Lecture 4, Hyperbolic groups

Olga Kharlampovich

WAM 2017

Definition: A metric space X is proper if closed balls of finiteradius in X are compact. The action of a group Γ on a metricspace X is cocompact if X/Γ is compact in the quotient topology.

Definition: If A,B ⊆ X are compact, thendHaus(A,B) = max{sup

a∈Ainfb∈B

d(a, b), supb∈B

infa∈A

d(a, b)} is the

Hausdorff distance between A and B.

The Svarc-Milnor Lemma: Let X be a proper geodesic metricspace. Let Γ act cocompactly and properly discontinuously on X .(Properly dicontinuous means that ∀ compact K ⊆ X ,|{γ ∈ Γ|γK ∩ K 6= ∅}| <∞). Then Γ is finitely generated and, forany x0 ∈ X , the map Γ→ X defined by γ 7→ γx0 is aquasi-isometry (where Γ is equipped with the word metric).

Proof: We may assume that Γ is infinite and X is non-compact.Let R be large enough that Γ-translates of B = B(X ,R) cover X .Set S = {s ∈ Γ \ 1|sB̄ ∩ B̄ 6= ∅}.Let r =inf{d(B̄, γB̄)|γ ∈ Γ, γ 6∈ S ∪ {1}}. Let λ = max

s∈Sd(x0, sx0).

We want to prove that:(a) S generates Γ,(b)∀γ ∈ Γ, λ−1d(x0, γx0)︸ ︷︷ ︸

i

≤ lS(γ) ≤ 1r d(x0, γx0) + 1︸ ︷︷ ︸

ii

,

(c) ∀x ∈ X , ∃γ ∈ Γ such that d(x , γx0) ≤ R.Note: dS(1, γ) = lS(γ) and dS(γ, δ) = lS(γ−1δ).(c) is obvious.(b-i) is also obvious.

To complete the proof we need to show (a) and (b − ii).Assume γ 6∈ S ∪ {1}. Let k be such thatR + (k − 1)r ≤ d(x0, γx0) < R + kr . As γ 6∈ S ∪ {1}, k > 1.Choose x1, ..., xk+1 = γx0, such that d(x0, x1) < R andd(xi , xi+1) < r for each i > 0. Choose 1 = γ0, γ1, ..., γk−1, γk = γsuch that xi+1 ∈ γi B̄ for each i . Let si = γ−1i−1γi , so γ = s1 · · · sk .

Now d(B̄, si B̄) ≤ d(γ−1i−1xi , siγ−1i xi+1) = d(xi , xi+1) < r , so

si ∈ S ∪ {1}. Therefore S generates Γ.Also, lS(γ) ≤ k ≤ 1

r d(x0, γx0) + r−Rr < 1

r d(x0, γx0) + 1 asrequired.Corollary: If K ≤ Γ is a finite index subgroup of finitely generatedgroup then K is quasi-isometric to Γ.

Quasi-geodesics

In the following, X is always a geodesic metric space.Recall: A quasi-geodesic is a quasi-isometrically embeddedinterval.

TheoremFor all δ ≥ 0, λ ≥ 1, ε ≥ 0, there exists R = R(δ, λ, ε) with thefollowing property: if X is a δ-hyperbolic metric space,c : [a, b]→ X is a (λ, ε)-quasi-geodesic, and [c(a), c(b)] is anygeodesic from c(a) to c(b),then dHaus(im c , [c(a), c(b)]) ≤ R(δ, λ, ε).

Quasi-geodesics

Corollary

A geodesic metric space X is hyperbolic if and only if for everyλ ≥ 1, ε ≥ 0, there exists an M such that every(λ, ε)-quasi-geodesic triangle is M-slim.

Corollary

If X is a δ− hyperbolic space, Y is quasi-geodesic space, and

f : Y → X

be a (k , c)− quasi-isometry, then Y is k (2R + δ) + c− hyperbolic.

Corollary

Hyperbolicity is a quasi-isometry invariant of geodesic metricspaces.

Quasi-geodesics

Proposition: Equivalent definition of a hyperbolic group: A groupΓ is hyperbolic if it acts properly discontinuously and cocompactlyby isometries on a proper hyperbolic metric space.Examples:a) Free groups;b) Zn is not hyperbolic for n > 1;c) Let M be any closed hyperbolic manifold. Then π1(M) isword-hyperbolic.

