Gluing Constructions and Local-to-Global

Gluing Constructions and Local-to-Global - ETH E

Research Collection
Doctoral Thesis
Gluing Constructions and Local-to-Global Results for
Hyperconvex Metric Spaces
Author(s):
Miesch, Benjamin
Publication Date:
2017
Permanent Link:
https://doi.org/10.3929/ethz-a-010867059
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ETH Library
Diss. ETH No. 23997
Gluing Constructions and
Local-to-Global Results
for
Hyperconvex Metric Spaces
A thesis submitted to attain the degree of
DOCTOR OF SCIENCES of ETH ZÜRICH
(Dr. sc. ETH Zürich)
presented by
Benjamin Raphael Miesch
MSc ETH Mathematik, ETH Zürich
born May 14, 1987
citizen of Titterten BL
accepted on the recommendation of
Prof. Dr. Urs Lang, examiner
Prof. Dr. Rafael Espı́nola, co-examiner
2017
Soli Deo Gloria.
Zusammenfassung
Hyperkonvexität beschreibt eine Schnitteigenschaft von Bällen in metrischen
Räumen und wurde 1956 von N. Aronszajn and P. Panitchpakdi für das Studium von Erweiterungen gleichmässig stetiger Abbildungen eingeführt.
In dieser Arbeit studieren wir nun verschiedene Eigenschaften von hyperkonvexen metrischen Räumen. Zuerst untersuchen wir Verklebungen von solchen
Räumen entlang isometrischer Teilmengen. Für schwach extern hyperkonvexe
Klebmengen können wir eine notwendige und hinreichende Bedingung angeben,
so dass der resultierende Raum wieder hyperkonvex ist. Als Konsequenz erhalten wir, dass das Verkleben von metrischen Räumen entlang extern hyperkonvexer oder stark konvexer Teilmengen die Hyperkonvexität erhaltet. Wir geben
auch eine Klassifizierung für Verklebungen von hyperkonvexen Vektorräumen
entlang eines linearen Unterraumes.
In einem zweiten Teil analysieren wir, wann ein lokal hyperkonvexer metrischer Raum hyperkonvex ist. Zu diesem Zweck beweisen wir einen CartanHadamard-Satz für metrische Räume, in denen wir lokal Geodäten mit einer
Konvexitätseigenschaft wählen können. Es folgt, dass ein vollständiger, einfach-zusammenhängender, lokal kompakter und lokal hyperkonvexer metrischer
Raum mit einer Längenmetrik, der lokal endliche kombinatorische Dimension
hat, ein hyperkonvexer metrischer Raum ist.
Des Weiteren beschäftigen wir uns mit schwächeren Formen der Hyperkonvexität. Wir zeigen, dass für n ≥ 3 jeder vollständige, fast n-hyperkonvexe
metrische Raum n-hyperkonvex ist, und beweisen dann, dass jeder vollständige,
4-hyperkonvexe metrische Raum bereits n-hyperkonvex für alle n ∈ N ist.
Anschliessend untersuchen wir Konvexitätseigenschaften von schwach extern hyperkonvexen Teilmengen und verbinden diese mit Lokal-global-Resultaten für diese Mengen. Dies führt unter anderem zu einem Helly-artigen Satz für
n .
schwach extern hyperkonvexe Teilmengen von l∞
Zum Schluss wenden wir uns noch kurz den Medianräumen und Würfelkomplexen zu. Wir beweisen, dass hyperkonvex metrisierte Würfelkomplexe mit
der entsprechenden Metrik CAT(0)-Räume sind. Zudem konstruieren wir eine
bi-Lipschitz-äquivalente, hyperkonvexe Metrik auf geodätischen Medianräumen
und zeigen, dass ein lokaler Medianraum mit einem geodätischen Bicombing ein
Medianraum ist.
iii
Abstract
Hyperconvexity is an intersection property of balls in metric spaces and was
introduced in 1956 by N. Aronszajn and P. Panitchpakdi to study extensions
of uniformly continuous transformations.
In this thesis, we study different properties of hyperconvex metric spaces.
First, we investigate gluings of such spaces along isometric subsets. For weakly
externally hyperconvex gluing sets, we can give necessary and sufficient conditions, so that the resulting space is hyperconvex as well. As a consequence, we
get that the gluing of metric spaces along externally hyperconvex or strongly
convex subsets preserves hyperconvexity. We then give a classification for gluings of hyperconvex vector spaces along linear subspaces.
