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Doctoral Thesis

The injective hull of hyperbolic groups

Author(s): Moezzi, Arvin

Publication Date: 2010

Permanent Link: https://doi.org/10.3929/ethz-a-006060216

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ETH Library DISS. ETH No. 18860

The Injective Hull of Hyperbolic Groups

A dissertation submitted to ETH ZURICH

for the degree of Doctor of Sciences

presented by Arvin Moezzi Dipl. Math. ETH Zurich

born 27.08.1976 citizen of Basel BS, Switzerland

accepted on the recommendation of Prof. Dr. Urs Lang, examiner Prof. Dr. Viktor Schroeder, co-examiner

2010

For my family

Abstract

The construction of the injective hull of a metric space goes back to Isbell in the sixties. In this thesis we investigate this construction in the setting of word metrics of ﬁnitely generated groups, hyperbolic groups in particular. There is a natural action of the group on its injective hull. We are mainly interested when this action is properly discontinuous and cocompact; we give a necessary condition. Furthermore, we introduce the concept of Axiom Y and show for a hyperbolic group satisfying Axiom Y that this action is cocompact and properly discontinuous. Moreover, in this case the injective hull can be given explicitly the structure of a polyhedral complex such that each cell in itself is injective and the group acts by cellular isometries on it. We give some simple examples of groups satisfying Axiom Y and a counter example of a Cayley graph of a hyperbolic group that does not satisfy Axiom Y.

Zusammenfassung

Die injektive Hülle von metrischen Räumen geht zurück auf Isbell in den Sechzigern. Wir betrachten in dieser Arbeit die injektive Hülle einer endlich erzeugten Gruppe versehen mit der Wort Metrik - hyperbolische Grup- pen aber im besonderen. Die Gruppe selbst operiert auf natürliche Weise isometrisch auf ihrer injektiven Hülle. Es stellt sich die Frage, wann diese Gruppenaktion eigentlich und cocompakt ist. Wir geben eine notwendige Bedingung an die Gruppe. Desweiteren führen wir die Eigenschaft Ax- iom Y ein und zeigen, dass jede hyperbolische Gruppe, die das Axiom Y erfüllt, eigentlich und cocompakt auf ihrer injektiven Hülle operiert. In diesem Fall lässt sich weiter zeigen, dass die injektive Hülle die Struktur eines polyedrischen Komplexes trägt, wobei jede Zelle für sich wieder injek- tiv ist und die Gruppe zellulär operiert. Wir geben einige einfache Beispiele von Gruppen, die das Axiom Y erfüllen, und auch ein Gegenbeispiel eines Cayley Graphen einer hyperbolischen Gruppe, welcher Axiom Y nicht erfüllt.

Danksagung

Die vorliegende Doktorarbeit wäre ohne die Hilfe und Unterstützung an- derer in dieser Form nicht möglich gewesen. Daher möchte ich gerne fol- genden Menschen an dieser Stelle danken; allen voran meinem Doktorvater, Professor Urs Lang. Er war es, der mich in die metrische Geometrie und geometrische Gruppentheorie einführte. Dieses faszienerende Teilgebiet der Mathematik wäre mir ansonsten wohl verschlossen geblieben. Auch nahm er sich stets die nötige Zeit für Fragen und hat die vorliegende Arbeit mit wichti- gen Ideen und Konzepten vorangetrieben; ohne seine grosse Mithilfe wäre die Doktorarbeit niemals zu dem geworden, was sie ist. Viel Zeit investierte er zudem, die Arbeit mit der von ihm gewohnten Genauigkeit durchzuar- beiten und half mir so, auch die letzten Fehler auszumerzen. Vorallem aber möchte ich ihm dafür danken, dass er mir vertraute und an meine Fähigkeiten glaubte - tausend Dank dafür! Ein grosses Dankeschön gilt auch dem Koreferenten, Professor Viktor Schroeder, der sich die Mühe machte, meine Arbeit zu lesen und für die mündliche Doktorprüfung kurzfristig nach Zürich zu reisen. Mein besonderer Dank gilt auch zwei guten Freunden von mir, Johnny Micic und Driton Komani. Beide haben grosse Teile der Arbeit mehrfach korrekturgelesen und haben mich bei der Prüfungsvorbereitung tatkräftig unterstützt. Sehr wichtig für mich waren auch die zahlreichen mathematischen Debatten mit ihnen zu allen möglichen Themen. Diese Diskussionen führten mir immer wieder vor Augen was ich am liebsten mache: Mathe- matik. Der lange Weg zum Abschluss der Doktorarbeit wäre sicherlich unendlich länger gewesen ohne die Freundschaft und die moralische Unterstützung vieler. Nennen möchte ich an dieser Stelle Fabian Roth, Yves Hauser, Bénédicte Gros, Thilo Schlichenmaier, Stefan Wenger, Michael Anderegg, Driton Komani, Johnny Micic, Leandra Simitovic, Theo Buehler, Beat Steiner, Kathrin Signer, Luzia Huggentobler, Demian Wismer, Kascha, Robin Krom, Ivo Kählin, Lorenz Reichel, Thomas Huber, Ralph Kirchhofer und Jan Oppliger. Vielen Dank, dass ihr für mich da wart. Und vielen Dank für eine sehr gelungene Doktorfeier. Danken möchte ich auch Tim Boin fürs Korrekturlesen der deutschen Passagen. Zuletzt und ganz besonders möchte ich meiner Familie danken. Ohne ihre Liebe und Unterstützung, die man leider allzu oft als selbstverständlich betrachtet, hätte ich dieses Ziel niemals erreicht. Danke für eure Stütze und eine Konstante in meinem Leben!

Contents

1 Isbell’s Injective Hull 5 1.1 Preliminaries ...... 5 1.2 isoM-Injectivity ...... 8 1.3 Bicombing and isoperimetric inequality ...... 13 1.4 Isbell’s injective hull ...... 15 1.5 Gromov–Hausdorffconvergence ...... 21 1.6 Compactmetricspaces ...... 26 1.7 Stronglyconvexsubspaces ...... 30 1.8 Normedvectorspaces...... 35 n 1.8.1 Affine, injective subspaces of R∞ ...... 41 n 1.8.2 Injective polyhedrons in R∞ ...... 46 1.9 Finitemetricspaces...... 51 2 Groups and their Injective Hull 57 2.1 Cayleygraphs...... 59 2.2 Conetypes...... 62 2.3 Combablegroups ...... 63 2.4 Hyperbolicgroups...... 69 2.5 AxiomY...... 72 2.6 Injectivehullofhyperbolicgroups ...... 73 3 Examples and a counter-example 83 3.1 Hyperbolicgroups...... 83 3.1.1 ACounterexample ...... 83 3.1.2 Small cancellation groups ...... 86 3.2 Abeliangroups ...... 90 A The Stone–Cechˇ compactification 95 B Group actions on geodesic metric spaces 99 CTheGrowthofaDehnFunction 101

Introduction

A metric space I is said to be isoM-injective if every 1-Lipschitz map, f : A → I, can be extended to B whenever A is isometrically embedded in B. In other words, there exists a 1-Lipschitz map f¯: B → I such that the diagram f A / I ? ¯ f B commutes. isoM-injective metric spaces share nice metric properties; they are contractible and complete geodesic metric spaces, and they all admit an isoperimetric inequality of euclidean type in the class of integral metric currents. One may wonder whether every X can be isometrically embedded into some isoM-injective metric space I and if there is a smallest such space, in the sense that whenever X can be isometrically embedded into some other isoM-injective metric space J, there exists an isometric embedding of I into J such that X / Io J commutes. In this case I is unique up to an isometry, and it is said to be the injective hull of X; we shall write EX for I. In [22] Isbell shows that every metric space X can indeed be embedded isometrically into a smallest isoM-injective metric space EX. He gives an explicit description of EX in terms of some function space on X. We will show in Theorem 1.55 that the assignment X → EX is continuous with respect to the Gromov–Hausdorﬀ distance; that is E E Xn →dGH X ⇒ Xn →dGH X. Furthermore, in Theorem 2.58 we are going to show that Gromov hyperbolicity for geodesic metric spaces is preserved under this assignment: 4

Theorem. The injective hull of a geodesic hyperbolic metric space is hyperbolic. However, in this thesis we are mostly interested in the following question: when does a finitely generated group (Γ,S) act nicely, i.e. properly discontinuous and cocompactly, on EΓS, where ΓS denotes the metric space Γ with the word metric; in particular what is the metric structure of EΓS in this case? We shall give in 2.39 a necessary condition for Γ to act nicely on EΓS: Proposition. Every group that acts properly discontinuous and cocompactly on an injective metric space is a combable group. But this condition is by no means sufficient. The finitely generated abelian group Zn, for example, with the standard generating set is a combable group but does not act nicely on EZn; compare page 68. In Section 2.5 we introduce the concept of Axiom Y(δ), a metric property which postulates that for any three points x, z, z′ with d(z, z′) ≤ δ one can find a point y such that d(z,y) ≤ R = R(δ) and

d(x, z) = d(x, y)+ d(y, z) d(x, z′) = d(x, y)+ d(y, z′).

The main result of this thesis, Theorem 2.71, can then be formulated as follows.

Theorem. Suppose (Γ,S) is a δ-hyperbolic group and ΓS satisfies Axiom Y(δ). Then, EΓS is a proper locally finite polyhedral complex with finitely many isometry types of cells, and Γ acts properly discontinuous and cocompactly on EΓS by cellular isometries.

The cell structure on EΓS can be given explicitly in terms of some admissible graphs on ΓS, and every cell endowed with the induced metric is itself isoM-injective; see Theorem 2.70. Moreover, we shall see in 2.72 that every tangent cone of EΓS is an isoM-injective metric space. For further study, we hope that this property would give some good link conditions on the cell structure of EΓS, allowing us to alter the metric on the cells in such a way that EΓS becomes a CAT(0) space. This would show that every hyperbolic group satisfying Axiom Y(δ) is a CAT(0) group. Finally we will see in Chapter 3 that a subclass of small cancellation groups and all ﬁnitely generated abelian groups satisfy Axiom Y(δ) for every δ > 0. We will also give a particular presentation of a δ-hyperbolic group which does not satisfy Axiom Y(δ). Chapter 1

Isbell’s Injective Hull

1.1 Preliminaries

We recall the notion of an injective object and injective hull (cf. [1]). Let C denote an arbitrary category and H a distinguished class of morphisms in C. We call all morphisms from H H-morphisms. An H-morphism i : A → B is called H-essential if for any g : B → C the following holds g ◦ i ∈H⇒ g ∈ H. 1.1 Definition. An object I in C is said to be H-injective if for every pair of objects A, B and every pair of morphisms f : A → I and i : A → B with i ∈ H there exists a morphism f¯ : B → I, such that the following diagram commutes f A / I . (1.2) ? i∈H ¯ f B The extension f¯ need not be unique. If f : A → I is H-essential then I is called the H-injective hull of A. 1.3 Lemma. The H-injective hull is unique up to a not necessarily unique isomorphism. Proof. Let I,I′ be two H-injective hulls of A with H-morphisms i: A → I and i′ : A → I′. From Definition 1.1 we have the following commuting diagram i idI A / I / I (1.4) ? O ? O ? ? idI′ ?? ¯ ′ ? i ? idI i ? ? ? ′ / ′ I ′ I idI 6 Chapter 1. Isbell’s Injective Hull where by definition of H-essential all the maps in the diagram are in H. In particular idI′ ◦ ¯i = idI′ and idI ◦ idI′ = idI . Hence idI′ is an isomorphism and therefore ¯i is also an isomorphism. 1.5 Definition. An H-retract R of I is an H-morphism i: R → I, such that i has a left-inverse r : I → R , i.e. r ◦ i = idR. The left-inverse r is called a retraction of I onto R.

1.6 Deﬁnition. Let {Xα}α∈I be a family of objects in C. An object α Xα together with a family of morphisms πα : α Xα → Xα is said toQ be the product of {Xα}α∈I if the following universalQ property is satisﬁed. For every object Y in C and any family of morphisms fα : Y → Xα there is a unique f : Y → α Xα such that πα ◦ f = fα for all α ∈ I. Q α Xα (1.7) w O πα wwQ ww ww {ww ∃!f Xα dHH HH HH fα HH HH Y

By the universal property the product of a family of objects is unique in C up to an isomorphism. The following fact will be needed later. 1.8 Lemma. The subcategory of H-injective objects is closed under retrac- tions and products. Proof. First of all we show that every H-retract R of an H-injective object I is again H-injective. Consider the commuting diagram

r u R i I O 6> O ~~ f ~~ i◦f ~~i◦f ~~ / A j∈H B where the extension i ◦ f exists since I is H-injective. Deﬁne f¯ := r ◦ i ◦ f. Since

f¯◦ j = r ◦ i ◦ f ◦ j = r ◦ i ◦ f ◦ j = r ◦ i ◦ f = idR ◦ f = f, it follows that f¯ is an extension of f in the sense of Deﬁnition 1.1. Therefore, R is H-injective. Preliminaries 7

Let α Xα and πα : α Xα → Xα be given as in Deﬁnition 1.6 and let all Xα beQ H-injective. WeQ need to show that for any f : A → α Xα and ¯ any H-morphism i : A → B there is an extension f : B → Qα Xα with ¯ f ◦ i = f. Since Xα is H-injective there is an extension πα ◦ f :QB → Xα of πα ◦ f : A → Xα, for every α ∈ I. In other words πα ◦ f ◦ i = πα ◦ f. Due to the universal property of the product α Xα in Deﬁnition 1.6 there is a ¯ ¯ unique morphism f : B → α Xα such thatQ πα ◦ f = πα ◦ f. We need to show that f¯ is an extensionQ of f, i.e. f¯◦ i = f. As

πα ◦ (f¯◦ i)= πα ◦ f ◦ i = πα ◦ f =: fα for all α ∈ I and since f is unique with this property, it follows that f¯◦i = f.

α Xα w O bFF πα wwQ F f ww FF ww FF {ww FF Xα f A dH ww H ww H ww πα◦f H ww i H {ww B

1.9 Deﬁnition. An absolute H-retract is an object A ∈ C with the property that every H-morphism i: A → B has a left-inverse in C. In other words there is a morphism r : B → A with r ◦ i = idA. 1.10 Lemma. Every H-injective object in C is an absolute H-retract.

Proof. Let I be H-injective and let i : I → B be an H-morphism. Then there exists a morphism idI : B → I such that

idI I / I ? i∈H idI B commutes. Therefore, r := idI is a left-inverse of i. Note that in general an absolute H-retract need not be H-injective. A counter-example is given next.

1.11 Example. Let C be the category consisting of {Z3, Z3 × Z5, Z3 ⋊ Z2} and of all group homomorphisms between them, and let H be the subclass 8 Chapter 1. Isbell’s Injective Hull

∼ of all injective homomorphisms in C; note that Z3 ⋊ Z2 = S3. Since 15 = |Z3 × Z5| > |Z3|, |Z3 ⋊ Z2| there is no H-morphism from Z3 × Z5 into Z3 or Z3 ⋊ Z2. On the other hand every H-morphism from Z3 × Z5 into itself is an isomorphism and hence invertible. It follows that Z3 × Z5 is an absolute H- retract. But on the other hand there is no group homomorphism ¯i: Z3⋊Z2 → Z3 × Z5 such that i Z3 / Z3 × Z5 8 qqq j∈H qqq qq ¯i qq Z3 ⋊ Z2 commutes, j(z) := (z, 0) ∈ Z3 ⋊ Z2 and i(z) := (z, 0) ∈ Z3 × Z5. Otherwise Z2 < Z3 ⋊ Z2 would be a normal subgroup, but it isn’t. Hence, Z3 × Z5 is not H-injective in C. However if a category has enough H-injective objects then the two notions, H-injectivity and absolute H-retract, are in fact equivalent. This is the subject of the next lemma. We say that a category C has enough H- injectives if every object in C can be embedded into an H-injective object by means of an H-morphism. 1.12 Lemma. Let C be a category that has enough H-injectives. Then, an object I ∈ C is H-injective if and only if it is an absolute H-retract in C. Proof. One implication directly follows from Lemma 1.10. We only need to show that every absolute H-retract R is H-injective. Let i: R → I be an H- morphism into an H-injective object I ∈ C. This exists since C has enough H-injectives. Then by assumption there is r : I → R with r◦i = idR. In other words R is an H-retract of I. Therefore, R is H-injective by Lemma 1.8.

1.2 isoM-Injectivity

Throughout this thesis Metr1 will denote the category of metric spaces and 1-Lipschitz maps, i.e. objects are metric spaces (X,dX ) and morphisms are maps f : (X,dX ) → (Y,dY ) such that

′ ′ dY (f(x), f(x )) ≤ dX (x, x ) for all x, x′ ∈ X. isoM will denote the class of all isometric embeddings, which we shall use as the distinguished class of morphisms H from Section 1.1 for the category Metr1. For convenience we shall write injective instead of isoM-injective unless otherwise stated. isoM-Injectivity 9

Injective objects in Metr1 can be characterized by the ball intersection property as Proposition 1.13 will show. This property is crucial in under- standing the idea of Isbell’s construction, [22], of the injective hull of a metric space X, which we shall introduce in Section 1.4.

1.13 Proposition (Ball Intersection Property). Let X =(X,d) be a metric space. The following three statements are equivalent:

(i) X is an injective object in Metr1. (ii) Whenever f : A → X is a 1-Lipschitz map deﬁned on a subset A of a metric space Y , then there exists a 1-Lipschitz extension f¯: Y → X of f.

(iii) Whenever {B(xi,ri)}i∈I is a nonempty family of closed balls in X with the property that ri +rj ≥ d(xi, xj) for all pairs of indices i, j ∈ I, then

the intersection i∈I B(xi,ri) is nonempty. Proof. The equivalenceT of (i) and (ii) is just a reformulation of Definition 1.1 where C is Metr1 and H is isoM. We first show (ii) ⇒ (iii). Let A denote the subset {xi}i∈I ⊂ X endowed with the metric inherited from X. Let Y = A ∐{y} be the extension of A by one more point with distance d(y, xi) := ri. By assumption d(y, xi)+ d(y, xj)= ri + rj ≥ d(xi, xj) for all i, j ∈ I, thus d is a metric on Y . Because X satisfies condition (ii) the inclusion map i : A ֒→ X can be extended to a 1-Lipschitz map ¯i : Y → X. Since

d (¯i(y), xi) ≤ d(y, xi)= ri ⇒ ¯i(y) ∈ B(xi,ri), ∀i ∈ I we see that i∈I B(xi,ri) = ∅. We proveT (iii) ⇒ (ii). Using Zorn’s lemma one can see that in order to prove (ii) it is enough to verify that every 1-Lipschitz map f : A → X can be extended by one more point y ∈ Y . In other words the domain of f¯ is the metric subspace Y¯ := A ∐{y} ⊂ Y where A is isometrically embedded. Therefore, we only need to deﬁne f¯(y) in a proper way such that f¯ remains 1-Lipschitz. Let ri := dY (xi,y). Since

ri + rj = dY (xi,y)+ dY (xj,y) ≥ dY (xi, xj) ≥ dX (f(xi), f(xj)) and condition (iii) holds, we know that the intersection F := ¯ i∈I B (f(xi),ri) is nonempty. So choose a point x ∈ F and set f(y) := x. TBy construction we then have ¯ dX f(y), f(xi) = dX (x, f(xi)) ≤ ri = dY (y, xi) and therefore f¯ is 1-Lipschitz. 10 Chapter 1. Isbell’s Injective Hull

1.14 Remark. There is also an intrinsic characterization of isoM-injective objects, among metric spaces with ﬁnite Assouad-Nagata dimension, in the much “larger” category M of all metric spaces with arbitrary Lipschitz maps as morphisms. See [24] for more details. In the literature, injective metric spaces are also known as hyperconvex spaces (cf. [2]) or tight spans (cf. [19], [20], [17], [11], [10]). As a ﬁrst example we will show that the real line is injective. Since every geodesic line in a metric space X is the isometric image of the real line, it follows that every geodesic in X is injective. 1.15 Corollary. The real line R with its usual metric induced by the euclidean norm, d(x, y)= |x − y|, is injective in Metr1.

Proof. Let {B(xi,ri)}i∈I be a nonempty family of closed balls in R such that xi − xj ≤|xi − xj| ≤ ri + rj. Hence

∀i, j ∈ I : xi − ri ≤ xj + rj ⇒ ∃x ∈ R : sup(xi − ri) ≤ x ≤ inf(xj + rj). i∈I j∈I

Therefore, xi − ri ≤ x ≤ xi + ri for all i ∈ I, which implies x ∈

i∈I B(xi,ri) = ∅. By the ball intersection property 1.13(iii) R must be injective.T We will list some properties of injective metric spaces in general. Recall that a metric space X is said to be geodesic if every two points in X can be joined by a geodesic segment. 1.16 Proposition. An injective metric space X is complete and geodesic. -Proof. Let X¯ denote the metric completion of X and i : X ֒→ X¯ the inclu ¯ sion map. Since X is injective there is an extension r := idX : X¯ → X of the identity map idX : X → X with r ◦ i = idX. Let (xn) be a Cauchy sequence in X and setx ¯n := i(xn). Because X¯ is complete (¯xn) converges to a point x¯ ∈ X¯. Therefore, we get that

d r(¯x), xn = d r(¯x),r ◦ i(xn) ≤ d x,¯ i(xn) = d(¯x, x¯n) → 0. Hence X is complete. On the other hand let x, y ∈ X be two arbitrary points and let A := {0,D} ⊂ [0,D] =: ID ⊂ R with D := d(x, y). Deﬁne a map f : A → X by setting f(0) := x and f(D) := y. The map f is an isometric embedding because d f(0), f(D) = d(x, y) = D = |0 − D|. In particular f is 1- ¯ Lipschitz. Therefore, there is an extension f : ID → X of f. Let s < t ∈ ID. D = d f¯(0), f¯(D) ≤ d f¯(0), f¯(s) + d f¯(s), f¯(t) + d f¯(t), f¯(D) (f¯ is 1-Lipschitz) ≤ s+(t − s)+( D − t)= D. isoM-Injectivity 11

Hence we have equality everywhere and d f¯(s), f¯(t) = t − s = |s − t|. 1.17 Proposition. Let (X,d) be an injective metric space. Every closed ball B(x, r) in X endowed with the inherited metric d is injective.

Proof. Let X′ := B(x, r). To distinguish balls in X′ and X we write B′ for balls in X′ instead of B. In other words B′(z,s)= B(z,s) ∩ X′. We have to verify the ball intersection property 1.13(iii) for X′. Let ′ ′ {B (xi,ri)}i∈I be a nonempty family of closed balls in X with d(xi, xj) ≤ ri + rj and let B := {B(xi,ri)}i∈I ∪ B(x, r). The family B satisﬁes the conditions of 1.13 (iii) because d(x, xi) ≤ r ≤ r + ri and d(xi, xj) ≤ ri + rj. Therefore, since X is injective and due to Proposition 1.13, the intersection of B is non-empty. And we get

′ B (xi,ri)= B(xi,ri) ∩ B(x, r) = B(xi,ri) ∩ B(x, r) = ∅. \i∈I \i∈I \i∈I Hence B(x, r) is injective.

1.18 Lemma. Let X be a proper metric space, that is every closed, bounded subset of X is compact, and let {An}n∈N be an exhausting, increasing family of subsets of X, i.e.

An ⊂ An+1 and An = X. n[∈N

Then X is injective if all the An’s are injective.

Proof. Let B := {B(xi,ri)}i∈I be a family of closed balls in X as in 1.13 (iii). By Proposition 1.13 we need to verify that the intersection of B is non-empty. n Let B := {B(xi,ri)∩An | xi ∈ An, i ∈ I}. Since X = n An and An ⊂ An+1 for every i ∈ I there is an Ni ∈ N such that xi ∈ ASn for all n ≥ Ni. In n particular B(xi,ri) ∩ An ∈ B for all n ≥ Ni. Let j ∈ I be arbitrary but ﬁxed. W.l.o.g. we may assume that xj ∈ An for all n, otherwise we take ′ n An := An ∪ ANj . As An is injective, the intersection of the family B is non- empty. Let an ∈ B∈Bn B. Since an ∈ B(xj,rj) for all n and X is proper, there is a convergentT subsequence ank → a. Note that ank ∈ B(xi,ri) for all nk ≥ Ni. Therefore, a ∈ B(xi,ri) for all i ∈ I and the claim follows.

1.19 Corollary. A proper metric space X is injective if and only if every closed ball in X is injective.

Proof. This is a direct consequence of Proposition 1.17 and Lemma 1.18. 12 Chapter 1. Isbell’s Injective Hull

1.20 Deﬁnition. Let (Xi, xi) i∈I be a family of pointed metric spaces. The ℓ∞-product is the metric space given by

(Xi, xi) := f ∈ Xi sup dXi (fi, xi) < ∞ , i∈I YI YI c ∞ endowed with the metric d (f,g) := supi∈I dXi (fi,gi). In the case where ∞ (Xi, xi)=(R, 0) we just write l (I) for I (R, 0). This is the Banach space of all bounded functions on the set I. Qb

The metric space (Xi, xi) depends on the choice of the family {xi}i∈I . Note that from the categoricalQb point of view satisﬁes the universal property of the product in the category of pointed metric spaces with 1-Lipschitz Qb maps. The following proposition shows that injectivity is preserved under products from Deﬁnition 1.20.

∞ ∞ 1.21 Proposition. The ℓ -product I (Xi, xi),d is injective in Metr1 whenever the X ’s are. In particular the Banach space ℓ∞(I) is injective. i Qb

Proof. Every closed ball in (Xi, xi) of radius r and with center {mi}i∈I is the product of the balls B(m ,r) ⊂ X . The claim follows from the ball Qb i i intersection property 1.13 in Xi. The real line R endowed with the euclidean metric is injective by 1.15. Therefore, the Banach space ℓ∞(I) is injective.

For our purposes ℓ∞(I) is an important class of examples of injective metric spaces. We shall show that every metric space X can be isometrically embedded into ℓ∞(I) for I = X. This is Kuratowski’s embedding theorem, which we prove next. In particular if X is injective the Banach space structure of ℓ∞(X) leads to good properties of X itself; see Section 1.3.

1.22 Theorem (Kuratowski Embedding Theorem). Every metric space (X,d) can be isometrically embedded into the Banach space ℓ∞(X) by means ∞ of ix0 : X ֒→ ℓ (X) given by

ix0 (x) := d(x, ) − d(x0, ), where x0 ∈ X.

The Kuratowski embedding ix0 is not canonical since it depends on the choice of the base point x0. Bicombing and isoperimetric inequality 13

Proof. We have to show that ix0 (x) is a bounded function on X.

sup | ix0 (x) (y)| = sup |d(x, y) − d(x0,y)| ≤ d(x, x0) < ∞. y∈X y∈X ∞ Hence ix0 (x) ∈ ℓ (X). Furthermore

This implies that ix0 is an isometric embedding.

Theorem 1.22 says that the category Metr1 has enough injectives, i.e. every metric space can be isometrically embedded into an injective one.

1.23 Deﬁnition. A metric space X is called an absolute 1-Lipschitz retract, or 1-ALR, if X is an absolute isoM-retract in Metr1 that is if every isometric embedding of X into another metric space Y is a 1-Lipschitz retract of Y .

1.24 Proposition. X is injective in Metr1 if and only if X is a 1-ALR. Proof. This is follows directly from Lemma 1.12.

1.3 Bicombing and isoperimetric inequality

In the following let I = [0, 1] ⊂ R. A bicombing on a metric space X is a choice of a, not necessarily continuous, path between any pair of points of X. The particular choice depends on the problem one is dealing with. In this thesis, for example, we use bicombings by quasi-geodesics (cf. Deﬁni- tion 2.24). Another important class are convex bicombings, see Deﬁnition 1.25 below. Metric spaces admitting a convex bicombing share some properties with normed vector spaces. They are for instance 1-Lipschitz contractible and a metric version of the Theorem of Krein–Mil’man holds (see [6]). Examples of such metric spaces are normed spaces, CAT(0) spaces (cf. [5]) and in fact all injective metric spaces. The latter is shown in Lemma 1.27.

1.25 Deﬁnition. A convex bicombing on a metric space X is a map

c : X × X → C(I,X) with the following properties. 14 Chapter 1. Isbell’s Injective Hull

(i) The path cxy := c(x, y): I → X is a geodesic segment from x to y parametrized with constant speed v = d(x, y), i.e.

′ ′ ′ d cxy(t),cxy(t ) = |t − t | d(x, y), ∀t, t ∈ I. (1.26) (ii) For every four points x, x′,y,y′ ∈ X the function D : I → R,

D(t) := d cxy(t),cx′y′ (t) , is a convex function on I.

