Fact. The image of a geodesic under a QI is a quasi-geodesic.

Definition. Let (X, d) be a proper geodesic metric space. If a, b, c ∈ X , then ageodesic triangle with vertices a, b, c is a choice of geodesics [a, b], [b, c] and [a, c].

Definition. Let δ ≥ 0. A geodesic triangle in X is δ-slim if each side of the triangleis contained in the δ-neighborhood of the union of the other two sides.

Definition. Let δ ≥ 0. X is δ-hyperbolic if all geodesic triangles in X are δ-slim.X is hyperbolic if it is δ-hyperbolic for any δ.

Example. E2 is not δ-hyperbolic for any δ.

Remark. If the diameter of {a, b, c} is less than δ, then any geodesic triangle ona, b, c can be “anything”. (For example, they could look “fat”, or even spherical,and still be slim.)

Proposition. X is 0-hyperbolic if and only if X is a tree.

Example. (1) E1 is 0-hyperbolic.

(2) Hn is hyperbolic. (δ = log(1 +√

2)? Independent of n?)

18 ANDY EISENBERG

10. October 7, 2010

Recall our setting from last time. X is a proper geodesic metric space. X ishyperbolic if there exists a δ > 0 such that all geodesic triangles are δ-slim.

Theorem. If X is a proper, cocompact CAT(0) space, then X is hyperbolic if andonly if X does not contain an isometrically embedded E2.

The single most important result about hyperbolic spaces is:

Theorem (Stability of quasi-geodesics). For all δ ≥ 0, λ ≥ 1, thhere existsR(λ, δ) = R such that if X is δ-hyperbolic and c : [0, T ] → X is a λ quasi-geodesicwith c(0) = p and c(T ) = q, then the Hausdorff distance satisfies:

dH(im c, [p, q]) < R,

where [p, q] denotes any geodesic from p to q.

Remark. You can think of Hausdorff distance as follows: every x ∈ im c is withinR of a point on [p, q], and vice versa.

We may prove this later, but first we’ll see why it’s important. An immediateconsequence:

Corollary. Suppose X,Y are geodesic metric spaces, and let f : X → Y be a λquasi-isometric embedding. If Y is δ-hyperbolic, then X is δ′ = δ′(δ, λ) hyperbolic.Hence hyperbolicity is a QI invariant.

Proof. Let a, b, c be the vertices of a triangle in X . Let c1 be a geodesic connectinga to b; c2 connects b to c; and c3 connects c to a. Then f(a), f(b), f(c) are connectedby the images of the ci, which are quasi-geodesics. Since Y is a geodesic space, wecan draw a geodesic triangle between these vertices as well.

Take p ∈ c1. By the theorem, there is some point z on f(c1) within R of f(p).Since the geodesic triangle in Y is δ-slim, there is some point z′, for example, onf(c3) within δ of z. Then, by the theorem again, there is a point f(q) on f(c3)within R of z′. By the quasi-isometric embedding inequality:

d(p, q) < λ(2R+ δ) + λ2.

Now, taking δ′ = λ(2R+ δ) + λ2, the geodesic triangle in X is δ′-slim. �

Corollary. If X ∼QI Y , then X is hyperbolic if and only if Y is hyperbolic.

Definition. A finitely generated group is hyperbolic (or Gromov hyperbolic, or wordhyperbolic) if any Cayley graph of Γ is hyperbolic.

Example. Free groups are hyperbolic, since trees are 0-hyperbolic. Γ = π1(Sg) forg ≥ 2.

Remark. In general, it’s difficult to tell if the Cayley graph is hyperbolic. Thehard part is really understanding the algebra of geodesics. To show a given groupΓ is hyperbolic, we hope to find a nice hyperbolic metric space on which Γ actsgeometrically. Then we apply Svarc–Milnor.

Example. Consider Z ⊕ Z. Geodesics from (0, 0) to (p, q) can be arbitrarily farapart (as p and q get large). So Z⊕Z is not hyperbolic. This is an example wherewe fully understand the geodesics in the graph.


Theorem. If Γ is hyperbolic, then Γ contains no Z ⊕ Z subgroup.

Remark. In the graph representing a RACG, a copy of Z⊕Z would be representedby a non-chordal cycle of length 4:

a d

c b

Theorem (Moussong). If you don’t see the above in the defining graph, then thegroup is hyperbolic.

If Γ is any finitely presented group that does not contain a Z ⊕ Z, then is Γhyperbolic? No. Consider B(m,n) = 〈a, b | b−1amb = an〉 for m 6= n ≥ 1. Whatif Γ does not contain Z ⊕ Z or B(m,n)? No. The examples are much harder, bywork of N. Brady (1999 or 2000).

Open Question. Suppose Γ admits a finite K(Γ, 1) and does not contain B(m,n)for m,n ≥ 1. Then is Γ hyperbolic?

We’ll discuss various other formulations of hyperbolicity. For example, hyper-bolic Cayley graphs behave a lot like trees. Let x, y, z be a triangle in X . Thereexist a, b, c ≥ 0 such that d(x, y) = a+ b, d(x, z) = a+ c, and d(y, z) = b + c. Nowwe construct the graph:

y

xa

v

b��

c @@@@

@@@

z

The map which sends the triangle to this graph, we will denote by χ△. We can de-

termine the “thinness” of the triangle x, y, z by looking at the diameter diamχ−1△ (v).

Definition. The triangle x, y, z is δ-thin if for all t ∈ T (a, b, c), diamχ−1△ (t) ≤ δ.

We call diamχ−1△ (v) the insize of the triangle.

Proposition. The following are equivalent:

(1) There exist δ0 such that all geodesic triangles in X are δ0-slim.(2) There exist δ1 such that all geodesic triangles in X are δ1-thin.(3) There exist δ2 such that the insize of any geodesic triangle is at most δ2.

Suppose Γ = 〈A | R〉 is a finite presentation.

Problem 1 (Word problem). Can it be decided in a finite number of steps whethera given word w ∈ {A ∪A}∗ has ε(w) = eΓ?

Suppose you could find a set of elements u1, v1, u2, v2, . . . , un, vn a set of wordssuch that

(1) ε(ui) = ε(vi),(2) |vi| < |ui|, and(3) if a word w ∈ {A ∪ A}∗ has ε(w) = eΓ, then w contains at least one ui as

a subword.

20 ANDY EISENBERG

If the answer is yes, then we have a very simple algorithm: look through w andcheck for one of the ui. If found, replace with vi. Since this reduces the length,applying this check recursively yields an algorithm that is linear in the length of w.

Definition. If we have a finite list of words as above, we say 〈A | R〉 is a Dehnpresentation for Γ.

Theorem. If Γ is hyperbolic, then Γ has a Dehn presentation.

Remark. In fact, this is if and only if.

11. October 12, 2010

Proposition. Suppose c, c′ : [a, b] → X are two geodesics from x to y. If X is aδ-hyperbolic geodesic space, then im(c) ⊂ Nδ(im(c′)) and im(c′) ⊂ Nδ(im(c)), whereNδ(U) denotes the δ-neighborhood of a set U .

Proof. If d(x, y) = ℓ(c) = ℓ(c′) < 2δ, then the claim is trivial. So suppose d(x, y) >2δ. Let z ∈ im(c) such that d(x, z), d(z, y) > δ. Let q be the point distance δalong the geodesic towards y. Now we have a triangle {q, x, y} which is δ-slim byhyperbolicity. So either d(z, im(c′)) < δ, or d(z, q′) < δ for q′ along the geodesicfrom q to y. But the latter is impossible by construction, hence d(z, im(c′)) < δ. �

Definition. Let X be a geodesic space. Fix k > 0. A path c : [a, b] → X in X iscalled a k-local geodesic if d(c(t), c(t′)) = |t− t′| for all t, t′ ∈ [a, b] with |t− t′| ≤ k.

Theorem. Let X be a δ-hyperbolic geodesic space. Fix k > 8δ. Suppose c : [a, b] →X is a k-local geodesic.

(1) im(c) ⊂ N2δ([c(a), c(b)]). (Recall the notation [c(a), c(b)] denotes any geo-desic from c(a) to c(b).)

(2) [c(a), c(b)] ⊂ N3δ(im(c)).(3) c is a λ-quasi-geodesic for λ = λ(k, δ).

Proof. (1) Fix a geodesic [c(a), c(b)], which we will denote [c(a), c(b)]. If b−a <8δ, then c is a geodesic, so that im(c) ⊂ Nδ([c(a), c(b)]).

Let t ∈ [a, b] such that d(c(t), [c(a), c(b)]) is maximal, and denote c(t) =x. Choose a subinterval [a′, t] of [a, b] with length at most k/2 and atleast 4δ, and let y = c(a′). Choose an analgous subinterval [t, b′] and takez = c(b′). Draw geodesics from y and z to [c(a), c(b)], and denote theimages by y′ and z′, respectively.

Now the restriction of c to [a′, b′] is a geodesic, so we have a geodesicquadrilateral {y, z, z′, y′}. Draw a diagonal from y′ to z to get two geodesictriangles. There exists a point w ∈ [y, y′]∪[y′, z′]∪[z′, z] such that d(w, x) ≤2δ. We want w ∈ [y′, z′] to complete the proof of the first claim.

Suppose w ∈ [y, y′]. Then we claim there would be a path from x to y′

via w which is shorter than [y, y′]. We have

d(x, y′) − d(y, y′) ≤ (d(x,w) + d(w, y′)) − (d(y, w) + d(w, y′))

= d(x,w) − d(y, w)

≤ d(x,w) − (d(y, x) − d(x,w))

= 2d(x,w) − d(y, x)

< 4δ − 4δ.


That is, d(x, y′) < d(y, y′), contradicting our choice of t. Hence w /∈ [y, y′].An analogous argument shows w /∈ [z, z′], hence w ∈ [y′, z′] and we aredone.

(2) Pick an arbitrary point p ∈ [c(a), c(b)]. Any point of im(c) lies in eitherN2δ([c(a), p]) or N2δ([p, c(b)]). There exists some x ∈ im(c) in both neigh-borhoods, since im(c) is connected. Let q ∈ [c(a), p] and r ∈ [p, c(b)] suchthat d(q, x) < 2δ and d(x, r) < 2δ. Now {q, r, x} forms a δ-slim triangle.Without loss of generality, p is within δ of [q, x], hence d(p, x) ≤ 3δ, andwe are done.

