Lecture Notes on Geometric Group Theory

Shengkui Yeshengkui.ye@xjtlu.edu.cn

July 28, 2018

Contents

I Lecture 1 : Lengths 3

1 Word lengths and Cayley graphs 3

2 The fundamental theorem of geometric group theory 4

3 Classifications of groups up to quasi-isometries 6

II Lecture 2: Curvatures I 9

4 CAT(0) spaces 94.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . 94.2 Local and global properties of CAT(0) spaces . . . . . . . . . . . 10

5 Gromov criterion 11

6 CAT(0) groups 13

III Lecture 3: Curvature II 18

7 Definitions and examples 18

8 Boundary of hyperbolic spaces 208.1 Visual topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208.2 Visual metric on ∂X . . . . . . . . . . . . . . . . . . . . . . . . . 22Some references:(1) M. Bridson and A. Haefliger, Metric spaces of non-positive curvature,

vol. 319, Grundlehren der Mathematischen Wissenschaften, Springer-Verlag,Berlin, 1999. (a comprehensive book containing almost everything in GGT.)

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(2) Bowditch, A course on geometric group theory, Mathematical Society ofJapan, Tokyo, 2006. (undergraduate basic textbook.)(3) J.Meier, Groups, graphs and trees: An introduction to the geometry of

infinite groups, Cambridge University Press, Cambridge, 2008. (undergraduatetextbook discussing many interesting examples.)

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Part I

Lecture 1 : LengthsThe geometric group theory studies finitely generated groups from the geometricpoint of view. In the first lecture, we will discuss the "lengths" on groups.

1 Word lengths and Cayley graphs

Notation: Let G be a group, S ⊂ A a subset, S−1 = {s−1 : s ∈ S}.

Definition 1.1 Let G be a group generating by a finite set S. Suppose thatS−1 = S and 1 ∈ S. For any element 1 6= g ∈ G, the word length

|g|S = min{n | g = s1s2 · · · sn,∀si ∈ S}.

For any two elements g, h ∈ G, the word distance dS(g, h) is defined as |g−1h|Swith the convention that |1|S = 0.

Example 1.2 When G = Z, S = {0, 1,−1}. Any g ∈ G, the length |g|S = |g|,the absolute value.

Lemma 1.3 The distance dS : G×G → R≥0 is a metric and invariant underleft G-translations, i.e. dS(gx, gy) = dS(x, y) for any x, y, g ∈ G.

Proof. By definition, dS(g, h) ≥ 0 and d(g, h) = 0 if and only if |gh−1| = 0and g = h. Since S = S−1 (symmetric), dS(x, y) = dS(y, x). It is easy thatdS(x, y) ≤ dS(x, z) + dS(z, y) for any x, y, z ∈ G.

Definition 1.4 Let G be a group generating by a finite set S. The Cayley graphΓ(G,S) is a graph with vertices G and edges {(g, gs) : g ∈ G, s ∈ S}.

Example 1.5 Let Fn be the free group on n letters S = {x1, x2, · · · , xn}. TheCayley graph Γ(G,S) is a a tree.

Note that a graph Γ has a path metric defined as d(v1, v2) = min{n : P is apath of length n connecting v1, v2} for any vertices v1, v2.

Lemma 1.6 The word metric dS is the same as the path metric d.

Proof. If g = s1s2 · · · sn is a word, then this gives a path from g to 1 (theidentity element) in the Cayley graph and vice versa. Note that both metricsare invariant under left G-actions.The metric dS clearly depends on the generating set S. We want to find the

relations between metrics defined by two different generating sets.

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Definition 1.7 Let X,Y be two metric spaces. A map f : X → Y is quasi-isometric if there exists constants A > 0, B ≥ 0, C ≥ 0 such that(1) (quasi-isometric ebedding) 1

AdX(x, x′)−B ≤ dY (f(x), f(x′)) ≤ AdX(x, x′)+B for any x, x′ ∈ X;(2) (cobound) ∀y ∈ Y,∃x ∈ X such that dY (f(x), y) < C.Two metric spaces X,Y are quasi-isometric if there exists such a quasi-

isometry f.

Example 1.8 Let X = Z, with word metric dS defined by S = {0, 1,−1} andY = R. The inclusion Z ↪→ R is quasi-isometric.

Lemma 1.9 For different generating sets S1, S2 of a group G, the two metricspaces (X, dS1) and (X, dS2) are quasi-isometric.

Proof. Let A1 = max{|s′|S : s′ ∈ S′} and A2 = max{|s|S′ : s ∈ S}. Forany element g ∈ G, we have |g|S ≤ A1|g|S′ and |g|S′ ≤ A2|g|S . Choose A =A1A2, B = 0, C = 0 and f the identity map to get a quasi-isometry.

Lemma 1.10 The quasi-isometries define an equivalence relation on the classof metric spaces.

Proof. Let f : X → Y with contants A,B,C and g : Y → Z with contantsD,E, F be quasi-isometries. Then the composite g ◦ f is a quasi-isometry withcontants AD and suffi ciently large M,N. An inverse f−1 of f is defined as thefollowing. For any y ∈ Y, choose x ∈ X such that d(f(x), y) < C. Definef−1(y) = x. Then

d(f−1(y1), f−1(y2)) = d(x1, x2) ≤ Ad(f(x1), f(x2))+B ≤ A(d(y1, y2)+2C)+B

and the other ineuqality is obtained similarly. For any x ∈ X, there existsf−1(f(x)) ∈ X such that d(x, f−1(f(x))) ≤ A(d(f(x), f(x)) + B) = AB. Thisproves that f−1 is also a quasi-isometry.

One of the basic problems in geometric group theory is:

Problem 1.11 Classify finitely generated groups up to quasi-isometries.

2 The fundamental theorem of geometric grouptheory

A geodesic in a metric space (X, d) is an isometric embedding I → X, whereI ⊂ R is an interval equipped with the metric induced from the standard metricof R.

Definition 2.1 A metric space (X, d) is geodesic if every two points x, y ∈ Xare connected by a path of length d(x, y). In other words, for any x, y ∈ X, thereis a continuous map f : [0, 1]→ X such that f(0) = x, f(1) = y and

d(x, y) = sup∑ni=1 d(f(xi), f(xi+1)),

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where the sup is taken over all partitions of [0, 1]. The path f([0, 1]) is called agedesic.

Example 2.2 X = Rn (the Euclidean space),Hn (the hyperbolic space) or Sn(the sphere) is a geodesic metric space.

Example 2.3 X = R2\{(0, 0)} is not geodesic, since there are no geodesicsconnecting (−1, 0) and (1, 0).

Definition 2.4 Let G be a group acting on a space X.(1) The group action is properly discontinuous if for any compact set K ⊂ X,

the set{g ∈ G | gK ∩K 6= ∅}

is finite.(2) The group action is cocompact, if there is a compact set K ⊂ X such

that∪g∈GgK = X.

(3) A metric space X is proper if any closed ball B(x, r) is compact.

