Spaces with Convex Geodesic Bicombings

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Doctoral Thesis

Spaces with convex geodesic bicombings

Author(s): Descombes, Dominic

Publication Date: 2015

Permanent Link: https://doi.org/10.3929/ethz-a-010584573

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ETH Library Diss. ETH No. 23109

Spaces with Convex Geodesic Bicombings

A thesis submitted to attain the degree of

Doctor of Sciences of ETH Zurich

presented by

Dominic Descombes

Master of Science ETH in Mathematics citizen of Lignières, NE and citizen of Italy

accepted on the recommendation of

Prof. Dr. Urs Lang, examiner Prof. Dr. Alexander Lytchak, co-examiner

2015

Life; full of loneliness, and misery, and suffering, and unhappiness — and it’s all over much too quickly. – Woody Allen

Abstract

In the geometry of CAT(0) or Busemann spaces every pair of geodesics, call them α and β, have convex distance; meaning d ◦ (α, β) is a convex function I → R provided the geodesics are parametrized proportional to arc length on the same interval I ⊂ R. Therefore, geodesics ought to be unique and thus even many normed spaces do not belong to these classes. We investigate spaces with non-unique geodesics where there exists a suitable selection of geodesics exposing the said (or a similar) convexity property; this structure will be called a bicombing. A rich class of such spaces arises naturally through the construction of the injective hull for arbitrary metric spaces or more generally as 1- Lipschitz retracts of normed spaces. In particular, the injective hulls of word hyperbolic groups admit bicombings. We are able to generalize a variety of classical results to this broader context of weak non-positive curvature including a flat torus theorem and an asymptotic rank theorem (an equivalence of several notions of embeddings of flats) — results first shown for manifolds of non-positive curvature. New tools are needed in this less restricted setting. We employ a well-known, elegant barycenter construction that will be the key to many proofs.

Zusammenfassung

In CAT(0)- oder Busemann-Räumen gilt für beliebige zwei Geodäten, nennen wir diese α und β, dass der Abstand d ◦ (α, β) eine konvexe Funktion I → R ist. Vorausgesetzt wird dabei nur, dass die beiden Geodäten auf demselben Intervall I definiert und proportional zur Bogenlänge parametrisiert sind. Geodäten sind daher eindeutig, was für viele interessante Räume eine zu restriktive Forderung darstellt. Wir untersuchen daher Räume, in welchen wir eine Auswahl an Geodäten treffen können, so dass die obige Konvexitätsbedingung (oder eine ähnliche) zumindest für diese Teilmenge gilt; diese Struktur nennen wir ein bicombing. Diese Klasse umfasst nun auch alle normierten Räume und allgemeiner alle 1-Lipschitz-Retrakte solcher. Weitere Beispiele von Räumen mit bicombings entstehen auf natürliche Weise durch die Konstruktion injektiver Hüllen metrischer Räume und im Speziellen besitzen injektive Hüllen hyperbolischer Gruppen diese Struktur. Wir beweisen zentrale klassische Resultate in diesem allgemeineren Rah- men schwacher, nichtpositiver Krümmung. Darunter einen Satz über flache Tori und ein Theorem über den asymptotischen Rang (eine Äquivalenz über die Einbettbarkeit flacher Räume) — Resultate welche ursprünglich für Mannigfaltigkeiten nichtpositiver Krümmung gezeigt wurden. Manchmal führen die klassischen Beweispfade zum Ziel, häufig aber auch nicht. Unter den neuen Hilfsmitteln, die nun benötigt werden, befindet sich auch eine Schwerpunktkonstruktion, von welcher wir wiederholt und an kritischen Stellen Gebrauch machen.

vii

Acknowledgments

Most of all, I want to express my gratitude to my advisor Urs Lang. Without his patience, gentle guidance, and firm support through these years, I would not have made it. I could always stop by his office to discuss mathematics and always left it with new ideas and renewed motivation for doing my research. I also thank the Swiss National Science Foundation for supporting me financially. And of course, I am also indebted to Alexander Lytchak for agreeing to act as co-examiner. The time I spent at the department of mathematics would not have been so rewarding without the wonderful people of group 1 & 4. Thank you all! And in particular, I am grateful for the time and fun I had with my office mates Dante Bonolis, Maël Pavón and Michael Th. Rassias. Finally, I thank my family for the support during all of my studies and most of all my girlfriend Julia.

Contents

1 Introduction ...... 1

2 Basic concepts ...... 6 2.1 Metric spaces ...... 6 2.2 Bicombings ...... 9 2.3 Group actions ...... 11

3 Improving bicombings ...... 12 3.1 From conical to convex bicombings ...... 12 3.2 Straight curves ...... 15 3.3 Combinatorial dimension ...... 18

4 Boundary at inﬁnity ...... 22

5 Flat planes ...... 28 5.1 Flat strips and half-planes ...... 28 5.2 The Flat Plane Theorem ...... 32

6 Barycenters ...... 36 6.1 The construction ...... 36 6.2 Linear barycenters ...... 42 6.3 Integration ...... 44

7 Semi-simple isometries ...... 45 7.1 Minimal displacement and axes ...... 45 7.2 σ-Axes ...... 48

8 Flat tori ...... 52

9 Asymptotic rank ...... 57

A Injective metric spaces ...... 64 A.1 Basic deﬁnitions and injective hulls ...... 64 n A.2 A characterization of injective subsets in `∞ ...... 68

Bibliography ...... 74

Chapter 1

Introduction

The geometry of spaces of global non-positive curvature is largely dominated by the convexity of the distance function. Thus a considerable part of the theory of CAT(0) spaces [Bal, BriH] carries over to Busemann spaces [Bus, Pap] (deﬁned by the property that d ◦ (α, β) is convex for any pair of constant speed geodesics α, β parametrized on the same interval). However, this larger class of spaces has the defect of not being preserved under limit processes. For example, among normed real vector spaces, exactly those with strictly convex norm satisfy the Busemann property, and a sequence of such n norms on R , say, may converge to a non-strictly convex norm. This motivates the study of an even weaker notion of non-positive curvature that dispenses with the uniqueness of geodesics but retains the convexity condition for a suitable selection of geodesics (compare Section 10 in [Kle]). In any normed space, the aﬃne segments t 7→ (1−t)x+ty (t ∈ [0, 1]) provide such a choice. In particular, the relaxed condition carries the potential for simultaneous generalizations of results for non-positively curved and Banach spaces. A selection of geodesics for some metric space (X, d) we call a geodesic bicombing σ and thereby mean a map σ : X × X × [0, 1] → X that singles out a constant speed geodesic σxy := σ(x, y, ·) from x to y (that is, σxy(0) = 0 0 0 x, σxy(1) = y, and d(σxy(t), σxy(t )) = |t − t |d(x, y) for all t, t ∈ [0, 1]) for every pair (x, y) ∈ X × X. Because we work exclusively with bicombings that are composed of geodesics, we suppress the term “geodesic” throughout; a bicombing is thus implicitly understood to be a geodesic bicombing. We are interested in bicombings satisfying one of the following two conditions, each of which implies that σ is continuous and, hence, X is contractible via X ×[0, 1] → X, (x, t) 7→ σox(1−t) for an arbitrary choice of basepoint o ∈ X. We call σ convex if

the function t 7→ d(σxy(t), σx0y0 (t)) is convex on [0, 1] (1.1) for all x, y, x0, y0 ∈ X, and we say that σ is conical if

0 0 d(σxy(t), σx0y0 (t)) ≤ (1 − t) d(x, x ) + t d(y, y ) for all t ∈ [0, 1] (1.2) 2 CHAPTER 1. INTRODUCTION and x, y, x0, y0 ∈ X. Obviously every convex bicombing is conical, but the reverse implication does not hold in general (see Example 3.2). For this, in order to pass condition (1.2) to subsegments, one would need to know in addition that σ is consistent in the sense that σpq(λ) = σxy((1 − λ)s + λt) (1.3) whenever x, y ∈ X, 0 ≤ s ≤ t ≤ 1, p := σxy(s), q := σxy(t), and λ ∈ [0, 1]. Thus, every conical and consistent bicombing is convex. Since we always assume our bicombings to be conical at least (for requiring a space to admit a bicombing is nothing more than it being geodesic), we adopt the convention that bicombing shall mean conical (and geodesic) bicombing from now on. Consequently, a consistent bicombing is automatically convex. Furthermore, we will say that a bicombing σ is reversible if

σxy(t) = σyx(1 − t) for all t ∈ [0, 1] (1.4) and x, y ∈ X. Consistent and reversible bicombings have also been employed in [HitL, FoeL]. We arrive at the following hierarchy of global non-positive (but non-coarse) curvature conditions (A) ⇒ (B) ⇒ (C) ⇒ (D) ⇒ (E) for a geodesic metric space X:

(A) X is a CAT(0) space,

(B) X is a Busemann space,

(D) X admits a convex bicombing,

(E) X admits a (conical) bicombing.

Clearly, if X is uniquely geodesic, then (E) ⇒ (B). When X is a normed real vector space, (A) holds if and only if the norm is induced by an inner product (compare Proposition II.1.14 in [BriH]), (B) holds if and only if the norm is strictly convex (see Proposition 8.1.6 in [Pap]), and (C) is always satisﬁed. In the general case, (C) is stable under limit operations, whereas (B) is not (see, for instance, Section 10 in [Kle] again). It is unclear whether (E) ⇒ (D) and (D) ⇒ (C) without further conditions. The goal of Chapter 3 is to establish these implications under suitable assumptions and to address questions of uniqueness. Our ﬁrst result refers to (E) and (D).

Theorem 3.4. Let X be a proper metric space with a bicombing. Then X also admits a convex bicombing.

The idea of the proof is to resort to a discretized convexity condition and then to gradually decrease the parameter of discreteness by the “cat’s cradle” construction from [AleB]. The passage from (D) to (C) appears to be more subtle. We ﬁrst observe that the linear segments t 7→ (1 − t)x + ty in an arbitrary normed space X may be characterized as the curves α: [0, 1] → X with the property that t 7→ d(z, α(t)) is convex on [0, 1] for every z ∈ X (see Theorem 3.8). We therefore term curves with this property 3 straight (in a metric space X). Since the geodesics of a convex bicombing are necessarily straight, the above observation shows that normed spaces have no such bicombing other than the linear one. By contrast, there are instances of non-linear straight geodesics in compact convex subspaces of normed spaces as well as multiple convex bicombings in compact metric spaces (see Examples 3.9 and 3.10). Nevertheless, we prove a strong uniqueness property of straight curves (Proposition 3.13) in spaces satisfying a certain ﬁnite dimensionality assumption, introduced by Dress in [Dre] and explained further below. This gives the following result regarding items (D) and (C).

Theorem 3.14. Let X be a metric space of ﬁnite combinatorial dimension in the sense of Dress, or with the property that every bounded subset has ﬁnite combinatorial dimension. Suppose that X possesses a convex bicombing σ. We then have that σ is consistent, reversible, and unique, that is, σ is the only convex bicombing on X.

Our interest in Theorems 3.4 and 3.14 comes from the fact that property (E) holds in particular for all injective metric spaces (equivalently absolute 1-Lipschitz retracts). X is injective if for every isometric inclusion A ⊂ B of metric spaces and every 1-Lipschitz map f : A → X there exists a 1-Lipschitz extension f : B → X of f. Basic examples include the real line, all `∞ spaces, and all metric (R-)trees. Furthermore, by a result of Isbell [Isb1], every metric space X possesses an essentially unique injective hull E(X), which is the smallest injective space into which X embeds isometrically. See Appendix A n for a short survey where we also derive a characterization of injective subsets of `∞. E(X) yields a finite polyhedral complex with `∞ metrics on the cells in case X is finite. Isbell’s explicit construction was rediscovered and further explored by Dress [Dre]. The combinatorial dimension dimcomb(X) of a general metric space X is the supremum of the dimensions of the polyhedral complexes E(Y ) for all finite subspaces Y of X. In case X is already injective, every such E(Y ) embeds isometrically into X, so dimcomb(X) is bounded by the supremum of the topological dimensions of compact subsets of X. Now by a theorem in [Lan], for every word hyperbolic group Γ the injective hull E(Γ) satisfies the prerequisites of the theorem above and, moreover, Γ acts properly and cocompactly on E(Γ).

Theorem 3.16. Every word hyperbolic group Γ acts properly and cocompactly by isometries on the proper, finite-dimensional metric space E(Γ) admitting a consistent bicombing σ that is furthermore reversible and unique. Hence σ is equivariant with respect to the isometry group of E(Γ), i.e. for every isometry L: E(Γ) → E(Γ) and x, y ∈ E(Γ) we have L ◦ σxy = σL(x)L(y). In Chapter 4 we discuss the asymptotic geometry of complete metric spaces X with a consistent bicombing σ. We define the geometric boundary ∂σX in terms of geodesic rays consistent with σ, and we equip Xσ = X∪∂σX with a natural metrizable topology akin to the cone topology in the case of CAT(0) spaces. If X is locally compact, Xσ is compact. In view of Theorem 3.16, this unifies and generalizes the respective constructions for CAT(0) or Busemann spaces and for hyperbolic groups. By virtue of the bicombing one can then give a rather direct proof of the following result on Xσ. 4 CHAPTER 1. INTRODUCTION

Theorem 4.4. Let X be a complete metric space with a consistent bicombing σ. Then Xσ is an absolute retract; in particular, Xσ is contractible and locally contractible. Moreover, ∂σX is a Z-set in Xσ, that is, for every open set U in Xσ the inclusion U \ ∂σX,→ U is a homotopy equivalence. Note that we do not assume X to be proper or finite dimensional. Bestvina and Mess [BesM] proved the analogous result for the Gromov closure P (Γ) of the (contractible) Rips complex of a hyperbolic group Γ. Theorem 3.16 and Theorem 4.4 together thus provide an alternative to their result, which has a number of important applications (see Corollaries 1.3 and 1.4 in [BesM]). The results so far have been published in [DesL1] together with Urs Lang. Then in Chapters 5–9 we carry the analogy with CAT(0) and Busemann spaces further with regard to existence results for flat subspaces, that is, for subspaces isometric to a normed space. Our first main result is the following generalization of the hyperbolicity criterion for cocompact CAT(0) spaces stated on p. 119 in [Gro]. A detailed proof of Gromov’s result was given in [Bri]. For the case of Busemann spaces, both Theorem 5.5 and Theorem 8.3 below were shown by Bowditch [Bow]. Theorem 5.5 (Flat plane). Let X be a proper metric space with a consistent and reversible bicombing σ and with cocompact isometry group. Then X is hyperbolic if and only if X does not contain an isometrically embedded normed plane. En route to this theorem we discuss a generalized Flat Strip Theorem. Unlike for Busemann spaces, the σ-convex hull of a pair of parallel σ-lines may be “thick” and the two lines may span different, though pairwise isometric, flat (normed) strips. We also give a criterion for the existence of an embedded normed half-plane. This is then used in Section 5.2 for the proof of the above result. A variant of this theorem for injective metric spaces is also shown. Another well-known result from the theory of spaces of non- positive curvature is the Flat Torus Theorem, originally proven for smooth manifolds in [GroW, LawY]. A detailed account of this result and its applications in the context of CAT(0) spaces is given in Chap. II.7 of [BriH]. We have: Theorem 8.3 (Flat torus). Let X be a proper metric space with a consistent and reversible bicombing σ. Let Γ be a group acting properly and cocompactly by isometries on X, and suppose that σ is Γ-equivariant. If Γ has a free abelian subgroup A of rank n ≥ 1, then X contains an isometrically embedded n-dimensional normed space on which A acts by translation.

Here, σ being Γ-equivariant means that γ ◦ σxy = σγ(x)γ(y) for all isometries γ ∈ Γ and (x, y) ∈ X × X. In Section 7 we establish basic properties of semi-simple isometries of spaces with bicombings. We employ a barycenter map for ﬁnite subsets which was introduced in the context of Busemann spaces in [EsSH]. Section 7.2 addresses the question whether a hyperbolic (axial) isometry of a metric space with a consistent bicombing σ also possesses an axis that is at the same time a σ-line. This is false in general but holds true for the hyperbolic elements of a group Γ as in Theorem 8.3. As an auxiliary tool we use a ﬁxed point theorem for non-expansive mappings originally proven for 5

Banach spaces in [GoeK]. Finally, Chapter 8 is dedicated to the proof of Theorem 8.3. Chapters 5–8 with the exception of the chapter about barycenters are a publication in preparation [DesL2] and will be co-authored with Urs Lang. In that paper some of the results (but not the main results) are stated in a somewhat more general setting. Here we decided not to do so since this facilitates the presentation without any drawback for the main results. The ﬁnal Chapter 9 will revolve around the following theorem which generalizes a result of Kleiner for Busemann spaces; see Proposition 10.22 and the more comprehensive Theorem D in [Kle]. Also compare [Wen].

Theorem 9.1. Let X be a proper metric space with a bicombing σ and cocompact isometry group. Suppose there are sequences Rk ∈ (0, ∞), Sk ⊂ X, and a normed vector n 1 n space (R , k · k), so that Rk → ∞, and R Sk converges to the unit ball B ⊂ (R , k · k) in k n the Gromov-Hausdorff topology. Then (R , k · k) can be isometrically embedded in X. Not that the bicombing is not assumed to be consistent nor convex or even reversible. With this we are able to present Kleiner’s Theorem D in our setting. In particular, this then shows that for every proper cocompact metric space X with a bicombing the following is true: X contains a flat (normed) n-plane whenever there is a quasi-isometric n embedding of R into X. Such a result was first shown for manifolds of non-positive curvature in [AndS]. The proof of Proposition 10.22 in [Kle] is performed in the spirit of CAT(0) geometry whereas our generalized statement we prove by a quite different argument. One ingredient is a Riemannian integral defined by means of the barycenter maps. The central steps then are a convolution like operation and a kind of metric differentiation argument. We also devote a whole chapter to introduce the barycenters needed. As a byproduct, we show that a space X is isometric to a convex subset of a normed space if it admits a barycenter satisfying a simple extra property. The final chapter about Theorem 9.1 and the auxiliary results about barycenters form another publication in preparation [Des1]. Chapter 2

Basic concepts

We use this chapter to ﬁx some notation and recall basic deﬁnitions and facts concerning metric spaces. For the proofs we omit and for a comprehensive introduction we refer to Part I in [BriH].

2.1 Metric spaces

Let X be a set. By a metric we mean a map d: X × X → R satisfying (i) d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 if and only if x = y, (ii) d(x, y) = d(y, x) for all x, y ∈ X and (iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X (triangle inequality). Then, when endowed with a metric, we call X a metric space; and in case we need to stress which metric is meant, we shall write (X, d). Some notation: For arbitrary x ∈ X; A, B ⊂ X; r ∈ R, we set diam(A) := supa,b∈A d(a, b) the diameter of A, d(x, A) := infa∈A d(x, a), d(A, B) := infa∈A,b∈B d(a, b), and denote by B(x, r) := {y ∈ X | d(x, y) ≤ r} the closed ball around x of radius r and by Ur(A) := {x ∈ X | d(x, A) < r} the open neighborhood of A. A map f : X → Y between metric spaces (X, dX ), (Y, dY ) is an isometry by definition, if it is surjective and dY (f(p), f(q)) = dX (p, q) for all p, q ∈ X. If such an f exists, then X and Y are called isometric. If the requirement of surjectiveness for f is dropped, we call this an isometric embedding of X into Y . The isometry group of a metric space X is the set of all isometries f : X → X with concatenation as group operation. A map f with dY (f(p), f(q)) ≤ c dX (p, q) for all p, q ∈ X is called c-Lipschitz continuous or just Lipschitz in case the value of the constant is irrelevant. A metric space is called proper if every closed ball B(x, r) is compact and this implies completeness. In metric spaces there is the following notion of length for paths (also called curves or sometimes segments) — i.e. for continuous maps α: I → X where I is an interval in R: k X l(α) := sup d(α(ti−1), α(ti)) t0≤t1≤···≤tk i=1 2.1. METRIC SPACES 7 where the supremum runs over all k ∈ N and tagged points t0 ≤ t1 ≤ · · · ≤ tk in I. We adopt the notation N = {1, 2, 3,...} and N0 = {0, 1, 2,...} throughout. When I = [a, b] we say that α is connecting (or joining) the points α(a) and α(b), and (whenever [a, b] ⊂ I) we always have l(α|[a,b]) ≥ d(α(a), α(b)) by the triangle inequality. In case that equality occurs for all a, b provided [a, b] ⊂ I, then α is commonly called a geodesic. But it is useful to adopt a slightly more restricted version of this definition. A continuous curve α: I → X for some interval I ⊂ R is called a geodesic if there is a constant c ∈ R such that d(α(t), α(t0)) = c|t − t0| for all t, t0 ∈ I, (2.1) thus the geodesics are required to be parametrized proportional to arc length (or parametrized by arc length in case of c = 1). We also refer to this property (2.1) by saying that the geodesics have constant speed (or unit speed if c = 1). Nevertheless, both definitions of geodesics coincide up to monotonic (but not necessarily continuous) reparametrizations. From now on geodesic shall implicitly mean constant speed geodesic when not mentioned otherwise. Note that we allow for c = 0, giving stationary geodesics, and that geodesics are always global geodesics (in contrast to the wording in Riemannian geometry where geodesics are usually only required to be local geodesics). In case we need a local geodesic we will explicitly add the adjective and thereby mean — to be precise — that for a path α: I → X and every t ∈ I there is an ε > 0 such that α|[t−ε,t+ε]∩I is a geodesic. So with our denomination there can be no closed geodesics (i.e. a periodic geodesic) like 2 2 the boundary of an `1 ball in `∞ which is a local geodesic. Throughout this thesis we n n understand `p , for n ∈ N and 1 ≤ p ≤ ∞, to be the Banach space R endowed with the Lp-norm 1 p p p kxkp := (|x1| + ··· + |xn| ) when p < ∞ and kxk = max{|x1|,..., |xn|} for p = ∞. Just for later use we also define the Banach spaces `p(X) for arbitrary index sets X as follows. For a real valued function X f : X → R (the set of all such we denote by R ), the Lp-norm shall be as usual

1 ! p X p kfkp := |f(x)| for p < ∞ and kfk∞ := sup |f(x)| for p = ∞; x∈X x∈X

X quantities which may be infinite. `p(X) then is the subset of R where the said norm is finite. From (2.1) we see that geodesics are just isometric embeddings of intervals where the metric on the interval is rescaled by a factor c. In case I = R and c = 1 we call a geodesic α a line (thus a line is just an isometric embedding of R), and if I = R+ := [0, ∞) and c = 1 we refer to α as a ray and α is said to issue from α(0). Note that there are spaces where no geodesics besides the stationary ones exist. A simple 1 2 example is S := {x | kxk2 = 1} ⊂ R with the restricted Euclidean metric since the arc connecting two distinct points is always longer than their distance. Spaces were there is a geodesic connecting any pair of points are denoted geodesic spaces. If, in addition, for every pair of points there is exactly one geodesic joining them (up to affine reparametrization), the space is called uniquely geodesic. Basic examples of geodesic 8 CHAPTER 2. BASIC CONCEPTS spaces are normed spaces where the linear segment t 7→ (1 − t)x − ty joins x to y realizing the distance. Moreover, they are uniquely geodesic if and only if their norm k · k is strictly convex, i.e. k(1 − t)v + twk < 1 for all t ∈ (0, 1) and distinct vectors v, w n of norm 1. For instance, this is the case in all `p spaces except for p = 1 or p = ∞ (and n ≥ 2) where there are infinitely many geodesics connecting any pair of distinct points. To give an example, for every 1-Lipschitz function f : I → R the paths t 7→ (t, f(t)) and 2 t 7→ (f(t), t) are geodesics in `∞, and in fact every geodesic therein can be described that way. Concerning whether a complete geodesic space is proper or not, we remark that local compactness is sufficient for such spaces to be proper by virtue of the Hopf-Rinow theorem. Nevertheless, we will always write proper instead of locally compact even in these cases. We turn to the notion of Hausdorff and Gromov-Hausdorff distance. Let X be any metric space, then for two non-empty subsets A, B the Hausdorff distance is the real value dH (A, B) := inf{ε | A ⊂ Uε(B),B ⊂ Uε(A)}. Clearly, this is not a metric on the set 2X of all subsets of X; the value can be infinite if no ε with A ⊂ Uε(B),B ⊂ Uε(A) exists and the distance dH (A, B) is zero if and only if the closures A, B coincide. However, triangle inequality holds and dH is a metric on the set of all non-empty, bounded and closed subsets of some metric space X. Often we want to compare spaces which are not subsets of a common ambient space. This leads to the following definition. Let X,Y be two metric spaces, then their Gromov-Hausdorff distance is dGH (X,Y ) := inf dH (ϕ(X), ψ(Y )) ϕ,ψ,Z where the infimum runs over all metric spaces Z with isometric embeddings ϕ: X → Z, ψ : Y → Z. Alternatively, and to circumvent set theoretic anomalies, one may replace Z by the disjoint sum of X,Y and take the Hausdorff distance with respect to all the pseudometrics which restrict to the given metrics on X and Y . Here pseudometric means that we allow distinct points to have zero distance. Now for compact spaces X,Y we have that dGH (X,Y ) = 0 if and only if they are isometric. And a sequence of spaces Xi converges to a limit X in Gromov-Hausdorff distance if any only if there is a sequence of (typically non-continuous) maps fi : X → Xi such that supx,y∈X |d(fi(x), fi(y))−d(x, y)| and dH (fi(X),Xi) converge to zero for i → ∞. (The index set of a sequence is always understood to be N = {1, 2, 3,...}.) Finally, since we use the well-known theorem of Arzelà-Ascoli rather frequently, we state a version adapted to our needs. Recall that a family of maps fi : X → Y from a metric space into another is called equicontinuous if for every ε > 0 there is a δ > 0 such that d(x, y) < δ implies d(fi(x), fi(y)) < ε uniformly for all i and x, y ∈ X. Theorem 2.1 (Arzelà-Ascoli). Let X be a separable metric space and Y a proper metric spaces. Then every sequence fi : X → Y such that

(i) the family fi|C is equicontinuous for every compact set C ⊂ X,

(ii) {fi(x) | i ∈ N} is bounded for every x ∈ X, 2.2. BICOMBINGS 9 then has a subsequence converging uniformly on compact subset to a continuous limit f. Moreover, if X happens to be geodesic, then {fi(x) | i ∈ N} being bounded for one x implies it for every x.