(3) We will not prove the third claim today.�

Corollary. Let X be a δ-hyperbolic geodesic space, and let k be a fixed positiveconstant greater than 8δ. Let c : [a, b] → X be a k-local geodesic. Then either c isa constant map at c(a) = c(b), or c(a) 6= c(b).

Proof. Without loss of generality, let a = 0. Suppose c(a) = c(b). Then im(c) ⊂N2δ(c(a)) by the theorem. If there exists t ∈ [0, b] with 8δ < t ≤ k, thend(c(a), c(t)) > 8δ, a contradiction. This implies c is geodesic, and therefore con-stant. �

Theorem. Any hyperbolic group has a Dehn presentation.

Proof. Consider ∆ = ∆(Γ, S) for some δ-hyperbolic group Γ. (Suppose ∆ is theδ-hyperbolic Cayley graph.) Fix k > 8δ. Any edge loop of length greater than 8δin ∆ will have a subpath of length in (8δ, k] that is not geodesic. This will allow usto build a Dehn algorithm.

Let U be the set of all reduced words in {S ∪ S}∗ of length at most k. Observethat U is a finite set. We need to construct a finite list of words u1, . . . , um anddv1, . . . , vm such that:

(1) ε(ui) = ε(vi),(2) |vi| < |ui|, and(3) if w is a word with ε(w) = eΓ, then w contains some ui.

If R = {uiv−1i }, then 〈S | R〉 is a presentation for Γ.

For each ui ∈ U , choose vi to be a geodesic with ε(ui) = ε(vi). If ui is a geodesic,we leave it out of the list. This gives a Dehn algorithm. �

Remark. Let p ∈ E2, and let c1, c2 be geodesics leaving p. Fix t, and connect c1(t)and c2(t) by a path outside B(p, t) of length at most πt.

Now consider p, c1, c2 in H2. How long is a path from c1(t) to c2(t) that liesoutside B(p, t)? It turns out it’s exponential in t.

Definition. Let X be a geodesic space. We say e : N → R is a divergence functionfor X if for all R, r ∈ N and all geodesics c1 : [0, a1] → X , c2 : [0, a2] → X satisfying

(1) c1(0) = c2(0) =: x,(2) R+ r ≤ min{a1, a2},(3) d(c1(R), c2(R)) > e(0),

any path connecting c1(R+ r) to c2(R+ r) outside B(x,R+ r) has length at leaste(r).

22 ANDY EISENBERG

12. October 14, 2010

The following are some consequences of having a Dehn presentation.

Proposition. Suppose Γ = 〈S | R〉 has a Dehn presentation. Then Γ has onlyfinitely many conjugacy classes of finite order elements.

Proof. Suppose g ∈ Γ has order n. Let [g] denote the conjugacy class of g. Let〈S | R〉 be a Dehn presentation of Γ. Let w ∈ {S∪S}∗ be shortest among all wordsv with ε(v) ∈ [g]. We know that ε(wn) = eΓ.

Because we have a Dehn presentation, there exists a relator r ∈ R such that thepath wn contains a subword that consists of more than half of r.

Claim. ℓ(w) < ℓ(r).

Proof. HW. �

Given the claim, take k = max{ℓ(r) | r ∈ R}. Then every element of finite orderis conjugate to some element of length at most k. Since there are finitely manysuch elements, this completes the proof. �

Definition. Let Γ = 〈S | R〉 be a finite presentation. If w is a freely reduced wordin {S ∪ S}∗ of length ℓ(w) and ε(w) = eΓ, then we can write

w =

N∏

i=1

pirǫi

i p−1i

where ǫ = ±1, and pi ∈ {S ∪S}∗. If there exists a constant k such that for all suchw, N < kℓ(w), then we say that this presentation satisfies a linear isoperimetricinequality.

Proposition. If Γ has a Dehn presentation 〈S | R〉, then this presentation satisfiesa linear isoperimetric inequality with k = 1.

Remark. In fact, the converse is true. A group is word hyperbolic if and only ifit has a Dehn presentation if and only if it has a presentation satisfying the linearisoperimetric inequality.

Example. Let Γ = 〈x, y | r1, r2〉 ∼= Z × Z3, where r1 = x3 and r2 = xyx−1y−1.Let w = y2x−1y−1x−1y−1x−1. Then we can draw w as

y //

x

��

y //

y//

x

OO

//

OO

We can fill in this picture in the Cayley graph:

y //

x

��

y //OOOO

//y

//

OOOO

x

OOOO

//

OOOO OOOO


The three squares are r2, and the half circle is r1. So

w = (x−1r2x)(x−1yr2y

−1x)(x−2r2x2)(r−1

1 ).

Definition. Let X be a meteric space. A path is a continuous map c : [a, b] → X .The length of the path is

ℓ(c) = supa=t0≤···≤tn=b

n−1∑

i=0

d(c(ti), c(ti+1)).

Note that this could be infinite. If it is finite, we say that c is rectifiable.

Fact. If c : [a, b] → X is rectifiable with ℓ(c) = ℓ, then we can reparametrize c asc : [0, ℓ] → X in such a way that ℓ(c|[0,t]) = t.

Remark. We really want to think of this reparametrization as [0, 1]λ−→ [0, ℓ]

c−→ X ,where λ is linear. By an abuse of notation, we will call the composition c. Thiscomposition should be parametrized proportional to arc length.

Proposition. Let X be a δ-hyperbolic geodesic space, let c : [0, 1] → X be a recti-fiable path parametrized proportional to arc length, and take p = c(0), q = c(1). If[p, q] is a geodesic between p and q in X, then for all x ∈ [p, q],

d(x, im(c)) ≤ δ| log2 ℓ(c)| + 1.

Proof. If ℓ(c) ≤ 1, then this is trivial: if ℓ(c) ≤ 1, then d(p, q) ≤ 1, and the righthand side of the inequality can never be less than 1.

Suppose ℓ(c) > 1. Let N be the integer such that ℓ(c)/2N+1 < 1 ≤ ℓ(c)/2N .Consider the δ-thin triangle {p, c(1/2), q}. There exists y1 ∈ [p, c(1/2)]∪ [c(1/2), q]such that d(x, y1) < δ. Without loss of generality, suppose y1 ∈ [p, c(1/2)]. Nowdraw the δ-thin triangle {p, c(1/4), c(1/2)]. Then, without loss of generality, thereis some y2 ∈ [c(1/4), c(1/2)] such that d(y1, y2) < δ.

Proceed by induction so that at the n+1 stage, we have yn ∈ [c(tn), c(t′n)]. To getyn+1, we build triangle △n+1 as {c(tn), c(t′n), c((tn +t′n)/2)}. Then d(yn, yn+1) < δ.

At the Nth stage, we have yN with d(yN , x) ≤ δN . Now yN is along a geodesicfrom c(tN ) to c(t′N ). Suppose without loss of generality that yN is closest to c(tN ).Then

d(x, c(tN )) ≤ d(x, yN ) + d(yN , c(tN )) ≤ δN + ℓ(c)/2N+1 ≤ δ| log2 ℓ(c)| + 1.

�

Proposition. If X is a δ-hyperbolic geodesic space, then X has an exponentialconvergence function.

Proof. Let

e(r) = max{3δ, 2 r−1

δ }.Suppose d(c1(R), c2(R)) > 3δ for geodesics c1, c2 starting at some point x0. Con-sider the δ-thin triangle {x0, c1(R + r), c2(R + r)}. From the definition of δ-thin,we can find i1 ∈ [x0, c1(R+ r)], i2 ∈ [x0, c2(R+ r)], and i3 ∈ [c1(R+ r), c2(R+ r)]such that

d(x0, c1(R+ r)) = d(x0, i1) + d(i1, c1(R+ r))

d(x0, c2(R+ r)) = d(x0, i2) + d(i2, c2(R+ r))

d(c1(R + r), c2(R+ r)) = d(c1(R + r), i3) + d(i3, c2(R + r))

24 ANDY EISENBERG

Also note that, in particular, d(ik, ij) ≤ δ for i, j ∈ {1, 2, 3}.

Claim. d(x0, i1) < d(x0, c1(R)) (and similarly for i2 and c2).

Proof. Otherwise, when we collapse the triangle to a tripod, we have:

c1(R+ r)

x0 c∗(R) v

wwwwwwwwww

GGGG

GGGG

GG

c2(R+ r)

where the preimage of v under the collapsing map is {i1, i2, i3} and the preimageof c∗(R) is {c1(R), c2(R)}. But then by the definition of δ-thin, we must have thatthe preimage of c∗(R) has diameter at most δ, contradicting the assumption thatd(c1(R), c2(R)) ≥ 3δ. �

Claim. d(i1, x0) = d(i2, x0) < R − δ.

Proof. Now suppose d(x0, i1) = d(x0, i2) ≥ R− δ. We have:

d(x0, c1(R)) = d(x0, i1) + d(i1, c1(R))

⇒ R ≥ R − δ + d(i1, c1(R))

⇒ δ ≥ d(i1, c1(R)).

Similarly, δ ≥ d(i2, c2(R)). By the triangle inequality:

d(c1(R), c2(R)) ≤ d(c1(R), i1)︸︷︷︸≤δ

+ d(i1, i2)︸︷︷︸<δ

+ d(i2, c2(R))︸︷︷︸≤δ

< 3δ,

contradicting the assumption that d(c1(R), c2(R)) ≥ 3δ. This completes the proofof the claim. �

Claim. d(x0, i3) < R.

Proof. By the triangle inequality: d(x0, i3) ≤ d(i3, i1) + d(i1, x0). �

Now we have B(i3, r) ⊂ B(x0, R + r). Let γ be a rectifiable path outside ofB(x0, R + r) from c1(R + r) to c2(R + r). (If the path is not rectifiable, theconclusion is trivial.) By the proposition,

r < d(i3, im(γ)) ≤ δ| log2 ℓ(γ)| + 1.