Example 2.5 (1) The translation action of Z on R is properly discontinuousand cocompact.(2) The irrational rotation of R2 is neither properly discontinuous nor co-

compact.

The following is the Minor-Schwarz theorem, usually called the fundamentaltheorem of geometric group theory.

Theorem 2.6 Let (X, d) be a proper geodesic metric space and let G be a groupacting properly discontinuously and cocompactly by isometries on X. Then

(1) The group G is finitely generated.(2) For any word metric on G and any point x ∈ X, the orbit map G→ X

given byg 7−→ gx

is quasi-isometric.

Proof. (1) Fix a closed ball B = BD(x0) with the center x0 and the radius Dsuch that X = ∪g∈GgB. Let S = {g ∈ G | gB ∩ B 6= ∅}, which is a symmetricfinite set by the proper discontinuity. If S = G, then nothing needs to prove.Suppose that S 6= G.

Let 2d = inf{d(B, gB) : g ∈ G\S}. The real number d > 0 as the following.For any R > 0 and h ∈ G such that d(B, hB) < R, we have d(x0, hB) ≤ D+R.Thus {h ∈ G : d(B, hB) < R} ⊂ {h ∈ G : hBD+R(x0) ∩ BD+R(x0) 6= ∅} is afinite set. This implies that

inf{d(B, gB) : g ∈ G\S} = inf{d(B, gB) < R : g ∈ G\S} = d(B, h0B) > 0,

for a suffi cienly large R > 0 and some h0 ∈ G\S.

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We check that G is generated by S. Consider a geodesic [x, gx] ⊂ X andk = [d(x,gx)

d ], the integer part. There exists a sequence of points yi ∈ [x, gx]such that y0 = x, y1, ..., yk+1 = gx and d(yi, yi+1) < d. Choose hi ∈ G such thatyi ∈ hiB. We take h0 = 1 and hk+1 = g. Then

d(B, h−1i hi+1B) = d(hiB, hi+1B) ≤ d(yi, yi+1) ≤ d,

which implies that si := h−1i hi+1 ∈ S. Therefore, we have hi+1 = hisi and

g = hk+1 = s0s1 · · · sk.(2) From the proof of (1), we know that

|g|S ≤ k + 1 ≤ d(x, gx)

d+ 1.

Let L = max{d(x, sx) : s ∈ S} and g = a1a2 · · · a|g|S , ai ∈ S. Then

d(x, gx) ≤ d(x, a1x) + d(a1x, gx) = d(x, a1x) + d(x, a2 · · · a|g|Sx)

≤∑|g|Si=1 d(x, aix) ≤ L|g|S .

Therefore,d|g|S − d ≤ d(x, gx) ≤ L|g|S

and d·dS(g1, g2)−d ≤ d(g1x, g2x) ≤ LdS(g1, g2) since both metrics are invariantof left translations.

Corollary 2.7 Let M be a compact Riemannian manifold. Then the funda-mental group π1(M) is quasi-isometric to the universal cover M. For example,Zn is quasi-isometric to Rn. Any surface group π1(Σg) (for a closed surface ofgenus g ≥ 2) is quasi-isometric to the upper-half plane.

3 Classifications of groups up to quasi-isometries

Lemma 3.1 A group homomorphism f : G1 → G2 between two finitely gen-erated groups with word metrics is quasi-isometric if and only if ker f and thecoset G2/ Im f are both finite.

Proof. =⇒ Suppose that for some constants A ≥ 1 and B ≥ 0, we have

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Ad(x, y)−B ≤ d2(f(x), f(y)) ≤ Ad1(x, y) +B

for any x, y ∈ G1. For any x, y ∈ ker f, since d(x, y) ≤ A(d2(f(x), f(y)) +B) =AB, a bounded number, we see that ker f is finite by choosing y = 1. Since thereexists a constant C ≥ 0 such that for any y ∈ G2, the distance d2(y, f(G1)) < C,we know that the coset G2/ Im f is finite.⇐= Note that the subgroup Im f acts on the Cayley graph Γ(G2, S2) by

left multiplications, isometrically and properly discontinuously. Since the cosetG2/ Im f is finite, the action is also cocompact. The fundamental theorem

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implies that Im f and G2 are quasi-isometric. It is enough to prove that G1

is quasi-isometric to Im f. Fix some finite generating sets S for Im f. Denoteby S1 ⊂ G1 the set consisting of representatives of S, viewed as elements incosets of ker f in G1. Then S1∪ker f is a generating set for G1. Thus d1(1, g) ≤d2(1, f(g)) + 1 and d2(1, f(g)) ≤ d1(1, g). Choose A = 1, B = 1 and C =max{d2(g, Im f) : g ∈ G2/ Im f}.

Example 3.2 All the non-abelian finitely generated free groups are quasi-isometric.

Proof. It is enough to note that the free group Fn of rank n (n ≥ 2) is afinite-index subgroup of F2 and apply the previous lemma. For example, letϕ : F2 � Z2 be the abelization and p : Z2 → Z→Z/n be the surjection inducedby projection to the first component. Let X be the Cayley graph of F2 withrespect to the standard generating set, which is a tree and thus contractible.Note that F2 acts on X freely by left translations. The quotient space X/F2 ishomotopy equivalent to the wedge of 2 circles. The quotient space X/ ker(p◦ϕ)is a degree-n covering space of X/F2. By considering the Euler characteristics,the group ker(p ◦ ϕ) = π1(X/ ker(p ◦ ϕ)) is a free group of rank n+ 1.

Example 3.3 Any two surface groups π1(Σg) and π1(Σg′) are quasi-isometric,where Σg,Σg′ are two closed surfaces of genus g, g′ > 1.

Proof. We will find an integre h such that Σh → Σg is a degree-d coveringand Σh → Σg′ is a degree-d′ covering. If this is done, π1(Σh) is a finite-indexsubgroup in both π1(Σg) and π1(Σg′) and then claim follows. Actually, theEuler characteristics implies that 2 − 2h = d(2 − 2g) = d′(2 − 2g′). Thus wecould choose d = g′ − 1 and d′ = g − 1 to finish the proof.

Now we introduce some quasi-isometric invariants.

Definition 3.4 Let G be a group generated by a finite set S. Let βS(S) = #{g ∈G : |g|S ≤ n} the number of elements in the closed ball of radius n. The growthfunction with respect to S is n 7−→ βn(S).

Example 3.5 Let G = Zn with the standard generating set S determined by thestandard basis. Then βk(S) = #{(a1, a2, · · · , an) ∈ G : |a1|+ |a2|+ · · ·+ |an| ≤k}, which grows like the function kn.

One says that G is of polynomial growth of degree ≤ d if there exists aconstant C such that βn(S) ≤ Cnd for any n. The infimum d0 of such d’s iscalled the order of polynomial growth.

Lemma 3.6 If S′ is another finite generating set, then there exists a contantk ≥ 1 such that

βn/k(S′) ≤ βn(S) ≤ βkn(S′).

