Introduction to Geometric Group Theory

Home , Brian Bowditch, Dehn function, Geometric group theory, Karen Vogtmann, Tietze transformations

Math 697: Introduction to Geometric Group Theory

Notes from course given by Dick Canary, Fall 2014. August 23, 2016

0 Overview 0.1 A Rambling Geometric group theory is really “a collection of things put together by a state of mind.” These things defi- nitely include hyperbolic group theory, but also various other things such as CAT(0) groups, topological group theory, and so on ... A little more concretely, geometric group theory is the study of groups by looking at their actions on spaces. Given a [finitely-presented] group G = hx1, . . . , xn | r1, . . . , rmi, we can define the word metric distance by setting d(g, h) to be the minimal length of word in {x1, . . . , xn} (and their inverses) over all representations −1 of g h. We define the Cayley graph ΓG of G by taking the vertices to be the elements of G and drawing edges labelled xi joining each g to gxi ∼ Example 0.1. • If G = Z = ha |i, then ΓG is a line (= R.) a a a a ··· ··· a−2 a−1 1 a a2

2 • If G = Z = ha, b | [a.b]i, then ΓG is a 2-dimensional lattice (“what is this? It is graph paper.”)

ab = ba b 0 a

1 1 • If G = F2 = ha, b |i, then ΓG is an inﬁnite 4-valent tree (the universal cover of S ∨ S )—see below

If M is a compact manifold, then π1(M) acts properly discontinuously and co-compactly on M˜ .

2 2 Example 0.2. If M = T , then π1(M) = Z , which acts properly discontinuously and cocompactly on M˜ = R2; we note that R2 “looks like” Z2 (coarsely speaking.) ˜ If M is a compact 3-manifold (e.g. a solid genus-2 torus), M is a thickening of ΓF2 , the inﬁnite 4-valent tree.

We make precise the notion of “looks like” by using the notion of quasi-isometry. Note that our spaces X and Y are usually assumed to be proper (i.e. closed balls are compact) and geodesic (i.e. distance between two points is given by the length of a shortest path.)

1 a−1 ba

b a−1 ba−1 a ba b−1 b−1 a−1 ba a−1 ba b b b a−1 ba−1 a−1 a a ba b−1 b−1 b−1 b−1 a−1 ba a−1 a a−1 a a−1 ba b−1 b−1 b b a−1 ba−1 a ba a−1 ba−1 a ba b b−1 b−1 b−1 b−1 a−1 ba a−1 ba b b b b a−1 ba−1 a−1 a−1 a a a ba b−1 b−1 b−1 b−1 a−1 a a−1 a b−1 b−1 b−1 b−1 a−1 ba−1 a ba a−1 ba−1 a ba b−1 b−1 b−1 b−1 b−1 b−1 b−1 a−1 a a−1 ba a−1 ba a−1 a b−1 b−1 b b a−1 ba−1 a−1 a a ba b−1 b−1 b−1 b−1 a−1 a a−1 a b−1 b−1 b−1 a−1 ba−1 a ba b−1 b−1 b−1 a−1 a b−1

Figure 1: Part of a Cayley graph of a nonabelian free group

1 Deﬁnition 0.3. h : X → Y is a quasi-isometric embedding if there exist k, c s.t. k d(x, z) − c ≤ d (h(x), h(z)) ≤ kd(x, z) + c for all x, z ∈ X. h is a quasi-isometry if in addition it is “coarsely surjective”, i.e. there exists D ≥ 0 s.t. for all y ∈ Y , there exists x ∈ X s.t. d(h(x), y) ≤ D.1 A key result in geometric group theory is the Milnor-Svarcˇ lemma, which states that if G acts properly discontinuously and co-compactly on a space X, then X is quasi-isometric to G with the word metric. Note that this notion of quasi-isometry, though seemingly loose and geometric, can capture algebraic aspects of structure (or aspects of algebraic structure.) e.g. G quasi-isometric to a free group implies G virtually free.2 There is also Gromov’s theorem, which states that any group with polynomial growth is virtually nilpotent. In fact, we can also use algebra to clarify geometry, as in the following [re-]formulation of hyperbolic

1Think of the plane, and graph paper. 2“Virtually” or “almost” in this context means true up to (true for a) ﬁnite-index subgroup. These adjectives will appear a lot, because quasi-isometry fails to distinguish between the whole group and ﬁnite-index subgroups.

2 spaces by Gromov: Fact. Suppose X is simply-connected with (sectional) curvature ≤ −k < 0. Then there exists δ = δ(k) such that any geodesic triangle in X is δ-slim, meaning that if T has vertices x, y, z, then [x, z] ⊂ Nδ([x, y]∪[y, z]), i.e. ∀t ∈ [x, z]: ∃s ∈ [x, y] ∪ [y, z] s.t. d(s, t) < δ. Deﬁnition 0.4 (Gromov, possibly Alexander). X is δ-hyperbolic if every geodesic triangle in X is δ-slim. e.g. trees are 0-hyperbolic. A space is hyperbolic if it is δ-hyperbolic for some δ ≥ 0. A group is hyperbolic if it acts properly discontinuously and co-compactly on a hyperbolic space. Hyperbolic groups have some very nice properties: • Hyperbolic groups have solvable word problem (which we might formulate as “can I build the Cayley graph [algorithmically]?”) by Dehn’s algorithm. • They also have solvable conjugacy problem and isomorphism problem.3

• The Tits Alternative: if a hyperbolic group G is not virtually cyclic, it contains a F2.

• A hyperbolic group cannot contain a copy of Z2. • A hyperbolic group has only finitely many conjugacy classzes of finite subgroups. • Being a hyperbolic space / group is a condition invariant under quasi-isometry. Hyperbolic spaces are those with constant negative curvature (-1); we can generalize this notion to look at spaces X with constant non-positive curvature, and this brings us to the theory of CAT(0) groups: Definition 0.5. A simply-connected space X is CAT(0) if every geodesic triangle in X is at least as thin as a Euclidean triangle with the same edge lengths. G is CAT(0) if it acts properly and cocompactly on a CAT(0) space. Definition 0.6. CAT(0) cube complexes are formed by gluing Euclidean unit cubes along faces. e.g. R2 is a CAT(0) cube complex

Example 0.7. Right-Angled Artin Groups (RAAGs) are CAT(0) groups AΓ formed as follows: start with a ﬁnite graph Γ; deﬁne (0) (1) AΓ = hΓ | vw = wv ⇐⇒ (v, w) ⊂ Γ i

i.e. the elements of AΓ are the vertices of Γ, and we add a commutator relation between two elements iﬀ there is an edge between the corresponding vertices. e.g. if Γ is the empty graph on n vertices, then AΓ = Fn. n If Γ = Kn, then AΓ = Z . 2 2 If Γ consists of two disjoint edges on 4 vertices, then AΓ = Z ∗ Z . These groups are “interpolating between the free group and the free abelian group.” Theorem 0.8 (Agol, Wise). If a hyperbolic group G acts properly discontinuously and cocompactly on a CAT(0) cube complex (in this case we say G is a cubulation), then a ﬁnite-index subgroup of G embeds in a RAAG. Theorem 0.9 (Haglund-Wise, Kahn-Markovic). If M is a closed hyperbolic 3-manifold, then M is cubulated. Finally, there is topological group theory. The prototypical result here is the proposition that states that every subgroup of a free group is free, proven using covering spaces. There is also the following generalisation of this:

Theorem 0.10 (Kurosh). If G = G1 ∗ G2 and H ⊂ G, then H = F ∗λ∈Λ Hλ, where F is free, and each HΛ is conjugate to a a subgroup of G1 or G2. (See Figure 2.)

3Something involving Tietze transformations was name-dropped here.

3 Figure 2: A free product may be thought of as the fundamental group of a wedge of spaces.

This has the following corollary:

Corollary. If G is ﬁnitely-generated, then G has a unique free decomposition G = G1 ∗ · · · ∗ Gn where each Gi is freely indecomposable (i.e. does not split as a free product.) Other topics that might come under this last heading include JSJ decompositions and Bass-Serre theory.

0.2 Deﬁnitions and quasi-isometry Deﬁnition 0.11. Given X a metric space and α : [0, 1] → X a path,

n−1 X `X (α) := sup d(α(ti), α(ti+1)) 0=t

Deﬁnition 0.12. α is rectiﬁable if `X (α) < ∞.

Deﬁnition 0.13. α is geodesic if `X (α) = d(α(0), α(1)). Exercise. (1) A subpath of a geodesic is a geodesic, (2) If 0 ≤ s ≤ t ≤ u ≤ 1, then

d(α(s), α(t)) + d(α(t), α(u)) = d(α(s), α(u))

if α is a geodesic. Definition 0.14. A metric space is geodesic if any two points are joined by a (by some) geodesic. e.g. Cayley graphs, Riemannian manifolds, curve complexes, etc. Definition 0.15. X is proper if all closed metric balls are compact. Our spaces will be proper geodesic metric spaces (mostly.) We always assume that G acts by isometries on X, i.e. d(x, y) = d(s(x), s(y) for all x, y ∈ X and s ∈ G. This is a common (and apparently commonly unstated!) assumption in geometric group theory. Definition 0.16. An action is properly discontinuous (or proper) if whenever K is compact {g ∈ G | g(K) ∩ K 6= ∅} is finite. Definition 0.17. An action of G is co-compact if X/G is compact.

Remark. X/G is a metric space, with the metric given by dX/G([x], [y]) := dX (Gx, Gy). Hence X/G compact =⇒ ∃R s.t. G(BR(x0)) = X—we can take e.g. R = diam(X/G). Fact. If G acts properly and co-compactly on a geodesic proper metric space, then G is ﬁnitely generated.

Proof. Choose D = B2 diam(X/G)(x), U = int(D). Note U ⊃ Bdiam(X/G)(x). Then G(U) = X. Let S = {g ∈ G | g(D) ∩ D 6= ∅}. S is ﬁnite. Claim. S is a generating set. Let H = hSi ⊂ G. If H 6= G, then let V = Hu and W = (G \ H)u: V and W are non-empty and open, and X is connected (since path-connected, since geodesic), so V ∩ W 6= ∅. Let x ∈ V ∩ W . x ∈ V =⇒ ∃h ∈ H, p ∈ U s.t. h(p) = x. x ∈ W =⇒ ∃g ∈ G \ H, q ∈ U s.t. g(q) = x. Then hg−1(p) = q and so hg−1 ∈ S, which implies g ∈ H. Contradiction.

4 Recall that h : X → Y is a (k, c)-quasi-isometric embedding if ∀x, z ∈ X a d(x, z) − c ≤ d(h(x), h(z)) ≤ kd(x, z) + c k (k is a bi-Lipschitz constant, disregard scale ≤ c.) Fact. If h : X → Y is a quasi-isometry, then ∃ quasi-isometry j : X → Y s.t. d(j(h(x)), x), d(h(j(y)), y) ≤ R + for all x ∈ X, y ∈ Y , and some R ∈ R0 . j is called a quasi-inverse. Proof. If y ∈ Y , then ∃x ∈ X s.t. d(y, h(x)) ≤ C. Choose j(y) s.t. d(y, h(j(y)) ≤ C. Then

d(j(y), j(z)) ≤ k(d(h(j(y)), h(j(z))) + c ≤ k(d(y, z) + 2c) + c ≤ kd(y, z) + (2kc + c)

Now repeat in the other direction, and take R = max{2kc + c, −}. Exercise. A composition of quasi-isometries is a quasi-isometry. Corollary. Quasi-isometric equivalence of metric spaces is an equivalence relation.

0.3 The Milnor-Svarcˇ lemma Lemma 0.18 (Milnor-Svarc)ˇ . Suppose G acts properly and co-compactly by isometries on a (proper geodesic metric) space X, and R is a finite generating set for G; then (G, dR) is quasi-isometric to X (here dR is the word metric associated to R.) Corollary. If G acts properly, co-compactly, and by isometries on two proper geodesic metric spaces X and Y , then X is quasi-isometric to Y . Remarks. (1) The assumptions imply that G is finitely-generated, so that we can choose a finite generating set R and put the word metric dR on G. (2) A Riemannian manifold is proper and geodesic iff it is complete.

Proof of Lemma. We construct a quasi-isometry τ : G → X. Pick x0 ∈ X and let τ : G → X be the orbit map given by g 7→ g(x0).

hx0 gx0

Figure 3: Balls of radius diam(G\X) cover X. Orbit/s is/are coarsely dense, and the coarse transitivity of the action means the Cayley graph supplies the linear upper bound constants.

If D = diam(X/G), then if x ∈ X ∃g ∈ G s.t. d(x, g(x)) ≤ D, and so τ is coarsely D-surjective.

5 To obtain the upper bound, let R = {r1, . . . , rg} and Q = maxi∈[g] d(x0, ri(x0)) for some ﬁxed x0 = τ(1) ∈ X. If g = r1 ··· rn and n = dR(1, g) (i.e. we have a minimal word representation), then d(x0, g(x0)) ≤ Qn, i.e. d(τ(1), τ(n)) ≤ QdR(1, g), i.e. we can extend G,→ X into a map CayG,X ,→ X. Since G acts by isometries on both G and on X, if h ∈ G, then

d(τ(h), τ(hg)) = d(τ(1), τ(g)) ≤ Qd(1, g) = Qd(h, hg)

and so if g1, g2 ∈ G then d(τ(g1), τ(g2)) ≤ QdR(g1, g2). To obtain the lower bound, let S = {g ∈ G | d(x0, g(x0)) ≤ 3D}. S is ﬁnite; let p = maxs∈S dR(1, s).

g4x0

x1 x2 x4 g3x0 g2x0 g1x0 x3

Figure 4: We obtain the lower bound by approximating our geodesic (black) with a broken curve (red) that takes hops of ≤ 3D between orbit points. Hops between black points are ≤ D apart by construction.

j d(x0,g(x0) k Choose a geodesic [x0, g(x0)] and divide it into N = D + 1 segments. For each i, choose gi ∈ G s.t. d(xi, gi(x0)) ≤ D. WMA g0 = id and gN = g. −1 −1 Then d(gi(x0), gi+1(x0)) ≤ 3D =⇒ d(x0, gi gi+1(x0)) ≤ 3D =⇒ gi gi+1 ∈ S. −1 −1 −1 −1 Nu g = g0(g0 g1)(g1 g2) ··· (gN−2gN1 )(gN−1gN ) and so

d(x , g(x )) d (1, g) ≤ Np ≤ p 0 0 + 1 . R D

1 Then D( p dR(1, g) − 1) ≤ d(x0, g(x0)). D Hence τ is a (max{Q, p },D)-quasi-isometry. The Milnor-Svarcˇ lemma makes it possible to talk about the growth of groups in a meaningful well-deﬁned way.

0.4 Growth functions of groups

If G is generated by a ﬁnite set S, deﬁne βG,S(n) = #{g ∈ G | dS(1, g) ≤ n},

Example 0.19. • βZ,{1} = 2n + 1 (linear) 2 • β 2 = 2n + n + 1 (quadratic) Z ,{e1,e2} Pn i−1 • βF2,{a,b} = 1 + i=1 4(3 ) (exponential)

6 (Note e.g.) if T is a (finite) generating set for Z2, then every element in T can be written as a word of length ≤ Q in S (for some Q ∈ Z+.) n 2 2 2 Then βZ ,S( Q ) ≤ βZ ,T (n) ≤ βZ ,S(Qn); changing the choice of generating set does not change the order of growth of βG,S. 2 ∼ Corollary. Z =6 F2. Theorem 0.20 (Gromov). G is virtually nilpotent (i.e. has a nilpotent4 subgroup of finite index) iff it has polynomial growth. Example 0.21. The Heisenberg group H may be represented as the group of unipotent upper-triangular matrices     1 a b   0 1 c  | a, b, c ∈ Z  0 0 1   1 1  or abstractly as hx, y, z | [x, y] = z, [x, z] = [y, z] = 1i (concretely we may take x =  1 , y = 1  1   1 1   1 1 , and z =  1 .) 1 1 4 It is a result of Milnor that βH = O(n ).

Given f, g : N → N, we say that f 4 g if f(n) ≤ Cg(Cn + c) for some C. We say f g (f and g are “quasi-equal”) if g 4 f and f 4 g, and f ≺ g (f is “quasi-less-than” g) if f 4 g and g 46 f. Example 0.22. (1) For 1 < a < b, na ≺ nb. (2) For 1 < α < β, αn βn. (3) na ≺ 2n for all a > 0.

Observation. If G is quasi-isometric to H, then βG βH . 2 Corollary. Z is not quasi-isometric to Z or to F2.

Proof. Let j : G → H be a (k, c)-quasi-isometry. Then j(BG(1, n)) ⊂ BH (j(1), kn + c), which has size βH,kn+c. −1 1 If h ∈ H, j (h) ⊂ ball in G of radius kc + 1 since d(g1, g2) > kc + 1) =⇒ d(j(g1), j(g2)) ≥ k (kc + 1) − c > 0, and hence βG(n) ≤ βH (kn + c)βG(kc + 1), and βG(kc + 1 is constant w.r.t. n. Hence βG 4 βH . Similarly, βH 4 βG. Therefore βG βH .

Which maps are quasi-isometrically equivalent? (1) Finite-index inclusions (2) Quotients by ﬁnite-index normal subgroups (3) No other criteria! (“Quasi-isometric rigidity”)

1 Hyperbolic groups 1.1 A whirlwind tour of the upper half-plane

The upper half-plane model of hyperbolic geometry takes as the hyperbolic plane H2 = {(x, y) | y > 2 dx2+dy2 2 v·w 0} = {z ∈ C | =(z) > 0} with the metric dshyp = y2 ; so given v, w ∈ Tz(H ), hv, wihyp = =(z)2 and kvk 2 kvkhyp = =(z)2 . This is a conformal model for H : ∠hypv, w = hv, w.

4 A group is nilpotent if its lower central series terminates, i.e. ∃R ∈ N s.t. [g1, [g2,..., [gR−1, gR]] ... ] = 1.

7 0 2 R 1 R 1 |γ (t)| Given a path γ : [0, 1] → H , `hyp(γ) := 0 kγ(t)khypdt = 0 =(γ(t)) dt. dhyp(z1, z2) := infγ:z1 z2 `hyp(γ). |z−w|2 There is an explicit formula given by cosh dhyp(z, w) = 1 + 2=(z)=(w) . Now we want to find what the geodesics in H2 are ... “first we find a geodesic”. Fact. The y-axis is a geodesic in H2. 0 0 Proof. Let p : 2 → {x = 0} be the projection, i.e. p(z) = =(z). d = , and kd (v)k ≤ kvk H p 0 1 p hyp hyp with equality iff v is vertical. Thus (equivalently) `(p ◦ γ) ≤ `(γ) with equality iff γ0x) is vertical for all x, which is equivalent to the assertion that shortest paths joining points on the y-axis are vertical.

az+b 2 Fact. z 7→ cz+d , where a, b, c, d ∈ R with ad − bc = 1, is an isometry of H .

dh 0 Proof. The “eﬃcient way to do this” is to compute it and show this acts as an isometry on the tangent bundle: dz = h (z) = (cz + d)−2, and =(h(z)) = =(z) , so kdh(v)k = kvk . However, this may be considered a “bad proof”, for “being a good |cz+d|2 hyp hyp Thurston student, I5 believe that no one every learned anything from a computation.” An alternative proof idea is to observe that M¨obiustransformations are products of inversions in circles perpendicular to the x-axis, i.e. to ∂H2. Fact. Every circle (or line) perpendicular to ∂H2 is a geodesic. These are the only geodesics.

Proof. These are all images of the y-axis under (appropriate) M¨obiustransformations. Corollary. Geodesics are unique Proof. Any 2 points can be moved onto the y-axis, and geodesics on the y-axis are unique.

Fact. Isom(H2) = PSL(2, R) consists of the M¨obiustransformations preserving H2. Idea of proof. PSL(2, R) acts transitively on T 1(H2) and an isometry is determined by the image of one vector in T 1(H2.

The Poincar´edisk model

2 2 iz+1 Take a Möbius transformation U : H → D and push forward the geometry. If we take U(z) = z+i , we 2|dz| obtain dsD2 = 1+|z|2 . Since U is a Möbiustransformation, we have that Fact. Geodesics in D2 are circles or lines perpendicular to ∂D2. Isom(D2, hyp) is the group of Möbius 2 1 R r 2 1+r transformations preserving D . If ξ ∈ S , d(0, rξ = 0 1−s2 ds = log 1−r .