So (r−1)/δ ≤ | log2 ℓ(γ)|. If ℓ(γ) ≥ 1, then by exponentiating we have 2r−1

δ ≤ ℓ(γ).It remains to show that ℓ(γ) ≥ 1.


The triangle {x0, c1(R+ r), c2(R + r)} collapses to the following tripod:

c1(R + r)

x0a1

a2wwwwwwwwww

a3 GGGG

GGGG

GG

c2(R + r)

Note that since c1 and c2 are geodesics, we have a2 = a3. Furthermore, by the firstclaim, we know that the image of c1(R) under the collapsing map must lie along theedge marked a2, and the image of c2(R) must lie along the edge marked a3. Sinced(c1(R), c1(R + r)) = r, and similarly for c2, we have that a2 = a3 > r. It followsthat d(c1(R + r), c2(R+ r)) ≥ 2r > 1, hence ℓ(γ) > 1, completing the proof. �

13. October 19, 2010

From last time, you can show a group is not hyperbolic by showing that theDehn function is not linear. But this isn’t much easier than showing that theCayley graph is not hyperbolic.

Proposition. Suppose Γ is a hyperbolic group. Let γ ∈ Γ have infinite order,∆ = ∆(Γ, S) be any Cayley graph, and α a geodesic from eΓ to γ. Then the bi-infinite path

(. . . , γ−1α, α, γα, γ2α, . . . )

is a quasi-geodesic line. As a consequence, the map Z → γ given by n 7→ γn is aQI embedding.

Example. Consider Γ = BS(1, 2) = 〈a, b | aba−1 = b2〉. We claim that 〈b〉 doesnot determine a quasi-geodesic line in ∆(Γ, S) for S = {a, b}.

First notice that anba−n = b2n

. In the subgroup 〈b〉, the element b2n

has length2n. But in the full group, it has length at most 2n+ 1, so 〈b〉 does not determinea quasi-geodesic line. It follows that BS(1, 2) is not a hyperbolic group.

Fact. If Γ is hyperbolic and γ ∈ Γ has infinite order, then |p| = |q| when γp isconjugate to γq.

Proof. Proof by picture (draw the brick wall). Finish on your own. (Hint: qm−1

copies of γq are conjugate via t to qm−1 copies of γp.) �

Conjecture. If Γ has a finite K(π, 1) and does not contain B(p, q) = 〈a, b |abpa−1 = bq〉, then Γ is hyperbolic.

Definition. Recall that a K(Γ, 1) is a CW complex with fundamental group iso-morphic to Γ, and contractible universal cover.

Fact. A finitely generated Abelian group is hyperbolic if and only if it contains acyclic subgroup of finite index.

Proof. If Γ contains a cyclic subgroup of finite index, then it is either finite or2-ended. In either case, it is hyperbolic.

Conversely, if Γ is finitely generated and Abelian, then by the structure theoremeither Γ is finite or virtually Zn. The former case is trivial. In the latter case,Γ ∼QI En, which is hyperbolic only when n = 1. So Γ is virtually Z. �

26 ANDY EISENBERG

Theorem (N. Brady). Not every finitely presented subgroup of a hyperbolic groupis hyperbolic.

Remark. However, every subgroup of a free group is free.

Fact. If N is a finitely generated, nontrivial, normal subgroup of a finitely generatedfree group F , then [F : N ] <∞. (Original proof due to Greenberg.)

Definition. Let X be a geodesic space. A subset C ⊂ X is called quasi-convex ifthere exists a k > 0 such that for all x, y ∈ C, each geodesic satisfies [x, y] ⊂ Nk(C).

Lemma. Suppose Γ = 〈A〉 is any finitely generated group and H ≤ Γ. If H isquasi-convex in ∆ = ∆(Γ, A), then H is finitely generated and H → Γ is a QI-embedding.

Proof. Let h ∈ H , and suppose h = a1 · · ·an is a geodesic for h ∈ ∆(Γ, A). Byquasi-convexity, this geodesic is in Nk(H). For each i, choose geodesics ui of lengthat most k connecting ai to hi ∈ H . Then hi = ui−1hiu

−1i . Then we can write

h = h1 . . . , hn, and for each i, |hi|A ≤ 2k + 1. Taking S = B(eΓ, 2k + 1) in ∆ ∩H ,we have a finite generating set for H .

From the above, we have dS(eΓ, h) ≤ dA(eΓ, h), since for any path of length nin the A metric, we have constructed a path of length n in the S metric. On theother hand, if dA(eΓ, h) ≤ 2k + 1, then dS(eΓ, h) = 1 by choice of S. So

dS(eΓ, h) ≤ dA(eΓ, h) ≤ (2k + 1)dS(eΓ, h).

This shows that (H,S) → (Γ, A) is a QI embedding. �

Proposition. If (Γ, A) is hyperbolic and (H,B) is a finitely generated subgroup,then H is quasi-convex if and only if H is QI embedded in Γ.

Corollary. If H is quasi-convex in (Γ, S), then H is quasi-convex in (Γ, T ) for anyfinite generating sets S, T .

14. October 21, 2010

Recall the lemma from last time:

Lemma. Let (Γ, A) be a finitely generated group and H ≤ Γ a subgroup. If H isquasi-convex in ∆(Γ, A), then H is finitely generated and H → Γ is a QI embedding.

Corollary. If (Γ, A) is hyperbolic with finite generating set A and (H,B) is afinitely generated subgroup of Γ, then:

(1) if H is quasi-convex with respect to one finite generating set, then it isquasi-convex with respect to all finite generating sets, so that we can say His quasi-convex in Γ; and

(2) H is quasi-convex in Γ if and only if H is QI embedded in Γ via the inclusionmap.

Proof. We will prove the second claim first. By the lemma, if H is quasi-convexthen it is QI embedded. Conversely, suppose H is QI embedded. We have aninduced inclusion map ∆H = ∆(H,B) → i∆Γ = ∆(Γ, A). Let [h1, h2]H be ageodesic in ∆H . Then, since the inclusion is a QI embedding, the image [h1, h2]Γis a quasi-geodesic in ∆Γ. From the stability of quasi-geodesics, we know thereexists R > 0 such that [h1, h2]Γ ⊂ NR(i([h1, h2]H)). Taking k = R + λ

2 , we have[h1, h2]Γ ⊂ Nk(H), and H is quasi-convex.


The first claim follows from the second, since all Cayley graphs are QI to eachother. �

Proposition. Suppose Γ is hyperbolic. If H ≤ Γ is quasi-convex, then H is finitelygenerated and QI embedded. Furthermore, H is hyperbolic.

Proof. It only remains to show that H is hyperbolic. We have seen previously thatif X → X ′ is a QI embedding and X ′ is hyperbolic, then X is hyperbolic. TakingH = X and Γ = X ′, H is hyperbolic. �

Exercise. Every finitely generated subgroup of a finitely generated free group isquasi-convex. (For HW.)

Exercise (Much harder, but fun). Every finitely generated subgroup of a hyper-bolic surface group (that is, π1(Sg) for g ≥ 2) is quasi-convex.

Remark. This is not true in general for hyperbolic groups. In fact, there exist ahyperbolic group Γ and finitely presented subgroupH such thatH is not hyperbolic.(N. Brady)

Example (Cannon–Thurston). It is also possible to have hyperbolic subgroups ofa hyperbolic group that are not quasi-convex (that is, that are not QI embedded).Consider Γ = π1(M

3) for a compact, hyperbolic 3-manifold M3. Take H = π1(Sg)to be a surface subgroup with g ≥ 2. In fact, H is normal.

Lemma. If H1, H2 are quasi-convex in Γ, then H1 ∩H2 is quasi-convex in Γ.

Proof. Left to the reader. �

Proposition. If Γ is hyperbolic and γ ∈ Γ, then CΓ(γ) is quasi-convex.

Corollary. If Γ is hyperbolic and γ ∈ Γ has infinite order, then Z → Γ given byn 7→ γn is a QI embedding.

Proof. It is sufficient to show that 〈γ〉 is quasi-convex in Γ. Suppose the orderof γ is infinite. CΓ(γ) is quasi-convex by the proposition, and therefore finitelygenerated by, say, S. Now

Z(CΓ(S)) =⋂

s∈S

CΓ(s)

is finitely generated, Abelian, and hyperbolic. So Z(CΓ(γ)) is virtually cyclic. Since〈γ〉 ⊂ Z(CΓ(γ)), so it is finite index in this center. (In fact, 〈γ〉 and Z(CΓ(γ)) areboth 2-ended, and by the inclusion, they are QI.) Therefore

〈γ〉 → Z(CΓ(γ)) → C(γ) → Γ

is a chain of QI embeddings, and is therefore a QI embedding. �

Corollary. Let Γ be hyperbolic and γ ∈ Γ have infinite order. There exists aconstant L such that for all x on a geodesic [γi, γj ] there exists k such that d(x, γk) <L.

Proof. Let α be a geodesic from eΓ to γ. Connect γi to γj by a quasi-geodesic bytaking geodesics γi ·α, γi+1 ·α, and so on. Fix a geodesic [γi, γj] and take x on thegeodesic. Project x to a point on the quasi-geodesic at distance at most R (fromthe stability of quasi-geodesics). This image is between two consecutive powers ofγ. Take L = R+ ℓ(γ)/2. �

28 ANDY EISENBERG

Corollary. If Γ is hyperbolic and γ ∈ Γ has infinite order, then |CΓ(γ)/〈γ〉| isfinite. In particular, this shows that there are no Z ⊕ Z subgroups in CΓ(γ), henceCΓ(γ) is 2-ended.

Proof. Let s ∈ CΓ(γ). It suffices to show that every coset s〈γ〉 intersectsB(eΓ, rL+ 2δ),where the L comes from the previous corollary.

Let s ∈ CΓ(γ). Choose m such that d(eΓ, γm) > 2ℓ(s) + 2δ. (We can do this

because γ is infinite order. We have the geodesic quadrilateral:

s a sγm

eΓ b γm

We claim that d(a, b) is less than 2δ. If there exists p ∈ [eΓ, s] and q ∈ [γm, sγm]with d(p, q) < 2δ, then d(eγ , γ

m) < 2δ + 2ℓ(s), a contradiction. So we can choosea, b as in the picture.