In particular, the order of polynomial growth of a group is an invariant of quasi-isometries.

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Proof. Let k1 = max{|s|S′ : s ∈ S}. Then {g ∈ G : |g|S ≤ n} ⊂ {g ∈ G :|g|S′ ≤ k1n} and βn(S) ≤ βk1n(S′) Similarly, we have βn(S′) ≤ βk2n(S) forsome k2 ≥ 1. Choose k = k1k2 to finish the first part. If βn(S) ≤ Cnd, thenβm(S′) ≤ βkm(S) ≤ Ckdmd, which means the polynomial growth of βm(S′) isstill of degree ≤ d. Therefore, the order is an quasi-isometric invariant.

Example 3.7 Let G = Zn, the order of polynomial growth of G is n. Therefore,Zn is not quasi-isometric to Zm when m 6= n.

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Part II

Lecture 2: Curvatures IIn the second lecture, we will discuss the curvatures on groups.

4 CAT(0) spaces

4.1 Definitions and examples

Let (X, d) be a geodesic metric space. A geodesic triangle ∆(x, y, z) consistsof three vertices x, y, z ∈ X and three geodesics [x, y], [y, z], [x, z] connectingthese vertices. A comparison triangle ∆(x, y, z) (or denoted by ∆(x, y, z)) isan Euclidean triangle in the plane R2 with three vertices x, y, z and edges oflengths d(x, y), d(y, z), d(x, z) respectively.

Definition 4.1 A geodesic metric space X is CAT(0) if for any geodesic tri-angle ∆(x, y, z) and any two points p, q ∈ ∆(x, y, z), we have

d(p, q) ≤ dR2(p, q),

where p, q are the corresponding points of p, q in the comparison triangle ∆(x, y, z).

Example 4.2 (1) A simplicial tree is CAT(0);(2) Euclidean spaces are CAT(0);(3) Upper-half plane is CAT(0).

Exercise 4.3 Being CAT(0) is not a quasi-isometry invariant. For example,R2 is CAT(0) but Z2 is not.

Definition 4.4 For any geodesics c : [0, a]→ X and c′ : [0, a′]→ X with c(0) =

c′(0), the Alexandrov angle is ∠c(0)(c, c′) := lim

t,t′→0sup arccos t

2+t′2−d(c(t),c′(t))2

2tt′ ,

the limit of angles in the comparison triangles.

Example 4.5 In a simplicial tree, the Alexandrov angle is either 0 or π.

Lemma 4.6 Let X be a geodesic metric space. The following are equivalent.(1) X is CAT(0);(2) For every geodesic triangle ∆([p, q], [q, r], [r, p]) and every point x ∈ [q, r],

we haved(p, x) ≤ d(p, x)

where p, x are the corresponding points of p, x in the comparison triangle;(3) For every geodesic triangle ∆([p, q], [q, r], [r, p]) and every point x ∈

[p, q], y ∈ [p, r] with x 6= p, y 6= p, the angles at the vertices corresponding top in the comparison triangles ∆(p, q, r) and ∆(p, x, y) satisfy

∠p(x, y) ≤ ∠p(q, r)

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(4) For any geodesics c : [0, a] → X and c′ : [0, a′] → X with c(0) = c′(0),

the (Alexandrov) angle ∠c(0)(c, c′) := limt,t′→0 sup arccos t

2+t′2−d(c(t),c′(t))2

2tt′ is nogreater than the angle between the corresponding sides of its comparison trianglein R2.

Proof. It is clear that (1) implies (2) and that (3) implies (4). For (2) im-plies (3), consider the comparison triangle ∆(p′, q′, y′) of the geodesic triangle∆(p, q, y). Note that d(q, y) ≤ dR2(q, y) in ∆(p, q, r). Apply the law of cosine toget that

∠p(q, r) ≥ ∠p′(q′, y′).Similarly, d(x, y) ≤ dR2(x, y′) in∆(p′, q′, y′).Once again the law of cosine impliesthat ∠p′(q′, y′) ≥ ∠p(x, y). Moreover, (1) and (3) are equivalent by the law ofcosine.Now we prove (4) implies (2). For a geodesic ∆([p, q], [q, r], [r, p]) and any

point x ∈ [q, r], consider the three comparison triangles∆(p′, q′, r′),∆(p′′, q′′, x′′)and ∆(p′′, r′′, x′′) of ∆(p, q, r),∆(p, q, x),∆(p, r, x), respectively. Note that theAlexandrov angles

π = ∠x([q, x], [x, r]) ≤ ∠x([q, x], [x, p]) + ∠x([x, p], [x, r])

≤ ∠x′′(q′′, p′′) + ∠x′′(p′′, r′′).

Choose the other point r′′′ in the line q′′x′′ such that dR2(x′′, r′′′) = dR2(x′′, r′′).

The law of cosine implies that dR2(p′′, r′′′) ≤ dR2(p′′, r′′) = dR2(p

′, r′) and∠q′′(p′′, x′′) ≤ ∠q′(p′, x′) and thus

d(p, x) = dR2(p′′, x′′) ≤ dR2(p′, x′).

4.2 Local and global properties of CAT(0) spaces

Lemma 4.7 Let (X, d) be a CAT(0) space. Then:(i) (X, d) is uniquely geodesic, i.e. any two points are joined by a unique

geodesic segment.(ii) X is contractible.

Proof. If there were two different geodesic segments [x, y]1, [x, y]2 connectingtwo points x, y, choose z1 ∈ [x, y]1, z2 ∈ [x, y]2 such that d(x, z1) = d(x, z2).In the comparison triangle of ∆(x, y, z1) with edges [x, z1], [z1, y] and [x, y]2degenerates, which implies that d(z1, z2) ≤ 0 and thus z1 = z2.Fix a point x ∈ X. For any other point y ∈ X, there exists a unique geodesic

[y, x]. The space shrinks to x along these geodesics.

Definition 4.8 A metric space X is non-positively curved if it is locally CAT(0),i.e. for every x ∈ X there exists rx > 0 such that the ball B(x, rx) with the in-duced metric is a CAT(0).

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Example 4.9 A smooth Riemannian manifold with non-positive sectional cur-vature is locally CAT(0).

Theorem 4.10 (Cartan-Hadamard) The universal cover of a complete con-nected locally CAT(0) space is CAT(0).

5 Gromov criterion

We now discuss how to construct non-positively curved spaces.A Euclidean cell is the convex hull of a finite number of points in Rn,

equipped with the standard Euclidean metric.

Definition 5.1 A Euclidean cell complex X is a space formed by glueing to-gether Euclidean cell-complexes via isometries of their faces. It has the piecewiseEuclidean path metric. Precisely, for any x, y ∈ X, let x = x0, x1, · · · , xn = ybe a path such that each successive xi, xi+1 is contained in a Euclidean simplexSi. Define the distance (called path metric) dX(x, y) = inf

∑n−1i=0 dSi(xi, xi+1),

where the infimum is taken over all such paths.