1.27 Lemma. Every injective metric space X admits a convex bicombing. Proof. The idea of the proof is to ﬁrst embed X isometrically into the Banach space l∞(X) and then push forward the natural convex bicombing on ℓ∞(X), as a normed vector space, onto X using a retraction. Here are the details. We can suppose that X is a subspace of ℓ∞(X). Otherwise we take a base ∞ point x0 ∈ X and embed X into ℓ (X) using the Kuratowski embedding ix0 from Theorem 1.22. Since X is injective there is by Proposition 1.24 a 1-Lipschitz retraction r : ℓ∞(X) → X. Deﬁne the bicombing on X by

cxy(t) := r (1 − t) x + ty , for all x, y ∈ X. We have to check that cxy is a geodesic and that 1.25(ii) holds for cxy. Let s, t ∈ I with t>s. It follows that

d(x, y) ≤ d x, cxy(s) + d cxy(s),cxy(t) + d cxy(t),y

(r is 1-lipschitz) ≤ sx − y∞+(t − s)x − y∞+ (1− t)x − y∞

= x − y∞ = d(x, y). Therefore, we have equality and in particular

d cxy(s),cxy(t) = |t − s|x − y∞, for all s, t ∈ I. Hence cxy is geodesic parametrized with constant speed v = x − y∞. Furthermore

′ ′ d cxy(t),cx′y′ (t) = r (1 − t) x + ty − r (1 − t) x + ty ∞ ′ ′ (ris 1-Lipschitz) ≤ (1 − t) x + ty − (1 − t) x + ty ∞ ′ ′ = (1 − t)(x − x)+ t(y − y)∞ ′ ′ ≤ (1 − t)x − x∞ + ty − y∞ = (1 − t) d(x, x′)+ td(y,y′), which proves that c is a convex bicombing. Isbell’s injective hull 15

1.28 Corollary. Every injective metric space X is contractible.

Proof. Choose a base pointx ¯ ∈ X. By Lemma 1.27 X admits a convex bicombing c: X × X → C(I,X). Let H(x, t) := cxx¯ (t). H is a contraction of X ontox ¯. We have

′ ′ ′ d H(x, t), H(x , t ) = d cxx¯ (t),cxx¯ ′ (t ) ′ ≤ dcxx¯ (t),cxx¯ ′ (t) + d cxx¯ ′ (t),cxx¯ ′ (t ) (c is a convex bicombing) ≤ td(x, x′)+ |t − t′| d(¯x, x′).

Hence H: X × I → X is a continuous map. Furthermore we have by deﬁni- tion that H(x, 0)=x ¯ and H(x, 1) = x.

In [26] S. Wenger has shown that every complete metric space admitting a convex bicombing has an isoperimetric inequality of euclidean type in the setting of integral metric currents. It would lead us too far astray to give the precise deﬁnitions and state the theorem here. But the interested reader is encouraged to read [26] for the details and proofs. Since every injective metric space admits a convex bicombing and is complete we get the following corollary.

1.29 Corollary. Every injective metric space X admits an isoperimetric inequality of euclidean type in the class of integral metric currents.

Proof. See [26].

1.4 Isbell’s injective hull

The aim of this section is to show that every metric space X has an isoM- injective hull in Metr1. The injective hull, denoted by EX, can be described explicitly in terms of a function space. This construction goes back to Isbell [22]. We are going to recall the details and give the proofs.

1.30 Deﬁnition. Let (E, ) be the poset of all functions g : X → R satisfying g(x)+ g(y) ≥ d(x, y) ∀x, y ∈ X, (1.31) where f g :⇔ f(x) ≥ g(x) for all x ∈ X. We call a function f : X → R extremal if f is a minimal element of the poset (E, ). In other words whenever a function g : X → R is given, satisfying condition (1.31) and with f g, we have g = f. 16 Chapter 1. Isbell’s Injective Hull

1.32 Remark. Let f ′ : X → R be a function that satisfies condition (1.31) ′ and let Ef ′ := {g ∈ E | f g} ⊂ E. Every linearly ordered subset of the poset Ef ′ has a minimum in Ef ′ - take the infimum of the subset for instance. Therefore, by Zorn’s lemma we obtain a minimal function in Ef ′ , which is by definition an extremal function on X. Hence for every function f ′ ∈ E there is an extremal function f with f ′ f. 1.33 Proposition. Every extremal function is non-negative and 1-Lipschitz. Proof. Let f be extremal. By definition we have 2f(x) = f(x)+ f(x) ≥ d(x, x) = 0. Hence f is non-negative. Furthermore let x, y ∈ X and define g : X → R by setting g(y) := f(x)+ d(x, y) and g(x′) := f(x′) for all x′ ∈ X \{y}. Note that either f g or g f, since g and f coincide everywhere except for y. Moreover g satisfies condition (1.31) because g(x′)+ g(y) = f(x′)+ f(x)+ d(x, y) ≥ d(x′, x)+ d(x, y) ≥ d(x′,y). Hence by the minimality of f we obtain that f g and in particular f(y) ≤ g(y)= f(x)+ d(x, y). Reversing the roles of x and y we get |f(x) − f(y)| ≤ d(x, y) for all x, y ∈ X. (1.34) Therefore, f is 1-Lipschitz. In the following we give an alternative characterization of extremal functions on X. 1.35 Proposition. A function f : X → R is extremal if and only if f satisfies condition (1.31) and for every x ∈ X inf f(x)+ f(y) − d(x, y) =0. (1.36) y∈X

Proof. Let f be an extremal function and suppose that there exists an x ∈ X such that f(x)+ f(y) − d(x, y) ≥ 2ε > 0 for all y ∈ X. Deﬁne g : X → R by setting g(x) := f(x) − ε ≥ 0 and g(y) := f(y) for y ∈ X \{x}. The function g satisﬁes condition (1.31), because by assumption g(x)+ g(y) = f(x) − ε + f(y) ≥ d(x, y) for all y ∈ X \{x}. But g f, which contradicts the minimality of f. On the other hand let f be a function satisfying condition (1.31) and (1.36). By Remark 1.32 there is an extremal function g with f g. Suppose that g = f, i.e. g(x)+ ε ≤ f(x) for some x ∈ X and some ε> 0. Since g is extremal the above implies that 0 = inf f(x)+ f(y) − d(x, y) ≥ inf g(x)+ ε + g(y) − d(x, y) = ε> 0. y∈X y∈X Isbell’s injective hull 17

The following two estimates will be needed later in this section.

1.37 Lemma. Let f1, f2 be two extremal functions on X. Then the following two inequalities hold for all x, y ∈ X:

|d(x, y) − f1(x)| ≤ f1(y) (1.38) and |f1(x) − f2(x)| ≤ f1(y)+ f2(y). (1.39) Proof. This is a direct consequence of (1.34) and (1.31). 1.40 Deﬁnition (Injective Hull). Let X be a metric space. Then EX denotes the set of all extremal functions on X equipped with the metric

d¯(f1, f2) := sup |f1(x) − f2(x)|. x∈X d¯ is well-deﬁned since the supremum is ﬁnite by inequality (1.39). ¯ We shall show that (EX, d) is the isoM-injective hull of (X,d) in Metr1. This is done in three steps. First of all we construct in Lemma 1.41 an isometric embedding i: X ֒→ EX. Then we verify that EX is injective and that the embedding i is isoM-essential, see Theorem 1.45. 1.41 Lemma. Every metric space X can be isometrically embedded into EX by z ∈ X → iz ∈ EX where iz : X → R denotes the distance function

iz(y) := d(z,y).

Furthermore, for all f ∈ EX the following identity holds:

d¯ f, iz = f(z). (1.42)

Proof. Let z ∈ X. By the triangle inequality it follows that iz(x)+ iz(y)= d(z, x)+ d(z,y) ≥ d(x, y). Therefore, iz satisﬁes condition (1.31). In order to show that iz is extremal we need to verify that iz is minimal. Using Remark 1.32 we get an extremal function f ∈ EX with the property that f iz. Then f(z) = 0 because 0 ≤ f(z) ≤ iz(z) = d(z, z) = 0. Hence iz(x)= d(z, x) ≤ f(x)+ f(z)= f(x) ≤ iz(x) for all x ∈ X. Since we have

d¯(iy,iz) d(y, z)= |d(y,y) − d(y, z)| ≤ sup |d(x, y) − d(x, z)| ≤ d(y, z), xz∈X }| { 18 Chapter 1. Isbell’s Injective Hull it follows that i is an isometric embedding of X into EX. On the other hand we get by (1.38)

d¯(iz,f) f(z)= |d(z, z) − f(z)| ≤ sup |d(x, z) − f(x)| ≤ f(z) xz∈X }| { for all f ∈ EX.

The following lemma will be needed for the main theorem of this section, Theorem 1.45, to prove that i: X ֒→ EX is in fact the injective hull of X in Metr1.

Lemma. Let i: X ֒→ EX be the isometric embedding from 1.43 Lemma 1.41. Then the following two statements are true.

(i) If s: EX → R is an extremal function on EX then s ◦ i: X → R is an extremal function on X.

(ii) If F : EX → EX is a 1-Lipschitz map that leaves i(X) pointwise ﬁxed then F is the identity on EX.

Proof. (i) Since s satisfies (1.31) and i is an isometric embedding, we get that s ◦ i(x)+ s ◦ i(y) = s(ix)+ s(iy) ≥ d¯(ix, iy) = d(x, y) for all x, y ∈ X. Thus s ◦ i fulfills condition (1.31). It remains to show that s ◦ i is minimal. Let g : X → R be a function on X with g ≤ s ◦ i and satisfying (1.31), and let x ∈ X. Define t: EX → R by setting t(ix) := g(x) and t(f) := s(f) for all f ∈ EX \{ix}. If suffices to prove that

t(ix)+ t(f)= g(x)+ s(f) > f(x) − ǫ (1.44) for all f ∈ EX \{ix} and ǫ> 0. Because by (1.42) we have f(x) = d¯(f, ix) and therefore (1.44) implies that t satisﬁes (1.31). By the minimality of s it follows that s ◦ i(x) ≤ t(ix) = g(x). Since x was arbitrary this proves that s ◦ i is minimal. Let f ∈ EX \{ix} and ǫ> 0. By Proposition 1.35 there is a y ∈ X such that f(x)+ f(y)

g(x)+ s(iy) ≥ g(x)+ g(y) ≥ d(x, y) > f(x)+ f(y) − ǫ.

On the other hand by (1.38) and (1.42), we get

s(f) − s(iy) ≥ −d¯(iy, f)= −f(y). Isbell’s injective hull 19

Adding these two estimates we obtain g(x)+ s(f) > f(x) − ǫ. (ii) Let f ∈ EX and g := F (f) ∈ EX. Since F : EX → EX is a 1-Lipschitz map and because of (1.42) we get that

g(x)= d¯(ix,g)= d¯ F (ix), F (f) ≤ d¯(ix, f)= f(x) for all x ∈ X. By minimality of f it follows F (f)= g = f. Hence F is the identity.

Theorem. The metric space (EX, d¯) is injective and i: X ֒→ EX is 1.45 an isoM-essential map. Therefore, EX is the isoM-injective hull of X in Metr1.

Proof. To prove the injectivity of EX we verify the ball intersection property 1.13(iii). Let B(fi,ri) i∈I be a nonempty family of closed balls in (EX, d¯) such that ri+ rj ≥ d¯(fi, fj) for all i, j ∈ I. Let r : EX → R be the function given by

r(f) := inf ri + d¯(fi, f) . i∈I In particular for f = fi we have that r(fi) = ri. Note that r satisﬁes ′ ¯ ¯ ¯ ′ condition (1.31) since r(f)+r(f ) ≥ infi,j∈I d(fi, fj)+d(fi, f)+d(fj, f ) ≥ d¯(f, f ′) for all f, f ′ ∈ EX. Therefore, by Remark 1.32 there is an extremal function s: EX → R with s ≤ r. Using Lemma 1.43 we conclude that s ◦ i is an extremal function on X. Let x ∈ X. By (1.42) and (1.38) it follows

¯ |f(x) − s ◦ i(x)| = |d(ix, f) − s(ix)| ≤ s(f) ≤ r(f).

¯ E Hence d(f,s ◦ i) = supx∈X |f(x) − s ◦ i(x)| ≤ r(f) for all f ∈ X. In other words s ◦ i is a point of intersection of the balls B(fi,ri) i∈I since

s ◦ i ∈ B f,r(f) ⊂ B(fi,ri). f∈\EX \i∈I

On the other hand let H and G be two maps given as in the following commutative diagram i X / EX D U) DD DD H ¯i G DD D! Y with G ∈ isoM. In order to prove that i is isoM-essential we need to verify that H is an isometric embedding. Because EX is injective and G ∈ isoM 20 Chapter 1. Isbell’s Injective Hull there is an extension ¯i: Y → EX of the map i. Let F := ¯i ◦ H : EX → EX. Then for all x ∈ X we have

F i(x) = (¯i ◦ H) ◦ i (x)= ¯i ◦ (H ◦ i)(x)= ¯i ◦ G(x)= i(x). Therefore, F ﬁxes i(X) pointwise and by Lemma 1.43 (ii). it follows that ¯i ◦ H = F = idEX . Now H is an isometric embedding because

d H(f),H(f ′) ≥ d¯ ¯i(H(f)),¯i(H(f ′)) = d¯(f, f ′) ≥ d H(f),H(f ′) for all f, f ′ ∈ EX.

The following corollary says that EX is the smallest injective metric space in which X can be isometrically embedded.

1.46 Corollary. Let j : X → Y be an isometric embedding of X into some injective metric space Y . Then every extension ¯j : EX → Y of j i.e. ¯j◦i = j, where i: X → EX, is an isometric embedding.

Proof. Note that an extension ¯j : EX → Y of j : X → Y always exists by the ,injectivity of Y . Since ¯j ◦ i = j ∈ isoM and i: X ֒→ EX is isoM-essential cf. 1.45, it follows that ¯j ∈ isoM.

1.47 Lemma. Let j : X → Y and h: Y → EX be isometric embeddings such that h ◦ j = i, where i: X → EX. Then EY and EX are isometric.

Proof. Consider the commuting diagram

i X / EX (1.48) D O bE DD EE h¯ DD h EE j DD EE D! ? E Y / EY iY where iY : Y → EY is the isometric embedding from Lemma 1.41 of Y into its injective hull and h¯ is an extension of h. Note that h¯ exists since EX is injective and iY ∈ isoM; it is an isometric embedding due to Corollary 1.46. ¯ ¯ ¯ Moreover, i(X) ⊂ h(EY ) because i = h◦iY ◦j. We claim that h(EY )= EX. Let r : EX → h¯(EY ) ⊂ EX be a 1-Lipschitz retraction of EX onto h¯(EY ), which by Lemma 1.10 exists since the latter is injective. Then r ﬁxes the subset i(X) ⊂ h¯(EY ) pointwise. Therefore, r is the identity on EX by 1.43; in particular EX = r(EX)= h¯(EY ). Gromov–Hausdorﬀ convergence 21

1.49 Remark. The isometric embedding h: Y → EX from 1.47 is uniquely determined by j and the condition h ◦ j = i since

h(y)(x)= d¯ h(y), i(x) = d¯ h(y), h j(x) = dY y, j(x) . ¯−1 ¯−1 It follows in particular that h = iY because h ◦ h = iY and iY ◦ h = iY ; compare diagram (1.48).

1.50 Corollary. If i(X) ⊂ Y ⊂ EX then EY is isometric to EX.

Proof. The claim follows directly from Lemma 1.47 with h: Y ֒→ EX the inclusion map and j = i: X → Y ⊂ EX.

1.5 Gromov–Hausdorﬀ convergence

In Section 1.4 we saw that every metric space X has an injective hull EX and that the latter can be given explicitly. In this section we are going to show that the assignment X → EX is continuous with respect to the Gromov–Hausdorﬀ pseudo-distance dGH in the sense that E E Xn →dGH X ⇒ Xn →dGH X. (1.51)

We give the definition of the Gromov–Hausdorff pseudo-distance dGH (X,Y ) of two metric spaces X and Y . There are several equivalent definitions, cf. [5], but the following one is appropriate for our purposes. Let

NZ (X, R) := {z ∈ Z | ∃x ∈ X : d(x, z) ≤ R}. (1.52)

Deﬁne the Gromov–Hausdorﬀ pseudo-distance between X and Y to be

Z dGH (X,Y ) := inf{ dH(X,Y ) | X,Y ⊂ Z isometrically embedded}, (1.53) Z where

Z dH (X,Y ) := inf{R> 0 |X ⊂ NZ (Y, R) and Y ⊂ NZ(X, R)} denotes the Hausdorﬀ distance between X and Y as subspaces of Z. However we can assume that the metric space Z in the deﬁnition (1.53) is injective. Otherwise we can embed Z isometrically into an injective metric space by Theorem 1.22. In order to prove (1.51) we need the following technical lemma. 22 Chapter 1. Isbell’s Injective Hull

Y 1.54 Lemma. Let X ⊂ Y be two metric spaces such that dH (X,Y ) ≤ ε, and let iX : X ֒→ EX and iY : Y ֒→ EY be deﬁned as in Lemma 1.41. Then for every isometric embedding i : EX → EY with i ◦ iX = iY |X we have

EY E E dH i( X), Y ≤ 4ε. Note that the condition i ◦ iX = iY |X is satisﬁed whenever i is an extension . of the map iY |X : X ֒→ EY

Proof. Let f ∈ EY . We shall show that there exists a gf ∈ i(EX) with d¯EY (f,gf ) ≤ 4ε. Y By assumption dH (X,Y ) ≤ ε. Therefore for each y ∈ Y there is a xy ∈ X with d(xy,y) ≤ ε. Consider the restriction f|X : X → R. Then f|X satisfies condition (1.31) since X ⊂ Y . By Remark 1.32 there is an extremal function g ∈ EX with g f|X . We show by contradiction that |g(x) − f|X (x)| ≤ 2ε must hold for all x ∈ X. Suppose that g(x) < f|X (x) − 2ε for some point x ∈ X. Define f ′ : Y → R by setting f ′(x) := g(x)+2ε and f ′(y) := f(y) otherwise. Then the function f ′ fulfills condition (1.31) since

f ′(x)+ f ′(y) = g(x)+2ε + f(y)

(f is 1-lipschitz and d(xy,y) ≤ ε) ≥ g(x)+ ε + f(xy)

(f(xy) ≥ g(xy) and g ∈ EX) ≥ d(x, xy)+ ε

(d(xy,y) ≤ ε) ≥ d(x, y).

But this is a contradiction to the minimality of f because f ′(x) = g(x)+ 2ε < f(x). Therefore we have that |f(x) − g(x)| ≤ 2ε for all x ∈ X. Let gf := i(g) ∈ EY . Then

gf (x)= d¯EY (gf , iY (x)) = d¯EY (i(g), i ◦ iX (x)) = d¯EX (g, iX(x)) = g(x) for all x ∈ X. Hence

d¯EY (f,gf ) = sup |f(y) − gf (y)| y∈Y

(f and gf are 1-Lipschitz) ≤ sup |f(xy) − gf (xy)| +2ε y∈Y

≤ sup |f(x) − gf (x)| +2ε x∈X = sup |f(x) − g(x)| +2ε ≤ 4ε. x∈X Gromov–Hausdorﬀ convergence 23

1.55 Theorem. Let X and Y be two metric spaces. Then,

dGH (EX, EY ) ≤ 8dGH (X,Y ).

In particular the assignment X → EX is continuous with respect to dGH in the sense of (1.51).

Proof. Let R>dGH (X,Y ). Then by the deﬁnition of dGH there is a metric Z space Z with X,Y ⊂ Z and such that dH (X,Y ) ≤ R, i.e. X ⊂ NZ (Y, R) and Y ⊂ NZ (X, R). In particular Y ⊂ NZ(X, R) ⊂ NZ (Y, 2R). Therefore we get that

Z Z dH Y, NZ(X, R) ≤ 2R, dH X, NZ (X, R) ≤ R. Let N := NZ (X, R) and let IX := iN |X : EX → EN denote the extension of .iN |X : X ֒→ EN where iN : N ֒→ EN is the embedding given by Lemma 1.41 Compare the commutative diagram.

⊆ iN X / N / EN (1.56) m6 m m iX m m mI :=i | m m X N X EX

By Corollary 1.46 it follows that IX is an isometric embedding. Since Z dH (X, N) ≤ R we get by Lemma 1.54 that EN E E dH IX ( X), N ≤ 4R. In the same way we deﬁne IY := iN |Y : EY → EN and conclude that EN E E dH IY ( Y ), N ≤ 8R. Therefore dGH (EX, EY ) ≤ 8R whenever dGH (X,Y ) < R and the claim follows. 1.57 Remark. The assignment X → EX is not an isometry. For example n n n n n consider Z ⊂ R1 where R1 := (R ,ℓ1) and Z is endowed with the induced n n 1 ℓ1-metric. Then it is easy to see that dGH (Z , R1 ) ≥ 2 > 0. But on the E n E n E n ∼ E n other hand dGH ( Z , R1 ) = 0 since Z = R1 ; see Example 1.99. 1.58 Corollary. Let Y be injective and X an arbitrary metric space with dGH (Y,X)=0. Then iX (X) ⊂ EX is dense, where iX : X ֒→ EX is the isometric embedding from Lemma 1.41. If in addition X is complete, then X is injective. 24 Chapter 1. Isbell’s Injective Hull

B Proof. First of all note that A ⊂ B is dense if and only if dH (A, B) = 0. This follows easily from the deﬁnition of the Hausdorﬀ distance, see (1.53). EX E Thus it is enough to verify that dH iX (X), X = 0. ∼ Since dGH (Y,X) = 0 and Y is injective, i.e. Y= EY , Theorem 1.55 gives us the following estimate

0 ≤ dGH (Y, EX)= dGH (EY, EX) ≤ 8dGH (Y,X)=0.

Hence dGH (Y, EX) = 0. We then obtain by the triangle inequality that dGH (X, EX) = 0, because

0 ≤ dGH (X, EX) ≤ dGH(X,Y )+ dGH (Y, EX)=0.

In other words, for every ε> 0 there is an injective metric space Z = Z(ε) in E Z E which X and X are isometrically embedded and such that dH (X, X) ≤ ε. W.l.o.g. we may assume that X, EX ⊂ Z. In order to distinguish the isometric image of EX in Z and EX itself, we shall write EZ X for the former. We have EZX ⊂ NZ(X,ε) and X ⊂ NZ (EZX,ε), thus X ⊂ NZ (EZX,ε) ⊂ NZ (X, 2ε). Let N := NZ (EZ X,ε). W.l.o.g. we may further assume that EZX ⊂ EN ⊂ Z. Otherwise we can embed EN isometrically into Z, using Corollary 1.46, and get N ⊂ EN ⊂ Z. Notice that the inclusion map i : EZX ֒→ EN obviously Z E E satisﬁes the conditions of Lemma 1.54 and so dH ( Z X, N) ≤ 4ε, since Z E E E dH ( Z X, N) ≤ ε. In particular we get N ⊂ NZ ( ZX, 4ε). On the other hand EZ X ⊂ NZ (X,ε) and hence

EN ⊂ NZ (X, 5ε).

Moreover, because X ⊂ N ⊂ EN ⊂ Z, there exists by corollary 1.46 an isometric embedding j : EX → EN such that j ◦ iX = idX . We conclude that for all ε> 0 X ⊂ j(EX) ⊂ EN ⊂ NZ (X, 5ε), or rather, since j ◦ iX = idX ,

iX (X) ⊂ EX ⊂ NEX iX (X), 5ε . EX E In other words dH iX (X), X = 0. As a Corollary we get the following. The completion of the Gromov– Hausdorﬀ limit of a sequence of injective metric spaces is injective. Gromov–Hausdorﬀ convergence 25

1.59 Corollary. Let {Xn}n∈N be a sequence of injective metric spaces converging to a metric space X with respect to the Gromov–Hausdorﬀ distance, i.e.

Xn →dGH X.

Then iX (X) lies dense in EX. If in addition X is complete, then X is injective. E E Proof. By Theorem 1.55 we know that Xn →dGH X. Since all Xn are in- E ∼ E E jective, i.e. Xn = Xn, we get that Xn →dGH X. Therefore dGH (X, X)= 0, because 0 ≤ dGH (X, EX) ≤ dGH (X,Xn)+ dGH (Xn, EX) → 0. The claim then follows using Corollary 1.58.

1.60 Remark. Recall that a sequence of pointed metric spaces (Xn, xn) is said to converge to (X, x) in the pointed Gromov–Hausdorﬀ topology, and we write Xn →pGH X, if for every r > 0 there is a sequence of numbers rn → r such that

B(xn,rn) →dGH B(x, r) (cf. [7] for a better definition; however the definition above is sufficient for our purposes). One may think that the assignment X → EX is also continuous with respect to this topology. But this is not true in general; take for instance 2 X = R2, the plane with the euclidean norm. Then the sequence of closed 2 balls Xn := B(0, n) ⊂ R2, n ∈ N, converges to X with respect to the pointed E 2 Gromov–Hausdorff topology. However R2 is an infinite dimensional Banach space and therefore not proper, see Theorem 1.103, whereas the completion of limpGH EXn must be a proper metric space due to Proposition 1.66 below and [5, I.5.44].

Anyhow the following is true:

1.61 Corollary. Let Xn be a family of injective metric spaces, such that (Xn, xn) converges to a proper metric space (X, x) in the pointed Gromov– Hausdorﬀ topology for some xn ∈ Xn and x ∈ X. Then X is injective.

Proof. This is an immediate consequence of Corollary 1.19, Theorem 1.55 and the deﬁnition of pointed Gromov–Hausdorﬀ convergence from Re- mark 1.60.

1.62 Remark. If X is a proper metric space and Xn ⊂ X a family of injective subspaces such that B(x, R) ∩ Xn = ∅, n ∈ N, for some x ∈ X and R > 0; let xn ∈ B(x, R) ∩ Xn. Then by an application of [7, 7.3.8], 26 Chapter 1. Isbell’s Injective Hull

Theorem of Blaschke, there is a subsequence of (Xn, xn) converging to a proper subspace (Y,y) of X in the Gromov–Hausdorﬀ topology. Therefore, Y is injective by Corollary 1.61.

1.63 Corollary. Suppose X is a proper metric space and An ⊂ X, n ∈ N, an increasing family of injective subspaces of X such that

An = X. n[∈N Then X is injective.

Proof. By Remark 1.62 we know that there is a subsequence of (An) converging to a proper injective metric subspace Y ⊂ X. Since n∈N An ⊂ Y and n∈N An = X it follows that Y = X. Hence, X is injective.S S 1.6 Compact metric spaces

For the sake of completeness we recall [22] regarding the injective hull of a compact metric space C. We shall show in Proposition 1.66 that the injective hull EC is compact. Furthermore, we are interested in the following question. What is the smallest closed subset EC ⊂ C, such that EEC and ∼ EC are isometric, i.e. EEC = EC? Is EC unique with this property? The uniqueness of EC may surprise, since it would follow that besides EC there is no other isometric copy of EC in C, compare Lemma 1.77.

1.64 Lemma. Let X be a metric space. Then diam(X) = diam(EX).

Proof. Recall that diam(X) := supx,y∈X d(x, y). Since X can be isometrically embedded into EX we have diam(X) ≤ diam(EX). If diam(X) = ∞ then it follows that diam(EX) = ∞ and the claim follows. Therefore, assume D := diam(X) < ∞. Let f ∈ EX and x ∈ X, and let ε > 0. By Proposition 1.35 there is an y ∈ X such that f(x)+ f(y) ≤ d(x, y)+ ε. In particular f(x) ≤ f(x)+ f(y) ≤ d(x, y)+ ε ≤ D + ε, or rather f(x) ≤ D since ε is arbitrary. Let g ∈ EX. We may further assume w.l.o.g. that g(x) ≤ f(x). Then |f(x) − g(x)| = f(x) − g(x) ≤ f(x) ≤ D. Therefore, ¯ E d(f,g) = supx∈X |f(x) − g(x)| ≤ D. It follows diam( X) ≤ diam(X). This proves the assertion.

1.65 Corollary. The injective hull of a bounded metric space is bounded.

Proof. This follows directly from Lemma 1.64. Compact metric spaces 27

1.66 Proposition. Let C be a compact metric space. Then EC is compact.

1 Proof. Let Cn be a maximal n -separated net in C. Because C is compact, Cn is finite. By Lemma 1.65 the injective hull ECn is a bounded and closed ∞ ∼ |Cn| subset of ℓ (Cn) = R . Thus ECn is compact for all n ∈ N. C 1 On the other hand by construction we have that dH (Cn,C) ≤ n . Hence the sequence of subsets Cn ⊂ C converges to C in the Gromov–Hausdorff topol- E E ogy. Therefore by Proposition 1.55 we obtain that Cn →dGH C. Since every complete Gromov–Hausdorff limit of a sequence of compact metric spaces is compact [5, 5.40], it follows that EC is compact. Proposition 1.66 can also be proved using Arzelà-Ascoli (cf. [22]), since extremal functions are 1-Lipschitz and as such uniformly continuous.

Before we can answer the first question, we need first to settle some notations and show a simple observation about extremal functions. 1.67 Definition. Let X be a metric space. By the defect of an extremal function f ∈ EX at x, y ∈ X, we mean the number ∆(f; x, y) := f(x)+ f(y)−d(x, y) ≥ 0. The defect ∆ : EX ×X ×X → R is a continuous function. Furthermore, we say that y lies between x andx ¯, when d(x, x¯)= d(x, y)+ d(y, x¯) or equivalently, when ∆(iy; x, x¯) = 0, where iy is the distance function defined in Lemma 1.41. We write x

1.69 Lemma. Let f ∈ EX be an extremal function on X. Whenever y lies between x and x¯, we have

∆(f; x, y) ≥ ∆(f; x, x¯).