By the previous corollar, there are some i, j such that d(a, sγj) < L and d(b, γi) <L. Now d(eΓ, sγ

j−i) = d(γi, sγj) ≤ 2L + 2δ. (The first equality follows from thefact that γi acts like an isometry.) This completes the proof. �

15. October 26-28, 2010

Remark. Let g ∈ Γ, and let ϕg be the automorphism given by conjugation by g.Then CΓ(g) = Fix(ϕg). By the following theorem, centralizers in hyperbolic groupsare quasi-convex.

Theorem (Neumann). If Γ is hyperbolic and ϕ ∈ Aut(Γ), then Fix(ϕ) is quasi-convex.

Theorem (Gersten, 70s). If F is a finitely generated free group and ϕ ∈ Aut(F ),then Fix(ϕ) is finitely generated.

Proof. Long and technical combinatorial group theory. �

Remark. Daryl Cooper gave a GGT proof of this last result that is much nicer.

Theorem. Let Γ be hyperbolic, let g ∈ Γ have infinite order, and let α = [e, g] bea choice of geodesic in a Cayley graph ∆. Define β = (. . . , g−1α, α, gα, . . . ) to be abi-infinite path. Then β is a quasi-geodesic.

Proof. We want to show:

1

λdβ(x, y) − λ ≤ d∆(x, y) ≤ λdβ(x, y) + λ.

We have d∆(x, y) ≤ dβ(x, y) trivially. It remains to find λ such that dβ(x, y) ≤λd∆(x, y) + λ.

Claim. Let R ∈ Z+. Let k ∈ Z such that d(e, gk) > 8R+ 2δ. Fix a geodesic frome to gk, and let y be the midpoint of this geodesic. Let I be the open subsegment ofthis geodesic of length 2R centered at y. Then for any p ∈ BR(e) and q ∈ BR(gk),the midpoint m1 of [p, q] lies within 2δ of I:

d(m1, I) < 2δ.


Proof. In the following, consult Figure 1 for a visual aid. We have a 2δ-thin geodesicrectangle {e, p, q, gk}. Fix a diagonal [p, gk] to form two δ-thin triangles {p, q, gk}and {e, p, gk}. If we draw the internal points a ∈ [q, gk], b ∈ [p, q], c ∈ [p, gk] for thetriangle {p, q, gk}, then we have d(a, b) < δ and d(a, c) < δ. So b, c ∈ BR+δ(g

k).Similarly, the internal points for the triangle {e, p, gk} lie in BR+δ(e).

Let m1 be the midpoint of [p, q]. Observe that d(p,m1) = d(q,m1) > 3R +δ. Let m′

1 ∈ [p, gk] such that d(p,m′1) = d(p,m1). Observe that, by δ-thinness,

d(m1,m′1) < δ. Let m2 be the midpoint of [p, gk]. We have d(m2, g

k) > 72R+ δ.

We now claim that d(m′1,m2) < R/2. By the reverse triangle inequality, we have

|d(p, q) − d(p, gk)| ≤ d(gk, q) < R. It follows that:

|d(p, q) − d(p, gk)| < R

⇒∣∣[d(p,m1) + d(m1, q)] − [d(p,m2) + d(m2, g

k)]∣∣ < R

⇒ |2d(p,m1) − 2d(p,m2)| < R

⇒ |d(p,m1) − d(p,m2)| < R/2

⇒ |d(p,m′1) − d(p,m2)| < R/2.

Let m′2 ∈ [e, gk] so that d(gk,m2) = d(gk,m′

2). By an argument analogous to theabove, d(m′

2, y) < R/2. Let m′′1 ∈ [e, gk] so that d(gk,m′

1) = d(gk,m′′1). Observe

that d(m′1,m

′′1) < δ and d(m2,m

′2) < δ by δ-thinness. Now:

d(m′′1 ,m

′2) = d(m′

1,m2) < R/2.

By the triangle inequality, d(m′′1 , y) < R. Now d(m1, I) ≤ d(m1,m

′′1) < 2δ, com-

pleting the proof of the claim. �

e

p

gk

q

y

m1

m′1

m′′1

m2

m′2

R

R+ δ

R

R+ δ

a

b

c

Figure 1. Picture to help with the proof of the first claim. Thispicture is not general.

Now let N be the number of vertices in B2δ(e). I has at most 2R vertices,and each 2δ-neighborhood of a vertex contains N vertices. Therefore N2δ(I) has atmost 2NR vertices. Consider the translates of [e, gk] by the elements e, g, . . . , g2NR.The midpoints of these segments must all be distinct, otherwise g would fix a point(which is impossible since g has infinite order). Since there are 2NR + 1 suchmidpoints, not all of them can be in N2δ(I). That is, there exists some powerp(R) ≤ 2NR such that gp(R) /∈ BR(e).

Observe that d(e, gp(R)) > R, and dβ(e, gp(R)) = |g|p(R). Hence p(R) > R/|g|.

30 ANDY EISENBERG

Claim. For all R ∈ Z, d(e, g2NR) ≥ R.

Proof. The claim is trivial for R ≤ 0. Suppose the claim is false. Then there existssome R0 > 0 such that d(e, g2NR0) < R0. Since the inequality is strict, there existsǫ > 0 such that d(e, g2NR0 < R0 − ǫ. (We will need this extra wiggle room later.)Let n ∈ Z+ be large and 0 ≤ R1 < R0, and define s = n(2NR0) +R1. We have

d(e, gs) = d(e, gn(2NR0)+R1)

≤ d(e, gn(2NR0)) + d(e, gR1)

≤ nd(e, g2NR0) + d(e, gR1)

< n(R0 − ǫ) + d(e, gR1)

= nR0 − nǫ+ d(e, gR1).

That is, for all n ∈ Z+ and some fixed R1 as above, d(e, gn(2NR0)+R1) < nR0 −nǫ+ d(e, gR1). Note that d(e, gR1) does not depend on n, and since ǫ is fixed theterm nǫ can be made as large as we want by raising n. It follows that there is somen0 ≫ 0 such that for all n ≥ n0,

d(e, gn(2NR0)+R1) < nR0.(1)

Take n > n0 and choose R > n|g|2NR0. By our previous discussion, there existsp(R) ≤ 2NR such that d(e, gp(R)) > R > nR0. But we have p(R) > R/|g| >n(2NR0), so d(e, gp(R)) < nR0 by Equation (1), a contradiction. This completesthe proof of the claim. �

We will now show that β is a quasi-geodesic. Pick x, y ∈ β. Recall that itremains to find λ such that dβ(x, y) ≤ λd(x, y) + λ.

There exists a ∈ Z such that d(x, g2aN ) ≤ N |g|. Similarly, there is some b ∈ Z

so that d(y, g2bN ) ≤ N |g|. Observe that, by the second claim, d(g2aN , g2bN ) =d(e, g2N |b−a|) ≥ 2|b− a|. We have

dβ(x, y) ≤ d(x, g2aN ) + d(g2aN , g2bN) + d(g2bN , y)

≤ N |g| + 2N |b− a||g| +N |g|= 2N |b− a||g| + 2N |g|.

We also have

2|b− a| ≤ d(g2aN , g2bN )

≤ d(g2aN , x) + d(x, y) + d(y, g2bN )

≤ N |g| + d(x, y) +N |g|= 2N |g|+ d(x, y).

Now

dβ(x, y) ≤ 2N |b− a||g| + 2N |g| (first calculation)

= N |g|(2|b− a|) + 2N |g|≤ N |g| [2N |g|+ d(x, y)] + 2N |g| (second calculation)

= 2N2|g|2 +N |g|d(x, y) + 2N |g|= N |g|d(x, y) + (2N2|g|2 + 2N |g|)

Choosing λ = 2N2|g|2 + 2N |g| completes the proof. �


Corollary. If γ has infinite order, 〈γ〉 ∼= Z is quasi-convex.

Remark. Recall that we have already used this to show that for infinite order γ,|C(γ)/〈γ〉| < ∞. This shows that C(γ) is 2-ended, and it follows that C(γ) isquasi-convex.

Remark. In the defining graph of a word hyperbolic right angled Coxeter group,if v represents some generator, the centralizer of v is the subgroup generated bySt(v). (Note: the star of v is the link of v and v itself.)

Theorem. Given any element g ∈ Γ of a hyperbolic group, CΓ(g) is quasi-convex.

One proof of this theorem shows that the fixed subgroup of an automorphismFix(ϕ) is quasi-convex. Then CΓ(g) = Fix(ϕg), where ϕg is conjugation by g.Another proof uses the following lemma, which also solves the conjugacy problem.

Proposition. Let (Γ, S) be a hyperbolic group with finite generating set. Thereexists a constant c0 = c0(δ, |S|) such that if a, b ∈ Γ are conjugate, then there existsx ∈ Γ with x−1ax = b and |x|S ≤ |a|S + |b|S + c0.

16. November 2, 2010

Remark (Philosophy). Let (Γ, A) be a finitely presented group. We can also thinkof Γ as π1(K) for some complex K. Algebraically, the word problem is aboutrecognizing a word w ∈ {A∪A}∗ has ε(w) = eΓ. Geometrically, the world problemis about recognizing whether a given loop is homotopically trivial.

Algebraically, the conjugacy problem is about recognizing when two words u,w ∈{A ∪ A}∗ have ε(u) = gε(w)g−1 for some g ∈ Γ. Geometrically, the conjugacyproblem is about recognizing when two loops are homotopic.

When K has some “nice” geometry, then there are well-known geometric tech-niques for solving the geometric versions of these problems. (“Nice” means thatK is either negatively curved or non-positively curved. For us, negatively curvedmeans δ-hyperbolic.)

We’ll see more of this philosophy on Thursday. Today’s goal is the following:

Proposition. Suppose Γ is a hyperbolic group and g ∈ Γ. Then CΓ(g) is quasi-convex.

Remark. We already know this is true for infinite order g, since 〈g〉 is quasi-convexand |CΓ(g)/〈g〉| is finite.