In the second part, we analyze under which conditions a locally hyperconvex
metric space is hyperconvex. For this purpose, we prove a Cartan-Hadamard
Theorem for metric spaces with a local geodesic bicombing. It follows that a
complete, simply-connected, locally compact and locally hyperconvex metric
space with finite combinatorial dimension, endowed with the length metric, is
hyperconvex.
Furthermore, we consider some relaxed notions of hyperconvexity. We show
that, for n ≥ 3, every complete, almost n-hyperconvex metric space is n-hyperconvex and prove that every complete, 4-hyperconvex metric space is n-hyperconvex for every n ∈ N.
Afterwards, we investigate convexity of weakly externally hyperconvex subsets and connect them with local-to-global results for these sets. This leads to
n .
a Helly-type theorem for weakly externally hyperconvex subsets of l∞
Finally, we turn our attention to median metric spaces and cube complexes.
We prove that cube complexes which are hyperconvex with respect to some
metric also possess a metric such that they become CAT(0)-spaces. Moreover,
we construct a bi-Lipschitz equivalent hyperconvex metric on geodesic median
metric spaces and show that a locally median metric space with a geodesic
bicombing is a median metric space.
v
Acknowledgments
First of all, I would like to thank Prof. Dr. Urs Lang for his encouragement
and guidance during the past years. He was always ready to take time for my
questions and could give me the advise needed. I am also indebted to Prof.
Dr. Rafael Espı́nola for agreeing to act as co-examiner. I joyfully remember
my visit to Sevilla.
Moreover, I really appreciated the companionship with my colleagues from
Assistant Groups 1 & 4 of the Department of Mathematics at ETH Zürich.
Especially, I would like to mention my office mate Christian, who always had a
helping hand. With Nicolas I could also conduct projects beyond mathematics.
Maël and Giuliano were excellent coworkers and travel companions.
Above all, I thank my parents for their unconditional love and support.
I gratefully acknowledge financial support from the Swiss National Science
Foundation.
vii
CONTENTS
Contents
Abstract
v
I
Introduction
1
II
Hyperconvex Metric Spaces
7
II.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . 7
II.2 An Intersection Property for Externally Hyperconvex Subsets 9
II.3 Weakly Externally Hyperconvex Subsets . . . . . . . . . . . . 13
II.4 Retracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
III
Gluing Hyperconvex Metric Spaces
23
III.1 Gluing along Weakly Externally Hyperconvex Subsets . . . . 23
III.2 Gluing Isometric Copies . . . . . . . . . . . . . . . . . . . . . 29
III.3 Gluing Hyperconvex Linear Spaces . . . . . . . . . . . . . . . 31
IV
Local to Global
IV.1 The Cartan-Hadamard Theorem for Metric
Spaces with Local Geodesic Bicombings . . . . . . . . . . . .
IV.2 Locally Hyperconvex Metric Spaces . . . . . . . . . . . . . .
IV.3 Absolute 1-Lipschitz Neighborhood Retracts . . . . . . . . .
43
43
49
51
V
Finite Hyperconvexity
55
V.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . 55
V.2 Almost n-Hyperconvex Metric Spaces . . . . . . . . . . . . . 56
V.3 4-Hyperconvex Metric Spaces . . . . . . . . . . . . . . . . . . 61
VI
Convexity of Weakly Externally Hyperconvex Subsets
VI.1 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . .
VI.2 σ-Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
VI.3 Locally Weakly Externally Hyperconvex Subsets . . . . . . .
VII Hyperconvex Metrics on Median Metric Spaces
VII.1 Median Metric Spaces . . . . . . . . . . . . . . .
VII.2 Cube Complexes . . . . . . . . . . . . . . . . . .
VII.3 Hyperconvex Metrics on Median Metric Spaces .
VII.4 Locally Median Metric Spaces . . . . . . . . . .
ix
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65
65
66
69
75
75
77
83
86
CONTENTS
A
Some Results on Geodesic Bicombings
87
A.1 Reversible Geodesic Bicombings . . . . . . . . . . . . . . . . 87
A.2 A Non-Consistent Convex Geodesic Bicombing . . . . . . . . 89
Bibliography
97
Index
101
x
Chapter I
Introduction
Hyperconvex metric spaces appear in various contexts. They were introduced
by N. Aronszajn and P. Panitchpakdi and later studied by J. Lindenstrauss
to receive extension results for linear operators between Banach spaces [AP56,
Lin64]. Furthermore, they occur as the injective hull or tight span of a metric
space [Isb64, Dre84], which has applications to geometric group theory [Lan13]
and theoretical computer science [CL91, CL94]. Hyperconvex metric spaces
also play an important role in metric fixed point theory [EK01].