Most often in this thesis the family of functions we are considering will have a uniform Lipschitz constant on every compact set, this clearly renders them equicontinuous on these set.

2.2 Bicombings

Recall the deﬁnition and various properties of bicombings deﬁned on page 1–2 in the introduction. Also recall that since a bicombing σ is a selection of one geodesic for every (ordered) pair of points, a space possessing such a σ is necessarily geodesic. Occasionally, when readability demands it, we will write [x, y] instead of σxy for the geodesics of a bicombing. And, when using this notation, we do not explicitly distinguish between [x, y] as a map and its image (or trace) as long as no confusion arises. The most basic of weak non-positive curvature conditions we are considering (and which every bicombing has by convention) is the conical property (1.2). Moreover, let us mention here that there is no loss of generality in assuming a space X with (conical) bicombing σ to be complete. As for two points x, y in the completion X, we may (and must) set

σxy := limk→∞ σxkyk given two sequences xk → x, yk → y with xk, yk ∈ X. For every t ∈ [0, 1], the sequence σxkyk (t) is then Cauchy and the convergence σxkyk → σxy is uniform and the limit independent of the chosen sequences by (1.2). Furthermore, and since we very often work with proper spaces, let us note:

Proposition 2.2. If a proper metric space X admits a (possibly non-reversible) bicombing σ, then X admits a reversible bicombing σ˜.

Proof. First we claim that the existence of a bicombing is equivalent to the existence of a midpoint assignment (x, y) 7→ x#y (i.e. x#y is a point such that d(x, x#y) = 1 d(x#y, y) = 2 d(x, y)) with the additional (conical) property 1 1 d(x#y, x0#y0) ≤ d(x, x0) + d(y, y0); (2.2) 2 2 and this equivalence holds for any complete space X (which every proper space is). Obviously (x, y) 7→ σxy(1/2) is a valid midpoint assignment with (2.2) given a bicombing σ. To define a bicombing from a midpoint assignment, set σxy(0) := x, σxy(1) := y, k k and Q0 := {0, 1}. Then for every k ≥ 1 let Qk := {m/2 | m odd , 0 < m < 2 }, and −k −k −k −k inductively define σxy(t) at t ∈ Qk to be σxy(t−2 )#σxy(t+2 ). See that t−2 , t+2 lie in ∪0≤i≤k−1Qi, and by induction it is easy to verify (1.2) for t ∈ Q := ∪k∈N0 Qk as well 0 0 0 as d(σxy(t), σxy(t )) = |t − t |d(x, y) for all t, t ∈ Q. So the maps σxy : Q → X defined so far are Lipschitz continuous for every pair (x, y), hence extend uniquely to geodesics σxy : [0, 1] → X (preserving (1.2)) as X is complete. 10 CHAPTER 2. BASIC CONCEPTS

In view of this it suﬃces to show that, starting from #, we can construct a symmetric 2 midpoint assignment , i.e. one with x y = y x for all x, y ∈ X. For a pair (x, y) ∈ X we deﬁne the sequences xi, yi by

x0 = x, y0 = y and xi = xi−1#yi−1, yi = yi−1#xi−1 for all i ≥ 1.

We have d(xi+1, yi+1) ≤ d(xi, yi) by applying (2.2) to the definition of the sequences as well as d(x , y ) d(x , x ) = d(x , y ) = i i = d(x , y ) = d(y , y ) for all j > i. i j j i 2 i j j i The latter results by induction over j and the fact that a, b ∈ M(x, y) implies a#b, 1 b#a ∈ M(x, y) where M(x, y) := {z | d(x, z) = d(z, y) = 2 d(x, y)} denotes the set of all midpoints between x and y. Consequently, if the monotone sequence d(xi, yi) would not converge to zero, then d(xi, xj) ≥ ε for all i, j and some ε > 0 thereby contradicting compactness. So both sequences converge to a common limit and we set x y = limi→∞ xi = limi→∞ yi = y x. The verification that obeys (2.2) follows when taking the limit i → ∞ of either of the inequalities 1 1 d(x , x0 ) ≤ d(x, x0) + d(y, y0), i i 2 2 1 1 d(y , y0) ≤ d(x, x0) + d(y, y0), i i 2 2 0 0 0 0 where xi, yi are the sequences in the construction of x y . And these inequalities are easily shown by mutual induction. Given a bicombing σ, we call a set C σ-convex if the image of every σ-geodesic (but not necessarily every geodesic) connecting two points of C is contained in C. Note that in the absence of reversibility for σ this requires both geodesics σxy, σyx to run entirely in C whenever x, y ∈ C. For any set D ⊂ X, we define the (closed) σ-convex hull to be the intersection of all closed σ-convex sets containing D. If we denote this hull by D, we have diam(D) = diam(D). This is a direct consequence of the following equivalent construction of the σ-convex S hull. Given D, let Dk be defined inductively as D0 = D and Dk = 2 σxy([0, 1]) (x,y)∈Dk−1 for k ≥ 1. Clearly Dk−1 ⊂ Dk for all k and the set ∪k∈N0 Dk is σ-convex. Now D is the closure of that set (which is still σ-convex as the bicombing is continuous). Since diam(Dk−1) = diam(Dk) from the conical inequality, taking the σ-convex hull does not increase the diameter. Also, observe that for a non-consistent or non-reversible bicombing the σ-convex hull of {x, y} may not just be σxy([0, 1]). In fact, the only bicombings for which the trace of every geodesic σxy is σ-convex are precisely the ones that are consistent and reversible. For arbitrary bicombings, examples of σ-convex sets include balls and neighborhoods Ur(A) of σ-convex sets A as well as M(x, y), the set of midpoints for any pair x, y.A midpoint for x, y is a point z with d(x, z) = d(z, y) = 1 2 d(x, y), and every geodesic α from α(0) = x to α(1) = y has to pass M(x, y) exactly at α(1/2) and only then. 2.3. GROUP ACTIONS 11

2.3 Group actions

Here we collect the definitions about group actions we need in the sequel. By an action by isometries of a group Γ on a metric space X we mean a group homomorphism ϕ:Γ → Isom(X) from Γ into the group of isometries on X. We do not distinguish (notation- wise) between γ as an element of Γ and the isometry ϕ(γ) assigned to it. Therefore we simply write γ(x) instead of (ϕ(γ))(x) for all x ∈ X. Equivalently to the definition above, we may define an action by isometries as a map Γ × X → X, (γ, x) → γ(x) such that γ(·): X → X is an isometry for all γ ∈ Γ and γ(δ(x)) = (γδ)(x), e(x) = x for all x ∈ X and γ, δ ∈ Γ where e ∈ Γ denotes the identity element. A proper action is one where {γ | γ(C) ∩ C 6= ∅} is finite for every compact subset C ⊂ X. We will use this mainly in the context of proper metric spaces X and there we can replace compact by bounded and have that {γ | γ(B(x, r)) ∩ B(x, r) 6= ∅} is finite for all x ∈ X, r ∈ R. An action by Γ on X is called cocompact if there exists a compact set C ⊂ X such that ΓC = X where ΓC = {γ(c) | γ ∈ Γ, c ∈ C}, thus the orbit of that compact set covers the whole space. This last property is especially useful in combination with the Arzelà-Ascoli Theorem 2.1. Let fi : X → Y be a sequence of maps and assume some group Γ acts cocompactly by isometries on the proper target space Y . If fi fulfills the requirements of Theorem 2.1 for a geodesic X but with the exception of part (ii) ({fi(x) | i ∈ N} is bounded for every x ∈ X), then there is a sequence γi ∈ Γ such that property (ii) holds for the sequence γi ◦ fi. Whenever the canonical action of the full isometry group Isom(X) on X is cocompact, we simply call X a cocompact metric space. Finally, given an action by Γ on a space with bicombing σ, we then say that σ is Γ-equivariant if

γ ◦ σxy = σγ(x)γ(y) for all x, y and γ ∈ Γ. (2.3)

If we want to express that (2.3) holds for a single isometry γ, we say that σ is γ- equivariant. Chapter 3

Improving bicombings

First we discuss some simple properties and examples of (conical) bicombings as deﬁned in (1.2). Lemma 3.1. Let X¯ be a metric space with a bicombing σ¯. If π : X¯ → X is a 1-Lipschitz ¯ retraction onto some subspace X of X, then σ := π ◦ σ¯|X×X×[0,1] deﬁnes a bicombing on X. Furthermore, if σ¯ is reversible so is σ.

Proof. Note that since π is a 1-Lipschitz retraction, σxy is indeed a geodesic for every pair (x, y) ∈ X2. Furthermore, since π is 1-Lipschitz and σ¯ is conical, we have

0 0 d(σxy(t), σx0y0 (t)) ≤ d(¯σxy(t), σ¯x0y0 (t)) ≤ (1 − t) d(x, x ) + t d(y, y ) for all t ∈ [0, 1] and x, y, x0, y0 ∈ X, so σ is conical as well. The last statement follows from σxy(t) = π(¯σxy(t)) = π(¯σyx(1 − t)) = σyx(1 − t). As mentioned earlier (also see Theorem A.5), a direct consequence of this lemma is that all injective metric spaces (or absolute 1-Lipschitz retracts) admit bicombings. An equally simple application gives an example of a bicombing that is not convex. 2 Example 3.2. Consider the set X of all points (u, v) ∈ `∞ with |u| ≤ 2 and b(u) := 2 |u| − 1 ≤ v ≤ |b(u)| endowed with the metric induced by the `∞ norm. Note that this is a geodesic metric space (even injective by Lemma A.8). With respect to this metric, the vertical retraction π from the triangle X¯ := {(u, v) | b(u) ≤ v ≤ 1} onto X that maps (u, v) to (u, min{v, |b(u)|}) is 1-Lipschitz. The linear bicombing σ¯ on X¯, deﬁned by σ¯xy(t) := (1 − t)x + ty, is convex. By Lemma 3.1, σ := π ◦ σ¯|X×X×[0,1] deﬁnes a bicombing on X. For x = (−2, 1) and y = (2, 1), we have σxy(1/4) = (−1, 0), σxy(3/4) = (1, 0), and σxy(1/2) = (0, 1). Hence, for z = (0, −1), the function t 7→ kσxy(t) − σzz(t)k∞ = kσxy(t) − zk∞ is clearly not convex.

3.1 From conical to convex bicombings

We now deﬁne a relaxed notion of convexity that will be useful for the proof of Theo- rem 3.4. We say that a bicombing σ on a metric space X is 1/k-discretely convex if, for 3.1. FROM CONICAL TO CONVEX BICOMBINGS 13 all x, y, x0, y0 ∈ X, the convexity condition t − s s − r d(σ (s), σ 0 0 (s)) ≤ d(σ (r), σ 0 0 (r)) + d(σ (t), σ 0 0 (t)) xy x y t − r xy x y t − r xy x y holds whenever the three numbers r < s < t belong to [0, 1] ∩ (1/k)Z. To check this condition, it clearly suﬃces to verify the “local” inequality

2d(σxy(s), σx0y0 (s)) ≤ d(σxy(s − 1/k), σx0y0 (s − 1/k))

+ d(σxy(s + 1/k), σx0y0 (s + 1/k)) for all s ∈ (0, 1) ∩ (1/k)Z. Note that every (conical) bicombing is 1/2-discretely convex. Proposition 3.3. Suppose that X is a complete metric space with a bicombing σ that is 1/k-discretely convex for some integer k ≥ 2. Then X also admits a bicombing that is 1/(2k − 1)-discretely convex.

Proof. Set l := 2k − 1. To construct the desired new bicombing σ˜, ﬁx x and y and deﬁne the sequences pi and qi recursively as q0 := σxy(k/l) and

pi := σxqi−1 (1 − 1/k), qi := σpiy(1/k), for i ≥ 1.

p σ˜xy q

p4 q3

y x

q1 = σp1y(1/k)

p1 = σxq0 (1 − 1/k)

q0 = σxy(k/l)

Since σ is conical, we get the inequalities

d(pi, pi+1) ≤ (1 − 1/k) d(qi−1, qi),

d(qi, qi+1) ≤ (1 − 1/k) d(pi, pi+1).

It follows that pi and qi are Cauchy sequences, so pi → p and qi → q. Then σxqi → σxq and σpiy → σpy uniformly, again because σ is conical. Note that p = σxq(1 − 1/k) and q = σpy(1/k); we can thus deﬁne σ˜xy(s) for s ∈ [0, 1] ∩ (1/l)Z so that ( σxq (l/k)s if s ≤ k/l, σ˜xy(s) = (3.1) σpy (l/k)s − (k − 1)/k if s ≥ (k − 1)/l. 14 CHAPTER 3. IMPROVING BICOMBINGS

0 To declare σ˜xy on all of [0, 1], we connect any pair of consecutive points x :=σ ˜xy(s) 0 and y :=σ ˜xy(s + 1/l), where s ∈ [0, 1) ∩ (1/l)Z, by the geodesic t 7→ σx0y0 (l(t − s)) for t ∈ [s, s + 1/l]. 0 0 0 0 Now if p , q and σ˜x0y0 result from the same construction for two points x and y , we want to show that for s ∈ (0, 1) ∩ (1/l)Z we have

2d(˜σxy(s), σ˜x0y0 (s)) ≤ d(˜σxy(s − 1/l), σ˜x0y0 (s − 1/l))

+ d(˜σxy(s + 1/l), σ˜x0y0 (s + 1/l)). In view of (3.1) this corresponds to the inequality

2d(α(t), β(t)) ≤ d(α(t − 1/k), β(t − 1/k)) + d(α(t + 1/k), β(t + 1/k)) where α = σxq, β = σx0q0 , t = (l/k)s if s ≤ (k − 1)/l; and α = σpy, β = σp0y0 , t = (l/k)s − (k − 1)/n if s ≥ k/l. However, these inequalities hold since σ is 1/k- discretely convex. This shows that σ˜ is 1/l-discretely convex. Now it follows easily from the construction that σ˜ is also conical.

From this result, we now obtain Theorem 3.4 by an application of the Arzelà-Ascoli Theorem 2.1, which requires X to be proper. We do not know whether the implication (E) ⇒ (D) holds in general without this assumption. Theorem 3.4. Let X be a proper metric space with a bicombing. Then X also admits a convex bicombing. Proof. Starting from the given bicombing σ1 := σ, we construct, by means of Proposi- i i tion 3.3, a sequence of bicombings σ on X such that σ is 1/ki-discretely convex, where i k1 = 2 and ki+1 = 2ki −1. This collection of maps σ is equicontinuous on every bounded set, and for every ﬁxed (x, y, t) in the separable domain X × X × [0, 1], the sequence i i(j) σxy(t) remains in a compact subset of X. One may thus extract a subsequence σ that converges uniformly on every compact set to a map σ¯, which is clearly a bicombing. Convexity t − s s − r d(¯σ (s), σ¯ 0 0 (s)) ≤ d(¯σ (r), σ¯ 0 0 (r)) + d(¯σ (t), σ¯ 0 0 (t)), xy x y t − r xy x y t − r xy x y

i(j) i(j) where r < s < t, follows from the corresponding inequality for σxy , σx0y0 and rl < sl < tl by choosing these numbers in [0, 1] ∩ (1/ki(j))Z such that rl → r, sl → s, and tl → t. The following observation regarding the above construction will be used in Exam- ple 3.10. Remark 3.5. Let σ be a bicombing on the proper metric space X, and suppose that for some pair of points x, y the consistency condition holds, i.e. σx0y0 (λ) = σxy((1−λ)s+λt) 0 0 for all 0 ≤ s ≤ t ≤ 1 and λ ∈ [0, 1], where x := σxy(s) and y := σxy(t). Then it is easily seen that σxy as well as all positively oriented subsegments are unaltered by the procedure in the above proof. In other words, the resulting convex bicombing σ¯ satisﬁes 0 0 σ¯x0y0 = σx0y0 for all x , y as above, in particular σ¯xy = σxy. 3.2. STRAIGHT CURVES 15

3.2 Straight curves

Whereas the preceding section dealt with the existence of convex bicombings, we now turn to the question of uniqueness. First we consider a property every geodesic from a convex bicombing necessarily shares. Let X be a metric space. We call a curve α:[a, b] → X straight (or a straight segment) if for every z ∈ X the function dz ◦ α is convex, where dz = d(z, ·). In particular, for ﬁxed s, t ∈ [a, b], taking z := α(s) one gets the inequality

d(α(s), α((1 − λ)s + λt)) ≤ λ d(α(s), α(t)) for all λ ∈ [0, 1], whereas for z := α(t) one obtains

d(α((1 − λ)s + λt), α(t)) ≤ (1 − λ) d(α(s), α(t)) for all λ ∈ [0, 1].

By taking the sum of these two inequalities one sees that straight curves are geodesics (of constant speed). The terminology is further justiﬁed by Theorem 3.8 below. In the proof of this result as well as in Example 3.10 we use Isbell’s injective hull construction [Isb1]. For convenience we review this in Appendix A. We denote the injective hull of a metric space X by E(X) and given a straight curve in X, we ﬁrst observe that this property persists when we pass from X to E(X).

Lemma 3.6. Let α:[a, b] → X be a straight curve in some metric space X. Regarding X as a subspace of its injective hull E(X) we then have that α is also straight in E(X).

Proof. By (A.2), the distance from an element f ∈ E(X) to a point x ∈ X ⊂ E(X) equals f(x). So we need to show that the function f ◦ α is convex. Given x := α(s), y := α(t), and q := α((1 − λ)s + λt), where λ ∈ [0, 1], let ε > 0 and choose (using the extremity of f, see (A.1)) a point p ∈ X such that f(p) + f(q) ≤ d(p, q) + ε. Since α is straight in X, we have

d(p, q) ≤ (1 − λ) d(p, x) + λ d(p, y).

Furthermore, d(p, x) ≤ f(p) + f(x) and d(p, y) ≤ f(p) + f(y). Combining these inequalities we get f(p) + f(q) ≤ f(p) + (1 − λ)f(x) + λf(y) + ε. Since ε was arbitrary, this gives f(q) ≤ (1 − λ)f(x) + λf(y), as desired.

Now let X = V be a normed real vector space. Isbell [Isb2] and Rao [Rao] (see also [CiaD]) showed that then the injective hull E(V ) has a Banach space structure with respect to which the isometric embedding e: V → E(V ) is linear. Since E(V ) is injective, collections of balls in E(V ) have the, so-called, binary intersection property, so the Banach space E(V ) is also injective in the linear category by [Nac]. Then a theorem of Nachbin, Goodner, and Kelley [Kel] implies that E(V ) is isometrically isomorphic to the space C(M) of continuous functions, with the supremum norm, on some extremally 16 CHAPTER 3. IMPROVING BICOMBINGS disconnected compact Hausdorﬀ space M. Summarizing, we may thus view V as a linear subspace of C(M), where M is such that E(V ) =∼ C(M). This fact will be used below. As usual, we call a curve α:[a, b] → V in a vector space V linear if it is of the form t 7→ p + tv for some p, v ∈ V . This is obviously a local property.

Proposition 3.7. Let M be a compact Hausdorﬀ space, and let α:[a, b] → U be a curve in an open subset U of C(M). Then α is straight (in U with the induced metric) if and only if it is linear.

Proof. Clearly every linear curve is straight. For the other direction, we assume that α: [0, 1] → U is a straight curve from 0 to y, where kyk∞ = l, and the closed 2l- neighborhood of α([0, 1]) is contained in U. We have to show that the two functions α(λ) and λy agree for every λ ∈ (0, 1). So ﬁx λ ∈ (0, 1) as well as m ∈ M. For an arbitrary ε > 0, choose an open neighborhood B of m such that

|α(λ)(m0) − α(λ)(m)| < ε/2, |y(m0) − y(m)| < ε/2 for all m0 ∈ B. By Urysohn’s lemma there is a non-negative function ϕ ∈ C(M) vanishing on M \ B and kϕk∞ = 2l. Put z± := α(λ) ± ϕ and note that these are elements of U. Since α is straight,

2l = kz± − α(λ)k∞ ≤ (1 − λ)kz±k∞ + λkz± − yk∞. (3.2)

0 Now if f ∈ C(M) is a function satisfying kfk∞ ≤ l as well as f(m ) < f(m) + ε for all 0 m ∈ B, then (f + ϕ)|B < 2l + f(m) + ε, (f + ϕ)|M\B ≤ l, −(f + ϕ) ≤ −f ≤ l, and l ≤ 2l + f(m), hence kf + ϕk∞ ≤ 2l + f(m) + ε.

Taking f = α(λ) and f = α(λ) − y we get kz+k∞ ≤ 2l + α(λ)(m) + ε and kz+ − yk∞ ≤ 2l + α(λ)(m) − y(m) + ε, respectively. (Note that kfk∞ ≤ l since α is a geodesic from 0 to y.) Together with (3.2), this shows that

α(λ)(m) − λy(m) + ε ≥ 0.

Similarly, for f = −α(λ) and f = y − α(λ) we obtain kz−k∞ ≤ 2l − α(λ)(m) + ε and kz− − yk∞ ≤ 2l + y(m) − α(λ)(m) + ε, respectively, which gives

λy(m) − α(λ)(m) + ε ≥ 0.

Letting ε → 0 we conclude that α(λ)(m) = λy(m), ending the proof as λ and m were arbitrary.

Combining these results we obtain the desired characterization of linear geodesics in normed spaces.

Theorem 3.8. A straight curve in an arbitrary normed space V is linear. Hence the bicombing of linear geodesics is the only convex bicombing on V . 3.2. STRAIGHT CURVES 17

Proof. As mentioned above, we can regard the normed space V as being a linear subspace of its injective hull E(V ), and we have E(V ) =∼ C(M) for some compact Hausdorﬀ space M. Now straight curves in V are straight in C(M) by Lemma 3.6 and linear by Proposition 3.7.

As we learned after developing the above result, on every normed space there is in fact no other (conical) bicombing than the linear one; see [GaeM]. One may ask whether the equivalence between straight and linear holds more generally for linearly convex subsets of normed spaces. The answer turns out to be negative.

Example 3.9. Consider the Banach space `∞([0, 1]) of bounded real valued functions on [0, 1], equipped with the supremum norm. The subset

C := {f ∈ `∞([0, 1]) | f(0) + f(1) = 1, f is convex, and f ∈ ∆1([0, 1])} is compact and linearly convex (see Appendix A for the deﬁnition of ∆1). Hence there is the convex bicombing of linear geodesics in C. But there are more straight segments: the curve α: [0, 1] → C, α(t) = dt, is non-linear, and by (A.2) we have t 7→ kf − α(t)k∞ = f(t) for every f ∈ C, which is a convex function by deﬁnition of C.