Lemma. There exists a constant K = K(δ) such that if σ, σ′ are geodesics fromx to y and x′ to y′, respectively, (and by reparameterizing one of them, we mayassume they are maps from the same interval) in ∆(Γ, S) that begin and end oneunit apart, then d(σ(t), σ′(t)) ≤ K for all t.

Proof. 1 For all t, there exists w ∈ [x′, y′] such that d(σ(t), w) ≤ 2δ + 1. Then

w = σ′(t) for some t. (We choose the smallest t with this property.) Without loss

of generality, t ≤ t (possibly by switching the roles of σ and σ′). We claim that

t− t ≤ 2δ + 2:

t = d(x′, w)

≤ d(x′, x) + d(x, σ(t)) + d(σ(t), w)

≤ 1 + t+ 2δ + 1.

32 ANDY EISENBERG

Now

d(σ(t), σ′(t)) ≤ d(σ(t), w) + d(w, σ′(t))

≤ 2δ + 1 + (t− t)

≤ 2δ + 1 + 2δ + 2

= 4δ + 3.

Taking K = 4δ + 3, we are done. �

Lemma. There exists a constant K = K(δ) such that if σ, σ′ are geodesics from x toy and x′ to y′, respectively, in ∆(Γ, S), then d(σ(t), σ′(t)) ≤ Kmax{d(x, x′), d(y, y′)}for all t.

Proof. This follows directly from the previous lemma. �

Proposition (Bridson and Howie). Let Γ = 〈S〉. There exists a constant c0 (de-pending on δ and |S|) such that if a, b ∈ Γ are conjugate, then there exists an x ∈ Γsuch that x−1ax = b and |x|S := |x| ≤ |a| + |b| + c0.

Proof. Since a, b are conjugate, choose some conjugating word x ∈ Γ (that is,x−1ax = b) of minimal length |x| = n. Let σx be a geodesic [e, x]. Then aσx is ageodesic [a, ax = xb]. Then we have a rectangle as in the second lemma:

a ax = xb

e

|a|

x

|b|

Each vertex σx(i) along σx lies within 2δ of one of the other three sides of thequadrilateral.

Claim. If i satisfies2δ + |a| < i < n− 2δ − |b|

then there exists v ∈ aσx such that d(σx(i), v) ≤ 2δ.

Proof. If there exists p ∈ [e, a] such that d(p, σx(i)) < 2δ, then

i = d(e, σx(i)) ≤ d(σx(i), p) + d(p, e) ≤ 2δ + |a|.Similarly, if p ∈ [x, xb] with d(σx(i), p) ≤ 2δ, then

n− i = d(σx(i), x) ≤ d(σx(i), p) + d(p, x) ≤ 2δ + |b|. �

For any i in this range, there exists a vertex v ∈ [a, ax] such that d(σx(i), v) ≤ 2δ.We can write v = aσx(j(i)).

Claim. |j(i) − i| ≤ 2δ.

Proof. Suppose without loss of generality, j(i) > i. If the claim is false, then

|ax| ≤ d(e, σx(i)) + d(σx(i), v) + d(v, ax)

≤ i+ 2δ + (n− j(i))

< n.

Since ax also conjugates a to b, this contradicts the minimality of the length ofx. �

Now we have d(σx(i), aσx(i)) ≤ 4δ.


Claim. The group elements σx(i)−1aσx(i) are distinct for i in the appropriaterange.

Proof. Suppose not. Then σx(i)−1aσx(i) = σx(j)−1aσx(j) for some j > i. Let

x′ = σx(i)σx(j)−1x ∈ Γ.

We will show that ax′ = x′b and |x′| < |x|, contradicting minimality of the lengthof x. We have

ax′ = aσx(i)σx(j)−1x

= σx(i)σx(j)−1 ax︸︷︷︸=xb

= σx(i)σx(j)−1x︸︷︷︸=x′

b

= x′b.

Write out x = s1s2 · · · sn, σx(i) = s1s2 · · · si, and σx(j) = s1 · · · sj . Then

x′ = (s1 · · · si)(s−1j · · · s−1

1 )s1 · · · sn

= s1 · · · sisj+1 · · · sn.

So |x′| ≤ n− (j − i). �

Recall that the word length of each σx(i)−1aσx(i) is at most 4δ, so the set{σx(i)−1aσx(i)} for i in the appropriate range is in B(4δ, eΓ). Hence the number ofpossible i’s, n− 2δ− |b| − (2δ+ |a|), is at most the number of vertices in B(4δ, eΓ).If V is the number of vertices in B(4δ, eΓ), then

n ≤ V + |a| + |b| + 4δ.

Taking c0 = V + 4δ, we are done. �

Proof of the first proposition. Let γ ∈ CΓ(g). We must show that there exists Qsuch that each vertex along [e, γ] is within Q of a point in CΓ(g). We build thefollowing geodesic quadrilateral:

ggσg

gσ = σg

e σ

There exists K such that d(σγ(i), gσγ(i)) ≤ K|g|. Let h = σγ(i)−1gσγ(i). Then hand g are conjugate, so there exists x ∈ Γ with hx = xg, and |x| ≤ |g| + |h| + c0 ≤|g| +K|g|+ c0 =: Q.

We claim that σγ(i)x ∈ CΓ(g). We have

σγ(i)xg = σγ(i)hx

= σγ(i)σγ(i)−1gσγ(i)x

= gσγ(i)x.

This completes the proof. �

34 ANDY EISENBERG


Proposition. If N is a non-trivial, finitely generated, normal subgroup of a finitelygenerated free group F , then [F : N ] <∞.

A generalization is Greenberg’s lemma:

Proposition. If Γ is hyperbolic and H is an infinite, quasi-convex subgroup, then[NΓ(H) : H ] <∞. In particular, if H is also normal, then [G : H ] <∞.

Theorem. If H is an infinite subgroup of a hyperbolic group Γ, then H containsan element of infinite order.

Open Question. Do all hyperbolic groups have a finite index, torsion-free sub-group?

Open Question. Can a CAT(0) group contain an infinite torsion subgroup?

Definition. Suppose (Γ, A) is a finitely generated group. We say that Γ has thequasi-monotone conjugacy property (we wiill abbreviate this as QMC) if there existsa constant K = K(A) such that, whenever u, v ∈ F (A) with ε(u) conjugate to ε(v)in Γ, there exists a word w = a1 · · ·an with w−1uw = v in Γ and

d(eΓ, w−1i uwi) ≤ Kmax{|u|, |v|},

where wi = a1 · · · ai for i = 1, . . . , n.

Proposition. Suppose (Γ, A) is a finitely generated group. If Γ has solvable wordproblem and the QMC property, then Γ has solvable conjugacy problem.

Proof. For each n ∈ Z+, consider B(n) = {w ∈ F (A) | ℓ(w) ≤ n}. Since Γ hasa solvable word problem, we can take v1, v2 ∈ B(n) and decide if there exists anelement a ∈ A±1 such that a−1v1a = v2 in Γ. If there does exist such an a, thenwe’ll say v1 ∼ v2.

We construct graphs G(n) with vertex set V = B(n) and edges determined bythe relation ∼. The QMC property tells us that u, v ∈ F (A) are conjugate in Γ ifand only if u, v lie in the same path component of G(n) for n = Kmax{|u|, |v|}.

The graph G(n) has at most (2|A|)n vertices. Any injective edge-path of lengthℓ has length at most (2|A|)n. An edge path of length ℓ between u and v gives aconjugating word. So u, v ∈ F (A) are conjugate in Γ if and only if there exists aconjugating word w with |w|A ≤ (2|A|)K max{|u|,|v|}. (We can also write this boundas µmax{|u|,|v|} where µ = (2|A|)K .) �

Remark. Recall: we already proved that for Γ hyperbolic, we can find a muchbetter bound on the length of the conjugating word. In fact, we showed there isa global constant c0 such that we can find a word conjugating u to v of length atmost |u| + |v| + c0. But the above proves the conjugacy problem for a larger classof groups.

Definition. w ∈ {A ∪A}∗ is cyclically reduced if ai 6= a−1i+1 and a1 6= a−1

n .

For example, if w = bab−1babb−1 can reduce to baa or aab, and these are conju-gate. Any word cyclically reduces to a cyclically reduced word that is unique up tocyclic permutation of the letters.

Let Fn be a free group. How do we solve the conjugacy problem? Take u, v ∈ Fn,and replace them with cyclically reduced words. Then check if the resulting wordsare cyclic permutations of each other.


Exercise. Figure out the running time. (Hint: linear?)

In a hyperbolic group, replace the notion of cyclically reduced words with wordswhose cyclic permutations are all (8δ + 1)-local geodesics. We will call words sat-isfying this property geometrically cyclically reduced. It turns out that it takes nomore than |u| + |v| steps to replace u and v with geometrically cyclically reducedwords. (This comes from the Dehn algorithm.)

Lemma. There exists a finite set of words Σ such that if u and v are geometricallycyclically reduced, then u and v are conjugate if and only if they are conjugateby a sequence of words from Σ where the length of the sequence is bounded bymax{|u|, |v|}.


Definition. Let Γ be a finitely generated group with finite generating set A. Letu ∈ F (A). Then the cone of u is

Cone(u) = {v ∈ F (A) | uv is geodesic in Γ}.Observe that if u is not geodesic in Γ, then Cone(u) = ∅. If u is geodesic in Γ, thenCone(u) 6= ∅ since it contains at least the empty string. Observe also that Cone(u)only depends on ε(u).

If γ ∈ Γ, then we define the cone of γ as Cone(γ) = Cone(u) where u is anygeodesic representative of γ.

Example. Let Γ = F2, A = {a, b}. Let γ = ba. Then Cone(γ) is the set of stringsthat do not begin with a−1. In fact, there are exactly five different cone types:

Cone(γ) = {all strings that do not begin with the inverse of the last letter of γ}.Note that in the case γ = e, Cone(γ) = F (A). In general, there are 2n+ 1 distinctcone types in Fn. In fact, there are 2n+ 1 cone types for Zn if we use a basis asour generating set.