Moreover, if ∆(f; x, y)=∆(f; x, x¯), it follows that f(¯x)= f(y)+ d(y, x¯). In other words the value of f at x¯ is determined by the value f(y). Note that, if ∆(f; x, y) = 0 and x ≤ y ≤ x¯, we get by the ﬁrst part of Lemma 1.69 that ∆(f; x, x¯)=∆(f; x, y) = 0. Hence f(¯x)= f(y)+ d(y, x¯). Proof. Let f ∈ EX. Then,

∆(f; x, x¯) = f(x)+ f(¯x) − d(x, x¯) (x ≤ y ≤ x¯) = f(x)+ f(¯x) − d(¯x, y) − d(y, x) (f is 1-lipschitz) ≤ f(x)+ f(y) − d(x, y) = ∆(f; x, y).

Hence if ∆(f; x, x¯)=∆(f; x, y), we have equality everywhere. By subtract- ing the third from the second line, we obtain that f(¯x) − d(¯x, y) − f(y) = 0. The claim follows. 1.70 Lemma. Let C be a compact metric space and f ∈ EC. Then for every c¯ ∈ C there exists a point c = c(f;¯c) ∈ C, such that

f(¯c)+ f(c)= d(¯c,c). (1.71)

In other words ic¯ ≤ f ≤ ic in EC, cf. Definition 1.67. Proof. This is a direct consequence of Lemma 1.35 and Lemma 1.66, and the fact that the defect ∆: EC × C × C → R is a continuous function. We now come back to the original question, what the smallest closed ∼ subset EC ⊂ C with the property EEC = C is. We shall make the following observation to motivate Definition 1.73 of extremal subsets in C. 1.72 Lemma. Let E ⊂ C be a closed subset such that EE is isometric to EC and let x, y ∈ C be arbitrary. Then there exists e ∈ E with the property that y ≤ x ≤ e. Proof. Let EC be isometric to EE. Therefore we shall identify EC with EE. Since E is compact and ix, iy ∈ EC = EE, we have by Definition 1.1 the following identity

d(x, y)= d¯(ix, iy)= |iy(¯e) − ix(¯e)| = |d(y, e¯) − d(x, e¯)| Compact metric spaces 29 for somee ¯ ∈ E. If d(y, e¯) − d(x, e¯)= d(x, y) then y ≤ x ≤ e¯. On the other hand if d(x, e¯) − d(y, e¯) = d(x, y), i.e. x ≤ y ≤ e¯, there is by Lemma 1.70, with f = ix ∈ EE andc ¯ =e ¯, a point e ∈ E such that e ≤ x ≤ e¯. Thus we get by Lemma 1.68 that y ≤ x ≤ e.

1.73 Deﬁnition. According to Lemma 1.72 we call a non-empty, closed subset E ⊂ C extremal, if for all x, y ∈ C there exists e ∈ E with x ≤ y ≤ e.

1.74 Lemma. Let E ⊂ C be an extremal subset. Then EE is isometric to EC.

Proof. We are going to embed C isometrically into EE. Let c ∈ C and let ic : E → R be the distance function ic(e) := d(c, e). By the triangle ′ ′ inequality we obtain the sub-additivity ic(e)+ ic(e ) ≥ d(e, e ). Since E is an extremal subset of C, there exists for all e ∈ E a pointe ¯ ∈ E such that e ≤ ′ ′ c ≤ e¯. Therefore infe′∈E ic(e)+ ic(e ) − d(e, e ) = ic(e)+ ic(¯e) − d(e, e¯) = 0. By Lemma 1.35 it follows that ic ∈ EE. Let c,c′ ∈ C. Since E is extremal there is a point e ∈ E with c ≤ c′ ≤ e. Hence

′ ′ ′ ′ d¯(ic, ic′ ) = sup |ic(e ) − ic′ (e )| = ic(e) − ic′ (e)= d(c, e) − d(c , e)= d(c,c ). e′∈E

Therefore the map i(c) := ic is an isometric embedding from C into EE. Thus we can assume by Corollary 1.46 that E ⊂ C ⊂ EE ⊂ EC. Since EE is injective it is a 1-Lipschitz retract of EC. Let r : EC → EE ⊂ EC denote the retraction. By deﬁnition r ﬁxes C pointwise, therefore due to Lemma 1.43(ii) r must be the identity. Hence EE = EC.

1.75 Lemma. The intersection EC of all extremal subsets of C is again extremal.

Proof. Let E, E′ be two extremal subsets of C and let x, y ∈ C. Recall that x ≤ y ≤ z if and only if ∆(iy; x, z) = 0. Since ∆(iy; x, ) is a continuous function on C and E is extremal, the set Ex,y := E ∩{c ∈ C | ∆(iy; x, c)=0} is non-empty and compact. Hence d(x, ) attains its maximum on Ex,y - ′ say at e ∈ Ex,y. Note that x ≤ y ≤ e. We shall show that e ∈ E . Because E′ is extremal there is a point e′ ∈ E′ with x ≤ e ≤ e′. Using the fact that E is extremal one more time we obtain a pointe ¯ ∈ E such that x ≤ e′ ≤ e¯. Hence it follows by Lemma 1.68 that x ≤ e ≤ e′ ≤ e¯. In particular d(x, e¯) = d(x, e)+ d(e, e′)+ d(e′, e¯) ≥ d(x, e). But d(x, e) is 30 Chapter 1. Isbell’s Injective Hull maximal by the choice of e. Hence d(e, e′) = 0, i.e. e = e′ ∈ E ∩ E′. Since E′ was arbitrary we get in particular

e ∈ EC := E E⊂C \extremal with x ≤ y ≤ e. Thus EC is extremal.

Note that EC is the set of all points in C that are extremal in the sense that there is no pair of points x, y ∈ C such that x

1.77 Lemma. There is no isometric copy of EC in C other than EC itself.

Proof. Suppose that there exists E ⊂ C and an isometry I : EC → E. Since EC is compact and EE = EEC = EC it follows by Lemma 1.72 that E is an extremal subset of C. But EC is by deﬁnition the smallest extremal subset, −1 hence EC E. But then by the same reason I (EC ) EC is extremal, which contradicts the deﬁnition of EC . Recall that a metric space C is said to be uniform if the isometry group Isom(C) operates transitively on C, i.e. for every pair of points x, y ∈ C there is an isometry T ∈ Isom(C) with T (x) = y. By Remark 1.77 we get that EC = C if C is a compact uniform metric space.

1.7 Strongly convex subspaces

A tripod T is a metric graph (V, E,ℓ) with four vertices V := {t0, t1, t2, m} and three edges E := {t0m, t1m, t2m} of a given length ℓi := ℓ(tim) ≥ 0. The vertex m is said to be the center of T . Note that the incidence relation in T is implicitly given by the labeling of the edges. For example the endpoints of t1m are t1 and m. dT will denote the length pseudo-metric on T induced by the length function ℓ: E → R+; cf. [5, 1.9]. It holds that

dT (ti, m)= ℓi and dT (ti, tj)= ℓi + ℓj, ∀ 0 ≤ i < j ≤ 2.

On the other hand, for a triple of points {x0, x1, x2} ⊂ X one obtains an unique tripod T = T (x0, x1, x2) with dT (ti, tj)= d(xi, xj) by solving the following system of equations for ℓi,

ℓi + ℓj = d(xi, xj) 0 ≤ i < j ≤ 2. Strongly convex subspaces 31

t2 t1

Figure 1.1: Tripod

1 In particular ℓi = 2 d(xi, xi+1)+ d(xi, xi+2) − d(xi+1, xi+2) ≥ 0 mod 3. T (x0, x1, x2) is said to be spanned by the subset {x0, x1, x2}. Furthermore, we say that {x0, x1, x2} can be joined by a tripod in X if there exists a geodesic triangle ∆ ⊂ X with vertex set {x0, x1, x2} such that ∆ is isometric to T (x0, x1, x2). 1.78 Lemma. Let X be an injective metric space. Every triple of points {x0, x1, x2} ⊂ X can be joined by a tripod in X.

Proof. Let A := {t0, t1, t2} ⊂ T (x0, x1, x2) and deﬁne f : A → X by f(tj) := xj. Then f is an isometric embedding because

d f(tj), f(tk) = d(xj, xk)= ℓj + ℓk = dT (tj, tk). Since X is injective there is a 1-Lipschitz extension f¯: T (x0, x1, x2) → X of f. Hence

d f¯(tj), f¯(tk) ≤ d f¯(tj), f¯(m) + d f¯(tk), f¯(m)

(f¯ is 1-Lipschitz) ≤ dT(tj, m)+ dT(tk, m )

= dT (tj, tk)

= d f¯(tj), f¯(tk) . ¯ ¯ Therefore, we have equality and in particular d f(tj), f(m) = dT (tj, m). 1.79 Remark. A Riemannian manifold M of dimension bigger or equal 2 can not be injective by Lemma 1.78. Since a geodesic in M is uniquely determined by its initial condition, three non collinear points in M can not be joined by a tripod in M.

Recall that a metric tree T is a geodesic metric space, where every geodesic triangle is a tripod. Or equivalently, T is a 0-hyperbolic geodesic metric space; compare Section 2.4 for a deﬁnition of hyperbolicity. 32 Chapter 1. Isbell’s Injective Hull

1.80 Corollary. Every uniquely geodesic, injective metric space X is a metric tree.

Proof. Let {x0, x1, x2} ⊂ X. By Lemma 1.78 there exists a tripod T in X joining them. Because X is uniquely geodesic, T is the only geodesic triangle with vertex set {x0, x1, x2}. Hence, X is a metric tree. 1.81 Deﬁnition. A subset C of a geodesic metric space X is strongly convex if every geodesic segment connecting two points of C is contained in C.

1.82 Lemma. Let C ⊂ X be a closed, strongly convex subset of an injective metric space X and x ∈ X. Then there is a unique point πC (x) ∈ C with d x, πC (x) = d(x, C). Furthermore, d(x, y)= d x, πC (x) + d πC (x),y , for all y ∈ C.

′ Proof. We ﬁrst show the uniqueness of πC (x). Let x ∈ X and z, z ∈ C with d(x, C)= d(x, z′)= d(x, z). By Lemma 1.78 there is a tripod T in X joining {x, z, z′}. Because m, the center of T , lies on a geodesic segment from z to z′ and C is strongly convex, m must lie in C. It follows that

d(x, z)+ d(m, z) ≤ d(x, m)+ d(m, z)= d(x, z), and hence d(m, z) = 0. In the same way one gets that d(m, z′) = 0. There- fore, 0 ≤ d(z, z′) ≤ d(z, m)+ d(m, z′) = 0. In other words z = z′. On the other hand, let (xn) be a sequence of points in C with d(x, xn) ≤ 1 d(x, C)+ n , and Tn,k be a tripod in X joining {x, xn, xk}. Using the same argument as above we see that the center mn,k of Tn,k lies in C, and that 1 d(x, x ) − + d(x , m ) ≤ d(x, m )+ d(m , x )= d(x, x ). n n n n,k n,k n,k n n

1 Therefore d(xn, mn,k) ≤ n . By exchanging xn with xk, we also obtain that 1 1 1 d(xk, mn,k) ≤ k . Since d(xn, xk) ≤ d(xn, mn,k)+d(mn,k, xk) ≤ n + k → 0, the sequence (xn) is a Cauchy sequence and converges to πC (x) := limn xn ∈ C. Let y ∈ C and let T be a tripod in X spanned by {x, πC (x),y}. Then the center m of T is in C and

d x, πC (x) + d m, πC (x) ≤ d(x, m)+ d m, πC (x) = d x, πC (x) . Therefore, d m, πC (x) = 0 and T is degenerated. In particular d(x, y) = d(x, m)+ d(m, y)= d x, πC (x) + d πC (x),y . Strongly convex subspaces 33

Note that by Lemma 1.82 the point x can be connected with every y ∈ C by a geodesic segment passing through πC (x).

1.83 Corollary. πC : X → C is a 1-Lipschitz retraction.

Proof. It is clear from the construction of πC that πC (x)= x for all x ∈ C. In other words πC : X → C is a left-inverse of the inclusion i: C ֒→ X. It remains to show that πC is a 1-Lipschitz map. Let x, y ∈ X. By Lemma 1.82 and the triangle inequality the following holds:

d x, πC (x) + d πC (x), πC (y) = d x, πC (y) ≤ d(x, y)+ d y, πC(y) ,

dy, πC(y) + dπC (y), πC(x) = dy, πC(x) ≤ d(y, x)+ dx, πC (x). Adding these two estimates we obtain the 1-Lipschitz condition for πC .

1.84 Corollary. Let X be an injective metric space. Every closed, strongly convex subset C ⊂ X is injective.

Proof. By Corollary 1.83 we know that C is a 1-Lipschitz retract of X in Metr1. The claim follows by Lemma 1.8.

2 In the following let ℘: R → R+ be a norm with the property

℘(a) < ℘(a + b) for all a, b ∈ P+, (1.85)

2 2 where P+ := {(0, 0) = (a1, a2) ∈ R | a1, a2 ≥ 0} is the positive cone in R . For instance every ℓp-norm with p = ∞ is of this form. Hence ℘ need not be uniquely geodesic as ℘ = ℓ1 shows.

1.86 Deﬁnition. The ℘-product of (X1,d1) and (X2,d2), denoted by X1 ×℘ X2, is the metric space X1 × X2,d℘ with d℘ (x1, x2), (y1,y2) := ℘ d1(x1,y1),d2(x2,y2) . Condition (1.85) gives the triangle inequality for d℘.

We shall show in Proposition 1.87 that an injective metric space can only be decomposed in this sense as an ℓ1-product of two subspaces. ∼ 1.87 Proposition. Let X be an injective metric space and X = X1 ×℘ X2. Then X is isometric to a τ-product X1 ×τ X2 where τ(x1, x2)= τ1|x1|+τ2|x2| and τ1 = ℘(1, 0), τ2 = ℘(0, 1) ∈ R+. 34 Chapter 1. Isbell’s Injective Hull

Proof. Let p2 ∈ X2 and Cp2 := X1 ×{p2}. The map πp2 : X → Cp2 deﬁned by πp2 (x1, x2) := (x1,p2) is 1-Lipschitz due to Condition (1.85). Thus Cp2 =

πp2 (X) is a closed subspace of X. We shall see that Cp2 is strongly convex.

Let p := (p1,p2), q := (q1,p2) ∈ Cp2 and let r := (r1,r2) ∈ X with d℘(p, q)= d℘(p,r)+ d℘(r, q). Suppose that r∈ / Cp2 or rather d2(p2,r2) > 0. Then

d℘(p, q) = d℘(p,r)+ d℘(r, q)

= ℘ d1(p1,r1),d2(p2,r2) + ℘ d1(q1,r1),d2(p2,r2)

(℘ is a norm) ≥ ℘d1(p1,r1)+ d1(q1,r1),d2(p2,r2)+ d2(p2,r2)

(Condition (1.85)) > ℘d1(p1,r1)+ d1(q1,r1), 0

(Condition (1.85)) ≥ ℘d1(p1, q1), 0

= d℘(p, q).

This is a contradiction to d℘(p, q)= d℘(p, q). Hence, r ∈ Cp2 and Cp2 ⊂ X is strongly convex. Note that π (x)= π (x), for all x =(x , x ) ∈ X, where Cp2 p2 1 2 π : X → C is the 1-Lipschitz retraction onto C deﬁned in Lemma 1.82. Cp2 p2 p2 This is a direct consequence of Condition (1.85) since

d℘(x, p)= ℘ d1(x1,p1),d2(x2,p2) ≥ ℘ 0,d2(x2,p2) = d℘ x, πp2 (x) , for all p = (p1,p2) ∈ Cp2 . Let x, y ∈ X with x = (x1, x2) and y = (y1,y2).

We have seen that Cy2 ⊂ X is a closed and strongly convex subset. Hence,

℘ d1(x1,y1),d2(x2,y2) = d℘(x, y) (Lemma 1.82) = d x, π (x) + d π (x),y ℘ Cy2 ℘ Cy2 = ℘ 0,d2(x2,y2) + ℘ d1(x1,y1), 0

(℘ is a norm) = ℘(1, 0) d1(x1,y1)+ ℘(0, 1) d2(x2,y 2)

= τ d1(x1, x2),d2(x2,y2)

= dτ(x, y).

1.88 Remark. From the proof above we have also learned that X1 and X2 are injective. In particular if X1 and X2 consist of more than one point, they both contain a non-trivial geodesic segment by Proposition 1.16. Therefore 2 X1 ×℘ X2 contains an isometric copy of an open subset in (R , ℘). We get the following corollary.

1.89 Corollary. An injective metric space can not have a non-trivial ℓp- decomposition for 1

1.8 Normed vector spaces

Our main focus here is the injective hull of normed vector spaces. By a n vector space we mean in the following a real vector space. Let Rp be the n Banach space R , ||p and let Norm1 denote the category of normed vector spaces and linear 1-Lipschitz maps, i.e. objects are normed vector spaces, X, ||X , and morphisms are linear maps T : X → Y with T := sup |T (z)|Y ≤ 1. |z|X ≤1

Furthermore, let isoL be the subclass of linear isometric embeddings. Since Norm1 ⊂ Metr1 and isoL ⊂ isoM it is natural to ask whether the following two statements are true:

1. X ∈ Norm1 is isoL-injective if and only if it is isoM-injective.

2. The isoL-injective hull of X ∈ Norm1 exists and is isometric to its isoM-injective hull.

A priori these two statements are not equivalent since the isoM-injective hull of a Banach space need not be a vector space anymore. However the first one follows from the second one. Isbell answers in [21] both questions in the affirmative. We shall recall the arguments of [21] first. Then, using a classification of isoL-injective Banach spaces due to Co- hen, cf. [9], we are going to prove that every separable injective Banach n space is isometric to R∞ for some n ∈ N; in particular it is finite-dimensional.

First we give some examples of isoL-injective objects in Norm1.

1.90 Lemma. In Norm1

(i) the real line R with the euclidean norm is isoL-injective.

(ii) the Banach space ℓ∞(S) is isoL-injective for any index set S.

Proof. (i) This is exactly the statement of the Hahn–Banach theorem; cf. [25]. (ii) The real line R is isoL-injective because of (i). By Lemma 1.8 it is therefore suﬃcient to verify that ℓ∞(S) along with the family of 1-Lipschitz maps ∞ ∞ πs : ℓ (S) → R, where πs(f) := f(s) with s ∈ S and f ∈ ℓ (S), corresponds to the product s∈S R in Norm1. But this is obviously true. Q 36 Chapter 1. Isbell’s Injective Hull

1.91 Lemma. Every normed vector space X can be embedded into a Banach space ℓ∞(S) by means of a linear isometry, for some index set S.

∗ Proof. Let (X , ) denote the dual space of (X, ||X ) and let

S := {x∗ ∈ X∗ |x∗ =1}.

It is well-known (cf. [25]) that for every z ∈ X,

∗ |z|X = sup |x (z)|. (1.92) x∗∈S

∞ ∗ Deﬁne T : X → ℓ (S) to be the linear map T (z) := x (z) x∗∈S. The right- hand side of (1.92) then corresponds to the supremum-norm T (z)∞, i.e. T is a linear isometry.

1.93 Corollary. Every isoL-injective vector space is isoM-injective.

Proof. Let X be isoL-injective. Using 1.91 we can embed X into some Ba- nach space ℓ∞(S) by an isoL-morphism j. There exists a linear 1-Lipschitz ∞ map r : ℓ (S) → X such that r ◦ j = idX ; see 1.10. Then X is an isoM- retract of ℓ∞(S) since r is 1-Lipschitz. The Banach space ℓ∞(S) on the other hand is isoM-injective because of 1.21. Hence, X is isoM-injective by Lemma 1.8.

Recall the proof of the Hahn–Banach theorem where it is shown that R with the euclidean norm is isoL-injective. The only essential property of R that is used in the proof is actually the isoM-injectivity of R. The next proposition is therefore not so surprising.

1.94 Proposition. A normed vector space X is isoL-injective if and only if X is isoM-injective in Metr1.

Proof. One implication has already been shown in 1.93. Suppose X is isoM-injective. Let L: A → X be a linear 1-Lipschitz map and i: A → B be a linear isometric embedding, where A, B ∈ Norm1. We need to verify that there is a linear extension L: B → X such that L◦i = L. Since i−1 : i(A) → A is a linear isometry the map L′ := L ◦ i−1 is a linear 1-Lipschitz extension of L to i(A) ⊂ B. We may therefore assume w.l.o.g. that A ⊂ B. Moreover, due to Lemma of Zorn it is suﬃcient to extend L to the linear subspace A¯(y) := {a + ty | a ∈ A, t ∈ R} for a y ∈ B \ A. So let ra := |a − y|B ∈ R, a ∈ A. Then

′ ′ ′ ra + ra′ = |a − y|B + |a − y|B ≥|a − a |B ≥|L(a) − L(a )|X . Normed vector spaces 37

isoM Since X is -injective, the family of balls B L(a),ra a∈A has a non- empty intersection; lety ¯ ∈ X be a point of intersection. Hence, |L(a)−y¯|X ≤ ra for all a ∈ A. Then L¯ : A¯(y) → X given by L¯(a + ty) := L(a)+ t y¯ is linear and satisﬁes ¯ L(a + ty) X = L(a)+ t y¯ X = |t| L(−a/t) − y¯ X

− a/t ∈ A ≤ | t| r−a/t = | t| − a/t − y B = a + ty B, for all a + ty ∈ A¯(y), t = 0. Hence, L¯ is 1-Lipschitz. We shall use the term injective simultaneously for both isoL-injective and isoM-injective. 1.95 Lemma. The isoM-injective hull of a normed vector space is a Banach space. Proof. Let X be a normed vector space. We embed X into an injective Banach space Z := ℓ∞(S) by means of a linear isometry using Lemma 1.91. For the sake of convenience we may therefore assume that X is a linear subspace of an injective Banach space Z. Let i: X → EX be as in 1.41 and let F be the poset of tuples (Y, f) where Y is a linear subspace of Z with X ⊆ Y and f : Y → EX is an isometric embedding such that f|X = i. The ′ ′ ′ ′ partial order on F is given by (Y, f) (Y , f ):⇔ Y ⊆ Y and f |Y = f. Note that F is non-empty since (X, i) ∈F. Every linearly ordered subfamily of F has a maximal element in F; by Lemma of Zorn there is a maximal one, (W, g) ∈ F. Because of 1.47 we know that EW is isometric to EX. Therefore, we may assume w.l.o.g. that g = iW where iW : W → EW is as in 1.41. We claim that iW (W ) = EW . Suppose this is not the case; takey ¯ ∈ EW \ iW (W ) and y := j(¯y) where j : EW → Z is the isometric embedding with j(iW (w)) = w, for all w ∈ W , from Corollary 1.46. Then y∈ / W . Let W ′ := {ty + w | t ∈ R,w ∈ W } ′ and let iw′ be the distance function iw′ (z) := |w − z|Z , z ∈ Z. Then

iy(w)= |y − w|Z = |j(¯y) − j(iW (w))|Z = d¯ y,¯ iW (w) =y ¯(w), w ∈ W ; hence iy|W =y ¯ ∈ EW . Note that if iy′ |W ∈ EW then iy′+w|W ∈ EW for all w ∈ W ; this follows from the fact that x → x − w is an isometry of Z leaving W invariant. Furthermore, because of Lemma 1.124 we have ity′ |W ∈ EW for every t ∈ R. In other words iw′ |W is an extremal function on ′ ′ ′ W for any w ∈ W . Let p = t1 y + w1 and q = t2 y + w2 be two points in W . 38 Chapter 1. Isbell’s Injective Hull

If t1 = t2 and either t1, t2 ≥ 0 or t1, t2 ≤ 0 then the straight line sp+(1−s) q, s ∈ R, through p and q meets W in exactly one point w := t1 q − t2 p; pq t1−t2 t1−t2 moreover, ip|W (wpq) − iq|W (wpq) = |p − q|Z. (1.96) ¯ Hence, d (ip|W , iq|W ) ≥ |ip|W (wpq) − iq|W (w pq)| = |p − q|Z. If on the other 1 hand t1 = t2 = t we may approximate q by the sequence qn := (t − n ) y + w2. We then obtain d¯(ip|W , iq|W ) ≥|p − q|Z, (1.97) ¯ since qn → q and d(ip|W , iqn |W ) ≥ |p − qn|Z for suﬃciently large n ∈ N. Note that by the triangle inequality we also have d¯(ip|W , iq|W ) ≤ |p − q|Z , ¯ ′ in other words d(ip|W , iq|W )= |p − q|Z. Let W+ := {ty + w | t ≥ 0,w ∈ W } ′ ′ ′ E and W− := {ty + w | t ≤ 0,w ∈ W }. Then f+ : W+ → W given by ′ E ′ f+(p) := ip|W is an isometric embedding. It follows by 1.47 that W+ is E ¯′ ¯′ ′ ′ isometric to W by means of f+ with the property that f+ ◦ iW+ = f+. ′ ′ E ′ ′ ′ ′ ′ We claim that f : W → W+ given by f (w ) := iw |W+ is an isometric ′ E ′ ′ ′ ′ embedding of W into W+. Let w ∈ W . For every x ∈ W+ we have

inf iw′ (x)+ iw′ (x + w) −|w|Z =0, w∈W since iw′ (x) = iw′−x(0), iw′ (x + w) = iw′−x(w) and iw′−x|W ∈ EW . Hence, ′ ′ E ′ ′ ′ ′ ′ ′ iw |W+ ∈ W+ by 1.35. Let w1,w2 ∈ W . If w1 ∈ W+, then

′ ′ ¯ ′ ′ ′ ′ ′ ′ ′ ′ |w2 − w1|Z ≥ d(iw1 |W+ , iw2 |W+ ) ≥|iw2 (w1)| = |w2 − w1|Z .

′ ′ ′ If on the other hand w1,w2 ∈ W−, then

′ ′ ¯ ′ ′ ′ ′ ¯ ′ ′ ′ ′ |w2 − w1|Z ≥ d(iw2 |W+ , iw1 |W+ ) ≥ d(iw2 |W , iw1 |W ) ≥|w2 − w1|Z, where the last estimate is because of (1.97). It follows that f ′ is an isometric ′ E ′ ¯′ ′ ′ E embedding of W into W+. Let f := f+ ◦ f : W → W . Then,

¯′ ¯′ ′ ′ f(w)= f+(f(w)) = f+(iW+ (w)) = f+(w)= iW (w),

′ ′ for all w ∈ W . Hence, (W , f) ∈F and (W, iW ) (W , f); this contradicts the maximality of (W, iW ). Cohen gives in [9] a complete classification of all injective normed vector spaces. They are up to an isomorphism exactly of the form C(M), the Banach space of all real-valued continuous functions on an extremely disconnected compact Hausdorff space M; a topological space is called extremely disconnected if the closure of every open subset is open. The Banach space ∞ ∼ ℓ (S) = Cb(S) for instance is injective and isomorphic to C(βS) where βS Normed vector spaces 39 is the Stone–Cechˇ compactification of S with the discrete topology; βS is an extremely disconnected compact Hausdorff space, see Appendix A and A.6 for more details. Cohen gives even an explicit description for M in the case of the injective hull EX ∼= C(M) of X. M corresponds to the projective hull, in the category of compact Hausdorff spaces and continuous maps, of ∗ ∗ ∗ ∗ a particular weak -closed subset S ⊂ cl (ext(B1 )); here cl (A) denotes the ∗ ∗ ∗ weak -closure of A ⊂ X and ext(B1 ) is the set of extremal points of the unit ball in X∗. It is not important for us how S or M can be constructed ∗ out of B1 ; the interested reader should look at [9],[16] for an exposition of this subject. However we need the following two simple facts from [9],[16]:

∗ ∗ ∗ (i) if N := |ext(B1 )| < ∞ then S = M ⊂ ext(B1 ) such that ext(B1 ) = N M ∪ −M and M ∩ −M = ∅; in particular |M| = 2 .

∗ (ii) if |ext(B1 )| = ∞ then |M| = ∞.