A metric space (X, d) is called hyperconvex if for every family {B(xi , ri )}i∈I
of closed balls with d(xi , xj ) ≤ ri + rj , we have
\
B(xi , ri ) 6= ∅.
i∈I
In other words, (X, d) is a metrically convex metric space whose closed balls
have the binary intersection property; compare [BL00, Definition 1.3]. Already
N. Aronszajn and P. Panitchpakdi showed that hyperconvex metric spaces are
the same as injective metric spaces and absolute 1-Lipschitz retracts [AP56].
From the construction of the injective hull by J. Isbell [Isb64] it follows that
hyperconvex metric spaces possess a geodesic bicombing which is invariant under isometries [Lan13, Proposition 3.8]. This means that we can select geodesics
in such a way that the distance between them fulfills a weak convexity property.
Therefore, we can look at hyperconvex metric spaces as some kind of weakly
negatively curved spaces. It turns out that many results for CAT(0) and Busemann spaces transfer to metric spaces with geodesic bicombings [DL16, Des16]
and hence also to hyperconvex metric spaces.
In this thesis, we continue in this spirit by providing gluing results for hyperconvex metric spaces and proving a local-to-global theorem for hyperconvex metric spaces which is based on a Cartan-Hadamard Theorem for metric
spaces with geodesic bicombings. Furthermore, we investigate finite hyperconvexity and study σ-convexity of weakly externally hyperconvex subsets. Then
a section on median metric spaces follows, where we construct a bi-Lipschitz
equivalent hyperconvex metric. In the appendix we finally include some further
results on geodesic bicombings.
1
CHAPTER I. INTRODUCTION
Gluing Hyperconvex Metric Spaces. It is a classical question if we can
glue two metric spaces along a common subset such that the properties of the
spaces are preserved. It is well known that gluing along a single point preserves
hyperconvexity [JLPS02, Lemma 2.1]. Recently, B. Piatek proved that if we
‘
glue two hyperconvex metric spaces along a unique interval, then the resulting
metric space is hyperconvex as well [Pia14].
This result can be generalized by studying externally hyperconvex and
weakly externally hyperconvex subsets. A subset A of a metric space (X, d)
is externally hyperconvex in X if for every family of closed balls {B(xi , ri )}i∈I
in X with d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri , we have
\
B(xi , ri ) ∩ A 6= ∅.
i∈I
Furthermore, a subset A of a metric space (X, d) is weakly externally hyperconvex in X if for every x ∈ X the set A is externally hyperconvex in A ∪ {x}.
As a first result we establish the binary intersection property for bounded
externally hyperconvex subsets. This will turn out to have great impact.
Proposition II.2.1. Let (X, d) be a hyperconvex metric space and {Ai }i∈I a
family of pairwise intersecting, externally
T hyperconvex subsets such that at least
one of them is bounded. Then we have i∈I Ai 6= ∅.
Eventually, we get the following characterization of gluings along weakly
externally hyperconvex subsets.
Theorem III.1.8. Let (X, d) be the metric space obtained by gluing a family
of hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A such that for each
λ ∈ Λ, A is weakly externally hyperconvex in Xλ . Then X is hyperconvex if and
only if for all λ ∈ Λ and all x ∈ X \ Xλ , the set B(x, d(x, A)) ∩ A is externally
hyperconvex in Xλ . Moreover, if X is hyperconvex, the subspaces Xλ are weakly
externally hyperconvex in X.
As a consequence, we obtain that gluing along strongly convex and gluing
along externally hyperconvex subsets preserves hyperconvexity. A subset A of
a metric space X is called strongly convex if for all x, y ∈ A, we have
I(x, y) := {z ∈ X : d(x, z) + d(z, y) = d(x, y)} ⊂ A.
We call I(x, y) the metric interval between x and y.
Corollary III.1.9. Let (X, d) be the metric space obtained by gluing a collection
(Xλ , dλ )λ∈Λ of hyperconvex metric spaces along some space A such that A is
closed and strongly convex in Xλ for each λ ∈ Λ. Then X is hyperconvex as
well.
Corollary III.1.10. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A such that A is
externally hyperconvex in each Xλ . Then X is hyperconvex. Moreover, A is
externally hyperconvex in X.