Example 5.2 Let X be the Euclidean cell complex by glueing together the threeunit squares in the coordinate planes in R3. Show that the distance

dX((0, 0, 1/2), (1/2, 1/2, 0)) =

√5

2

and there are two paths realizing this distance.

Definition 5.3 Let X be a Euclidean cell complex and v ∈ X. The (geometric)link Lk(x,X) is the set of unit tangent vectors ("directions") at x in X. Pre-cisely, let S be the set of all geodesics [x, y] with y in a simplex containing x.Two geodesics are called equivalent if one is contained in the other. The linkLk(x,X) is the set of equivalence classes of geodesics in S.

If X is one n-dimensional Euclidean cell, the link Lk(x,X) is part of Sn−1

and thus the topology on Lk(x,X) is defined as the "angle" topology. In general,the topology on Lk(x,X) is defined as the path metric coming from each cell.

Definition 5.4 A metric space (X, d) is CAT(1) if any geodesic triangle ∆(x, y, z)of perimeter less than 2π has the same inequality as in Definition 4.1 when thecomparison triangle in R2 (in the definition of CAT(0)) is replaced by the onein unit sphere S2.

Example 5.5 A circle of length less than 2π is never CAT(1), while a circleof length not less than 2π is always CAT(1).

Theorem 5.6 (Gromov criterion) Let X be a Euclidean cell complex with fi-nitely many isometric cells. The following are equivalent:(1) X is non-positively curved, i.e. locally CAT(0);

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(2) each vertex link Lk(x,X) is CAT(1);(3) each vertex link Lk(x,X) contains no closed geodesic loop of length less

than 2π.

Outline of the proof of Theorem 5.6. The previous example shows that(2) implies (3). Actually, (3) also implies (2) by a theorem of Gromov.We prove that (1) and (2) are equivalent. We choose a small neighborhood

U of x as the cone of Lk(x,X). In other words, for a small number a the setU = {ty : t ∈ [0, a], y ∈ Lk(x,X)} with the convention that 0y = 0y′ for anyy, y′ ∈ Lk(x,X). The metric on U is actually given by d(ty, t′y′)2 = t2 + t′2 −2tt′ cos(min{π, dLk(x,X)(y, y

′)}). Suppose that U is CAT(0). For any geodesictriangle ∆(y1, y2, y3) in Lk(x,X) of perimeter less than 2π, choose a comparisontriangle∆(y1, y2, y3)) in S2.We consider three points ty1, ty2, ty3 ∈ U (such thatthe geodesic triangle ∆(ty1, ty2, ty3) is still of perimeter less than 2π) and thethree points ty1, ty2, ty3 in the unit ball with boundary S2. For any y ∈ [y2, y3],the distance d(ty, ty1) ≤ d(ty, ty1) when U is CAT(0). But the definition of themetric d gives that dLk(x,X)(y, y1) ≤ dS2(y, y1). This proves that Lk(x,X) isCAT(1).Suppose that Lk(x,X) is CAT(1). For any three points xi = tiyi ∈ U, i =

1, 2, 3, we consider the comparison triangle ∆(x1, x2, x3) in R2. If one ti is zero,then the geodesic triangle ∆(x1, x2, x3) with induced metric is isometric to∆(x1, x2, x3), noting that the metric on U could be defined as d above. Wethus may assume ti 6= 0. We prove that U is CAT(0) in the following threecases.

Case 5.7 (a) d(y1, y2) + d(y2, y3) + d(y1, y3) < 2π. Choose ∆(y1, y2, y3) in S2

to be the comparison triangle of the geodesic triangle with vertices y1, y2, y3 inthe link Lk(x,X). The cone C of ∆(y1, y2, y3) in the unit ball has the met-ric defined by d(ty, t′y′)2 = t2 + t′2 − 2tt′ cos ds2(y, y

′) for any y, y′ ∈ S2.Note that the ∆(x1, x2, x3) could be ∆(t1y1, t2y2, t3y3). For any p = ty, q =t′y′ ∈ ∆(x1, x2, x3) ⊂ U, the distance d(p, q) ≤ dR2(p, q) since dLk(x,X)(y, y

′) ≤dS2(y, y

′), where p, q are the corresponding points in the comparison triangle of∆(x1, x2, x3) in the cone C.

Case 5.8 (b) d(y1, y2) + d(y2, y3) + d(y1, y3) ≥ 2π but all d(yi, yj) < π, i, j =1, 2, 3. Let ∆(O, x′1, x

′2),∆(O, x′1, x

′3) be the comparison triangles of the geodesic

triangles ∆(x, x1, x2),∆(x, x1, x3). Without loss of generality, we may assumethat x′2, x

′3 lie different sides of [O, x′1]. By the definition of the metric d on

the cone U, we have that ∠O(x′1, x′2) = d(y1, y2),∠O(x′1, x

′3) = d(y1, y3) and

∠x′1(O, x′2) = ∠x1(x, x2),∠x′1(O, x

′3) = ∠x1(x, x3). Since d(y1, y2) + d(y1, y3) >

π, we have

∠O(x′3, x′2) = 2π − ∠O(x′1, x

′2)− ∠O(x′1, x

′3)

= 2π − d(y1, y2)− d(y1, y3) ≤ d(y2, y3) = ∠x(x2, x3)

and thus dR2(x′2, x′3) ≤ d(x2, x3) by the law of cosine. Let ∆(x1, x2, x3) =

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∆(x′′1 , x′′2 , x′′3) be the comparison triangle. Then

∠x′′1 (x′′2 , x′′3) ≥ ∠x′1(x

′2, x′3) = ∠x′1(x

′2, O) + ∠x′1(O, x

′3)

= ∠x1(x, x2) + ∠x1(x, x3) ≥ ∠x1(x2, x3),

which implies that the Alexandrov angle is not greater and thus U is CAT(0).

Case 5.9 (c) d(yi, yj) ≥ π for some i, j. Without loss of generality, assumed(y2, y3) ≥ π. Note that x lies in the geodesic [x2, x3] by considering the de-fined metric d. Let ∆(O, x′1, x

′2),∆(O, x′1, x

′3) be the comparison triangles of the

geodesic triangles ∆(x, x1, x2),∆(x, x1, x3). Since ∠O(x′1, x′2) + ∠O(x′1, x

′3) =

d(y1, y2) + d(y1, y3) ≥ π, the same idea as the proof Lemma 4.6 implies that theAlexandrov angle ∠x2(x1, x3) = ∠x′2(x

′1, O) ≤ ∠x′′2 (x′′1 , x

′′3) for any comparison

triangle ∆(x′′1 , x′′2 , x′′3) of ∆(x1, x2, x3).

Corollary 5.10 A 2-dimensional Euclidean cell complex X is non-positivelycurved if and only if each link Lk(x,X) contains no injective loops of length lessthan 2π.

Example 5.11 The boundary of a 3-dimensional cube is not non-positivelycurved.