Quasiconvex subgroups

Definition A subspace Y of a geodesic metric space X isquasiconvex if there exists a K ≥ 0 such that, for all y1, y2 ∈ Yand for all x ∈ [y1, y2], d(x ,Y ) ≤ K .Definition: A subgroup H of a hyperbolic group is calledquasiconvex if it is a quasiconvex space of some (any) Cayleygraph of Γ.

Example: Consider Z2 with the l1-metric. Then the diagonalsubgroup ∆ ⊂ Z2 is not quasiconvex (though it isquasi-isometrically embedded).

Quasiconvex subgroups

Corollary: Suppose that Γ is a word-hyperbolic group and H is asubgroup. Then H is quasiconvex in some (any) Cayley graph of Γif and only if H is finitely generated and H ↪→ Γ is aquasi-isometric embedding.

Quasiconvex subgroups

Proof: (⇐) is immediate from the Theorem. For the otherdirection, fix a generating set S for Γ, assume H is quasiconvex inthe Cayley graph of Γ with constant K , and let h be in H.Consider a geodesic in the Cayley graph of Γ from 1 to h, which wecan take to be of the form s1...sn for si in S . Let vi be the verticesof this geodesic, so vi = s1...si .

1 g2 gi gn = h

v2

vi

g1

u1

u2ui

h1h2

hi hns1

s2

si

v1

Quasiconvex subgroups

1 g2 gi gn = h

v2

vi

g1

u1

u2ui

h1h2

hi hns1

s2

si

v1

∀vi there exists gi inH s. t. d(vi , gi ) ≤ K . Take g0 = 1 and gn = h. Let hi = g−1i−1gi ,

so h = h1...hn. Let ui = vig−1i , lS(ui ) ≤ K . For each i , we have

that hi = ui siu−1i+1 and so lS(hi ) ≤ 2K + 1. Therefore, H is

generated by T = B(1, 2K + 1) ∩ H, a finite set. Furthermore,lT (h) ≤ n = lS(h).lS(h) ≤ (2K + 1)lT (h) (as each element of T has S-length at most2K + 1) so H ≤q.i . Γ.

Quasiconvex subgroups

Definition: LetH≤ G . A homomorphism φ : GH is a retraction ifφ(h) = h for h ∈ H. H is a retract of G .

Exercise: If Gφ−→ H is a retraction and G is finitely generated,

then the inclusion H ↪→ G is a quasi-isometric embedding. (Hint:you can choose a generating set S for G such that φ(s) = s orφ(s) = 1 for all s ∈ S .Example: Marshall Hall’s Theorem implies that every finitelygenerated subgroup of a free group is a retract of a finite-indexsubgroup, so every finitely generated subgroup of a free group isquasiconvex.

Theorem1)Quasiconvex subgroups of hyperbolic groups are hyperbolic.2) Intersection of two quasi-convex subgroups is quasi-convex.3) Membership problem in a quasiconvex subgroup is decidable.

Quasiconvex subgroups

Example

In a hyperbolic group, if g ∈ G is infinite order, then thecentralizer of g is quasiconvex.

Example

No hyperbolic group contains Z2 as a subgroup. In fact, it cannotcontain a Baumslag-Solitar subgroup,⟨

a, b|b−1amb = an⟩.

Quasiconvex subgroups

DefinitionLet G = 〈S |R〉 . We say that the symmetric presentation is a Dehnpresentation if for any reduced word w with w = 1 in G , thereexists a relator r ∈ R so that r = r1r2,

l (r1) > l (r2) ,

andw = w1r1w2.

In other words, any word that represents the identity in G containsmore than one half of a relator, and so it can be shortened. A wordwhich cannot be further reduced or shortened by this method(replacing r1 by r−12 , a shorter word) is called Dehn reduced.