2 R Fact. A circle of hyperbolic radius R about 0 ∈ D has Euclidean radius tanh 2 .

1+tanh R/2 Proof. R = log 1−tanh R/2 . Corollary. This circle has hyperbolic length

Z 2π 2 Z π tanh ds = sinh 2R = 2π sinh R πeR 2 R 0 1 − tanh 2 0 and the disk it bounds has area ZZ 4 R 2 2 2 dx dy = 2π(cosh(R) − 1) πe . D (1 − x − y )

5i.e. Dick Canary

8 R θ R θ sinh R = 2 e

Figure 5: Hyperbolic circles have exponential circumference

and here we see the exponential divergence of geodesics and exponential growth which are two characteristics of negatively-curved spaces. Note also the constant isopermetric inequality obtained from the above: frac2π(cosh(R) − 1)2π sinh(R) → 1 as R → 0 or as R → ∞. Moreover there are unique geodesics (unlike e.g. in spherical geometry) and the sum of the angles in a triangle add up to < π: these are also characteristics of negatively-curved spaces.

Ideal triangles

Deﬁnition 1.1. An ideal triangle in H2 is a triangle spanned by geodesics with endpoints in ∂H2.

Figure 6: Two ideal triangles (in the upper half-space model.)

Fact. Ideal triangles are all congruent and have area π.

Proof. Every ideal triangle is congruent to the one with vertices at ±1 and ∞ since PSL(2, R) acts transitively on triples of points in ∂H2. R 1 R√∞ 1 The area of this particular ideal triangle is given by −1 1−x2 y2 dy dx = π.

Fact. All 2/3-ideal triangles (i.e. triangles in H2 with two of their vertices in ∂H2) with internal angle α are congruent and have area π − α.

Proof. All such triangles are congruent to the one obtained by truncating the ideal triangle with vertices at ±1 and ∞.

Figure 7: A 2/3-ideal triangle with a nonzero angle α (in the upper half-space model.)

R 1 R√∞ 1 The area of this latter 2/3-ideal triangle is given by cos(π−α) 1−x2 y2 dy dx = π − α. We note in passing that this is in fact (may in fact be obtained as) a special case of the Gauss-Bonnet theorem. Fact. A hyperbolic triangle with angles α, β, γ has area π − (α + β + γ). In particular, α + β + γ ≤ π. Proof. See Figure 8

9 a a c

c b b

Figure 8: Here a, b, c stand for α, β, γ resp.; they represent the angles of the small white triangle, and also the areas of the shaded 2/3-ideal triangles by the previous Fact.

−1 Fact. Triangles in H2 are cosh (2)-slim. −1 Proof. If T has sides S1,S2,S3, and z ∈ S1, d(z, S2 ∪ S3) ≤ cosh (2): if R = d(z, S2 ∪ S3), then half of B(z, R) ⊂ T , so

Area(B(z, R)) < 2 Area(T ) 2π cosh(R) − 2π < 2π cosh R < 2

Figure 9: That ball is at most half the triangle in area, and our desired inequality follows.

Hyperbolic sports! Let’s think about baseball in hyperbolic space. How many outfielders would you need on a hyperbolic baseball diamond? (Suppose the outfield is the area between 100 and 300 feet out from the pitcher.) Well, 2 π 2 2 2 in Euclidean space E , the outfield has area 4 (300 − 100 ) ≈ 62, 832 square feet. In H , the outfield would 2π 100 have area 4 (cosh 300 − cosh 100) 10 square feet. Even assuming each outfielder could cover an area of 104 square feet ... that doesn’t look good (due to exponential growth in this negatively-curved space.) 2 1 2 (You would also never see the ball coming: in E the visual size of a ball of radius R (e.g.) is πR . In H , 1 2 it is π sinh R ∼ πeR .) Due to this exponential growth, real estate is cheap in hyperbolic space: one could imagine “a sketchy real-estate agent in hyperbolic space” hawking spectacular (but relatively—or worse—worthless) timeshares. But back to sports—maybe we should choose another sport. Let’s try golf. Suppose you were taking a shot at 300 feet, and you were 1 degree off. In E2, that would translate into your being ∼ 5.24 feet off at 2 2π sinh(300) 300 feet. In H , you would be 300 feet off. Oh well. You would have the same problem/s with something like soccer: you would never see the ball coming, and the game would mostly involve passes to nowhere. Or, if one imagined a sort of rectangular field with

10 equally long goal lines on both sides, the center line would be extremely narrow, and then the game would mostly involve the goalies kicking the balls out of bounds. Similarly, hyperbolic space would be a terrible place to go for a walk: you would never ﬁnd your way home unless you were perfectly precise (due to the exponential divergence of geodesics.)

Some notes coming out of the tour ...

Deﬁnition 1.2. A (complete) n-manifold is hyperbolic if it is locally isometric to Hn. π Example 1.3. Σg is made from a regular hyperbolic 4g-gon with internal angles 2g and has area 2π(2g−2) = 2π|χ(Σg)|.

Figure 10: (Left) a torus formed by gluing the sides of a square; (right) a genus-2 surface formed by gluing π the sides of a regular hyperbolic octagon with angles 4 .

Recall a proper geodesic metric space is hyperbolic if ∃δ > 0 s.t. all triangles are δ-slim (i.e. if T is a geodesic triangle in X with edges S1,S2,S3 and x ∈ S1, then d(x, S1 ∪ S2) ≤ δ. A group is hyperbolic if it acts co-compactly and by isometries on a hyperbolic metric space.

Example 1.4. All finite groups are hyperbolic: in particular G finite =⇒ ΓG (a Cayley graph of G) is diam(ΓG)-hyperbolic. Similarly, all compact hyperbolic metric spaces are δ-hyperbolic. Fn is hyperbolic: it acts on a punctured (hyperbolic) surface. π1(Σg) (with g ≥ 2) is hyperbolic: it acts co-compactly and by isometries on Σg. Z2 is not ... why? (We will find out.)

1.2 The Fellow Traveller Property

Deﬁnition 1.5. If α : J → X is a (k, c)-quasi-isometric embedding, and J is an interval in R, then α is called a (k, c)-quasi-geodesic.

Note that geodesics need not be unique in hyperbolic metric spaces. 2 The universal cover of a hyperbolic surface Σg is H . Since Σg is a simply-connected Riemannian manifold 2 2 locally isometric to H , so π1(Σg) acts properly and co-compactly on H if g ≥ 2. If α :[a, b] → X is a (k, c)-quasi-geodesic and X is δ-hyperbolic, then ∃D = D(k, c, δ) s.t. if [α(a), α(b)] is a geodesic (or even quasi-geodesic) going from α(a) to α(b), then

α([a, b]) ⊂ ND[α(a), α(b)] ⊂ N2D(α([a, b])). The key fact used in the proof (which will be given below, along with a more precise statement of the Property) is the exponential divergence of geodesics. For now, we pause to state and prove an important corollary of the fellow traveller property: Corollary. If f : X → Y is a quasi-isometric embedding, and Y is hyperbolic, then X is hyperbolic. Proof. Suppose Y is δ-hyperbolic and f is a (k, c)-quasi-isometric embedding. Consider a (geodesic) triangle in X with sides s1, s2, s3. 0 Let si be a geodesic in Y with the same endpoints as f(si). By the fellow traveller property, there exists 0 0 0 0 0 0 0 p ∈ si s.t. d(f(0), p ) ≤ D = D(k, c, δ). Since s1, s2, s3 form a geodesic triangle, ∃q ∈ s2 ∪ s3 (WLOG s3) 0 0 00 0 00 s.t. d(p , q ) < δ and q = f(q) ∈ f(s2) s.t. d(q , q ) < D. 1 1 Then d(f(p), f(q)) < 2D + δ =⇒ d(n, q) < k (2D + δ) + C =⇒ T is k (2D + δ) + C -hyperbolic.

11 Figure 11: A quasi-isometry maps the (thin) black geodesic triangle on the right to the (thinnish) black quasi-geodesic triangle on the right, which can be straightened to (is uniformly close to) the red geodesic triangle on the right.

Fact (Exponential divergence of geodesics). Suppose X is δ-hyperbolic, p lies on a geodesic [x, y] ∈ X, and α : [0, 1] → X is a rectiﬁable path joining x to y. Then d(p, α(I)) ≤ δ log2(`(α)) + 1.

Figure 12: Any rectiﬁable curve (in particular, quasi-geodesic—grey) which is at least D away in Hausdorﬀ distance (red) from a geodesic (black) is exponentially (≥ 2(D−1)/δ) long.

Proof. “A picture I’m going to spend all sorts of time making precise.” Parametrize α proportional to arc-length. Choose N ∈ N s.t. 2−(N+1)`(α) < 1 ≤ 2−N `(α), i.e. 2N ≤ `(α) < 2N+1, i.e. (since we have parametrized k k+1 ∝ arc-length) 1 ≤ ` α 2N , 2N < 2. 1 1 Consider the (geodesic) triangle [x, y], [α(0), α( 2 )], [α( 2 ), α(1)]. 1 1 Pick y1 ∈ [α(0), α( 2 )] ∪ [α( 2 ), α(1)] s.t. d(y1, p) < δ 1 1 1 1 1 If y1 ∈ [α(0), α( 2 )] (say), consider the triangle [α(0), α( 2 )], [α(0), α( 4 )], [α( 4 ), α( 2 )]. 1 1 1 k k+1 Choose y2 ∈ [α(0), α( 2 )]∪[α( 4 ), α( 2 )] s.t. d(y1, y2) < δ. Continue on until you ﬁnd yN ∈ α 2N , α 2N with d(yN , yN−1) < δ.

Figure 13: Starting with p on the geodesic (red dots), we ﬁnd a sequence of points yi (successive red dots), each within δ of the previous, by δ-hyperbolicity. There are logarithmically many such points before we get within distance 1 of our rectiﬁable curve.

12 k k+1 Then d(yN , p) < Nδ, and d(yN , α([0, 1]) < 1 since ` α 2N , 2N < 2. (Consider the geodesic triangle with endpoints α k , α k+1 , and y .) 2N 2N N Hence we have d (p, α([0, 1])) ≤ Nδ + 1 ≤ δ log2(`(α)) + 1

We can use this to show, among other things, the following Theorem 1.6 (Fellow Traveller Property). If γ :[a, b] → X is a (k, c)-quasigeodesic, X is δ-hyperbolic and [γ(a), γ(b)] is a geodesic, then there exists D = D(k, c, δ) s.t. γ([a, b]) ⊂ ND([α(a), α(b)]) and [γ(a), γ(b)] ⊂ ND(γ([a, b])).

z' y' p z y

Figure 14: Sketch of argument that quasigeodesics are uniformly close to geodesics: the dotted red circle is distance D from p; we apply exponential divergence of geodesics to compare the blue rectiﬁable path with the orange geodesic path.

Proof. Let D be the Hausdorﬀ distance between γ and [γ(a), γ(b)], and let p ∈ [γ(a), γ(b)] be s.t. d(p, γ) = D. Let BD(p) be a ball of radius D around p (see diagram above.) Pick y 6= z ∈ [γ(a), γ(b)] outside BD(p), at distance D away from the boundary along [γ(a), γ(b)], and let y0, z0 be points on γ before and after (resp.) 0 0 it intersects BD(p). Note d(y, y ) ≤ d(y, γ) ≤ D and d(z, z ) ≤ d(z, γ) ≤ D. Then, by the triangle inequality, d(y0, z0) ≤ 6D; choosing γ to be a nice quasi-geodesic (and possibly changing the quasi-isometry constants to some other k0, c0 which depend only on k and c—see below), we 0 0 0 0 0 0 have `(γ([y , z ])) ≤ 6Dk + c . Then, since γ([y , z ]) avoids the ball BD(p), exponential divergence of geodesics says that we should have

0 d(p, BD(p)) = D ≤ δ log2(6Dk + c + 4D) + 1 For ﬁxed k, c, once we take large enough D, the LHS will be larger than the RHS; hence in particular there is some maximum D = D(k, c, δ) for which the required inequality will hold. Hence the any point on the quasi-geodesic γ lies relatively close to the geodesic between its endpoints, i.e. γ([a, b]) ⊂ ND([γ(a), γ(b)]). Conversely, to show that any point on the geodesic lies relatively close to the quasi-geodesic i.e.

[γ(a), γ(b)] ⊂ ND(γ([a, b]))

pick q ∈ γ([a, b]) maximising d(q, [γ(a), γ(b)]). Let p be the corresponding point on γ([a, b]). If d(q, p) ≤ D0 where D0 = D(k, c, δ) was the maximum we found above, we are done (choose D = D0.) If not, choose points q1 ∈ [γ(a), q) ⊂ γ and q2 ⊂ (q, γ(b)] ⊂ γ. Let

X = {p ∈ [γ(a), γ(b)] | d(p, q1) ≤ D0}

Y = {p ∈ [γ(a), γ(b)] | d(p, q2) ≤ D0}

13 and choose p1 ∈ X, p2 ∈ Y s.t. d(pi, qi) ≤ D0. If s and t are s.t. γ(s) = q1 and γ(t) = q2, then 0 0 `(γ([s, t])) ≤ k (2D0) + c and so 1 d(q, p) ≤ `(γ([s, t])) + max{d(p, q ), d(p, q )} 2 1 2 c ≤ k0D + + D 0 2 0

q2 q1

p p p1 2

Figure 15: Sketch of argument that geodesics are uniformly close to quasigeodesics: each of the blue segments has length ≤ D0.

0 c and if we now choose D = D1 = k D0 + 2 + D0 then we have our desired result. To show that we can take our quasi-geodesics to be nice (in particular, rectiﬁable), we will need the following slightly technical lemma: Lemma 1.7. Given a (k, c)-quasi-geodesic γ :[a, b] → X in a geodesic metric space X, there exist k0 and c0 (depending on k and c) and a (k0, c0)-quasi-geodesic β :[a.b] → X s.t. (1) β(a) = γ(a) and β(b) = γ(b). (2) The Hausdorﬀ distance between γ([a, b]) and β([a, b]) is ≤ k + c. (3) `(β[a, b]) ≤ k0d(β(a), β(b)) + c0.

Proof. Let S = ([a, b] ∩ Z) ∪ {a, b} = {a = t0 ≤ t1 ≤ · · · ≤ tn = b}. Choose β(t1) = γ(ti)∀ti ∈ S. Note that d(γ(ti), γ(ti+1) ≤ k(1)+c = k+c since γ is a (k, c)-quasi-geodesic.

Choose β|[ti,ti+1] to have image a geodesic segment [β(ti, β(ti+1)] parametrized proportional to arc-length.

1 k If p ∈ γ([ti, ti+1]), then d(p, γ(ti) ∪ γ(ti+1) ≤ k 2 + c = 2 + c. k+c If q ∈ β([ti, ti+1]), then d(q, γ(ti) ∪ γ(ti+1) ≤ 2 . k Thus the Hausdorﬀ distance between β and γ is < 2 + c (since γ(ti) ∪ γ(ti+1) ⊂ β([a, b]) ∪ γ([a, b]).) This proves (2). Given t ∈ [a, b], let r(t) be a point in S closest to t. Then k + c k + c d(β(s), β(t)) ≤ + d(β(r(s)), β(r(t)) + 2 2 = d(γ(r(s)), γ(r(t)) + (k + c) ≤ k|r(s) − r(t)| + c + k + c 1 1 ≤ k |s − t| + + + (k + 2c) γ arc-length parametrization 2 2 ∵ = k|s − t| + 2(k + c)

14 R r

Figure 16: Exponential divergence in the hyperbolic plane. The two rays span an angle θ; any path between their intersections with the outer arc that stays outside the inner sector has length ≥ θ sinh(R + r) (by hyperbolic trig) and k + c k + c + d(β(s), β(t)) + ≥ d(β(r(s)), β(r(t))) 2 2 d(β(s), β(t)) ≥ d(γ(r(s)), γ(r(t))) − (k + c) 1 ≥ |r(s) − r(t)| − c − (k + c) k 1 ≥ (|s − t| − 1) − c − (k + c) k 1 1 ≥ |s − t| − k + + 2c k k

1 and so β is k, max 2(k + c), k + k + 2c -quasi-geodesic, as desired. As defined it satisfies (1). Now let us show that it also satisfies (3): β is (k + c)-Lipschitz on [ti, ti+1) if i 6= 0 and i + 1 6= N, and

`(β([a, t1]), `(β([tn−1, b]) ≤ k + c

`(β[a, b]) ≤ (k + c)|tn−1 − t0| + 2(k + c) ≤ (k + c)|a − b| + 2(k + c)

as desired.

1.3 Divergence functions

A map e : N → R is a divergence function for X if ∀R, r ∈ N and for all geodesics α, β : [0, ∞) → X with α(0) = β(0) = 0 and d(α(R), β(R)) ≥ e(0), any path joining α(R + r) to β(R + r) outside of B(α(0),R + r) (a ball of radius R + r around α(0) = β(0)) has length ≥ e(r).6 ( e(0) = 1 Example 1.8. (0) If X is a tree, then works (since there are no geodesic paths outside e(N \{0}) = +∞ the requisite ball.)

(1) If X = H2, and we set e(0) = 1 (a normalisation constant; we can take this to be any positive number), then θ sinh R > 1 if d(α(R), β(R)) > 0, then

θ sinh(R + r) `(γ) ≥ θ sinh(R + r) > sinh r θ sinh R & where the last inequality holds up to some bi-Lipschitz (multiplicative) constant. 6“There are various deﬁnitions in the literature ... all of them awkward.”

15 Figure 17: The geodesic triangle considered here consists of the red side, plus the two black sides. The dotted geodesic is [α(R), β(R)].

Fact. A hyperbolic space X has an exponential divergence function. n−2 Speciﬁcally, if X is δ-hyperbolic, then e(n) = max{3δ, 2 δ } is a divergence function. It may also be shown that any non-hyperbolic space has a linear (or sublinear) divergence function

Proof. Suppose d(α(R), β(R)) > 3δ. Consider the geodesic triangle

α([0,R + r]), β([0,R + r]), [α(R + r), β(R + r)].

T is δ-slim, so there exists a point p ∈ [α(R + r), β(R + r)] s.t. d(p, α(R)) < δ. Suppose p ∈ β([0,R + r]). Then R − δ < d(p, α(0)) < R + δ (two applications of the triangle inequality), and so p = β(t) where t ∈ (R −δ, R +δ), and so d(p, β(R)) < δ, and so d(α(R), β(R)) < 2δ, which contradicts that d(α(R), β(R)) > 3δ. So p ∈ [α(R + r), β(R + r)]. Then d(p, α(0)) ≤ R + δ, and, applying the triangle equality, we ﬁnd that Br−δ(p) ⊂ BR+r(α(0)) (otherwise we obtain some point q outside BR+r(α(0)) with d(α(0), q) ≤ d(α(0), p) + d(p, q) ≤ R + δ + (r − δ) = R + r—contradiction.) Hence if γ is a path joining α(R + r) to β(R + r) outside BR+r(α(0)), we have d(p, γ) ≥ r − δ. r−δ−1 Then, by exponential divergence of geodesics, r −δ ≤ d(p, γ) ≤ δ log2(`(γ))+1, and so `(γ) ≥ 2 δ . There is also a converse which we will not prove here:

Theorem 1.9. If X is a proper geodesic metric space and has a superlinear divergence function e, i.e.

e(n) lim = +∞, n→∞ n then X is hyperbolic.

1.4 Dehn presentations A ﬁnite presentation Γ = hX; Ri is a Dehn presentation if whenever w is a word in X and w ∼ id, there exists a subword w0 of w which is “more than half a relator”, i.e. ∃ a word v0 s.t. `(v0) < `(w0) and v0w0 ∈ R. Fact. If Γ has a Dehn presentation, the word problem is solvable in linear time.7

Proof. Consider the following algorithm: • Given a word w in X, search w for subword which are more than half a relator • If you don’t ﬁnd any, w 6∼ id.

−1 0 • If you ﬁnd such a subword w0 (with v0w0 ∈ R and `(v0) < `(w0), then write w = aw0b ∼ av0 b = w and note `(w0) < `(w). • If w0 = id, stop (“and declare victory.”) If not, repeat above steps with w0 in the place of w. Note that this will terminate in ≤ `(w) steps (e.g. by induction on `(w).) 7Linear in the length of the word, although this is also somewhat dependent on how substring search is implemented and how operations are counted.

16 Figure 18: A k-local geodesic (for k up to 9, in fact) which is not a geodesic, in a [bi-inﬁnite] ladder.

Theorem 1.10. Hyperbolic groups have Dehn presentations (in particular, they are ﬁnitely presented and have solvable word problem.)

Remarks. • Conversely, if Γ has a Dehn presentation then Γ is hyperbolic: see Section 1.7

• e.g. Z ⊕ Z does not have a Dehn presentation (consider the word anbn; it does not contain more than half a relator.) A key step in the proof of the above theorem is a fact involving k-local geodesics:

Deﬁnition 1.11. A k-local geodesic is a path α :[a, b] → X s.t. if s, t ∈ [a, b] with |s − t| < k then α|[s,t] is geodesic. Fact. If k > 8δ, then k-local geodesics are not closed loops (or are constant.) Note that k-local geodesics need not be geodesics (in δ-hyperbolic spaces): e.g.

Proof of theorem. The idea is to construct a Dehn presentation as follows: let A be a ﬁnite generating set and let CΓ be the Cayley graph of Γ on A. Let R be the set of all words w ∼ id in A s.t. `(w) < 16δ + 2. More generally, we may choose any k > 8δ with k ∈ N and R to be the set of all words w ∼ id in A s.t. `(w) ≤ 2k. Claim. hA; Ri is a Dehn presentation.