6 CAT(0) groups

Definition 6.1 A group G is CAT(0) if it acts properly and cocompactly on aCAT(0) space by isometries.

Example 6.2 A finite group, fundamental groups of closed hyperbolic mani-folds, fundamental groups of compact non-positively curved Euclidean cell com-plexes, the finitely generated abelian group Zk are CAT(0).

Proof. Let G be a finite group. Then G acts transitively on the tree with avertices {G} ∪ O and edges {(O, g) : g ∈ G}. If G is the fundamental groupof a compact non-positively curved Euclidean cell complexes, then G acts geo-metrically on the universal covering X with a metric induced from X. TheCartan-Hadamard theorem implies that X is a CAT(0) space.

Example 6.3 Let Γ be a finite graph with the vertex set V and the edge set E.The right angled Artin group (RAAG) G(Γ) = 〈v, v ∈ V | vwv−1w−1, vw ∈ E〉.Prove that G(Γ) is a CAT(0) group.

Proof. Fix a point p. For each vetex v, attach a loop lv to p. For each edge vw,attach a square along the boundary v, w, v−1, w−1. The resulting complex Khas the fundamental group G(Γ). Consider the link Lk(p,K) and use Gromovcriterion to show that K is non-positively curved (any closed loop in Lk(p,K)has length at least 2π).We study the structure of CAT(0) groups in the following.

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Definition 6.4 Let X be a metric space and γ : X → X an isometry. Thetranslation length of γ is |γ| = inf{d(γx, x) : x ∈ X}. The set of points attainingthe infimum is denoted by Min(γ). The isometry γ is called semi-simple ifMin(γ) 6= ∅. If Γ is a group of isometries, then Min(Γ) := ∩γ∈ΓMin(γ).

Lemma 6.5 When X is CAT(0), the minimal set Min(γ) is convex and γ-invariant.

Proof. This is a consequence of the following fact: for any two geodesicsc : [0, 1] → X and c′ : [0, 1] → X, parametrized propositional to arc-length, wehave

d(c(t), c′(t)) ≤ (1− t)d(c(0), c′(0)) + td(c(1), c′(1)).

If c(0) = c′(0), consider the comparison triangle ∆(c(0), c(1), c′(1)).We haved(c(t), c′(t)) ≤ td(c(1), c′(1)) by the elementary geometry. If c(0) 6= c′(0), definea new linear parametrized curve c′′ : [0, 1] → X such that c′′(0) = c(0) andc′′(1) = c′(1). Then

d(c(t), c′(t)) ≤ d(c(t), c′′(t))+d(c′′(t), c′(t)) ≤ (1−t)d(c(0), c′(0))+td(c(1), c′(1)).

Definition 6.6 Let X be a metric space. An isometry γ : X → X is called:(1) elliptic if γ has a fixed point;(2) hyperbolic if the translation length 0 6= |γ| = d(γx, x) for some x ∈ X;(3) parabolic otherwise (i.e. not semi-simple).

Example 6.7 (1) Any isometry of the Euclidean space Rn is semi-simple.(2) Let H2 = {(x, y) : y > 0} be the upper half plane with metric dx2+dy2

y2 .

The translation τa(x) = x+ a, a > 0, is parabolic.

Lemma 6.8 Let γ : X → X be an isometry of a CAT(0) space X and C anγ-invariant convex, complete subspace.

(1) ∀x ∈ X,∃!p(x) ∈ C such that d(x, p(x)) = d(x,C) := infc∈C d(x, c). Thisgives a projection p : X → C, which is distance-non-increasing.(2) p(γx) = γp(x) for any x ∈ X. Moreover, Min(γ|C) = Min(γ) ∩ C.

Proof. Let D = d(x,C) and suppose that d(x, yn) → d(x,C) for a sequence{yn}.We claim that {yn} is a Cauchy sequence, which proves that p(x) exists bythe completeness of C and is also unique. Actually, for any ε > 0, there exists Nsuch that when n,m > N we have d(x, xn) < D+ ε and d(x, xm) < D+ ε. Con-sider the comparison triangle ∆(x, xn, xm) in R2. The edge [xn, xm] lies in theregion {v ∈ R2 : D ≤ d(x, v) < D + ε} (if ∃y ∈ [xn, xm] satisfying d(x, y) < D,then dX(x, y) ≤ d(x, y) < D by the CAT(0) inequality, a contradiction!). By el-ementary geometry, the longest segement contained in the region is 2

√ε2 + 2εD,

implying that d(xn, xm) ≤ 2√ε2 + 2εD.

We now prove d(p(x), p(y)) ≤ d(x, y) for any x, y ∈ X. Without loss ofgenerality, assume x, y ∈ X\C. Let ∆(x, px, py) be the comparison triangle.

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The angle ∠px(x, py) ≥ π/2 (otherwise, ∃z ∈ [px, py] such that xz⊥[px, py].But dX(x, z) ≤ d(x, z) < d(x, px) = d(x, px) with z ∈ C, a contradiction).Similarly, we have ∠py(px, y) ≥ π/2 for another comparion triangle ∆(x, y, py)by composing the two comparion triangles together along [x, py] (otherwise, wecould a point z ∈ [px, py] such that d(z, y) < d(py, y). Denote u = [z, y]∩ [x, py].Then dX(y, z) ≤ d(y, u)+d(u, z) ≤ d(y, u)+d(u, z) < d(py, y), a contradiction).We see that d(p(x), p(y)) ≤ d(x, y).Note that d(γx, γpx) = d(x, px) = d(x,C) = d(γx,C) and thus γpx = p(γx)

by the uniqueness of projection points. Therefore, pMin(X) = Min(γ) ∩ C =Min(γ|C).

Theorem 6.9 Let X be a CAT(0) space.(1) An isometry γ of X is hyperbolic if and only if there exists a geodesic

line c : R→ X such that γc(t) = c(t+ a) for some a > 0. The set c(R) is calledan axis of γ and the number a = |γ|.(2) If X is complete and γm is hyperbolic (resp. elliptic) for some m 6= 0,

then γ is hyperbolic (resp. elliptic).Let γ be a hyperbolic isometry of X.(3) The axes of γ are parallel to each other and their union is Min(γ);(4)Min(γ) is isometric to a product Y ×R and the restriction of γ toMin(γ)

is of the form (y, t) 7−→ (y, t+ |γ|);(5) Every isometry α that commutes with γ leavesMin(γ) = Y ×R invariant

and its restriction to Y × R is of the form (α′, α′′), where α′ is an isometry ofY and α′′ is a translation of R.