Quasiconvex subgroups

If G has a finite Dehn presentation (G is finitely generated, R isfinite) then you can check all subwords of length at most

N = max {l (r) |r ∈ R} ,

to see if a reduction can be made. This procedure for solving theword problem is called Dehn’s Algorithm, originally created by MaxDehn in 1910 to solve the word problem in surface groups. Its run

time is O(|w |2

)in its simplest iteration. There are at most

|w | − N subwords in each step, and at most |w | steps in thereduction.

Quasiconvex subgroups

Example

Let G = 〈a, b, c , d | [a, b] [c , d ]〉 . Let R be the symmetrized set ofgenerators, so that R contains all cyclic conjugates of [a, b] [c, d ]and its inverse.

What about other hyperbolic groups?

DefinitionA path γ in a metric space X is called a k− local geodesic if everysubpath of length k is a geodesic.

Example

On a sphere of radius 1, a great circle is a π−local geodesic.

Quasiconvex subgroups

LemmaLet G be δ−hyperbolic group and let γ be a 4δ− local geodesic.Let g be the geodesic between the endpoints of γ (called γ+ andγ−). Assume

l (g) > 2δ,

and let r and s be points on γ and g respectively, both distance 2δfrom γ+. Then

d (r , s) ≤ δ.

The proof here is by induction on the length of γ and uses the thintriangle property multiple times.

Quasiconvex subgroups

TheoremIf γ is a 4δ−local geodesic in a δ−hyperbolic group G , then γ iscontained in the 3δ neighborhood of the geodesic between itsendpoints.

Quasiconvex subgroups

We will use this theorem to create a shortening algorithm in ourhyperbolic group G .

TheoremLet G be a δ− hyperbolic group with generators S . Let R be equalto the set of words

R = {w : w = 1 ∈ G , |w | < 8δ and w represents the identity element in G} .

We aim to show that 〈S |R〉 is a Dehn presentation for G . Inparticular G is finitely presented and has a solvable word problem.

Quasiconvex subgroups

Proof.Take a word w in the generators so that w = 1 in G . If the loop win the Cayley graph is already a 4δ-local geodesic, then it is in the3δ neighborhood of the origin, so it is already an element of R andwe can see that it represents the identity. Now say that w is not a4δ which is not a geodesic between its endpoints. Replace it in wby the geodesic between its endpoints, call it w2. Then the pathw1w

−12 has length less than 8δ, so that word is in R. Further, we

note that w contained w1, the longer part of the relator.Now, we can continue this process until we reduced w to a4δ−local geodesic, a case we have already covered. Thus

G = 〈S |R〉 ,

is a Dehn presentation.

Conjugacy problem

TheoremIf Γ is a δ-hyperbolic group and u = γvγ−1 in Γ, then there existssuch an element γ with `(γ) ≤ M(`(u) + `(v)), where M dependsonly on Γ. Therefore the conjugacy problem in Γ is decidable.

Isomorphism problem

Decidable! (Dahmani, Guirardel)

Stallings-like graph

Theorem: (Khar., Miasnikov, Weil) There is an algorithm toconstruct a Stallings-like graph for a quasiconvex subgroup H of ahyperbolic group G (if we know that G is hyperbolic).

Previously: Kapovich, Arzhantseva-Olshanski.

Rips construction

There are f.g. non-quasiconvex subgroups of hyperbolic groupssuch that the membership problem is undecidable. (Ripsconstruction:Theorem Given any finitely presented group Qand any realnumber λ > 0, there is a C ′(λ) group G and a short exactsequence1→ N → G

p→ Q → 1with N finitely generated.Theorem (Rips) There is a small cancellation group G such that:G has finitely generated subgroups whose intersection is notfinitely generated.G has a finitely generated but not finitely presented subgroupThe subgroup membership problem in G is not solvable. Indeed,there is a subgroup (N) for which there is no algorithm decidingwhether a word in the generating set for G represents an elementof the subgroup.

Open questions

1. (Gromov) Given a hyperbolic group G with one topologicalend (i.e. a freely indecomposable hyperbolic group), does itcontain a surface subgroup?

2. Are hyperbolic groups residually finite?

3. Does every hyperbolic group have a proper subgroup of finiteindex?

4. (A.Miasnikov) Given a finite presentation of a hyperbolicgroup (which is not necessarily a Dehn presentation), is itpossible to find a Dehn presentation for this group inpolynomial time?