E n 1.98 Example. We would like to calculate the injective hull R1 , using [9]. n∗ n ∗ n Recall that the dual space R1 is isomorphic to R∞ and |ext(B1 )| = 2 ; ∗ n 2n n−1 B1 ⊂ R∞ is an n-dimensional cube of side length 2. Then |M| = 2 =2 by (i). Therefore, we get that

E n ∼ n−1 ∼ 2n−1 R1 = C {1,..., 2 } = R∞ . n n E n ∼ E n ∼ 2n−1 1.99 Example. Let Z ⊂ R1 . Then Z = R1 = R∞ . This can be seen n n as follows. Let x, y ∈ R1 ; deﬁne z =(z1,...,zn) ∈ Z by

⌈yi⌉ , if xi ≤ yi zi := ⌊yi⌋ , if yi ≤ xi.

|x − z|1 = |x − y|1 + |y − z|1. (1.100)

n E n n Let F : R1 → Z be given by F (y) := iy where iy(z) := |y − z|1, y ∈ R1 . n n n Note that iy ∈ EZ ; by (1.100) there exists for every x ∈ Z a point z ∈ Z such that |x−z|1 = iy(x)+iy(z), in other words iy satisﬁes Proposition 1.35. n We claim that F is an isometric embedding: let x, y ∈ R1 . Then by (1.100) n there is z ∈ Z with ix(z) − iy(z) = |x − z|1 −|y − z|1 = |x − y|1. Hence, d¯ F (x), F (y) ≥|x − y|1. But F , on the other hand, is 1-Lipschitz because of the triangle inequality; hence F ∈ isoM. It then follows by 1.47 that E n ∼ E n R1 = Z . 40 Chapter 1. Isbell’s Injective Hull

Using Cohen’s result we shall show in 1.103 that every separable injective n Banach space is isomorphic to some R∞. 1.101 Lemma. Let M be an extremely disconnected topological space and U, V ⊂ M two disjoint open subsets. Then, U ∩ V = ∅. Proof. Since U ∩ V = ∅ and V is open it follows that U ∩ V = ∅. On the other hand U is open; hence U ∩ V = ∅. 1.102 Lemma. An extremely disconnected Hausdorﬀ space M with a countable local base is discrete. Proof. Assume that M is not discrete; there is a point x ∈ M and a sequence {xn} in M converging to x with xn = x, for all n ∈ N. Let n1 := 1. Since

M is Hausdorﬀ there is an open subset U1 containing xn1 such that x∈ / U1.

Because x = limn xn ∈/ U1 there is an n2 > n1 with xn2 ∈/ U1. Then, we can ﬁnd an open subset U2 containing xn2 such that x∈ / U2 ∪ U1 and U1 ∩ U2 = ∅. By continuing this procedure we obtain a monotone increasing sequence of numbers {ni}i∈N and a family of pairwise disjoint, open subsets

{Ui}i∈N such that xni ∈ Ui. Let U := i∈N U2i and V := i∈N U2i+1; then U and V are disjoint open subsets of M.S Because of LemmaS 1.101 we have that

U ∩V = ∅. On the other hand x ∈ U ∩V = ∅ since x = limi xn2i = limi xn2i+1 ; this contradicts U ∩ V = ∅. 1.103 Theorem. Every injective separable Banach space is isomorphic to n some R∞.

Proof. Let X ∈ Norm1 be injective and separable. Then X is isomorphic to some C(M), cf. [9], where M is an extremely disconnected, compact Hausdorﬀ space. Since X is separable, C(M) is separable; we shall show that M has a countable local base. Let S ⊂ C(M) be a countable subset with S = C(M). Consider F : M → Y := f∈S If deﬁned by F (m) :=

f(m) f∈S, where If := −|f|∞, |f|∞ ⊂ R; byQ Tychonoff’s theorem Y is a compact Hausdorff space. Then Y has a countable local base 1 U (y ) := y ∈ Y |π (y) − π (y )| < , ∀1 ≤ i ≤ k , f1,...,fk,n 0 fi fi 0 n n o where k ∈ N, f1,...,fk ∈ S and y0 ∈ Y ; πf : Y → If is the projection of Y onto the f-th factor If . Furthermore, F is a continuous map since πf ◦ F = f : M → R is continuous for every f ∈ S. We claim that F is an embedding of M into Y ; it is enough to verify that F is injective, because M is compact and Y is Hausdorff. So let x = x′ ∈ M. By Lemma of Urysohn there is g ∈ C(M) such that g(x) − ε>g(x′)+ ε for some ε > 0. On the Normed vector spaces 41

other hand there is f ∈ S with |g − f|∞ < ε; hence, f(m) ≥ g(m) − ε and g(m)+ ε ≥ f(m) for all m ∈ M. Therefore, f(x) > f(x′). In other words F (x) = F (x′). We have just shown that M has a local base; so by 1.102 it is a discrete and compact space, in particular M is ﬁnite. Thus C(M) ∼= ∼ n ℓ∞ {1,...,n} = R∞ where n = |M| < ∞. n 1.8.1 Aﬃne, injective subspaces of R∞ n Let X be a k-dimensional linear subspace of R∞ endowed with the induced norm. Then the unit ball in X corresponds to the intersection of X with n n n n the unit ball in R∞ – the n-dimensional cube C := [−1, 1] ⊂ R . If X k is injective, then due to Theorem 1.103 it is isometric to R∞ by a linear transformation. Therefore, X ∩ Cn and the k-dimensional cube Ck ⊂ Rk must have the same number of extremal points, in particular

| ext(X ∩ Cn)| =2k.

3 n For instance, the subspace X6 := (x1, x2, x3) ∈ R | x1 +x2 +x3 =0 ⊂ R∞ 3 can not be injective, since X6 ∩ C is a polygon with 6 vertices whereas the square C2 has only 4. In the next proposition we shall use this observation n to give a characterization of all aﬃne, injective subspaces of R∞.

n 1.104 Proposition. Let X ⊂ R∞ be an aﬃne subspace of codimension n − k. Then, up to a permutation of the coordinates, X is injective if and k only if there are numbers ci ∈ R and λi,j ∈ R with j=1 |λi,j| ≤ 1 and P k

(x1,...,xn) ∈ X ⇔ xi = ci + λi,j xj, for all k < i ≤ n. (1.105) Xj=1

n In this case r : R∞ → X,

k k

r(x) := x1,...,xk,ck+1 + λk+1,j xj,...,cn + λn,j xj , (1.106) Xj=1 Xj=1

n is a 1-Lipschitz retraction of R∞ onto X. Proof. We may assume w.l.o.g. that X is a linear subspace of Rn. Otherwise ∗ ∗ we displace X by some vector −c, c ∈ X. Let {E1 ,...,En} be the dual basis n ∗ ∗ of the standard basis of R and let {e1,...,ek} denote the corresponding k n n dual basis of R . Recall that the dual space of R∞ is isometric to R1 and 42 Chapter 1. Isbell’s Injective Hull

∗ ∗ n that {±E1 ,..., ±En} is the set of extremal points of the unit ball B1 := n B(0, 1) ⊂ R1 . “⇒”: Suppose that X is injective. Then by 1.103 there is a linear iso- k ։ n metric embedding T : R∞ X ⊂ R∞. Using the Theorem of Hahn–Banach ∗ n k we get that T : R1 → R1, the dual transformation of T , is surjective and ∗ n k that T maps the ball B1 onto B1 . It follows in particular that

k ∗ n {±e1,..., ±ek} = ext(B1 ) ⊂ T ext(B1 ) . ∗ ∗ ∗ ∗ ∗ We may assume that T (Ej )= ej , for all 1 ≤ j ≤ k. Let k < i ≤ n. T (Ei ) ∗ ∗ ∗ ∗ k is a convex linear combination of {±e1,..., ±ek} since T (Ei ) ∈ B1 . In other words, there are non-negative numbers i,j, ¯i,j ∈ [0, 1], 1 ≤ j ≤ k, k ∗ ∗ k ∗ ∗ with j=1(i,j + ¯i,j) = 1 and T (Ei ) = j=1 i,jej − ¯i,jej . Let x ∈ X, k and letPy ∈ R∞ with x = T (y). Deﬁne λi,jP:= i,j − ¯i,j. By the triangle k k inequality we have that j=1 |λi,j| ≤ j=1 i,j + ¯i,j = 1. Furthermore, X satisﬁes condition (1.105)P since P

∗ ∗ ∗ ∗ xi = Ei (x)= Ei T (y) = T (Ei )(y) k ∗ ∗ = i,jej − ¯i,jej (y) Xj=1 k k ∗ ∗ ∗ = λi,jej (y)= λi,jT (Ej ) (y) Xj=1 Xj=1 k k k ∗ ∗ = λi,jEj T (y) = λi,jEj (x)= λi,j xj. Xj=1 Xj=1 Xj=1

“⇐”: Suppose on the other hand that X is of the form (1.105), and let k n T : R∞ → R∞ be the map

k k

T (y) := y1,...,yk,ck+1 + λk+1,j yj,...,cn + λn,j yj . (1.107) Xj=1 Xj=1

k Then the condition (1.105) is equivalent to T (R∞) = X. Hence, in order k to show that X = T (R∞) is injective, it is enough to verify that T is an ′ k isometric embedding. Let y,y ∈ R∞. Then

k k k ′ ′ ′ ′ | λi,j (yj − yj)| ≤ |λi,j||yj − yj| ≤ |λi,j| |y − y |∞ ≤|y − y |∞, Xj=1 Xj=1 Xj=1 Normed vector spaces 43

′ ′ for all k < i ≤ n. Therefore, |T (y)−T (y )|∞ ≤|y−y |∞. The lower estimate, ′ ′ |y − y |∞ ≤ |T (y) − T (y )|∞, also holds because of (1.107). Hence, T is an isometric embedding. n k n Let πk : R∞ → R∞ be the projection of R∞ onto the ﬁrst k coordinates, that is πk(x1,...,xn) := (x1,...,xk). Then πk and r = T ◦ πk are obviously 1-Lipschitz. Furthermore, by (1.105) we get x = T πk(x) = r(x) ⇔ x ∈ X. Hence, r is a 1-Lipschitz retraction onto X.

1.108 Lemma. Let S be a subset of {(i, j) | 1 ≤ i ≤ j ≤ n}, n ∈ N. Suppose that the system of equations in Rn,

ES : xi + xj = cij, (1.109)

(i, j) ∈ S and cij ∈ R, is solvable. Then ES is equivalent to one of the form (1.105) in the sense that they have the same solution space X.

Proof. The proof goes by induction over the number of equations, N := |S|. We may assume w.l.o.g. that the equations in (1.109) are linearly indepen- dent, in particular |S| ≤ n. We remove the linearly dependent equations, without changing X, otherwise. In the case of N = 1 the claim is true since ′ xi = cij − xj is of the form (1.105). Let |S| = N + 1 and S = S ∪{(a, b)}. Then by the induction hypothesis ES′ is equivalent to the system,

′ xi = ci + τi xji , 1 ≤ i ≤ N and N < ji ≤ n, (1.110) with τi ∈ {−1, 0, 1}; up to a permutation of {1,...,n}. Note that (1.110) is ′ of the form (1.105). If a ≤ N then by substituting xa with ca + τa xja one ′ ′ gets an equivalent system to ES of the form τa xja + xb = cab − ca and ES , where ja > N. We can do the same with b if b ≤ N. Hence, we may further assume that a, b > N, in particular that a = N +1. Whenever ji = N +1 = b ′ we shall replace the i-th equation of (1.110) with xi =(ci + τi cN+1b) − τi xb by substituting τi xN+1 with τi cN+1b − τi xb. If on the other hand ji = ′ τi N +1 = b we may replace the i-th equation with xi = (ci + 2 cN+1b) by cN+1b substituting τi xN+1 with τi 2 instead. Either way we obtain from (1.110) a new equivalent system ES¯′ such that ji > N + 1, for all 1 ≤ i ≤ N. In other words, ES¯′ and xN+1 + xb = cN+1b is of the form (1.110) and the claim follows.

Proposition 1.104 and Lemma 1.108 give the following corollary.

n 1.111 Corollary. The solution space X ⊂ R∞ of (1.109) endowed with the induced metric is injective for any choice of S and cij. 44 Chapter 1. Isbell’s Injective Hull

1.112 Deﬁnition. Let

c n X () := {x ∈ R | x := 1x1 + + nxn = c} be a hyperplane in Rn, where 0 = ∈ Rn and c ∈ R. For the sake of convenience, we just write X() for X0(). By the upper half-space of Xc() c n we mean the subset X+ () := {x ∈ R | x ≥ c}, whereas the lower half- c c n space of X () is X− () := {x ∈ R | x ≤ c}.

c c c c −c Note that X ()= X+ () ∩ X− () and X+ ()= X− (−). 1.113 Remark. c c X− () and X+ () are isometric with respect to any norm on n R . An isometry can be given for instance by the reﬂection Sp(x) := 2 p − x at p ∈ Xc().

c n 1.114 Corollary. A hyperplane X () ⊂ R∞ is injective if and only if ||1 ≤ 2 ||∞. Proof. We may assume w.l.o.g. that c = 0 since Xc() and X() are isometric aﬃne subspaces. Now suppose that X() is injective. By Propo- n sition 1.104 there is λ ∈ R with |λi| = 1, for some i ∈ {1,...,n}, and |λ|1 ≤ 2 such that X(λ) ≡ X(). Then and λ are collinear and in particular = ±||∞ λ. Thus, ||1 = ||∞|λ|1 ≤ 2 ||∞. n On the other hand, let ∈ R with ||1 ≤ 2 ||∞. Since X() ≡ X(t), for all t ∈ R, we may further assume w.l.o.g. that n = ||∞ = 1. Then it n−1 follows that j=1 |j| = ||1 −||∞ ≤ 2 ||∞ −||∞ =1 and (x1,...,xn) ∈ n−1 X() ⇔ xn =P j=1 −j xj. Hence, X() is injective by 1.104. P 1.115 Example. Let X6 be the linear subspace from the paragraph before Proposition 1.104. Then X6 = X() with = (1, 1, 1). Therefore, X6 can not be injective since 3 = ||1 > 2=2 ||∞. Because of Proposition 1.104 and Corollary 1.114 every aﬃne, injective n subspace X ⊂ R∞ of codimension k is the intersection of k injective hyperplanes. The converse is not true in general as the following example shows.

4 1.116 Example. Let Y := {(x1, x2, x3, x4) ∈ R | x1 + x2 + x3 +3 x4 = 0} ′ 4 ′ n and Y := {(x1, x2, x3, x4) ∈ R | x4 = 0}. The intersection Y ∩ Y ⊂ R∞ is ′ isometric to X6 from Example 1.115. Therefore, Y ∩ Y is not injective even though Y and Y ′ are both by 1.114.

The next lemma will be needed in Subsection 1.8.2.

1.117 Lemma. The following three statements are equivalent. Normed vector spaces 45

c n (i) The hyperplane X () ⊂ R∞ is injective.

c n (ii) The upper half-space X+ () ⊂ R∞ is injective. c n (iii) The lower half-space X− () ⊂ R∞ is injective.

Proof. The equivalence of (ii) and (iii) is obvious since the two half-spaces are isometric by Remark 1.113. c n n c (i) ⇒ (ii): suppose that X () ⊂ R∞ is injective and let π : R∞ → X () n c denote a 1-Lipschitz retraction of R∞ onto X (); see Proposition 1.24. We c shall verify the ball intersection property for X+ (). Let {B(xi,ri)}i∈I be a c n family of balls in X+ () as in Proposition 1.13 (iii). Since R∞ is injective n there is a point x ∈ R∞ with |x − xi|∞ ≤ ri, for all i ∈ I. Suppose that c c x∈ / X+ () or rather x ∈ X− (). Nothing has to be shown otherwise. Let c c i ∈ I. Since xi ∈ X+ () and x ∈ X− () there is an s ∈ [0, 1] such that c xs := s xi + (1 − s) x ∈ X (). Note that π(xs) = xs and |x − xi|∞ = |x − xs|∞ + |xs − xi|∞. Then

|π(x) − xi|∞ ≤ |π(x) − π(xs)|∞ + |π(xs) − xi|∞

≤ |x − xs|∞ + |xs − xi|∞ = |x − xi|∞ ≤ ri.

c In other words π(x) is a point of intersection of {B(xi,ri)}i∈I in X+ (). c (iii) ⇔ (ii) ⇒ (i): let {B(xi,ri)}i∈I be a family of balls in X () as in c c Proposition 1.13 (iii). Since X () ⊂ X+ () and the latter is injective, there c is a point x+ ∈ X+ () with |xi − x+ |∞ ≤ ri, for all i ∈ I. For the same c reason we also ﬁnd an x− ∈ X− () with |xi − x− |∞ ≤ ri, for every i ∈ I. Let c s ∈ [0, 1] be such that x0 := s x− + (1 − s) x+ ∈ X (), and let i ∈ I. Then

|x0 − xi|∞ ≤ s|x− − xi|∞ + (1 − s)|x+ − xi|∞ ≤ sri + (1 − s) ri = ri.

c c The point x0 ∈ X () is a point of intersection of {B(xi,ri)}i∈I in X (). Hence, Xc() is injective by Proposition 1.13.

c n 1.118 Remark. Let X () ⊂ R∞ be an injective, aﬃne subspace and n c c let π : R∞ → X () be any 1-Lipschitz retraction onto X (). A 1- Lipschitz retraction onto the upper half-space of Xc() can then be given n c by π+ : R∞ → X+ (),

c x if x ∈ X+ (), π+ (x) := n c π(x) if x ∈ R∞ \ X+ ().

This is implicitly shown in the proof of “(i) ⇒ (ii)” above. Let r+ denote in n c the following the retraction π+ in the case that π = r, where r : R∞ → X () is the retraction from (1.106). 46 Chapter 1. Isbell’s Injective Hull

n 1.8.2 Injective polyhedrons in R∞ We will carry out some preliminary work for Section 1.9 where it is shown that the injective hull of a finite metric space is a finite cell complex and that n its cells are injective convex polyhedrons in R∞. In Proposition 1.127 we are going to give a sufficient condition for a convex polyhedron to be injective which will be enough for our purposes. However, a complete classification of n injective convex polyhedrons in R∞ remains still open. First of all, we need to fix some terminology and recall some simple and well known facts from convex geometry in Rn; see for example [12], [5, 7.34]. In the following C denotes a closed convex subset of Rn. A subset F ⊂ C is a face of C if either F = C, or F is the intersection of C and a c c hyperplane X () with C ⊂ X+ (). Every face is a closed convex subset of Rn. The dimension of F , dim F , is the dimension of the smallest affine n subspace XF ⊂ R containing F . In this case XF is said to be generated by F . A point x ∈ F is an inner point of F if x lies in the interior of F as a subset of XF ; int(F ) denotes the set of all interior points of F . The solid tangent cone of C at x ∈ C is the closed subset

TxC := x + s (C − x), (1.119) s[∈N where x + s (C − x) := {x − s (y − x) ∈ Rn | y ∈ C} is the blow-up of C at x by the factor s ∈ N. The solid tangent cone is a conical subset of Rn with apex x, and TxC ⊂ XC for every x ∈ C. If TxC is a half-space in XC , say of a hyperplane HxC ⊂ XC , then HxC is said to be the supporting hyperplane, or the tangent space, of C at x. The corresponding half-space TxC is the supporting half-space of C at x. A closed convex subset is the intersection of its supporting half-spaces. 1.120 Remark. The subset (C − x) is convex and 0 ∈ (C − x) since x ∈ C. Then s (C −x) ⊂ (s+1)(C −x) for every s ∈ N. Hence, the family of subsets in (1.119), x + s (C − x) , is an ascending family of convex subsets in s∈N Rn containing x.

1.121 Remark. Let z be an inner point of a face F of C. Then TzF = XF .

N ci n If C is the intersection of ﬁnitely many half-spaces, C = i=1 X+ (i) ⊂ R , and dim C = n then every supporting hyperplane of C isT of the form HxC = cj X (j) for some 1 ≤ j ≤ N. Furthermore, every solid tangent cone TxC and every face F of C can be written as

cj cj TxC = X+ (j) and F = C ∩ X (j), (1.122) j\∈S j\∈S′ Normed vector spaces 47 for some S,S′ ⊂{1,...,N}. The half-spaces and hyperplanes in (1.122) can be chosen to be supporting half-spaces and supporting hyperplanes of C. A convex polyhedron, or just a polyhedron, is a compact convex subset n ci n P ⊂ R that is the intersection of ﬁnitely many half-spaces X+ (i) ⊂ R , 1 ≤ i ≤ N. Every proper face of P is a polyhedron, and the intersection of two polyhedrons is either empty or a polyhedron again.

In 1.125 we shall give a necessary condition for a closed convex subset to be injective in terms of its solid tangent cones.

1.123 Deﬁnition. Let (X,d) be a metric space and λ > 0. Then, λ X denotes the scaled metric space (X,dλ) where dλ(x, y) := λd(x, y).

1.124 Lemma. The subclass of injective metric spaces is closed under scal- ing.

Proof. This is a direct consequence of the deﬁnition 1.1 of injectivity and 1 the fact that a map f : X → λ Y is 1-Lipschitz if and only if f : λ X → Y is 1-Lipschitz.

n 1.125 Proposition. Let C be a convex subset of R∞ that is injective. Then (i) every solid tangent cone of C is injective, and

(ii) every supporting hyperplane of C is injective.

n Proof. (i) Let x ∈ C and let As := x + s (C − x)=(1 − s) x + sC ⊂ R∞ be the blow-up of C at x by the factor s ∈ N. Then, As is isometric to s C for every s. Furthermore, the family {As}s∈N is an ascending family of injective n subsets of R∞; compare Remark 1.120 and Lemma 1.124. It follows from 1.63 that TxC = s∈N As is an injective metric space. This proves the first part of the claim.S (ii) Let p be an inner point of C and let k := dim C. From Remark 1.121 we know that the affine subspace XC corresponds to TpC. The latter is injective because of (i). Hence, XC is also injective. In particular, XC is k isometric to R∞ by Theorem 1.103. Let HzC be a supporting hyperplane of C at some point z ∈ C, and let TzC denote the corresponding half- space. In order to show that HzC is injective we may restrict to XC since ∼ k HzC, TzC ⊂ XC = R∞. In other words we may assume w.l.o.g. that n = k. Note, by the definition of a supporting hyperplane, that dim TzC = k. Then TzC is injective due to (i). Therefore, by using Lemma 1.117 we get that HzC is also injective. 48 Chapter 1. Isbell’s Injective Hull

Condition 1.125 (ii) is not sufficient to ensure that C is injective; see 1.136 for a counter example. However, it is not known whether 1.125 (i) is a sufficient condition. n 1.126 Remark. Proposition 1.125 is not only true for subsets of R∞ but for convex subsets of any finite dimensional Banach space. The proof goes exactly the same way. 1.127 Proposition. n N cj Let C ⊂ R∞ be a convex subset and C = j=1 X+ (j). Suppose further that T cj XS := X (j) (1.128) j\∈S is injective, for all S ⊂{1,...,N}. Then C is injective. Proof. The proof goes by induction over k := dim C and N, the number of half-spaces representing C. In the case of N = 1 the claim follows immediately from Lemma 1.117. Hence, we may assume that N > 1. Let k = 0. In this case we don’t have to show anything either since C is a point, and ′ cj a point is obviously injective. Let k > 0. Define Ci := j=i X+ (j) and ci ′ Fi := X (i) ∩ C , 1 ≤ i ≤ N. If dim Fi = k for all 1 ≤ i ≤TN then N N ci ci C ⊂ X (i) ⊂ X+ (i)= C. i\=1 i\=1

N ci It follows that C = i=1 X (i); hence C is injective by (1.128). Therefore ′ ′ we may assume w.l.o.g.T that dim F1 ≤ k − 1. Let C := C1 and F := ′ F1. Then, C and F both obviously satisfy (1.128). Since dim F ≤ k − 1 and C′ is an intersection of N − 1 half-spaces, it follows by the induction hypothesis that C′ and F are both injective. Therefore, it is enough to verify that C ⊂ C′ is a 1-Lipschitz retract of C′; see Lemma 1.8. Note that ′ c1 C = C ∩ X+ (1). From 1.125 (i) and Remark 1.121 it follows that XF is n n n injective. Let π1 : R∞ → XF ⊂ R∞ denote a 1-Lipschitz retraction of R∞ onto XF and π2 : XF → F ⊂ XF a 1-Lipschitz retraction of XF onto F ; see Proposition 1.24. Deﬁne r : C′ → C ⊂ C′ to be the map x if x ∈ C ⊂ C′, r(x) := ′ π2 ◦ π1(x) if x ∈ C \ C. We need to verify that r is 1-Lipschitz. Let x ∈ C and y ∈ C′ \C. The other two cases, x, y ∈ C and x, y ∈ C′ \ C, are obvious. Since C′ is convex, and 1(x) ≥ c1 and 1(y)

′ c1 ′ y := s x + (1 − s) y ∈ X (1) ∩ C = F. Normed vector spaces 49

Therefore,

|r(x) − r(y)|∞ = |x − π2 ◦ π1(y)|∞ ′ ′ ≤ |x − y |∞ + |y − π2 ◦ π1(y)|∞ ′ ′ ′ (y ∈ F ⊂ XF ) = |x − y |∞ + |π2 ◦ π1(y ) − π2 ◦ π1(y)|∞ ′ ′ (π2 ◦ π1 is 1-Lipschitz) ≤ |x − y |∞ + |y − y|∞

= |x − y|∞.

Then C = r(C′) is injective.

1.129 Remark. Unlike 1.125, cf. Remark 1.126, it is essential for the proof n of 1.127 that C is actually a subset of R∞ because Lemma 1.117 is valid only n for R∞. 1.130 Example. The trivial metric space that is a point is obviously injec- 2 tive. Hence, condition (1.128) is satisfied by any family of halfspaces in R∞. 2 In particular every polyhedron in R∞ is injective by Proposition 1.127. 2 Moreover, every compact convex subset C ⊂ R∞ is the Gromov– Hausdorff limit of an ascending family of polyhedrons Cn ⊂ C; take for instance finite subsets Sn ⊂ ext(C) with Sn ⊂ Sn+1 and such that

S := n∈N Sn is dense in ext(C). Then, Cn := conv(Sn) ⊂ C are poly- hedronsS and since S ⊂ ext(C) is dense, Cn →dGH C. Then all Cn are injective and by 1.61 the limit C is also injective. 2 Now let C be an arbitrary closed convex subset of R∞; deﬁne Cn := C ∩ B(0, n), n ∈ N. Then {Cn} is an ascending family of compact convex 2 subsets of R∞ with C = n∈N Cn. Hence, using Lemma 1.18 we get that C 2 is injective. To sum up, everyS closed convex subset of R∞ is injective. The following example will be important later in section 1.9.

′ 2 ′ n 1.131 Example. Let S,S ⊂ {1,...,n} . Deﬁne CS,S ⊂ R∞ to be the + ′ intersection of the family {Hij | (i, j) ∈ S}∪{Hab | (a, b) ∈ S }, where

+ n Hij := {x ∈ R∞ | xi + xj ≥ cij } n Hab := {x ∈ R∞ | xa + xb =¯cab} and cij, c¯ab ∈ R. Note that xa + xb =c ¯ab ⇔ xa + xb ≥ c¯ab, −xa − xb ≥ −c¯ab. By Proposition 1.127 and Corollary 1.111 the convex subset CS,S′ is injective ′ for any choice of S,S and cij, c¯ab ∈ R.

n 1.132 Lemma. Let C be a convex subset of R∞ as in Proposition 1.127. Then every proper face of C satisﬁes the requirements of 1.127 too. 50 Chapter 1. Isbell’s Injective Hull

N cj Proof. Let C = j=1 X+ (j) be as in 1.127 and let F be a proper face of C. ck By (1.122) the faceT F can be written as F = C ∩ k∈S′ X (k), for some S′ ⊂{1,...,N}. Then T

N cj ck F = X+ (j) ∩ X (k) j\=1 k\∈S′ N cj ck ck = X+ (j) ∩ X+ (k) ∩ X− (k) j\=1 k\∈S′ N cj ck −ck = X+ (j) ∩ X+ (k) ∩ X+ (−k). (1.133) j\=1 k\∈S′ In other words F is the intersection of finitely many half-spaces. Fur- thermore, the representation (1.133) satisfies condition (1.128) since by cj N assumption the family {X+ (j)}j=1 satisfies (1.128) and we have that −ck ck ′ X (−k)= X (k), for all k ∈ S . n 1.134 Corollary. Let C be a convex subset of R∞ as in Proposition 1.127. Then every face of C with the induced metric is injective. Proof. This follows immediately from 1.127 and 1.132. In the next counter-example we construct an injective, 4-dimensional 4 polyhedron P1 ⊂ R∞ in such a way that one of its faces is not injective. Therefore, P1 can not be of the form (1.128) since otherwise this would contradict 1.134. 1.135 Counter-Example. 4 0 Let = (1, 1, 1, 3) and r+ : R∞ → X+ () be the 1-Lipschitz retraction from 1.118. Note that X() is injective since 4 ||1 = 6 ≤ 6=2 ||∞; compare Corollary 1.114. Let m := (1, 1, 1, 1) ∈ R∞ and

4 P1 := {x ∈ R∞ ||x − m|∞ ≤ 2, (x)= x1 + x2 + x3 +3 x4 ≥ 0}.

4 Then P1 = r+ (B) with B := B(m, 2) ⊂ R∞ since r+ (B) ⊂ B. By Proposi- tion 1.17 the ball B is injective, and therefore P1 is injective; see 1.8. Let F1 be the face P1 ∩ X() of P1. It is easy to see that the subspace 4 H := X() ∩ {x ∈ R∞ | x4 = −1} is a supporting hyperplane of F1. 4 ∼ 3 Note that H in {x ∈ R∞ | x4 = −1} = R∞ is given by the equation ′ (x1, x2, x3) := x1 + x2 + x3 = 3. Hence, H is not injective because of ′ ′ 1.114 since | |1 = 3 > 2=2 | |∞. Therefore, F1 can not be injective; compare Proposition 1.125. In other words, P1 is an injective polyhedron that is not of the form (1.128). Finite metric spaces 51

1.136 Counter-Example. Let , P1 and F1 be as in Example 1.135 and let

1 (k) 4 1 k P := x ∈ R∞ ||x − m|∞ ≤ 2, 0 ≤ (x) ≤ = P1 ∩ X− (), n k o k ∈ N. Then dim P (k) = 4 and every supporting hyperplane of P (k) is isometric to either X() or to one of the supporting hyperplanes of the (k) cube B(m, 2); thus they are all injective. But since the sequence (P )k∈N converges to the face F1 in the Gromov–Hausdorff topology and the latter is not injective, cf. Example 1.135, it follows by 1.59 that all but finitely many of the P (k)’s are not injective. In other words we have constructed a 4 family of convex subsets in R∞ that are not injective but whose supporting hyperplanes are all injective. This shows that condition 1.125 (ii) is not sufficient.