Proof. (1) The if part is proved in Lemma 6.8. When γ is hyperoblic, for anyx ∈Min(γ) the union of geodesic segments [γnx, γn+1x]n∈Z is a geodesic (choosem to be the middle point of [x, γx]. Then d(m, γm) = d(x, γx) = 2d(m, γx) =d(m, γx) + d(γx, γm) implying the union is locally geodesic and thus geodesic).(2) When γ is hyperbolic, this follows (1). If γn is elliptic, γn has fixed point

x. Then {γix : i = 1, 2, ...} is bounded and thus has a center y, which is a fixedpoint of γ.(3) For two axes c, c′ : R → X, consider the projections of X to these two

axes. The equivariance of projections (see Lemma 6.8 ) and the convexity (seethe proof of Lemma 6.5) proves that c and c′ are parallel.(4) From the proof of (1), Min(γ) is a union of axes. From the convexity,

Min(γ) is a CAT(0) subspace. By (3), all these axes are parallel. Let p :Min(γ) → c(R) be the projection onto the image of an axe. Choose Y =p−1(c(0)).(5) Since α maps axes of γ onto axes, the product structure is preserved.

Since α commutes with a translation, itself is a translation.The following is usually called the flat torus theorem.

Theorem 6.10 (Flat torus theorem) Let A = Zn act properly on a CAT(0)space X by semi-simple isometries. Then(1) Min(A) = ∩γ∈AMin(γ) is non-empty and splits as a product Y × Rn.

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(2) Every element α ∈ A leaves Min(A) invariant and respects the productdecomposition; α acts as the identity on the first factor Y and as a translationon the second factor Rn.(3) The quotient of each y × Rn by the action of A is an n-torus.(4) If an isometry of X normalizes A, then it leaves Min(A) invariant and

preserves the product decomposition.(5) (virtual splitting of centralizers) If a subgroup Γ < Isom(X) normalizes

A, then a subgroup of finite index in Γ centralizes A. Moreover, if Γ ⊃ A isfinitely generated, then Γ has a subgroup of finite index that contains A as adirect factor.

Proof. The proof of (1),(2),(3) is based on the previous theorem by induction onn. Choose the standard basis e1, e2, ..., en of Zn. The previous theorem impliesthat Min(e1) = Z × R. Then the subgroup 〈e2, ..., en〉 = Zn−1 act invariantlyon Z and trivially on z0×R (note that the subgroup acting on R by non-trivialtranslations is generated by e1). (4) is proved by noting that γ maps the convexhull of A-orbits to the convex hull of A-orbits.For any fixed real number r, the set {a ∈ A : |a| = r} is finite. Since the

translation length |γaγ−1| = |a|, the image Γ→ Aut(A) given by conjugations isfinite (because the number of possible images for each basis lement ei is finite).The kernel Γ0 is a subgroup of finite index in Γ and its elements centralize A.Any element γ ∈ Γ0 acts on Min(A) = Y × Rn by γ(y, x) = (γ1y, γ2x) by

(5) of the previous theorem, where γ2 is a translation. This gives a grouphomomorphism f : Γ0 → Rn(< Isom(Rn)). When Γ is finitely generatedand Γ ⊃ A, the image Im f ∼= Zn. After changing of basis, we may assumeA = 〈e1, ..., en〉 and f(e1) lies in the first component of Zn. Projecting Zn to thefirst component, we have a surjection f1 : Γ0 → Z with f1(e1) 6= 0. This surjec-tion gives f−1

1 (〈f1(e1)〉) → 〈f1(e1)〉. By maping f1(e1) to e1, we get a splitingf−1

1 (〈f1(e1)〉) = 〈e1〉 × Γ1 and Γ1 ∩ A = 〈e2, ..., en〉. Note that f−11 (〈f1(e1)〉) is

of finite index in Γ0 (and thus in Γ). The proof of (5) is finished by an inductiveargument on n.

Theorem 6.11 If the group Γ acts properly and cocompactly by isometries ona CAT(0) space X, then(1) Γ is finitely presented.(2) Γ has only finitely many conjugacy classes of finite subgroups.(3) Every solvable subgroup of Γ is finitely generated and virtually abelian

(i.e. there exists an abelian subgroup Γ0 of finite index).(4) Every abelian subgroup of Γ is finitely generated.(5) If Γ is torsion-free, then it is the fundamental group of a compact cell

complex whose universal cover is contractible.

Proof. (3) (outline) It is enough to prove that if 1→ A→ Γ→ B → 1 is exact,finitely generated A = [Γ,Γ] is virtually abelian, B is abelian, then Γ is virtuallyabelian (after this, an inductive argument will prove the general solvable case).Let A0 < A be a subgroup of index n. Since A acts on the left cosets of any

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index-n subgroup H, there is a group homorphism fH : A → Sym(n) (thesymmetric group). Since A is finitely generated, the number of all such fH isfinite. Therefore, we could choose a finite-index subgroup A1 = ∩ ker fH < Awhich is characteristic (i.e. any φ ∈ Aut(A) we have φ(A1) = A1). Then Γnormalizes A1 and thus a finite-index subgroup Γ0 contains A1 as direct factorby the virtual splitting of centralizers. Note that [Γ0,Γ0] < (A∩Γ0)/A1 is finiteand thus Γ0 is virtually abelian.

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Part III

Lecture 3: Curvature IIIn this lecture, we study hyperbolic spaces and groups

7 Definitions and examples

Definition 7.1 Let δ ≥ 0.(1) A geodesic metric space X is Gromov δ-hyperbolic if for every geodesic

triangle ∆(x, y, z), each geodesic edge is contained in the δ-neighborhood of theunion of the other two sides. A geodesic metric space X is hyperbolic if it isGromov δ-hyperbolic for some δ > 0.

(2) A finitely generated group G is hyperbolic if its Cayley graph is hyperbolic.

Example 7.2 (1) Bounded metric spaces are δ-hyperbolic, where δ can be thediameter. Thus finite groups are hyperbolic.(2) Trees are 0-hyperbolic. Thus free groups are hyperbolic.(3) Real hyperbolic spaces are δ-hyperbolic for δ = (log3)/2. The Euclidean

plane is not hyperbolic. Therefore, surface groups (π1(Σg), g > 1) are hyperbolic.(4) A CAT(0) space is not necessary hyperbolic, eg. R2; and hyperbolic space

is also not necessary CAT(0), eg. bounded spaces.(5) In a certain statistical sense, almost all finitely presented groups are

hyperbolic.

Hyperbolicity is a large-scale property of metric spaces.

Definition 7.3 Let X be a metric space and p ∈ X. The Gromov product(, )p : X ×X → R is defined as

(x, y)p =1

2(d(x, p) + d(y, p)− d(x, y)).

For a geodesic triangle∆(x, y, z), consider the comparison triangle∆(x, y, z) ⊂R2. The inscribed circle of ∆(x, y, z) has three tangent points ix, iy, iz (withoutconfusions, we also denote the corresponding points in ∆(x, y, z) by ix, iy, iz).It is easy to see that (x, y)z = d(z, ix) = d(z, iy). By shrinking the inscribed cir-cle to its center O∆, the triangle ∆(x, y, z) becomes four-point tree T∆ (calledtripod) with vertices x, y, z, O∆ and edges xO∆, yO∆, zO∆ whose lengths are(y, z)x, (x, z)y, (x, y)z respectively. Let f∆ : ∆(x, y, z) → T∆ be the naturalmap.