Definition. Given a constant k > 0, the k-tail is the set

k -tail(γ) = {h ∈ Γ | d(e, γh) < d(e, γ), d(e, h) ≤ k}.Remark. Observe that there are only finitely many possible k-tails.

Theorem. If Γ is hyperbolic, then Γ has finitely many cone types (for any Cayleygraph).

Proof. Given γ ∈ Γ, the idea is that to determine Cone(γ), we need to know whichstrings to avoid locally at γ.

Let k = 2δ + 2. Then the k-tail of γ determines Cone(γ). That is, if γ, γ′ havethe same k-tail, then they have the same cone type.

Suppose γ, γ′ ∈ Γ with k -tail(γ) = k -tail(γ′). Let v ∈ Cone(γ). If |v| = 0, thenv = e, and v ∈ Cone(γ′). If |v| = 1, then v ∈ Cone(γ′) by construction. (Followthe definitions.)

We will now induct on the length of v. We claim that if v ∈ Cone(γ) andv ∈ Cone(γ′) and a ∈ A±1 with va ∈ Cone(γ), then va ∈ Cone(γ′). Supposeva /∈ Cone(γ′). Then there exists a word w ∈ F (A) with |w| < d(e, γ′)+ |v′|+1 andε(w) = ε(γ′va). Write w = w1w2 where |w1| = d(1, γ)− 1, and then |w2| ≤ |v|+ 1.Now the edge paths u′v and w1w2 are geodesics that end 1 unit apart. By a lemma

36 ANDY EISENBERG

from last week, u′v and w1w2 are (2δ+1)-uniformly close. Hence d(w1, γ′) < 2δ+2.

Notice that (γ′)−1w1 ∈ k -tail(γ′). So (γ′)−1w1) ∈ k -tail(γ) by assumption.Let p be a geodesic path to γ(γ′)−1w1. Since (γ′)−1w1 ∈ k -tail(γ), d(γ(γ′)−1w1, e) <

d(e, γ). Now notice that the path pw2 is a path from e to γva:

γ(γ′)−1w1w2 = γ(γ′)−1γ′va = γva.

Moreover, the length of pw2 is shorter than d(e, γ) + |v| + 1. This contradicts thefact that uva is geodesic. Hence Γ has finitely many cone types. �

Proposition. If Γ is an infinite group with finitely many cone types, then Γ con-tains an element of infinite order.

Proof. Since Γ is infinite, there exists a geodesic edge path in a Cayley graph ofΓ that begins at e and has length greater than the number of cone types. Let wlabel such a path. Break up w = u1u2u3, where the shared end of u1 and u2 is γ1

and the shared end of u2 and u3 is γ2, so that Cone(γ1) = Cone(γ2). (These twovertices must exist by the pidgeonhole principle.)

Since w is geodesic, u2u3 ∈ Cone(γ1) = Cone(γ2). Then u22u3 ∈ Cone(γ1), so

u1u22u3 is geodesic. Iterating, for all n > 0 we have u1u

n2u3 is geodesic. Since

subwords of geodesics are geodesics, u2 is an infinite order element of Γ, and we aredone. �

Proposition. If Γ is hyperbolic and H ≤ Γ is infinite and quasi-convex, then[NΓ(H) : H ] <∞. In particular, if H ⊳Γ, then H is finite index.

Remark. In the case of free groups, this is called Greenberg’s lemma.

To prove the above, we will need the following ingredient:

Proposition. If Γ is hyperbolic, then for all R > 0, there exist only finitely manyconjugacy classes [γ] with translation length τΓ(γ) ≤ R.

19. November 16, 2010

Definition. Let (Γ, A) be a finitely generated group. For γ ∈ Γ, define the trans-lation number of γ as follows:

τΓ,A(γ) = limn→∞

d(eΓ, γn)

n= lim

n→∞

|γn|n.

Remark. This limit always exists, because word length is subadditive. It is a clas-sical fact that if f : N → N is subadditive (that is, f(m+ n) ≤ f(m) + f(n)), thenlimn→∞ f(n)/n exists.

Question. We know that Γ acts on ∆(Γ, A) geometrically. Define

T (γ) = inf{d(x, γx) | x ∈ ∆}.Is T (γ) = τ(γ)?

Proposition. Some properties of the translation number:

(1) τΓ(γ) only depends on the conjugacy class of γ.(2) τΓ(γm) = |m| · τΓ(γ).(3) If H < Γ is finitely generated and QI embedded, then there exists K such

that1

KτΓ(h) ≤ τH(h) ≤ KτΓ(h).


Proposition. If Γ is hyperbolic with finite generating set A, then for all R > 0there exist only finitely many conjugacy classes [γ] such that τΓ,A(γ) ≤ R.

Proof. Let γ ∈ Γ, and let u be a minimal length word in [γ]. If |u| > 8δ + 1 := k,then the path β determined by the powers un is a quasi-geodesic. Recall thatk-local geodesic implies quasi-geodesic for k > 8δ + 1.)

Observe that n|u| = dβ(e, un). We have

1

λ|un| − λ ≤ n|u| ≤ λ|un|

(Recall from our proof that k-local geodesics are quasi-geodesics, we do not needthe extra λ on the right.) Now |u|/λ ≤ |un|/n, so

τΓ(u) = limn→∞

|un|n

≥ |u|λ.

Since there are only finitely many words of length at most a fixed R, this completesthe proof. �

Theorem (Delzant). If (Γ, A) is a hyperbolic group, then {τΓ,A(γ) | γ ∈ Γ} hasthe following property. There exists N ∈ N such that for all γ ∈ Γ, NτΓ,A(γ) ∈ N.

Proposition. Let Γ be a hyperbolic group and H ≤ Γ infinite and quasi-convex.Then [NΓ(H) : H ] < ∞. In particular, if H ⊳Γ, then [Γ : H ] < ∞. (This is ageneralization of Greenberg’s lemma for free groups.)

Proof. We will show there exists a constant D ≥ 0 such that d(γ0, H) ≤ D for allγ0 ∈ NΓ(H).

Since H is quasi-convex, H is finitely generated, hyperbolic, and QI embedded.Since H is an infinite order hyperbolic group, there exists an α ∈ H of infiniteorder. There exists a constant k1 such that if g ∈ CΓ(α), then d(g, α) < k1

(since [CΓ(α), 〈α〉] < ∞). And there exists a constant k2 such that if γ−1αγ ∈ H ,then τH(γ−1αγ) ≤ k2τΓ(γ−1αγ) = k2τΓ(α). Also noticce that if γ−1αγ ∈ H ,then there are only finitely many H-conjugacy classes with τH < k2τΓ(α). Letc−11 αc1, . . . , c

−1n αcn be the conjugacy classes.

Suppose γ0 ∈ NΓ(H). Then γ−10 αγ0 ∈ H , so there exists h ∈ H and i such that

γ−10 αγ0 = h−1(c−1

i αci)h. Now

α = γ0h−1(c−1

i αci)hγ−10 ,

so γ0h−1c−1

i ∈ CΓ(α). It follows that d(γ0h−1c−1

i , 〈α〉) < k1.Since γ0 ∈ NΓ(H), we can write γ0h

−1 = h′γ0 for some h′. So

d(γ0, H) = d(h′γ0, H)

≤ d(g′γ0, h′γ0c

−1i ) + d(h′γ0c

−1i , H)

= |ci| + d(γ0h−1c−1

i , H)

≤ |ci| + d(γ0h−1c−1

i , 〈α〉)≤ max{|ci|} + k1. �

Theorem (Gromov, Delzant). If Γ is a hyperbolic group, then for any finite set{h1, . . . , hr} ⊂ Γ there exists n ∈ N such that 〈hn

1 , . . . , hnr 〉 is free of rank at most r.

Theorem (E. Rips). Every hyperbolic group Γ acts on a simplicial complex P withthe following properties:

38 ANDY EISENBERG

(1) P is finite-dimensional, locally finite, and contractible.(2) Γ acts simplicially with compact quotient Γ \ P and finite stabilizers.(3) Γ acts freely and transitively on the vertices.(∗) If Γ is torsion free, then the quotient Γ \ P is a K(Γ, 1).

Proof. We’ll only describe the complex. Fix a generaing set A for Γ. For eachR > 0, we construct a simplicial complex PR(Γ) as follows. V (PR) = Γ for all R.{x0, . . . , xn} ⊂ Γ determines an n-simplex in PR if diamΓ,A{x0, . . . , xn} ≤ R. If Ris large enough, then PR is contractible. �

20. November 18, 2010

Let H be a finitely generated subgroup of Γ generated by a subset S′ ⊂ S ofthe generating set for Γ. When we erase the edges in S \ S′, then we’re left withconnected components in bijective correspondence with the cosets of H in G. Thecoset containing the identity is a Cayley graph for H .

Example. Consider S3 = 〈ρ, τ〉, and N = 〈p〉. The Cayley graph of S3 is givenby:

(1 2)

e

(1 3 2) (1 2 3)

(2 3) (1 3)

ρ

ρ

ρρ ρ

ρ

τ

τ τ

Then N \ ∆ is an edge with loops on the ends. (The vertices are in bijectivecorrespondence with the cosets of N in Γ.)

Example. Now let H = 〈τ〉. Deleting the ρ edges:

(1 2)

e

(1 3 2) (1 2 3)

(2 3) (1 3)

τ

τ τ

The quotient ∆(S3) \H is just a triangle.

Proposition. Suppose 1 → N → Γ → B → 1 is an exact sequence of finitelygenerated infinte groups with Γ hyperbolic. Then N cannot be ǫ-quasi-convex.


Proof. Suppose N is ǫ-quasi-convex. N acts on any ∆(Γ). Consider the quotientΛ = N \ ∆. This is infinite and locally finite (it’s a covering space action, so thevertices in the quotient will have the same valence as in the cover). Because B isinfinite, we can find arbitrarily long edge paths in the quotient. Let p : ∆ → Λ bethe quotient map, and let β be a path from p(e) to some point b in Λ of length atleast k = 2ǫ+ 2δ. Choose u ∈ N with |u| > 2k + 2δ.