1.9 Finite metric spaces

The purpose of this section is to study the structure of the injective hull of a ﬁnite metric space X. In this case EX carries a natural cell structure; it can be given explicitly in terms of graphs in X which correspond to the cells. This fact was already known to Isbell, cf. [22], and has been used in recent years; see for instance [10], [19], [20], [17]. We shall show that the cells of N this decomposition are injective polytopes in R∞. Let X = {1,...,N} be in the following a ﬁnite metric space and i: X → EX the isometric embedding from 1.41. We are going to write fj,1 ≤ j ≤ N, instead of f(j) whenever f : X → R; in other words we think of EX as a subspace of RN .

1.137 Deﬁnition. Let f ∈ EX ⊂ RN . The graph generated by f is the non-directed graph G(f)=(V, E) with vertex set V = X and edges E where

{i, j}∈E :⇔ fi + fj = dX (i, j). (1.138)

A path in G(f) of length k ∈ N is a (k+1)-tuple of vertices p =(v0,...,vk) ∈ k V with {vi, vi+1}∈E for all 0 ≤ i

1.139 Remark. Let f ∈ EX. For every j ∈ X there exists by Lemma 1.70 a point k ∈ X such that fj + fk = dX (j, k); in other words every j ∈ X is the vertex of an edge {j, k}∈E of G(f). Furthermore, we have {j, j}∈E if and only if fj = 0 since fj + fj = dX (j, j) = 0 by (1.138). 1.140 Deﬁnition. We say that a non-directed graph G =(V, E) with vertex set V = X is X-admissible if there is an extremal function fG ∈ EX such that G = G(fG); let (A(X), ≺) denote the poset of all X-admissible graphs N with (X, E1) ≺ (X, E2):⇔ E1 ⊂ E2. The convex subset P (G) ⊂ R is by deﬁnition the set of solutions of the system of relations, f ∈ RN ,

fi + fj =c ¯ij := dX (i, j), ∀{i, j}∈E (1.141)

fk + fl ≥ ckl := dX (k,l), ∀{k,l} ∈E/ . (1.142)

Note that G ≺ G′ ∈A(X) if and only if P (G) ⊃ P (G′).

1.143 Lemma. If G ∈A(X) then P (G) ⊂ EX ⊂ RN . Proof. Let f ∈ EX be such that G = G(f), and let g ∈ P (G). Then by (1.141) and (1.142) we obtain gi + gj ≥ dX (i, j) for all i, j ∈ X. Let j ∈ X. By Remark 1.139 there is k ∈ X such that {j, k}∈E, hence gj +gk = dX (j, k) due to (1.141). Therefore, g ∈ EX by 1.35.

1.144 Remark. Let f,g ∈ P (G) and v ∈ X, and let δ := g(v)−f(v). Then by (1.141) it follows that f(v)+ f(w) = dX (v,w) = g(v)+ g(w) whenever {v,w}∈E, hence g(w)= f(w) − δ. Therefore,

g(w)= f(w)+(−1)k δ, for all w ∈ [v], (1.145) where k is the length of a path joining v and w. In other words the function g is completely determined on the path component [v] by the value g(v). Furthermore, if [v] contains a k-cycle (v0,...,vk−1, v0), k = 2n + 1 odd, n n+1 then g(vn) = f(vn)+(−1) δ = f(vn)+(−1) δ; since (v0,...,vn) and (v0, v2n, v2n−1,...,vn) are paths of length n and n + 1 joining v and vn. Therefore, δ = g(v) − f(v) = 0, and g, f coincide on [v]. If on the other hand [v] does not contain any cycle of odd length then one has one degree of freedom to alter f on [v] without leaving P (G); this is the subject of the next lemma.

1.146 Lemma. Let f ∈ EX, and let {[v1],..., [vs]} be the path components of G(f) that do not contain any cycle of odd length. Then there is ε > 0 s s such that for every (t1,...,ts) ∈ [−ε,ε] ⊂ R there is g ∈ P (G(f)) with g(vj) − f(vj)= tj, for all 1 ≤ j ≤ s. Finite metric spaces 53

Proof. Let G(f)=(X, E) and set 1 ε := min fj + fk − dX (j, k). 2 {j,k}∈E/

Then ε> 0, since by definition if {j, k} ∈E/ then fj + fk − dX (j, k) > 0. Let s l (t1,...,ts) ∈ [−ε,ε] . Furthermore, δ(x) := (−1) where x ∈ [vi] and l is the length of a path from vi to x; note that δ(x) is well-defined since [vi] does not contain any cycle of odd length. Define

f(x) if x∈ / s [v ] g(x) := i=1 i f(x)+ δ(x) ti if x ∈ S[vi]. Then g(x) ≥ f(x) − ε, for all x ∈ X. Therefore, if {j, k} ∈/ E we have gj + gk ≥ fj + fk − 2ε ≥ dX (j, k). On the other hand if {j, k}∈E and j ∈ [vi] for some 1 ≤ i ≤ s, then k ∈ [vi] and δ(k) = −δ(j). Hence g(j)+ g(k) = f(j)+ δ(j) ti + f(k) − δ(j) ti = f(j)+ f(k) = dX (j, k); the case where j, k ∈ [x] = [vi], for all 1 ≤ i ≤ s, is trivial. It follows that g ∈ P (G(f)) with g(vi) − f(vi)= ti.

1.147 Deﬁnition. For an extremal function f ∈ EX we deﬁne n(f) ∈ N0 to be the number of path components of G(f) that do not contain any cycle of odd length.

1.148 Remark. If f ∈ i(X), that is f = ix for some x ∈ X, then fx + fy = dX (x, x)+ dX (x, y) = dX (x, y), for all y ∈ X. Therefore {x, y} ∈ G(f) for all y ∈ X; in other words there is only one path component, [x], in G(f). Furthermore, [x] has a cycle, (x, x), of length 1 since {x, x}∈E; hence n(f) = 0.

1.149 Proposition. Let (V, E)= G = G(fG) ∈A(X).

N (i) P (G) ⊂ R∞ is a polyhedron of dimension n(fG) and fG is an inner point of P (G). On the other hand if g is an inner point of P (G) then G(g)= G; in particular P (G(g)) = P (G).

(ii) The aﬃne subspace

N XE := g ∈ R∞ | gi + gj = dX (i, j), {i, j}∈E corresponds to XP (G), the hyperplane generated by P (G). Furthermore, we have that XP (G) ∩ EX = P (G). (iii) All faces of P (G) are exactly of the form P (G′) where G′ ∈A(X) and G ≺ G′. 54 Chapter 1. Isbell’s Injective Hull

N (iv) P (G) ⊂ R∞ with the induced metric is injective.

Proof. (i) We know from Lemma 1.143 that P (G) ⊂ EX; thus by 1.64 it N is a bounded subset of R∞ and therefore compact. On the other hand, by deﬁnition, P (G) is the intersection of a family of half-spaces, compare (1.141) and (1.142); hence P (G) is a polyhedron. Let s := n(fG), and let {[v1],..., [vs]} be the path components of G which do not con- s tain any cycle of odd length and Xs := X \ i=1[vi]. Furthermore, let N s N πs : R∞ → R∞ denote the projection πs(f) := Sf(v1),...,f(vs) , f ∈ R∞. s Then, π := πs|P (G) : P (G) → R∞ is injective by 1.144 and so π is an aﬃne homeomorphic embedding. Note that π(P (G)) ⊂ C1 ∩ C2 where

s s C1 := {t ∈ R∞ | tk + tl ≥ dX (vk, vl)} k,l\=1 and s s C2 := {t ∈ R∞ | tk ≥ dX (vk, x) − fG(x)}; k\=1 x\∈Xs the former follows directly from (1.142) whereas the latter uses (1.142) and the fact that every g ∈ P (G) coincides with fG on Xs ⊂ X by Remark 1.144. However π(P (G)) and C1 ∩ C2 do not coincide in general. Because of Lemma 1.146 we have B(fG,ε) ⊂ π(P (G)) for some ε > 0; s B(fG,ε) is the closed ball in R∞. Hence, fG is an inner point of π(P (G)) and therefore of P (G). Let on the other hand g be an inner point of P (G). Then by (1.141) we have that G ≺ G(g); we need to show that G(g) ≺ G, i.e. {k,l} ∈/ G ⇒{k,l} ∈/ G(g). But since g is also an inner point of π(P (G)) and π(P (G)) ⊂ C1 ∩ C2, it follows that g is an inner point of C1 and C2. In other words g(vk)+g(vl) >dX (vk, vl) and g(vk)+g(x)= g(vk)+fG(x) >dX (vk, x) for 1 ≤ k,l ≤ s and all x ∈ Xs; thus {vk, x}, {vk, vl} ∈/ G(g). Furthermore, because g|Xs = fG|Xs , we have for x, y ∈ Xs that {x, y} ∈/ G(g) whenever {x, y} ∈/ G. But, on the other hand the arguments above do not depend on the particular choice of the representative vi of [vi]; therefore G(g) ≺ G and in particular G(g)= G. s (ii) Let πs be the projection onto R∞ defined in the proof of (i). Then it follows by Remark 1.144 that πs|XE is injective. The map πs|XE is also s surjective since P (G) ⊂ XE and πs(P (G)) ⊂ R∞ has non-empty interior; compare the proof of (i). Therefore, dim P (G) = dim XE and in particular XP (G) = XE since XP (G) is the smallest affine subspace containing P (G). Let g ∈ EX ∩ XE . Then by the definition of XE we have gi + gj = dX (i, j) Finite metric spaces 55

for all {i, j}∈E. On the other hand g ∈ EX; hence gk + gl ≥ dX (k,l) for {k,l} ∈E/ . In other words g ∈ P (G) and we get that P (G)= EX ∩ XE . (iii) Let F ⊂ P (G) be a proper face and g ∈ F . Then, G G(g) and P (G(g)) ∩ int P (G) = ∅; otherwise using (i) we would have G = G(g) and g ∈ int P (G) which contradicts the assumption that g ∈ F . It follows that P (G(g)) ⊂ F for every g ∈ F . Since there are only ﬁnitely many m X-admissible graphs we get F = i=1 P (G(gi)), gi ∈ F . Hence, there is a P (G(gj)), 1 ≤ j ≤ m, with non-emptyS interior in F . Therefore, XF =

XP (G(gj )) and

E E (ii) F ⊂ X ∩ XF = X ∩ XP (G(gj )) = P (G(gj)).

Hence, F = P (G(gj)) and the claim follows. (iv) This follows directly from 1.131 and the definition of P (G). We have seen that the injective hull EX of a finite metric space X can N be decomposed into finitely many polyhedral cells P (G) ⊂ R∞, G ∈A(X), that is EX = P (G). (1.150) G∈A[(X) The cell structure on (1.150) is given by the relation, P (G′) is a face of P (G) if and only if G ≺ G′. Moreover, all the cells P (G) are isometrically embedded into EX, and EX is isometric to the quotient metric structure given by the polyhedral decomposition (1.150) as defined in [5, I.7.4]. The latter follows easily since EX is a geodesic metric space.

Chapter 2

Groups and their Injective Hull

In this thesis we are basically interested in the question when a group with a distinguished set of generators acts nicely on the injective hull of the according Cayley graph in the sense that it acts properly discontinuous and cocompactly by isometries; cf. Appendix B. We are going to give a necessary but not sufficient algebraic condition by means of the Dehn function of the group; the latter measures the complexity of the word problem in the group. But first we need to fix some notations and definitions, and state some well known facts from geometric group theory; compare [4] for a more comprehensive introduction.

Let P ≡S|R be a presentation of a group Γ, i.e. there is a short exact sequence 0 → N(R) −→i F (S) −→π Γ → 0 (2.1) where F (S) is the free group generated by S ⊂ Γ, R is a set of words in the alphabet S¯ := S ∪ S−1 and N(R) is the normal closure of R in F (S). The elements of S are called generators and the words in R are called relators. We say that Γ is ﬁnitely generated if |S| < ∞; it is said to be ﬁnitely presented if |S|, |R| < ∞. A word in the alphabet S¯ is said to be reduced if it doesn’t contain any subword of the form ss−1 or s−1s, s ∈ S. The free group F (S) is the group of all reduced words in S¯. Let w1 and w2 be two reduced words in S¯; we write Γ w1 ≡ w2 :⇔ π(w1)= π(w2). (2.2) Γ A word w in S¯ is said to be null-homotopic if w ≡ 1. Since the sequence (2.1) is exact we have kern(π)= N(R); therefore

N F (S) −1 w is null-homotopic ⇔ w ≡ ci rici ∈ N(R) (2.3) Yi=1 58 Chapter 2. Groups and their Injective Hull

±1 for some ri ∈ R and ci ∈ F (S), 1 ≤ i ≤ N. Note that the equation on the right-hand side just means that both words have the same freely reduced form in the sense that, after successively canceling subwords of the form aa−1 and a−1a on either side, we get the same words. 2.4 Remark. If P ≡S|R is a presentation for Γ, then

−1 P1 ≡S1|R1 :≡ S ∪{g}| R ∪{g a1 ...as} ,

Γ where g ∈ Γ with g ≡ a1 ...as and aj ∈ S¯ for all 1 ≤ j ≤ s, and

P2 ≡S2|R2 :≡ S | R ∪ R¯ , where R¯ is a set of null-homotopic words in S¯, are presentations for the same group Γ. 2.5 Deﬁnition. Let P ≡S|R be a presentation of Γ. The word length of a word w = a1 ...as over the alphabet S¯ is

ℓ(w) := s.

The combinatorial area of a null-homotopic word w in P is given by

N F (S) −1 ± AreaP (w) := min N ∈ N | w ≡ ci rici, ci ∈ F (S) ,ri ∈ R . (2.6) Yi=1

Note that by equation (2.3) we always have AreaP (w) < ∞. The Dehn function of the presentation P is the function δP : [0, ∞) → [0, ∞) given by

δP (x) := max AreaP (w) | w is null-homotopic and ℓ(w) ≤ x . Hence, δP (x)= δP (n) whenever n ≤ x < n + 1 since ℓ(w) ∈ N.

Although the Dehn function δP depends on the selected set of generators and relators, the growth of δP however doesn’t; it is an invariant of the group Γ itself. We are going to make this precise in the following. Let f1, f2 : [0, ∞) → [0, ∞) be two monotonic increasing functions. Then f2 is said to grow faster than f1, we write f1 f2, if there exists a constant C > 0 such that f1(x) ≤ Cf2(Cx + C)+ Cx + C, ∀x ≥ 0. (2.7)

The functions f1 and f2 are said to have the same growth if f1 f2 and f2 f1; in this case we write f1 ≃ f2. An easy calculation shows that ≃ is an equivalence relation on the set of monotonic increasing functions on Cayley graphs 59

[0, ∞). The linear function x is a minimal element since x f for every f; take for instance C = 1. Note that the last linear term, Cx + C, in (2.7) is needed to ensure that Dehn functions of two presentations of a group are equivalent in this sense; see (C.4). For instance P = a| and P ′ = a, b|b are both presentations of Z whereas δP (x) = 0 and δP ′ (x)= x.

2.8 Remark. In order to verify that f1 f2 it is suﬃcient to ﬁnd constants A, B, C, D and E ∈ N such that

f1(x) ≤ Af2(Bx + C)+ Dx + E.

This is true since f2 and x → Bx + C are both monotonic increasing.

2.9 Proposition. Let P ≡ S|R and P ′ ≡ S′|R′ be two ﬁnite presentations of a group Γ. Then δP ≃ δP ′ .

Proof. See [4] or Appendix C for a proof.

By δΓ we mean the equivalence class [δP ] of a finite presentation P of Γ with respect to the relation ≃. We say that Γ satisfies a quadratic isoperi- 2 metric inequality if δΓ [x ]. We shall see in Section 2.3 that the geometry of an injective metric space enforces any group acting nicely on it to be combable by quasi-geodesics. In this case the group itself must be finitely presentable and must satisfy a quadratic isoperimetric inequality. However, our main focus will be on groups with a linear Dehn function in the sense that δΓ = [x]; hyperbolic groups.

2.1 Cayley graphs

There is a natural metric graph associated to every ﬁnitely generated group, on which the group acts nicely. The graph depends on the set of generators S and is called the Cayley graph of the group with respect to S. The aim of this section is to introduce the Cayley graph and list some facts.

2.10 Definition. Let Γ be a finitely generated group with a distinguished set of generators S ⊂ Γ; we write (Γ,S). The Cayley graph CΓ(S) with respect to S is the simplicial metric graph (V, E) with vertex set V := Γ and −1 edges, (g, h) ∈ E ⊂ Γ × Γ:⇔ g h ∈ S; edges in CΓ(S), by definition, have all length one. Let dS denote the simplicial metric on CΓ(S). Then CΓ(S) 60 Chapter 2. Groups and their Injective Hull is a proper geodesic metric space since all edges have length one and S is finite; see [5, 1.9]. By ΓS we mean the 0-skeleton of CΓ(S), i.e. the vertex set Γ, endowed 1 with the induced metric dS; then dH (ΓS,CΓ(S)) = 2 in CΓ(S). Moreover if γ ∈ ΓS and g ∈ CΓ(S) with dS(γ,g) ∈ N, then g ∈ ΓS; this can be easily seen for instance by induction over the length dS(γ,g).

2.11 Remark. The restriction of the metric dS on ΓS can be given in a simple algebraic way. Let

|γ|S := min{ℓ(w) | π(w)= γ,w ∈ F (S)}, (2.12)

γ ∈ Γ, where π is the projection from F (S) onto Γ deﬁned in (2.1). Then −1 for any γ, γ¯ ∈ ΓS the identity dS(γ, γ¯)= |γ γ¯|S holds; in particular |γ|S = dS(e, γ) where e is the neutral element of Γ. Note that

−1 −1 |γγ¯|S = dS(γ , γ¯) ≤ dS(γ , e)+ dS(e, γ¯)= |γ|S + |γ¯|S and −1 |γ |S = dS(γ, e)= dS(e, γ)= |γ|S, −1 for all γ, γ¯ ∈ Γ. Furthermore, by (2.12) we have |γ|S =1 ⇔ γ ∈ S¯ := S∪S .

If it is apparent we shall drop the index S and just write |γ| for |γ|S.

2.13 Deﬁnition. A discrete geodesic segment in ΓS, parametrized by arc- length, is an isometric embedding c: {0,...,N}→ ΓS; note that −1 −1 ¯ c(i − 1) c(i) S = dS c(i − 1),c(i) = 1 and si := c(i − 1) c(i) ∈ S.

Furthermore, we deﬁnec ¯ to be the word s1s2 ...sN ∈ F (S) over the alphabet ¯ −1 S. Then c0 π(¯c) = c0 s1 ...sN = c0c0 c1s2 ...sN = c1s2 ...sN = = cN , hence −1 cN = c0π(¯c), c0 = cN π(¯c ).

Moreover, ℓ(¯c)= N = dS c(0),c(N) . 2.14 Lemma. Let c: [0, N] → CΓ(S) be a geodesic segment parametrized by arc-length in CΓ(S) with endpoints c(0),c(N) ∈ ΓS. Then the restriction of c to [0, N] ∩ N is a discrete geodesic in ΓS. In particular, ΓS is a discrete geodesic metric space.

Proof. First of all N = dS c(0),c(N)) ∈ N since c(0),c(N) ∈ ΓS. We shall show that c(i) ∈ ΓS for every 0 ≤ i ≤ N and i ∈ N. But this is true since dS c(0),c(i) = i ∈ N and c(0) ∈ ΓS. Cayley graphs 61

′ 2.15 Lemma. Let S,S ⊂ Γ be two ﬁnite sets of generators. Then, ΓS and ΓS′ are bilipschitz equivalent by means of the identity map idΓ : ΓS → ΓS′ . ′ ′ ′ Proof. Let N := max{|s |S | s ∈ S }. Then |γ|S ≤ N |γ|S′ for all γ ∈ Γ. We show this assertion by induction over the length n := |γ|S′ . The case where n = 0 is trivial; suppose |γ|S′ = n + 1. Then by (2.12) it follows that ′ ′ ′ ′ ′ ′ ′ ′ ′ γ = s γ with |γ |S′ = n and s ∈ S¯ . Hence, |γ|S = |s γ |S ≤ |s |S + |γ |S ≤ ′ ′ N + N |γ |S′ = N(1 + |γ |S′ )= N |γ|S′ and in particular ′ −1 ′ −1 ′ ′ dS(γ,γ )= |γ γ |S ≤ N|γ γ |S′ = NdS′ (γ,γ ), for all γ,γ′ ∈ Γ. The identity on Γ is therefore an N-Lipschitz map from ′ ΓS′ into ΓS. By reversing the roles of S and S we then get that idΓ is a bilipschitz equivalence between ΓS and ΓS′ .

2.16 Lemma. Γ operates on ΓS by isometries via Lg : ΓS → ΓS where Lg(h) := gh. Lg is called the left multiplication on Γ by g. −1 Proof. The map Lg : ΓS → ΓS is surjective since h = Lg(g h) for every h ∈ Γ. Furthermore, −1 −1 −1 dS Lg(γ), Lg(¯γ) = dS(gγ,gγ¯)= |γ g gγ|S = |γ γ¯|S = dS(γ, γ¯), for γ, γ¯ ∈ Γ; hence Lg is an isometry. Moreover, L: Γ → Isom(ΓS) is a group ′ ′ homomorphism since Lgg′ (γ)=(gg )γ = g(g γ)= Lg(Lg′ (γ)).

2.17 Remark. Γ operates transitively on ΓS by left multiplication since for any two elements, γ, γ¯ ∈ Γ, we have Lγγ¯−1 (¯γ) = γ; hence ΓS is a uniform metric space, see page 30.

2.18 Lemma. The isometry Lg : ΓS → ΓS can be extended in a unique way to an isometry L¯g : EΓS → EΓS. Furthermore, −1 L¯g(f) (h)= f(g h), f ∈ EΓS and h ∈ ΓS.

Proof. Let i: ΓS → EΓS denote the isometric embedding from 1.41 and g ∈ Γ. Then i ◦ Lg : ΓS → EΓS is an isometric embedding and hence there is ¯ ¯ ¯ − ¯ a 1-Lipschitz extension Lg : EΓS → EΓS, Lg ◦ i = i ◦ Lg. Since Lg 1 ◦ Lg ◦ i = ¯ − − ¯ − ¯ Lg 1 ◦ i ◦ Lg = i ◦ Lg 1 ◦ Lg = i it follows that the map Lg 1 ◦ Lg ﬁxes i(ΓS) ¯ −1 ¯ ¯ pointwise; hence Lg ◦ Lg = idEΓS by 1.43 (ii). In other words Lg is an isometry for every g ∈ Γ. On the other hand, let L¯ : EΓS → EΓS be another extension, i.e. L¯ ◦ ¯−1 ¯ i = i ◦ Lg. Then by the same argument as above we see that Lg ◦ L = −1 ¯ −1 ¯ ¯ ¯ Lg ◦ L ﬁxes the subset i(ΓS) pointwise; therefore Lg ◦ L = idEΓS and ¯ ¯ ¯ L¯ = L¯g. Furthermore, L¯g(f) (h)= d L¯g(f), ih = d f, L¯g−1 (ih) = d f, i ◦ ¯ −1 Lg−1 (h) = d f, ig−1h = f(g h) for all h ∈ Γ. 62 Chapter 2. Groups and their Injective Hull

2.19 Proposition. Γ acts on EΓS properly discontinuous via L¯.

Proof. Let L¯g : EΓS → EΓS be the isometry from 2.18 and i: ΓS → EΓS. We need to verify that L¯ : Γ → Isom(EΓS) is a group homomorphism. Let ′ g,g ∈ Γ; then L¯g′ ◦L¯g ◦i = i◦(Lg′ ◦Lg)= i◦Lg′g. In other words the map L¯g′ ◦ L¯g is an extension of Lg′g. Hence, by the uniqueness of the extension it follows that L¯g′g = L¯g′ ◦ L¯g. Let R> 0 and deﬁne

NR(e) := {g ∈ Γ | B(ie, R) ∩ L¯g(B(ie, R)) = ∅}; note that L¯g(ie)=(i ◦ Lg)(e) = ig and L¯g(B(ie, R)) = B(ig, R). Let g ∈ NR(e) and f ∈ B(ie, R) ∩ B(ig, R) = ∅. Then |g|S = d¯(ig, ie) ≤ d¯(ig, f)+ d¯(f, ie) ≤ 2 R; hence g ∈ B(e, 2 R) in ΓS. In particular

2R |NR(e)|≤|B(e, 2 R)| ≤ 2|S| < ∞, and the action L¯ is properly discontinuous; compare B.2.

2.2 Cone types

2.20 Deﬁnition. Let (Γ,S) be a ﬁnitely generated group. By the cone type of γ ∈ ΓS with respect to S we mean

coneS(γ) := a ∈ Γ |γa| = |γ| + |a| ; in other words γ lies in CΓ(S) on a geodesic segment from e to γa, for every a ∈ coneS(γ); e ≤ γ ≤ γa (recall 1.67). ′ ′ ′ 2.21 Lemma. Let a ∈ coneS(γ) with a = a c and |a| = |a | + |c|, a ,c ∈ Γ. ′ Then a ∈ coneS(γ). Proof. We have the following estimates, |γa′|≥|γa|−|c| = |γ| + |a|−|c| = |γ|+|a′|≥|γa′|. Therefore, equality holds and in particular |γa′| = |γ|+|a′|; ′ hence, a ∈ coneS(γ). 2.22 Proposition. Suppose that (Γ,S) is a ﬁnitely generated group with only ﬁnitely many cone types {Ci := coneS(γi) | 1 ≤ i ≤ n}. Let f ∈ EΓS. Then for every γ ∈ ΓS there is a cone type Cγ such that

f(γa)= f(γ)+ |a|, ∀a ∈ Cγ. In particular, on the subset

γCγ := {γa | a ∈ Cγ} ⊂ ΓS f is completely determined by the value f(γ). Combable groups 63

Proof. Let n ∈ N; then by 1.35 there is a point γn ∈ Γ such that 1 d (γ,γ ) ≤ f(γ)+ f(γ ) ≤ d (γ,γ )+ . (2.23) S n n S n n Since (Γ,S) has only ﬁnitely many cone types we may assume that −1 coneS(γn γ) is constant, say Cγ, for all n; otherwise we choose a constant −1 subsequence. For a ∈ Cγ we have the identity dS(γa,γn) = dS(γn γa, e) = −1 −1 |γn γa| = |γn γ| + |a| = dS(γ,γn)+ dS(γ,γa). Therefore

dS(γ,γn)+ dS(γ,γa) = dS(γa,γn)

≤ f(γa)+ f(γn)

(f is 1-lipschitz) ≤ f(γ)+ dS(γ,γa)+ f(γn) 1 (2.23) ≤ d (γ,γa)+ d (γ,γ )+ ; S S n n hence 1 d (γ,γ )+ d (γ,γa) ≤ f(γa)+ f(γ ) ≤ d (γ,γa)+ d (γ,γ )+ . S n S n S S n n By combining this with (2.23) we then obtain 1 1 d (γ,γa) − ≤ f(γa) − f(γ) ≤ d (γ,γa)+ . S n S n

Since n ∈ N was arbitrary it follows that |a| = dS(γ,γa) = f(γa) − f(γ). This proves the claim.

2.3 Combable groups

It turns out that isometries and geodesics are too narrow notions to be useful in context of ﬁnitely generated groups. We need a more ﬂexible concept: that of quasi-isometries and quasi-geodesics; see [5],[13] for more details.

2.24 Deﬁnition. A not necessarily continuous map f : X1 → X2 between two metric spaces X1 and X2 is called a (λ,ε) quasi-isometric embedding, with λ ≥ 1 and ε ≥ 0, if 1 d (x, y) − ε ≤ d f(x), f(y) ≤ λd (x, y)+ ε, (2.25) λ 1 2 1 for all x, y ∈ X1. For the sake of convenience we shall omit the constants λ and ε if their values are not essential. A quasi-isometric embedding f is said to be a quasi-isometry if there is a constant D ≥ 0 such that

NX2 (f(X1),D)= X2; (2.26) 64 Chapter 2. Groups and their Injective Hull

(compare Deﬁnition (1.52)). A quasi-geodesic in X is a quasi-isometric embedding c : J → X of an interval J ⊂ R with the euclidean metric into X.