Proposition 7.4 Let X be a geodesic space. The following are equivalent:(1) (slim-triangle property) X is δ0-hyperbolic: there exists δ > 0 such that

each geodesic edge is contained in the δ0-neighborhood of the union of the othertwo sides, for every geodesic triangle ∆(x, y, z).

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(2) (thin-triangle property) There exists δ1 > 0 such that any two pointsp, q with f∆(p) = f∆(q), the distance d(p, q) ≤ δ1 for every geodesic triangle∆(x, y, z).(3) There exists δ2 > 0 such that the diameter of f−1(O∆) is less than δ2,

for every geodesic triangle ∆(x, y, z).

Proof. It is obvious (2) implies (1). For (1) implies (3), choose p ∈ [x, y] (thegeodesic segment) (or [x, z]) such that d(p, ix) ≤ δ0. By the triangle inequalities,d(p, iz) = |d(y, p) − d(y, iz)| = |d(y, p) − d(y, ix)| ≤ d(p, ix) ≤ δ0 and thusd(ix, iz) ≤ d(p, iz) + d(p, ix) ≤ 2δ0. Thus the diameter of f−1(O∆) is less than4δ0.For (3) implies (2), suppose without loss of generality that p ∈ [y, z] and

d(y, p) < d(y, ix) and q is the other point in f−1∆ (f∆(p)). If we can show that

p, q are the tangent points of another geodesic triangle, the proof is finished.Actually, let c(t) : [0, 1]→ X be a monotone parametrization of [y, z]. Note thatthe tangent points of ∆(x, y, c(t)) move continuously as a function of t. Whent = 0, the tangent point on [y, c(t)] is y while t = 1 gives the tangent point ix.Therefore some point t0 ∈ (0, 1) gives the tangent point on [y, c(t0)] is y. Sinced(y, p) = d(y, q), the point q is also a tangent point.

Corollary 7.5 If X is δ-hyperbolic, then

(x, y)w ≥ min{(x, z)w, (y, z)w} − 2δ

for any x, y, z, w ∈ X.

Proof. We will use the thin-trangle property to prove the claim. Without loss ofgenerality, we assume that S := d(x, z)+d(y, w) ≤M := d(x, y)+d(z, w) ≤ L :=d(x,w) +d(y, z). Note that L ≤M + 2δ if and only if (x, y)w ≤ (y, z)w + 2δ. Let∆(x,w, y),∆′(x,w, z) be geodesic triangles with the tangent points (ix, iw, iy)and (i′x, i

′w, i′z). Considering the path from y to z that proceeds via ix, iy, i′z

and i′w, we have that d(y, z) ≤ d(y, ix) + d(ix, iy) + d(iy, i′z) + d(i′w, z) which is

b+ d(ix, iy) + l + d(i′z, i′w) + d in the notation of the following diagram:

Also d(x,w) = a+ c− l. Thus L ≤ a+ b+ c+ d+ 2δ = M + 2δ.

Lemma 7.6 Let x, y, z ∈ X, a δ-hyperbolic space and α a geodesic connectingx, y. Then

(x, y)z ≤ d(z, α) ≤ (x, y)z + 8δ.

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Proof. If a ∈ α, then xy = xa + ax, xz ≤ xa + az, yz ≤ ya + az and thusaz ≥ (x, y)z. For the other inequality, let a ∈ α, p ∈ [x, z] be the tangent points.Then xa+az ≤ xp+δ1 +pz+δ1 = xz+2δ1. Similarly, ya+az ≤ yz+2δ1. Sumthese two inequalies together to get that az ≤ (x, y)z+2δ1. The proof is finishedby noting that δ1 ≤ 4δ as shown in the proof of the previous proposition.

Corollary 7.7 If α, β are two geodesics connecting the same pair of points,then α ⊂ N(β, 8δ) and β ⊂ N(α, 8δ).

Lemma 7.8 (Stability of Quasi-Geodesics) Let X be a δ-hyperbolic space, c :[0,+∞) → X be a (A,B)-quasi-geodesic (i.e. 1

A |t − t′| − B ≤ d(c(t), c(t′)) ≤

A|t − t′| + B for all t, t′) and [p, q] a geodesic segment joining endpoints of c.Then there exists a constant R = R(A,B, δ) such that [p, q] is contained in anR-neighborhood of c and c is contained in an R-neighborhood of [p, q] (in otherwords, [p, q] ⊂ N(c,R), c ⊂ N([p, q], R) or the Hausdorff distance between c and[p, q] is less than R).

Theorem 7.9 Let f : X → X ′ be a quasi-isometric embedding between geodesicmetric spaces. If X ′ is hyperbolic, then X is hyperbolic as well.

Proof. Let ∆(x, y, z) be a geodesic triangle in X. Choose ∆(fx, fy, fz) tobe another geodesic triangle in X ′ connecting the image of the three vertices.Since X ′ is hyperbolic, each edge of ∆(fx, fy, fz) is contained in a boundedneighborhood of the other twos. The previous lemma implies that the image ofeach edge in ∆(x, y, z) is of bounded distance from the corresponding edge in∆(fx, fy, fz). Therefore, ∆(x, y, z) is also a slim triangle.

8 Boundary of hyperbolic spaces

Let X be a geodesic space. A geodesic ray is an isometric embedding c :[0,+∞) → X such that c([0, t]) is a geodesic connecting c(0) and c(t) for anyt ≥ 0. Two geodesic rays c1, c2 are defined to be equivalent or asymptotic ifd(c1(t), c2(t)) ≤ K for some constant K and any t ≥ 0.

Definition 8.1 The boundary ∂X is the set of equivalence classes of geodesicrays.

Example 8.2 The boundary of the real line R consists of two points.

8.1 Visual topology

We now define a (visual) topology on ∂X. For techniqical reasons, define ageneralized ray to be a geodesic c : I → X, where I = [0, R] for some R ≥ 0or I = [0,+∞). In the case I = [0, R] it is convient to define c(t) = c(R) whent ≥ R. Then X = X ∪∂X = {c(∞) : c is a generalized gedesic ray}, where c(∞)denotes the equivalence class of c.

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Definition 8.3 Let X be a proper geodesic space. Fix a base point p ∈ X.Define xn → x in X if there exists generalized ray cn, c starting from p withcn(∞) = xn, c(∞) = x such that every subsequence of {cn} contains a subse-quence converging (uniformly on compact subsets) to c.

Lemma 8.4 (Arzela-Ascoli theorem) Let {ci : [0,∞)→ X, i ∈ I} be a sequenceof generalized geodesic rays with {ci(0), i ∈ I} contained in a bounded set B ⊂ X.Then any subsequence cij of {ci} has a convergent subsequence cijk → c, wherec is also a generalized geodesic ray.