Let z ∼ id be a word in X. Think of z as a loop based at id ∈ CΓ. By our fact, z is not a k-local geodesic; 0 hence there exists a subsegment z0 of length ≤ k which is not geodesic. If z0 is a geodesic in CΓ joining the 0 endpoints of z0, then `(z0) < `(z0). 0 Now z0 corresponds to a subword u0 of z, and z0 describes a word v0 s.t. u0 ∼ v0 and `(v0) < `(u0) ≤ k. −1 −1 −1 Then u0v0 ∼ id and `(u0v0 ) < 2k =⇒ u0v0 ∈ R, so z has a subword which is more than half a relator in R. So now we have this nice result ... but need to do the hard work on k-local geodesics: Lemma 1.12. If X is δ-hyperbolic, k > 8δ and γ :[a, b] → X is a k-local geodesic, then

(1) γ([a, b]) ⊂ N2δ(γ(a)γ(b)).

(2) γ(a)γ(b) ⊂ N3δ(γ([a, b]))

k+4δ (3) γ is (L, C)-quasi-geodesic, where L = k−4δ and C = 2δ Note that, in particular, this implies Corollary. k-local geodesics are not nontrivial closed loops

Proof. If |a − b| ≤ k, γ is geodesic, and hence not a nontrivial closed loop. Else d(γ(a+4δ), γ(a)) = 4δ since γ|[a,a+4δ] is geodesic, but then this implies γ(a+4δ) 6⊂ N2δ(γ(a)γ(b)) = B2δ(γ(a)) if γ(a) = γ(b), which contradicts (1) above. Or, in short (and less precisely), “local geodesics make steady progress.”

Proof of lemma. (1) Choose x = γ(t0) s.t. D := d(x, γ(a)γ(b) = supt d(γ(t), γ(a)γ(b). We wish to show D < 2δ. There are several cases to consider:

17 x

x y 4 z 4 < < y z x 4 D - 3 D < < D y

y' z' <

Figure 19: The setup (on the left), and why we don’t go oﬀ to the sides (two contradictions, on the right.)

(a) t0 − 4δ > a and t0 + 4δ < b “And now all we have to do is to make that incomprehensible by turning it into notation.”

Let y = γ(t0 − 4δ) and z = γ(t0 + 4δ). Choose y0, z0 ∈ γ(a)γ(b) s.t. d(y, y0) ≤ D and d(z, z0) ≤ D. Consider the geodesic triangle 0 0 zz , yz , γ([t0 − 4δ, t0 + 4δ]). Since X s δ-hyperbolic, there exists x0 ∈ zz0 ∪ yz0 s.t. d(x, x0) < δ. Claim. x0 ∈/ zz0, so x0 ∈ yz0. If x0 ∈ zz0, then d(x0, z) > 3δ by the triangle inequality; then d(x0, z0) < D−3δ, and so d(x, γ(a)γ(b)) < D − 2δ < D, which contradicts the deﬁnition of D. Now choose x00 ∈ yy0 ∪ y0z0 s.t. d(x0, x00) < δ. As above, x0 ∈ y0z0 ⊂ γ(a)γ(b). Then D ≤ d(x, x00) ≤ d(x, x0) + d(x0, x00) < 2δ. The other two cases (pictured below) are similar.

(b) t0 − 4δ ≤ a but t0 + 4δ < b (or vice versa.)

z 4 x <4 x D <4 <

<4 < (b) z' (b)

(a) (a)

Figure 20: Schematic setups for the degenerate cases—(b) on the left, (c) on the right.

Now N2δ(γ(a)x)∩γ([a, b]) and N2δ(xγ(b))∩γ([a, b]) are open in γ([a, b]) and nonempty and cover γ([a, b]). Since γ([a, b]) is connected, ∃y in their intersections. ∃p, q ∈ γ(a)x, xγ(b) (resp.) s.t. d(y, q), d(y, p) < 2δ. Now ∃x0 ∈ yp ∪ yq s.t. d(x, x0) < δ by δ- hyperbolicity, so by the triangle inequality d(x, y) < 3δ.

18 y

< x p q

(a) (b)

Figure 21: Set-up for part (2).

(3) Roughly speaking, the idea is that“quasi-geodesics make [roughly] linear progress.” 0 k 0 Let k = 2 + 2δ > 6δ. Partition the interval [a, b] as a = t0 < t1 < ··· < tm ≤ b s.t. |ti − ti−1| = k and 0 η = |tn − b| < k . 0 Notice that each γ|[ti−1,ti] is geodesic, since k < k. For each i, choose xi ∈ γ(a)γ(b) s.t. d(xi, γ(ti)) < 2δ. 0 By the triangle inequality, d(xi, xi+1) > k − 4δ > 2δ > 0.

(t2)

(t1) <2 <2

x2 x1 >2

(a) (b)

Figure 22: Crux of the argument: we choose projections which are suﬃciently separated.

k We wish to obtain something of the form d(γ(a), γ(b)) > m( 2 − 2δ) + (η − 2δ), and this we would obtain if we can show that the xi proceed monotonically along γ(a)γ(b).

Claim. x0, x1, . . . , xm appear monotonically on γ(a)γ(b).

Fix i ∈ {0, 1, . . . , m}. Choose s0 = γ(ti+1 + 2δ) and s1 = γ(ti+1 − 2δ). We have corresponding geodesic triangles T0,T1.

γ(ti)

2δ 2δ

γ(ti−1) γ(ti+1)

T0 T1 < 2δ < 2δ

xi m00 xi−1 xi+1

00 00 Figure 23: xi and m are both within 2δ of γ(ti), and then xi−1 ∈ xim puts xi−1 within 3δ of γ(ti)— contradiction.

Since T0 is δ-thin, every point of T0 lies within 3δ of γ(ti−1).; similarly, every point on T1 lies within 3δ of γ(ti+1).

Since d(γ(ti), γ(ti+1) > 6δ, d(γ(ti),T1) > 3δ, and similarly d(γ(ti),T0) > 3δ.

Look at the [geodesic] quadrilateral with vertices s0, s1, xi−1, xi+1. Using the same argument as yesterday 00 00 involving δ-slim triangles and lower bounds on d(ti, −), ∃m ∈ xi−1xi+1 s.t. d(m , γ(t)) < 2δ.

19 Suppose xi lies before xi−1 on γ(a)γ(b). Now d(xi−1, xi) > 2δ; consider the triangle with vertices 00 00 00 xi, m , γ(ti). Note xi−1 ∈ xim and d(x, γ(ti)), d(γ(ti), m ) < 2δ. 0 00 0 Since ∃xi−1 ∈ xiγ(ti) ∪ γ(ti)m with d(xi−1, xi−1) < 2δ, we may conclude d(xi−1, γ(ti)) < 3δ, which contradicts the earlier statement that γ(ti) stays at least 3δ away from T0 and T1.

Now

d(γ(a), γ(b)) ≥ m(k0 − 4δ) + (η − 2δ) b − a k ≥ − 1 ( − 2δ) + η − 2δ k0 2 k − 4δ k = (b − a) − ( − η) k + 4δ 2 ≥ L(b − a) − 2δ

and after a similar bound in the other direction (d(γ(a), γ(b)) ≤ `(γ) ≤ mk0 + η) we are done.

1.5 Dehn functions

(Sometimes also known as isoperimetric functions.) A Dehn function for a group Γ = hX; Ri is a function f : N → R s.t. if w ∼ id is a freely-reduced word Qn i −1 in X, then w can be written as a product of n ≤ f(`(w)) conjugates of relations, i.e. w = i=1 piri pi , where ri ∈ R, pi ∈ Γ, and i ∈ {±1}. i −1 Consider the Cayley graph CΓ and add in a (2-)cell for each piri pi to form a 2-dimensional CW complex with π1(KfΓ) = Γ. Assign each edge length 1 and each cell area 1; then every closed loop α ⊂ KfΓ bounds a [singular] disk Dα of area ≤ f(`(α)). 2 2 e.g. KZ⊕Z = T and KeZ⊕Z = R . `(α) If M is a simply-connected Riemannian manifold of sectional curvature ≤ − < 0, then Area(D) < .

Example 1.13. (0) For Γ = Z = ha; i, or more generally Γ = Fm = ha1, . . . , am; i, f(n) = 0 is a Dehn function (since there are no relations.)

2 2 (1) For Γ = Z = ha, b | [a, b]i, CΓ is “graph paper”, KfΓ = R .

w ∼ id =⇒ w a closed loop in the 1-skeleton of KfΓ, so it bounds a disc Dw; by the (Euclidean) 2 `(w) `(w)2 n2 isoperimetric inequality, the area of Dw is bounded above by π 2π = 4π l hence fn = 4π is a Dehn function. k k −k −k To see that any Dehn function here must be at least quadratic, look at wk := a b a b ∼ id. Now 2 2 `(wk) = 4k and the area of Dwk is k ; hence if g is any Dehn function for Z = ha, b | [a, b]i, we must n2 have g(n) ≥ 16 . 2 ◦ (2)Γ= π1(Σ2) = ha, b, c, d | [a, b][c, d]i acts on H with fundamental domain a regular (all-45 ) octagon. 2 This gives a tessellation of H by such regular octagons; CΓ is the nerve of the tessellation (and is 2 2 quasi-isometric to H .) KfΓ = H ; KΓ has a tessellation by octagons dual to the ﬁrst tessellation.

Let L be the length of each edge in CΓ and R be the area of an octagon in the dual tessellation.

w ∼ id gives rise to a loop α ∈ CΓ of length L · `(w); by the hyperbolic isometric inequality, Rn = L Area(Dα) < `(α) = L · `(w)) if it takes n conjugates of relations to reduce w to id. So n < R `(w), and L so f(t) = R t is a Dehn function for Γ.

“Now seems as good a time as any to introduce ... ”

20 1.6 Baumslag-Solitar groups and other counter-examples Or, down the rabbit-hole of a family of standard counter-examples in combinatorial group theory. The Baumslag-Solitar group with parameters m and n is deﬁned as BS(m, n) = ha, b | a−1bma = bni. 2 Note that these are HNN extensions of Z . Some examples

• BS(1, 1) =∼ Z2 • BS(1, 2) = ha, b | a−1ba = b2i has exponential Dehn function b

Relations look like a a

b b A Cayley tree for the group (with the given presentation) consists of inﬁnitely branched copies of cross-sections which look like

n Figure 24: One sheet of a Cayley graph for BS(1, 2). Note we have a−nban = b2 .

Now consider w = b−1anb−1a−nbanba−n (go over one, go all the way down, go over one downstairs, go all the way back up now in a different sheet, go back one over upstairs in the different sheet, come down again, then go back one in the intersection of the two sheets and go all the way back up in the first sheet.)

n n an a an a

b b

`(w) = 4n + 4, but we may verify that the disk it spans has exponential area 2(2n − 1) (this is an obvious upper bound; to make this argument fully rigorous we need to show this is the best possible disk for this word, i.e. that it is a lower bound as well.)

• BS(2, 3) is not Hopﬁan, i.e. it is a proper quotient of itself

Other fun facts: the metabalian group G = ha, t | (t−1at)a(t−1at) = a2i has super-exponential Dehn function (i.e. its Dehn function grows faster than any ﬁnite tower of exponentials.)

21 1.7 Back to Dehn functions and hyperbolicity We say that Γ = hX; Ri has a linear isoperimetric inequality (LIP) if it has a Dehn function of the form f(n) = kn. Observation. If Γ = hX; Ri = hY ; Si, and hX; Ri has a linear Dehn function, so does hY ; Si. Proof. Suppose f(n) = kn is a Dehn function for hX; Ri. ∃A s.t. for all y ∈ Y `X (y) ≤ kA. QB j −1 ∃B s.t. for all r ∈ R, r = j=1 pjsj pj (after rewriting r as a word in Y .) If w is a word in Y with w ∼ id, we can rewrite w in the alphabet X to obtain a freely-reduced z ∼ id Q|S| i −1 with `X (z) ≤ A`Y (w) and z = i=1 qiri qi . Then

|S|  B  Y Y ij −1 −1 w = qi  pijsij pij  qi i=1 j=1

P|S| and so |S| ≤ k`X (z) ≤ kA`Y (w); hence i=1 B ≤ kAB`Y (w), and so g(n) = kABn is a Dehn function for hY ; Si.

We may generalise the argument to obtain that: if f is a Dehn function for hX; Ri =∼ hY ; Si, then ∃A, B s.t. g9n) = Bf(An) is a Dehn function for hY ; S. Proposition 1.14. If Γ = hX; Ri is a Dehn presentation, then Γ has a LIP.

Proof. If w is a word in X with w ∼ id, we get a sequence id = w0, w1, . . . , ws = w, where s ≤ `(w) and wj = ajujbj where ujvj ∈ R with `(uj) < `(vj).

aj vj

Figure 25: One of linearly many relators obtained from a Dehn presentation.

−1 Note that wj−1 = ajvj bj = aj−1ujbj−1, and that our sequence of wj’s reduces w to id in s ≤ `(w) steps. Theorem 1.15. If Γ has a LIP, then Γ is hyperbolic.

Proof. The proof makes use of Van Kampen diagrams on putatively c-thick triangles in CΓ to obtain a contradiction. Hence all triangles in CΓ are δ-slim for some δ > 0, i.e. CΓ must be δ-hyperbolic. Choose p ∈ xy s.t. d(p, yz ∪ xz) > c. WLOG (up to an additive fudge factor) p is a vertex of the Cayley graph. Let λ ∈ (0, 1) (a concrete value for which can be backed out of the subsequent argument.) Choose q to be the point in xp closest to p s.t. d(q, yz ∪ xz) = λc. Choose r to be the point in py closest to p s.t. d(r, yz ∪ xz) = λc. Choose q0, r0 ∈ xz ∪ yz s.t. d(q, q0) = d(r, r0) = λc. In the non-degenerate case, q0 ∈ xz and r0 ∈ yz. Choose s to be the closest point to q0 on q0z s.t. d(s, yz) = λc and s0 ∈ yz s.t. d(s, s0) = λc.

22 s s'

q' r'

q r

Figure 26: A schematic illustration of the argument: ﬁlling a [hexagon inscribed inside a] thick triangle with quadratically many relators.

“By a small (≤ 2) fudge of these constants,” we can assume all these points are vertices. 0 0 0 0 Write `1 := `(qr), `2 = `(q s), and `3 = `(r s ). WLOG ` = `1 ≥ `2 + `3. By the triangle inequality, `1 > 2(1 − λ)c. 0 0 0 0 Consider the hexagon H with vertices q, q , s, s , r , r. `(H) = `1 + `2 + `3 + 3λc < 6` . H bounds a (singular) disk D. Let L0 = qr. Let ?(L0) be the union of all disks in H which touch L0. If K is the maximum length of 0 `(L0) ` any relator in R, then the Area(?(L0)) (i.e. the number of 2-cells in ?(L0)) is ≥ K = K . 0 0 0 ∃ a path L1 in ∂(?(L0)) joining qq to rr s.t. `(L1) ≥ ` − 2K by the triangle inequality (as long as c `0−2K 10 K.) Then Area(?(L1)) ≥ K . Suppose f(n) = ρn is a Dehn function for Γ = hX; Ri; then Area(D) ≤ 6`0ρ. 0 0 λc Since qq , rr have length distance λc, we can iterate this process of taking the star ≥ K times. Then Area(D) ≥ Area ?(L0) ∪ · · · ∪ ?(L λc ) K `0 `0 − 2K `0 − 2λC ≥ + + ··· + K K K λc `0 − 2λC ≥ K K λc ≥ `0 2K2

2K2ρ This gives a contradiction if we choose c > λ . In the degenerate cases, [it is an exercise to check that] a simpliﬁed version of this argument similarly gives us the desired result.

In fact, we can prove: Fact. If Γ has a Dehn function which is subquadratic, then Γ is hyperbolic.

Fact. If G and H are hyperbolic, then the free product G ∗ H is hyperbolic. Proof. G has a Dehn presentation hX; Ri and H has a Dehn presentation hY ; Si.

23 r'

q r

Figure 27: A degenerate case: a schematic illustration

Then hX ∪ Y ; R ∪Si is a Dehn presentation for G∗ H: if w is freely reduced, then w = w1w2 ··· wn where wi ∈ G ⇐⇒ wi+1 ∈ H; then w ∼ id =⇒ ∃i : wi ∼ id =⇒ wi contains more than half a relator either in R or in S.

1.8 The conjugacy problem for hyperbolic groups Theorem 1.16. The conjugacy problem is solvable for hyperbolic groups. Let us first introduce a bit of notation, and a key lemma: given Γ = hX; Ri, a word w in X is fully reduced (or cyclically reduced) if all cyclic conjugates of w (i.e. “take some stuff off front and stick it to the back”) are freely reduced. Lemma 1.17. If u and v are fully reduced and conjugate, then either (1) `(u), `(v) ≤ 8δ + 1, or (2) ∃ cyclic conjugates u0, v0 of u, v (resp.) and a word w of length ≤ 2δ + 1 s.t. wu0w−1 = v0. Proof. Let w be a minimal-length word s.t. wu0w−1 = v0.

u0 q 0 0 u1 u2 w w 0 0 v1 v2 v0 p

0 1 Observation. If p is a vertex on Sv0 (the side associated to v ), then d (p, Su0 ) > `(w) − 2 , for if not then ∃q ∈ Su0 s.t. d(p, q) < `(w). 0 0 0 0 If now we write u1 and u2 to denote the parts of u up to and after q resp., and similarly write v1 and 0 0 0 0 0 0 −1 0 0 v2 to denote the parts of v before and after p resp. (see swag picture above), then w (u2u1)(w ) = v2v1, 0 0 0 0 0 and u2u1, v2v1 are cyclic conjugates to u and v (resp.), and `(w ) < `(w), which contradicts that w was a minimal-length conjugator between any pair of cyclic conjugates for u0 and v0. Now suppose p is a midpoint of Sv0 and that `(w) > 2δ + 1, so ∃q on a vertical side of the rectangle s.t. d(p, q) < 2δ. x

q < 2δ y p `(v) 2

24 `(v) `(v) Then d(q, y) > 2 − 2δ and d(q, x) = `(w) − d(q, y) < `(w) − 2 + 2δ, so 1 `(v) `(w) − ≤ d(p, x) ≤ d(x, q) + 2δ ≤ `(w) + 4δ − 2 2

`(v) 1 and hence 2 ≤ 4δ + 2 , i.e. `(v) ≤ 8δ + 1. Symmetrically, `(u) ≤ 8δ + 1. So either `(w) ≤ 2δ + 1, or `(v), `(u) ≤ 8δ + 1, as desired.

Solution to conjugacy problem. (0) Let B1 = {w | `(w) ≤ 2δ + 1}. B1 is finite. −1 If u, v are short (`(u), `(v) ≤ 8δ+1 and are conjugate, then let wuv be s.t. wuvuwuv = v. Let B2 = {wuv}. Since there are finitely many pairs of such short words, B2 is finite.

Let B = B1 ∪ B2. (1) If u and v are words in X, ﬁrst conjugate and reduce until you ﬁnd u0, v0 conjugate to u, v (resp.) fully reduced. Note `(u0) ≤ `(u) and `(v0) ≤ `(v). (2) Check if wu00w−1 = v00 ∀w ∈ B where u00 and v00 range over all cyclic conjugates of u0 and v0 (resp.)

1.9 Finiteness in hyperbolic groups We remark also that stronger results holds:

Theorem 1.18. Γ hyperbolic =⇒ there are only ﬁnitely many conjugacy classes of ﬁnite subgroups of Γ. Remark. We note that this result can be generalised as follows: if Γ acts properly and co-compactly on a simply-connected negatively (non-positively) curved manifold, the same conclusion holds.

Proof. The idea is this: pick x0 ∈ X. If H ⊂ Γ is finite, consider H(x0). Then H(x0) has a well-defined barycenter (center of mass), because the distance function is convex. Since H(x0) is preserved by H, so is its barycenter; hence H has a fixed point h0 ∈ X. Now ∃ compact K ⊂ X which contains a point Γ-equivalent to any point in X (i.e. which contains a fundamental domain.) Hence (implicitly assuming −1 X = Cay(Γ)) ∃γ0 s.t. γ0(h0) ∈ K =⇒ γ0Hγ0 ⊂ {γ ∈ Γ | γ(K) ∩ K 6= ∅}, which is finite (since −1 −1 γ0Hγ0 (γ0h0) = γ0Hh0 = γ0h0 ∈ (γ0Hγ0 )K ∩ K. To make this rigorous we will use the following

Lemma 1.19. If X is δ-hyperbolic and Y ⊂ X is non-empty and bounded, let rY = inf{ρ > 0 | ∃x ∈ X s.t. Y ⊂ B(x, ρ)} be the radius of Y . Then for all > 0, the set C(Y ) := {x ∈ X | Y ⊂ B(x, rY + )} (the set of “-candidates for barycenters”) has diameter ≤ 4δ + 2.