2.27 Example. Let (Γ,S) be a ﬁnitely generated group. The inclusion 1 e: ΓS → CΓ(S) is a (1, 0) quasi-isometry with D = 2 .

2.28 Remark. A quasi-isometric embedding f : X1 → X2 is a quasi- isometry if and only if there is a quasi-inverse to f in the sense that there is a map g : X2 → X1 with the property that

(i) d1 x, g ◦ f(x) ≤ C for all x ∈ X1, (ii) d2 y, f ◦ g(y) ≤ C for all y ∈ X2, for some constant C ≥ 0. Note that g is in this case a quasi-isometric embedding. One may define the quasi-inverse g : X2 → X1 of f as follows: Pick for every y ∈ X2 a point x ∈ X1 with d2 y, f(x) ≤ D, and set g(y) := x. Then g satisfies (i) and (ii) with C = max {D, λ D + λε}. 2.29 Remark. An easy calculation shows that the composition of two quasi- isometric embeddings is a quasi-isometric embedding, with different constants λ and ε. In particular if F : X1 → X2 is a (λ,ε) quasi-isometric embedding and c : J → X1 is a geodesic parametrized by arc-length, i.e. d1 c(s),c(t) = |t − s|, then F ◦ c : J → X2 is a (λ,ε) quasi-geodesic; this follows directly from (2.25). 2.30 Definition. Let (Γ,S) be a finitely generated group. A combing by (λ,ε) quasi-geodesics on ΓS is a family of (λ,ε) quasi-geodesics cγ : [0, Tγ] → ΓS, γ ∈ ΓS and Tγ ∈ N, with the property that

(i) cγ(0) = e and cγ(Tγ)= γ, we shall put cγ(t) := γ for t ≥ Tγ ,

(ii) there are constants k1 ≥ 1 and k2 ≥ 0 such that

′ dX cγ (t),cγ′ (t) ≤ k1 dS(γ,γ )+ k2 for all t ≥ 0. (2.31)

(Γ,S) is said to be combable if there is a combing by quasi-geodesics on ΓS. We say that a ﬁnitely generated group Γ is combable if ΓS is combable for some ﬁnite generating set S of Γ.

Even though Definition 2.30 depends on the generating set S one easily verifies that if (Γ,S) is combable then any other (Γ,S′) is combable for any finite generating set S′. For more details on combable groups see [13]. Combable groups 65

2.32 Remark. The requirement Tγ ∈ N is not a restriction; since whenever ′ Tγ ≥ 0 is not an integer one can extend cγ to the domain [0, Tγ] by setting ′ cγ(t) := γ for Tγ ≤ t ≤ Tγ := ⌈Tγ ⌉. One easily veriﬁes that the extended cγ is a (λ,λ + ε) quasi-geodesic and that condition (2.31) remains valid with the same constants k1 and k2. One may even relax the two requirements to

dS cγ (Tγ),cγ ≤ D, for all γ ∈ ΓS, and

dX cγ(t),cγ′ (t) ≤ k1 dS(cγ(Tγ ),cγ′ (Tγ′ )) + k2 for all t ≥ 0. (2.33) In this case cγ may be altered toc ¯γ (t) := cγ(t), for t < Tγ , andc ¯γ (t) := γ otherwise. Thenc ¯γ are all (λ, D + ε) quasi-geodesics and they satisfy (2.31) ¯ ¯ with new constants k1 := k1 and k2 := (k1 +1)2D + k2.

2.34 Example. Every ﬁnite group is combable. For instance take Tγ = 0 and cγ(0) := e for all γ ∈ Γ. 2.35 Theorem. Let Γ be a combable group. Then,

2 (i) Γ satisﬁes a quadratic isoperimetric inequality, δΓ [n ]. (ii) Γ is ﬁnitely presentable.

Proof. Let S ⊂ Γ be a ﬁnite generating set and π : F (S) → Γ as in (2.1). Suppose w := s1 ...sn ∈ F (S) is a null-homotopic word of length n and put γi := π(s1 ...si), for 0 ≤ i ≤ n; then γ0 = e and γn = π(w) = e, moreover

γi si+1 = γi+1. Let ci := cγi be the (λ,ε) quasi-geodesic from the combing on 1 ΓS from e to γi, and Ti := Tγi its domain. From λ Ti −ε ≤ dS ci(0),ci(Ti) = dS(e, γi) ≤ n we obtain an upper bound Ti ≤ λn + ε. For every ci and every integer 0 ≤ j ≤ Ti − 1 we choose a discrete geodesic segment di,j in ΓS from ¯ ci(j) to ci(j + 1); the word di,j ∈ F (S) is the word presentation of di,j, see Deﬁnition 2.13. Moreover, for every 0 ≤ k ≤ Ti set

k−1

c¯i(k) := d¯i,j ∈ F (S). (2.36) Yj=0 ¯ Then ℓ(di,j)= dS ci(j),ci(j + 1) ≤ λ + ε and by 2.13, k−1 k−1 −1 −1 π(¯ci(k)) = π(d¯i, j)= ci(j) ci(j +1) = ci(0) ci(k)= ci(k). Yj=0 Yj=0 66 Chapter 2. Groups and their Injective Hull

Deﬁne bi,j to be a discrete geodesic segment in ΓS from ci(j) to ci+1(j), 1 ≤ i ≤ n − 1 and 0 ≤ j ≤ Ti − 1, and let d¯i,j be the corresponding word ¯ ¯ presentation in F (S); for di,Ti we set si+1. For the length of di,j we get by (2.31) the estimate ℓ(d¯i,j) = dS ci(j),ci+1(j) ≤ k1 + k2. Now consider ′ −1 ′ the word w :=c ¯i(Ti) si+1 c¯i+1(Ti+1) ; it is null-homotopic since π(w ) = −1 γi si+1γi+1 = e, and can be written in F (S) as

0 0 ′ F (S) ¯ ¯ ¯−1 ¯−1 −1 −1 w ≡ c¯i(j) di,j bi,j+1 di+1,jbi,j c¯i(j) =: c¯i(j) ri,j c¯i(j) (2.37) jY=Ti jY=Ti

−1 ¯ ¯−1 ¯ This is true sincec ¯i(j) c¯i(j − 1) di,j−1 = di,j−1di,j−1 by (2.36); they cancel each other. In other words, most of the terms of (2.37) vanish, and in the end the right-hand side corresponds to

¯ ¯−1 ¯−1 ′ c¯i(Ti) di,Ti si di+1,Ti di+1,0 = w .

Note that

¯ ¯ ¯−1 ¯−1 ℓ(ri,j)= ℓ(di,j)ℓ(bi,j+1)ℓ(di+1,j)ℓ(bi,j ) ≤ 2(λ + ε + k1 + k2) =: K.

Furthermore, using 2.13 and the deﬁnitions of the geodesic segments di,j and bi,j we get

¯ ¯ ¯−1 ¯−1 ci(j)π(ri,j) = ci(j)π(di,j)π(bi,j+1)π(di+1,j)π(bi,j ) ¯ ¯−1 ¯−1 = ci(j + 1)π(bi,j+1)π(di+1,j)π(bi,j ) ¯−1 ¯−1 = ci+1(j + 1)π(di+1,j)π(bi,j ) ¯−1 = ci+1(j)π(bi,j )

= ci(j); hence π(ri,j) = 1. The words ri,j are null-homotopic and uniformly bounded by K. So set R′ := {r ∈ F (S) | π(r)= e, ℓ(r) ≤ K}; note that R′ is ﬁnite since |R′| ≤ (2|S|)K. We conclude from (2.37) that −1 every word of the formc ¯i(Ti) si+1 c¯i+1(Ti+1) can be written in F (S) as the ′ product of Ti + 1 conjugates of words in R . Sincec ¯0(T0) andc ¯n(Tn) are by deﬁnition the empty words we can rewrite

n−1 F (S) −1 w = s1 ...sn ≡ c¯i(Ti) si+1 c¯i+1(Ti+1) . Yi=0 Combable groups 67

n−1 Hence, w can be represented as the product of i=0 (Ti + 1) conjugates of ′ n−1 n−1 2 R . On the other hand i=0 (Ti + 1) ≤ i=0 (λnP+ ε + 1) ≤ Mn for some constant M ≥ 0. P P Suppose that P ≡ S|R is an arbitrary presentation of Γ; then S|R ∪ R′ is also a presentation of Γ by 2.4. On the other hand, every r ∈ R is null-homotopic and can be written as a product of conjugates of R′. We conclude by 2.4 that S|R = S|R ∪ R′ = S|R′; in other words P ′ ≡S|R′ is a ﬁnite presentation of Γ. Furthermore, every null-homotopic word w can be written as a product of at most Mℓ(w)2 conjugates of relators ′ 2 of R ; it follows that δΓ [x ].

2.38 Lemma. Suppose that c1 and c2 are two geodesics in a metric space X, parametrized by arc-length, both issuing from e ∈ X and with endpoints x1 and x2. If dH (c1,c2) ≤ Kd(x1, x2) for some K ≥ 0 then

d(c1(t),c2(t)) ≤ 2Kd(x1, x2), for all t ≥ 0.

Note that we have set c1(t) := x1, for t ≥ d(e, x1), and that dH (c1,c2) is + + by deﬁnition the Hausdorﬀ distance of c1(R0 ) and c2(R0 ) in X.

′ ′ Proof. Let t ≥ 0. There is t ≥ 0 with d(c1(t),c2(t )) ≤ Kd(x1, x2). Then

′ ′ ′ |t − t | = |d(e, c1(t)) − d(e, c2(t ))| ≤ d(c1(t),c2(t )) ≤ Kd(x1, x2),

′ ′ and we obtain d(c1(t),c2(t)) ≤ d(c1(t),c2(t )) + |t − t | ≤ 2Kd(x1, x2).

2.39 Proposition. A group Γ acting properly discontinuous and cocompactly on an injective metric space is combable.

Proof. Let X be the injective metric space on which Γ acts nicely. We may assume that X has more than one point; otherwise Γ would be finite and thus combable by 2.34. From Lemma B.4 and Proposition B.5 we know that Γ is finitely generated by some S ⊂ Γ and that ΓS is quasi-isometric to X; say by q : X → ΓS with constants λ,ε and D as in Definition 2.24. For every γ ∈ ΓS we choose a point xγ ∈ X such that dS γ, q(xγ) ≤ D, and set Tγ := dX (xe, xγ ). We may further assume that Tγ >0, for all γ = e. If this ′ ′ is not the case we may choose another xγ = xe with d(xe, xγ) ≤ 1; note that this can be done since X is geodesic and consists of more than one point. ′ ′ ′ Then dS(g(xγ),γ) ≤ dS(g(xγ),g(xγ)) + dS(g(xγ),γ) ≤ λ + ε + D =: D is again a uniform bound. 68 Chapter 2. Groups and their Injective Hull

Let c be a convex bicombing on X, cf. 1.27, and deﬁnec ¯γ : [0, Tγ] → X to bec ¯γ (t) := cxexγ (t/Tγ ). Thenc ¯γ is a geodesic segment, parametrized by arc-length, from xe to xγ . By the convexity of c,

′ ′ ′ dH (¯cγ, c¯γ ) ≤ max d cxexγ (t), xxex ′ (t) ≤ max td(xγ, xγ )= d(xγ, xγ ). t∈I γ t∈I

We conclude from Lemma 2.38 that d(¯cγ(t), c¯γ′ (t)) ≤ 2d(xγ, xγ′ ), for all t ≥ 0. Now set cγ : [0, Tγ] → ΓS,

cγ(t) := q c¯γ (t) . By 2.29 the cγ are all (λ,ε) quasi-geodesics. Furthermore, dS(γ,cγ(Tγ )) = ′ dS(γ, q(xγ )) ≤ D. Hence it remains to show that (2.33) holds. Let γ,γ ∈ ΓS. Then

dS cγ (t),cγ′ (t) = dS q c¯γ(t) , q c¯γ′ (t)

≤ λdS c¯γ(t),c¯γ′ (t) + ε

≤ 2λd(xγ, xγ′ )+ ε 2 2 ≤ 2λ d q(xγ ), q(xγ′ ) +2λ ε + ε 2 2 = 2λ dcγ(Tγ ),cγ′ (Tγ′ ) +2λ ε + ε. Therefore, cγ is a combing by quasi-geodesics on (Γ,S) by 2.32. 2.40 Corollary. A ﬁnitely generated group (Γ,S) acting properly discontinuous and cocompactly on EΓS must be ﬁnitely presentable and must satisfy a quadratic isoperimetric inequality.

2.41 Example. By a result of Gersten, Holt and Riley, cf. [14], a nilpotent group Γ of degree c has a polynomial Dehn function of degree c + 1, δΓ = [xc+1]. For instance the Heisenberg group H, the group of 3 × 3 upper triangular matrices over Z with ones on the diagonal, has degree 2 and 3 therefore δH = [x ]. As a consequence, a nilpotent group of degree c ≥ 2 can not act properly discontinuous and cocompactly on its injective hull. However an abelian group satisﬁes a quadratic isoperimetric inequality because every abelian group is nilpotent of degree c = 1.

2.42 Example. The special linear group Γ = SL(n, Z) is not combable for n ≥ 3, cf. [13]. Therefore, Γ does not operate nicely on ΓS for any ﬁnite generating set S ⊂ Γ.

Though 2.40 is a necessary condition for a group to act nicely on its injective hull, it is not suﬃcient. Take for instance the abelian group Γ = Hyperbolic groups 69

n Z with the standard generating set S := {e1,...,en}. Then Γ satisﬁes a n n n quadratic isoperimetric inequality and ΓS = Z1 ⊂ R1 , the lattice in R1 E ∼ 2n−1 with the induced ℓ1-metric; note that ΓS = R∞ due to Example 1.99. n Moreover, Γ acts nicely on R1 by means of Lz(x) := (x1 + z1,...,xn + zn), n n n z ∈ Z and x ∈ R1 ; hence ΓS and R1 are quasi-isometric by B.5. If on E n the other hand Γ acts properly discontinuous and cocompactly on Z1 then E n E ∼ 2n−1 ΓS, ΓS are quasi-isometric. We conclude that R1 and ΓS = R∞ must n be quasi-isometric. But this is not possible for n > 2 since asdim R1 = n, 2n−1 n−1 whereas asdim R∞ = 2 . The asymptotic dimension is a concept due to Gromov and is a quasi-isometry invariant; it is a coarse version of the covering dimension in topology, cf. [3].

2.4 Hyperbolic groups

Although Gromov hyperbolicity can be deﬁned on a broader class of metric spaces, cf. [5], the one for geodesic metric spaces will be suﬃcient for our purposes. So let X be a geodesic metric space. For x, y, z ∈ X we put

1 (x y)z := 2 (d(x, z)+ d(y, z) − d(x, y)); note that 0 ≤ (x y)z ≤ min{d(x, z),d(y, z)} and (x y)z +(z y)x = d(x, z). We write again x ≤ y′ ≤ y when y′ lies on a geodesic segment in X from x to y. Then X is said to be δ-hyperbolic if there exists a constant δ ≥ 0 such that d(x′,y′) ≤ δ ′ ′ ′ ′ whenever d(x , z)= d(y , z) ≤ (x y)z; z ≤ x ≤ x and z ≤ y ≤ y. The property to be hyperbolic is a quasi-isometry invariant among geodesic metric spaces in the sense of Proposition 2.43. This makes the concept of hyperbolicity suitable in the context of ﬁnitely generated groups; compare Deﬁnition 2.44.

2.43 Proposition. Suppose that X and Y are geodesic metric spaces and quasi-isometric. Then, X is hyperbolic if and only if Y is hyperbolic.

Proof. See [5, III.1.9].

2.44 Definition. We call a finitely generated group (Γ,S) hyperbolic if the Cayley graph CS(Γ), as a geodesic metric space, is hyperbolic. From 2.27 ′ and Proposition 2.43 we conclude that in this case CΓ(S ) is hyperbolic for any other finite generating set S′ of Γ. We may thus say that Γ is hyperbolic. 70 Chapter 2. Groups and their Injective Hull

The next proposition shows that hyperbolic groups satisfy the necessary condition 2.40 in order to act properly discontinuous and cocompactly on their injective hull.

2.45 Proposition. Every hyperbolic group is combable.

Proof. Suppose that CΓ(S) is δ-hyperbolic. For γ ∈ ΓS choosec ¯γ to be a geodesic segment in CΓ(S), parametrized by arc-length, from e to γ; we set ′ Tγ := dS(e, γ). Let 0 ≤ t ≤ Tγ. If0 ≤ t ≤ T := (γ γ )e then by the δ-hyperbolicity of CΓ(S) we obtain that dS(¯cγ(t), c¯γ′ (t)) ≤ δ. Now suppose ′ T ≤ t ≤ Tγ ; choose a geodesic d from γ to γ . Since CΓ(S) is δ-hyperbolic ′ and Tγ − t ≤ Tγ − T =(γ e)γ we get that

−1 dS c¯γ(t),d(Tγ − t) = dS c¯γ (Tγ − t),d(Tγ − t) ≤ δ; ′ ′ ′ hence dS(¯cγ (t),γ ) ≤ dS(¯cγ(t),d(Tγ − t)) + dS(d(Tγ − t),γ ) ≤ δ + dS(γ,γ ). ′ ′ We conclude that dH (¯cγ , c¯γ′ ) ≤ (δ + 1) dS(γ,γ ) for all γ,γ ∈ ΓS. Then, by 2.38 it follows that

′ dS(¯cγ(t), c¯γ′ (t)) ≤ 2(δ + 1) dS(γ,γ ) for all t ≥ 0.

Let q : CΓ(S) → ΓS bea(λ,ε) quasi-inverse to the inclusion e: ΓS → CΓ(S); compare 2.27. Deﬁne a combing cγ on ΓS by cγ(t) := q(¯cγ(t)), 0 ≤ t ≤ Tγ . Then cγ is a (λ,ε) quasi-geodesic by Remark 2.29 and dS cγ (Tγ),γ = dS(q(γ),γ) ≤ D for some constant D. Furthermore,

dS(cγ (t),cγ′ (t)) = dS q(¯cγ(t)), q(¯cγ′ (t))

≤ λdS(¯cγ(t), c¯γ′ (t)) + ε ′ ≤ 2λ(δ + 1) dS(γ,γ )+ ε, for t ≥ 0. This proves that c is a combing on ΓS. As a corollary of Theorem 2.35 we obtain:

2.46 Corollary. Every hyperbolic group is ﬁnitely presentable and satisﬁes a quadratic isoperimetric inequality.

One can even say more. The next remarkable fact is due to M. Gro- mov [18].

2.47 Theorem. A ﬁnitely generated group Γ is hyperbolic if and only if Γ has a linear Dehn function i.e. δΓ = [x]. Proof. See [5], [18] for a proof. Hyperbolic groups 71

The following important property of hyperbolic groups is due to J. Can- non; see [8], [5]. We will use it in Section 2.6 in order to show that the injective hull of a subclass of hyperbolic groups is a locally ﬁnite polyhedral complex.

2.48 Deﬁnition. The k-tail of γ ∈ ΓS is deﬁned as the set k-tail(γ) := a ∈ Γ |a| ≤ k and |γa| < |γ| . 2.49 Theorem. Let (Γ,S) be a δ -hyperbolic group. The cone type of an element γ ∈ ΓS is completely determined by its (δ + 1)-tail(γ).

′ ′ Proof. Let γ,γ ∈ ΓS with (δ + 1)-tail(γ)=(δ + 1)-tail(γ ). We prove the claim by induction over the length of elements of coneS (γ). The case n = 0 is trivial; the only element of length 0 is the neutral element ′ e and e ∈ coneS(γ) ∩ coneS(γ ). Now suppose

′ ′ ′ a ∈ coneS(γ) ⇔ a ∈ coneS(γ ), (2.50)

′ ′ for all |a | ≤ n. Let a ∈ coneS (γ) with |a| = n + 1; then a = a s with s ∈ S ′ ′ and |a | = n; compare (2.12). Due to Lemma 2.21 we have a ∈ coneS (γ) ′ ′ and therefore a ∈ coneS(γ ) by induction hypothesis (2.50). Suppose that ′ ′ ′ ′ ′ ′ ′ ′ a∈ / coneS (γ ); then |γ a |≥|γ a|≥|γ a | − 1 = |γ | + n − 1. Let c be a ′ ′ ′ geodesic in CΓ(S) from e to γ a, and c be a geodesic from e to γ a with c(|γ′|) = γ′; both parametrized by arc-length. Note that the latter exists ′ ′ ′ ′ ′ since a ∈ coneS (γ ) and therefore e ≤ γ ≤ γ a . Because 1 (γ′a′ γ′a) = (|γ′a′| + |γ′a| − d (γ′a′,γ′a)) ≥|γ′| + n − 1 ≥|γ′| e 2 S ′ ′ ′ ′ ′ ′ and Γ is hyperbolic we have dS c(|γ |),c (|γ |) ≤ δ. Letγ ¯ := c (|γ | − 1); ′ thenγ ¯ ∈ ΓS and

′ ′ ′ ′ ′ dS(¯γ ,γ ) ≤ dS c (|γ |),c(|γ |) +1 ≤ δ +1 |γ¯′| = |γ′| − 1 < |γ′|.

In other words d := γ′−1γ¯′ ∈ (δ + 1)-tail(γ′). Hence, d ∈ (δ + 1)-tail(γ) and in particular |γd| < |γ|. Leta ¯ := d−1a =γ ¯′−1γ′a; then

|γ¯′| +1+ n = |γ′| + n = |γ′a′|≥|γ′a| = |γ¯′| + |γ¯′−1γ′a| = |γ¯′| + |a¯|.

Therefore, |a¯| ≤ n + 1 and

|γ| + n +1= |γa| = |γda¯|≤|γd| + |a¯| < |γ| + n +1.

′ This gives a contradiction to the assumption; hence a ∈ coneS (γ ). 72 Chapter 2. Groups and their Injective Hull

2.51 Corollary. Let Γ be a hyperbolic group and S ⊂ Γ a ﬁnite generating set. Then, ΓS has only ﬁnitely many cone types.

Proof. Because of Theorem 2.49 the cone type of an element γ ∈ ΓS is determined by its (δ + 1)-tail(γ). But on the other hand (δ + 1)-tail(γ) ⊂ |B(e,δ+1)| B(e, δ + 1) in ΓS. Hence, there are at most |P(B(e, δ +1))| =2 < ∞ diﬀerent cone types in ΓS.

2.5 Axiom Y

We are going to do some preliminary work for the next section. We shall introduce the concept of Axiom Y, which plays a crucial role in our main result. 2.52 Definition. We say that a metric space X satisfies Axiom Y(s), s> 0, if there is R(s) > 0 with the property that for any triple of points x, z′, z ∈ X with d(z, z′) ≤ s there exists y = y(x, z, z′) ∈ X such that d(z,y) ≤ R(s) d(x, z) = d(x, y)+ d(y, z) d(x, z′) = d(x, y)+ d(y, z′), i.e. x ≤ y ≤ z and x ≤ y ≤ z′; compare Definition 1.67. We say that X satisfies Axiom Y if it satisfies Axiom Y(s) for every s> 0. The next technical lemma is an iterated version of the property of Axiom Y. 2.53 Lemma. Suppose that X satisfies Axiom Y. Then for every natural number N ≥ 2 and every r > 0 there is an R(N,r) > 0 with the property that whenever x, z ∈ X and F ⊂ B(z,r) is a finite set with z ∈ F and |F | ≤ N, there exists a point y ∈ X such that d(y, z) ≤ R(N,r) and

d(x, z′)= d(x, y)+ d(y, z′), i.e. x ≤ y ≤ z′, for all z′ ∈ F . Proof. The proof goes by induction over N. If N = 2, the result holds with R(2,r) := R(r) since X satisﬁes Axiom Y. Now let N ≥ 2, and suppose that the claim holds for N. Let r > 0 and x, z ∈ X, and let F ⊂ B(z,r) be a ﬁnite set with z ∈ F and |F | ≤ N + 1. By Axiom Y, for every z′ ∈ F \{z}, there is a point yz′ such that d(yz′ , z) ≤ R(r),

′ x ≤ yz′ ≤ z and x ≤ yz′ ≤ z . (2.54) Injective hull of hyperbolic groups 73

′ ′ ′ Let E := yz′ | z ∈ F \{z} and choose y ∈ E; note that E ⊂ B(y , 2R(r)) and |E| ≤N. Hence, by the induction hypothesis, there is a point y ∈ X ′ ′ such that d(y,y ) ≤ R(N, 2R(r)) and x ≤ y ≤ yz′ for all z ∈ F \{z}. It follows that

d(y, z) ≤ d(y,y′)+ d(y′, z) ≤ R(N, 2R(r))+ R(r) =: R(N +1,r).

Moreover, by 1.68 and (2.54) we get x ≤ y ≤ z′ and x ≤ y ≤ z. This proves the claim.

2.55 Lemma. Let (X,d) be a geodesic δ-hyperbolic metric space. If X satisﬁes Axiom Y(δ) then X satisﬁes Axiom Y.

Proof. Let s> 0 and R′(s) := R(δ)+ s where R(δ) is as in Deﬁnition 2.52. Suppose we have x, z, z′ ∈ X given with d(z, z′) ≤ s. Let c and c′ be geodesic segments from x to z and z′ respectively, parametrized by arc-length. Let ′ ts := (z z )x and note that

1 d(x, z) − t = (d(x, z) − d(x, z′)+ d(z, z′)) ≤ d(z, z′) ≤ s. s 2

′ ′ ′ As X is δ-hyperbolic we get that d(¯y , y¯) ≤ δ withy ¯ := c(ts) andy ¯ := c (ts). Since X satisﬁes Axiom Y(δ) there is yδ ∈ X such that

d(¯y,yδ) ≤ R(δ)

x ≤ yδ ≤ y¯ ′ x ≤ yδ ≤ y¯ .

′ Now d(z,yδ) ≤ d(z, y¯)+ d(¯y,yδ) ≤ s + R(δ) = R (s). Since x ≤ yδ ≤ y¯ and ′ x ≤ y¯ ≤ z, we have x ≤ yδ ≤ z (cf. Lemma 1.68). Similarly, x ≤ yδ ≤ z . In ′ other words, X satisﬁes Axiom Y(s) with y = yδ and R (s) = R(δ)+ s for every s> 0.

2.6 Injective hull of hyperbolic groups

In this section we are going to study to what extent a hyperbolic group acts nicely on EΓS, for some finite generating set S ⊂ Γ. We show that hyperbolic groups (Γ,S) satisfying Axiom Y act nicely on EΓS. Moreover, in this case EΓS can be given the structure of a finite-dimensional and locally finite polyhedral complex, in a way similar to Section 1.9, on which Γ acts by cellular isometries. 74 Chapter 2. Groups and their Injective Hull

2.56 Remark. If a group Γ acts nicely on EΓS then by B.3 there is an R > 0 such that Γ B(ie, R) = EΓS; e ∈ ΓS is the neutral element of Γ. Hence for every extremal function f ∈ EΓS there is g ∈ ΓS such that f ∈ g B(ie, R) = B(ig, R), in other words d¯ f, i(ΓS) ≤ R. Thus a necessary condition for Γ to act nicely on EΓS is that dH (EΓS, i(ΓS)) < ∞ in EΓS. The next proposition proves this for Gromov hyperbolic groups.

2.57 Proposition. Suppose X is a geodesic metric space and δ-hyperbolic. Then, d¯ f, i(X) ≤ 2δ for all f ∈ EX. Proof. Let f ∈ EX. We deﬁne r ≥ 0 as the inﬁmum of all r′ ≥ 0 for which there exists a z ∈ X such that f(x) ≥ d(x, z) − r′ for all x ∈ X. Our aim is to show that r ≤ 2δ. Let ε> 0. Then we choose z ∈ X such that

f(x) ≥ d(x, z) − r − ε for all x ∈ X, and set E := {x ∈ X : f(x)

f(y) ≥ d(y, z) − r − ε ≥ d(y, z¯) − r + ε.

On the other hand, for all w ∈ X \ E,

f(w) ≥ d(w, z) − r + δ +3ε ≥ d(w, z¯) − r + ε.