Proof. Recall that the classical Arzela-Ascoli theorem is that if a sequencefi : [a, b] → X is uniformly bounded and equi-continuous (i.e. ∀x ∈ [a, b]∀ε >0∃δ > 0,such that d(fi(x), fi(y)) < ε for any i), then any subsequence has asubsequence converging to a continuous function f : [a, b]→ X. In our situation,for any K > 0 the restricted sequence ci : [0,K] → X is uniformly bounded( ci([0,K]) ⊂ N(B,K) the K-neighborhood of B) and equi-continuous by iso-metric embedding. Therefore, any subsequence cij |[0,K] has a converging subse-quence cijk |[0,k] → cK , which is a generalized ray. When K →∞, ck → c notingthat cK is a restriction of cK′ when K ′ > K.

Lemma 8.5 Fix c0(∞) ∈ ∂X and ε > 0. For any integer n, define Vn(c0, ε) ={c(∞) ∈ X : d(c(n), c0(n)) < ε}. Then {Vn(c0, ε) : n ∈ N, ε > 0} is a funda-mental system of neighborhoods of c0(∞), i.e. any open set containing c0(∞)contains some Vn(c0).

Proof. It is enough to prove that ci(∞)→ c0(∞) if and only if for every n > 0there exists Nn > 0 such that ci ∈ Vn(c0) when i > Nn. Suppose that ci → c0and for some n0 there is a sequence cij /∈ Vn0(c0). By Arzela-Ascoli theoremthere is a subsequence cijk → c′ /∈ Vn0(c0), a contradiction. Conversely, if forevery n, ε we have ci ∈ Vn(c0, ε) when i > Nn for some Nn, then ci(∞) = c0(∞).

Example 8.6 LetM be a non-positively curved manifold. Fix p ∈M, each unittangent vector v ∈ TpM determines a geodesic rany rv. In the visual topology,∂M is identified with the sphere Sn−1 as the set of all unit tangent vectors.In particular, the boundary of n-dimensional Euclidean space Rn or hyperbolicspace Hn is Sn−1.

Theorem 8.7 (1) The topology on X = X ∪ ∂X is independent of p;(2) X ↪→ X is a homeomorphism onto its image. Moreover, X is compact

and ∂X is closed;(3) If f : X → X ′ is a quasi-isometry between proper δ-hyperbolic spaces.

Then f induces a homeomorphism f∂ : ∂X → ∂X ′.

Proof. (1) Given any point q ∈ X and any geodesic ray c with c(0) = p, thereexists a ray c′ with c′(0) = q and c′(∞) = c(∞).

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(2) X is obvious homeomorphic to its image and X is open in X. Theprevious lemma shows that X satisfies the first axiom of countability and X issequentially compact by the Arzela-Ascoli.(3) For any geodesic ray c in X, the map f ◦ c is a quasi-geodesic in X ′. Let

c′(n) be the geodesic from f(c(0)) to f ◦c(n). By Arzela-Ascoli theorem, {c′(n)}has a subsequence convergent to a geodesic ray c′. By Lemma 7.8, c′ lies in abounded neighborhood of f ◦ c. Define f∂(c(∞)) = c′(∞)). It is direct to checkthat f∂ is well-defined and continuous.

8.2 Visual metric on ∂X

In this subsection, let X be a δ-hyperbolic space and p ∈ X.

Definition 8.8 The Gromov product (, )p : X ×X → R is defined as

(x, y)p =1

2(d(x, p) + d(y, p)− d(x, y)).

In general, for x, y ∈ X ∪ ∂X define

(x, y)p = sup lim infi,j→∞

(xi, yj)p

where the supremum is taken over all sequences xi → x, yj → y in X.

Lemma 8.9 For x 6= y ∈ ∂X, there exists a geodesic c : (−∞,∞) → X con-necting these two points, i.e. c(−∞) = x and c(∞) = y. Moreover, (x, y)p <∞for any p ∈ X. In particular, (x, y)p =∞ if and only if x = y ∈ ∂X.

Proof. Fix p ∈ X, let c1, c2 be two geodesics starting from p to x and y respec-tively. Choose T > 0 such that d(c1(T ), c2) > δ. Let γn be the geodesic segmentconnecting c1(n) and c2(n) for each n > T. By the thin triangle property, thereis a point pn ∈ γn such that d(c1(T ), pn) < δ. the distance d(p, γn) is boundedby a constant D(x, y). The Arzela-Ascoli theorem implies that a subsequence of[pn, c2(n)] ⊂ γn has a limit c of bounded distance from c1(T ), which is a geo-desic connecting x and y. We actually proved that d(p, c) ≈ limn(c1(n), c2(n))pis also finite.

Let ε > 0 and define ρε(x, y) = e−ε(x,y)p for x, y ∈ ∂X and

dε(x, y) = inf Σni=0ρε(xi, xi+1)

where the infimum is taken over all chains x = x0, x1, ..., xn+1 = y, no boundon n.

Theorem 8.10 Let ε > 0 and ε′ = e2δε − 1. If ε′ ≤√

2− 1, then dε is a visualmetric on ∂X, indeed

(1− 2ε′)ρε(x, y) ≤ dε(x, y) ≤ ρε(x, y)

for all x, y ∈ ∂X. If X is a proper geodesic space, the topology induced by dε isthe same as the visual topology.

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Proof. It is obvious that dε(x, y) ≤ ρε(x, y). The other inequality is proved byinduction on the size n in the chain x = x0, x1, ..., xn+1 = y and proving thefact that

(1− 2ε′)ρε(x, y) ≤ Σni=0ρε(xi, xi+1) := S(n).

When n = 0 or S(n) ≥ (1 − 2ε′), this inequality is obvious. We suppose thatn ≥ 1 and S(n) < (1 − 2ε′). Choose p = max{p : S(p) ≤ S(n)/2} (so S(n) −S(p+ 1) ≤ S(n)/2). By inductive hypothesis, both ρε(x0, xp) and ρε(xp+1, xn)are no greater than S(n)/(2 − 4ε′). Moreover, ρε(xp, xp+1) ≤ S(n). Note thatρε(x, y) ≤ (1 + ε′) max{ρε(x, z), ρε(z, y)} for any x, y, z ∈ ∂X, by Corollary 7.5.Hence (1− 2ε′)ρε(x, y) ≤ (1 + ε′)2S(n) max{1− 2ε′, 1/2}. In order to completethe induction, it remains to note that (1 − 2ε′)(1 + ε′)2 ≤ 1 and (1 + ε′)2 ≤ 2for all ε′ ≤

√2− 1.

If X is a proper geodesic space, xn → x if and only if (xn, x)p → ∞. Tosee this, choose geodesic cn such that cn(∞) = xn and define pn = cn(n). Notethat xn → x if and only if pn → x, which is equivalent to (pn, x)p → ∞.Since (pn, xn)p →∞, we have that (xn, x)p ≥ min{(xn, pn)p, (pn, x)p}− 2δ and(pn, x) ≥ min{(xn, pn)p, (xn, x)p} − 2δ.

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