Proof. Pick x1, x2 ∈ C(Y ), let m be the midpoint of x1x2, and picky ˆ ∈ Y s.t. d(ˆy, m) ≥ rY .

< r +

≥ r m yˆ

25 Then ∃p ∈ xiyˆ s.t. d(p, m) < δ. WLOG we let xi = x1. d(x1,x2) Now d(p, x1) ≥ d(x, m) − δ = 2 − δ, and so

d(p, yˆ) = d(x1, yˆ) − d(x1, p) d(x , x ) ≤ d(x , yˆ) + δ − 1 2 1 2 d(x , x ) < r + + δ − 1 2 Y 2 d(x , x ) d(m, yˆ) ≤ d(p, yˆ) + δ ≤ r + + 2δ − 1 2 Y 2

d(x1,x2) but d(m, yˆ) ≥ rY , and so rY ≤ rY + + 2δ − 2 , or d(x1, x2) ≤ 2(2δ + ) = 4δ + 2.

Hence the C(Y ) (the -coarse barycenters of Y ) are coarsely well-defined. Suppose H ⊂ Γ is finite; identify H with the set of vertices H(id) ⊂ CΓ. Let C1(H) be the set of 1-coarse barycenters. Since diam(C1(H)) ≥ 1, ∃ vertex γ ∈ C1(H). −1 −1 −1 Since H leaves H invariant, H(C1(H)) = C1(H), so γ Hγ(γ (C1(H))) = γ (C1(H)). But id ∈ −1 −1 γ (C1(H)) since γ ∈ C1(H), so γ (C1(H)) ⊂ B4δ+2(id) =: R. Hence ∀β ∈ γ−1Hγ, β(R) ∩ R 6= ∅, i.e. γ−1Hγ ⊂ {α ∈ Γ | α(R) ∩ R 6= ∅}, which is finite. Since every finite subgroup is conjugate into R, and R is finite, we conclude that there are only finitely many possibilities for the conjugacy class of H. Corollary. If A ⊂ Γ is abelian, then A is virtually cyclic.

Proof. If A contains an infinite-order element α, then A ⊂ Z(α), and hence is virtually infinite cyclic (see following section.) If A is abelian, not virtually cyclic, and does not contain an infinite-order element, then it contains infinitely many isomorphism classes of finite subgroups, which is impossible since conjugate subgroups are isomorphic.

Fact. Γ hyperbolic =⇒ Γ contains only finitely many conjugacy classes of finite-order elements. In particular, there exists an upper bound on the order of a finite-order element.

Proof. Morally, the result holds because in a hyperbolic space the distance function is convex, and so we have uniquely-defined barycenters. Let hX; Ri be a Dehn presentation for Γ. Suppose γ ∈ Γ s.t. γn = id, n > 1, and γ has minimal length of any element in its conjugacy class. Let w be a minimal-length word in X s.t. w ∼ γ. Then wn ∼ id and hence wn contains more than half n n a relator in R, i.e. ∃r = r1r2 ∈ R of w with `(r1) > `(r2) and r1 is a subword of w . −1 If r1 is contained in a copy of w, then w = ar1b ∼ ar2 b ∼ γ, and hence w was not a minimal-length representation of γ: contradiction. Otherwise, if `(w) > `(r1), then w = utv where vu = r1, since r1 is a subword of ww but not of w. Now −1 −1 −1 take a conjugate u wu = tvu = tr1 ∼ tr2 ; `(tr1) > `(tr2 ), which contradicts that γ was a minimal-length representative of its conjugacy class. Hence `(w) ≤ R := max{`(r) | r ∈ R}, and so every finite-order element of Γ is conjugate to a word of length ≤ R. There are only finitely many such words, and we are done. One moral of this story is that “there’s a lot of juice in the Dehn presentation.”

Remark. There exists a ﬁnitely-presented group which contains an isomorphic copy of every other ﬁnitely- presented group (see e.g. de la Harpe for references.)

26 1.10 Cyclic subgroups in hyperbolic groups

Theorem 1.20. If Γ is hyperbolic and γ ∈ Γ has infinite order, then the map Z → Γ given by n 7→ γn n n is a quasi-isometric embedding, i.e. ∃k, c s.t. k − c ≤ dΓ(1, γ ) ≤ kn + c (note that n = dZ(1, n) and n n dΓ(1, γ ) = `(γ ).) We also describe this by saying that “cyclic subgroups [of hyperbolic groups] are undistorted.” Remarks. (1) `(γn) ≤ `(γ), since we can take a word w = γ; then `(w) = `(γ) and `(wn) = n`(w) = n`(γ). (2) This property is not exclusive to hyperbolic groups. e.g. in Z ⊕ Z = ha, b | [a, b]i, `((arbs)n) = n(r + s) = n(`(arbs)). n (3) hai is distorted in BS(1, 2) = ha, b | bab−1 = a2i: we may show (inductively on n) that a2 = bnab−n n (e.g. b2ab−1 = b(bab−1)b−1 = ba2b−1 = (bab−1)2 = (a2)2 = a4) and so `(a2 ) ≤ 2n+1. This is k2n −c for any fixed choice of k and c as long as n is large enough. Moreover, BS(1, 2) is not isomorphic to a subgroup of a hyperbolic group. In fact, BS(m, n) is never a subgroup of a hyperbolic group (e.g. BS(1, 1) =∼ Z ⊕ Z.) (4) On the other hand, there do exist finitely-presented subgroups of hyperbolic groups which are not hyperbolic. Gromov asked the question: does there exist a finitely-presented group Γ s.t. Γ is not hyperbolic and does not contain any BS(m, n) as a subgroup? (Or, put differently, are there other obstructions to hyperbolicity than the Baumslag-Solitar groups?) This was answered in the affirmative by Brady in 1999—in fact there exist torsion-free finitely-presented counter-examples; however his examples have no finite K(Γ, 1). ∼ Definition 1.21. A finite K(Γ, 1) is a finite cell complex X s.t. π1(X) = Γ and X˜ is contractible (i.e. πn(X˜) = 1 for all n, which is equivalent to πn(X) = 1 for all n > 1 by Whitehead’s theorem.) It is known that all torsionfree hyperbolic groups have finite K(Γ, 1). We may also speak of virtually finite K(Γ, 1). New question: What if we assume finite K(Γ, 1)? Corollary. If Γ is hyperbolic, and γ ∈ Γ has infinite order, then Z(γ)/hγi is finite, where Z(γ) denotes the centralizer {β ∈ Γ | βγβ−1 = γ}

Proof. Suppose Z → Γ given by n 7→ γn is a (k, c)-quasi-isometric embedding. n n Pick a geodesic word w s.t. w = γ; then w = γ is a (k, c)-quasigeodesic path in CΓ. n By the fellow traveller property, ∃L = L(k, c) s.t. 1γ ⊂ NL(Pwn ) where Pwn is the path determined by wn. Suppose β ∈ Z(γ). Choose n > 0 s.t. d(1, γn) > 2`(β) + 4δ + 4. c β βγn = γnβ β(1γn) 2δ 1γn γ id a

Supposing d(1, γn) `(γ), we may choose a vertex a which is almost a midpoint, i.e. a ∈ 1γn and n n d(1,γ ) 1 d(a, id), d(a, γ ) ≥ 2 − 2 . n n 1 Then d(a, id), d(a, γ ) > 2δ implies that there exists a vertex c ∈ β(1γ ) s.t. d(a, c) ≤ 2δ + 2 . Now every point on the path determined by wn lies within `(w) of the form γj for some 0 ≤ j ≤ n, so, by the fellow traveller property (as applied above) ∃j ∈ N s.t. d(a, γj) ≤ L + `(w); similarly ∃k ∈ N s.t. d(c, βγk) ≤ L + `(w). j k 1 k−j 1 Hence d(γ , βγ ) ≤ 2(δ + L + `(w)) + 2 and so `(βγ ) ≤ 2(δ + L + `(w)) + 2 =: C. Thus every coset in Z(γ)/hγi has a representative of length ≤ C. Since there are only ﬁnitely many words of length ≤ C, there are only ﬁnitely many cosets.

27 We may compare the above corollary to Preissman’s Theorem, which states that if M is a negatively- curved closed manifold, then the only abelian subgroups of π1(M) are (isomorphic to) Z or the trivial subgroup. From this corollary we may deduce

Corollary. No hyperbolic group contains an isomorphic copy of Z ⊕ Z. Hence no BS(n, n) is contained in a hyperbolic Γ. Another corollary of Theorem 1.20 is the following: Corollary. If Γ is hyperbolic, γ ∈ Γ has inﬁnite order, and hγth−1 = γs for some h ∈ Γ then s = ±t.

n n Proof. If γs = hγth−1 for some h ∈ Γ and s > t, then inductively γs = hnγt h−n e.g.

2 2 h2γt h−2 = h(hγt h−1)h−1 = h(hγth−1) ··· (hγth−1)h−1 = hγs ··· γsh−1 = hγsth−1 = (hγth−1) ··· (hγth−1) 2 = γs ··· γs = γs

Suppose n 7→ γn is a (k, c)-quasi-isometric embedding. Then

n s n n − c ≤ `(γs ) = `(hnγt h−n) ≤ 2n`(h) + tn`(γ) k for all n, but this is impossible if s > t. Combining the above two corollaries, we may conclude

Corollary. H ⊂ Γ hyperbolic =⇒ H =6∼ BS(m, n) for m 6= ±n.

Corollary. No hyperbolic group contains an isomorphic copy of Z ⊕ Z (so no BS(n, n) are contained in Γ either.) And now, after exploring these consequences, we turn to the proof of the theorem itself:

nR Proof of Theorem 1.20. It suﬃces to prove that ∃R ∈ Z>0 s.t. d(1, g ) > n for all n, since this would imply k k k d(1, g ) ≥ b R c − (R − 1)`(g), where `(g) is bounded by d(1, g), but on the other hand d(1, g ) ≤ `(g)k, so we would have k − (R − 1)`(g) − 1 ≤ d(1, gk) ≤ `(g)k. R By equivariance, |s − t| < (R − 1)`(g) + 1 ≤ d(γs, γt) ≤ `(g)|s − t| R n 1 so n 7→ γ would be a (k, c)-quasi-isometric embedding (with k = max{`(g), R } and c = (R − 1)`(g) + 1.) To proceed with this we will need the following

Lemma 1.22. Given δ > 0, ∃C > 0 s.t. if n ∈ N, CΓ is δ-hyperbolic, p and q are vertices in CΓ with `(pq) > 8n + 2δ, In is a subinterval of pq of radius n about y, u ∈ Bn(p) and v ∈ Bn(q), and m is a midpoint of uv, then d(m, In) ≤ C. “It’s always a mistake to state the lemma before drawing the picture.”

28 u v m () p y q In

Figure 28: We have some control over midpoints of long geodesics.

Proof. This is (by now) a standard argument: d(m, pq ∪ pu) < 2δ by considering the two δ-slim triangles making up our quadrilateral, but d(m, pu) > 2δ (similarly to why d(m, qv) > 2δ), so ∃z ∈ pq s.t. d(m, z) < 2δ, and then

d(z, p) ≤ d(m, u) + d(u, p) + d(m, z) `(pq) + 2n + 2 ≤ + n + 2δ 2 `(pq) = + 2n + 2δ + 1 2

so d(z, y) ≤ 2n + 2δ + 1, and hence d(z, In) ≤ 2δ + 1, and d(m, In) ≤ 4δ + 1. We have now proven our lemma with C = 4δ + 1.

1 Now, armed with this lemma, we proceed. Let R be the number of vertices in BC (1); this is at least n the number of vertices in NC (In). Given n, pick k s.t. d(1, gk) > 8n + 2δ. Let y be a midpoint of 1gk. Note gsgs+k = gs(1gk), and let m be the midpoint of gsgs+k given by gs(y). s Claim. #{s ∈ N | g (1) ∈ Bn(1)} ≤ nR. s s s Suppose g ∈ Bn(1). Then, by our Lemma, d(g (y),In) ≤ C. All the g (y) are distinct, since the action of Γ on CΓ is free. Hence

s #{s ∈ N | g (1) ∈ Bn(1)} ≤ # vertices in NC (In) ≤ nR

Claim. d(1, gnR) ≥ n ∀n ∈ N.

n0R For suppose not; then ∃n0 s.t. d 1, g = n0 − < n0. For all s > 0, we can write s = asn0R + bs where 0 ≤ bs < n0R. Then

d(1, gs) ≤ d(1, gasn0R) + d(gasn0R, gs) = d(1, gasn0R) + d(1, gbs )

≤ as(n0 − ) + bs`(g)

=: as(n0 − ) + D

s If as(/2) > D then d(1, g ) ≤ as(n0 − 2 ), so ∃s0 s.t. whenever s > s0 we have s d(1, gs) < R + δ

n0−2 for some δ > 0 (satisfying R + δ ≈ n − R.) s 0 Then #{s ∈ N | d(1, g ) < n} ≥ n(R + δ) − s0, which is eventually bigger than nR, contradicting what was argued above. The proof of this last claim establishes the theorem, as was argued at the beginning of this proof.

29 1.11 Quasiconvexity

Deﬁnition 1.23. If X is a proper geodesic metric space, we say A ⊂ X is k-quasiconvex if for all a1, a2 ∈ A, a1a2 ∈ Nk(A) (for all geodesics a1a2.

Figure 29: An example of a k-quasiconvex set in the plane

If Γ is a group with presentation Γ = hX; Ri, and H ⊂ Γ is a subgroup, then H is quasiconvex (i.e. k-quasiconvex for some k ≥ 0) if H is a quasiconvex subset of CΓ. Example 1.24. • Cyclic groups are quasiconvex subsets of hyperbolic groups. • Finite-index subgroups are quasiconvex (since they are coarsely dense.) • Note that quasiconvexity is dependent on the presentation. e.g. in Z ⊕ Z = ha, b | [a, b]i, hai is quasiconvex but habi is not

Figure 30: Paths of the form anbn are geodesics in Z ⊕ Z = ha, b|[a, b]i, but can get arbitrarily far from habi ⊂ ha, b|[a, b]i.

However in hb, ab | [b, ab]i =∼ Z ⊕ Z, habi is quasiconvex but hai is not. Exercise. In the hyperbolic setting, quasiconvexity of a subgroup is independent of the choice of generating set. Hint: fellow traveller property.

Theorem 1.25. If Γ is hyperbolic and H ⊂ Γ is quasiconvex, then H is hyperbolic. In particular, H is finitely presented. Remarks. (1) (Brady) There exists a hyperbolic group with non-hyperbolic subgroups (i.e. the above result is not trivial.) (2) There is a general construction, due to Rips, which, given a finitely-generated group G, builds a short exact sequence 1 N Γ G 1 where Γ is hyperbolic. If G is infinite, then N is finitely generated but not finitely presented. (3) The converse does not hold: hyperbolic subgroups of hyperbolic groups need not be quasiconvex. The counterexamples here are rather canonical—they come from hyperbolic 3-manifolds. e.g. take a surface Σ of genus g ≥ 2 and define S = Σ × [0, 1], let φ : S → S be a psuedo-Anosov 3 (“generic”) self-homeomorphism. Then the mapping torus Mφ = H /Γ, but π1(Γ) ⊂ Γ is not quasiconvex.

30 To prove this we will first need a Lemma 1.26. If Γ = hX; Ri is finitely generated and H ⊂ Γ is quasiconvex, then H is finitely generated, and the inclusion map ι : H,→ Γ is a quasi-isometric embedding.

Remark. The converse is not true in general. e.g. in Γ = Z ⊕ Z = ha, b | [a, b]i, habi is not quasiconvex, but it is quasi-isometrically embedded in Γ. Proof. Suppose H is k-quasiconvex in Γ = hX; Ri. This is equivalent to ι(h1)ι(h2) ⊂ Nk(H) ⊂ CΓ whenever h1, h2 ∈ H. 8 Given h ∈ H, let w be a geodesic word in Γ represented by w = xn ··· x1.

hi-1 hi ui-1 ui ... x2 ai-1 ai x1 1

−1 ∀i : ∃hi ∈ H s.t. dCΓ (hi, ai) ≤ k. Hence there exists a word ui in X s.t. `(ui) ≤ k and hi = ui ai, so −1 −1 hihi−1 = ui xiui−1. Choose u0 = un = 1, h0 = 1, hn = h. Then

h = xn ··· x1 −1 −1 −1 = (un xnun−1)(un−1xn−1un−2) ··· (u1 x1u0) −1 −1 −1 = (hnhn−1)(hn−1hn−2) ··· (h1h0 ) = hn = h

−1 and `X (hihi−1) ≤ 2k + 1 ∀i. Now let S = {h ∈ H | `x(h) ≤ 2k + 1}. By the above, S is a ﬁnite generating set for H. Note moreover that dS(1, h) ≤ dX (1, h) ≤ (2k + 1)dS(1, h) so that ι : H,→ Γ is a quasi-isometric embedding, as desired, Proof of Theorem 1.25. Recall that if A and B are proper geodesic metric spaces and h : A → B is a quasi-isometric embedding, and B is hyperbolic, then A is hyperbolic (see Corollary to Theorem 1.6.) Since ι : H,→ Γ extends to a quasi-isometric embedding between CH and CΓ (by equivariance), we see that CH is hyperbolic. 8Note we seem to be reversing, in this proof, the sign conventions we have been using all along, for some reason.

31 Corollary. If Γ is hyperbolic, then H ⊂ Γ quasiconvex ⇐⇒ H,→ Γ is a quasi-isometric embedding. In particular, for hyperbolic groups quasiconvexity of subgroups is independent of the choice of generators. Proof. The Lemma gives us the =⇒ direction. For the other direction, suppose Γ is δ-hyperbolic, and H,→ Γ is a (k, c)-quasi-isometric embedding. If h1h2 is a geodesic in CH , ι(h1h2) is a (k, c)-quasi-geodesic in CΓ. The fellow traveller property now tells us that ∃L = L(k, c, δ) s.t. NL(ι(h1h2)) ⊃ ι(h1)ι(h2) where ι(h1)ι(h2) is a geodesic in CΓ joining h1 to h2. Thus every point in ι(h1h2) lies within k + c of an element of H =⇒ ι(h1)ι(h2) lies within k + c + L of H =⇒ H is (k + c + L)-quasiconvex. Quasiconvexity can also be used to formulate a common generalization (for hyperbolic groups) of the following classical theorem/s

Theorem 1.27 (Howson’s Theorem). If H1,H2 ⊂ Fn are finitely-generated, then H1 ∩ H2 is finitely- generated. To state the other result we first require some terminology:

n 9 n n Deﬁnition 1.28. H ⊂ Isom(H ) is convex co-compact if ∃x0 ∈ H (or ∀x0 ∈ HH : these formulations n are equivalent by equivariance [?]) s.t. the orbit map τ :Γ → H given by γ 7→ γ(x0) is a quasi-isometric embedding10.

n To see why the name “convex co-compact”: let Λ(Γ) := Γx0 ∩ ∂H . (In fact, Γ also has a boundary ∂Γ and τ extends to a mapτ ˆ :Γ ∪ ∂Γ → Hn ∪ ∂Hn.) Now the convex hull CH(Λ(Γ)), i.e. the smallest closed Γ-invariant convex subset of Hn containing Λ(Γ) in its closure, is formed by the union of all ideal simplices with endpoints in Λ(Γ). Hence k-quasiconvexity =⇒ CH(Λ(Γ)) ⊂ NL(Γx0) =⇒ CH(Λ(Γ))/Γ is compact, and convex as a subset of Hn/Γ. n Theorem 1.29 (Susskind). If Γ ⊂ Isom(H ) is discrete, and H1,H2 ⊂ Γ are convex co-compact, then H1 ∩ H2 is convex co-compact. The common generalisation is this:

Theorem 1.30. If Γ is hyperbolic and H1,H2 ⊂ Γ are quasiconvex, then H1 ∩ H2 is quasiconvex. Note that

Theorem 1.31 (Thurston). Γ convex co-compact but not co-compact and H ⊂ Γ ﬁnitely generated =⇒ H convex co-compact

Definition 1.32. Γ has the finitely-generated intersection property (FGIP) if H1,H2 ⊂ Γ finitely- generated =⇒ H1 ∩ H2 finitely-generated. From these results of Thurston and of Susskind we may conclude

Corollary. If Γ ⊂ Isom(Hn) is convex co-compact but not compact, then Γ has the FGIP. Example 1.33. Let Σ be a surface of genus g ≥ 2 and φ :Σ → Σ be a psuedo-Anosov self-homeomorphism. 3 Let M = Mφ be the mapping torus of φ. Note φ pseudo-Anosov implies Mφ = H /Γ (for some hyperbolic group Γ), by geometrization.