This contradicts the deﬁnition of r and proves that there exist x, y ∈ E with (x y)z < δ +2ε. Then

f(x)+ f(y) 0 is arbitrary, we have r ≤ 2δ. Now let r′ > 2δ and choose z′ ∈ X such that f(x) ≥ d(x, z′)−r′ for all x ∈ X. We deﬁne g : X → R so that g(z′)= r′ and g(x)= f(x) for x ∈ X\{z′}; then g(x)+g(z′)= f(x)+r′ ≥ d(x, z′). Hence, g satisﬁes (1.31), thus f(z′) ≤ g(z′)= r′ by the minimality of f. Finally, (1.42) gives d¯(f, i(z′)) = f(z′) ≤ r′. We conclude that d¯ f, i(X) ≤ 2δ. 2.58 Theorem. The injective hull of a geodesic hyperbolic metric space is hyperbolic. Injective hull of hyperbolic groups 75

Proof. Let X be δ-hyperbolic. By 2.43 it is enough to show that EX and X are quasi-isometric. We claim that the embedding i: X → EX from 1.41 is a quasi-isometry. It obviously satisﬁes (2.25) with constants λ = 1 and ε = 0. We know on the other hand from the previous proposition that dH X, i(X) ≤ 2δ; hence i also satisﬁes (2.26). 2.59 Remark. Let X be δ-hyperbolic, and let f ∈ EX and n ∈ N. By ¯ 1 Proposition 2.57 there is zn ∈ X with f(zn)= d f, i(zn) ≤ 2δ + n . Further- ¯ ¯ ¯ more, d(z1, zn)= d i(z1), i(zn)) ≤ d f, i(z1) + d f, i(zn) ≤ 4δ +2. If X is proper, then by taking a convergent subsequence of {zn} ⊂ B(z1, 4δ + 2) we obtain a new point z := lim zn ∈ X with f(z) ≤ 2δ. Let ε> 0. For any pair of points x, y ∈ X with f(x)+ f(y) ≤ d(x, y)+ ε, it holds 1 ε ε (x y) = d(x, z)+ d(y, z) − d(x, y) ≤ f(z)+ ≤ 2δ + . z 2 2 2 ′ ′ ′ ′ Now take x ≤ z ≤ z and x ≤ y ≤ y with d(x, z ),d(x, y )=(z y)x; then d(z′,y′) ≤ δ since X is δ-hyperbolic. Moreover, d(z, z′)= d(x, z) − d(x, z′)= (x y)z +(z y)x − (z y)x =(x y)z. Hence ε d(z,y′) ≤ d(z, z′)+ d(z′,y′)=(x y) + d(z′,y′) ≤ 3δ + , z 2 and in particular we get that d z, [x, y] ≤ 3δ + ε for any geodesic segment [x, y].

From these two observations we can conclude, in the case of ΓS, the following.

2.60 Lemma. Suppose that (Γ,S) is δ-hyperbolic and let f ∈ EΓS. Then there is a point z = z(f) ∈ ΓS such that

f(z) ≤ 2δ +1,

′ ′ ′ with the property: whenever f(γ)+ f(γ ) ≤ dS(γ,γ )+ ε, γ,γ ∈ ΓS and ε ≥ 0, we have ′ dS(z, [γ,γ ]d) ≤ 3δ +1+ ε ′ ′ for every discrete geodesic segment [γ,γ ]d in ΓS from γ to γ .

1 Proof. Recall that dH (ΓS,CΓ(S)) ≤ 2 . Since ΓS ⊂ CΓ(S), we may assume by 1.46 for the sake of convenience that ΓS ⊂ EΓS ⊂ ECΓ(S) and ΓS ⊂ ′ ′ CΓ(S) ⊂ ECΓ(S). So let f ∈ EΓS and let γ,γ ∈ ΓS with f(γ)+ f(γ ) ≤ ′ dS(γ,γ )+ ε. Then f ∈ ECΓ(S); therefore by Remark 2.59 there is a point 76 Chapter 2. Groups and their Injective Hull

′ ′ ′ ′ z ∈ CΓ(S) with f(z ) ≤ 2δ and such that dS(z , [γ,γ ]) ≤ 3δ + ε. Now take ′ 1 a point z ∈ ΓS with dS(z, z ) ≤ 2 . Then, since f is 1-Lipschitz, f(z) ≤ ′ 1 f(z )+ 2 < 2δ + 1. Furthermore, 1 1 d (z, [γ,γ′] ) ≤ d (z′, [γ,γ′]) + + ≤ 3δ +1+ ε. S d S 2 2

Recall that Proposition 2.22 states that whenever ΓS has ﬁnitely many cone types, as in the hyperbolic case, and f ∈ EΓS, there is in every point γ ∈ ΓS a cone Cγ = Cγ(f) attached such that f(γa) = f(γ)+ |a|S for all a ∈ Cγ. The following lemma says that if ΓS is hyperbolic then every extremal function on ΓS grows with gradient one away from a neighborhood of its minimum.

2.61 Lemma. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisﬁes Axiom Y(δ). There exists R1 = R1(δ) ∈ N such that for all f ∈ EΓS one can ﬁnd a point z = z(f) ∈ ΓS with f(z) ≤ 2δ +1 and the property that for every γ ∈ ΓS with dS(γ, z) ≥ R1 there exists a γ¯ ∈ B(z, 3δ + 2) such that

−1 Cγ = coneS(¯γ γ).

Proof. Set R1 := R +3⌈δ⌉ + 2 where R := ⌈R |B(e, δ + 1)|, δ +1 ⌉ is as in Lemma 2.53; here B(e, δ + 1) denotes the closed ball in ΓS. Letz = z(f) be from Lemma 2.60 and γ ∈ ΓS with dS(γ, z) ≥ R1. Moreover, let γ1 ∈ ΓS −1 with f(γ)+ f(γ1) ≤ dS(γ,γ1) + 1 and coneS (γ1 γ)= Cγ ; compare the proof of 2.22. Since B(γ, δ +1) = γB(e, δ + 1), using Lemma 2.55 and Lemma 2.53 we get a y ∈ ΓS such that dS(γ,y) ≤ R and

′ γ ≤ y ≤ γ1, for all γ′ ∈ B(γ, δ +1). By Lemma 2.60 there isγ ¯ ∈ B(z, 3δ + 2) such that γ¯ ∈ [γ,y]d ∪ [y,γ1]d. As dS(γ, γ¯) ≥ dS(γ, z) − dS(¯γ, z) ≥ R1 − (3δ + 2) ≥ R ≥ dS(γ,y), we have in factγ ¯ ∈ [y,γ1]d and so y ≤ γ¯ ≤ γ1. It follows that γ′ ≤ γ¯ ≤ γ, for all γ′ ∈ B(γ, δ + 1). This implies

′ ′ dS(¯γ,γ )

−1 −1 (δ + 1)-tail(¯γ γ)=(δ + 1)-tail(γ1 γ).

−1 −1 Hence coneS(¯γ γ) = coneS(γ1 γ)= Cγ. Injective hull of hyperbolic groups 77

2.63 Proposition. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisﬁes Axiom Y(δ). Then there exists R2 = R2(δ) ∈ N such that for all f ∈ EΓS there exists z = z(f) ∈ ΓS with f(z) ≤ 2δ +1 and the property that

f|B ∈ EB, for any B ⊂ ΓS with B(z, R2) ⊂ B. Furthermore, every element γ¯ ∈ ΓS lies in a cone γCγ, for some γ ∈ B.

′ ′ Proof. Set R2 := max{R1, R }, where R := ⌈R |B(e, 3δ + 2)|, 3δ +2 ⌉, and let z be as in Lemma 2.61. Let x ∈ ΓS with dS(x, z) > R2. Using Lemma ′ ′ ′ 2.53 we obtain a new point x ∈ ΓS such that dS(z, x ) ≤ R and

x ≤ x′ ≤ z′,

′ ′ for all z ∈ B(z, 3δ + 2). We may assume that R2 ≥ dS(z, x ) ≥ R1; other- ′ wise we may take another point on the geodesic segment [x, x ]d such that ′ ′ dS(z, x ) = R1. Due to Lemma 2.61 there isx ¯ ∈ B(z, 3δ + 2) with Cx′ = ′−1 ′ coneS(¯x x ). But since x ≤ x′ ≤ x¯′, ′−1 ′−1 ′ it follows that x x ∈ coneS (¯x x )= Cx′ and

′ ′−1 ′ x = x x x ∈ x Cx′ .

′ ′ In other words, x lies in the cone x Cx′ for some x ∈ B(z, R2) ⊂ B; this proves the last part of the assertion. Furthermore,

f(x)= f x′(x′−1x) = f(x′)+ |x′−1x|, by Proposition 2.22. We shall verify 1.35 in order to show that f|B ∈ EB. Let y ∈ B and ǫ> 0. Then there exists x ∈ ΓS with f(y)+f(x) ≤ dS(x, y)+ǫ ′ because of 1.35. Due to the previous paragraph, there is x ∈ B(z, R2) ⊂ B with f(x)= f(x′)+ |x′−1x|. Therefore, since

′ ′−1 f(y)+ f(x )+ |x x| = f(y)+ f(x) ≤ dS(y, x)+ ǫ, we get by the triangle inequality

′ ′−1 ′ f(y)+ f(x ) ≤ dS(y, x) −|x x| + ǫ ≤ dS(y, x )+ ǫ.

From 1.35 we conclude that f|B ∈ EB.

We are going to give EΓS the structure of a locally ﬁnite, polyhedral complex in a way similar to Section 1.9. 78 Chapter 2. Groups and their Injective Hull

2.64 Corollary. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisfies Axiom Y(δ). Then for every f ∈ EΓS and every γ ∈ ΓS there is γ¯ ∈ B(z, R2) such that f(γ)+ f(¯γ)= dS(γ, γ¯), (2.65) where z = z(f) and R2 are as in Proposition 2.63. If G(f), A(ΓS) and P (G(f)) are defined as in the finite case (cf. 1.137 and 1.140), then P (G(f)) ⊂ EΓS.

Proof. Let f ∈ EΓS and γ ∈ ΓS, and set B := B(z, R2) ∪{γ}. Then, by Proposition 2.63 we get that f|B ∈ EB. Since B(z, R2) ⊂ ΓS is compact there exists a pointγ ¯ ∈ B with f(γ)+ f(¯γ) = dS(γ, γ¯); see Lemma 1.70. Ifγ ¯ = γ, then we are done. Suppose thatγ ¯ = γ. Then f(γ) = 0 and ′ ′ ′ ′ f(γ)+ f(¯γ )= iγ (¯γ )= dS(γ, γ¯ ) for everyγ ¯ ∈ B(z, R2). Let g ∈ P (G(f)). By deﬁnition we have g(γ1)+ g(γ2) ≥ dS(γ1,γ2) for all γ1,γ2 ∈ ΓS. Because of (2.65) and by Proposition 1.35 we get therefore g ∈ EΓS. 2.66 Remark. From Lemma 1.69 we know that

′ ′ f(γ )= f(γ)+ dS(γ,γ ),

′ −1 for all γ ∈ γ coneS(¯γ γ). In other words, Cγ (f) can be chosen to be the −1 subset coneS (¯γ γ) whenever we have f(¯γ)+ f(γ) = dS(¯γ,γ). Because of the corollary above we may thus assume w.l.o.g. that Cγ(g) = Cγ(f) for every g ∈ P (G(f)) and every γ ∈ ΓS since G(f) ≺ G(g).

2.67 Corollary. Let f ∈ EΓS, and let B(z, R2) ⊂ B where z = z(f) and R2 are as in Proposition 2.63. If we have g ∈ EΓS and g|B = f|B then f = g.

′ ′ Proof. Let γ ∈ ΓS. By Proposition 2.63 there is γ ∈ B with γ ∈ γCγ, Cγ = −1 Cγ(f). Using Remark 2.66 we can assume w.l.o.g. that Cγ = coneS(¯γ γ) for someγ ¯ ∈ B with f(¯γ)+ f(γ) = dS(¯γ,γ). Since g|B = f|B we have g(¯γ)+ g(γ) = dS(¯γ,γ); we may thus further assume that Cγ(g) = Cγ. It follows that

′ ′ ′ ′ g(γ )= g(γ)+ dS(γ ,γ)= f(γ)+ dS(γ,γ )= f(γ ).

2.68 Lemma. Let f ∈ EΓS and B(z, R2) ⊂ B; compare Proposition 2.63. ΓS The convex subset P (G(f)) ⊂ EΓS ⊂ R is isometric to P (G(f|B)) ⊂ EB ΓS B ΓS by means of the linear map TB : R → R given by TB(h) := h|B, h ∈ R . Injective hull of hyperbolic groups 79

Proof. First of all, note that TB P (G(f)) ⊂ P (G(f|B)) ⊂ EB because of ′ Lemma 1.143 and since f|B ∈ EB. Let g,g ∈ P (G(f)). By 2.66 we may ′ assume w.l.o.g. that Cγ(g) = Cγ (g ) = Cγ(f) =: Cγ for every γ ∈ ΓS. Letγ ¯ ∈ ΓS; using Proposition 2.63 we get a point γ ∈ B withγ ¯ ∈ γCγ . ′ ′ Hence g(¯γ) = g(γ)+ dS(γ, γ¯) and g (¯γ) = g (γ)+ dS(γ, γ¯); in particular ′ ′ |g(¯γ) − g (¯γ)| = |g(γ) − g (γ)|. It follows that TB|P (G(f)) is an isometric embedding since

¯ ′ ¯ ′ ¯ ′ ¯ ′ ¯ ′ d(TB(g), TB(g )) = d(g|B,g |B) ≤ d(g,g ) ≤ d(g|B,g |B)= d(TB(g), TB(g )).

We have to show that TB|P (G(f)) is surjective. For everyγ ¯ ∈ ΓS we choose in the following a point γ = γ(¯γ) ∈ B such thatγ ¯ ∈ γCγ; ifγ ¯ ∈ B we set γ =γ ¯. Letg ¯ ∈ P (G(f|B)) and deﬁne g : ΓS → R to be g(¯γ) :=g ¯(γ)+ dS(γ, γ¯), γ¯ ∈ ΓS. Then, TB(g)=¯g and

′ ′ ′ ′ g(¯γ)+ g(¯γ )=¯g(γ)+ dS(γ, γ¯)+¯g(γ )+ dS(γ , γ¯ ) ′ ′ ′ ′ ≥ dS(γ, γ¯)+ dS(γ,γ )+ dS(γ , γ¯ ) ≥ dS(¯γ, γ¯ ),

′ ′ for allγ, ¯ γ¯ ∈ ΓS. Now let {γ,¯ γ¯ }∈E, where G(f)=(ΓS, E). Then,

′ ′ ′ ′ ′ dS(¯γ, γ¯ )= f(¯γ)+ f(¯γ )= f(γ)+ dS(γ, γ¯)+ f(γ )+ dS(γ , γ¯ ) ′ ′ ′ ′ ≥ dS(γ, γ¯)+ dS(γ,γ )+ dS(γ , γ¯ ) ≥ dS(¯γ, γ¯ ).

′ ′ Hence we have equality, in particular f(γ)+ f(γ )= dS(γ,γ ) and

′ ′ ′ ′ dS(γ, γ¯)+ dS(γ,γ )+ dS(γ , γ¯ )= dS(¯γ, γ¯ ).

′ ′ Therefore, {γ,γ }∈E, and since γ,γ ∈ B andg ¯ ∈ P (G(f|B)) we have that ′ ′ g¯(γ)+¯g(γ )= dS(γ,γ ). Thus,

′ ′ ′ ′ g(¯γ)+ g(¯γ )=¯g(γ)+ dS(γ, γ¯)+¯g(γ )+ dS(γ , γ¯ ) ′ ′ ′ ′ = dS(γ, γ¯)+ dS(γ,γ )+ dS(γ , γ¯ )= dS(¯γ, γ¯ ).

In other words we have shown that g ∈ P (G(f)) and TB(g)=¯g; therefore TB|P (G(f)) is surjective.

2.69 Corollary. The subset P (G(f)) is an injective polyhedron in RΓS and f ∈ P (G(f)) is an inner point. Moreover, (i) g ∈ P (G(f)) is an inner point if and only if G(g)= G(f), and

(ii) every face of P (G(f)) is of the form P (G) for some ΓS-admissible graph G ∈A(ΓS) with G(f) ≺ G. 80 Chapter 2. Groups and their Injective Hull

Proof. Let R ≥ R2 and z = z(f) as in Proposition 2.63. Because of 2.68 we know that P (G(f)) is isometric to P (G(f|B)) ⊂ EB by means of the linear map TB with B := B(z, R) ⊂ ΓS ﬁnite. It follows that P (G(f)) is an injective polyhedron in RΓS and that f is an inner point; see 1.149 (i). (i) Let g ∈ P (G(f)) be an inner point, then TB(g)= g|B is an inner point of P (G(f|B)). Therefore, G(g|B) = G(f|B) by 1.149 (i). Since R ≥ R2 was arbitrary we get that G(g)= G(f). The other implication is obvious. (ii) Let g ∈ P (G(f)) be an inner point of a face F of P (G(f)), and choose ′ R ≥ R2 big enough such that B(z(g), R2) ⊂ B := B(z(f), R). Then, TB′ (g) is an inner point of the face TB′ (F) of P (G(f|B′ )). From 1.149 (i) and (iii) we know that 2.68 TB′ (F)= P (G(g|B′ )) = TB′ (P (G(g))), hence F = P (G(g)) since TB′ |P (G(f)) is injective. Moreover, G(f) ≺ G := G(g) because g ∈ P (G(f)).

2.70 Theorem. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisfies Ax- iom Y(δ). Then EΓS is isometric to the locally finite polyhedral complex ′ ′ P := {P (G)}G∈A(ΓS ) where P (G ) is a face of P (G) if and only if G ≺ G . Moreover, every cell of P is injective and there are only finitely many isometry types of cells in P.

Proof. P is a cell decomposition of EΓS because the interior of two distinct ′ ′ cells are disjoint by 2.69 (i), and every cell is of the form P (G ), G ∈A(ΓS), see 2.69 (ii). Furthermore, P (G)= P (G(f)) is isometric to P (G(f|B)) ⊂ EB for some f ∈ EΓS and with B := B(z(f), R2); see Lemma 2.68. The latter, on the other hand, is isometric to the ball B(e, R2) since Lz(f) B(e, R2) = ¯ B(z(f), R2). It follows that P (G) is isometric to P (G) ⊂ EB(e, R2) where G¯ ∈A(B(e, R2)); e ∈ ΓS is the neutral element in Γ. Since |B(e, R2)| < ∞ there are only ﬁnitely many isometry types of cells and P (G) is an injective polyhedron. Let R> 0. For every f ∈ B(ie, R) ⊂ EΓS we have z(f) ∈ B(e, R+2δ+1) ¯ ¯ since dS(e, z(f)) ≤ d(ie, f)+ d(f, iz(f)) ≤ R + f(z(f)) ≤ R +2δ +1. In other words B(z(f), R2) ⊂ B(e, R2 + R +2δ + 1) =: B, for every f ∈ B(ie, R). Therefore, it follows by 2.67 and Lemma 2.68 that P is a locally ﬁnite cell decomposition of EΓS, since

P (G(f)) ∈P| f ∈ B(ie, R) ≤ |A(B)| < ∞. Because P is a locally ﬁnite cell decomposition of EΓS and the latter is a geodesic metric space, we get in particular that EΓS is in fact isometric to the metric polyhedral complex P = {P (G)}G∈A(ΓS). Injective hull of hyperbolic groups 81

Every ball B(ie, R) in EΓS, R > 0, can be covered by ﬁnitely many P (Gj) ∈ P, 1 ≤ j ≤ N. Since these are compact we obtain that EΓS is a proper metric space. We get our main result.

2.71 Theorem. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisfies Axiom Y(δ). Then, EΓS is a proper locally finite polyhedral complex with finitely many isometry types of cells, and Γ acts properly discontinuous and cocompactly on EΓS by cellular isometries.

Proof. Γ acts on EΓS by isometries via L¯ from 2.18. It is clear that this action is cellular on EΓS since

¯ ′ − ′ Lg P (G(f )) = P Lg 1 (G(f )) , ′ for all f ∈ EΓS and g ∈ Γ. Due to Theorem 2.70 and Proposition 2.19 we only need to show that Γ acts cocompactly on EΓS. So let f ∈ EΓS and put ¯ z := z(f) as in Proposition 2.63. Then, d(f, iz) = f(z) ≤ 2δ + 1; in other words f ∈ B(iz, 2δ +1) = Lz B(ie, 2δ + 1) . Let C := B(ie, 2δ + 1). Then C is compact and Γ(C)= EΓS.

2.72 Corollary. Suppose (Γ,S) is δ-hyperbolic and CΓ(S) satisﬁes Axiom Y(δ). Then, the tangent cone TgEΓS is injective at every point g ∈ EΓS.

Proof. Since EΓS is a locally ﬁnite polyhedral complex, the claim follows immediately using Proposition 1.17 and Corollary 1.50.

Chapter 3

Examples and a counter-example

3.1 Hyperbolic groups

3.1.1 A Counterexample Let Γ be the group Γ= a, b|a4, a2ba−2b−1 ≡: S|R. (3.1)

We shall see in Corollary 3.8 and 3.10 that Γ is hyperbolic but ΓS does not satisfy Axiom Y. Let w be a word in the alphabet S¯. Then n(w,s′) denotes in the following the number of occurrences of the letter s′ ∈ S¯ in w and N(w,s) := n(w,s) − n(w,s−1), s ∈ S. ′ 3.2 Lemma. Let Γ be given as in (3.1). Let M, M , nj, mk ∈ Z with nj = ±1 and mk =0 , 1 ≤ j, k ≤ s. ′ Γ (i) If bM an1 bm1 ans bms aM ≡ 1 then M ′ = s =0 and M ≡ 0 mod4.

Γ (ii) If w ≡ w′ then N(w, b)= N(w′, b).

Proof. (i) The idea of the proof is to map Γ into a free product where the ′ situation is much simpler. For that purpose let Γ be the free product Z2 ∗ Z = A, B|A2, and f : Γ → Γ′ be the group homomorphism induced by f(a) := A and f(b) := B. Note that f is well deﬁned since

Γ′ f a4 = A2A2 ≡ 1 Γ′ f a2ba−2b−1 = A2 BA−2B−1 ≡ 1. 84 Chapter 3. Examples and a counter-example

We have ′ ′ ′ Γ BM ABm1 ABms AM = f(bM an1 bm1 ans bms aM ) ≡ f(1) = 1, ′ ′ Γ that is BM ABm1 ABms AM ≡ 1. By the definition of the free product this ′ Γ implies that M ′ = 0 and s = 0. Therefore bM an1 bm1 ans bms aM = aM ≡ 1. 4 In the same way we can define a group homomorphism g : Γ → Z4 = t|t by setting g(a) := t and g(b) := 1; this shows that M = 0 mod 4. (ii) We define a group homomorphism g′ : Γ → Z by g′(a) := 0 and g′(b) := 1. Note that g′ is well defined and that g′(w)= N(w, b). The claim Γ now follows since w ≡ w′ ⇒ N(w, b)= g(w)= g(w′)= N(w′, b). 3.3 Lemma. Let Γ ≡ S|R be as in (3.1). Then every word w over the alphabet S¯ can be written in the form N F (S) ′ M ±1 m1 ±1 ms M −1 w ≡ b a b a b a cj rjcj, (3.4) Yj=1 ′ −1 where M ,M,mk ∈ Z, cj ∈ F (S) and rj ∈ R¯ := R ∪ R . Furthermore, 0 ≤|M| ≤ 3, mk =0 for all 1 ≤ k ≤ s, and N ≤ ℓ(w).

Proof. We prove the claim by induction over the number of blocks of a’s and b’s in w. To illustrate what we mean by blocks of a’s and b’s, the word babab3a for instance will have 6 blocks whereas a3ba−1b5 will only have 4.

Case I: The case where w consists of only one ”b” block is trivial. So let w = am; we may assume w.l.o.g. that m > 0. Then m = 4t + M for some m M t 4 0 ≤ M ≤ 3 and t ∈ N. Hence a = a i=1 a where t ≤ m = ℓ(w). Q Case II: w ends with a ”b” block; that is w = w′bj where w′ does not end with a ”b”. Then by the induction hypothesis N F (S) ′ ′ M ±1 m1 ±1 ms M −1 w ≡ b a b a b a ci rici, (3.5) Yi=1 with N ≤ ℓ(w′)= ℓ(w) − j. Therefore N F (S) ′ M ±1 m1 ±1 ms M −1 j w ≡ b a b a b a ci rici b Yi=1 N F (S) ′ M ±1 m1 ±1 ms M j −j −1 j ≡ b a b a b a b b ci ri cib . Yi=1 Hyperbolic groups 85

If |M| ≤ 1 then we are done. W.l.o.g. we may assume that 2 ≤ M ≤ 3. In this case, by using the identity

F (S) a2bj ≡ b(a2bj−1) b−ja−2(a2ba−2b−1)a2bj (3.6) j times, one obtains

N F (S) ′ M ±1 m1 ±1 ms M j −j −1 j w ≡ b a b a b a b b ci ri cib Yi=1 j N F (S) ′ M ±1 m1 ±1 ms M−2 j 2 −1 2 −2 −1 −j −1 j ≡ b a b a b a b a di (a ba b ) di b ci ri cib . Yi=1 Yi=1 In other words w is of the form

N+j F (S) ′ M ±1 m1 ±1 ms M−2 j 2 −1 w ≡ b a b a b a b a c¯i ric¯i, (3.7) Yi=1 and N + j ≤ ℓ(w′)+ j = ℓ(w) − j + j = ℓ(w). Note that even if M = 2 and ms = −j the presentation (3.7) is of the form (3.4).

Case III: Suppose that w = w′ak where w′ does not end with an ”a”. We may assume w.l.o.g. that k > 0. By the induction hypothesis w′ is of the form (3.5), hence

N F (S) ′ M ±1 m1 ±1 ms M k −k −1 k w ≡ b a b a b a a (a ci ) ri (cia ). Yi=1 If M + k ≤ 0 then −3 ≤ M ≤ M + k ≤ 0; hence nothing has to be shown. Suppose that M + k > 0. Then M + k =4t + M¯ for some 0 ≤ M¯ ≤ 3 and 1 t ∈ N. We have t ≤ 4 (M + k) and M ≤ 3 ≤ 3k, hence t ≤ k. We can thus write w in the following way

t N F (S) ′ M ±1 m1 ±1 ms M¯ 4 −k −1 k w ≡ b a b a b a a (a ci ) ri (cia ), Yi=1 Yi=1 where N + t ≤ ℓ(w′)+ k = ℓ(w). 3.8 Corollary. The group Γ= a, b | a4, a2ba−2b−1 is hyperbolic. Proof. By using 3.2 and 3.3 one gets immediately that Γ has a linear Dehn function. Hence, Γ is hyperbolic by 2.47. 86 Chapter 3. Examples and a counter-example

In the next paragraph we are going to show that ΓS doesn’t satisfy Axiom Y. 3.9 Proposition. Let Γ be as in (3.1). Then the following is true for

± ±1 N zN := a b ∈ ΓS, N ∈ N. + − + − (i) The elements z0 , z0 , z1 , z1 ,... are pairwise distinct. − + ± (ii) d(zN , zN ) ≤ 2 and |zN | = N +1 (iii) The sequence

± ±1 ±1 ±1 2 ±1 N cN := e, a , a b, a b ,...,a b is the only discrete geodesic segment in ΓS from the neutral element e ± + − to zN . In particular cN and cN have no segment in common. Proof. (i) This follows directly from Lemma 3.2. By the identity (3.6) we obtain successively

− + −N 2 N −N+1 2 N−1 2 d(zN , zN )= |b a b | = |b a b | = = |a | ≤ 2.

Γ ± ±1 N This proves the ﬁrst part of (ii). Now let w ≡ zN = a b be a shortest ¯ ± word in the alphabet S representing zN . By Lemma 3.2 (ii) it then follows that w must be of the form w = bsa±1bN−s for some 0 ≤ s ≤ N. Suppose Γ s> 0 then bs−N a±1b−sa±1bN ≡ 1 which contradicts 3.2 (i). Therefore s =0 ± ± and |zN |S = ℓ(w) = N + 1. This proves the rest of the claim since cN is a ± discrete path of length N +1 from e to zN .

3.10 Corollary. The metric space ΓS doesn’t satisfy Axiom Y. + − Proof. Let R ≥ 0 and take N ∈ N to be N ≥ R. Then, d(zN , zN ) ≤ 2 ± + − and |zN |S = N +1 > R. Furthermore, the geodesics cN , cN are the only + − one issuing from e to zN and zN respectively. Since they have no common segment at all, we conclude that ΓS doesn’t satisfy Axiom Y(2).

3.1.2 Small cancellation groups We ﬁrst settle some deﬁnitions and concepts from the theory of small cancellation groups and state some well known facts about geodesic triangles in the Cayley graph of such groups. Then, Axiom Y is proved for a small subclass with the objective of showing Axiom Y for groups of the form

Tg := a1, b1,...,ag, bg | [a1, b1][a2, b2] [ag, bg] ≡: Sg|Rg, g ∈ N, (3.11) Hyperbolic groups 87

−1 −1 where [x, y] denotes the commutator xyx y of x and y. Note that Tg is the fundamental group of the orientable surface of genus g. In the following S|R denotes a presentation of a group Γ; we may assume that whenever r ∈ R then r−1 and any cyclic permutation of r are in R. This is not a restriction since any set of relators can be made in this form by adding the inverse and cyclic permutations without changing the group Γ nor the Cayley graph CΓ(S). 3.12 Definition. A word u in the alphabet S¯ := S ∪ S−1 is called a piece of R if it is a common prefix of two distinct relators r, r¯ ∈ R. For example the word a2 is a piece of R := {a2ba, aba2, ba3, a3b} whereas ab is not. 3.13 Definition. The set of relators R is said to satisfy condition C′(λ) if every piece u of r ∈ R has length |u| < λ |r|. Let q ∈ N. We say that R satisfies condition T (q) if for any finite subset {r1,...,rl} ⊂ R, 3 ≤ l < q, −1 −1 −1 with r1 = r2 ,...,rl−1 = rl ,rl = r1 , one of the products

r1r2 ,...,rl−1rl , rlr1 must be freely reduced.