Let us mention here that we do not know whether the non-linear straight curve α belongs to a convex bicombing, thus whether there is a bicombing σ on C with σα(0)α(1) = α. It may very well be that there is no convex bicombing on C other than the linear one. In contrast to this we conclude this section with an example of a compact metric space that admits at least two distinct convex bicombings.

Example 3.10. Consider two geodesics α, β : [0, 1] → `∞([0, 1]) from d0 to d1: the linear geodesic α(s) = (1 − s)d0 + sd1, and the Kuratowski embedding β(t) = dt of [0, 1]. Let B ⊂ `∞([0, 1]) be the bigon composed of these two geodesic segments. By (A.2) we have

kα(s) − β(t)k∞ = α(s)(t) = s + t − 2st. (3.3)

Since this last term is symmetric in s and t, there is an isometric involution ι: B → B that interchanges α and β. Furthermore, as `∞([0, 1]) is injective, we can embed the injective hull E(B) of B into `∞([0, 1]) (and identify it with its image), so that B ⊂ E(B) ⊂ `∞([0, 1]). Now E(B) is not linearly convex in `∞([0, 1]), but by retracting the linear geodesics with endpoints in E(B) to E(B) we obtain a bicombing σ on E(B)

(compare Lemma 3.1). Note that since α was already linear, we have σd0d1 = α. Note further that E(B) is compact, because B is compact. Theorem 3.4 then yields a convex bicombing σ¯ which, by Remark 3.5, still satisﬁes σ¯d0d1 = α. Finally, the involution ι extends uniquely to an isometry I of E(B) (see Proposition A.4 which is Proposition 3.7 in [Lan]). Mapping σ¯ by I we get a convex bicombing τ¯ on E(B) with τ¯d0d1 = β, distinct from σ¯. 18 CHAPTER 3. IMPROVING BICOMBINGS

3.3 Combinatorial dimension

The example just described contrasts with Theorem 3.14, which we will prove in this section. First we discuss the structure of injective hulls of finite metric spaces and the notion of combinatorial dimension. X ∼ n Let X be a finite metric space, with |X| = n ≥ 1, say. The set ∆(X) ⊂ R = R is an unbounded polyhedral domain, determined by the finitely many linear inequalities f(x) + f(y) ≥ d(x, y) for x, y ∈ X (in particular f ≥ 0). As X is finite, a function f ∈ ∆(X) is extremal if and only if for every x ∈ X there exists a point y ∈ X such that f(x) + f(y) = d(x, y). So the injective hull is a polyhedral subcomplex of ∂∆(X) of dimension at most n/2. (It is not difficult to see that E(X) consists precisely of the bounded faces of ∂∆(X).) For n ≤ 5, the various possible combinatorial types of E(X) are depicted in Section 1 of [Dre] (where ∆(X) and E(X) are denoted PX and TX , respectively). To describe the structure of E(X) further, one may assign to every f ∈ E(X) the undirected graph with vertex set X and edge set

A(f) = {x, y} | x, y ∈ X, f(x) + f(y) = d(x, y) .

Note that this graph has no isolated vertices (because f is extremal), but may be disconnected, and there is a loop {x, x} if and only if f(x) = 0, which occurs if and only if f = dx (by (A.2)). Call a set A of unordered pairs of (possibly equal) points in X admissible if there exists an f ∈ E(X) with A(f) = A, and denote by A (X) the collection of admissible sets. The family of polyhedral faces of E(X) is then given by {P (A)}A∈A (X), where P (A) = {f ∈ ∆(X) | A ⊂ A(f)}, and where P (A0) is a face of P (A) if and only if A ⊂ A0. We define the rank rk(A) of A as the dimension of P (A). This number can be read off as follows. If f, g are two elements of P (A), then f(x) + f(y) = d(x, y) = g(x) + g(y) for {x, y} ∈ A, so f(y) − g(y) = −(f(x) − g(x)). Thus the difference f − g has alternating sign along all edge paths in the graph (X,A). It follows that there is either no or exactly one degree of freedom for the values of f ∈ P (A) on every connected component of (X,A), depending on whether or not the component contains a cycle of odd length. We call such components (viewed as subsets of X) odd or even A-components, respectively. The rank rk(A) = dim(P (A)) is then precisely the number of even A-components of X. (Here we have adopted the notation from [Lan], whereas [Dre] uses Kf = {(x, y) ∈ X × X | f(x) + f(y) = d(x, y)} in place of A(f).) Now let again X be a general metric space. We recall that the combinatorial dimension of X, introduced by Dress, is the possibly infinite number

dimcomb(X) = sup{dim(E(Y )) | Y ⊂ X, |Y | < ∞}, see Theorem 90 on p. 380 in [Dre]. This theorem contains in particular a characterization of the inequality dimcomb(X) ≤ n in terms of a 2(n + 1)-point inequality, which may be rephrased as follows. 3.3. COMBINATORIAL DIMENSION 19

Theorem 3.11 (Dress). Let X be a metric space, and let n ≥ 1 be an integer. The inequality dimcomb(X) ≤ n holds if and only if for every set Z ⊂ X with |Z| = 2(n + 1) and every ﬁxed point free involution i: Z → Z there exists a ﬁxed point free bijection j : Z → Z distinct from i such that

X d(z, i(z)) ≤ X d(z, j(z)). (3.4) z∈Z z∈Z

The case n = 1 corresponds to the much simpler fact that dimcomb(X) ≤ 1 if and only if X is 0-hyperbolic in the sense of Gromov [Gro] or tree-like in the terminology of [Dre], that is, for every quadruple of points x, x0, y, y0 ∈ X,

d(x, x0) + d(y, y0) ≤ max{d(x, y) + d(x0, y0), d(x, y0) + d(x0, y)}.

Theorem 3.11 follows from more general considerations in Section (5.3) of [Dre]. We will not use this result in the sequel. By the results of [Lan], every proper metric space with integer valued metric that is discretely geodesic and δ-hyperbolic has finite combinatorial dimension. By contrast, the unit circle S1 in the Euclidean plane with either the induced 1 (chordal) or the induced inner metric satisfies dimcomb(S ) = ∞, as is seen by looking at the constant extremal function f = diam(S1)/2, restricted to the vertices of a regular 2n-gon. Similarly, the metric bigon B constructed in Example 3.10 has dimcomb(B) = ∞: consider the function f defined by f(α(s)) = f(β(1 − s)) = kα(s) − β(1 − s)k∞/2 for s ∈ [0, 1]. Among the finite dimensional normed spaces, only those with a polyhedral norm have finite combinatorial dimension, equal to the number of pairs of opposite facets of the unit ball. The following proposition is the key observation for the proof of Theorem 3.14.

Proposition 3.12. Let X be a metric space of ﬁnite combinatorial dimension. Then for every pair of points x0, y0 ∈ X there exists a δ > 0 such that

d(x0, y0) + d(x, y) ≤ d(x, y0) + d(x0, y) for all pairs of points x ∈ B(x0, δ) and y ∈ B(y0, δ).

Proof. When x0 = y0, the triangle inequality in X gives the result. Now assume that x0 6= y0. Denote by F the collection of all real valued functions with ﬁnite support spt(f) ⊂ X such that f ∈ E(spt(f)), x0, y0 ∈ spt(f), and {x0, y0} ∈ A(f), that is,

f(x0) + f(y0) = d(x0, y0).

Since dimcomb(X) < ∞, there exist an integer n and a function f ∈ F such that rk(A(g)) ≤ n = rk(A(f)) for all g ∈ F . Since x0 6= y0, we have n ≥ 1, thus f > 0. By restricting f to a smaller set if necessary, we can arrange that A(f) is the collection of n disjoint pairs {x0, y0}, {x1, y1},..., {xn−1, yn−1}. There exists a δ > 0 such that for all p ∈ {x0, y0} and q ∈ {x1, y1, . . . , xn−1, yn−1} we have d(p, q) > δ, f(p) ≥ δ, and

f(p) + f(q) ≥ d(p, q) + 2δ. (3.5) 20 CHAPTER 3. IMPROVING BICOMBINGS

Note that d(x0, y0) = f(x0) + f(y0) ≥ 2δ. Let x ∈ B(x0, δ) and y ∈ B(y0, δ). If x = x0 or y = y0 or x = y, the desired inequality holds. So assume that x0 6= x 6= y 6= y0. Then x, y, x0, y0, . . . , xn−1, yn−1 are pairwise distinct. Put a := d(x, y0) − f(y0) and b := d(x0, y) − f(x0). We have

a ≥ d(x0, y0) − d(x, x0) − f(y0) = f(x0) − d(x, x0) ≥ f(x0) − δ, (3.6) b ≥ d(x0, y0) − d(y, y0) − f(x0) = f(y0) − d(y, y0) ≥ f(y0) − δ.

Since f(x0), f(y0) ≥ δ, this gives in particular a, b ≥ 0 and hence

a + f(x0) ≥ δ ≥ d(x, x0), (3.7) b + f(y0) ≥ δ ≥ d(y, y0).

Furthermore, for every q ∈ {x1, y1, . . . , xn−1, yn−1}, combining (3.6) and (3.5) we obtain

a + f(q) ≥ f(x0) + f(q) − δ ≥ d(x0, q) + δ ≥ d(x, q), (3.8) b + f(q) ≥ f(y0) + f(q) − δ ≥ d(y0, q) + δ ≥ d(y, q).

Now, in the case that a + b < d(x, y), we could deﬁne a function g with ﬁnite support by putting g(q) := f(q) for q ∈ {x0, y0, . . . , xn−1, yn−1} and by choosing g(x) > a and g(y) > b such that g(x) + g(y) = d(x, y). In view of (3.7) and (3.8), this function would satisfy A(g) = {x0, y0},..., {xn−1, yn−1}, {x, y} , so that g ∈ F and rk(A(g)) = n + 1, in contradiction to the maximality of n. We conclude that d(x, y) ≤ a + b = d(x, y0) + d(x0, y) − d(x0, y0).

Resuming the discussion of straight curves, we can now prove the following.

Proposition 3.13. Suppose that X is a metric space of ﬁnite combinatorial dimension, and α, β : [0, 1] → X are two straight curves. Then the function s 7→ d(α(s), β(s)) is convex on [0, 1]. In particular, every pair of points in X is joined by at most one straight segment, up to reparametrization.

Proof. For s, t ∈ [0, 1], put h(s, t) := d(α(s), β(t)). Fix s0 ∈ (0, 1). Then it follows from Proposition 3.12 that there exists an ε > 0 such that [s0 − ε, s0 + ε] ⊂ [0, 1] and, for all s, t ∈ [s0 − ε, s0 + ε],

h(s0, s0) + h(s, t) ≤ h(s, s0) + h(s0, t). (3.9)

Now suppose that s0 − ε ≤ s < s0 < t ≤ s0 + ε, and let λ ∈ (0, 1) be such that s0 = (1 − λ)s + λt. Since h(s, ·) and h(·, t) are convex functions on [0, 1], we have

h(s, s0) ≤ (1 − λ) h(s, s) + λ h(s, t), (3.10) h(s0, t) ≤ (1 − λ) h(s, t) + λ h(t, t). 3.3. COMBINATORIAL DIMENSION 21

Combining (3.9) and (3.10) we conclude that

h(s0, s0) ≤ (1 − λ) h(s, s) + λ h(t, t).

Note that this holds whenever s0 − ε ≤ s < s0 < t ≤ s0 + ε and s0 = (1 − λ)s + λt, where ε > 0 depends on s0. Since s 7→ h(s, s) = d(α(s), β(s)) is continuous on [0, 1], it follows easily that this function is convex.

Now the next theorem is an immediate corollary of Proposition 3.13. Note that a bicombing σ can be restricted to any ball because balls are σ-convex.

Theorem 3.14. Let X be a metric space of ﬁnite combinatorial dimension in the sense of Dress, or with the property that every bounded subset has ﬁnite combinatorial dimension. Suppose that X possesses a convex bicombing σ. We then have that σ is consistent, reversible, and unique, that is, σ is the only convex bicombing on X. Since straight segments are unique, σ is also equivariant with respect to the isometry group.

Recall that equivariant with respect to the isometry group means that for every isometry γ : X → X and all x, y ∈ X we have γ ◦ σxy = σγ(x)γ(y). The assumptions of the theorem are, in particular, satisﬁed for proper injective metric spaces of ﬁnite combinatorial dimension.

Theorem 3.15. Let X be a proper injective metric space of ﬁnite combinatorial dimension. Then X possesses a unique convex bicombing, which furthermore is consistent, reversible, and equivariant with respect to the whole isometry group.

Proof. Such spaces admit a bicombing by the remark after Lemma 3.1 and a convex bicombing by Theorem 3.4.

As addressed before, by a theorem of [Lan] we know that if Γ is a Gromov hyperbolic group, endowed with the word metric with respect to some ﬁnite generating set, then the injective hull E(Γ) is proper, has ﬁnite combinatorial dimension, and Γ acts properly and cocompactly on E(Γ) by isometries. Therefore the above theorem is applicable, and we obtain the last result of this section mentioned in the introduction.

Boundary at inﬁnity

We now consider a complete metric space X with a consistent bicombing. Note that completeness is no restriction as any bicombing may be extended to the completion of the underlying space. We define the geometric boundary and the closure of X by means of geodesic rays that are consistent with the given bicombing, and we equip the closure with a simple explicit metric. (Some general references for the analogous constructions in the case of CAT(0) spaces or Gromov hyperbolic spaces are [Bal, BriH, BuyS, Gro].) Then we prove Theorem 4.4. Recall that by a ray in X we mean an isometric embedding of R+ := [0, ∞). Two rays ξ, η in X are asymptotic if the function t 7→ d(ξ(t), η(t)) is bounded or, equivalently, the Hausdorff distance between the images of ξ and η is finite. In the presence of a consistent bicombing σ on X we call a ray ξ : R+ → X a σ-ray if

ξ((1 − λ)s + λt) = σxy(λ) (4.1) whenever 0 ≤ s ≤ t, x := ξ(s), y := ξ(t), and λ ∈ [0, 1]. It follows that for any two 0 0 0 0 σ-rays ξ, η the map t 7→ d(ξ(a + a t), η(b + b t)) is convex on R+ for all a, a , b, b ≥ 0. The geometric boundary ∂σX is the set of equivalence classes of mutually asymptotic σ-rays in X, and we write

Xσ := X ∪ ∂σX.

For a uniﬁed treatment of the two parts of Xσ it is convenient to associate with every pair (o, x) ∈ X × X the eventually constant curve %ox : R+ → X, ( σox(t/d(o, x)) if 0 ≤ t < d(o, x), %ox(t) := (4.2) x if t ≥ d(o, x).

As a preliminary remark we note that for any basepoint o ∈ X and r ∈ R+, the radial retraction ϕr : X → B(o, r) deﬁned by ϕr(x) := %ox(r) satisﬁes

2r d(ϕ (x), ϕ (y)) ≤ d(x, y) (4.3) r r d(o, x) 23

0 whenever d(o, x) ≥ d(o, y) and d(o, x) > r. To see this, let s := r/d(o, x) and y := σoy(s). 0 We have ϕr(x) = σox(s), and since σ is conical, d(ϕr(x), y ) ≤ s d(x, y). Further- 0 0 0 0 more, d(y , ϕr(y)) = d(o, ϕr(y)) − d(o, y ) ≤ d(o, ϕr(x)) − d(o, y ) ≤ d(ϕr(x), y ) and so 0 0 d(ϕr(x), ϕr(y)) ≤ d(ϕr(x), y )+d(y , ϕr(y)) ≤ 2s d(x, y). In particular, ϕr is 2-Lipschitz. 2 The constant 2 is optimal, as one can see by looking at the space `∞ with 0 as the basepoint, the map ϕ1, and the points (1, 1), (1 + ε, 1 − ε). In order to prove that for every pair (o, x¯) ∈ X × ∂σX there is a σ-ray issuing from o and representing the class x¯, we shall need the following estimate (compare Lemma II.8.3 in [BriH] for the case of CAT(0) spaces).

Lemma 4.1. Let X be a metric space with a consistent bicombing σ, and let o, p ∈ X. Then for any σ-ray ξ with ξ(0) = p we have

2t d(o, p) d(% (t),% (t)) ≤ ox oy T − d(o, p) whenever T > 2d(o, p), x, y ∈ ξ([T, ∞)), and 0 ≤ t ≤ T − 2 d(o, p).

0 Proof. Assume d(o, x) ≤ d(o, y) and let s := d(p, x)/d(p, y) and y := σoy(s). Since ξ is a σ-ray, we have x = σpy(s). As σ is conical,

d(x, y0) ≤ (1 − s) d(p, o) ≤ d(o, p). (4.4)

Now d(o, x) ≥ d(p, x) − d(o, p) ≥ T − d(o, p) and so d(o, y0) ≥ d(o, x) − d(x, y0) ≥

o %oy(t)

%ox(t) ∂B(o, t)

0 y = σoy(s) p ξ T x y

0 T −2 d(o, p). Hence, for 0 ≤ t ≤ T −2 d(o, p), we have ϕt(x) = %ox(t) and ϕt(y ) = %oy(t), and (4.3) gives the result.

The next result now follows by a standard procedure.

Proposition 4.2. Let X be a complete metric space with a consistent bicombing σ. Then for every pair (o, x¯) ∈ X × ∂σX there is a unique σ-ray %ox¯ with %ox¯(0) = o that represents the class x¯. Furthermore, if r ≥ 0 and x = %ox¯(r), then %xx¯(t) = %ox¯(r + t) for all t ∈ R+.

Proof. Let ξ be a σ-ray in the class x¯, and let p := ξ(0). For n = 1, 2,... , put %n := %oξ(n). It follows from the preceding lemma that for every ﬁxed t ≥ 0 the sequence %n(t) is Cauchy. In the limit one obtains a σ-ray %ox¯ issuing from o. As in (4.4), we have 24 CHAPTER 4. BOUNDARY AT INFINITY d(ξ(t),%n[t d(o, ξ(n))/n]) ≤ d(o, p) for 0 ≤ t ≤ n, hence d(ξ(t),%ox¯(t)) ≤ d(o, p) for all 0 t ≥ 0. In particular, %ox¯ is asymptotic to ξ. Finally, if % is another σ-ray issuing 0 from o and asymptotic to ξ, then t 7→ d(%oξ(t),% (t)) is a non-negative, bounded, convex 0 function on R+ that vanishes at 0, so % = %oξ. From this uniqueness property, the last assertion of the proposition is clear.

A natural topology on Xσ = X ∪ ∂σX may be described in diﬀerent ways. First we ﬁx a basepoint o ∈ X and consider the set

Rσ,o := {%ox¯ | x¯ ∈ Xσ} = {%ox | x ∈ X} ∪ {%ox¯ | x¯ ∈ ∂σX} of generalized σ-rays based at o, given by (4.2) and Proposition 4.2. We equip Rσ,o with the topology of uniform convergence on compact subsets of R+. Clearly Rσ,o is compact if and only if X is proper, as a consequence of the Arzelà-Ascoli theorem. The topology of Rσ,o agrees, under canonical identiﬁcation, with the cone topology on Xσ, a basis of which is given by the sets

Uo(¯x, t, ε) := {y¯ ∈ Xσ | d(%ox¯(t),%oy¯(t)) < ε} (4.5)

∂σX x¯

%ox¯(t)

ε Uo(¯x, t, ε)

Figure 4.1: Uo(¯x, t, ε) for an x¯ ∈ ∂σX for x¯ ∈ Xσ and t, ε > 0. Note that since ϕr is 2-Lipschitz, we have

d(%ox¯(r),%oy¯(r)) ≤ 2 d(%ox¯(t),%oy¯(t)) for all r ∈ [0, t]. (4.6)

Note also that Uo(x, t, ε) is just the open ball U(x, ε) in case t ≥ d(o, x) + ε. It follows readily from the following lemma that this topology on Xσ is independent of the choice of basepoint o.

Lemma 4.3. Given o ∈ X, x¯ ∈ ∂σX, ε, t > 0, and p ∈ X, there exists T > 0 such that Up(¯x, ε/4,T ) ⊂ Uo(¯x, ε, t). 25

Proof. It follows from Lemma 4.1 and the construction of the ray %ox¯ in Proposition 4.2 that if T is chosen suﬃciently large, depending on d(o, p) and ε, t, and if x := %px¯(T ), then d(%ox(t),%ox¯(t)) ≤ ε/4.

Likewise, for any point y¯ ∈ Xσ, if y := %py¯(T ) and d(p, y) is large enough, then

d(%oy(t),%oy¯(t)) ≤ ε/4.

Now if y¯ ∈ Up(¯x, T, ε/4), that is, d(x, y) < ε/4, then

d(%ox(t),%oy(t)) ≤ 2 d(x, y) < ε/2 by (4.3), provided d(o, x), d(o, y) > t. We conclude that d(%ox¯(t),%oy¯(t)) < ε and thus y¯ ∈ Uo(¯x, t, ε) for suﬃciently large T .

Next we equip Xσ with the metric deﬁned by Z ∞ −s Do(¯x, y¯) := d(%ox¯(s),%oy¯(s)) e ds (4.7) 0

(compare Section 8.3.B in [Gro]). We have Do(o, x¯) ≤ 1, with equality if and only if x¯ ∈ ∂σX. For d(%ox¯(t),%oy¯(t)) = a, (4.6) yields Z ∞ Z t Z ∞ a −s −s −s e ds ≤ Do(¯x, y¯) ≤ 2a e ds + 2s e ds, t 2 0 t and it follows easily from these estimates that the metric (4.7) induces the cone topology. Observe also that if d(o, x) = R and y = ϕr(x) for some 0 ≤ r ≤ R, then Z R Z ∞ −s −s −r −R Do(x, y) = (s − r) e ds + (R − r) e ds = e − e . r R In particular, for any σ-ray ξ issuing from o, the curve λ 7→ ξ(− log(1 − λ)), λ ∈ [0, 1), is a unit speed geodesic with respect to Do. Accordingly, for λ ∈ [0, 1], we deﬁne the radial retraction ψλ : Xσ → BDo (o, λ) = {y¯| Do(o, y¯) ≤ λ} such that ψ1 = id and ψλ(¯x) := %ox¯(t) for λ < 1 and t := − log(1 − λ). In this latter case, the generalized ray %ox with endpoint x := ψλ(¯x) agrees with ϕt ◦ %ox¯, and since ϕt is 2-Lipschitz we obtain Z ∞ −s Do(ψλ(¯x), ψλ(¯y)) = d(ϕt(%ox¯(s)), ϕt(%oy¯(s))) e ds ≤ 2 Do(¯x, y¯) (4.8) 0 for all x,¯ y¯ ∈ Xσ. Thus ψλ is 2-Lipschitz with respect to Do. Finally, we note that if x¯ ∈ Xσ and λ, µ ∈ [0, 1], then clearly

Do(ψλ(¯x), ψµ(¯x)) ≤ |λ − µ|, (4.9) with equality when Do(o, x¯) ≥ max{λ, µ}. Now we turn to the result mentioned in the introduction. Recall that a metrizable space is an absolute retract if it is a retract of every metrizable space containing it as a closed subspace. 26 CHAPTER 4. BOUNDARY AT INFINITY

Proof. To start, we wish to show that Xσ is contractible and locally contractible. The map (¯x, λ) 7→ ψ1−λ(¯x) is continuous on Xσ ×[0, 1] by (4.8) and (4.9) and contracts Xσ to o, so Xσ is contractible. As for the local property, we prove that in fact each of the sets Uo(¯x, t, ε) for x¯ ∈ Xσ and t, ε > 0 (see (4.5)) is contractible. The same map as above, but −t restricted to Uo(¯x, t, ε) × [0, e ], contracts Uo(¯x, t, ε) to the subset U(%ox¯(t), ε) ∩ B(o, t), which as an intersection of balls is σ-convex and hence itself contractible. To prove that ∂σX is a Z-set in Xσ, let an open set U in Xσ be given, and assume that U 6= ∅, Xσ. We want to find a homotopy H : U × [0, 1] → U from the identity on U to a map into U \ ∂σX such that the restriction of H to (U \ ∂σX) × [0, 1] takes values in U \ ∂σX. To this end, we note that the function h: U → R defined by 1 h(¯x) := D (o, x¯) − inf{D (¯x, y¯) | y¯ ∈ X \ U} o 2 o σ is Lipschitz continuous with respect to Do and satisfies h(¯x) < Do(o, x¯) for all x¯ ∈ U because U is open. It is then easy to see that H(¯x, λ) := ψmax{1−λ,h(¯x)}(¯x) serves the purpose. It remains to show that Xσ is an AR (absolute retract). In case X has finite topological dimension, we have

dim(Xσ) = dim(X) < ∞.