Consider a geodesic β from e to some preimage c := p−1(b). Then {e, u, uc, c}forms a geodesic quadrilateral. Let r be the midpoint of [e, u] and s the midpointof [uc, c]. Then d(r, s) < 2δ. Since N is ǫ-QC, there is an n1 within ǫ of r and n2

within ǫ of s, with n1 ∈ N , n2 ∈ cN . Now d(n1, n2) < 2δ + 2ǫ. Since p(n1) = p(e)and p(n2) = b, we have found a shorter path than β from p(e) to b, contradictingour choice of β. �

We’ll now define the boundary of a proper, geodesic, δ-hyperbolic metric space.Here are the properties we want:

(1) ∂(Fn) ∼= Ends(Fn)(2) ∂(H2) ∼= S1

(3) ∂(E2) ∼= S1

(4) X := X ∪ ∂(X) should be compact, with X open and dense in X(5) π : ∂(X) ։ Ends(X), where the point preimages should be connected com-

ponents(6) any isometry of X extends to a homeomorphism of ∂(X)(7) the homeomorphism type of ∂(X) is a QI invariant (true for hyperbolic

spaces, but false for CAT(0) spaces)

Let X be a geodesic metric space with fixed base point x0 ∈ X . We mightdefine the boundary of X as the set of geodesic rays in X based at x0 with atopology so that two rays are close if they stay close for a long time. It is clear that∂(Fn) ∼= Ends(Fn) with this definition.

Definition. Let X be a proper, geodesic, δ-hyperbolic metric space. Supposeγ1, γ2 : [0,∞) → X are geodesic rays. We say that they are asymptotic, denotedγ1 ∼ γ2, if there exists k > 0 such that for all t ≥ 0, d(γ1(t), γ2(t)) ≤ k. (Equiva-lently, γ1 ∼ γ2 if their images in X have finite Hausdorff distance.)

Definition. Let X be a proper, geodesic, δ-hyperbolic metric space. Let w ∈ Xbe a fixed base point. The boundary of X is ∂w(X) = {[γ] | γ(0) = w}, where [γ]is the asymptotic class of geodesic rays.

We write ∂(X) = {[γ] | γ geodesic}, where γ is a geodesic ray with unspecifiedbase point. It can be shown that ∂w(X) = ∂(X), so this is an equivalent definitionof the boundary.

Definition. In any metrix space X with fixed base point w ∈ X , we can definethe Gromov inner product :

(x · y)w =1

2[d(x,w) + d(y, w) − d(x, y)].

If we draw the triangle w, y, x and collapse to a tripod, the Gromov inner productis the distance from w to the center vertex.

If X is a geodesic, δ-hyperbolic metric space, then for all x, y, z:

(x · y)w ≥ min{(x · z)w, (z · y)x} − δ.

40 ANDY EISENBERG

In a δ-hyperbolic metric space, (x · y)w measures how long it takes [w, x] and [w, y]to get more than δ apart.

Definition. Let {xi} ⊂ X be a sequence. We say that it converges to ∞, denotedxi → ∞, if lim infi,j(xi · xj)w = ∞. Two sequences {ai} and {bi} are related iflim infi(ai · bi)w = ∞. We write {ai}R{bi}. This is in fact an equivalence relation.The boundary of X is

∂(X) = {[{xn}] | xn → ∞}.

21. November 23, 2010

Throughout today, X is a proper, geodesic, δ-hyperbolic metric space. Recallfrom last time:

∂gX = {[γ] | γ : [0,∞) → X geodesic}∂g

pX = {[γ] | γ : [0,∞) → X geodesic, γ(0) = p}∂qX = {[r] | r : [0,∞) → X quasi-geodesic}

Recall the notation γ(∞) := [γ].

Lemma. The natural map f : ∂gX → ∂qX is a bijection. For each p ∈ X andz ∈ ∂gX there exists a geodesic ray c : [0,∞) → X with c(0) = p and c(∞) = z.

Proof. f is trivially an injection (since any geodesic ray is a quasi-geodesic ray).Now let p ∈ X and r : [0,∞) → X a quasi-geodesic ray. Let cn = [p, r(n)]. Since Xis proper we can use the Arzela–Ascoli theorem to get a subsequence of (cn) thatconverges to a ray c : [0,∞) → X . We have c(0) = p. Moreover, there exists Ksuch that for all n, the image of cn is K-close to the image of r. So im(c) is alsoK-close to im(r), hence c(∞) = r(∞). �

In fact, this shows that all three of the point sets above are equal. So we willwrite only ∂X for the boundary from now on. We will describe a topology on theboundary to satisfy the following. Fix a basepoint p ∈ X and let c1 and c2 begeodesic rays based at p. Then c1 and c2 are “close” if they stay (δ-)close for a longtime.

Lemma (Reverse triangle inequality). If a, b, c are the lengths of the sides of ageodesic triangle, then |b− c| ≤ a.

Lemma. Suppose c, c′ : [0, T ] → X are geodesics with c(0) = c′(0). If there existsK > 0 and t0 ∈ [0, t] such that d(c(t0), im(c′)) ≤ K, then d(c(t), c′(t)) ≤ 2δ for allt ≤ t0 −K − δ.

Proof. Let t ∈ [0, T ] such that d(c(t0), c′(t)) ≤ K. For any a < t0−K−δ, c(a) is not

δ-close to [c′(t), c(t0)]. But for any such a there exists a′ such that d(c(a), c′(a)) ≤ δ.So |a− a′| ≤ δ by the reverse triangle inequality. So d(c(a), c′(a)) ≤ 2δ. �

Lemma. If c1, c2 : [0,∞) → X are geodesic rays with c1(0) = c2(0) and c1(∞) =c2(∞), then d(c1(t), c2(t)) ≤ 2δ for all t ≥ 0.

Proof. Choose K such that im(c1) and im(c2) are K-Hausdorff close. Given t,choose t0 > t+K + δ. Since c1(∞) = c2(∞) �

Lemma. If c1, c2 : [0,∞) → X are geodesic rays with c1(∞) = c2(∞), then thereexist T1, T2 such that d(c1(T1 + t), c2(T2 + t)) ≤ 5δ for all t ≥ 0.


Sketch of proof. Using Arzela–Ascoli, build c′1 with c′1(0) = c1(0) and c′1(∞) =c1(∞) as the limit of sequences of geodesics from c1(0) to c2(n). �

Lemma. If z 6= w ∈ ∂X, then there exists a geodesic line c : R → X with c(∞) = zand c(−∞) = w.

Proof. We’ll prove this in detail later. Brief sketch: fix a point p. Pick a geodesicc1 = [p, w] and another geodesic c2 = [p, z]. Go out along each geodesic far enoughto be more than δ apart, and connect c1 and c2 be a geodesic from c1(t) to c2(t).Any geodesic from w to z will have to go through a δ-ball around c1(t). Use Arzela–Ascoli twice to show a sequence of geodesics converges to a geodesic line. �

We’ll topologize X = X ∪ ∂X as follows.

Definition. A generalized ray is a geodesic c : I → X with either I = [0,∞) orI = [0, R]. In the latter case, we extend c to [R,∞) by the constant map at c(R).

Definition. X = {c(∞) | c is a generalized ray} (where c(∞) means the endpointfor a generalized ray of the form c : [0, R] → X).

Definition. A sequence {ai} ⊂ X converges to ∞ if

limi,j→∞

(ai · aj)p = ∞.

We say that two sequences which converge to ∞ are related, denoted {ai}R{bi}, if

limi→∞

(ai · bi) = ∞.

We write ∂sX = {[{ai}] | {ai} → ∞}.Recall the Gromov inner product:

(a · b)p =1

2[d(a, p) + d(b, p) − d(a, b)].

This measure how long the geodesics [p, a] and [p, b] stay δ-close. It also measuresd(p, [a, b]), since d(p, [a, b]) ≤ (a · b)p + δ. (Note that this last inequality explainswhy we care about sequences converging to ∞. If they don’t, then any geodesicfrom ai to bi will have to go through some particular δ-ball at distance sup(a · b)p

away from p. If the sequences do converge to ∞, then lim d(p, ai) = ∞.)

Example. The relation R needs δ-hyperbolicity to be transitive. Consider theCayley graph ∆(Z ⊕ Z, S = {x, y}), with an = xn, bn = yn, and cn = xnyn. Letw = (0, 0). We have {ai} → ∞, {bi} → ∞, and {ci} → ∞. It is easy to calculate

(ai · bi)w = 0

(ai · ci)w = i

(bi · ci)w = i

so {ai}R{ci}, {bi}R{ci}, but {ai}�R{bi}.Recall that in a δ-hyperbolic space, for all x, y, z,

(x · y)z ≥ min{(x · z)w, (z · y)w} − δ.

We define a basis for a topology on X as follows. If x ∈ X , use the B(x, ǫ). Ifx ∈ ∂X , define

(x · y)p = infxi→xyi→y

{lim infi

(xi · yi)p},

42 ANDY EISENBERG

and let N(x, k) = {y ∈ X | (·y) > k} for x ∈ ∂X .

Example. To see why we need both the liminf and the inf, consider the Cayleygraph of Z ⊕ Z2.

22. November 30, 2010

Recall X = X ∪ ∂(X). We wrote ∂sX = {[{xn}] | xn → ∞}, where xn → ∞means (xi · xj)w → ∞ for some base point w as i, j → ∞.

We will extend the Gromov inner product to ∂X as follows. Let x, y ∈ X . Letw ∈ X be a fixed base point. Define:

(x · y)s := infxi→xyi→y

{lim infi

(xi · yi)w}.

Proposition. If x, y ∈ X, then (x · y)s = (x · y)w = (x · y), where we drop the basepoint in the right hand side since the basepoint will not matter.

Proof. Suppose xi → x and yi → y in X . We have

|(x · y) − (xi · y)| =1

2|d(x,w) + d(y, w) − d(x, y) − d(xi, w) − d(y, w) + d(xi, y)|

≤ 1

2[|d(x,w) − d(xi, w)| + |d(xi, y) − d(x, y)|]

≤ 1

2[d(x, xi) + d(x, xi)]

= d(x, xi).