There exists a short exact sequence 1 π1(Σ) Γ Z 1 and we may write Γ = −1 hπ1(Σ), t | tat = φ(a) ∀a ∈ π1(Σ)i, “the standard form for a HNN extension”. Notice that the limit sets agree: Λ(π1(Σ)) = Λ(Γ), since π1(Σ) C Γ and Λ(Γ) is the smallest closed non-empty Γ-invariant subset.

9 2 3 Fun facts: Isom(H ) =∼ PSL(2, R); Isom(H ) =∼ PSL(2, C). 10By the Lemma above, this is equivalent to quasiconvexity of Γ.

32 Now CH(Λ(π1(Σ))) = CH(Λ(Γ)), so CH(Λ(π1(Σ)))/π1(Σ) is not compact, i.e. π1(Σ) is not convex co-compact. Pick γ ∈ [π1(Σ), π1(Σ)]. Let H = hγ, ti; “by 3-manifold theory” (or by passing to high enough powers), ∼ H = F2. Suppose α ∈ H ∩ π1(Σ). Then

α = tm1 γn1 tm2 γn2 ··· γnr tmr+1

and m1 + ··· + mr+1 = 0. Now since tm1 γntm2 = (φm1 (γ))n1 t−m1 tm2 , we may inductively write

α = tm1 γn1 tm2 γn2 ··· γnr tmr+1 = tm1 γn1 t−m1 tm1+m2 γn2 t−m1−m2 ··· γnr t−m1−m2−···−mr = (tm1 γt−m1 )n1 (tm1+m2 γt−m1−m2 )n2 ··· (tm1+···+mr γt−m1−···−mr )nr

n −n so G ∩ π1(S) = ht γt | n ∈ Ni = F∞ γ γ γ γ γ

γ t ··· ··· t−2 t−1 id t t2

Note that this example is not at all esoteric, in fact it is rather generic: Theorem 1.34 (Agol). If M is a closed hyperbolic 3-manifold, then there exists a ﬁnite cover Mˆ of M which ﬁbers over S1. This result was known as the Virtually Fibered Conjecture.

Corollary. π1(M) does not have the FGIP, since π1(Mˆ ) does not have the FGIP. Thus we have this stark contrast:

Corollary. If Γ is convex co-compact in Isom(H3) , then Γ has the FGIP iﬀ Γ is not co-compact (i.e. H3/Γ is not closed. Remark. By applying geometrization and some other technical arguments, we may ﬁnd that the same holds for all discrete subgroups Γ ⊂ Isom(H3).

Taking inspiration from negatively-curved spaces 2 In H , if B is a ball disjoint from a geodesic γ, and πγ is nearest-point projection onto γ, then πγ (B) has uniformly bounded diameter.

horoball

Nearest-point projection onto geodesics (and even quasi-geodesics) is coarsely well-deﬁned in a hyperbolic metric space. Moreover, balls suﬃciently far from the geodesic project to uniformly bounded sets. Similarly, we can hope to transfer (coarsely) properties of negatively-curved spaces to hyperbolic groups or spaces.

33 1.12 The Tits Alternative

Theorem 1.35 (Tits Alternative for SLn(k)). If Γ ⊂ SLn(k) with char(k) = 0, then either (1) Γ is virtually solvable, or (2) Γ contains a free group. Exercise. Solvable subgroups of hyperbolic groups are virtually cyclic.

th Hint. The idea is that if HR (the R term in the commutator series) is trivial, then HR−1 is abelian and hence virtually cyclic; since HR−1 is a normal subgroup of HR−2, HR−2 sits inside the normalizer Z(HR−1); inductively (there is some work to be done there) we may show that H0 = Γ is also virtually cyclic. There is an analogous Theorem 1.36 (Tits Alternative for Hyperbolic Groups). Γ hyperbolic =⇒ either (1) Γ is virtually cyclic, or (2) Γ contains a free group of rank ≥ 2. which we will work towards next.

The SL2(R) case ∼ 2 Consider γ ∈ PSL2(R) = Isom+(H ). One of three cases holds: λ (1) γ hyperbolic, i.e. γ conjugate to with γ > 1. λ−1

1 1 (2) γ parabolic, i.e. conjugate to 1

cos θ sin θ (3) γ elliptic, i.e. conjugate to . − sin θ cos θ In case (1), γ may be represented (is conjugate to) z 7→ λ2z, which preserves the y-axis (call this the axis of γ, ﬁxes 0, ∞ ∈ ∂H2, and translates along the axis by 2 log λ. After conjugating back, we obtain that γ is translation along some invariant geodesic (translation axis) by 2 log λ.

∂−γ ∂+γ

In case (2), γ may be represented by z 7→ z + 1: the translation distance → 0 as =(z) → ∞, and we have one fixed point in ∂H2 After conjugating back, we obtain that γ is a flow along horocycles based at some point (the fixed point) in ∂H2.

34 In case (3), γ ﬁxes a point inside H2 and rotates around it (in the case of our chosen conjugacy representative, a rotation of 2θ about i.) Now we play

Ping-Pong Lemma 1.37. If Γ contains 2 hyperbolic elements with distinct ﬁxed point sets, then Γ contains a copy of F2.

− + − + ∂ γ1 ∂ γ1 ∂ γ2 ∂ γ2

− − + + ∂ γ1 ∂ γ2 ∂ γ1 ∂ γ2

Figure 31: Ping-pong (two possible Schottky group pictures.)

n n Proof. We work in the upper-half-plane. Choose n large enough s.t. C1, γ (C1),C2, γ (C2) are all disjoint. n n n n Let H be hγ1 , γ2 i, and pick x0 above C1, γ (C1),C2, γ (C2). n sr n tr n s1 n t2 n Pick w = (γ1 ) (γ2 ) ··· (γ1 ) (γ2 ) freely-reduced. We can show, using ping-pong (i.e. since γ2 (x0) lies under one of the four arcs, and then we keep bouncing between the spaces underneath four arcs, so W (x0) lies under one of the four arcs), that W (x0) 6= x0. 2 ∼ Hence every freely-reduced word acts non-trivially on H , so H = F2. And, after more argument, we may ﬁnd that, if the hypotheses of the ping-pong lemma are not satisﬁed, then Γ is virtually solvable, i.e. we have the Tits Alternative. Now we wish to replicate the outlines of this proof in δ-hyperbolic spaces. To do so, we will need a notion of the boundary of such a space, which will play the role of ∂H2.

The boundary of a δ-hyperbolic space We may think of ∂ 2 as T 1 ( ), the unit tangent bundle at a distinguished point. But we wish to remove H x0 H2 the dependence on the distinguished point, and then we obtain asymptotic classes of geodesic rays, where the asymptotic equivalence is given by bounded Hausdorff distance (not vanishing Hausdorff distance—think of the bi-infinite ladder)

Let X be a proper geodesic δ-hyperbolic metric space, and α, β : [0, ∞) → X be geodesic rays. Deﬁnition 1.38. We say α is asymptotic to β (denoted α ∼ β) if the Hausdorﬀ distance

dHaus (α([0, ∞)), β([0, ∞))) < +∞

Equivalently, we may say

35 Lemma 1.39. α ∼ β ⇐⇒ ∃k : d(α(t), β(t)) < k ∀t. Proof. The ⇐ direction is obvious. To prove the =⇒ direction, let r = d(α(0), β(0)) and k = dHaus(im(α), im(β)). Fix t, and choose s so that α(s) is exactly the Hausdorﬀ distance k away from β(t).

α(t) α(0) k r β(s) β(0)

Notice that s = d(α(s), α(0)), so, applying the triangle inequality to β([0, t]) and to α([0, s]), we obtain

t − (r + k) ≤ d(α(s), α(0)) ≤ r + t + k

Hence s ∈ [t − (r + k), t + (r + k)], i.e. d(α(s), α(t)) ≤ r + k, and so d(α(t), β(t)) ≤ (r + k) + k = 2k + r for all t. Proposition 1.40 (Visibility Properties). (1) If p ∈ X and z ∈ ∂X, then there exists a geodesic ray α s.t. [α] = z and α(0) = p. (2) If z1 6= z2 ∈ ∂X, then ∃ geodesic γ : R → X s.t. γ|[0,∞) = z2 and γ(−x)|[0,∞) = z1. Thus we may define ∂X as the set of all geodesic rays in X modulo the equivalence relation given by asymptoticity. Remark. The geodesics in (2) need not be unique: consider again (e.g.) the bi-infinite ladder. Proof. There are two main ingredients: δ-hyperbolicity, and Arzelà-Ascoli.

(1) Suppose z = [β] and β(0) 6= p.

We will construct a geodesic ray α ∼ β with α(0) = p, as follows: deﬁne αn : [0, tn] → X s.t. αn is geodesic and αn(0) = p, αn(tn) = β(n).

β β(n) β(n + 1) ···

αn αn+1 p = α(0)

Figure 32: Constructing rays with prescribed asymptotic class and starting point, by taking successive approximations αn from the given starting point to points further and further out along an arbitrary geodesic ray representative.

By Arzel`a-Ascoli, αn converges (passing to a subsequence if needed) to some α : [0, ∞) → X a geodesic ray (note that tn → ∞ since tn = d(α(0), β(n)) ≥ d(β(0), β(n)) − d(α(0), β(0)) = n − d(α(0), β(0)).)

By considering the [geodesic] triangle with sides α([0, tn]), β([0, tn]), α(0)β(0), we see that d(α(t), β([0, n])) ≤ r + δ for all t ∈ [0, tn], where r := d(α(0), β(0)).

So dHaus(αn([0, tn]), β([0, n])) ≤ r + δ and, since also αn → α, we have

dHaus (α([0, ∞)), β([0, ∞))) ≤ r + δ

and hence [α] = [β] (the Hausdorﬀ distance between the two geodesic rays is bounded.)

36 z1 z2

β(n) α(n) γn β α

Figure 33: Constructing rays with prescribed asymptotic classes by taking successive approximations γn.

(2) Pick p ∈ X,[α] = z2, [β] = z1 with z1 6= z2 s.t. α(0) = β(0) = p. Notice that ∃T s.t. whenever t > T , d (α(t), β([0, ∞))) > 2δ and d (β(t), α([0, ∞))) > 2δ.

Claim. ∃qn ∈ β(n)α(n) s.t. d(qn, p) ≤ T + δ.

Proof. Let A = {q ∈ β(n)α(n) | d(q, α([0, ∞))) < δ} and B = {q ∈ β(n)α(n) | d(q, β([0, ∞))) < δ}.

Since X is δ-hyperbolic, α(n)β(n) ⊂ A ∪ B. If n > T , then β(n) ∈/ A and α(n) ∈/ B, so ∃qn ∈ A ∩ B, since A and B are open and β(n)α(n) is connected.

α(n) qn β(n)

< δ

0 0 0 0 Hence there exist t, t s.t. d(qn, α(t)) < δ and d(qn, β(t )) < δ but one of t, t < T (since d(α(t), β(t )) < 0 2δ, t, t ∈ T ) and so d(an, p) ≤ T + δ.

Now choose γn :[sn, tn] → X with sn ≤ 0 ≤ tn s.t. γn(sn) = beta(n), γn(0) = qn, and γn(tn) = α(n).

Since γn(0) lies in a compact (closed and bounded) set, γn is a geodesic, sn → −∞, and tn → +∞, γn converges to a geodesic γ : R → X (by Arzel`a-Ascoli.) We can check that γ|[0,∞) = [α]: using δ-hyperbolicity, dHaus(γn([0, tn]), α([0, n])) ≤ (T + δ) + δ, and so dHaus (γ([0, ∞)), α([0, ∞))) ≤ T + 2δ < ∞, and we are done. We may similarly verify that γ|[0,−∞ = [β].

We now proceed to deﬁne a topology on X ∪ ∂X as follows: ﬁx p ∈ X. Identify x ∈ X with the geodesic segment/s joining p to x and z ∈ ∂X with the geodesic rays (from p.) We say yn → y if any sequence of geodesic segments (or rays) αn s.t. [αn] = yn has a subsequence converging to α (in the compact-open topology) and [α] = y. Notice this gives the usual topology on X.

Deﬁnition 1.41. C ⊂ X ∪ ∂X is closed if whenever (yn) ⊂ C and yn → y, then y ∈ C. Exercise. This topology is independent of the choice of p. Theorem 1.42. X ∪ ∂X, with the topology deﬁned above, is compact.

Proof. If yn ∈ X ∪ ∂X, then one gets a sequence αn : [0, rn] → X of geodesic segments or rays s.t. α(0) = p and either α(rn) yn ∈ X or rn = ∞,[αn] = yn. We can pass to a subsequence s.t. rn converges to r ∈ R or to +∞. If rn → r, then up to subsequence yn → y ∈ X (since X is proper); if rn → +∞, then up to subsequence αn converges to α : [0, ∞) → X a geodesic ray, and so [yn] converges (up to [the same] subsequence) to [y] ∈ X.

37 Hence X ∪ ∂X is sequentially compact. We may also ﬁnd that X ∪ ∂X is second-countable, and hence also compact. Corollary. ∂X is compact (since it is closed in X ∪ ∂X.) Fact. Any quasi-geodesic ray is asymptotic to a geodesic ray, i.e. if α : [0, ∞) → X is a (k, c)-quasi-geodesic, then ∃γ : [0, ∞) → X geodesic s.t.

dHaus(α([0, ∞)), γ([0, ∞))) < ∞.

Proof. For any n ∈ N, let γn be a geodesic segment s.t. γn(0) = γ(0) and γn(rn) = α(n). Since α is quasi-geodesic, rn = d(α(0), α(n)) → ∞. Therefore ∃L s.t. dHaus(α([0, rn]), γn([0, rn])) ≤ L; up to subsequence, γn → γ : [0, ∞) → X a geodesic ray, and dHaus(γ, α) ≤ L. Corollary. Quasi-geodesic rays converge to a well-deﬁned point in ∂X. Thus we could also have deﬁned ∂X as the set of quasi-geodesic rays modulo asymptotic equivalence. We can now also extend quasi-isometric embeddings to the boundary—and there they can become home- omorphisms!

Quasi-isometry and δ-thinness on ∂X Fact. If X and Y are hyperbolic metric spaces and f : X → Y is a quasi-isometric embedding, then ∃ a topological embedding ∂f : ∂X → ∂Y s.t. ∂f([α]) = [f ◦ α]. Proof. ∂f is well-deﬁned since α is a geodesic ray implies f ◦ α a quasi-geodesic ray, and if α ∼ β, then f ◦ α ∼ f ◦ β. ∂f is injective, since [α] 6= [β] implies dHaus(α, β) = +∞ and so dHaus(f ◦ α, f ◦ β) = +∞, and so f ◦ α 6∼ f ◦ β, and hence, ﬁnally, ∂f([α]) 6= ∂f([β]). Claim. ∂f is continuous.

Proof. Suppose zn → z in ∂X. Then ∃ geodesic rays αn : [0, ∞) → X s.t. αn(0) = p and [αn] = zn.

Every subsequence of αn has a subsequence αnk s.t. αnk → α∞ where [α∞] = z.

Then [f ◦ αnk ] = ∂f([zn]); ∃L s.t. ∀n ∃ geodesic ray γn s.t. dHaus(γn, f ◦ αn) ≤ L.

Passing to a further subsequence if required, γnk converges to a geodesic ray γ, and dHaus(γ, f ◦α∞) ≤ L.

So, up to subsequence, ∂f(znk ) → [γ] = [f ◦ α∞] = ∂f(z), so ∂f(zn) → ∂f(z), and hence ∂f is continuous.

Corollary. If X and Y are quasi-isometric, then ∂X is homeomorphic to ∂Y .

Exercise. (1) Hn quasi-isometric to Hn ⇐⇒ n = m (since ∂Hn is homeomorphic to the n-sphere. ∼ 2 ∼ 1 (2) Fn is not quasi-isometric to π1(Σ) for any compact surface Σ, since ∂ Cay(π1(S), Γ) = ∂H = S , but ∂Fn is homeomorphic to a Cantor set.

Fact. If α, β : [0, ∞) → X are geodesic rays with α(0) = β(0) =: p and α ∼ β, then dHaus(α, β) < δ. Proof. Here’s a sketch of one:

> k + δ α(n)

p < δ k := dHaus(α, β) β(n)

Fact. If α, β : R → X are geodesics with the same endpoints in ∂X, then dHaus(α, β) < 2δ.

38 Proof. Apply the above argument twice:

(-n) (n) (0)

r k+r k+r (0) (-n) (n)

Fact. If α, β, γ are 3 bi-inﬁnite geodesics in X which form an ideal triangle11, then if x ∈ α, d(x, β ∪γ) < 2δ.

Proof. d(α(t), α(n)β(n) ∪ α(−n)β(n)) < δ for each n ∈ N by δ-thinness. WLOG d(α(t), α(n)β(n)) < δ for inﬁnitely many n.

(-n) (n)

(n)

Up to subsequence, α(n)β(n) → γˆ, a geodesic with the same endpoints as γ. Then d(α(t), γˆ) < δ and dHaus(γ, γˆ) < 2δ =⇒ d(α(t), γ) < 3δ. Other true facts which we may not prove in this course (their proof might require the use of heavier machinery such as the Gromov product):

• ∂X admits a “visual metric” (in fact a class of visual metrics)—e.g. the visual metric on ∂Hn is the 1 n standard spherical metric on Tp H . • ∂f is quasi-conformal, i.e. ∃k > 1 s.t.

sup {d(∂f(x), ∂f(y)) | d(x, y) = r} lim sup x,y∈∂X ≤ k. r→0 infx,y∈∂X {d(∂f(x), ∂f(y)) | d(x, y) = r}

n • If Γ1, Γ2 ⊂ Isom+(H ) with n ≥ 3, and Γ1 and Γ2 are co-compact, then Γ1, Γ2 conjugate by a quasi- n n conformal map of ∂H iﬀ Γ1, Γ2 are conjugate in Isom(H ) (i.e. by a conformal map.)

• (Paulin) If Γ1 and Γ2 are hyperbolic groups, then Γ1 and Γ2 are quasi-isometric iﬀ ∂CΓ1 is quasi-

conformally homeomorphic to ∂CΓ2 .

Isometries of a hyperbolic space Deﬁnition 1.43. An isometry of a hyperbolic space X is

n (a) elliptic if {γ (x0)} is bounded

n (b) parabolic if {γ (x0)} has exactly one accumulation point.

n (c) hyperbolic if the orbit map/s τ : Z → X given by n 7→ γ (x0) is an / are quasi-isometric embedding/s. 11 α, β, γ form an ideal triangle if ∂+α = ∂−β, ∂+β = ∂−γ, and ∂+γ = ∂−α.

39 Remarks. • If γ is an inﬁnite-order element of a hyperbolic group, we know (have proved) that the action of γ on CΓ is hyperbolic. • If γ is ﬁnite-order, it is elliptic.

• In H2, a hyperbolic element γ has an axis A, i.e. a geodesic s.t. γ(A) = A. The same is not necessarily true in a δ-hyperbolic space, although “you could get spot-on invariance by passing to a power, or live with coarse invariance”—∃ 2 ﬁxed points on boundary, and this gives us a quasi-ﬁxed [quasi-]geodesic. e.g. consider X = S1 × R with action given by γ(z, t) = (e2πiαz, t + 1). Geodesics in X have the form n {z0} × R, and are not preserved by γ or by γ for any n > 0. On the other hand, if Γ = Zn ⊕ Z, CΓ is the regular n-gon ×R, and γ = (1, 1) acts by rotation by 2π n n and translation by 1, then there are no γ-invariant geodesics, but γ = (0, n) does have invariant geodesics.

n Lemma 1.44. If {γ (z0)} accumulates at a ∈ ∂X, then ∂γ(a) = a.

n n Proof. Suppose a = lim γ (x0). If y ∈ X, then γ (y) (we claim) converges to a. n n ni ni dHaus(pγ i (y), pγ i (x0)) ≤ r + δ where r := d(x0, y), since d(γ (x0), γ (y) = r. ni Moreover d(p, γ (x0)) → ∞ in X. n n Hence pγ i (x0) → α and pγ i (y) → αˆ geodesic rays. Now dHaus(ˆα, α) ≤ r + δ =⇒ [ˆα] = [α] = a; hence lim γni (y) = a, and so ni ni+1 ∂γ(a) = γ(lim γ (x0)) = lim γ (x0) = a.

So if γ is not elliptic or parabolic, then ∂γ has ≥ 2 fixed points, so ∃ geodesic A s.t. ∂A is fixed by ∂γ, and so dHaus(A, γ(A)) < 2δ. Fact. If γ is not elliptic or parabolic, then γ is hyperbolic and ∂γ (in fact γ¯) has North-South dynamics, i.e. ∃ fixed points a and b of ∂γ and γn(z) → a if z ∈ ∂X − {b} (uniformly on compact subsets of ∂X − {b}.) Proof. Omitted (see Gromov.) Fact. If γ ∈ Γ has infinite order and Γ is hyperbolic, then ∃m > 0 s.t. γm has an invariant geodesic axis (not necessarily unique.) Proof. We construct a lexicographically-least (lex-least) geodesic joining the 2 fixed points of ∂γ. Since ∃ only finitely many lex-least geodesics between any 2 points, γm must leave one of them invariant for some m > 0. 0 0 0 Let X = {x1, . . . , xn} order the generating set. Given w, w words in X, w < w if either (a) `(w) < `(w ) 0 0 or (b) `(w) = `(w ) and w occurs before w in dictionary order (e.g. x1x2x4 < x1x3x4.) There exists a unique lex-least path joining any two vertices in CΓ.

x1 h

Figure 34: The top path from h to g is lex-least; the bottom is not.