Indeed, for every ε ∈ (0, 1), any open covering of the Do-ball BDo (o, 1 − ε) with mesh −1 ≤ ε gives rise, via ψ1−ε, to an open covering of Xσ with mesh ≤ 3ε and the same mul- tiplicity (see, for instance, [BuyS] for the definitions). It is then a standard result that contractible and locally contractible metrizable spaces of finite dimension are absolute retracts (see [Dug1]). However, finite dimension is not needed. Every metric space X with a bicombing is strictly equiconnected, as defined in [Him], so X is an AR by Theorem 4 in that paper (see also [Dug2]). Now, by Corollary 6.6.7 in [Sak] (a result attributed to O. Hanner and S. Lefschetz), Xσ is an ANR (absolute neighborhood retract) as it contains X as a homotopy dense subset; alternatively, one can give a short direct argument along the lines of Lemma 1.4 in [Tor]. Thus Xσ is a contractible ANR or, equivalently, an AR (see Corollary 6.2.9 in [Sak]).

As a concluding remark, we first note that ∂σX may be strictly smaller than the set ∂X of asymptote classes of (general) rays in X. For instance, any 1-Lipschitz function 2 f : R+ → R determines a ray t 7→ (t, f(t)) in `∞, thus there are plenty of such rays that are not asymptotic to linear (σ-)rays. Presumably, if X admits different consistent bicombings σ, then ∂σX also depends on the choice of σ. On the other hand, ∂σX agrees with ∂X if, for instance, X has roughly unique geodesics, that is, if there exists a constant δ ≥ 0 such that any two geodesics α, β : [0, 1] → X connecting the same points 27 are within uniform distance at most δ from each other. Indeed, a simple modification of Lemma 4.1 and Proposition 4.2 first shows that for every ray ξ in X there is a σ-ray asymptotic to ξ and issuing from the same point. As in the proof of Lemma 4.3, one can then conclude that distinct bicombings yield homeomorphic boundaries. Clearly every Gromov hyperbolic geodesic metric space X has roughly unique geodesics, and if X is proper, it is well-known that there is also a natural bijection between ∂X and the boundary ∂∞X defined in terms of sequences converging to infinity. This latter fact fails in general for non-proper spaces but remains true, in the presence of a consistent bicombing σ, when X is complete. In fact, for any complete Gromov hyperbolic metric space X with such a bicombing, ∂σX is homeomorphic to ∂∞X with the topology induced by any visual metric (however, the above Do rarely induces a visual metric on the boundary). Chapter 5

Flat planes

0 0 Two lines ξ, ξ in X are called parallel if sups∈R d(ξ(s), ξ (s)) < ∞. And, as before, two rays η, η0 in X are asymptotic if sup d(η(s), η0(s)) < ∞. We use the deﬁnition of s∈R+ σ-ray from the previous chapter. As for lines, ξ : R → X will be called a σ-line if its trace is σ-convex; equivalently,

[ξ(s), ξ(t)](λ) = ξ((1 − λ)s + λt) (5.1) for all s, t ∈ R and λ ∈ [0, 1]. Here we used the notation [x, y](t) for σxy(t), and by [x, y] we shall denote the map as well as its trace and will continue to do so without further comment. For two σ-lines ξ, ξ0, the function s 7→ d(ξ(s), ξ0(s)) is convex and non- negative, hence constant in case ξ, ξ0 are parallel. Note that there is a subtle diﬀerence between the otherwise identical deﬁnitions of σ-ray and σ-line. For the condition (4.1) we restrict to s ≤ t, whereas equation (5.1) must hold for all s, t ∈ R. But since we work with reversible bicombings throughout this chapter, the two properties are equivalent.

5.1 Flat strips and half-planes

Proposition 5.1 (Flat strip). Let X be a metric space with a consistent and reversible 0 bicombing σ. Suppose that ξ, ξ : R → X are two parallel σ-lines with disjoint images. Then the map 0 f : R × [0, 1] → X, f(s, t) = [ξ(s), ξ (s)](t) is an isometric embedding with respect to the metric on R × [0, 1] induced by some norm 2 on R .

If X is CAT(0), the norm is Euclidean. If X is a Busemann space, the norm is strictly convex. If X has ﬁnite combinatorial dimension, the norm is polyhedral.

0 0 Proof. For r ∈ R, put h(r) := d(ξ(0), ξ (r)). We have d(ξ(R), ξ (R + r)) = h(r) for every R ∈ R since the left hand side is a convex non-negative bounded function of R, hence 5.1. FLAT STRIPS AND HALF-PLANES 29 constant. We claim that for every pair of points p = (s, t) and p0 = (s + ∆s, t + ∆t) in R × [0, 1] we have (|∆t| h(∆s/∆t) if ∆t 6= 0, d(f(p), f(p0)) = (5.2) |∆s| if ∆t = 0.

There is no loss of generality in assuming ∆t ≥ 0. Suppose ﬁrst that ∆t > 0, and put r := ∆s/∆t. Let q := (s − tr, 0) and q0 := (s + (1 − t)r, 1) denote the points where the 0 line through p, p intersects R × {0} and R × {1}. Then d(f(q), f(q0)) = d(ξ(s − tr), ξ0(s − tr + r)) = h(r). (5.3) We put η := [ξ(s), ξ0(s)] and η0 := [ξ(s + ∆s), ξ0(s + ∆s)]. By convexity, we get

f(q0) ξ0

f(p0) = η0(t + ∆t)

η(1 − ∆t) ∆t η(t) = f(p) ∆s η0(∆t)

ξ f(q) = ξ(s − tr) that d(f(q), f(p)) = d(ξ(s − tr), η(t)) ≤ t d(ξ(s − r), η(1)) = t h(r). (5.4) Likewise, we have d(f(p0), f(q0)) ≤ (1 − t − ∆t) h(r) (5.5) as well as d(η(0), η0(∆t)) ≤ ∆t h(r) and d(η(1 − ∆t), η0(1)) ≤ ∆t h(r). Hence, by the convexity of λ 7→ d(η(λ), η0(λ + ∆t)) on [0, 1 − ∆t], also d(f(p), f(p0)) = d(η(t), η0(t + ∆t)) ≤ ∆t h(r). (5.6) From (5.3)–(5.6) and the triangle inequality we see that all inequalities derived so far are in fact equalities. In view of (5.6), this shows in particular the ﬁrst part of (5.2).

The second case follows by continuity from the ﬁrst, since |r| − h(0) ≤ h(r) ≤ h(0) + |r| for all r ∈ R and hence lim |∆t| h(∆s/∆t) = |∆s|. ∆t→0 0 Now, to conclude the proof, note that h(r) > 0 for all r ∈ R, as ξ and ξ have disjoint 2 images. It then follows readily from (5.2) that there is a norm k · k on R such that 0 0 0 d(f(p), f(p )) = kp − pk for all p, p ∈ R × [0, 1]. Note that the triangle inequality for k · k is just inherited from X. 30 CHAPTER 5. FLAT PLANES

The following example shows that, in general, if we replace ξ by s 7→ ξ(s + a) for some a 6= 0, we may get a different strip in X. Example 5.2. Define piecewise affine functions g, h: R × [0, 1] → R such that  1 t if s ≤ 0,  2  1 g(s, t) = 2 |s − t| if 0 ≤ s ≤ 1,  1  2 (1 − t) if s ≥ 1, 1 2 and h(s, t) = 2 − g(s, 1 − t). Note that g = h outside of (0, 1) , whereas the graphs of g 2 3 and h over [0, 1] bound a simplex Σ in R . Consider the space 3 X := {(s, t, u) ∈ `∞ | g(s, t) ≤ u ≤ h(s, t)} 3 equipped with the metric induced from `∞. X is injective by Lemma A.8 and therefore

ξ0(0)

ξ0(1)

ξ(1) ξ(0)

Figure 5.1: the simplex Σ in X

0 possesses a unique consistent and reversible bicombing σ. The geodesics ξ, ξ : R → X given by ξ(s) := (s, 0, g(s, 0)) and ξ0(s) := (s, 1, g(s, 1)) are two (parallel) σ-lines. It is then not difficult to see (compare the reasoning in Example 8.4) that the strip formed by the segments [ξ(s), ξ0(s + 1)] corresponds to the graph of g, whereas the segments [ξ(s + 1), ξ0(s)] trace out the graph of h. We also see that in Proposition 5.1, for fixed t ∈ (0, 1), the lines s 7→ f(s, t) need not be σ-lines in general: Clearly the lines s 7→ [ξ(s), ξ0(s + 1)](1/2) and s 7→ [ξ(s + 1), ξ0(s)](1/2) in the above example cannot both be σ-lines, and the strip there is never σ-convex as the hull always equals the whole space. For a proper X, however, one may 0 always obtain an embedding f : R × [0, 1] → X such that f(·, 0) = ξ, f(·, 1) = ξ , and s 7→ f(s, t) is a σ-line parallel to ξ and ξ0 for every fixed t ∈ (0, 1). Every accumulation point (w.r.t. compact-open topology) of the family of embeddings fa, a ∈ N, defined 0 through fa(s + ta, t) = [ξ(s), ξ (s + a)](t) for all (s, t) ∈ R × [0, 1], yields such an f. We now proceed to an existence result for embedded flat half-planes, which will be instrumental in the proof of Theorem 5.5. We need the following analogue of the Tits cone in the case of CAT(0) spaces. Let ∂σX denote the set of equivalence classes of mutually asymptotic σ-rays like we defined them at the beginning of Chapter 4. For (a, ξ), (b, η) ∈ R+ × ∂σX, we put 1 d∞((a, ξ), (b, η)) := lim d(ξ(aλ), η(bλ)). λ→∞ λ 5.1. FLAT STRIPS AND HALF-PLANES 31

Note that the limit exists by convexity, and |a − b| ≤ d∞((a, ξ), (b, η)) ≤ a + b. This deﬁnes a pseudometric d∞ on R+ × ∂σX, and the respective quotient metric space (shrinking {0}×∂σX to a single point) is a metric cone over ∂σX (compare [Bal], p. 38). In particular, for a > 0, d∞((a, ξ), (a, η)) = a d∞((1, ξ), (1, η)), and this is zero if and only ξ and η are asymptotic. The following result should now be compared with Proposition II.4.2 in [Bal] and Proposition II.9.8 and Corollary II.9.9 in [BriH].

Proposition 5.3 (Flat half-plane). Let X be a metric space with a consistent and reversible bicombing σ. Suppose that ξ : R → X is a σ-line and for every s ∈ R, ηs : R+ → X is the σ-ray with ηs(0) = ξ(s) asymptotic to η := η0. Then, for all a, b > 0, the function s 7→ d(ξ(s + a), ηs(b)) is non-decreasing on R with limit

lim d(ξ(s + a), ηs(b)) = d∞((a, ξ), (b, η)). (5.7) s→∞

Furthermore, if for every a ∈ R the function s 7→ d(ξ(s + a), ηs(1)) is constant on R and non-zero, the map f : R × R+ → X, f(s, t) := ηs(t), is an isometric embedding with respect to the metric on R × R+ induced by some norm 2 on R . Proof. Let a, b > 0. First we show that for all 0 < r ≤ λ ≤ r + 1, 1 d(ξ(ar + a), η (b)) ≥ d(ξ(aλ), η(bλ)). (5.8) ar λ

Since ηar and the ray t 7→ η(br + t) are asymptotic, we have

d(ηar(b), η(br + b)) ≤ d(ηar(0), η(br)) = d(ξ(ar), η(br)). (5.9)

It follows that

d(ξ(ar + a), ηar(b)) ≥ d(ξ(ar + a), η(br + b)) − d(ξ(ar), η(br)) r + 1 r ≥ − d(ξ(aλ), η(bλ)), λ λ which is (5.8). Putting λ = 1 we get d(ξ(ra + a), ηra(b)) ≥ d(ξ(a), η0(b)). Likewise, for all s ∈ R and 0 < r ≤ 1,

d(ξ(s + ra + a), ηs+ra(b)) ≥ d(ξ(s + a), ηs(b)), so s 7→ d(ξ(s + a), ηs(b)) is non-decreasing. Furthermore, for all s ∈ R and λ ≥ 1, we have 1 d(ξ(s + a), η (b)) ≤ d(ξ(s + aλ), η (bλ)). s λ s Together with (5.8), this gives (5.7). 32 CHAPTER 5. FLAT PLANES

For the second part of the proposition we have that s 7→ d(ξ(s+a), ηs(1)) is constant for every a ∈ R and that these values h(a) := d(ξ(a), η0(1)) are all positive. We ﬁrst claim that d(ξ(s + ta), ηs(t)) = t h(a) (5.10) for all t ≥ 0. The left hand side is convex as a function of t, thus it suﬃces to show this equality for 0 ≤ t ∈ Z. For t = 0, 1, (5.10) clearly holds. Consequently, by convexity, d(ξ(s + ta), ηs(t)) ≥ t h(a) for all t > 1. The reverse inequality for 1 < t ∈ Z follows by the triangle inequality since

d(ηs+ka(t − k), ηs+ka−a(t − k + 1)) ≤ d(ηs+ka(0), ηs+ka−a(1)) = h(a) for k = t, t − 1,..., 1 (compare (5.9)). Next, we claim that for every pair of points 0 p = (s, t) and p = (s + ∆s, t + ∆t) in R × R+ we have

(|∆t| h(−∆s/∆t) if ∆t 6= 0, d(f(p), f(p0)) = |∆s| if ∆t = 0, similarly as in the proof of Proposition 5.1. To show this, suppose without loss of generality that ∆t ≥ 0. Let ﬁrst ∆t > 0, and put a := ∆s/∆t and q := (s − ta, 0). Then (5.10) yields

0 d(f(p), f(p )) = d(ηs(t), ηs+∆s(t + ∆t))

≤ d(ηs(0), ηs+∆s(∆t)) = ∆t h(−a) as well as d(f(q), f(p)) = t h(−a) and d(f(q), f(p0)) = (t + ∆t) h(−a). This gives d(f(p), f(p0)) = ∆t h(−a), as claimed. The rest of the proof follows as in Proposi- tion 5.1.

5.2 The Flat Plane Theorem

We now turn to Theorem 5.5. Recall that a metric space X is δ-hyperbolic, for some constant δ ≥ 0, if for every quadruple (w, x, y, z) ∈ X4,

d(w, y) + d(x, z) ≤ max{d(w, x) + d(y, z), d(w, z) + d(x, y)} + 2δ.

If such a δ exists, X is said to be hyperbolic. As is well known, for a geodesic metric space this is equivalent to saying that geodesic triangles are slim, in an appropriate sense. It also suﬃces to consider triangles whose sides are given by a ﬁxed bicombing.

Lemma 5.4. Let X be a metric space with a map that selects for every pair of points x, y ∈ X a geodesic segment [x, y] = [y, x] ⊂ X connecting them. If for every triple 3 δ (x, y, z) ∈ X the segment [x, z] is contained in the closed 2 -neighborhood of [x, y]∪[y, z], then X is δ-hyperbolic. 5.2. THE FLAT PLANE THEOREM 33

4 δ Proof. Let (w, x, y, z) ∈ X . The union of the closed 2 -neighborhoods of [x, y] and [y, z] δ covers [x, z], and also the union of the closed 2 -neighborhoods of [z, w] and [w, x] covers [x, z]. It follows that there is either a pair of points x0 ∈ [x, y] and z0 ∈ [z, w] with d(x0, z0) ≤ δ or a pair of points y0 ∈ [y, z] and w0 ∈ [w, x] with d(y0, w0) ≤ δ. In the ﬁrst case,

d(w, y) + d(x, z) ≤ d(w, z0) + δ + d(x0, y) + d(x, x0) + δ + d(z0, z) = d(w, z) + d(x, y) + 2δ.

Similarly, in the second case, d(w, y) + d(x, z) ≤ d(w, x) + d(y, z) + 2δ.

In particular, a non-hyperbolic X with a bicombing σ contains a sequence of fatter and fatter σ-triangles. The following argument then uses a ruled surface construction together with the cocompact isometric action to produce a collection of mutually asymptotic rays as in Proposition 5.3. This diﬀers from the strategy in [Bow] and is inspired by the proof for CAT(0) spaces in [Bri, BriH], although we make no use of angles.

Theorem 5.5 (Flat Plane). Let X be a proper metric space with a consistent and reversible bicombing σ and with cocompact isometry group. Then X is hyperbolic if and only if X does not contain an isometrically embedded normed plane.

Proof. If X contains an isometrically embedded normed plane, then clearly X cannot be hyperbolic. Suppose now that X is not hyperbolic. By Lemma 5.4 there are sequences 1 2 3 1 3 of points yk, yk, yk ∈ X and pk ∈ [yk, yk] such that

1 2 2 3 B(pk, k) ∩ ([yk, yk] ∪ [yk, yk]) = ∅ (5.11)

i i i+1 for all integers k ≥ 1. Put rk(·) := d(pk, ·). For i = 1, 2, let xk be a point in [yk, yk ] with i i minimal distance to pk, and let ξk : [0, rk(xk)] → X be a unit speed parametrization of the i i segment [pk, xk] from pk to xk. Then, for every pair (i, j) ∈ {(1, 1), (1, 2), (2, 2), (2, 3)}, we deﬁne the “ruled surface”

i,j i j ∆k : [0, rk(xk)] × [0, rk(yk)] → X

i,j i i i,j so that ∆k (·, 0) = ξk and, for each s ∈ [0, rk(xk)], ∆k (s, ·) is a constant speed parame- i j i j trization of the segment [ξk(s), yk] from ξk(s) to yk. Thus

i j i,j i,j 0 d(ξk(s), yk) 0 0 d(∆k (s, t), ∆k (s, t )) = j |t − t | ≤ 2|t − t |, rk(yk)

i j i j j i because d(ξk(s), yk) ≤ rk(xk) + rk(yk) ≤ 2rk(yk) by the choice of xk. Note also that by convexity, i,j i,j 0 i i 0 0 d(∆k (s, t), ∆k (s , t)) ≤ d(ξk(s), ξk(s )) = |s − s |. (5.12) 34 CHAPTER 5. FLAT PLANES

i,j 2 It follows that each ∆k is 2-Lipschitz, where here and below we equip R with the 0 i 0 l1-metric. Furthermore, putting s := rk(xk), we notice that for 0 ≤ r ≤ s ≤ s and j 0 ≤ t ≤ rk(yk), i i,j i,j 0 i,j i,j 0 d(ξk(r), ∆k (s, t)) ≥ rk(∆k (s , t)) − r − d(∆k (s, t), ∆k (s , t)) i 0 ≥ rk(xk) − r − (s − s) = s − r (5.13)

i by the triangle inequality, the choice of xk, and (5.12). Now we choose a sequence of isometries γk of X so that γk(pk) ∈ K for all k and i j for some ﬁxed compact set K. By (5.11), rk(xk), rk(yk) > k. Since X is proper, we i,j can extract a sequence k(l) so that each of the four sequences γk(l) ◦ ∆k(l) converges uniformly on compact sets, as l → ∞, to a 2-Lipschitz map

i,j f : R+ × R+ → X i i,j j i,j with boundary rays ξ := f (·, 0) and η := f (0, ·). Furthermore, for every s ∈ R+, i,j i,j j i,j ηs := f (s, ·) is a ray asymptotic to η , so f is in fact 1-Lipschitz. From the construction we also have that d(η1(t), η3(t)) = 2t for all t ≥ 0, in particular η1, η3 are non-asymptotic. Hence, there is at least one pair (i, j) such that ξi, ηj are non- i,j i j i,j asymptotic. We put f := f , ξ := ξ , η := η , and ηs := ηs for some such pair. We claim that for all a ∈ R and b > 0, the limit

L(a, b) := lim d(ξ(s + a), ηs(b)) s→∞ exists and is strictly positive. Clearly L(0, b) = b. If a>0, then L(a, b) = d∞((a, ξ), (b, η)) > 0 by the ﬁrst part of Proposition 5.3 and since ξ, η are non-asymptotic. If a < 0, the same result still shows that s 7→ d(ξ(s + a), ηs(b)) is non-increasing on [|a|, ∞), so the limit exists, and L(a, b) ≥ |a| as a consequence of (5.13). Next, for every integer l ≥ 1 we deﬁne the 1-Lipschitz map

fl :[−l, ∞) × R+, fl(s, t) := f(l + s, t) = ηl+s(t).

Then we choose isometries γ¯l of X so that (¯γl ◦ fl)(0, 0) ∈ K for all l and for some ﬁxed compact set K. As above, there exists a subsequence l(m) such that the sequence γ¯l(m) ◦ fl(m) converges uniformly on compact sets to a 1-Lipschitz map ¯ f : R × R+ → X ¯ ¯ ¯ with boundary line ξ := f(·, 0) and mutually asymptotic rays η¯s := f(s, ·) for s ∈ R. For every a ∈ R and b > 0, we now have that

d(ξ¯(s + a), η¯s(b)) = L(a, b) > 0 for all s ∈ R. Hence, by Proposition 5.3, f¯ is an isometric embedding with respect to 2 some norm on R . Using once more that X is cocompact, we then conclude that X contains an isometrically embedded normed plane. 5.2. THE FLAT PLANE THEOREM 35

It is clear that if X is a CAT(0) or a Busemann space, then this property is inherited by any isometrically embedded normed plane, thus the corresponding norm must be Euclidean or strictly convex, respectively. We brieﬂy discuss another variant of Theo- rem 5.5, which happens to have a very short proof, without reference to bicombings. Recall the notion of injective metric space from Appendix A. Given any metric space Q = {w, x, y, z} of cardinality four, suppose that c := d(w, y)+d(x, z) is not less than the maximum of a := d(w, x) + d(y, z) and b := d(w, z) + d(x, y). The injective hull (or the tight span [Dre]) of Q is isometric to the (possibly degenerate) rectangle [0, (c − a)/2] × 2 [0, (c − b)/2] in `1 with four segments of appropriate lengths attached at the corners, where the terminal points of these segments correspond to Q (see Fig. A1 on p. 336 in [Dre]). Now the δ-hyperbolicity of Q means precisely that the width (the minimum of the two side lengths) of this `1-rectangle is not bigger than δ. This has the following easy consequence.

Theorem 5.6. A proper, cocompact injective metric space X is hyperbolic if and only 2 2 if X does not contain an isometric copy of `1 or, equivalently, of `∞. Proof. Suppose that X is not hyperbolic. Then, by the above discussion, for arbitrarily large δ > 0 there exists a quadruple Q ⊂ X whose injective hull contains an isometric 2 copy of [0, δ] × [0, δ] ⊂ `1. Since X is injective, this `1-square embeds isometrically into X by the respective property of the injective hull. From a sequence of such squares with side lengths tending to inﬁnity we obtain an isometric embedding of the entire `1-plane, using the fact that X is proper and cocompact. Chapter 6

Barycenters

Here we develop the tool that forms a crucial component of many proofs to come. In [EsSH], Es-Sahib and Heinich introduced an elegant barycenter construction for Buse- mann spaces, which was reviewed and partly improved in a recent paper by Navas [Nav].

6.1 The construction

The construction and proofs translate almost verbatim to spaces with reversible bicombings. As such spaces may lack unique midpoints, the only modiﬁcation required is to set bar2(x, y) := σxy(1/2). For ﬁnite subsets, the result is as follows.

Theorem 6.1. For every complete space X with reversible bicombing σ and every n ∈ N n there is a barycenter map barn : X → X with the following properties:

(i) barn(x1, . . . , xn) lies in the σ-convex hull of {x1, . . . , xn},

(ii) barn is permutation invariant, i.e. for any permutation π ∈ Sn we have barn(x1, . . . , xn) = barn(xπ(1), . . . , xπ(n)),

(iii) γ(barn(x1, . . . , xn)) = barn(γ(x1), . . . , γ(xn)) for every isometry γ of X provided σ is γ-equivariant,

1 Pn (iv) d barn(x1, . . . , xn), barn(y1, . . . , yn) ≤ min n i=1 d(xi, yπ(i)). π∈Sn

Note that for n = 2, bar2 as above is exactly what we call a symmetric midpoint assignment. And (in view of Proposition 2.2) if the given space is proper, even a (possibly non-reversible) bicombing (or midpoint assignment) is enough to obtain these barycenters. Also observe that (i) implies barn(x, . . . , x) = x for every n and x as we would expect.