Now

|(x · y) − (xi · yi)| ≤ |(x · y) − (xi · y)| + |(xi · y) − (xi · yi)|≤ d(x, xi) + d(y, yi)

→ 0. �

Remark. {xn} → ∞ is equivalent to d(w, [xi, xj ]) → ∞. If we have two sequences{xn} and {yn} converging to different points on the boundary, then we can drawgeodesics [xi, yi] and take the limit (using AA arguments) to get an ideal triangle{w, x, y}, where xn → x ∈ ∂X and yn → y ∈ ∂X . Then |(x ·y)s−d(w, [x, y])| < 2δ.So the extended Gromov inner product still measures distance from the third sideof the triangle.

We define a basis for a topology on X as follows.

(1) For x ∈ X , r > 0, take B(x, r) = {y ∈ X | d(x, y) < r}.(2) For x ∈ ∂X , k > 0, take Nx,k = {y ∈ X | (x · y)s > k}.

Let B be the collection of open sets given above.

Remark. Consider a geodesic ray c, and let c(∞) = x ∈ ∂X . Consider the 2δ ballaround c(n). Define a set

Vn(c(∞)) = {c′(∞) | c′ : I → X geodesic, c′(0) = w, d(c′(n), c(n)) < 2δ}.These are not quite a basis for the point x. But every Vn contains some Nx,k, andthe sets Vn can be more useful in proofs.

Proposition. The following are properties of the extended Gromov inner producton X. Let x, y ∈ X.


(1) (x · y)s = ∞ if and only if x, y ∈ ∂X and x = y.(2) If x ∈ ∂X and {xi} ⊂ X, then (xi · xs → ∞ if and only if xi converges to

∞ and [{xi}] = x.(3) If x, y ∈ X, then there exist sequences {xi} → x, {yi} → y such that

(x · y)s = limi→∞(xi · yi). If x, y ∈ X, then the constant sequences work.(4) If x, y ∈ ∂X and xi → x, yi → y, then

(x · y)s ≤ lim infi

(xi · yyi) ≤ (x · y)s + 2δ.

(5) Let x, y ∈ X and yi → y. Then lim infi(x · yi)s ≥ (x · y)s.(6) For x, y, z ∈ X,

(x · y)s ≥ min{(x · z)s, (y · z)s} − δ.

Proof. Will be sent out. �

Proposition. B forms a basis for a topology on X.

Proof. It is trivial that B covers X. Let B1, B2 ∈ B. Then we must show thereexists B3 ∈ B with B3 ⊂ B1 ∩ B2. If B1 and B2 are both of type (1), thenthis is trivial. If B1 is of type (1) and B2 is of type (2), then B1 = B(x, ǫ) andB2 = N(z, k). Let y ∈ B1 ∩B2. Since y ∈ B1, y ∈ X . So there exists ǫ1 such thatB(y, ǫ1) ⊂ B(x, ǫ) = B1. Since y ∈ B2, we know (y · z) > k. So there exists ǫ2 suchthat (y · z) > k + ǫ2 > k. Let ǫ′ = min{ǫ1, ǫ2}.

Claim. B(y, ǫ′) ⊂ B1 ∩B2.

Proof. It is clear that B(y, ǫ′) ⊂ B1. Choose p ∈ B(y, ǫ′). Choose a sequence zi → zsuch that limi→∞(p · zi) = (p · z). (We can choose such a sequence by property 3.)Then

|(zi · p) − (zi · y)| ≤ d(p, y) < ǫ′ ≤ ǫ2.

Then

−ǫ2 ≤ lim infi

(zi · p) − lim infi

(zi · y) ≤ ǫ2

⇒ −ǫ2 ≤ (z · p) − (z · y)

(Think about this.) It will follow that

(z · p) = (z · p) − (z · y) + (z · y) ≥ −ǫ2 + (z · y) > −ǫ2 + k + ǫ2 = k. �

The case where B1 and B2 are both of type (2) will be sent out. �

To prove X is compact, we will do the following.

(1) Prove X is metrizable: we will show separable and first countable (hencesecond countable) and regular (hence metrizable by Urysohn metrizationtheorem). Proving regularity requires case by case checking.

(2) Prove X is sequentially compact. (Hint: AA.)

Remark. There is a map ∂X → Ends(X) which is continuous and surjective, andthe point preimages are the connected components of ∂(X). In particular, we candetect when a group is 1-ended by checking whether the boundary is connected.

44 ANDY EISENBERG

23. December 2, 2010

Theorem. If X,X ′ are proper, geodesic, δ-hyperbolic metric spaces and f : X →X ′ is a QI-embedding, then there exists a map ∂f : ∂X → ∂X ′ that is an (topolog-ical) embedding. If f is a QI, then ∂f is a homeomorphism.

Proof. Pick base points p ∈ X and p′ = f(p) ∈ X ′. Let c1, c2 : [0,∞) → X begeodesic rays with c1(0) = c2(0) = p and c1(∞) = c2(∞). Then f ◦ c1 and f ◦ c2are quasi-geodesics in X ′.

Pick geodesic rays c′1 ∈ [f ◦ c1] and c′2 ∈ [f ◦ c2] (that is, (f ◦ ci)(∞) = c′i(∞)).Then c1(∞) = c2(∞) if and only if c′1(∞) = c′2(∞).

We define a map ∂f : ∂X → ∂X ′ by ∂f(c) = c′, where c′ is a geodesic rayasymptotic to f ◦ c. The above shows we have a well-defined injective function.

We now show continuity of the map. With ci and c′i as above, we will show that ifc1 ∈ Vn(c2) (that is, c1(n) ∈ B(c2(n), 2δ)), then c′1 ∈ Vn′(c′2). Since f ◦c1 is a quasi-geodesic, there is some constant K and some t0 such that d(c′1(t0), f ◦ c1(n)) ≤ K.We now have

d(c′1(t0), im c′2) ≤ 2K + λ · 2δ + λ =: L.

It follows that d(c′1(t), c′2(t)) ≤ 2δ for all t ≤ t0 − L− δ. (We proved a lemma that

said this.)Now since d(p′, f ◦ c1(n)) ≥ n

λ− λ, we have t0 >

nλ− λ − K. Now for n′ >

nλ− λ−K − L− δ, c′1 ∈ Vn′(c′2).

If f is a QI, then there is a quasi-inverse f−1, and the above argument showscontinuity of the map ∂f−1. This shows that ∂f is a homeomorphism. �

Let G be a word hyperbolic group. Then ∂G makes sense, and we may aswell work in a Cayley graph. If f : G → G is a QI, then ∂f : ∂G → ∂G is ahomeomorphism. In particular, G acts by isometries on any Cayley graph, so anyg ∈ G acts as a homeomorphism on ∂G.

Let g ∈ G be an element of infinite order. We have already seen that {gn | n ∈ Z}determines a quasi-line. That is, there is a QI embedding Z → G, hence there isan injection of ∂Z, a discrete pair of points, into ∂G. Picking e ∈ G as our basepoint, we can pick a geodesic line that is asymptotic to {gn}, and this line goes totwo distinct points in ∂G. Call them

g+ = limi→∞

gi

g− = limi→∞

g−i.

Since g leaves {gn} invariant, ∂g (the induced homeomorphism on the boundary)will fix {g±}.Remark. No parabolics.

Now let z ∈ ∂G, z 6= g±.

Claim. If z 6= g−, then

limi→∞

(∂g)i · z = g+.

Sketch of proof. z 6= g− implies that 〈g−, z〉 < ∞, and 〈g−, ∂g · z〉 < 〈g−, z〉. (Wewill use angled brackets for the Gromov inner product for now, since we used dotfor the action.) Similarly, 〈(∂g)i · z, g+〉 → ∞ as i→ ∞. �


Theorem. For all U, V open with g+ ∈ U and g− ∈ V with U ∩V = ∅, there existsN such that for all n ≥ N :

(∂g)n(∂G− V ) ⊂ U.

Proof. Given the above claim, this is just compactness. �

Theorem. If G is hyperbolic with |∂G| > 2, then G contains F2.

Sketch of proof. Ingredients:

(1) G is not infinite torsion, so there exists g ∈ G with infinite order.(2) G is not 2-ended (since |∂G| > 2) and “no parabolics”, so there exists an

infinite order h ∈ G with {h±} ∩ {g±} = ∅.(3) Use previous theorem to get sets appropriate for ping pong.

�

Theorem. A much stronger version of the above theorem (same conditions asbefore): given a set {g1, . . . , gr} there exists an N such that {gN

1 , . . . , gNr } ∼= Fr′ for

some r′ ≤ r.

Definition. A word hyperbolic group G is non-elementary if G contains an F2.

Remark. If G is a word hyperbolic group, then ∂G has 0, 2, or infinitely manypoints. If it has 0, it is finite. If it is nonempty, then G contains an infinite orderelement, hence ∂G has at least two points. If it has more than 2 points, then itcontains an F2, which has infinitely many points in its boundary.

Definition. If H is a subgroup of a word hyperbolic group G, then the limit setof H is

Λ(H) = ∂G ∩H,where the closure of H is taken in G ∪ ∂G.

Theorem. The set {g± | g ∈ G, o(g) = ∞} is dense in ∂G.

Proof. A geodesic ray in ∆(G,S) is some infinite geodesic word in {S∪S}∗. Let z bethe boundary point the ray approaches, and pick a sequence of open sets (indexedby i) around it. We claim there is some g+

i in each open set, and lim g+i = z.

Take some beginning chunk of the infinite word z. There exists some M suchthat if aM is the first M letters of this chunk and bi is the rest, then aM (bi)

r isa geodesic string for all r. (This is the ultra pumping lemma.) We will denotelimr→∞ aM (bi)

r := (aM bia−1M )+. Then take gi = aM bia

−1M . �

Corollary. If H is an infinite normal subgroup of a word hyperbolic group G, thenΛ(H) = ∂G.

Proof. Use ∂g · h+ = (ghg−1)+. (Details left to the reader.) �

September 7, 2010 - Tufts Universitymath.tufts.edu/documents/ruane/GGTNotes.pdfGEOMETRIC GROUP THEORY NOTES 5 Example. Consider F 2 = ha,bi. (Picture of inﬁnite 4-valent tree.) Consider

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