Claim. Call a (bi-infinite) geodesic I special if for every r, s ∈ Z the segment I([r, s]) is lex-least. We claim that there exist only finitely many special bi-infinite geodesics joining ∂−γ to ∂+γ (and there exists at least one such geodesic.)

40 n Proof. Existence: Consider τ : Z → Cγ given by n 7→ γ , extended to a quasi-geodesicτ ˆ : R → CΓ. −m m Let Im be a lex-least geodesic joining γ to γ . Since τ is a quasi-isometric embedding, ∃L s.t. dHaus(Im, τˆ([−m, m])) < L. Let Im,n be a minimal geodesic segment in Im whose endpoints lie within L of n γ . There exist ﬁnitely many choices for Im,n.

Up to subsequence Im,n is constant as m → ∞. We can diagonalize to produce a subsequence s.t. Imj ,n is constant for all mj.

Let I = lim Imj . Then dHaus(I, τ(Z)) < L, so ∂I = ∂τ(Z) = ∂+γ ∪ ∂−γ. Finiteness: If I and I0 are 2 special geodesics with the same endpoints and I(0) = I0(0), then, for all 0 n ∈ Z, d(I(n),I (n)) < 2δ; hence there are ≤ #{B2δ(1)} =: N choices for I(n). But I(n),I(0) determine I([0, n]) due to lex-leastness, so ∃ ≤ N choices for I([0, n]) for all n, and hence exists ≤ N choices for I([0, ∞)). Similarly, there are ≤ N choices for I((−∞, 0]). 2 Hence there are ≤ N choices for I(R); but ∃ ≤#B1+δ(1) =: R choices for I(0) (up to reparametrization), and so there are ≤ N 2R < ∞ choices for I. Now observe the image of a special geodesic under γ is special. Since γ fixes ∂+γ and ∂−γ, the image under γ of a special geodesic joining ∂+γ and ∂−γ is a special geodesic joining ∂+γ to ∂−γ. Since there are finitely many special geodesics joining ∂+γ to ∂−γ, there exists n > 0 and some such special geodesic I s.t. γn(I) = I. Definition 1.45. The stable translation length (often just called the translation length) of a group element γ is defined by

dCΓ,X (1, γn) τΓ,X (γ) := lim n→∞ n Remarks. (1) τ is well-defined. n n+m m m n+m m dCγ,X (1, γ ) is sub-additive as a function of n, since d(1, γ ) ≤ d(1, γ ) + d(γ , γ ) = d(1, γ ) + d(1, γn). Now for any subadditive function the limit above is well-defined: for any fixed d ∈ N, we can write n = dq + r where 0 ≤ r < d.Then

f(n) f(dq + r) qf(d) + f(r) f(d) f(r) = ≤ ≤ + n dq + r dq + r d n

f(n) f(d) and so lim supn→∞ n ≤ d for all d ∈ N. Then f(n) f(d) f(n) lim sup ≤ lim inf = lim inf n→∞ n d→∞ d n

f(n) and so limn→∞ n exists.

n n f(rn) nf(rn) (2) τ(γ ) = nτ(γ), since τ(γ ) = lim r = lim rn = nτ(γ). (3) τ is conjugacy-invariant: consider α = xγx−1. `(γn) − 2`(x) ≤ `(αn) = `(xγnx−1) ≤ 2`(x) + `(γn); taking the limit of the ratio over n as n → ∞, we obtain τ(xγx−1) = τ(γ).

Corollary. τ(γ) ∈ Q. n Proof. If γ has ﬁnite order, then dCΓ,X (1, γ ) is bounded and so τ(γ) = 0. If γ has inﬁnite order, then there exists m > 0 and some geodesic I s.t. γm(I) = I. Pick x a vertex of I. Then α = x−1γnx leaves invariant the geodesic x−1I. α acts by translation by d ∈ Z on x−1I, and so τ(α) = d ∈ Z (since d(1, αn) = nd. m m τ(γ ) Hence τ(γ ) = d ∈ Z, and so τ(γ) = m ∈ Q.

41 Figure 35: The geodesic segment between the red markers on the left is projected to the closed, not [necessarily] simple geodesic loop on the quotient surface (on the right.)

Hyperbolic ping-pong

Now remember our model setting from the hyperbolic plane: if Γ ⊂ Isom(H2) is torsion-free, discrete, and co-compact, let Σ = H2/Γ; then τ(γ) is the length of the closed geodesic on Σ in the homotopy class of γ Now—this is an aside—we can look at things like the number of geodesics of (bounded by) some fixed length; the growth function of this quantity is known as the entropy of the manifold (?) We will not study this, but only say one thing about this subject (“the very first thing that should be said, really”): Observation. Given R > 0, there exist only finitely many conjugacy classes of elements with translation length ≤ R. “Really something that belongs to an earlier section (1.9).”

Proof. If u is geodesic and all of its cyclic conjugates are geodesic12 and `(u) > 8δ + 1, then un is a (8δ + 1)- local geodesic. n n n`(u) We showed that there exist k, c (depending on δ) s.t. u is a (k, c)-quasi-geodesic; hence `(u ) ≥ k −c, `(u) and so τ(u) ≥ k . Now every conjugacy class determines (contains) a fully-reduced word u, and there exist only ﬁnitely many fully-reduced words of length < max{Rk, 8δ + 1}. Hence there exist only ﬁnitely many conjugacy classes of elements of translation length ≤ R.

Now we are in a position to begin working towards ping-pong in δ-hyperbolic spaces. We ﬁrst begin with a very general version of the ping-pong lemma (of which Lemma 1.37 was a more speciﬁc version.)

Lemma 1.46. Let Ω be a set and {h1, . . . , hr} be a collection of bijections of Ω, and A1+,A1−,...,Ar+,Ar− −1 ∼ be non-empty mutually disjoint sets s.t. hi(Ω−Ai−) ⊂ Ai+ and hi (Ω−Ai+) ⊂ Ai−. Then hh1, . . . , hri = Fr. Proof. Let hn ··· h1 be a reduced word in {h , . . . , h } with = ±1. Pick v ∈ Ω − S (A ∪ A ). in i1 1 r i i i+ i−

+ h1

A1,+ - h2

A2,- A2,+

+ h2 A1,-

- h1

h1 (v) ∈ A . Notice that h2 6= h−1 since we are reduced, and so h1 ∈/ A . Hence h2 (h1 (v)) ⊂ i1 i11 i2 i1 i1 i2(−2) i2 i1

Ai22 .

12∃ ≥ 1 such u in every conjugacy class.

42 Iteratively, hn ··· h1 (v) ∈ A . in i1 inn In particular, w(v) 6= v, and so v is non-trivial in Γ := hh1, . . . , hri. ∼ Since this is true for any fully-reduced w, we may conclude Γ = Fr.

Theorem 1.47. If Γ is hyperbolic and γ1, . . . , γr ∈ Γ are infinite-order and have distinct fixed points (i.e. N N ∼ ∂i γi 6= ∂j γj unless i = j and i = j), then there exists N s.t. the subgroup hγ1 , . . . , γr i = Fr. M Proof. There exists M s.t. Ii is fixed by γi . Choose αi(0) ∈ Ii; choose αi : R → X (for all i) s.t. M γ (αi(R)) = αi(R) = Ii for some fixed n > 0. Given R > 0, let

Ai+ := {x ∈ CΓ | ∃ti > R : d(x, αi(ti)) = d(x, Ii)}

Ai− := {x ∈ CΓ | ∃ti < −R : d(x, Ii) = d(x, αi(ti))}

Claim. ∃R 0 s.t. A1+,A1−,...,Ar+,Ar− are mutually disjoint. To establish this we will ﬁrst need a Lemma 1.48. Projection onto a geodesic is coarsely well-deﬁned in a δ-hyperbolic space. More precisely: if X is δ-hyperbolic, I is a geodesic in X, and x ∈ X, y1, y2 ∈ I s.t. d(y1, x) = d(y2, x) = d(x, I), then d(y1, y2) < 4δ.

Proof. Let m be a midpoint of y1y2 ⊂ I. I

y1 < δ p

m < δ x

Since X is δ-hyperbolic, there exists p ∈ xy1 s.t. d(p, m) < δ; then, since y1 is a closest point to x on I, and xy1 is geodesic, d(p, y1) ≤ d(p, m) < δ. Then d(m, y1) ≤ d(p, m) + d(p, y1) < δ + δ = 2δ. Hence, since m is a midpoint, d(y1, y2) < 4δ.

Proof of Claim. Now, by the above lemma, if R > 4δ then Ai+ ∩ Ai− = ∅ for all i. To establish the rest of the claim: letρ ˆ = maxi{d(αi(0), 1}, and let ρ > 0 be some constant s.t. if xi ∈ Ii and xj ∈ Ij with i 6= j, then xixj ∩ Bρ(1) 6= ∅.

Such a ρ must exist: if not, ∃xin , xjn with i 6= j s.t. d(xin xjn , 1) → ∞. But, up to subsequence,

xin xjn → ab with a ∈ ∂Ii and b ∈ ∂Ij. Since d(1, ab) < ∞, but d (1, xin xjn ) → d(1, ab), this gives a contradiction. We now claim that R > 2ρ + 2ˆρ + 4δ works. The idea behind why is that if x ∈ CΓ, then either the projection of x to Ii lies within R of αi(0) or the projection of x to Ij lies within R of αj(0). More precisely: suppose x ∈ CΓ. Let xi ∈ Ii and xj ∈ Ij be s.t. d(xi, x) = d(x, Ii) and d(xj, x) = d(x, Ij). ∃p ∈ Bρ(1) ∩ xixj. By considering the triangle xixj ∪ xxi ∪ xxj, we ﬁnd that ∃q ∈ xxi (say) s.t. d(p, q) < δ. Since xi is a closest point on Ii to q,

d(xi, q) ≤ d(q, αi(0)) ≤ d(q, p) + d(p, 1) + d(1, αi(0)) < δ + ρ +ρ. ˆ

But d(xi, αi(0)) ≤ d(xi, q)+d(q, αi(0)) ≤ (δ+ρ+ρ ˆ)+(δ+ρ+ρ ˆ) < R. Hence x∈ / Ai+ ∪Ai−, since xj = αj(tj) for some |tj| < R.

So if R > 4δ + 2ρ + 2ˆρ, then Ai+ ∩ Ai− = ∅ for all i, and Aii ∩ Ajj = ∅ whenever i 6= j.

43 Ii

Figure 36: Why ρ must exist, a schematic illustration

x j x

p q

Figure 37: Why R > 2ρ + 2ˆρ + 4δ works, a schematic illustration.

m Now choose n s.t. nτ(γi ) > 8R for all i. Suppose x ∈ CΓ − Ai−. Then if xi ∈ Ii and d(x, xi) = d(x, Ii), then xi = αi(t) for t ≥ −R;

nM d(γi (x),Ii) = d(αi(t + nMτ(γi)))

nM implies αi(t + nMτ(γi)) is a projection of γi (x). nM By assumption, t + nMτ(γi) > 7R, so any projection of γi (x) onto Ii has coordinate ≥ 7R − 4δ > 6R, nM and so γi (x) ∈ Ai+. (Or, roughly speaking, the projection onto Ii is not just coarsely well-deﬁned, but N also coarsely equivariant.) So if N = nM, then γi (CΓ − Ai−) ⊂ Ai+ for all i. −N Similarly, γi (CΓ − Ai+) ⊂ Ai− for all i. N N ∼ By Lemma 1.46 (ping-pong), hγ1 , . . . , γr i = Fr.

Recall the Tits Alternative Theorem 1.49. If Γ is hyperbolic, then OOTFH: (a) Γ is ﬁnite,

(b) Γ is virtually inﬁnite cyclic, or

44 Recall: if we let β(n) = #{γ ∈ Γ | `(γ) ≤ n}, then [we showed that] β(n) is well-deﬁned up to quasi- equivalence. In case (a) above, β is bounded; in case (b), β is linear; in case (c), β is exponential.

P n Theorem 1.50 (Gromov, Cannon). B(x) = n β(n)x is a rational function for Γ hyperbolic (for any choice of generators of Γ.)

Then, combined with the discussion above, we have

Corollary. If Γ is quasi-isometric to Z, then Γ is virtually cyclic. Proof. Quasi-isometric groups have quasi-equivalent growth functions β.

Proof of Tits Alternative. (1) If Γ is inﬁnite, then Γ has an inﬁnite-order element (this part we will not prove unless we develop more machinery about ∂X.)

(2) If Γ contains two infinite-order elements α, β with distinct fixed points, then it contains F2 (from Theorem 1.47.) (3) If [there exist infinite-order elements and] all infinite-order elements share ≥ 1 fixed point/s, then Γ is virtually cyclic. Claim. If α, β infinite-order share one fixed point, then they share both fixed points.

Figure 38: tl;dr: the situation pictured leads to nondiscreteness.

M M Proof. Choose M s.t. α and β have invariant axes A and B. WMA M = 1. WLOG ∂+α = ∂+β. Parametrize A, B by a : R → CΓ, b : R → CΓ (resp.) s.t. α(a(t)) = a(t + τ(α)) and β(b(t)) = b(t + τ(β)). Let R = d(a(0), b(0)); then d(a(t), b(t)) < R + 3δ for all t > 0. (Since ideal triangles in a δ-hyperbolic space are 3δ-thin, we have s < R+t+3δ and t < R+s+3δ. Combining the two, obtain |s−t| < R+3δ.) Then, for all n > 0,

d(α−nβαn(a(0)), b(τ(β))) = d (β(a(nτ(α))), αn(b(τ(β))) ≤ d (β(a(nτ(α))), αn(a(τ(β)))) + (R + 3δ) = d (β(a(nτ(α))), a(τ(β) + nτ(α))) + (R + 3δ) ≤ d (β(b(nτ(α))), a(τ(β) + nτ(α))) + 2(R + 3δ) = d (b(nτ(α) + τ(b)), a(τ(β) + nτ(α))) + 2(R + 3δ) ≤ 3(R + 3δ)

−n n −n1 n1 −n2 n2 Therefore there are ∃#B3(R+3δ)(1) choices for α βα . Hence ∃n1 6= n2 s.t. α βα = α βα , i.e. αn2−n1 βαn1−n2 = β, i.e. αn2−n1 ∈ Z(hβi), which is virtually cyclic; so some power of α is the same power of β, i.e. ∂±α = ∂±β. Claim. If all inﬁnite-order elements of Γ share both of their ﬁxed points, then Γ is virtually cyclic.

45 Proof. Let a, b be the fixed points of all the infinite-order elements of Γ. Let S be the set of special geodesics. S is finite. Let α ∈ Γ have infinite order. −1 −1 If γ ∈ Γ, then γαγ has infinite order and ∂±(γαγ ) = ∂±α = {γ(a), γ(b)}.

So ∃ subgroup Γ0 of index ≤ 2 s.t. γ(a) = a and γ(b) = b for all γ ∈ Γ0, i.e. −γ(S) = S.

Let φ :Γ0 → ΣS (the permutation group on S), and Γ1 = ker(φ) which has ﬁnite index in Γ0.

∼ Now Γ1 acts freely and properly on I ∈ S, and hence Γ1 = Z; alternatively, we may say that if α ∈ Γ1 has inﬁnite order, then Γ1 ⊂ Z(hαi) and so Γ1 is virtually cyclic.

1.13 The Rips Complex Theorem 1.51 (Rips). If Γ is hyperbolic, it acts on a simplicial complex P s.t. 1. P is ﬁnite-dimensional, contractible, and locally ﬁnite.

2. Γ acts simplicially and co-compactly with ﬁnite stabilizers. 3. Γ acts freely and transitively on P 0. Corollary. Γ torsion-free =⇒ P/Γ is a ﬁnite K(π, 1).

Corollary. For all large enough n, Hn(Γ, Q) is trivial. This proof of this one uses spectral sequences.

0 Proof. Consider the Cayley graph CΓ with R > 0. Then PR(CΓ) is the simplicial complex s.t. PR(CΓ) =

Γ and {γ0, . . . , γn} span an n-simplex if diamCΓ ({γ0, . . . , γn}) ≤ R. PR(CΓ) is locally ﬁnite and ﬁnite-

dimensional since BCΓ (R, 1) is ﬁnite.

#BCΓ (R, 1) = #BCΓ (R, γ) for all γ ∈ Γ. γ({γ0, . . . , γn}) = (γγ0, . . . , γγn) determines a simplicial action on PR(CΓ). 0 Since Γ acts freely on Γ, it acts freely on PR(CΓ) . 0 0 The action of Γ is co-compact since it acts transitively on PR(CΓ) and PR(CΓ) is locally ﬁnite. γ stabilizes a simplex {γ0, . . . , γn} iﬀ it permutes the vertices, so the stabilizer of a n-simplex has order ≤ (n + 1)!

If R > 1, PR(CΓ) = CΓ,Y , where Y = {γ ∈ Γ | dCΓ (1, γ) ≤ R}; in particular, PR(CΓ) is connected (i.e. has trivial π0.) The proof of the theorem then reduces to the following

Claim. If R > 4δ + 6 then PR(CΓ) is contractible.

To establish this last result, it suffices to show that πn(P ) is trivial for all n > 0, i.e. any map of a n-sphere into P can be contracted to a point. Any image of a sphere lies in a finite subcomplex, so it suffices to show any subcomplex L can be contracted to a point in P . Case 1: diam(L) ≤ R. In this case, the vertices in L0 have diameter ≤ R, so the vertices of L0 span a simplex in P ; hence L ⊂ some simplex of P , and so L can be contracted to a point. Case 2: diam(L) > R. 0 0 0 Fix v0 ∈ L , and choose w ∈ L s.t. dCΓ (v0, w) is maximal (over all w ∈ L ), so dCΓ (v0, w) > R. R R Choose a vertex y ∈ v0w s.t. d(y, w) ∈ [ 2 , 2 + 1]. Claim. If u ∈ L0 with d(u, w) ≤ R, then d(u, y) ≤ R.

Proof. Consider the triangle T with edges v0w, v0u and uw. Notice that `(uw) ≤ R and `(v0u) ≤ `(v0w).

46 v0 y R/2 w

Since CΓ is δ-hyperbolic, ∃p ∈ uw ∪ v0u s.t. d(y, p) < δ. R If p ∈ uw: since d(w, y) ≥ 2 > 2δ, hence we have d(w, y) > d(p, y), and thus d(u, y) < d(u, w) ≤ R. Otherwise, if p ∈ v0u: if d(u, y) > R, then d(u, p) > R − δ; hence d(v0, w) ≤ d(v0, p) + d(p, y) + d(y, w) ≤ R R d(v0, p) + δ + 2 + 1. But d(v0, u) = d(v0, p) + d(p, u) ≥ d(v0, p) + R − δ, since R − δ > 2 + δ + 1 by our choice of R > 4δ + 6. Hence we would have d(v0, u) > d(v0, w), contradicting the maximality of d(v0, w)—hence we must conclude d(u, y) ≤ R.

0 Let L be obtained from L by replacing every simplex (w, γ1, . . . , γn) by the simplex (y, γ1, . . . , γn). Our 0 claim above guarantees that (y, γ1, . . . , γn) is a simplex. One can homotope L to L within P by linearly deforming (w, γ1, . . . , γn) to (y, γ1, . . . , γn) through (w, y, γ1, . . . , γn). This either (a) reduces max{d(v0, w) | w ∈ L0} by ≥ 1, or (b) reduces # {w ∈ L0 | d(v, w) maximal}, so after ﬁnitely many steps we arrive at Lˆ s.t. diam(Lˆ) < R and L can be homotoped to Lˆ; and then we may apply the arguments of Case 1 above.

1.14 The Mostow Rigidity Theorem

Or, a sketch of a pretty cool result

3 3 13 Theorem 1.52. If M = H /ΓM and N = H /ΓN are closed hyperbolic 3-manifolds, and f : M → N is a homotopy equivalence, then ∃ an isometry g : M → N homotopic to f.

Sketch of proof. (1) f : M → N lifts to a quasi-isometry fe : H3 → H3 (essentially, using the Milnor-Svarcˇ lemma.)