Proof. First, property (i) forces us to set b1(x1) = x1. Secondly, we deﬁne bar2(x1, x2) =

σx1x2 (1/2). This deﬁnitions are in accord with all the required properties since the 6.1. THE CONSTRUCTION 37

0 0 0 bicombing is assumed the be reversible. For n ≥ 3, let x1 = x1, x2 = x2, . . . , xn = xn be the initial points and recursively

i i−1 di−1 i−1 xj := barn−1(x1 ,..., xj , . . . , xn ) for i ≥ 1,

di−1 where xj means that this point is omitted. We proceed by induction over n — assum- i i ing the properties for barn−1 — and claim that diam({x1, . . . , xn}) converges to zero i i for i → ∞. This diameter equals the one for the σ-convex hull of {x1, . . . , xn}, and i−1 i−1 this set is clearly contained in the hull of {x1 , . . . , xn }. The resulting monotonic sequence converges to a single point which we declare to be the value of barn(x1, . . . , xn); i equivalently let it be the limit of xj when i → ∞ (the choice of j is obviously ir- i i 1 i−1 i−1 relevant). For the claim, observe that d(xk, xl) ≤ n−1 d(xk , xl ) from (iv). Hence 0 0 x3 x3

1 2 x1 x3 1 x2 x1 0 2 x4

To inﬁnity... and beyond! - Buzz Lightyear 2 1 x2 x4 = bar3(x1, x2, x3) 2 x1 0 0 x2 x2

1 x3 = bar2(x1, x2)

0 0 x1 = x1 x1

Figure 6.1: construction of bar3(x1, x2, x3) and bar4(x1, x2, x3, x4)

i i 1 i−1 i−1 diam({x1, . . . , xn}) ≤ n−1 diam({x1 , . . . , xn }) and barn is well-deﬁned. Since all i points xj lie in the σ-convex hull of {x1, . . . , xn}, so does the barn(x1, . . . , xn). (ii) results from the permutation invariant nature of the construction, and since — starting i i from the xj for barn(x1, . . . , xn) — γ(xj) are the points that arise in the construction of barn(γ(x1), . . . , γ(xn)), we obtain (iii) as well. Finally, (iii) reduces (iv) to 1 Pn d barn(x1, . . . , xn), barn(y1, . . . , yn) ≤ n i=1 d(xi, yi). By the inductive assumption we have 1 n d(xi , yi ) ≤ X d(xi−1, yi−1) j j n − 1 k k k=1 k6=j and taking the sum over j thereof, we arrive at

1 n 1 n 1 n X d(xi , yi ) ≤ X d(xi−1, yi−1) ≤ · · · ≤ X d(x0, y0). n j j n j j n j j j=1 j=1 j=1 38 CHAPTER 6. BARYCENTERS

The leftmost term converges to the distance we are about to estimate (when i → ∞) and thereby completes the proof.

i It is clear from the construction of the points xj in the proof above that

1 1 barn(x1, . . . , xn) = barn(x1, . . . , xn)

1 for xj = barn−1(x1,..., xcj, . . . , xn). This relation leads to the subsequent estimate we will use in Proposition 6.4, which is the key step on route to Theorem 6.5.

Lemma 6.2. For the barycenters of Theorem 6.1 above and any x ∈ X and integers 1 ≤ k ≤ n we have

!−1 n d(x, bar (x , . . . , x )) ≤ X d(x, bar (x| )). n 1 n k k I I⊂{1,...,n} |I|=k

Here bark(x|I ) denotes the barycenter of the k-tuple whose entries are the k (not necessarily distinct) values xi for i ∈ I (and the tuples order is irrelevant as shown before). Proof. Fixing k, we apply induction over n. For k = n the statement is trivial and our base case; hence assume n > k and the statement to hold for n − 1 ≥ 1. By means of Theorem 6.1(iv) together with x = barn(x, . . . , x) and the previously mentioned relation, we obtain 1 n d(x, bar (x , . . . , x )) ≤ X d(x, x1) n 1 n n j j=1 1 n = X dx, bar (x ,..., x , . . . , x ). n n−1 1 cj n j=1

We may estimate the last expression by

!−1 1 n − 1 n X X d(x, bar (x| )) n k k I j=1 I⊂{1,...,n}\{j} |I|=k using the induction hypothesis. For a ﬁxed I ⊂ {1, . . . , n}, |I| = k the double sum counts 1 n−1−1 n−1 d(x, bark(x|I )) exactly n − k times. Since we have the identity n k (n − k) = k for binomial coeﬃcients, this concludes the proof.

One very natural property may not hold for the barycenters constructed so far. If for an n-tuple x = (x1, . . . , xn) we write k · x for the kn-tuple where every entry is repeated k times, e.g. 3 · x = (x1, x1, x1, x2, x2, x2, . . . , xn, xn, xn), then it may not be the case that barn(x) = barkn(k · x). (6.1) 6.1. THE CONSTRUCTION 39

bar6(x, x, y, y, z, z) mx bar3(x, y, z)

Figure 6.2: the K of Example 6.3

Example 6.3. Let K be the tree that emerges from gluing two copies of [0, 1] to [0, 2] identifying all three zeros to become the single point m. It is easy to see that the geodesics are unique and form a consistent bicombing (in fact this is a CAT(0) space). Now if y, z are the endpoints of the branches of length 1 and x the endpoint of the remaining branch of length 2, then bar3(x, y, z) is at distance 1/3 from m on the geodesic to x, but bar6(x, x, y, y, z, z) lies at distance 13/45 from m on the same geodesic.

See that one can facilitate many computations of barycenters in a space X with bicombing by detecting the following pattern. Whenever there is a subset Y ⊂ X being σ-convex, isometric to a convex subset of a normed vector space via ϕ: Y → −1 V , and such that ϕ ((1 − λ)ϕ(x) + λϕy) equals the restricted bicombing on Y ; then the above construction yields a barycenter (on Y ) behaving like the linear barycenter 1 Pn barn(x1, . . . , xn) := n i=1 xi on ϕ(Y ). Thus confronted with the task of computing barn(x1, . . . , xn) in X when x1, . . . , xn ∈ Y , we may as well compute the linear barycenter of ϕ(x1), . . . , ϕ(xn) and take the preimage. This remark is of course rather trivial since a map ϕ as above transports one bicombing into the other, and therefore Y and ϕ(Y ) are indistinguishable from our point of view. The fact that (6.1) may not hold prevents us to define the barycenter for probability measures with, say, finite support. To address this defect, one is tempted to define

∗ barn(x) := lim barkn(k · x) k→∞ and hope this limit always exists. One of the central observations of [EsSH, Nav] is that this in fact works, and we present a streamlined version of a proof found in [Nav], which we were able to simplify considerably by use of elementary statistics.

Proposition 6.4. Still in the setting of Theorem 6.1, let the barycenters there be given. Then for any n and n-tuple x we have that k 7→ barnk(k · x) is a Cauchy sequence in X and thus convergent.

Proof. Let n, k be positive integers and x ∈ Xn an n-tuple. We want to estimate d barkn(k · x), bar(k+l)n((k + l) · x) for arbitrary l ≥ 1; by Lemma 6.2 this is less or 40 CHAPTER 6. BARYCENTERS equal to

!−1 (k + l)n X d bar (k · x), bar (k + l) · x . kn kn kn I I⊂{1,...,(k+l)n} |I|=kn

For an I, over which the above sum runs, let i1, . . . , in be deﬁned as follows: ij counts how many times xj appears in [(k + l) · x]I . So [(k + l) · x]I starts with i1 copies of x1 followed by i2 copies of x2 and so on. Now the continuity property 6.1(iv) of the barycenters leads to

d barkn(k · x), barkn (k + l) · x I D ≤ (|i − k| + |i − k| + ··· + |i − k|) . 2kn 1 2 n

For every speciﬁc assignment of the i there are exactly k+lk+l ··· k+l sets I that j i1 i2 in produce these values. Consequently our upper bound now reads

k+l k+l ··· D X i1 in (|i − k| + |i − k| + ··· + |i − k|) . 1 (k+l)n 2kn 1 2 n 0≤i1,...,in≤k+l kn i1+···+in=kn Pn By symmetry of this expression we may replace j=1 |ij − k| by n|i1 − k|. Running the sum over i2, . . . , in and exploiting Vandermonde’s identity we simplify further to

D k+l k+l(k+l)(n−1) X i kn−i |i − k|. 2k (k+l)n i=0 kn The fraction of binomial coeﬃcients is the probability mass function of a hypergeometric distribution with parameters kn (draws without replacement), (k +l)n (balls in the urn) and k + l (balls that count). Let Y be a random variable distributed accordingly; the mean value is E[Y ] = k and the variance may be estimated by Var[Y ] ≤ k. The above D expression is therefore equal to 2k E[|Y − E[Y ]|] and by Jensen’s inequality D D q D E[|Y − E[Y ]|] ≤ E[(Y − E[Y ])2] ≤ √ . 2k 2k 2 k So we have a bound for d barkn(k · x), bar(k+l)n((k + l) · x) independent of l and going to zero for k → ∞ as sought.

Now we have everything in place to define barycenters for probability measures with finite support (at this point the barycenters could be easily defined for a much wider

1This is (essentially) the first absolute centric moment of a multivariate hypergeometric distribution, and a simplified expression for this exists which would end the proof here. Since that expression is rather hard to find in the literature, we choose not to use it. 6.1. THE CONSTRUCTION 41 class of measures, see again [EsSH, Nav]; we skip this since we have no use for it in the sequel). Every such measure µ on a metric space X can be written uniquely as a sum Pk i=1 aiδxi of Dirac measures (point measures) where x1, . . . , xk are pairwise distinct Pl points and a1, . . . , ak > 0, a1 + ··· + ak = 1. Given another measure ν = j=1 bjδyj , a mass transport from µ to ν shall be a function T : {1, . . . , k} × {1, . . . , l} → [0, ∞) Pk Pl provided i=1 T (i, j) = bj for all j and j=1 T (i, j) = ai for all i. By the cost c(T ) P of a mass transport we understand the real value i,j T (i, j)d(xi, yj). We define the Wasserstein distance dW (µ, ν) to be the infimum over the costs of all mass transports from µ to ν, and it is not hard to verify that this is indeed a metric and that the infimum is always attained for some transport T which we call an optimal transport. (There are of course much more sophisticated definitions of this distance but this one is enough for our purpose and relatively easy to handle.) Note that the bound in estimate Pn 1 Pn 1 6.1(iv) is exactly the Wasserstein distance for µ = i=1 n δxi and ν = j=1 n δyj . To see this, fix an optimal transport T and consider the bipartite graph on the disjoint union {1, . . . , n}∪{˙ 1, . . . , n} where i in the first set is connected to j in the second if T (i, j) > 0. This graph fulfills the premises of Hall’s marriage theorem, and hence there is a permutation π such that T (i, π(i)) > 0 for every i. If we define the transport T˜ to be equal to T except for the points (i, π(i)) where we subtract D := mini=1,...,n T (i, π(i)), we obtain n ˜ X c(T ) = c(T ) + Dd(xi, yπ(i)). i=1 T˜ is an (optimal) transport from (1 − nD)µ to (1 − nD)ν so c(T˜) = (1 − nD)c(T ) and therefore the distances coincide.

Theorem 6.5. For every complete space X with reversible bicombing there is a barycenter map bar assigning each probability measure µ with ﬁnite support | spt(µ)| < ∞ a barycenter bar(µ) ∈ X. This map has the properties

(i) bar(µ) lies in the σ-convex hull of spt(µ),

(ii) γ(bar(µ)) = bar(γ∗µ) for every isometry γ of X provided σ is γ-equivariant (where Pl Pl γ∗µ := i=1 aiδγ(xi) denotes the push-forward of µ = i=1 aiδxi ),

(iii) d(bar(µ), bar(ν)) ≤ dW (µ, ν).

∗ Proof. First let barn be the maps provided by Theorem 6.1 and deﬁne barn by virtue of the previous proposition:

∗ barn(x1, . . . , xn) := lim barkn(k · x), where x = (x1, . . . , xn). k→∞

∗ It is straightforward to verify properties (i) through (iv) of 6.1 for all barn. Moreover, ∗ ∗ now clearly barn(x1, . . . , xn) = barkn(k · x) for every k ≥ 1. Because of this equality, we Pl may now unambiguously deﬁne bar(µ) provided the coeﬃcients ai of µ = i=1 aiδxi all lie in Q. Let m be a positive integer such that mai ∈ N for all i, then bar(µ) := barm(x) where x is a tuple starting with ma1 copies of x1, then ma2 copies of x2, et cetera. 42 CHAPTER 6. BARYCENTERS

(i) and (ii) are clear from this definition and (iii) follows from the remark preceding this proof. Finally, measures with rational coefficients are dense among all probability measures with finite support2 and the properties for the uniquely extended map are once again not hard to verify.

6.2 Linear barycenters

One may ask if the barycenters constructed so far may be reﬁned further to expose even nicer properties. Bearing in mind the prototype of all barycenters — barn(x1, . . . , xn) := 1 Pn n i=1 xi on normed spaces — a reasonable and mild additional assumption may be

bar2(bar2(x1, x2), bar2(x3, x4)) = bar2(bar2(x2, x3), bar2(x4, x1)) (6.2) to hold for every selection of points x1, x2, x3, x4. Here bar2 may originate from any barycenter constructed so far or even just a reversible bicombing by σx1x2 (1/2). To- gether with bar2(x1, x2) = bar2(x2, x1) this implies that bar2(bar2(x1, x2), bar2(x3, x4)) is permutation invariant. If (6.2) holds, we call this a linear barycenter; this denotation will be justified by Theorem 6.7 below. Given this property, it is considerably easier than before to construct a barycenter like in Theorem 6.5. But first let us demonstrate that (6.2) fails in general. If K is the tree (hence with unique bicombing) composed by three copies of [0, 1] glued along their initial points 0 (similar to Example 6.3) and x1, x2, x3 are the ends opposite to 0, then with x4 := x3 (6.2) is violated. Lemma 6.6. Given a linear barycenter on a complete space X, there is a barycenter on probability measures with finite support satisfying (i)–(iii) of Theorem 6.5 as well as

(iv) bar((1 − λ)µ + λν) = bar((1 − λ)δbar(µ) + λδbar(ν)) for all µ, ν and λ ∈ [0, 1].

Moreover, X then admits the consistent and reversible bicombing σ˜ deﬁned by σ˜xy(λ) := bar((1 − λ)δx + λδy).

Proof. Assume we have a barycenter map bar2 with property (6.2) above. we recursively set

bar2k (x1, . . . , x2k ) := bar2(bar2k−1 (x1, . . . , x2k−1 ), bar2k−1 (x2k−1+1, . . . , x2k )) for all k ≥ 2. bar2, bar4 are permutation invariant. We proceed by induction on k ≥ 3 to show the permutation invariance of all just deﬁned barycenters. It is enough to show bar2k (x1, . . . , x2k ) = bar2k (x2k , x2, . . . , x2k−1, x1) since bar2k−1 is assumed to have the invariance. Notated suggestively the left hand side reads

bar2(bar2(bar2k−2 (x1,...), bar2k−2 (...)), bar2(bar2k−2 (...), bar2k−2 (. . . , x2k ))).

Because bar4 is permutation invariant this leads to

bar2(bar2(bar2k−2 (...), bar2k−2 (...)), bar2(bar2k−2 (x1,...), bar2k−2 (. . . , x2k ))).

2 Pl Pl 1 Pl dW ( i=1 aiδxi , i=1 biδxi ) ≤ 2 diam({x1, . . . , xl}) i=1 |ai − bi| 6.2. LINEAR BARYCENTERS 43

Now x1 and x2k may be interchanged, and then reverting the above step leads to the desired result. So properties (i)–(iv) of 6.1 follow easily for these maps. Moreover, l permutation invariance implies bar2k (x) = bar2k+l (2 · x) by induction over l. Next let k Pl Q := {m/2 | m ∈ Z, k ≥ 0}, a set dense in R. Then for µ = i=1 aiδxi where all k coefficients are in Q\{0} we set bar(µ) := bar2k (x) where k is chosen such that 2 ai ∈ N for i = 1, . . . , l and x is arranged like in the proof of Theorem 6.5. Like there, the map extends to all probability measures with finite support and 6.5(i)–(iii) follow easily and we are left to show (iv). It is enough to restrict to µ, ν, λ such that all coefficients of µ, ν, (1 − λ)µ, λν are in Q — hence there is k such that all coefficients multiplied by k 2 yield an integer. Let x, y be tuples such that bar(µ) = bar2k (x), bar(ν) = bar2k (y). Now the right hand side of (iv) equals

bar2k (z,...,z,w,...,w) for z = bar(µ), w = bar(ν);

k k z appearing 2 (1 − λ) times and w appearing 2 λ times. If we put in bar2k (x) for z and bar2k (y) for w, we obtain a bar22k expression for the left hand side of (iv). This terminates the proof since checking the statement about σ˜ is a straightforward computation.

Theorem 6.7. Let X be a complete space with a linear barycenter. Then X is isometric to a convex subset of a normed space. Proof. Let bar denote the barycenter map provided by the previous lemma. We set n o X V := µ: 2 → R | spt(µ)| ≤ ∞, µ(X) = 0

(a subset of all signed measures on X w.r.t. the power set 2X as σ-algebra) endowed with the usual scalar multiplication and addition. From elementary measure theory we know that for every µ ∈ V there are unique non-negative (and ﬁnitely supported) measures µ+, µ− such that µ = µ+ − µ−. Deﬁne |µ| = µ+(X) = µ−(X) and the semi-norm kµk := |µ|d(bar(µ+/|µ|), bar(µ−/|µ|)) for µ 6= 0 and k0k = 0. Absolute homogeneity is obvious and since (µ + ν)+ = µ+ + ν+, (µ + ν)− = µ− + ν− and excluding the trivial cases µ = 0 or ν = 0 or µ + ν = 0 we have ! !! µ+ + ν+ µ− + ν− kµ + νk = |µ + ν| d bar , bar |µ + ν| |µ + ν| ! !! µ+ ν+ µ− ν− = |µ + ν| d bar |µ| + |ν| , bar |µ| + |ν| . |µ+ν| |µ| |µ+ν| |ν| |µ+ν| |µ| |µ+ν| |ν|

Now + + ! |µ| µ |ν| ν |µ| |ν| bar |µ+ν| + |µ+ν| = bar |µ+ν| δ µ+ + |µ+ν| δ ν+ , |µ| |ν| bar |µ| bar |ν| − − ! |µ| µ |ν| ν |µ| |ν| bar |µ+ν| + |µ+ν| = bar |µ+ν| δ µ− + |µ+ν| δ ν− |µ| |ν| bar |µ| bar |ν| 44 CHAPTER 6. BARYCENTERS by Lemma 6.6(iv) and the two terms on the right have distance less or equal to

|µ| + − |ν| + − d bar µ , bar µ + d bar ν , bar ν |µ + ν| |µ| |µ| |µ + ν| |ν| |ν| following from 6.6(iii). Therefore the triangle inequality holds. If now W denotes the quotient space of V by the equivalence µ ∼ ν if kµ − νk = 0, then — picking a basepoint o ∈ X — the map ϕ: X → W, x 7→ δx − δo is an isometric embedding.

6.3 Integration

In preparation for the second step of the proof of Theorem 9.1 we define a Riemannian integral for maps with values in a space with reversible bicombing. Let f : M → X be a map from a compact metric space with Borel probability measure µ to a complete space with reversible bicombing. We call f Riemann integrable if it is bounded and continuous outside a set D of points of discontinuity for which we have µ(D) = 0 (equivalently µ(Uδ(D)) → 0, where Uδ stands for the open δ-neighborhood). For a finite Borel k partition Ai, i = 1, . . . , k of M (that is ∪i=1Ai = M and Ai ∩ Aj = ∅ whenever i 6= j) and a selection of tagged points ai ∈ Ai, we define the Riemann sum to be

k ! X bar µ(Ai)δf(ai) . i=1

The mesh of a partition is defined to be maxi=1,...,k diam(Ai). We claim that for every ε there is a δ such that if the mesh of two partitions Ai,Bj does not exceed δ, then their Riemann sums are not more than ε apart. If K is the diameter of the image of f, take δ small enough to ensure µ(Uδ(D))K ≤ ε/4 and (from uniform continuity on compact subsets) d(f(x), f(y)) ≤ ε/4 for all x, y ∈ M \Uδ(D) with d(x, y) ≤ δ. Let Cij := Ai ∩Bj (dropping empty sets) be the common refinement of the two partitions and subdivide it further into Cij ∩ Uδ(D) and Cij \ Uδ(D). Now a short computation shows that the Riemann sum of the last partition has distance less or equal ε/2 to both, the Riemann sum of Ai and Bj. This gives the desired bound. Hence we may define the integral (or barycenter) of f as the limit of Riemann sums when the mesh of the partitions becomes zero (and such exist as M is compact). Furthermore, it is now straightforward to verify the formula Z Z Z d f dµ, g dµ ≤ d(f(t), g(t)) dµ(t). (6.3) M M M Chapter 7

Semi-simple isometries

In preparation for the proof of Theorem 8.3 we now discuss semi-simple isometries of a metric space X with a (not necessarily consistent) bicombing σ. The main purpose is to establish basic properties regarding sets of minimal displacement, analogous to those in the case of CAT(0) spaces. Whereas in the latter case a key role is played by the projection onto convex subspaces, we make use of the barycenter maps, introduced in the preceding chapter, instead. Here and in the subsequent Flat tori Chapter the less elaborate barycenter of Theorem 6.1 serves our needs. Nevertheless, we would lose nothing in using the reﬁned barycenter of Theorem 6.5. We use either of them but, for notational convenience, adopt the notation of the former theorem and suppress the subscript n; thus we write bar(x1, . . . , xn).

7.1 Minimal displacement and axes

We turn to the discussion of isometries and begin by recalling some standard terminology and basic facts. First, let X be an arbitrary metric space. For any map γ : X → X we denote by dγ(x) := d(x, γ(x)) the displacement at a point x ∈ X, and we put

|γ| := inf dγ(x) and Min(γ) := {x ∈ X | dγ(x) = |γ|}. x∈X An isometry γ of X is called parabolic if Min(γ) is empty and semi-simple otherwise. In the latter case, γ is elliptic if |γ| = 0 (that is, γ has a ﬁxed point) and hyperbolic if |γ| > 0. For an isometry γ, a line ξ : R → X will be called an axis of γ if there exists a t > 0 such that γ(ξ(s)) = ξ(s + t) for all s ∈ R. (7.1) Then, for x := ξ(0) and any y ∈ X, the triangle inequality gives

k k k d(x, γ (x)) ≤ d(x, y) + k dγ(y) + d(γ (y), γ (x))

= k dγ(y) + 2 d(x, y) (7.2)

k for all k ≥ 1, where d(x, γ (x)) = kt, thus t ≤ dγ(y) and so dγ(x) = t = |γ|. Hence every isometry γ with an axis is hyperbolic, and all axes of γ are contained in Min(γ). For the 46 CHAPTER 7. SEMI-SIMPLE ISOMETRIES converse, let γ be a hyperbolic isometry of X with |γ| =: t, let x ∈ Min(γ), and suppose there is a geodesic β : [0, t] → X from x to γ(x). Then the curve ξ : R → X satisfying

k ξ(kt + s) = γ (β(s)) for all k ∈ Z and s ∈ [0, t] (7.3) is a local geodesic (in fact it preserves all distances less than or equal to t), because ξ is parametrized by arc length and d(ξ(kt + s), ξ(kt + t + s)) = dγ(β(s)) ≥ t. This curve ξ also satisﬁes (7.1), hence it is an axis of γ if it happens to be a line. This is the case, for example, if X is a Busemann space, as then every local geodesic in X is a geodesic. (If α:[a, b] → X is the geodesic from ξ(a) to ξ(b), then the non-negative function s 7→ d(ξ(s), α(s)) is locally convex, hence convex on [a, b], hence identically zero as it vanishes at the endpoints.) Thus every hyperbolic isometry of a Busemann space is axial (compare Chap. 11 in [Pap]). The following result shows in particular that this last fact remains true in the more general context of this thesis. Recall that a bicombing σ of X is γ-equivariant, for an 2 isometry γ of X, if γ ◦ σxy = σγ(x)γ(y) for every pair (x, y) ∈ X . Proposition 7.1. Let γ : X → X be an isometry of a complete metric space X with a γ-equivariant reversible bicombing σ. Then:

1 k 2 (i) For all x, y ∈ X and k ≥ 1, |γ| ≤ k d(x, γ x) ≤ dγ(y) + k d(x, y). 1 k (ii) For all x ∈ X, limk→∞ k d(x, γ x) = |γ|. (iii) If γ is hyperbolic, then for every x ∈ Min(γ) there exists an axis of γ through x.