3.14 Definition. Let p, q ∈ N. A(p, q)-presentation for Γ is a presentation ′ 1 Γ ≡S, R such that R satisfies the conditions C ( p ) and T (q). A(p, q)-group is a group which permits a (p, q)-presentation. 3.15 Remark. (6, 3)-groups and (4, 4)-groups are known to be hyperbolic, cf. [15]. 3.16 Definition. Let (X,d) be a metric space. A simple geodesic digon in X is a union of two geodesic segments with a common start and end point, but which don’t have any further points in common. A simple geodesic triangle is a geodesic triangle ∆ with the property that any two different edges of∆ meet in exactly one point, a corner of ∆. 3.17 Remark. On can easily verify that every geodesic triangle in a metric space X is composed of a finite number of segments, simple geodesic digons and of at most one simple geodesic triangle; compare Figure 3.1. Furthermore, if x is an endpoint of a simple digon one could chose either one of its two segments to join x with z and z′. In other words, the only situation where Axiom Y may fail is at a corner of a simple geodesic triangle; hence in order to prove Axiom Y for X it is enough to investigate only simple geodesic triangles in X. 88 Chapter 3. Examples and a counter-example

z z′

Figure 3.1: A geodesic triangle with one simple geodesic triangle, and with a couple of geodesic segments and simple digons

We need the following classiﬁcation theorem for simple geodesic triangles in the Cayley graph of (6, 3) and (4, 4)-groups in order to prove Axiom Y for such groups.

3.18 Theorem. Let Γ = S|R be either a (6, 3) or (4, 4)-presentation for Γ. Then any simple geodesic triangle ∆ in CΓ(S) must be one of the form a) − d) listed in Figure 3.2; however ∆ may be degenerated in the sense that

A2i−1 = A2i for one or more 1 ≤ i ≤ 3. Moreover, a 2-face in Figure 3.2 corresponds to a relation in R and an internal edge is a common piece of two adjacent relators. The hatched 2-face, on the other hand, is the only face which touches all three external segments of the border of ∆.

Proof. See [15] for a proof. Hyperbolic groups 89 a) x b) x

r r z¯ z¯′ z¯ z¯′

A1 A6 A2 A5

′ z ′ z z z A3 A4

c) x d) x z¯ r z¯′ z¯ r z¯′

A1 A6 A1 A6 A2 A5 A2 A5

A23

z z′ z z′ A3 A4 A3 A4

Figure 3.2: Classiﬁcation of Simple Triangles

3.19 Proposition. Let Γ = S|R be a (6, 3) or (4, 4)- presentation of Γ such that ℓ(r) is even for all r ∈ R and ℓ(u)=1 for every piece of R. Then ΓS satisﬁes Axiom Y with a linear function R(s)= s + D for some constant D = D(R). Note that this is a huge restriction. For instance there can’t be any subword of the form aa . . . a with k > 2 in any r ∈ R. k-times Proof. Note that it is| enough{z } to consider simple geodesic triangles; compare ′ Remark 3.17. So let x, z and z be the corners of a simple triangle ∆ in ΓS and suppose w.l.o.g. that d(¯z, x) ≥ d(¯z′, x); compare Figure 3.2. If x is not a vertex of the hatched 2-face in ∆ then ℓ(r)= d(¯z, x)+ d(x, z¯′)+ d(¯z′, z¯)= d(¯z, x)+ d(x, z¯′) + 1; 90 Chapter 3. Examples and a counter-example note that d(¯z, z¯′) = 1 since it is a common piece of two relators. The left- hand side is even by assumption; hence d(¯z, x)=1+ d(¯z′, x). Therefore, the path (z, z,¯ z¯′, x) from z to x is also a geodesic which has the segment (¯z′, x) in common with the geodesic (z′, x). Thus we may further assume that x is a vertex of the hatched 2-face. In this case we get

′ d(z, x) ≤ d(z, A3)+ d(A3, x) ≤ d(z, A3)+ D ≤ d(z, z )+ D with D := max{ℓ(r) | r ∈ R} since in each case b) - d) we can get from D A3 to x along paths of length smaller than 2 by crossing at most two 2- faces. We conclude that if d(x, z) > d(z, z′)+ D then there is a point y with d(z,y) ≤ d(z, z′)+ D such that (z,y,x) and (z′,y,x) are geodesic with common segment (y, x).

3.20 Corollary. The presentation Tg = Sg|Rg satisﬁes Axiom Y for every g ≥ 2.

Proof. Let g ≥ 2. We shall show that Rg satisﬁes 3.19. Note that relators in Rg are up to a cyclic permutation of the indices of the following form:

[a1, b1] ... [ag, bg] [bg, ag] ... [b1, a1] −1 −1 −1 −1 b1a1 b1 [a2, b2] ... [ag, bg]a1 a1 [bg, ag] ... [b2, a2]b1a1b1 −1 −1 −1 −1 . a1 b1 [a2, b2] ... [ag, bg]a1b1 b1 a1 [bg, ag] ... [b2, a2]b1a1 −1 −1 −1 −1 b1 [a2, b2] ... [ag, bg]a1b1a1 a1b1 a1 [bg, ag] ... [b2, a2]b1

Hence, words of length one are the only pieces of Rg. Then, since every ′ 1 relator in Rg has length 4g it follows that Tg is C ( 4g−1 ). On the other hand any group presentation is by deﬁnition T (3). Therefore, Tg is a (6, 3)-group and satisﬁes Axiom Y.

3.2 Abelian groups

The aim of this section is to prove that Axiom Y is valid in abelian groups with respect to any ﬁnite generating set S. Since in the abelian case there are a lot of geodesics between two given points, in fact any permutation of a given length minimizing word is again minimal, we need a way to ﬁx one. This is done in the following by introducing the lexicographical order on the set of geodesics joining two points and choose the unique minimal one. Most of the ideas and arguments are out of [13]; we are going to recall them for the sake of completeness. Let S := {a1,...,ak} be an arbitrary Abelian groups 91 symmetric1 set of generators of an abelian group Γ. Every element g ∈ Γ can be written as Γ n n1 n2 nk g ≡ a := a1 a2 ak k with n := (n1,...,nk) ∈ N . We say that m < n if either ℓ(m) := j mj < j nj or if there exists an index i ≤ k such that mj = nj, ∀j

k n Γ Wg := {n ∈ N | a ≡ g} exists and is unique. The minimal element corresponds to a shortest word presentation for g and represents a discrete geodesic segment in ΓS from the neutral element 0 ∈ Γ to g; compare Deﬁnition 2.13. Thus

|g| = ℓ(n(g)) = ℓ(x)+ ℓ(n(g) − x)= |ax| + d(ax,g) (3.21)

k for all x n(g) where m n ∈ N :⇔ mi ≤ ni for all i ≤ k. Note that for every x ∈ Nk one has m < n ⇔ x + m < x + n. Following [13] we show first that the minimal presentations of two adjacent group elements in ΓS only differ by a minimal relation in the sense of the definition below.

Γ 3.22 Deﬁnition. A tuple (u, v) ∈ N2k is called a relation if au ≡ av. A relation (u, v) = (0, 0) is called minimal if it is minimal with respect to the order on N2k.

3.23 Remark. Let (u, v) ∈ N2k be a relation with (u, v) ≺ (n(g), n(h)) for some g, h ∈ Γ. Then u = v; otherwise one would have either u < v or v < u. Suppose w.l.o.g. that u < v. Since (n(h) − v + u) ∈ Wh one would have a shorter element in Wh than n(h) which contradicts the minimality of n(h).

3.24 Lemma. There are only ﬁnitely many minimal relations in N2k.

Proof. This is a corollary of the following Lemma.

3.25 Lemma. Let Nl be endowed with the partial order and let M ⊂ Nl be a subset of minimal elements. Then M is ﬁnite.

Proof. Note since all elements in M are minimal that

m(j) m(k) (3.26)

1S ⊂ Γ is called symmetric if a ∈ S ⇔ a−1 ∈ S 92 Chapter 3. Examples and a counter-example for all m(j) = m(k) ∈ M. The proof of the lemma goes by induction over the dimension l. The case where l = 1 is trivial, since in this case is a well ordering on N and hence M can only consist of exactly one element. So let l> 1 and suppose that M is infinite. Because of condition (3.26) there must (1) (j) be an index s such that ms > ms ≥ 0 for infinitely many j’s. Therefore, (ji) by taking a subsequence we obtain an infinite sequence ji with ms = const. The new set (ji) (ji) (ji) l−1 {(m1 ,..., ms ,...mk )} ⊂ N is then a minimal subset in Nl−1 and by the induction hypothesis it is finite. This contradicts the assumption that M is infinite.

Γ 3.27 Lemma. Let g, h be two adjacent elements in ΓS, that is, h − g ≡ ai ∈ S. Then there is a minimal relation (u, v) ∈ N2k and x ∈ Nk such that (i) (x, x) (n(h), n(g))

(ii) n(h)= x + u and n(g)+ n(ai)= x + v.

Proof. If n(h) = n(g) + n(ai) take x := n(g) and (u, v) := (n(ai), n(ai)). Γ We may suppose that n(h) = n(g) + n(ai). Since h ≡ g + ai the tuple (n(h), n(g) + n(ai)) is a relation; therefore it is the sum of ﬁnitely many minimal ones. Furthermore, since n(h) = n(g) + n(ai) there must be a minimal relation (u, v) (n(h), n(g) + n(ai)) among them such that u = v. Then, by Remark 3.23 it follows that (u, v) (n(h), n(g)); hence we obtain ′ ′ that n(ai) v. Let u := n(h)−u and v := n(g)+n(ai)−v. By construction it follows that (u′, v′) (n(h), n(g)) and that (u′, v′) is a relation; therefore, ′ ′ u = v =: x by 3.23. Then, n(h) = x + u and n(g) + n(ai) = x + v, and (u, v) is a minimal relation. Now we can state and prove our claim regarding Axiom Y in abelian groups. 3.28 Lemma. There is a constant r > 0 such that, whenever two elements g, h in ΓS are adjacent and |g|, |h| >r, there is (x, x) (n(g), n(h)) with x x dS(a ,g), dS(a , h) ≤ r and x x |g| = |a | + dS(a ,g) x x |h| = |a | + dS(a , h).

In other words ΓS satisﬁes Axiom Y(1). Abelian groups 93

Proof. Because of Lemma 3.24 we know that there exists only ﬁnitely many 2k minimal relations m := {(u1, v1) ..., (uL, vL)} in N . Set

r := max{ℓ(ui)+1,ℓ(vi)+1|(ui, vi) ∈ m}, where m is the finite set of all minimal relations. Then the claim follows by choosing x to be as in Lemma 3.27 and using the identity (3.21); note that Γ (n(h) − x, n(g) + n(ai) − x) ∈ m whenever h ≡ g + ai. 3.29 Corollary. Every finitely generated abelian group with an arbitrary finite set of generators satisfies Axiom Y with a linear function R(s) := rs.

Proof. Since ΓS is a uniform metric space, see Remark 2.17, it is suﬃcient to show Axiom Y at the neutral element 0 ∈ Γ. Let r be deﬁned as in Lemma 3.28 and z, z′ ∈ Γ be any two points with |z|, |z′| > rs, where ′ s := dS(z, z ). We are going to show by induction over the length s that there is a x(s) = x(s)(z, z′) ∈ Nk with (x(s), x(s)) (n(z), n(z′)) and the following properties:

x(s) ′ x(s) dS(z, a ),dS(z , a ) ≤ R(s)= rs (3.30) and x(s) x(s) |z| = |a | + dS(a , z) (3.31) ′ x(s) x(s) ′ |z | = |a | + dS(a , z ). (3.32)

′ x(s) Note that if y(z, z ) := a ∈ ΓS then (3.30) - (3.32) are exactly the at- tributes of Axiom Y. The case s = 1 is true due to Lemma 3.28. ′ Now let dS(z, z ) = s + 1. There is an intermediate pointz ¯ between z and ′ ′ z such that dS(z, z¯)= s and dS(¯z, z ) = 1. Applying the induction hypothesis to the two pairs of points (z, z¯) and (¯z, z′), one obtains two new points, x(s) = x(s)(z, z¯) and x(1) = x(1)(¯z, z′), with the properties listed above. So let

(s+1) k (s) (1) (s) (1) x := max{n ∈ N | n x and n x } = g.c.d.(x , x ),

(s+1) (s) (1) in other words xi = min{xi , xi }. Thus we have

(s) (s+1) (s) (1) (1) x(1) ℓ(x − x )= xi − xi ≤ z¯i − xi = dS(¯z, a ) ≤ r. (sX) (1) Xi xi >xi A similar argument shows that

(1) (s+1) x(s) ℓ(x − x ) ≤ dS(¯z, a ) ≤ r s. 94 Chapter 3. Examples and a counter-example

Therefore

x(s+1) x(s) x(s) x(s+1) dS(z, a ) ≤ dS(z, a )+ dS(a , a ) ≤ rs + r = R(s + 1) ′ x(s+1) ′ x(1) x(1) x(s+1) dS(z , a ) ≤ dS(z , a )+ dS(a , a ) ≤ r + rs = R(s + 1).

The identities (3.31) and (3.32) are then a direct consequence of the identity (3.21) since (x(s+1), x(s+1)) (x(s), x(1)) (n(z), n(z′)). Appendix A

The Stone–Cechˇ compactiﬁcation

In the following X denotes a topological space. Let βX be a compact Haus- dorﬀ space and i: X → βX a continuous map with the property that every continuous map j : X → K into a compact Hausdorﬀ space K can be uniquely extended to βX, i.e. there is an unique ¯j : βX → K such that

i X / βX (A.1) C CC CC ∃!¯j j CC C! K commutes. It is easy to see that βX is uniquely determined up to a homeo- morphism by property (A.1). In this case βX is said to be the Stone–Cechˇ compactification of X. Note that βX is not a compactification in the sense that i: X → βX is a homeomorphic embedding; in fact i is not injective in general. However if X is Tychonoff then i(X) ⊂ βX and X are homeomorphic by means of the map i; compare Proposition A.10. A.2 Proposition. The Stone–Cechˇ compactification exists for every X.

Proof. By Cb(X) we mean in the following the set of all continuous and bounded functions f : X → R. Let

T (X) := I(f), (A.3)

f∈YCb(X) where I(f) := [−c,c] ⊂ R and c := supx∈X |f(x)|. And let πf : T (X) → I(f) be the projection onto I(f) i.e. πf (rg) := rf , (rg) ∈ T (X). Recall that by the deﬁnition of the product topology a map g : Y → T (X) is continuous 96 Chapter A. The Stone–Cechˇ compactiﬁcation

if and only if πf ◦ g : Y → I(f) is continuous for every f ∈ Cb(X). Let i: X → T (X) be the map

i(x) := f(x) . (A.4) f∈Cb(X) Then i is continuous since πf ◦ i = f ∈ Cb(X) for all f ∈ Cb(X). Define βX := i(X) ⊂ T (X). Then βX with the induced topology is compact and Hausdorff since by Tychonoff’s Theorem (cf. [23]) the product T (X) is a compact Hausdorff space. We need to verify that βX satisfies the universal property (A.1). We shall show the uniqueness of ¯j first. Suppose that ¯j′ : βX → K and ¯j′ ◦ i = j. Then ¯j(i(x)) = j(x) = ¯j′(i(x)), for all x ∈ X. Hence, ¯j and ¯j′ coincide on i(X). But since βX = i(X) the continuous maps, ¯j and ¯j′, must coincide everywhere. In order to show the existence of an extension we consider the case where K is a compact interval I ⊂ R first. So let f : X → I be continuous, i.e. f ∈ Cb(X). Define f¯: βX → I by f¯ := πf |βX . Then f¯ is a continuous ¯ ¯ function and f ◦ i = πf |βX ◦ i = πf ◦ i = f. Hence, f is an extension of f. Now let K be an arbitrary compact Hausdorff space. By Urysohn’s Lemma, cf. [23], there exists for every pair of points x, y ∈ K a function g ∈ Cb(K) with g(x) = g(y). It follows that iK : K → T (K) is injective, where iK is defined as in (A.4). Therefore, iK is a homeomorphic embedding since K is compact and iK is injective. In other words we may assume w.l.o.g. that K ⊂ T (K). Let j : X → K ⊂ T (K). From the above we know that the function πf ◦ j : X → I(f) can be extended to πf ◦ j : βX → I(f) for every f ∈ Cb(K). The map j : βX → T (K) given by

j (t) := πf ◦ j (t) , f∈Cb(K) t ∈ βX, is continuous. Moreover, we have by deﬁnition

j ◦ i (x)= πf ◦ j (x) = f(x) = j (x), f∈Cb(K) f∈Cb(K) for all x ∈ X. Thus, j is an extension of j to βX. If ¯j(βX) ⊂ K we are done. But this is true since ¯j−1(K) ⊂ βX is a closed subset and i(X) ⊂ ¯j−1(K).

A.5 Remark. From the proof above we also learn that i(X)= βX.

A.6 Corollary. The Banach spaces Cb(X) and C(βX) are isometric.

∗ Proof. Let i: X → βX be as in (A.1) and let i : C(βX) → Cb(X) be the ∗ linear map given by i (g) := g ◦ i, g ∈ C(βX). Let f ∈ Cb(X). Using (A.1) 97 with K = I(f) we get an f¯ ∈ C(βX) with i∗(f¯) = f¯ ◦ i = f. Therefore, i∗ is surjective. It remains to show that i∗ is an isometry. This is true because

∗ g∞ = g|i(X)∞ = g ◦ i∞ = i (g)∞, where the first equality holds since i(X)= βX and g is continuous. A.7 Remark. From A.6 and the Gelfand–Naimark theorem it follows imme- ∗ diately that βX corresponds to the set of characters of Cb(X), the C -algebra of all bounded functions on X, endowed with the weak∗-topology; compare [23] for more details. A.8 Definition. A Hausdorff space X is said to be Tychonoff if any closed subset A ⊂ X and any point x∈ / A can be strictly separated by Cb(X) in the sense that there is a function g ∈ Cb(X) such that g(x) = 0 and g(A) = 1, and 0

i(O)= Ug ∩ i(X), where Ug = {(rh) ∈ T (X) ||rg| < 1} ⊂ T (X) is an open subset of T (X). Hence, i(O) is open in i(X). 98 Chapter A. The Stone–Cechˇ compactiﬁcation

Recall that a topological space X is said to be extremely disconnected if the closure of every open subset is open.

A.11 Lemma. The Stone–Cechˇ compactiﬁcation of a discrete topological space is extremely disconnected.

Proof. Let X be a discrete topological space and O ⊂ βX an open subset of βX. We need to verify that O is open. By A.10 and A.9, we may assume that X ⊂ βX. Let O1 := O ∩ X and O2 := X \ O1. Then, βX = O1 ∪ O2 since βX = X. Deﬁne a map ξ : X →{0, 1} by ξ(O1) := 0 and ξ(O2) := 1. Then ξ is continuous, and by (A.1) there is an extension ξ¯: βX →{0, 1}. We ¯−1 ¯ ¯−1 claim that ξ (0) = O. Note, by the continuity of ξ, that O1 = ξ (0) and −1 O2 = ξ¯ (1); in particular O1 and O2 are disjoint. Furthermore, O2 ∩ O = ∅ since O2 ∩ O = ∅. Hence, O ⊂ O1 and O ⊂ O1. The inclusion O1 ⊂ O on −1 the other hand follows directly from O1 ⊂ O. Therefore, O = O1 = ξ¯ (0) is an open subset of βX. . Appendix B

Group actions on geodesic metric spaces

Let Γ be an arbitrary group and X a geodesic metric space. Γ is said to operate on X by isometries if there is a group homomorphism L: Γ → Isom(X) of Γ into Isom(X), the group of isometries of X. For convenience we shall write g(x) instead of L(g)(x) where g ∈ Γ and x ∈ X. B.1 Deﬁnition. A group action L: Γ → Isom(X) is called properly discontinuous if for every x ∈ X and R> 0, the set

NR(x) := {g ∈ Γ | g B(x, R) ∩ B(x, R) = ∅} is finite. Γ is said to act cocompactly on X if there exists a compact subset C ⊂ X such that Γ(C) := g(C)= X. g[∈Γ Furthermore, we say that a group Γ acts nicely on X if Γ acts properly discontinuous and cocompactly on X. Our definition of a properly discontinuous group action is different from the usual one, cf. [5, 8.2]. However, in the context of geodesic metric spaces the two coincide. B.2 Remark. In order to show that a group acts properly discontinuous on X it is enough to verify that NR(x0) is finite for some x0 ∈ X and all R> 0, ′ since NR(x) ⊂ NR′ (x0) with R = R + d(x, x0), x ∈ X. B.3 Remark. Suppose Γ operates cocompactly on X and x ∈ X. Then, Γ B(x, R) = X with R := d(x, x0) + diam(C), x0 ∈ C; this is obvious since C⊂ B(x, R ). 100 Chapter B. Group actions on geodesic metric spaces

B.4 Lemma. If Γ acts nicely on a geodesic metric space X, then Γ is ﬁnitely generated.

Proof. Let x ∈ X and O := B˙ (x, 2R) be the open ball of radius 2R, where ′ R is given as in B.3; then X = Γ(O). Moreover, let S := N2R(x) ⊂ Γ and ′ ′ ′ Γ := {s1 ...sn ∈ Γ | sj ∈ S , n ∈ N} be the subgroup generated by S . We claim that Γ is generated by S′. We have that

Γ′(O) ∩ (Γ \ Γ′)(O)= ∅.

Otherwise we would have g(O) ∩ g′(O) = ∅ for some g′ ∈ Γ′ and g ∈ Γ \ Γ′. ′−1 ′−1 ′ Therefore, g g(O) ∩ O = ∅ and in particular g g = s ∈ N2R(x) = S . Hence, g = g′s ∈ Γ′ since g′ ∈ Γ′; this contradicts the assumption, g∈ / Γ′. It follows that {Γ′(O), (Γ \ Γ′)(O)} is an open partition of X. Since X is connected and Γ′(O) is non-empty, O ⊂ Γ′(O), we have (Γ \ Γ′)(O) = ∅. Thus, Γ \ Γ′ = ∅ or in other words Γ = Γ′.

B.5 Proposition (The Svarc–Milnorˇ Lemma). Suppose Γ acts nicely on a geodesic metric space X and x0 ∈ X. Then, ΓS is quasi-isometric to X by means of Q(γ) := γ(x0) ∈ X, γ ∈ ΓS, for any ﬁnite generating set S ⊂ Γ. Proof. See [5]. Appendix C

The Growth of a Dehn Function

We shall give here the proof of the fact that the growth of a Dehn function δP does not depend on the particular choice of the presentation P for the group Γ, but is in fact an invariant of the group itself. For the deﬁnitions and notations see Section 2 of this thesis or [4]. C.1 Proposition. Let P ≡S|R and P ′ ≡S′|R′ be two ﬁnite presentations of a group Γ. Then δP ≃ δP ′ . (C.2)

Proof. Let P1 and P2 denote new presentations of Γ derived from P by adding a new generator and new relators; see Remark 2.4. In order to show (C.2) it is suﬃcient to verify that

δP ≃ δP1 ≃ δP2 ; by adding one by one a generator of S′ \ S to S and a generator of S \ S′ to S′ we can ﬁrst assume that P and P ′ have the same set of generators since

δP ≃ δP1 . Then (C.2) follows immediately from δP ≃ δP2 .

δP ≃ δP1 : We always have AreaP1 (w) ≤ AreaP (w) for every null-homotopic word w in S, since w is also a word in S1, because S ⊂ S1, and since

R ⊂ R1. We claim that AreaP (w) ≤ AreaP1 (w). By the deﬁnition of the combinatorial area AreaP1 (w) we have the following representation of w:

AreaP1 (w) F (S1) −1 ± w ≡ τi r¯iτi =:w, ¯ r¯i ∈ R1. (C.3) Yi=1 ′ Let in the following Nx(w ) ∈ Z denote the number of the letters x and −1 ′ x in the word w ; note that Nx(w) = Nx(¯w) for every letter x because 102 Chapter C. The Growth of a Dehn Function

−1 of (C.3). Suppose that rg := g a1 ...as =r ¯l for some l. First of all we have that 0 = Ng(w) = Ng(¯w) because w is a word in S. Therefore, since −1 −1 Nx(τiτi ) = 0 for all i, there must be rg among the remaining relatorsr ¯i −1 to cancel the letter g ofr ¯l. We conclude thatw ¯ must be of the form

−1 −1 −1 −1 ′−1 −1 −1 ′ −1 τi r¯iτi τ (g a1 ...asτ) τj r¯jτj τ (as ...a1 g)τ τk r¯kτk Yi Yj Yk with −1 −1 ′−1 −1 −1 F (S1) g a1 ...asτ τj r¯jτj τ as ...a1 g ≡ 1. Yj

−1 ′ F (S1) −1 ⇔ τ τ ≡ τj r¯jτj Yj Hence

F (S1) −1 −1 ′ −1 w ≡ τi r¯iτi τ τ τk r¯kτk Yi Yk F (S1) −1 −1 −1 ≡ τi r¯iτi τj r¯jτj τk r¯kτk Yi Yj Yk In other words we have canceled two conjugated relators from the representation (C.3); this contradicts the minimality of (C.3). Hence,r ¯i ∈ R for all i; by further substituting every occurence of g± in (C.3) with the corre- ± sponding word (a1 ...as) we obtain a representation of w in P ≡ S|R.

Therefore, AreaP (w) ≤ AreaP1 (w), and in particular AreaP (w) = AreaP1 (w) for every null-homotopic word w in S. On the other hand let w be a null-homotopic word in the alphabet S1. If −1 Ng(w) ≥ 1 then we can reduce Ng(w) by extracting a conjugate of rg from w as the following argument shows: let w = τ ′gτ. Then

F (S1) ′ −1 −1 −1 −1 w ≡ (τ a1 ...asτ)(τ rg τ) =: w1 (τ rg τ), with Ng(w1)= Ng(w)−1 and ℓ(w1)= ℓ(w)+s−1 <ℓ(w)+s. Furthermore, w1 is null-homotopic since π(w1) = π(w) = 1, where π : F (S1) → Γ is the quotient map of the presentation P1; see (2.1). By proceeding we get therefore that Ng(w) F (S1) ′ −1 −1 w ≡ w (τi rg τi), Yi=1 ′ ′ where w is null-homotopic word in S with ℓ(w ) ≤ ℓ(w)+ Ng(w) s ≤ (s + 1) ℓ(w). Note that the same holds if Ng(w) ≤ −1; we then have to extract rg 103

−1 ′ instead of rg . Hence w can be written as a product of AreaP (w )+ |Ng(w)| conjugated relators from R1 where |Ng(w)| ≤ ℓ(w). That is

′ AreaP1 (w) ≤ AreaP (w )+ ℓ(w).

We have shown that

δP (n) ≤ δP1 (n), δP1 (n) ≤ δP (1 + s) n + n. (C.4)

Hence, δP1 ≃ δP . ¯ δP ≃ δP2 : Since R ⊂ R2 = R ∪ R and S = S2, we get immediately the estimate AreaP2 (w) ≤ AreaP (w) for every null-homotopic word w in P . On ¯ the other hand let N := maxr¯∈R¯{ℓ(¯r)}; then everyr ¯ ∈ R can be written as a product of at most C := δP (N) conjugated relators from R sincer ¯ is null-homotopic. It follows that AreaP (w) ≤ C AreaP2 (w). Hence,

δP2 (n) ≤ δP (n), δP (n) ≤ C δP2 (n),

and we get that δP2 ≃ δP .

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Curriculum Vitae

Geboren am 27.08.1976 in Tehran, Iran. Die ersten zwei Schuljahre in Iran besucht. 1984 Ankunft in der Schweiz. Von 1985 bis 1987 Besuch der Pri- marschule an der St. Johann Schule, Basel. Danach am Mathematisch- Naturwissenschaftlichen Gymnasium Basel bis 1995: Eidgenössische Ma- turität Typus C. Von 1995 bis 1998 Physik Studium an der ETH Zürich. 1998 dann Wechsel zum Mathematik Studium, ebenfalls an der ETH Zürich. Diplom als Mathematiker im April 2003, Diplomarbeit betreut durch Prof. Dr. Urs Lang. Von Oktober 2003 bis 2009 Doktorarbeit an der ETH Zürich unter der Aufsicht von Prof. Dr. Urs Lang. Angestellt als Assistent am Mathematik Departement von 2003 bis September 2009.