(2) fe extends to a quasi-conformal map g : ∂H3 → ∂H3 (recall a quasi-conformal map is a map which distorts the conformal structure by a bounded amount.)14

(3) g is diﬀerentiable almost everywhere and dg(z) 6= 0 a.e. (4) Either (“Sullivan”) ˆ ˆ Theorem 1.53 (Tukia). If g : C → C conjugates ΓM to ΓN (both co-compact) and dg(z) 6= 0, then g is conformal (or anti-conformal.)

or (“Mostow”) If g is not conformal, then there exists dg a measurable ΓM -invariant line ﬁeld (direction of maximal stretch); now apply the ergodicity of the action of Γ on Cˆ × Cˆ to show this is impossible. The proof of (1) proceeds, morally, by noting that fe = f∗ which is a quasi-isometry. More precisely, notice that f∗ :ΓM → ΓN is an isomorphism, and so f∗ is a quasi-isometry w.r.t. any word metric; then ˇ fe :ΓM (x0) → ΓN (fe(x0)) is a (k, c)-quasi-isometry (for some k, c) by Milnor-Svarc. WMA f is Lipschitz and let D = diam(M); then fe : H3 → H3 is a (k, c + 2D)-quasi-isometry. 13We may generalize this to ﬁnite-volume 14 3 3 Note further that g : ∂H → ∂H induces a map ∂ΓM → ∂ΓN .

47 For the proof of (2), recall that a homeomorphism g is k-quasiconformal if for all z,

iθ maxθ d(g(z), g(z + re ) lim sup iθ ≤ k r→0 minθ (d(g(z), g(z + re ))

Claim. Let g = fe : Cˆ → Cˆ. Then g is quasiconformal. In fact this follows from the slightly more general

Proposition 1.54. Given k, c > 0 ∃L > 1 s.t. if h : H3 → H3 is a (k, c)-quasi-isometry, then h¯ : ∂H3 → ∂H3 is L-quasiconformal. Proof. The proof uses a key Lemma 1.55. ∃D > 0 s.t. if L is a geodesic, P is a totally geodesic hyperplane s.t. L ⊥ P , L0 = 0 h(∂+L)h(∂−L) is a geodesic asymptotic to h(L), and πL0 denotes nearest-point projection, onto L , then

diam(πL0 (fe(P ))) ≤ D

P z L ~ h

Figure 39: Illustration of Lemma 1.55

3 To reduce our result to this lemma, suppose z ∈ C; let L = z∞ and Pr be a plane in H with ∂Pr a ¯ ˆ circle Sr of radius r about z. Now h(Sr) = h(Pr) ∩ C. WLOG h(z) = 0, h¯(∞) = ∞, and so (we may assume) L0 is the t-axis. D ¯ D ˆ Now let s = min(πL0 (h(Pr))); then πL0 (h(Pr)) ⊂ [s, e s]. Hence h(Pr) ⊂ B(0, e s)−B(0, s) ⊂ C. Hence

max d(h(z), h(z + reiθ) eDs ≤ = eD min (d(h(z), h(z + reiθ)) s and so the lim sup of this quantity as r → 0 is also ≤ eD; hence h¯ is now eD-quasiconformal.

Figure 40: Lemma 1.55 implies quasiconformality

48 M1 M2

L P

Figure 41: Proof of Lemma 1.55: the circle is x; and then we mess around with the fellow traveller property (see actual text of proof for an explanation ...

Proof of Lemma. This involves, essentially, a lot of argument using the fellow traveller property. ¯ ¯ Let z ∈ ∂P and x ∈ L ∩ P . Let L = z1z2, M1 = zz1, and M2 = zz2. Letx ¯ = h(x), L¯ = h(z1)h(z2), and ¯ ¯ M i = h(zi)h(z). Choose R ≥ 0 s.t. a (k, c)-quasi-geodesic (always) lies in a R-neighborhood of its associated geodesic. d(¯x, L¯) ≤ R; d(x, Mi) = T =⇒ d(¯x, h(Mi)) ≤ kT + c =⇒ d(¯x, M¯ i) ≤ R + kT + c. (It is an exercise to show that) πL¯ is 1-Lipschitz. ¯ ¯ Then d(πL¯ (¯x), πL¯ (Mi)) ≤ R + kT + c and πL¯ (z) = πL¯ (M1) ∩ πL¯ (M2) imply d(πL¯ (¯x), πL¯ (h(z)) ≤ D0 := R + kT + c. ¯ ¯ ¯ But h(xz) ⊂ NR(x¯h(z)), and so if y ∈ h(x¯h(z)), then d(πL¯ (y), πL¯ (¯x)) ≤ R + D0 since πL¯ (x¯h(z)) lies ¯ between πL¯ (¯x) and πL¯ (h(z)). Since this does not depend onz ¯, if y ∈ h(P ), then d(πL¯ (y), πL¯ (¯x)) ≤ R + D0 =⇒ diam πL¯ (h(P )) ≤ 2(R + D0) The proof of (3) and (4) ... we will leave to the reader to ﬁll in (warning: requires substantial additional material / background!)

We may compare this to more general / other [quasi-isometric] rigidity theorems, e.g.

3 Theorem 1.56 (Gromov, Cannon). Γ quasi-isometric to H =⇒ ∃Γ0 ⊂ Γ of ﬁnite index s.t. Γ0 acts co-compactly in H3. Note that this can be extended to semisimple Lie groups, solvable Lie groups, and even general symmetric spaces. There is also

Theorem 1.57 (Tukia, Gabai, Casson-Jungreis, Mess). Γ quasi-isometric to H2 =⇒ Γ virtually a surface group. and also the simpler Theorem 1.58. Γ quasi-isometric to a free group =⇒ Γ virtually free.

49 2 CAT(0) groups

Let X be a proper geodesic metric space. If ∆ is a geodesic triangle in X, there exists a comparison triangle ∆ in Euclidean space E2 with the same side-lengths:

y y X

z z x x in comparison (Euclidean) space

If a, b ∈ ∆, there exist corresponding points a, b ∈ ∆. Deﬁnition 2.1. X is CAT(0) if d(a, b) ≤ d(a, b) for all ∆, a, b ∈ X.

We may think of this as a strengthening of (δ-)hyperbolicity. We may replace E2 in the above with H2 to obtain the deﬁnition of CAT(-1) spaces; sometimes we use comparison triangles on the sphere—here we obtain CAT(1) spaces.

Example 2.2. Some examples include

• Euclidean space En • Trees

• H2 • M a Riemannian manifold with sectional curvature ≤ 0. Fact. X is uniquely geodesic Proof. If not, consider the degenerate triangle formed by two geodesics between distinct points x, y ∈ X (and a side of length 0):

x y z

Some choice of points in this triangle (or the triangle with x, y, z where z is on one of the geodesics, very close to x) would violate the CAT(0) condition. Fact. Local geodesics are global geodesics i.e. if γ : [0,R] → X has the property that ∀t ∈ [0,R] ∃ > 0 s.t. γ|[t−,t+] is geodesic, then γ is geodesic.

Proof. Let T = {t | γ|[0,t] geodesic}. T is closed and contains a neighborhood of 0. We now claim that T is also open: let s = sup T and suppose s < R.

50 Then d(γ(s − 2 ), γ(s + 2 )) = , but has to be less than in the comparison triangle—contradiction. Corollary. If X,Y are CAT(0), and f : X → Y is a locally-isometric immersion, then f : X → Y is an isometric embedding. Fact. X is contractible

Proof. Fix x0 ∈ X. For any x ∈ X, ∃ geodesic γX : [0, rX ] → X s.t. γx(0) = x0 and γx(rx) = x. Define H : X × [0, 1] → X by H(X, t) = γx(trx). H is a homotopy from id to the constant map with image x0. Note that we may define such a map H for any space, but that here we can sue the CAT(0) condition to show that it is locally Lipschitz / continuous, and so indeed a homotopy as desired. Definition 2.3. Γ is a CAT(0) group if it acts properly and co-compactly on some CAT(0) space X. Example 2.4. • Free abelian groups Zn

• Free groups Fn

• Surface groups π1(Σ)

• π1(M) of closed non-positively-curved Riemannian manifolds M Fact. If X and Y are CAT(0) spaces, then X × Y (with the product metric) is CAT(0).

Corollary. If Γ1 and Γ2 are CAT(0) groups, then Γ1 × Γ2 is CAT(0).

2.1 Isometries of CAT(0) space Here (and for the rest of this section of the notes) let X be a proper geodesic CAT(0) space. Suppose γ ∈ Isom(X).

Deﬁnition 2.5. τ(γ) := infx∈X d(x, γ(x)) is the translation length. min(γ) := {x ∈ X | d(x, γ(x)) = τ(γ)}. Deﬁnition 2.6. γ is • elliptic if τ(γ) = 0 and min(γ) 6= ∅. • hyperbolic if τ(γ) > 0 and min(γ) 6= ∅ • parabolic if min(γ) = ∅ It may be shown that any γ ∈ Isom(X) is of one of these three types. Fact. If γ is hyperbolic, it has an invariant axis A, i.e. a geodesic A s.t. γ(A) = A and γ acts on A by translation by τ(γ).

S n n+1 Proof. Deﬁne A := γ (x0)γ (x0) where x0 ∈ min(γ). n∈Z

2 3 γ (x0) γ (x0) ··· x 0 γ(x0)

We only need to observe that A is locally geodesic, and by equivariance it suﬃces to check this at γ(x0). Suppose not; then we have

(x0)

m (m)

x0 (x0)

51 2 Now let m be the midpoint of x0γ(x0); γ(m) will then be the midpoint of γ(x0)γ (x0). Then

d(m, γ(m)) < d(m, γ(x0)) + d(γ(x0), γ(m)) = τ(γ)

which contradicts that τ(γ) is the minimum translation distance. Fact. If Γ acts on X properly and co-compactly, then all the elements of Γ are elliptic or hyperbolic

Proof. This follows once we prove that min(γ) 6= ∅. Let x1, . . . , xn be a sequence in X s.t. d(xn, γ(xn)) → τ(γ). Let K ⊂ X be a compact set s.t. Γ · K = X. −1 For all n, there exists αn ∈ Γ s.t. αn(xn) =: yn ∈ K. Since d(αnγαn (yn), yn) = d(γ(xn), xn) → τ(γ), −1 we have finitely many choices for αnγαn . −1 Now pass to a subsequence s.t. αnγαn =γ ˆ and yn → y. Then, since τ(γ) = τ(ˆγ), we have yn → y =⇒ d(ˆγ(yn), yn) → τ(ˆγ), and so y ∈ min(ˆγ) 6= ∅. But now min(ˆγ) = αn(min(γ)) and hence we may conclude min(γ) 6= ∅. Corollary. Finite order elements of Γ (as above) have fixed points. Corollary. If γ ∈ Γ has infinite order, and α ∈ Γ with αγα−1 = γp, then p = ±1. In other words, if H ⊂ Γ is isomorphic to BS(m, n) := ha, t | tamt−1 = ani, then |m| = |n|. Proof. Since αγα−1 = γp, τ(γp) = τ(γ). Since γ has an invariant axis, τ(γp) = |p|τ(γ) (from the below arguments.) Hence we must have |p| = 1. Let us close this section by proving some useful facts about nearest-point retraction in CAT(0) spaces (which we need also to complete the above.) Fact. If X is CAT(0) and C ⊂ X is closed and convex (i.e. if x, y ∈ C, then xy ∈ C), then the nearest-point retraction π1 : X → C is well-defined and 1-Lipschitz. Proof. To show that it is well-defined, we look at comparison triangles: if yz ∈ C, ∃t¯ ∈ y¯z¯ s.t. d(t, x) ≤ d(t,¯ x¯) ≤ d(¯x, y¯) = d(x, y). x

z¯ y¯ t¯

To show that it is 1-Lipschitz, for any x, y ∈ X form the (comparison) quadrilateralx ¯y¯π1(x)π1(y).

y x

(x) (y)

Figure 42: The [quadrilateral we form the] comparison quadilateral [to].

π If the angle α < 2 , then ∃ a point on πC (x)πC (y) closer tox ¯ than πC (x), and hence ∃ a point on π π πC (x)πC (y) ⊂ C closer to x than πC (x). This is impossible, so α ≥ 2 , and similarly β ≥ 2 . Since the comparison quadrilateral must be Euclidean, we have

d(πC (x), πC (y)) = d(πC (x), πC (y)) ≤ d(¯x, y¯) = d(x, y).

52 Corollary. If A is an invariant axis of a hyperbolic element γ, then it is an invariant axis for γp. Proof. A is preserved by γp and γp acts on A by translation by |p|τ(γ); hence it suﬃces to show that there is no smaller translation distance. p p If z ∈ X, then πA is γ -equivariant, since A is γ -invariant. Then

p p d(z, γ (z)) ≥ d(πA(z), πA(γ (z))) p = d(πA(z), γ (πA(z))) = |p|τ(γ) and we are done.

2.2 Abelian subgroups of CAT(0) groups We can perform, starting from relatively elementary arguments, a number of fun [quite strong] results about abelian subgroups of CAT(0) groups. Here we present some of the key results and lemmas without proof:

Theorem 2.7 (Flat Strip Theorem). If A1 and A2 are asymptotic geodesics, then they bound an isometrically embedded (“ﬂat”) copy of R × [0,D] where D > 0. We say that A1 and A2 are parallel.

Recall that A1,A2 : R → X are asymptotic if supt d(A1(t),A2(T )) < +∞.

Corollary. min(γ) = Y × R, where Y ⊂ X is convex, γ acts as the identity on Y and by translation by τ(γ) on R. ∼ n T Theorem 2.8 (Flat Torus Theorem). If A ⊂ Γ CAT(0) is s.t. A = Z , then min(A) := α∈A min(α) 6= ∅, and in particular min(A) =∼ Z × Rn where Z ⊂ X is convex; A preserves this splitting, acts trivially on Z and as a co-compact translation group on Rn; min(A)/A := T n × Z where T n denotes the n-torus S1 × · · · × S1.

Corollary. Every abelian subgroup of a CAT(0) group is ﬁnitely-generated. Corollary. S ⊂ Γ solvable =⇒ S virtually abelian.

2.3 Solvability of CAT(0) word problem D Fact. Suppose Γ acts on a CAT(0) space X properly and co-compactly, and Γ · B x0, 4 = X. Then S = {γ ∈ Γ | d(x0, γ(x0)) ≤ D + 1} generates Γ. If γ ∈ Γ s.t. d(x0, γ(x0)) ≤ 2D + 1, then γ = s1s2s3s4 where si ∈ S.

D Proof. Write B := B x0, 4 and let H = hSi ⊂ Γ, V = H(B), and W = (Γ · H)(B). Now V and W are open; V ∪ W = X since Γ · B = X, and V ∩ W = ∅, since if gB ∩ hB = ∅ then h−1gB ∩ B 6= ∅ and hence h−1g ∈ S. If h ∈ H, then h(h−1g) = g ∈ H; hence V 6= ∅ =⇒ W = ∅; hence H = Γ, i.e. S generates. 1 D 1 Now choose x1, x2, x3 ∈ x0γ(x0), with d(xi, xi+1) = 4 d(x0, γ(x0)) ≤ 2 + 4 .

x3 (x0)

x2 x1

53 D Since X = Γ · B, for each i, there exists γi ∈ Γ s.t. d(γi(x0), xi) < 4 . Choose γ0 = id and γ4 = γ. Then D D 1 d(γ (x ), γ (x )) < 2 · + + < D + 1 i 0 j 0 4 2 4

−1 and so γi γi+1 ∈ S. −1 Then γ = s1s2s3s4 where si = γi−1γi ∈ S.

D Theorem 2.9. If Γ acts properly co-compactly on a CAT(0) space X, and X = Γ · B x0, 4 , let

S = {γ ∈ Γ | d(x0, γ(x0)) ≤ D + 1}

R = {w ∈ F (S) | `S(w) ≤ 10, w ∼ id}

QN −1 2 Then w = F (S), w ∼Γ id =⇒ w = i=1 xirjxi in F (S), where N ≤ (D + 1)`S(w) and `S(xi) ≤ (D + 1)`S(w) for i = 1,...,N.

Proof. Given γ ∈ Γ, we will produce a word σγ ∈ F (S) s.t. γ = σγ in Γ. Let n be s.t. 0 ≤ n − d(x0, γ(x0)) < 1 and let c : [0, n] → X be a geodesic with c(0) = x0 and D −1 c(d(x0, γ(x0))) = γ(x0). Choose σγ (i) s.t. d(σγ (i)(x0), c(i)) < 4 and σγ (i)σγ (i + 1) ∈ S. −1 Deﬁne s1 = σγ (i) and si = σγ (i − 1) σγ (i) for i = 2, . . . , n; then let σγ = s1 ··· sn. 0 Now let us ask: how much does it cost to add b ∈ S to our word σγ ? Consider γ = γb.

si+1

id b

s'i+1

Then

0 d(c(i)(x0), c (i)(x0)) ≤ d(γ(x0), γb(x0)) = d(x0, b(x0)) ≤ D + 1

D 0 0 D but d(c(i)(x0), σγ (i)(x0) ≤ 4 (and d(c (i)(x0), σγ (i)(x0) ≤ 4 ) imply D D d(σ (i)(x ), σ 0 (i)(x )) ≤ + D + 1 + < 2D + 1 γ 0 γ 0 4 4

−1 and hence ∃ a word α(i) ∈ F (S) s.t. `S(α(i)) ≤ 4, α(i) = σγ0 (i)σγ (i) ; then

n−1 −1 Y −1 0 −1 −1 σγ bσγ0 = σγ0 (i) α(i) si+1α(i + 1)(si+1) σγ0 (i) . i=0

Note `(σγ ) ≤ d(x0, γ(x0)) + 1 and we may choose, in the above, n = max{`S(σγ ), `S(σγ0 )}. Now let w := b1 ··· bm ∈ F (S). Deﬁne γi := b1 ··· bi a subword of w; then

m Y w = σ b σ−1 γj−1 j γj j=1

m nj −1 Y Y −1 = σγj (i)rijσγj (i) j=1 i=0

where the rij ∈ R and σγm = id; hence R is a full set of relations.

54 Moreover, the number of relations used N is given by n1 + ··· + nm. Note that we have nj ≤ `S(γj) ≤ j(D + 1) + 1, and `S σγj (i) ≤ i(D + 1) + 1, and so

m X N = n1 + ··· + nm ≤ j(D + 1) + 1 j=1

 m  X ≤ (D + 1)  j + m j=1 m + 1 = (D + 1) + m 2 2 2 ≤ (D + 1)m = (D + 1)`S(w)

and `S(σj(i)) ≤ i(D + 1) + 1 ≤ m(D + 1) = `S(w)(D + 1) for all i. Corollary. If Γ is a CAT(0) group, then deﬁne S and R as above; then Γ = hS; Ri is ﬁnitely-presented, has solvable word problem, and has a quadratic Dehn function.

2.4 Cube Complexes Deﬁnition 2.10. A cube complex is a space constructed from unit Euclidean cubes by identifying faces by isometries

Figure 43: A cube complex (black / grey) and a link of a vertex (blue).

We will always assume that our cube complexes are locally ﬁnite. Deﬁnition 2.11. The link of a vertex v in a cube complex K, denoted Lk(K, v), is a simplicial complex s.t. a neighborhood of v is isometric to a neighborhood of the cone point in the cone on Lk(K, v). Cube complexes need not be CAT(0): consider, for instance

Figure 44: (The pictured red triangle on the left = ﬁlled yellow in the middle) ∆ is thicker than the comparison triangle ∆¯ ⊂ ∆; the right diagram is a diﬀerent example of a non-CAT(0) cube complex.

We can however ﬁnd relatively simple / concise conditions s.t. cube complexes satisfying these conditions will be CAT(0): We say that K is locally CAT(0) if ∀x ∈ K ∃ a convex neighborhood U of x in K s.t. all triangles in U satisfy the CAT(0) criterion; we also say that K has non-positive curvature.

55 Theorem 2.12 (Cartan-Hadamard). K simply-connected, complete, locally CAT(0) =⇒ K CAT(0).

Definition 2.13. A simplicial complex K is a flag complex if whenever w0, . . . , wn are vertices in K with [wi, wj] an edge in K for all 0 ≤ i 6= j ≤ n,[w0, . . . , wn] span an n-simplex in K. Theorem 2.14 (Gromov’s Link Criterion). K locally CAT(0) ⇐⇒ for all vertices v ∈ K, Lk(K, v) is a flag complex.

In 2 dimensions this rules out (for instance) the things we see in Figure 44. The advantage of this criterion is that it is “incredibly checkable.” Here is one case where we can use Gromov’s Link Criterion (together with Cartan-Hadamard) to show that a space is CAT(0). Recall the deﬁnition of a Right-Angled Artin Group (RAAG): start with G a (ﬁnite) 0 1 graph. Form the group Γ = AG = hG | [v, w] = 1 ⇐⇒ [v, w] ∈ G i. Fact. RAAGs are CAT(0).

Proof. We will build a locally CAT(0) cube complex K (the Salvetti complex) s.t. π1(K) = AG; then we may consider the action of AG on K˜ (which is CAT(0) by Cartan-Hadamard.) We construct K as follows: ﬁrst let K(1) be the bouquet of circles (with one circle for each vertex of G), i.e. K(0) consists of a single point and K(1)“ =00 G(0). −1 −1 Whenever vi, vj are joined by an edge in G, append the square with boundary vivjvi vj to K.