(iv) If C ⊂ X is non-empty, σ-convex, and γ-invariant, then |γ| = γ|C . Proof. The second inequality in (i) is just (7.2). For the ﬁrst, we employ the barycenter construction. Given x ∈ X and k ≥ 1, put x := (x, γx, . . . , γk−1x) and γx := (γx, . . . , γkx). Then 1 |γ| ≤ d(bar(x), γ bar(x)) = d(bar(x), bar(γx)) ≤ d(x, γkx), k where the last two steps use parts (iii) and (iv) of Theorem 6.1, respectively. The limit formula (ii) is an immediate consequence of (i). As for (iii), let γ be a hyperbolic isometry with |γ| =: t, and let x ∈ Min(γ). Then, for y = x, (i) shows that d(x, γnx) = nt for all n ≥ 1, thus any curve ξ as in (7.3) is a line. Finally, given any set C as in (iv), its closure C is still σ-convex and γ-invariant, and furthermore complete. Now ﬁx any x ∈ C and apply (ii) for both γ and γ|C to conclude that |γ| = γ|C . Clearly γ|C = γ|C . The following standard result will be used several times in the sequel.

Lemma 7.2. Let X be a proper metric space, and let Γ be a group acting properly and cocompactly by isometries on X. Suppose that Φ:Γ × X → R is a function such that (i) Φ(δγδ−1, δx) = Φ(γ, x) for all γ, δ ∈ Γ and x ∈ X; and 7.1. MINIMAL DISPLACEMENT AND AXES 47

(ii) Φ(γ, ·) is continuous on X for every γ ∈ Γ.

For every ﬁnite subset {γ1, . . . , γn} ⊂ Γ; if there is a sequence xk ∈ X along which the displacement functions dγ1 , . . . , dγn stay bounded, then it exists an x ∈ X such that Φ(γ, x) = limk→∞ Φ(γ, xk) for every γ ∈ hγ1, . . . , γni for which this limit exists.

The assumptions on Φ are satisﬁed, in particular, for Φ(γ, x) := dγ(x). The lemma then shows that every element of Γ is a semi-simple isometry of X (cf. Proposition II.6.10 in [BriH]).

Proof. Let xk be the presumed sequence. Since the action is cocompact, we may assume, by passing to a subsequence if necessary, that there exist elements δk ∈ Γ such that δkxk −1 converges to some point z ∈ X. First let j = 1: Since d(δkxk, δkγjδk (δkxk)) = |γj|xk −1 is bounded, it follows that d(z, δkγjδk z) is bounded too. Hence, because the action is −1 proper, we may pass to a further subsequence in order to arrange that δkγjδk is equal to the same element γ¯j ∈ Γ for every k. Repeating the argument for j = 2, . . . , n we arrive at a map γj 7→ γ¯j which extends to a homomorphism γ 7→ γ¯ on all of hγ1, . . . , γni −1 and with the property that δkγδk =γ ¯ independent of k. Finally, by the properties of Φ and for γ for which the limit exists, we have

−1 Φ(γ, δ z) = Φ(¯γ, z) = lim Φ(¯γ, δkxk) = lim Φ(γ, xk); k k→∞ k→∞

−1 so setting x := δk z — for any (ﬁxed) value of k — concludes the proof.

From the above results we obtain a crucial fact for the proof of Theorem 8.3.

Proposition 7.3. Let X be a proper metric space with a reversible bicombing σ. Let Γ be a group acting properly and cocompactly by isometries on X, and suppose that σ is Γ-equivariant. Then for every ﬁnitely generated abelian subgroup A of Γ the set

Min(A) := \ Min(α) α∈A is non-empty (and σ-convex, α-invariant for every α ∈ A, and closed).

Proof. For an individual α ∈ A the set Min(α) is non-empty, as noted after Lemma 7.2. T Now suppose that B ⊂ A is a finite set such that Min(B) := β∈B Min(β) 6= ∅, and let α ∈ A \ B. Note that Min(B) is σ-convex and furthermore α-invariant, as α commutes with every element of B. Using Proposition 7.1(iv) we find a sequence xk such that dα(xk) → α|Min(B) = |α| and dβ(xk) = |β| for all β ∈ B. Applying Lemma 7.2 for the set B ∪{α} ⊂ Γ we get a point y ∈ Min(B ∪{α}). This shows that Min(B) 6= ∅ for every finite set B ⊂ A. Exhausting A by an increasing sequence of finite subsets we obtain a sequence xk in X such that for every α ∈ A, the sequence dα(xk) is eventually constant with value |α|. Applying Lemma 7.2 again, for generators α1, . . . , αn of A, we conclude that Min(A) is non-empty. 48 CHAPTER 7. SEMI-SIMPLE ISOMETRIES

7.2 σ-Axes

In Proposition 7.1 we showed that every hyperbolic isometry γ of a complete metric space X with a γ-equivariant bicombing σ is axial. It is natural to ask whether γ also admits an axis that is at the same time a σ-line. Such an axis will be called a σ-axis. It turns out that the answer to this question is negative in general, see Example 7.7. However, we shall prove in Proposition 7.8 that any group Γ satisfying the assumptions of Theorem 8.3 acts by “σ-semi-simple” isometries, that is, every element has either a ﬁxed point or a σ-axis. We start with an auxiliary result which will be useful in the proof of Proposition 7.6. In [GoeK], Goebel and Koter proved a ﬁxed point theorem for “rotative” non-expansive mappings in closed convex subsets of Banach spaces. The argument can easily be adapted to the present context.

Theorem 7.4. Let Y be a complete metric space with a bicombing σ. Then every 1-Lipschitz map ϕ: Y → Y for which there exist an n ≥ 2 and 0 ≤ a < n such that

d(y, ϕn(y)) ≤ a d(y, ϕ(y)) for all y ∈ Y has a ﬁxed point. Furthermore, the ﬁxed point set of ϕ is a 1-Lipschitz retract of Y .

Proof. Let λ ∈ (0, 1) (an appropriate value depending only on a, n will be determined at the end of the proof). For every x ∈ Y , the map sending y ∈ Y to [x, ϕ(y)](λ) is λ- Lipschitz and thus has a unique ﬁxed point fλ(x) ∈ Y by Banach’s contraction mapping theorem. This yields a map fλ : Y → Y with the property that fλ(x) = [x, ϕ(fλ(x))](λ) for all x ∈ Y . By the conical property,

d(fλ(x), fλ(y)) ≤ (1 − λ) d(x, y) + λ d(ϕ(fλ(x)), ϕ(fλ(y)))

≤ (1 − λ) d(x, y) + λ d(fλ(x), fλ(y)), so fλ is 1-Lipschitz. Furthermore, fλ has the same ﬁxed points as ϕ, since [x, ϕ(x)](λ) = x if and only if ϕ(x) = x. By the assumption on ϕ,

n n d(y, ϕ(fλ(y))) ≤ d(y, ϕ (y)) + d(ϕ (y), ϕ(fλ(y))) n−1 ≤ a d(y, ϕ(y)) + d(ϕ (y), fλ(y)). (7.4)

To estimate the last term, note that, again by the conical property,

m m m d(ϕ (y), fλ(y)) ≤ (1 − λ) d(ϕ (y), y) + λ d(ϕ (y), ϕ(fλ(y))) m−1 ≤ (1 − λ)m d(ϕ(y), y) + λ d(ϕ (y), fλ(y)), for m ∈ {1, . . . , n − 1}. It follows that

n−1 n−1 d(ϕ (y), fλ(y)) ≤ (1 − λ)cλ(n) d(ϕ(y), y) + λ d(y, fλ(y)), (7.5) 7.2. σ-AXES 49

n−2 where cλ(n) := (n − 1) + (n − 2)λ + ··· + λ . Combining the fact that d(y, fλ(y)) = λ d(y, ϕ(fλ(y))) with (7.4) and (7.5) we get

a + (1 − λ)c (n) d(y, f (y)) ≤ λ λ d(y, ϕ(y)). λ 1 − λn

Now let x ∈ Y , and put y := fλ(x). Then y = [x, ϕ(y)](λ) and therefore λ d(y, ϕ(y)) = (1 − λ) d(x, y), thus we obtain

a + (1 − λ)c (n) d(f (x), f 2(x)) ≤ λ d(x, f (x)). λ λ 1 + λ + ··· + λn−1 λ

Since the factor on the right converges to a/n < 1 for λ → 1, there is a λ ∈ (0, 1) making k it strictly less than 1. Then, for every x ∈ Y , the sequence k 7→ fλ (x) is Cauchy. Since fλ is 1-Lipschitz, it follows that the limit point %(x) of this sequence is a fixed point of fλ, hence a fixed point of ϕ, and % is a 1-Lipschitz retraction of Y onto the fixed point set of ϕ.

Now let γ be any isometry of a metric space X with a bicombing σ. We associate with γ the map −1 1 ϕγ : X → X, ϕγ(x) = [γx, γ x]( 2 ). 1 1 −1 −1 Note that d(ϕγ(x), ϕγ(y)) ≤ 2 d(γx, γy) + 2 d(γ x, γ y) = d(x, y), thus ϕγ is 1- Lipschitz. The interest of this map comes from the following simple fact.

Lemma 7.5. Let γ be an isometry of a metric space X with a γ-equivariant consistent and reversible bicombing σ, and let x ∈ X be such that γ(x) 6= x. Then there exists a σ-axis of γ through x if and only if x is a ﬁxed point of the associated map ϕ = ϕγ.

Proof. If ξ : X → R is a σ-axis of γ through x, then clearly ϕ(x) = x. Conversely, suppose that x is a ﬁxed point of ϕ. Put t := dγ(x). Let β : [0, t] → X be deﬁned by β(s) = [x, γx](s/t), and consider the corresponding unit speed curve ξ : R → X satisfying (7.3). Since ϕ(x) = x and σ is consistent, it follows that ξ is a “local σ-line”, in fact every subsegment of length t is σ-convex. Then, as in the case of Busemann spaces, it follows that ξ is a (global) σ-line and hence a σ-axis of x.

We now show that the translation length |ϕγ| = infx∈X dϕγ (x) of ϕγ is always zero, provided the bicombing σ is γ-equivariant.

Proposition 7.6. Let γ be an isometry of a metric space X with a γ-equivariant bicombing σ. Then for all x ∈ X and n ≥ 1,

n √ d(x, ϕγ (x)) ≤ n dγ(x), and |ϕγ| = 0. 50 CHAPTER 7. SEMI-SIMPLE ISOMETRIES

Proof. We assume without loss of generality that X is complete. We write ϕ := ϕγ. Let k l x ∈ X. For all l ∈ Z and 0 ≤ k ∈ Z, put xk,l := ϕ (γ (x)) and dk,l := d(x, xk,l). Note that ϕ and γ commute because σ is γ-equivariant. In particular, for k ≥ 1, we have xk,l = [xk−1,l−1, xk−1,l+1](1/2) and hence

−1 dk,l ≤ 2 (dk−1,l−1 + dk−1,l+1). By induction on k this yields ! k k d ≤ 2−k X d . k,l i 0,l−k+2i i=0

Since d0,l ≤ |l| dγ(x), we obtain ! ! k k k k d ≤ 2−k X d ≤ 2 d (x) · 2−k X i − k . k,0 i 0,2i−k γ i 2 i=0 i=0

−kk As 2 i is the probability mass function of a binomial distribution with parameters k and 1/2 (number of trials and probability of success, respectively), let Z be a random variable distributed accordingly. Recall that the mean and variance are E[Z] = k/2, Var[Z] = k/4, hence d q k,0 ≤ E[|Z − E[Z]|] = E (Z − E[Z])2 2 dγ(x) √ q q k ≤ E[(Z − E[Z])2] = Var[Z] = 2 √ k by Jensen’s inequality. Thus d(x, ϕ (x)) = dk,0 ≤ k dγ(x). For any c > |γ|, Y := {x ∈ X | dγ(x) ≤ c} is a non-empty, complete and σ-convex set with γ(Y ) = Y and, consequently, ϕ(Y ) ⊂ Y . Now if |ϕ| was positive, then for some √ suﬃciently large n and for some a < n we would have d(x, ϕn(x)) ≤ c n ≤ a|ϕ| ≤ a d(x, ϕ(x)) for all x ∈ Y , and Theorem 7.4 would provide a ﬁxed point y = ϕ(y), in contradiction to |ϕ| > 0. This shows that |ϕ| = 0.

The following example shows that, in general, the inﬁmum |ϕγ| = 0 need not be attained. The isometry γ we construct is axial, but has no σ-axis.

Example 7.7. Let X := `∞(Z) be the Banach space of bounded functions x: Z → R, with the supremum norm, and consider the usual affine bicombing (x, y, λ) 7→ (1 − λ)x + λy. Let %: X → X be the shift map satisfying %(x)(k) = x(k − 1) for all x ∈ X and k ∈ Z, and let p ∈ X be defined by p(k) = 1 for k ≥ 1 and p(k) = 0 otherwise. k The isometry γ : X → X, γ(x) := %(x) + p, satisfies kγ (0)k∞ = k for all k ≥ 1, so |γ| = dγ(0) = 1 by Proposition 7.1. Thus γ is hyperbolic and hence axial. The associated map ϕ = ϕγ : X → X is given by 1 ϕ(x) = (%(x) + %−1(x) − z), 2 7.2. σ-AXES 51 where z := %−1(p) − p is the indicator function of 0. Now an x ∈ X with ϕ(x) = x 1 would have to fulfill x(k) = 2 (x(k − 1) + x(k + 1)) for k 6= 0 and hence be constant (as 1 it has to remain bounded), but this violates the condition x(0) = 2 (x(−1) + x(1) − 1). (Nevertheless, we still have |ϕ| = 0 and points with arbitrary small ϕ-displacement are not hard to compute.)

In contrast to this example, the following holds.

Proposition 7.8. Let X be a proper metric space with a consistent and reversible bicombing σ. Let Γ be a group acting properly and cocompactly by isometries on X, and suppose that σ is Γ-equivariant. Then every isometry γ ∈ Γ has either a ﬁxed point or a σ-axis.

Proof. Let γ ∈ Γ. In view of Lemma 7.5 we just need to show that the associated map ϕ := ϕγ has a fixed point. Let r > |γ|. Applying Proposition 7.6 to the complete, σ-convex and γ-invariant set Xr := {x ∈ X | dγ(x) ≤ r}, we find a sequence of points in Xr along which the displacement function dϕ tends to zero. Now, as a consequence of Lemma 7.2, there exists a point y ∈ X such that dϕ(y) = 0. Indeed, for Φ(γ, x) := −1 1 dϕ(x) = d(x, [γ x, γx]( 2 )), assumption (i) in the lemma is satisfied because

−1 −1 −1 1 d(δx, [δγ δ (δx), δγδ (δx)]( 2 )) −1 1 −1 1 = d(δx, δ ◦ [γ x, γx]( 2 )) = d(x, [γ x, γx]( 2 )) by the δ-equivariance of σ.

Let us mention that, for a given isometry γ in the setting above, the union (of the traces) of all axes need not be σ-convex — see Example 8.4 at the end of the next chapter. 0 0 Nevertheless, if ξ, ξ are two axes of γ and d(ξ(R), ξ (R)) =: c > 0, then for every λ ∈ [0, 1] 0 there is an axis η with d(ξ(R), η(R)) = (1−λ)c, d(η(R), ξ (R)) = λc. To see this, observe that when we adjust the parametrization of the σ-lines such that d(ξ(0), ξ0(0)) = c, then B(ξ(0), (1 − λ)c) and B(ξ0(0), λc) intersect in a non-empty, compact, and σ-convex set C. This set is ϕγ-invariant for the associated 1-Lipschitz map ϕγ and hence contains a ﬁxed point thereof. Chapter 8

Flat tori

We now prove Theorem 8.3. Thus, throughout this section, X denotes a proper metric space with a consistent and reversible bicombing σ, equivariant with respect to a group Γ that acts properly and cocompactly by isometries on X, and A is a free abelian subgroup of Γ of rank n. We retain the multiplicative notation for A ⊂ Γ, but we ﬁx once and for n n all an isomorphism ι:(Z , +) → (A, ·). For generic points a, b ∈ Z , the corresponding elements of A will be denoted by α := ι(a), β := ι(b) without further comment. We n write b1 = (1, 0,..., 0, ), . . . , bn = (0,..., 0, 1) for the canonical generators of Z and put βi := ι(bi). With this convention, we can state the assertion of Theorem 8.3 as follows: n n There is a norm on R and an isometric embedding f :(R , k · k) → X such that

n n αf(p) = f(p + a) for all p ∈ R and a ∈ Z . (8.1)

This implies that d(f(p), αkf(p)) = kkak for all k ≥ 1, therefore kak must be equal to the translation length |α| by Proposition 7.1(ii). We ﬁrst show that a norm with this latter property indeed exists. Notice that we already know from Proposition 7.3 that Min(A) is non-empty.

n n Lemma 8.1. There is a unique norm k · k on R such that kak = |α| for every a ∈ Z . n With respect to the metric on Z induced by this norm, the map a 7→ αx is an isometric n embedding of Z into X for every x ∈ Min(A).

n n Proof. Deﬁne kak := |α| for all a ∈ Z . Then, for every x ∈ Min(A) and a, b ∈ Z ,

kb − ak = |α−1β| = d(x, α−1βx) = d(αx, βx), and this is non-zero if α 6= β, for otherwise α−1β would have a ﬁxed point by Proposi- tion 7.8 above and inﬁnite order as A is free, in contradiction to the action being proper. m Furthermore, kmak = |m|kak for m ∈ Z because |α | = |m||α| by Proposition 7.1(ii). It n n follows that k · k extends uniquely to a norm on Q and then also to a norm on R .

n n n In the following, R (and Z , Q ) are always equipped with the metric induced by this norm k · k. The next result will constitute the last step of the proof of Theorem 8.3. 53

n Proposition 8.2. Assume that there exists a sequence of 1-Lipschitz maps fk : R → X n such that for all p ∈ R and i ∈ {1, . . . , n},

lim d(βifk(p), fk(p + bi)) = 0. k→∞ n Then there is an isometric embedding f : R → X satisfying (8.1).

Proof. First note that the displacement of βi along the sequence fk(0) is bounded by d(βifk(0), fk(bi)) + d(fk(bi), fk(0)), where the ﬁrst term goes to zero by assumption and the second is bounded by kbik. Arguing as for Lemma 7.2 we may assume, after passing to a subsequence, that there are isometries δk ∈ Γ such that δkfk(0) converges to a point −1 in X and δkαδk =: α ∈ Γ is constant for every α ∈ A. By the Arzelà-Ascoli theorem we may further assume that the sequence of 1-Lipschitz maps δk ◦ fk converges uniformly n n n on compact sets to a 1-Lipschitz map f∞ : R → X. Now, for all p ∈ R and a ∈ Z , we have that

−1 −1 d(αδk f∞(p), δk f∞(p + a)) = d(αf∞(p), f∞(p + a)) = lim d(αδkfk(p), δkfk(p + a)) k→∞

= lim d(αfk(p), fk(p + a)). k→∞

By assumption this last term is zero if a ∈ {b1, . . . , bn}. Thus, for any ﬁxed k, the map −1 n f := δk ◦ f∞ satisﬁes αf(p) = f(p + a) for all p ∈ R and for all generators a = bi, n hence for all a ∈ Z . This property then forces the 1-Lipschitz map f to be isometric n on Z because kak = |α| ≤ d(f(p), αf(p)) = d(f(p), f(p + a)) ≤ kak

n n n for all p, a ∈ Z . Furthermore, every line segment in R connecting two points in Z is embedded isometrically. Since the set of all pairs of points which lie on a common such n n segment is dense in R × R , we conclude that f is in fact an isometric embedding. n Now we proceed as follows. First we construct a 1-Lipschitz map g : R → Min(A) n that sends every ray R+a = {λa | λ ≥ 0} with a ∈ Z \{0} isometrically to a (σ- )ray asymptotic to a σ-axis of α. Then we use a discrete averaging process based on barycenters (Theorem 6.1 or 6.5) to find a sequence of maps satisfying the assumptions of Proposition 8.2. Theorem 8.3 (Flat torus). Let X be a proper metric space with a consistent and reversible bicombing σ. Let Γ be a group acting properly and cocompactly by isometries on X, and suppose that σ is Γ-equivariant. If Γ has a free abelian subgroup A of rank n ≥ 1, then X contains an isometrically embedded n-dimensional normed space on which A acts by translation. (The norm of the embedded plane is the one of Lemma 8.1.) n Proof. We fix a point x ∈ Min(A) and define a map g : R → X as follows. First, for n a ∈ Z \{0} and λ ∈ [0, 1], put g(λa) := lim [x, αkx] λ . k→∞ k 54 CHAPTER 8. FLAT TORI

The limit exists by Lemma 4.1 since the orbit hαix stays within ﬁnite distance of some σ-axis of α by Proposition 7.8, and the deﬁnition is clearly consistent for distinct repre- n sentations of the same point. For a, b ∈ Z \{0} and λ ∈ [0, 1] we have

d(g(λa), g(λb)) = lim d[x, αkx] λ , [x, βkx] λ k→∞ k k λ ≤ lim d(αkx, βkx) k→∞ k = λka − bk,

n in particular g is 1-Lipschitz on Q \{0}. It follows that g extends uniquely to a 1- n Lipschitz map g : R → X. n n Next, for all integers k ≥ 1, put Ik := [−k, k] ∩ Z . We want to establish the following estimate of sublinear growth:

e(k) := sup d(g(a), αx) = o(k)(k → ∞). (8.2) a∈Ik

n n Given an ε > 0, there is a ﬁnite set B ⊂ Z and a constant C such that for every a ∈ Z there exist a point b ∈ B and a positive integer m such that ka − mbk ≤ εkak + C. For each b ∈ B we pick a point yb ∈ Min(A) on a σ-axis of β. Then

m mk 1 mk 1 d(g(mb), β yb) = lim d [x, β x] , [yb, β yb] ≤ d(x, yb), k→∞ k k

m hence d(g(mb), β x) ≤ 2 d(x, yb). So let D := max{2 d(x, yb) | b ∈ B}. Now, for every n a ∈ Z , if b and m are as above, we have

d(g(a), αx) ≤ d(g(a), g(mb)) + d(g(mb), βmx) + d(βmx, αx) ≤ 2ka − mbk + D ≤ 2(εkak + C) + D.

This clearly yields (8.2). To conclude the proof we now construct maps that meet the requirements of Propo- n sition 8.2. Deﬁne fk : R → X by

−1 fk(p) := bar({α g(p + a) | a ∈ Ik}).

Since g is 1-Lipschitz, it follows from Theorem 6.1(iv) that fk is 1-Lipschitz as well. For the generators bi we have

−1 βifk(p) = bar({α g(p + bi + a) | a ∈ Ik − bi}), −1 fk(p + bi) = bar({α g(p + bi + a) | a ∈ Ik}), the ﬁrst equality being a consequence of Theorem 6.1(iii) and a change of variable. In order to estimate d(βifk(p), fk(p + bi)) we need a pairing of points in Ik with points 55

in Ik − bi. We match a ∈ Ik ∩ (Ik − bi) with itself and a ∈ Ik \ (Ik − bi) with a˜ := a − (2k + 1)bi ∈ (Ik − bi) \ Ik. For a pair of the latter type we have

−1 d(α g(p + bi + a), x) = d(g(p + bi + a), αx)

≤ d(g(p + bi + a), g(a)) + d(g(a), αx)

≤ kp + bik + e(k)

−1 as well as d(˜α g(p + bi +a ˜), x) ≤ kp + bik + e(k + 1), since Ik − bi ⊂ Ik+1. Thus

−1 −1 d(α g(p + bi + a), α˜ g(p + bi +a ˜)) ≤ 2(kp + bik + e(k + 1)).

n−1 n Since there are (2k + 1) such pairs (a, a˜) out of |Ik| = (2k + 1) pairs in total, we conclude that 2(kp + b k + e(k + 1)) d(β f (p), f (p + b )) ≤ i → 0 (k → ∞) i k k i 2k + 1 by Theorem 6.1(iv) and (8.2).

One may ask if the embedded ﬂats, respectively their image, can be made σ-convex. This is not the case as the conclusive next example shows.

Example 8.4. Let f : R → R be the 2-periodic extension of x 7→ |x| deﬁned on [−1, 1]. 2 Now we set r(x) = f(x2) and r¯(x) = max{f(x1), f(x2)}, a pair of maps `∞ → R. The premises of Lemma A.8 are met, and

3 Q := {x ∈ `∞ | r(x1, x2) ≤ x3 ≤ r¯(x1, x2)} is therefore an injective space with unique consistent and reversible bicombing by Theo- rem 3.15. Observe that for every geodesic α in Q, the projection (α1, α2, 0) is necessarily

(0, 0, 0)

(−1, −1, 1)

Figure 8.1: Part of the space considered in Example 8.4.

2 linear in `∞. To see this, remember that α is straight so dα(t0)+(2,0,0) ◦ α (which equals

2 − α1 locally) and dα(t0)+(−2,0,0) ◦ α (equal to −2 + α1 locally) are convex implying that 56 CHAPTER 8. FLAT TORI

α1 is affine around t0. Likewise, α2 is affine so the projection is (locally) linear. Now 2 Z acts properly and cocompactly by isometries when we assign x 7→ x + (2z1, 2z2, 0) to 2 every z ∈ Z , and the Flat Torus Theorem applies. Let H be the plane through (0, 0, 0) with normal vector (1, 1, 0). H intersects Q in a one dimensional zigzag line (which is a geodesic line of Q in fact). So do all planes H + k(1, 1, 0) for k ∈ Z. We now have that the embedded flat provided by the Flat Torus Theorem contains all these zigzag lines. This either by geometric reasons (one can not embed a plane in a half-plane essentially) or more simply since the flat is invariant under the given action. We conclude that the σ-convex hull of every embedded flat coincides with Q and in particular includes all the 3-dimensional tetrahedral cells. 2 Finally, we briefly discuss the σ-axes of (1, 0), (0, 1) ∈ Z and claim that the former run through the graph of r (the bottom margin of Q) and the latter through the graph of r¯ (the top margin). This determines the axes since their projection is linear. To circumvent lengthy proofs, we exemplary show this for the dotted (and non-linear) σ- 2 axis of (0, 1) ∈ Z depicted in Figure 8.4 only. First the points with maximal third coordinate form an orbit and hence the axis needs to include them all. Moreover, this also determines its projection and therefore the route on the ascending and descending 3 parts as depicted. The remaining linear segments are clearly straight (even in `∞), therefore σ-geodesics and part of the axis as the bicombing is consistent. Chapter 9

Asymptotic rank

The wording asymptotic rank was coined by Wenger in [Wen]. Given a metric space X, the general idea is to take the supremum over all n ∈ N for which there is a n- dimensional normed space embedding in some way into X; the notion of embeddings adopted may be as restricted as isometric embeddings or much more relaxed like quasi- isometric embeddings. Then we typically want to show that for a certain class of spaces all these ranks coincide. For Busemann spaces such an equivalence was shown by Kleiner [Kle] — see Theorem D. Theorem 9.1 below forms the missing piece to carry over (most of) Kleiner’s Theorem D to our setting and generalizes the analogous Proposition 10.22 in [Kle] (with a slight improvement even in the case of Busemann spaces). The original proof relies in a crucial way on properties of Busemann functions that are no longer available in our context. Therefore we take a quite different road and make use of the bicombing geodesics only to provide us with a barycenter map. After the proof of Theorem 9.1, which we do in three steps, we finally state the analogue to Kleiner’s Theorem. This result shows, among other things, that a proper cocompact metric space X with a bicombing contains a flat (normed) n-plane whenever there is a quasi-isometric n embedding of R into X — a result which was first shown for manifolds of non-positive curvature in [AndS]. For the definition of Gromov-Hausdorff topology see the first section of Chapter 2. Theorem 9.1. Let X be a proper metric space with a bicombing σ and cocompact isometry group. Suppose there are sequences Rk ∈ (0, ∞), Sk ⊂ X, and a normed vector n 1 n space (R , k · k), so that Rk → ∞, and R Sk converges to the unit ball B ⊂ (R , k · k) in k n the Gromov-Hausdorff topology. Then (R , k · k) can be isometrically embedded in X. Observe that we may assume σ to be reversible by Proposition 2.2 and can therefore utilize the barycenter maps. For this section, k · k always refers to the norm given above n and B(r) denotes the norm-ball of radius r centered at 0 ∈ R . The assumptions of the theorem may be restated as follows. There are maps Φk : B(Rk) → X such that

d(Φk(x), Φk(y)) − kx − yk ≤ Rkεk (9.1) for a sequence Rk as above and a null sequence εk → 0. For brevity, we call such a sequence Φk an RHA; a relative Hausdorﬀ approximation. Also note that a subsequence 58 CHAPTER 9. ASYMPTOTIC RANK of an RHA is still an RHA. A priori the maps of an RHA need not be continuous which leads to the ﬁrst step of the proof of Theorem 9.1.

Lemma 9.2 (Step 1). Given an RHA for a space X with reversible bicombing (and thus a barycenter by means of Theorem 6.1), one may construct an RHA whose maps are all L-Lipschitz continuous. More generally, one may require X to be merely Lipschitz n − 1-connected and the lemma still holds.

n Proof. Set λk = Rkεk and restrict every Φk to those points of λkZ that lie in B(Rk). On this grid (which is a δλk-separated set for δ := min kx − yk > 0) we have x6=y∈Zn 1 d(Φ (x), Φ (y)) ≤ kx − yk + λ ≤ 1 + kx − yk. (9.2) k k k δ

n n For every x ∈ λkZ the set x + λk[0, 1] is a cube with vertices in the grid. Let Ck be ˜ the union of those cubes contained in B(Rk). We extend every restricted map Φk to Ck n through the following procedure: To y ∈ x + λk[0, 1] we assign a probability measure ˜ n n 1 supported on Φk(x + λk{0, 1} ) having, for every e ∈ {0, 1} , weight

n y − x Y(1 − e ) − (1 − 2e ) i i i i λ i=1 k ˜ ˜ at Φk(x + λke). The barycenter thereof shall then be the value for Φk(y). Observe that where the cubes intersect the values coincide. These extensions are L-Lipschitz for some n constant independent of k. With ∆ := diamk·k([0, 1] ) we have B(Rk − ∆λk) ⊂ Ck and on these sets we get ˜ ˜ d(Φk(x), Φk(y)) − kx − yk ≤ 2∆λkL + 2∆λk + λk when comparing x, y to two ∆λk-close points on the grid together with the ﬁrst inequality ˜ of (9.2). Since the right hand side divided by Rk − ∆λk converges to zero, the Φk form an RHA of L-Lipschitz maps. A space X is Lipschitz n−1-connected if there is a constant l such that for every m ∈ {0, 1, . . . , n − 1}, every c-Lipschitz map from the standard sphere Sm into X possesses a lc-Lipschitz extension to Bm+1 (the unit ball whose boundary is Sm). Given this, the ˜ restricted maps Φk may be extended directly through Theorem 1.5 in [LanS]. Since the existence of a bicombing easily implies that a space is Lipschitz (n − 1)-connected (even for arbitrary n), this yields a shorter proof of this lemma (and without requiring the bicombing to be reversible).

In both Steps 2 and 3 we apply Fubini’s theorem several times to real valued functions 1 n h ∈ L (Ω, Ln) deﬁned on (reasonably nice) subsets Ω ⊂ (R , k · k) of the normed space

1The chosen weights are one example of barycentric coordinates in an affine space. Taking the linear combination of these weights with the points x + λke would recover y. 59 at hand. For a direction v ∈ ∂B, let C be the orthogonal projection of Ω to v⊥, the Euclidean orthogonal complement of v. We then have Z Z Z n 1 n−1 h dL = h(t) dLv(t) dLv⊥ (c). Ω C (c+Rv)∩ Ω n n 1 L denotes the n-dimensional Lebesgue measure on R . Lv shall be the push-forward of 1 n (R, L ) by an isometry R → (c + Rv) ⊂ (R , k · k), thus it measures the length w.r.t. the given norm k · k (which may differ from the Euclidean norm k · k2). Finally, in order for the product of the right hand side measures to yield Ln, we have to rescale Ln−1 on ⊥ v appropriately. The correct factor is kvk2/kvk and the resulting measure denoted by n−1 Lv⊥ . The next step employs a smoothing procedure by means of integration. Recall from Section 6.3 the definition of Riemannian integrals for maps with values in spaces with reversible bicombings. The definition (9.3) can be seen as a sort of convolution operation that evens out the defects of the given RHA. Lemma 9.3 (Step 2). From an L-Lipschitz RHA for a space with reversible bicombing one may construct a 1-Lipschitz RHA.

Proof. Let Φk,Rk, εk be as presumed and choose sequences µk, λk such that Rkεk ≺ µk ≺ λk ≺ Rk, where ak ≺ bk means ak/bk → 0. Observe that from Rkεk ≺ µk we obtain a sequence δk → 0, such that kx − yk ≥ µk implies d(Φk(x), Φk(y)) ≤ (1 + δk)kx − yk. So on this intermediate scale, so to speak, we still have RHA-like estimates; we will use this later in the proof. We define a new RHA Ψk on B(Rk − λk) by Z n Ψk(x) := Φk dL . (9.3) B(x,λk) In Section 6.3 we defined the integral for probability measures only, so the above integral is understood to be taken with respect to the measure appropriately normalized. Comparing Φk to Ψk by means of (6.3) yields d(Φk(x), Ψk(x)) ≤ Lλk, thus Ψk is in fact an RHA. In order to compare Ψk(x) and Ψk(y) we define a map ϕk from B(x, λk) to B(y, λk). For every p ∈ B(x, λk), the (possibly non-unit speed) line t 7→ p + t(y − x) intersects both B(x, λk)\Interior(B(y, λk)) and B(y, λk)\Interior(B(x, λk)) in a segment of equal length. Let ϕk map every such first segment onto the corresponding second by appropriate (and unique) translation in direction of y − x. On the remaining part of B(x, λk), which is the interior of B(x, λk)∩B(y, λk), ϕk shall be the identity. Elementary n n geometry tells us that ϕk is measure preserving (i.e. L (ϕk(A)) = L (A)) and Φk ◦ ϕk is integrable on B(x, λk). Moreover, by a direct comparison of the Riemann sums involved, Z n Ψk(y) = Φk ◦ ϕk dL B(x,λk) and consequently Z 1 n d(Ψk(x), Ψk(y)) ≤ n d(Φk(t), Φk(ϕk(t))) dL (t). (9.4) L (B(λk)) B(x,λk) 60 CHAPTER 9. ASYMPTOTIC RANK

l (c0) c0 v,k z ϕk(z)

∂B(y, λk)

x sv y c lv,k(c) z0

s 0 ϕk(z ) Cv,k

Figure 9.1: case analysis

Now we ﬁx a direction v ∈ ∂B and estimate (9.4) for y = x + sv and s ≤ µk. Let Cv,k ⊥ be the orthogonal projection of B(x, λk) to v . We can rewrite the last integral as Z Z 1 1 n−1 n d(Φk(t), Φk(ϕk(t))) dLv(t) dLv⊥ (c). L (B(λk)) Cv,k (c+Rv)∩B(x,λk)

Let lv,k : Cv,k → R be the map that assigns the length of (c + Rv) ∩ B(x, λk) to every c. We consider two cases. Case 1: lv,k(c) ≥ µk(≥ s). Then d(Φk, Φk ◦ ϕk) is non-zero on an initial segment of (c + Rv) ∩ B(x, λk) of length s at most. There ϕk has displacement exactly lv,k(c) hence d(Φk, Φk ◦ ϕk) ≤ (1 + δk)lv,k(c) where δk is the sequence mentioned in the beginning. Case 2: lv,k(c) < µk. Again, the integrand is non-zero only on the initial segment of length at most s and we estimate the displacement of ϕk — which is max{lv,k(c), s} — by µk. Consequently d(Φk, Φk ◦ϕk) ≤ Lµk from the assumed Lipschitz continuity of Φk. Now we split the integral according to these two cases and get for the ﬁrst one Z 1 n−1 n s(1 + δk)lv,k(c) dLv⊥ (c) = (1 + δk)s = (1 + δk)kx − yk. L (B(λk)) Cv,k The second part can be estimated generously by

n−1 n−1 n−1 Lv⊥ (Cv,k)Lµk λk Lv⊥ (Cv)Lµk n s = n n s, L (B(λk)) λk L (B) where Cv is the orthogonal projection of the unit ball B. The measure of the cross n−1 sections Lv⊥ (Cv) is bounded by a constant independent of v, and therefore the above 61 fraction (i.e. everything except the s) converges uniformly to zero for k → ∞. Adding this to δk we arrive at d(Ψk(x), Ψk(y)) ≤ (1 + δk)kx − yk whenever kx − yk ≤ µk. Since the domains B(Rk − λk) are convex, the Lipschitz estimate extends to all pairs x, y. To conclude Step 2, we concatenate every Ψk with the 1/(1 + δk)-Lipschitz contraction ˜ z 7→ σΨk(0)z(1/(1+δk)) towards Ψk(0) producing 1-Lipschitz maps Ψk. And the estimate ˜ d(Ψk(x), Ψk(x)) ≤ δk/(1+δk)d(Ψk(0), Ψk(x)) together with, say, d(Ψk(0), Ψk(x)) ≤ 2Rk ˜ for all but finitely many k ensures that Ψk is still an RHA. We can now conclude the proof of Theorem 9.1 through the next and final Step 3. We use a kind of metric differentiation argument. Since we already made the maps of the RHA 1-Lipschitz, we must now deal with the fact that they may collapse locally. But, intuitively, at points where the differential is large this should not be the case. Keep in mind the following example of an RHA, which may justify why we√ have to chose n a density point a √in the proof below: Φk : B(k) →√ R , x 7→ max{kxk − k, 0} x/kxk, thus Φk maps B( k) to 0 and every other x a k-step towards 0. Now a = 0 is no appropriate choice to complete Step 3 as Φk converges uniformly on compact sets to the constant function Φ = 0. Lemma 9.4 (Step 3). From a 1-Lipschitz RHA for a proper space with cocompact isom- n etry group, we may construct an isometric embedding of the normed space (R , k · k).

Proof. Let Φk : B(Rk) → X be the presumed RHA of 1-Lipschitz maps and pick any direction v ∈ ∂B. For every x ∈ B(Rk) there is a minimal t ∈ R such that pk(x) := x + tv ∈ B(Rk) defining a map pk : B(Rk) → B(Rk). Furthermore, set Hk : B(Rk) → R, x 7→ d(Φk(pk(x)), Φk(x)). This map is 1-Lipschitz along straight segments [pk(x), qk(x)] ⊂ B(Rk), where qk(x) := x + tv ∈ B(Rk) for a maximal t, and locally Lipschitz at interior points of B(Rk). The latter claim holds since pk is Lipschitz there too, which may be seen by looking at the convex hull of an appropriate ball around an interior point x and its projection pk(x). Therefore, and by virtue of Rademacher’s theorem, we may ∞ n differentiate Hk in direction of v yielding an L (B(Rk), L ) derivative ∂vHk : B(Rk) → [−1, 1] defined almost everywhere with respect to the n-dimensional Lebesgue measure. From the fundamental theorem of calculus for Lipschitz functions we deduce

Z d(pk(x),qk(x)) ∂vHk pk(x) + tv dt = Hk(qk(x)). 0

For the maps hv,k(x): B → [−1, 1], x 7→ ∂vHk(Rkx) — that are just rescaled versions of the ∂vHk — this translates to Z 1 d Φk(pk(Rkx)), Φk(qk(Rkx)) hv,k dLv = . (x+Rv)∩B Rk The right hand side is less or equal to the length of (x + Rv) ∩ B and at least that quantity minus εk (the εk being the one provided by the RHA). Let Cv be the orthogonal projection of B to v⊥. Fubini’s theorem gives Z Z Z n 1 n−1 hv,k dL = hv,k dLv dLv⊥ (x), B Cv (x+Rv)∩B 62 CHAPTER 9. ASYMPTOTIC RANK thus we reach Z n n−1 n n L (B) − εkLv⊥ (Cv) ≤ hv,k dL ≤ L (B). B n Since hv,k is bounded by 1, we necessarily have L ({x ∈ B | hv,k(x) ≥ 1 − δ}) converging to Ln(B) for every δ > 0 when k → ∞. And this holds for arbitrary directions v. Next we pick a countable dense subset D ⊂ ∂B enumerated by vi for i ∈ N. From the above (and constructing first the values k(1, j), then k(2, j), and so on) we deduce the following: There is a double sequence k : N × N → N such that for every i ∈ N the sequence j 7→ k(i, j) is strictly increasing, j 7→ k(i + 1, j) is a subsequence of j 7→ k(i, j), and n 1 n −(i+j+1) L {x ∈ B | hvi,k(i,j)(x) ≥ 1 − j } ≥ L (B) 1 − 2 . Note that this implies Ln(T {x ∈ B | h (x) ≥ 1 − 1 }) ≥ Ln(B)/2. So possibly i,j∈N vi,k(i,j) j replacing the initial RHA by m 7→ Φk(m,m) (but again writing Φk for this new RHA), we may assume that \ A := {x ∈ B | hv,k(x) ≥ cv,k} k∈N,v∈D has positive n-dimensional Lebesgue measure for an appropriate choice of real numbers cv,k converging to 1 for every fixed v ∈ D. By virtue of Lebesgue’s density theorem we next choose a density point a ∈ A in the interior of B; by definition this is a point such that Ln(A ∩ B(a, δ)) lim = 1. δ→0 Ln(B(a, δ)) Recall that X was assumed to be proper. We make use of the cocompact isometry group as usual and can assume that Φk ◦ (x 7→ x + Rka) converges uniformly on compacta to n a 1-Lipschitz map Φ:(R , k · k) → X. It remains to verify that d(Φ(x), Φ(y)) ≥ kx − yk and it suffices to consider pairs where x − y is a multiple of some v ∈ D. To obtain a contradiction, assume there is an l > 0 with

d(Φk(x + Rka), Φk(y + Rka)) ≤ (1 − 2l)kx − yk for all but ﬁnitely many k ∈ N. Since the maps at hand are 1-Lipschitz, a suﬃciently small r provides

d(Φk(x + s + Rka), Φk(y + s + Rka)) ≤ (1 − l)kx + s − y − sk for s in G := {s | ksk ≤ r, s⊥x − y}. Let Z be the convex hull of (x + G) ∪ (y + G); a cylindric shape. From Fubini’s theorem and the fundamental theorem of calculus we deduce Z Z Z n 1 n−1 ∂vHk dL = ∂vHk dLv dLv⊥ (g) Z+Rka x+G+Rka (g+Rv)∩(Z+Rka) Z n−1 n ≤ (1 − l)kx − yk dLv⊥ (g) = (1 − l)L (Z) x+G+Rka 63 for k large enough. This translates to Z n n hv,k dL ≤ (1 − l)L ((1/Rk)Z) a+(1/Rk)Z and, for R > 0 (and k large enough) such that a + (1/Rk)Z ⊂ B(a, R/Rk) ⊂ B, ﬁnally Z n n n hv,k dL ≤ L (B(a, R/Rk)) − lL ((1/Rk)Z) B(a,R/Rk) lLn(Z) = Ln(B(a, R/R )) 1 − . k Ln(B(a, R))

But this contradicts the fact that a is a density point of A. As for every α ∈ (0, 1), we have

n n L B(a, R/Rk) ∩ {x ∈ B | hv,k(x) ≥ cv,k} ≥ L (B(a, R/Rk) ∩ A) n ≥ αL (B(a, R/Rk))

n n eventually, so αcv,kL (B(a, R/Rk)) − (1 − α)L (B(a, R/Rk)) becomes a lower bound for the above integral at last. Therefore d(Φ(x), Φ(y)) ≥ kx − yk which completes the proof.

These three steps compose a proof of Theorem 9.1, and as a consequence we obtain the previously mentioned generalization of a part of Theorem D in [Kle]. We refer to that paper for the deﬁnition of asymptotic cones.

Theorem 9.5. Let X be a proper metric space with bicombing and cocompact isometry group. Then for every positive integer n the following are equivalent:

(1) There exists an isometric embedding of some n-dimensional normed space into X.

n (2) There is a quasi-isometric embedding of R into X. n (3) There is a bi-Lipschitz embedding of R into some asymptotic cone Xω of X.

(4) There is a sequence of subsets Zk of some asymptotic cone Xω of X, a sequence n −1 0 < rk → 0, and a norm k · k on R such that rk Zk converges to the unit ball n B ⊂ (R , k · k) in the Gromov-Hausdorﬀ topology.

(5) There exist a sequence of sets Sk ⊂ X, a sequence 0 < Rk → ∞, and a norm n −1 n k · k on R such that Rk Sk converges to the unit ball B ⊂ (R , k · k) in the Gromov-Hausdorﬀ topology.

Proof. The implications (1) ⇒ (2) ⇒ (3) are clear, whereas (5) ⇒ (1) is just Theo- rem 9.1. The implication (4) ⇒ (5) is shown as in the last paragraph on p. 455 in [Kle]. The implication (3) ⇒ (4) is proved by a metric diﬀerentiation argument, compare Proposition 10.18 in [Kle] and Corollary 2.2 in [Wen]. Appendix A

Injective metric spaces

First we swiftly develop the basics about injective metric spaces needed in this thesis. We follow the outline in Section 2 and 3 of [Lan] very closely (often to the extent of copying word for word) and refer thereon and to the original papers [AroP, Isb1, Dre] for a more comprehensive picture and further results. In the subsequent section we will n characterize all injective subsets of `∞.

A.1 Basic deﬁnitions and injective hulls

A metric space X is injective (as an object in the metric category with 1-Lipschitz maps as morphisms) if for every metric space B and every 1-Lipschitz map f : A → X defined on a set A ⊂ B there is 1-Lipschitz extension f : B → X. Basic examples include the real line (f(b) := infa∈A(f(a) + d(a, b)) yields a maximal 1-Lipschitz extension) and all `∞ spaces for arbitrary index sets (as the maps may be extended component-wise) and (R-)trees. Also, every 1-Lipschitz retract X ⊂ Y of an injective space Y is itself injective since f : A → X may first be extended to f : B → Y and then concatenated with the retraction π : Y → X. A metric space X is called an absolute 1-Lipschitz retract if for every isometric embedding i: X → Y into another metric space Y there exists a retraction of Y onto i(X). Clearly, every injective space X is an absolute 1-Lipschitz retract since the identity on Y restricted to i(X) extends to a 1-Lipschitz retraction. Conversely, every space X embeds isometrically into `∞(X) via the Kuratowski embedding x 7→ dx − dz and is therefore injective if it is a 1-Lipschitz retract. Thus these two notions coincide, and every injective space admits a reversible bicombing by Lemma 3.1 and the Kuratowski embedding. So injective spaces are geodesic, contractible, and also never empty (as we allow A = ∅= 6 B in the definition above). By a remarkable result of Isbell [Isb1], every metric space X has an injective hull E(X) (unique up to isometry) which is minimal among injective spaces containing an isometric copy of X in the following sense. There is an isometric embedding e: X → E(X) with the following property. Whenever there is a metric space Z and a 1-Lipschitz map h: E(X) → Z such that h ◦ e is an isometric embedding, then h is an isometric embedding as well. We call such an isometric embedding e essential. This already implies A.1. BASIC DEFINITIONS AND INJECTIVE HULLS 65 uniqueness up to isometry (see Theorem A.3(ii)) and shall be the definition of injective hull adopted here. Also note that in case X is injective already, then e is an isometry. To see this, take h: E(X) → E(X) to be the 1-Lipschitz retraction onto the absolute 1-Lipschitz retract e(X). Now h ◦ e is an isometric embedding and so is h, therefore h must be the identity on E(X). We now review Isbell’s construction X 7→ E(X). Given X a metric space X, consider the vector space R of arbitrary real valued functions on X and put X ∆(X) := {f ∈ R | f(x) + f(y) ≥ d(x, y) for all x, y ∈ X}. Call f ∈ ∆(X) extremal if there is no g ≤ f in ∆(X) distinct from f. The set E(X) of extremal functions is equivalently described as

X E(X) = f ∈ R | f(x) = supy∈X (d(x, y) − f(y)) for all x ∈ X , thus f is extremal if and only if for every x and ε > 0 there is a y such that

f(x) + f(y) ≤ d(x, y) + ε. (A.1)

Furthermore, from

f(x) − d(x, x0) = sup(d(x, y) − d(x, x0) − f(y)) ≤ f(x0) y∈X for f ∈ E(X) and all x, x0 ∈ X it follows that E(X) is a subset of

∆1(X) := {f ∈ ∆(X) | f is 1-Lipschitz}.