Figure 45: A graph for a RAAG (left) and its associated Salvetti complex (right.)

(0) By construction, π1(K) = AG; now make K a ﬂag complex (if v0, . . . , vm ∈ G are s.t. vi, vj are joined by an edge for all i 6= j, then glue in a n-cube.) The resulting complex has all Lk(K, v) ﬂag complexes, and hence is CAT(0) by Gromov’s Link Criterion.

∼ 2 2 Examples 2.15. • If G consists of four vertices and two disjoint edges, then Γ = AG = Z ∗ Z , the Salvetti complex K = S(Γ) is a wedge of two (2-)tori, and the link of the join vertex v in K consists of two squares joined at a point.

∼ 3 • If G is a triangle, then Γ = AG = Z , K = S(Γ) is a 3-torus, and the link of any vertex in K is homeomorphic to a 2-sphere.

2.5 Cartan-Hadamard Let us prove the Cartan-Hadamard Theorem. We will do so in slightly more generality:

Deﬁnition 2.16. X is CAT(k) if every geodesic triangle ∆ in X (of perimeter < √2π if k > 0) is at least k ¯ 2 as thin as the comparison triangle ∆ in Mk (the unique simply-connected Riemannian manifold of constant curvature k.)

Deﬁnition 2.17. X is locally CAT(k) if for all x ∈ X there exists a convex neighborhood Ux 3 x s.t. every triangle in Ux satisﬁes the CAT(k) condition. We sometimes also say that X has curvature ≤ k.

56 Lemma 2.18. If (a0, b, c0) and (a1, b, c1) are geodesic triangles in X both satisfying the CAT(k) condition and c0 ∪ c1 is a geodesic in X, then (a0, a1, c0 ∪ c1) is CAT(k).

c0 A c1 B0 B1

b a0 a1

¯ ¯ ¯ ¯ ¯ ¯ Proof. Let ∆1 = (a ¯1, b, c¯1) and ∆0 = (a ¯0, b, c¯0), and consider ∆0 ∪¯b ∆1.

c¯0 c¯1 A¯

a¯0 b a¯1

¯ ¯ ¯ Claim. The internal angle of ∆0 ∪¯b ∆1 at A is ≥ π.

Proof. If not, pick p0, p1 near A. We know that, since c0 ∪c1 is geodesic in X, d(p0, p1) = d(p0,A)+d(A, p1) = d(p ¯0, A¯) + d(A,¯ p¯1) > d(p ¯0, p¯1). On the other hand, by the CAT(k) condition in each triangle we have d(p ¯0, p¯1) ≥ d(p0, p1). Contradiction.

And now

A¯

p¯0 p¯1

a¯0 a¯1

c¯0 c¯1

Corollary. If X1 and X2 are CAT(k) spaces, Ai ⊂ Xi (i = 1, 2) are convex subsets, and we have an isometry j : A1 → A2, then X1 ∪j X2 is also CAT(k). Sketch of proof. Quite literally:

X2 X1 = X2

Theorem 2.19. If X is locally CAT(k), and given any p, q ∈ X s.t. d(p, q) < √π there exists a unique k geodesic from p to q (i.e. X is locally geodesic), then X is CAT(k).

57 Proof. Fix p ∈ X and R < √π . B (p) is compact, so there exists > 0 s.t. for all x ∈ B (p), B (x) ⊂ U , k R R x For all q ∈ BR(p) there exists a unique geodesic cq : [0, 1] → X parametrized proportional to arc-length. cq is a continuous function of q (by Arzel`a-Ascoli)and so ∃δ > 0 s.t. if x, y ∈ BR(p) and d(x, y) < δ then d(cx(t), cy(t)) < for all t. Claim. If d(x, y) < δ, then the triangle (p, x, y) satisﬁes the CAT(k) condition.

i i+1 i i+1 Proof. Choose n s.t. d cx n , cx n < δ and d cy n , cy n < δ. x δ δ p y

1 1 (p, cx n , cy n ) is contained in B(p) ⊂ Up, and hence is CAT(k). i i i i+1 i Now, inductively (on i) assume (p, cx n , cy n is CAT(k). Add ﬁrst cx n , cx n , cy n and i+1 i i+1 then cx n , cx n , cy n , each of which is CAT(k); the union is CAT(k) by Lemma 2.18.

Claim. If x, y ∈ BR(p) then the triangle (p, x, y) is CAT(k). Proof. Once again, we glue inductively using our Lemma 2.18: x < δ < δ < δ < δ < δ < δ < δ < δ < δ < δ y

(This construction is sometimes known as Alexandrov’s patchwork.)

Proposition 2.20. X simply-connected and locally CAT(0) imply X locally uniquely geodesic.

Proof. Given c : [0, 1] → X a rectiﬁable path joining x to y parametrized proportional to arc-length, let 1 R = `(c). Choose = n > 0 s.t. if x ∈ BR(p), B(x) ⊂ Ux. First replace each c| i i+1 by a geodesic segment (straighten piecewise) and reparametrize by arc-length [ n , n ] 0 0 0 2i+1 2i+3 to get c ; then D(c) is obtained from c by replacing each c 2n , 2n with a geodesic segment.

Figure 46: Birkhoﬀ curve shortening, a schematic representation

m Claim. D (c) → c∞, a geodesic joining x to y, up to subsequence Ideas behind proof. Arzel`a-Ascoliand some sort of energy minimization.

58 n If c0, c1 are distinct geodesics, let H = {cs} be an isotopy by rectiﬁable curves from c0 to c1. D (cs) → cs is a family of geodesics interpolating between c0 and c1. Now d(c0(t), cs(t)) < (for all s > 0) implies our space is locally convex, and hence (by compactness of [0, 1]) globally convex. Then we may use the argument that appeared much early for why CAT(0) spaces are uniquely geodesic.

Note that this Proposition, together with the preceding Theorem, together imply the Cartan-Hadamard theorem.

2.6 Gromov’s link criterion Theorem 2.21. If K is an Euclidean cube complex, then K is locally CAT(0) if lk(K, x) is CAT(1) for all vertices v. (Note that this implies K˜ is CAT(0) by Cartan-Hadamard.) Proof. If p lies in the interior of a cube, one is clearly locally CAT(0) at p (i.e. some neighborhood of p is CAT(0).) If p lies on a face but is not a vertex: if v is a vertex of that face and U is a neighborhood of v, then ∃ V a neighborhood of p which isometrically embeds in U; hence we can reduce to the case where p = v is a vertex. Let U0 = int(C1 ∪· · ·∪Cn), where the Ci are the cubes with v as a vertex. By dividing cubes into smaller cubes, WMA v is only one of the vertices of each Ci. If q ∈ U0, ∃ unique geodesic segment γq joining v to q; let r : U0 → lk(K, v) be the projection map. Metrize lk(K, v) as a union of all-right spherical simplices, i.e. if p, q ∈ Ci, d(r(p), r(q)) = ∠(γp, γq). Consider a geodesic α : [0, 1] → X lying in U0 which does not pass through p = v. Also assume r ◦ α is not constant. Consider the partition Pα = {t0 = 0 < t1 < t2 < ··· < tn+1 = 1} s.t. α([ti, ti+1]) ⊂ Cj(i) and j(i) 6= j(i + 1): we may think of this as “understanding our geodesic cube-by-cube.” 2 Consider the triangles ⊂ Cj(i) and glue the comparison triangles Γi together in E . The total angle atv ¯ is ≤ π, since otherwise γα(0) ∪ γα(1) is shorter than α, which contradicts α geodesic. Now the total angle atv ¯ is equal to `(r ◦ α); r ◦ α must be geodesic, or we could use the geodesic joining r(α(0)) to r(α(1)) to construct a shorter path joining α(0) to α(1). Notice that ∃ > 0 s.t. B(x) is convex, since if is small, any geodesic joining points in B(x) lies in U0, and so has the form above, and in this picture distance to x is locally convex, hence convex.

Claim. If (a, b, c) is a geodesic triangle in B(x), then it satisﬁes the CAT(0) inequality.

Proof. Case 1: Suppose our triangle passes through p. WLOG p ∈ a.(γB, c, γA) is CAT(0) since its comparison triangle pathwise-isometrically embeds; in particular ∃ a 1-Lipschitz map of the comparison triangle into K. Similarly (γC , b, γA) is CAT(0). By Lemma 2.18, (a, b, c) is CAT(0).

γB

a c

γC γA C A b

Case 2: d(r(A), r(B)) + d(r(B), r(C)) + d(r(C), r(A)) < 2π (equivalently, r(a, b, c) is a triangle with perimeter < 2π.) In this case the CAT(1) condition will come into play straight away: make a comparison tetrahedron in E3:

59 A A c c B B b b a a

v C v C

Figure 47: The tetrahedron (left) and its comparison tetrahedron (right.)

Now r(a, b, c) is a geodesic triangle in lk(K, x) which has comparison triangle r(¯a, ¯b, c¯) in S2 ∈ E3. Since the link is CAT(1), d(r(p), r(q)) ≤ d(r(¯p), r(¯q)) and so θ¯ > θ =⇒ d(p, q) ≤ d(¯p, q¯). This works even if one of the edges is radial: a C B

b c A

Case 3:The perimeter of r(a, b, c) is ≥ 2π: “things are about to get vaguer.”

A A c

B b

a v

v C C B

Figure 48: The triangle only gets thicker when we correct.

Corollary. If lk(K, v) is ﬂag for all v, then K is locally CAT(0). Idea of proof. The same argument will show (inductively) that lk(K, v) is locally CAT(1), and then we can show that locally CAT(1) =⇒ globally CAT(1) in this case. This concludes the proof of the reverse (right-to-left as we stated it) direction of Gromov’s Link Criterion; the forward direction is an easy exercise.

2.7 Special Cube Complexes

Guest lecture by Mark Hagen, December 3, 2014.

Definition 2.22. A group G is residually-finite if ∀g ∈ G \{1}: ∃H ≤ G of finite index s.t. g∈ / H. Theorem 2.23. Let F be a finite-rank free group; then F is residually finite.

60 Remark. SLn(Z) is residually ﬁnite, and so RAAGs are residually ﬁnite since (by a result of Davis- Januskiewicz) any RAAG embeds into some SLn(Z).

Proof. Let B be a wedge of ﬁnitely many circles with π1(B) = F . Choose g ∈ F \{1} and an edge-path γ : [0, n] → B representing g; we require γ to be locally injective.

Figure 49: Given a locally-injective edge-path γ (solid arrowed lines), we can canonically complete it (dashed lines) to a covering space of B.

Make a cover Bˆ → B s.t. deg(Bˆ → B) < ∞ and the lift of γ at basepoint is not closed—then g∈ / π1(Bˆ) ,→ π1(B). Remark. (1) It didn’t matter that the domain of γ was an interval. Theorem 2.24. If B is a wedge of circles, Γ a ﬁnite graph, ϕ :Γ → B an immersion (i.e. a locally- injective graph homomorphism), then ∃ a ﬁnite cover Bˆ → B and an embedding ϕˆ :Γ → Bˆ s.t.

Bˆ B ϕ ϕˆ Γ

(2) On the other hand, ∃ CAT(0) groups with no proper ﬁnite-index subgroups (see examples by Burger- Mozes and Wise.)

Deﬁnition 2.25. Let X and Y be non-positively-curved cube complexes. The combinatorial map15 f : X → Y is a local isometry if (1) f is locally injective, and

(2) if e1, . . . , en are 1-cubes of X with a common 0-cube and f(e1), . . . , f(en) are 1-cubes of Y which span 0 0 an n-cube C of Y , then ∃ n-cube C of X spanned by e1, . . . , en with f(C ) = C.

Figure 50: If the blue edges map to the red edges, and there’s a grey cube in the target space, there’s a corresponding yellow cube in the domain which maps to the grey cube.

Deﬁnition 2.26 (Haglund-Wise 2008). A cube complex X is special if ∃ a simplicial graph Γ and a local isometry X → SΓ, where SΓ denotes the Salvetti complex of AΓ.

Exercise. Basepoint-preserving local isometries induce monomorphisms on π1. Now let’s talk about why it’s good to be special.

15k-cubes go to k-cubes, and the interiors map isometrically.

61 Figure 51: Examples of non-special (left) and special (right) cube complexes.

Theorem 2.27 (Canonical Completion and Retraction). Let X be a special cube complex, and let Y be a compact cube complex and φ : Y → X a local isometry. Then ∃ a ﬁnite cover Xˆ → X and an embedding Xˆ X ˆ ˆ φ ˆ ˆ φ : Y → X s.t. φˆ and ∃ a retraction r : X → im(φ). Y

We may also deﬁne specialness in terms of hyperplanes (in fact this was how special cube complexes were originally deﬁned): for each cube C = [−1, 1]n in our cube complex, a midcube in C is a subspace made by restricting one of the dimensions to {0}. The hyperplanes are then built “by radiating out the midcubes as far as possible.”

Figure 52: The red, blue, and green are hyperplanes.

If X is a non-positively-curved cube complex, let Γ be the graph whose vertex set is the set of hyperplanes, and edge set is given by E(Γ) := {(H1,H2) | H1 ∩ H2 6= ∅}. (0) (0) Deﬁne φ : X → SΓ as follows: for x ∈ X , set φ(x) = {∗} ∈ SΓ ; if e is a 1-cube of X, ∃ hyperplane He intersecting e. Let E be the 1-cube of SΓ labelled by He and send the interior of e isometrically to the interior of E (this can be done if the hyperplanes of X are 2-sided. φ is a local isometry, provided

(1) hyperplanes in X are embedded (2) no self-osculations (see above) (3) hyperplanes are 2-sided (4) no inter-osculations (see above.)

62 Figure 53: The four hyperplane pathologies (from left to right): non-embedded hyperplanes, one-sided hyperplanes, self-osculations, and inter-osculations. Illustration taken from Haglund-Wise.

2.8 A mad dash for the ﬂat torus theorem Lemma 2.28 (Flat Triangles Lemma). If ∆ = (p, q, r) is a geodesic triangle in a CAT(0) space X and ¯ 2 ∠ppq, pr = ∠pp¯q,¯ p¯r, then p, q, r spans a Euclidean 2-simplex isometric to the 2-simplex D in E spanned by ∆¯ . Proof. (1) If s ∈ qr, then d(p, s) = d(¯p, s¯). (See diagrams for Theorem 2.19.)

We observed earlier that α1 + α2 ≥ π =⇒ α¯1 +α ¯2 ≥ π. ¯ ¯ ¯ Claim. θ1 + θ2 ≤ θ. ¯ Since d(p,¯ s¯) = d(p, s) ≤ d(¯p, s¯), we have θi = θi for i = 1, 2 and soα ¯1 +α ¯2 = π; thus ∆¯1 ∪ ∆¯2 = ∆,¯ and hence d(p, s) = d(¯p, s¯.

¯ S (2) Construct a ruled surface D = s∈qr ps. (See diagrams for Theorem 2.19, particularly the second one for Alexandrov’s patchwork.) j : D¯ → D takes p¯s¯ to ps, is an isometry along each p¯s¯.

Claim. Ifx ¯i ∈ p¯s¯i for i = 1, 2, then d(j(x ¯1), j(x ¯2)) = d(x ¯1, x¯2). ¯ By the same argument, as above, δ1 = δ1 and so on, and we obtain our claim about distance.

Lemma 2.29 (Flat Quadrilateral Lemma). If (p, q, r, s) span a quadrilateral in X and θp +θq +θr +θs ≥ 2π, then θp + θq + θr + θs = 2π and (p, q, r, s) span a ﬂat ﬁlled-in quadrilateral in X Proof. Apply the Flat Triangle Lemma twice

ﬂat gluing is ﬂat

ﬂat

Figure 54: Arguing as in the Flat Triangle Lemma, each triangle is ﬂat; and the gluing is also ﬂat.

Theorem 2.30 (Flat Strip Theorem). If γ , γ : → X are asymptotic geodesics (i.e. sup d(γ (t), γ (t)) < 1 2 R t∈R 1 2 2 +∞), then γ1, γ2 span a ﬂat strip isometric to [0,D] × R ⊂ E for some D > 0 (i.e. γ1, γ2 are parallel in a Euclidean sense.)

63 Idea of proof. Look at projections πi : X → γi(R) for i = 1, 2. As nearest-point retractions, these are 1-Lipschitz. WLOG π1(γ2(0)) = γ1(0); let D = d(γ1(0), γ2(0)). d(γ1(t), γ2(t)) is locally convex as a function of t, hence convex, and is bounded non-negative; hence it is constant. So d(γ1(t), γ2(t)) = D for all t. Similarly, d(γ1(t), γ2(t + a)) is convex for all a ∈ R. π Hence we may infer that all angles are ≥ 2 , since γ1(0) is the closest point on γ1(R) to γ2(0), etc. By the Flat Quadrilateral Lemma, (γ1(0), γ2(0), γ2(t), γ1(t)) span a ﬂat quadrilateral for all t; we can now pass to the limit t → ±∞ to obtain the desired result.

Theorem 2.31 (Decomposition Theorem). If X is CAT(0), γ1 ⊂ X is a geodesic, and Xγ1 := {γ(R) ⊂ γ parallel to γ }, then X = X0 × , where X0 is a convex subset of X. 1 γ1 γ1 R γ1

Proof. Since every pair of geodesics in Xγ spans a ﬂat strip, Xγ is convex. Consider π1 : Xγ1 → γ1(R) and let X0 = π−1(0). γ1 1 If x ∈ X0 , ∃γ : → X parallel to γ s.t. γ (0) = x. γ1 x R 1 x Deﬁne j : X0 × → X by j(x, t) = γ (t). We can check that X0 is convex. γ1 R Γ1 x γ1 Some facts that now follow about hyperbolic isometries: if X is CAT(0), and γ is a hyperbolic isometry, let Min(γ) = {x ∈ X | d(x, γ(x)) = τ(γ)}. Then

(1) γ has an invariant axis in Min(γ). Moreover, every point in Min(γ) lies on an invariant axis. (2) All axes are parallel

(3) Min(γ) is convex and homeomorphic to Y × R where Y is convex in X, and the action of γ on the ﬁrst factor is trivial.

(4) α ∈ Z(Γ) =⇒ α(Min(γ)) = Min(γ).

Theorem 2.32 (Flat Torus Theorem). If A =∼ Zn acts on a CAT(0) group Γ properly, then T n (1) Min(A) := α∈A Min(α) 6= ∅, and splits as Y × E . (2) α ∈ A preserves the splitting and acts by translation on En. n (3) {y0} × E /A is a ﬂat torus

“There is also some stuﬀ about the normaliser ... ”

Idea of proof. Induct on the rank of A. In the n = 1 case we obtain the picture of the individual hyperbolic isometry. In general, suppose we have a splitting Y × En which behaves well for Zn−1 ⊂ Zn. Consider α ∈ A which n−1 n−1 acts trivially on E ; consider the action of α on Y , which splits as Y2×R. Then Min(A) = Y2×E×E . Some consequences of the Flat Torus Theorem: suppose Γ acts on X properly and co-compactly. Then

(1) A ⊂ Γ virtually abelian =⇒ A ﬁnitely-generated. (2) A ⊂ Γ virtually solvable =⇒ A virtually abelian. (3) If X does not contain a ﬂat 2-plane, then Γ is a hyperbolic group.

64 3 Sketches of student talks

Yubo gave talk on the following Theorem 3.1. Let f : F → F be an automorphism of a finitely-generated free group. Then the fixed group Fix(f) := {α ∈ F | f(α) = α} is finitely-generated. Proof. See e.g. S. M. Gersten, “On Fixed Points of Automorphisms of Finitely Generated Free Groups” (Bulletin of the AMS, Vol. 8 No. 3, May 1983.)

This generalizes to Theorem 3.2 (Paulin, Cooper). Let Γ be a hyperbolic group and f :Γ → Γ be a group automorphism. Then Fix(f) is ﬁnitely-generated. This is used to prove (rather than going in the other direction)

Theorem 3.3 (Neumann). Fix(f) is quasiconvex You gave a talk on Terence Tao’s proof of Gromov’s theorem. Both were somewhat muddled, but hey, student talks.

References

• Alessandro Sisto, Lecture notes on Geometric Group Theory • Brian Bowditch, A course on geometric group theory

• Karen Vogtmann’s lecture notes for a course in geometric group theory • Cornelia Drutu and Misha Kapovich, Lectures on Geometric Group Theory • An excellent published source for much of the material is: Martin Bridson and Andre Haeﬂiger, Metric Spaces of Non-positive curvature, Springer-Verlag, see chapters III.H and III.Gamma for hyperbolic groups and sections II.1, II.5, II.6, II.7, II.8 and III.Gamma.1 for CAT(0) groups. • Another published source for material on hyperbolic groups is; Sur les groups hyperboliques d’apr`es Mikhael Gromov, Etienne´ Ghys and Pierre de la Harpe, Birkhauser.