X Note that a function f ∈ R belongs to ∆1(X) if and only if

kf − dxk∞ = f(x) for all x ∈ X. (A.2)

The metric (f, g) 7→ kf − gk∞ on ∆1(X) is thus ﬁnite as kf − gk∞ ≤ kf − dxk∞ + kg − dxk∞ = f(x) + g(x) for any x ∈ X. The set E(X) ⊂ ∆1(X) is equipped with the induced metric, and one has the canonical isometric embedding

e: X → E(X), e(x) = dx, (A.3) where dx denotes the map y 7→ d(x, y). In order to show that E(X) is indeed the injective hull of X, we need the following two propositions.

Proposition A.1. For every metric space X there is a map p: ∆(X) → E(X) such that

(i) p(f) ≤ f for all f ∈ ∆(X), thus p(f) = f for all f ∈ E(X);

(ii) kp(f) − p(g)k∞ ≤ kf − gk∞ for all f, g ∈ ∆(X). The right hand side of (ii) may be inﬁnite for f, g ∈ ∆(X) but is ﬁnite if they belong to ∆1(X). So p yields a 1-Lipschitz retraction of ∆1(X) onto E(X). 66 APPENDIX A. INJECTIVE METRIC SPACES

∗ X Proof. For every f in the domain of p we deﬁne f ∈ R by

f ∗(x) = sup(d(x, y) − f(y)) for all x ∈ X. y∈X

∗ ∗ ∗ ∗ ∗ Clearly, f ≤ f and |f (x) − g (x)| ≤ kf − gk∞ hence kf − g k∞ ≤ kf − gk∞. We deﬁne 1 q(f) := (f + f ∗) 2 and have kq(f) − q(g)k∞ ≤ kf − gk∞, q(f) ≤ f and q(f) ∈ ∆(f) for f ∈ ∆(X). To see the last fact, add up f ∗(x) + f(y) ≥ d(x, y) and f(x) + f ∗(y) ≥ d(x, y). Now we deﬁne p(f) to be the pointwise limit of the monotonically decreasing sequence of maps k 7→ qk(f). Clearly p(f) ∈ ∆(X) and (i) and (ii) hold. We are left to show p(f) ∈ E(X). For all k ≥ 1; p(f) ≤ qk(f) and hence p(f)∗ ≥ qk(f)∗, so

0 ≤ p(f) − p(f)∗ ≤ qk(f) − qk(f)∗ = 2(qk(f) − qk+1(f)).

The last term converges pointwise to zero, thus p(f)∗ = p(f).

Proposition A.2. For every metric space X the metric space E(X) is injective.

Proof. Since E(X) is a 1-Lipschitz retract of ∆1(X) by the previous proposition, we may as well just prove that ∆1(X) is injective. Given metric spaces ∅ 6= A ⊂ B and a 1-Lipschitz map F : A → ∆1(X), a 7→ fa, put

f b(x) := inf (fa(x) + d(a, b)) a∈A for all x ∈ X. Clearly f b is a non-negative 1-Lipschitz function as the inﬁmum of a 0 family of such. For a, a ∈ A and y ∈ X, we have fa(y) − fa0 (y) ≤ kfa − fa0 k∞ = 0 0 kF (a) − F (a )k∞ ≤ d(a, a ) and so

0 f b(x) +f b(y) ≥ inf (fa(x) + fa0 (y) + d(a, a )) a,a0∈A

≥ inf (fa(x) + fa(y)) ≥ d(x, y). a∈A

0 This shows that f b ∈ ∆1(X). For b, b ∈ B and x ∈ X,

0 0 f b(x) − d(b, b ) = inf (fa(x) + d(a, b) − d(b, b )) ≤f b0 (x), a∈A

0 hence kf b−f b0 k∞ ≤ d(b, b ). If b ∈ A, then f b(x) ≤ fb(x) and fb(x) ≤ fa(x)+kfa−fbk∞ ≤ fa(x) + d(a, b) for all x ∈ X and a ∈ A, so that f b = fb. Thus F : b 7→f b is a 1-Lipschitz extension of F .

We now state Isbell’s result about E(X).

Theorem A.3. For every metric space X, the following hold. A.1. BASIC DEFINITIONS AND INJECTIVE HULLS 67

(i) If L: E(X) → E(X) is a 1-Lipschitz map that ﬁxes e(X) pointwise, then L is the identity on E(X) (here e denotes the canonical embedding (A.3));

(ii) E(X) is an injective hull of X;

(iii) if Y is another injective hull of X, then E(X) and Y are isometric.

Proof. For (i) we use (A.2). The map L takes f ∈ E(X) to some g ∈ E(X) such that

g(x) = kg − dxk∞ = kL(f) − L(dx)k∞ ≤ kf − dxk∞ = f(x) for all x ∈ X, so g = f since f is extremal. We already know that E(X) is injective, so for (ii) it remains to show that e is essential. Suppose h: E(X) → Z is 1-Lipschitz and h ◦ e: X → Z is an isometric embedding. Since E(X) is injective, e: X → E(X) extends to a 1-Lipschitz map e: Z → E(X), thus e ◦ h ◦ e = e. The map L := e ◦ h must be the identity on E(X) by (i). As both h and e are 1-Lipschitz, h is in fact an isometric embedding. As for (iii), if Y is another injective hull and i: X → Y an essential isometric embedding, then i extends to a 1-Lipschitz map I : E(X) → Y , so I ◦ e = i. Likewise, there is a 1-Lipschitz map e: Y → E(X) with e ◦ i = e. Since i is essential, e is an isometric embedding; furthermore, e ◦ I ◦ e = e, thus e ◦ I = idE(X) by (i). Hence e is an isometry onto E(X) and I its inverse.

Of course, if X is injective, then e: X → E(X) is an isometry, and therefore an arbitrary space X is injective if and only if {dx | x ∈ X} are the only extremal functions on X. Moreover, if X is compact so is E(X) as a consequence of the Arzelà-Ascoli theorem. We conclude this section with some results involving isometries of X. The isometry group of X will be denoted by Isom(X).

Proposition A.4. Let X be a metric space. Then:

(i) For every γ ∈ Isom(X) there is a unique isometry γ¯ : E(X) → E(X) with the property that γ¯ ◦ e = e ◦ γ. One has γ¯(f) = f ◦ γ−1 for all f ∈ E(X), and (γ, f) 7→ γ¯(f) is an action of Isom(X) on E(X).

−1 X (ii) The linear isomorphism f 7→ f ◦ γ of R maps ∆(X) onto itself, and the map p: ∆(X) → E(X) constructed in the proof of Proposition A.1 has the additional property that γ¯(p(f)) = p(f ◦ γ−1) for all γ ∈ Isom(X) and f ∈ ∆(X).

Proof. Clearly there can not be several extensions of γ by A.3(i). Concerning existence, observe that e◦γ is an essential isometric embedding; now with i := e◦γ and Y := E(X) one can argue as for A.3(iii). If f ∈ E(X) and x ∈ X, then

−1 −1 (¯γ(f))(x) = kγ¯(f) − dxk∞ = kf − γ¯ (dx)k∞ = kf − dγ−1(x)k∞ = f(γ (x)).

Obviously (γ, f) 7→ γ¯(f) = f ◦γ−1 is an action of Isom(X) on E(X). It is straightforward −1 X to check that the linear isomorphism f 7→ f ◦ γ of R maps ∆(X) onto itself and that 68 APPENDIX A. INJECTIVE METRIC SPACES it commutes with the operators deﬁned in the proof of Proposition A.1, thus f ∗ ◦ γ−1 = (f ◦ γ−1)∗, q(f) ◦ γ−1 = q(f ◦ γ−1), and

γ¯(p(f)) = p(f) ◦ γ−1 = p(f ◦ γ−1).

As an application of the above result we obtain:

Theorem A.5. Every injective metric space X admits a reversible bicombing σ, equivariant with respect to the whole isometry group (γ◦σxy = σγ(x)γ(y) for every γ ∈ Isom(X) and x, y ∈ X).

Proof. Since X is injective, the canonical map e: x 7→ dx is an isometry of X onto E(X). Let p: ∆(X) → E(X) be the map from the proof of Proposition A.1. For all x, y ∈ X and t ∈ [0, 1], we have (1 − t)dx + tdy ∈ ∆1(X), and we set

−1 σxy(t) := (e ◦ p)((1 − t)dx + tdy), which is a reversible bicombing by Lemma 3.1. Now let γ ∈ Isom(X), and recall that −1 −1 γ¯(p(f)) = p(f ◦ γ ) for all f ∈ ∆(X), by Proposition A.4. Since dv ◦ L = dL(v) for all v ∈ X, we have

γ¯(p((1 − t)dx + tdy)) = p((1 − t)dγ(x) + tdγ(y)) for x, y ∈ X and t ∈ [0, 1]. As e−1 ◦ γ¯ = γ ◦ e−1, this gives the equivariance.

n A.2 A characterization of injective subsets in `∞ We move Proposition A.10 to the appendix since we make use of it in examples only. And we released it as a short note on arXiv [Des2]. Because the combinatorial dimension n (see Section 3.3 for the deﬁnition) of `∞ is n, the combinatorial dimension of every subset is bounded by n. Thereby all injective subsets provide a space with unique consistent and reversible bicombing (see Theorem 3.15). n For the rest of this appendix k · k denotes the usual maximum norm on `∞ and for n−1 every element x therein let xbi := (x1, . . . , xi−1, xi+1, . . . , xn) ∈ `∞ denote the vector x but with the i-th coordinate omitted. In the following we are using the 1-Lipschitz 3 map (a, b, x) 7→ min{max{a, x}, b} from `∞ to R quite frequently where a ∈ {−∞} ∪ R, b ∈ R ∪ {∞} but we always assert that a ≤ b. We write p([a, b], x) for the value of this projection, i.e. the unique real number in the closed interval {y ∈ R | a ≤ y ≤ b} closest to x.

n−1 Lemma A.6. Let r, r¯ : `∞ → R be two 1-Lipschitz maps with r ≤ r¯ (at every point in n−1 `∞ ). Then the map

n n ϕ : `∞ → `∞, ϕ(x) := p([r(xb1), r¯(xb1)], x1), x2, . . . , xn n A.2. A CHARACTERIZATION OF INJECTIVE SUBSETS IN `∞ 69 is a 1-Lipschitz retraction to the hence injective and non-empty subset

n Q := {x ∈ `∞ | r(xb1) ≤ x1 ≤ r¯(xb1)} with the property that ϕd(x)1 = xb1. The lemma remains valid if we allow the (constant) function r = −∞ as the lower or r¯ = ∞ as the upper bound or both. Proof. Every component of ϕ is 1-Lipschitz, hence the whole map is and the rest is simple to prove as well.

n−1 Lemma A.7. Let λ ∈ [0, 1) and for every i = 1, . . . , n let ri, r¯i : `∞ → R be a pair of λ-Lipschitz maps with ri ≤ r¯i. Then the subspace

n Q := {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi) for i = 1, . . . , n} is injective — in particular every such set of inequalities is solvable. The lemma remains valid if any of the lower or upper bounds take the constant value −∞ or ∞, respectively.

Proof. Denote by ϕ1 the map from the previous lemma (applied to r1, r¯1) and anal- ogously ϕ2, . . . ϕn for the other components. We deﬁne the following sequence of 1- Lipschitz maps

Φm := ϕk ◦ · · · ◦ ϕ2 ◦ ϕ1 ◦ (ϕn ◦ · · · ◦ ϕ2 ◦ ϕ1) ◦ · · · ◦ (ϕn ◦ · · · ◦ ϕ2 ◦ ϕ1), where m denotes the number of concatenated maps on the right hand side. Formally

n Φ0 = Id`∞ and Φm = ϕ[m] ◦ Φm−1, where [m] := ((m − 1) mod n) + 1. n We claim that the sequence Φm converges pointwise and for every x ∈ `∞ we have limm→∞ Φm(x) ∈ Q. Then, since every (1-Lipschitz) Φm ﬁxes Q, the pointwise limit Φ is a 1-Lipschitz retraction to Q concluding the proof. n In order to prove the claim, pick an arbitrary x ∈ `∞ and deﬁne dk to be a sequence of real numbers such that m X Φm(x) = x + dke[k] and consequently |dm| = kΦm(x) − Φm−1(x)k, k=1 where e[k] denotes the [k]-th vector of the standard basis. This is possible since Φm(x) − Φm−1(x) = ϕ[m](Φm−1(x)) − Φm−1(x) and the map ϕ[m] alters nothing but the [m]-th coordinate. Now let m > n and set i := [m] as well as

a = Φm−n(x), b = Φm−1(x), c = Φm(x).

We want an estimate for |dm| = kc − bk. Since a = ϕi(Φm−n−1(x)), we have ri(abi) ≤ Pm−1 ai ≤ r¯i(abi) and also b = a + k=m−n+1 dke[k], hence bi = ai as well as kb − ak = max{|dm−n+1|, |dm−n+2|,..., |dm−1|}. We compare the following two lines:

|dm| = |p([ri(bi), r¯i(bi)], bi) − bi| = |ci − bi|, 0 = |p([ri(abi), r¯i(abi)], ai) − ai|. 70 APPENDIX A. INJECTIVE METRIC SPACES

2 As (p, q) 7→ | min{p, max{ai, q}}−ai| is a 1-Lipschitz map from `∞ to R and furthermore

ri(bi) − ri(abi), r¯i(bi) − r¯i(abi) ≤ λkb − ak, because ri, r¯i are λ-Lipschitz by assumption, this yields

|dm| ≤ λkb − ak = λ max{|dm−n+1|, |dm−n+2|,..., |dm−1|}. (A.4)

To end the proof we set D := max{|d1|,..., |dn|} and get

(m−1)%n |dm| ≤ Dλ

(where % denotes integer division, the ﬂoored value of the real division) by induction using (A.4). So the convergence of Φm(x) is like that of a geometric series. It remains to check that limm→∞ Φm(x) ∈ Q. This follows immediately from the fact that the sets n n ϕi(`∞) = {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi)} are closed and the subsequences k 7→ Φi+kn(x) (convergent to the same limit as Φm(x)) lie completely in these sets. So the limit lies in n the intersection of all the sets ϕi(`∞) and this is exactly Q.

The proof does not work with λ = 1 since the sequence Φm(x) does not has to converge as the example n = 2, r1(x2) =r ¯1(x2) = x2, r2(x1) =r ¯2(x1) = 1 + x1 shows. Moreover, Q can be empty. But even assuming Q 6= ∅ (change 1 + x1 to −x1 in the example) the sequence can be divergent, and it is of no help to subtract a convergent subsequence (which, in the case Q 6= ∅, always exists). Nevertheless, we can show the lemma to hold for λ ≤ 1 assuming Q 6= ∅.

n−1 Lemma A.8. For every i = 1, . . . , n let ri, r¯i : `∞ → R be a pair of 1-Lipschitz maps with ri ≤ r¯i. Then the subspace

n Q := {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi) for i = 1, . . . , n} is injective assuming either that all the maps ri, r¯i are bounded (hence again the system of inequalities is solvable automatically) or that Q is non-empty (requiring the existence of a solution). Again, the lemma remains valid if any of the lower or upper bounds take the constant value −∞ or ∞, respectively.

Proof. The bounded case first. Let λk = 1 − 1/k and l be a lower bound for the maps ri k k and u an upper bound for the maps r¯i. Define r¯i = λk(¯ri − u) + u and ri = λk(ri − l) + l k k and observe that the maps ri , r¯i for some fixed k are all λk-Lipschitz. Moreover, for k k every fixed i the sequence r¯i (ri ) is monotonically converging down (up) to r¯i (ri) pointwise. Define the sets

n n k k o Qk := x ∈ `∞ | ri (xbi) ≤ xi ≤ r¯i (xbi) for i = 1, . . . , n which are injective by the first proposition. And we have Q1 ⊃ Q2 ⊃ Q3 ⊃ · · · as well as T k Qk = Q (this already implies Q 6= ∅ as all the Qk are compact). If we can show that Qk converges to Q w.r.t. Hausdorff distance (implying Gromov-Hausdorff convergence), then Q is injective — either see Section 1.5 in [Moe] or derive a direct proof of this easy n A.2. A CHARACTERIZATION OF INJECTIVE SUBSETS IN `∞ 71 special case here. So assume for some > 0 we would have dH(Qk,Q) > for every k (first only for infinitely many k but then for all by monotonicity of Qk). Then taking T a convergent sequence xk ∈ Qk \ U(Q) 6= ∅ one immediately gets limk→∞ xk ∈ Qk leading to a contradiction. (In fact, this could also be derived from well-known theorems about Hausdorff distance.) This ends the proof in the case of bounded maps u ≤ ri, r¯i ≤ l. To reduce the case assuming Q 6= ∅ to the previous one we refer to Corollary 1.19 in [Moe]. There it is shown that every proper metric space Q is injective if and only if every closed ball in Q is injective. Thereby we only need to verify that Q ∩ B(q, r) is injective for all q ∈ Q, may assume q = 0 ∈ Q and have

Q ∩ B(0, r) = {x | p([−r, ∞], ri(xbi)) ≤ xi ≤ p([−∞, r], r¯i(xbi)) for all i} = {x | p([−r, r], ri(xbi)) ≤ xi ≤ p([−r, r], r¯i(xbi)) for all i} . The last set above is injective by the bounded case, and we are left to show the second equality since the ﬁrst one is clear. Clearly, all three sets are contained in B(0, r). For a point x in that ball, we have kqbi − xbik ≤ r and ri(qbi) ≤ 0 ≤ r¯i(qbi) by our choice of q. Consequently, ri(xbi) ≤ r, −r ≤ r¯i(xbi) and from this it is now obvious that the sets coincide.

n Can every injective subset of `∞ be written as a set of 2n 1-Lipschitz inequalities in the sense of the lemma above? The answer turns out to be positive. We need a last deﬁnition before turning to the proof. By a cone C(p, x) in a metric space X we mean the set {q ∈ X | d(p, q) = d(p, x) + d(x, q)}. n We mainly need this deﬁnition for cones in `∞ of the following special form: C(x−ei, x) or n C(x + ei, x), where ei, i = 1, . . . , n are vectors of the standard basis of R . In these cases we shorten the notation further to C(x, +i) := C(x − ei, x) and C(x, −i) := C(x + ei, x), where the second parameter intuitively indicates the direction as, e.g.

C(x, +i) = {x + bei + y | b ≥ 0, y⊥ei, kyk ≤ b}

(⊥ being the Euclidean orthogonality relation).

n Lemma A.9. For every injective subset Q ⊂ `∞ there are n pairs of 1-Lipschitz maps n−1 ri, r¯i : `∞ → R (i = 1, . . . , n) with ri ≤ r¯i such that

n Q = {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi) for i = 1, . . . , n}.

We allow for ri = −∞ or r¯i = ∞ (or both) for any i which is equivalent to drop some of the inequalities.

1 Proof. The injective subsets of `∞ = R are exactly the closed intervals. Thereby the 0 case n = 1 is trivial or more a matter of declaring the convention xbi = 0 ∈ `∞, and so we assume n ≥ 2. Since Q is injective, only the distance functions dq for q ∈ Q are 72 APPENDIX A. INJECTIVE METRIC SPACES extremal (see the remark after Theorem A.3). Therefore, given any point x outside Q, the function dx|Q is non-extremal. So we may assign to every such point the positive quantity

ε(x) := sup{ε | there is p ∈ Q with kx − pk + kx − qk ≥ kp − qk + ε for all q ∈ Q}

(from choosing q ∈ Q such that kx−qk = d(x, Q) we see that ε(x) ≤ 2d(x, Q)). Moreover, for every x let px be such that kx−pxk+kx−qk ≥ kpx −qk+ε(x)/2 for every q ∈ Q. Next n we select a cone Cx for every x ∈ `∞ \Q. To that end, let a ∈ R be some positive number, the exact value will be determined in the course of the proof. Assume that the i-th coordinate of x−px has maximal absolute value among all coordinates (if there are several such coordinates we simply choose one). Now set Cx = C(x − aε(x)ei, +i) if that value is positive and Cx = C(x + aε(x)ei, −i) if it is negative. Observe that x ∈ Interior(Cx) always. Assume that Q ∩ Cx contains a point q and Cx := C(x − aε(x)ei, +i) (the case Cx := C(x + aε(x)ei, −i) works the same way). A straightforward computation yields C(x, +i) ⊆ C(px, x) and hence

kpx − (q + aε(x)ei)k = kpx − xk + kx − (q + aε(x)ei)k and consequently kpx − qk ≥ kpx − xk + kx − qk − 2aε(x).

But this violates the definition of px if we choose a < 1/4. We do so and have Q∩Cx = ∅ n for all x ∈ `∞ \ Q. For every i, we define r¯i to be the pointwise infimum over the family n−1 of 1-Lipschitz functions `∞ → R; y 7→ kxbi − yk + xi − aε(x) where every x such that Cx = C(x − aε(x)ei, +i) contributes exactly one member. If there is no such x, we let r¯i := ∞. Similarly, ri := −∞ if there is no x with Cx = C(x + aε(x)ei, −i) or otherwise the supremum over all functions y 7→ kxbi − yk + xi + aε(x) for x with Cx = C(x + aε(x)ei, −i). It is now not hard to deduce

n Q = {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi) for i = 1, . . . , n} from x ∈ Interior(Cx) and Q ∩ Cx = ∅. It remains to show ri ≤ r¯i for all pairs. n−1 n First notice that ri > r¯i at some point in `∞ implies there are points x, y ∈ `∞ \ Q with Cx := C(x − aε(x)ei, +i), Cy := C(y + aε(y)ei, −i) such that the intersection Interior(Cx)∩Interior(Cy) is not empty. To show that this can not happen for appropriate choice of a, we assume otherwise and start with the easy observation that the apex x − aε(x)ei of Cx lies in Interior(Cy). Therefore x˜ := x − aε(x)ei − aε(y)ei lies in Interior(C(y, −i)) and the same holds for p˜x := px − aε(x)ei − aε(y)ei as px ∈ C(x, −i). So we have

kx − pxk + kx − pyk ≤ kx˜ − p˜xk + kx˜ − pyk + a(ε(x) + ε(y))

= kp˜x − pyk + a(ε(x) + ε(y))

≤ kpx − pyk + 2a(ε(x) + ε(y)), n A.2. A CHARACTERIZATION OF INJECTIVE SUBSETS IN `∞ 73

C(y, −1) C y py

C(py, x˜) x˜ x

Cx C(x, +1) p˜x px ∂C(x, −1)

Figure A.1: In this illustration px, py are chosen in ∂C(x, −1), ∂C(y, +1), respectively, since these are the critical cases.

hence by deﬁnition of px and py this leads to ε(x) ≤ 4a(ε(x)+ε(y)) and ε(y) ≤ 4a(ε(x)+ ε(y)), respectively. Now take a < 1/8. The sum of the last two inequalities involving ε(x), ε(y) then yields a contradiction proving that Interior(Cx) ∩ Interior(Cy) is in fact empty.

All in all we arrive at the goal of this section which summarizes the previous four lemmas.

n Proposition A.10. A non-empty subset Q ⊂ `∞ is injective if and only if it can be written as a solution set of (at most 2n) 1-Lipschitz inequalities like

n Q = {x ∈ `∞ | ri(xbi) ≤ xi ≤ r¯i(xbi) for i = 1, . . . , n}

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