2
Furthermore, we apply this theorem to give a full characterization for gluings
of finite dimensional hyperconvex vector spaces along linear subspaces. In the
n denotes the vector space Rn endowed with the maximum norm.
following, l∞
n and X = lm . Moreover, let V be a linear
Theorem III.3.13. Let X1 = l∞
2
∞
subspace of both X1 and X2 such that V 6= X1 , X2 . Then X = X1 tV X2 is
k ×X 0 , X = lk ×X 0 ,
hyperconvex if and only if there is some k such that X1 = l∞
2
∞
1
2
k
0
0
0
n−k
m−k .
V = l∞ × V , and V is strongly convex in both X1 = l∞ and X20 = l∞
With the exception of Section II.4, the results so far have been published
in two papers, one of them joint with M. Pavón [Mie15, MP17]. Section II.4 is
part of the preprint [MP16].
Local to Global. The classical Cartan-Hadamard Theorem was generalized by W. Ballmann [Bal90] for metric spaces with non-positive curvature, and
by S. Alexander and R. Bishop [AB90] for locally convex metric spaces. We
now prove the Cartan-Hadamard Theorem in a more general setting, namely for
spaces which are not uniquely geodesic but locally possess a suitable selection
of geodesics.
In a metric space (X, d), a geodesic bicombing is a selection of a geodesic
between each pair of points. This is a map σ : X × X × [0, 1] → X such that
for all x, y ∈ X, the path σxy := σ(x, y, ·) is a geodesic from x to y. Moreover,
we say that the geodesic bicombing σ is consistent if
σσxy (s1 )σxy (s2 ) (t) = σxy ((1 − t)s1 + ts2 )
for all 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1]. A geodesic bicombing σ is called convex if
the function t 7→ d(σxy (t), σx̄ȳ (t)) is convex for all x, y, x̄, ȳ ∈ X. Furthermore,
we say that σ is reversible if σyx (t) = σxy (1 − t) for all x, y ∈ X and t ∈ [0, 1].
Theorem IV.1.2. Let X be a complete, simply-connected metric space with a
convex local geodesic bicombing σ. Then the induced length metric on X admits
a unique consistent, convex geodesic bicombing σ̃ which is consistent with σ. As
a consequence, X is contractible. Moreover, if the local geodesic bicombing σ is
reversible, then σ̃ is reversible as well.
This especially applies to certain locally hyperconvex metric spaces and
results in the following local-to-global theorem for hyperconvex metric spaces
and the analogue for absolute 1-Lipschitz retracts.
Theorem IV.2.1. Let X be a complete, locally compact, simply-connected, locally hyperconvex length space with locally finite combinatorial dimension. Then
X is a hyperconvex metric space.
Theorem IV.3.1. Let X be a locally compact absolute 1-Lipschitz uniform
neighborhood retract with locally finite combinatorial dimension. Then X is an
absolute 1-Lipschitz retract.
Chapter IV is contained in a paper in preparation [Mie16].
3
CHAPTER I. INTRODUCTION
Finite Hyperconvexity. For the study of extensions of uniformly continuous functions and compact linear operators a weaker form of hyperconvexity is
considered. In [Lin64], J. Lindenstrauss characterizes all Banach spaces B with
the property that any compact linear operator with target B possesses an ”almost” norm preserving extension in pure metric terms, namely as the Banach
spaces which are n-hyperconvex for every n ∈ N. A counterpart for uniformly
continuous maps between metric spaces was later proven by R. Espı́nola and
G. López, see [EL02]. This motivates a closer look on results concerning nhyperconvexity in general metric spaces. Note that the following definitions are
slightly different from the ones given in [AP56].
Let A be a subset of a metric space (X, d). The subset A is . . .
• n-hyperconvex if for every family {B(xi , ri )}ni=1
T of n closed balls with
centers xi ∈ A and d(xi , xj ) ≤ ri + rj , we have ni=1 B(xi , ri ) ∩ A 6= ∅;
• almost n-hyperconvex if for every family {B(xTi , ri )}ni=1 of n closed balls
with xi ∈ A and d(xi , xj ) ≤ ri + rj , we have ni=1 B(xi , ri + ) ∩ A 6= ∅,
for every > 0;
• externally n-hyperconvex in X if for every family {B(xi , ri )}ni=1 of n closed
balls with xiT∈ X, d(xi , A) ≤ ri and d(xi , xj ) ≤ ri + rj , we have that the
intersection ni=1 B(xi , ri ) ∩ A 6= ∅ is non-empty;
• weakly externally n-hyperconvex in X if for every x ∈ X, the set A is
externally n-hyperconvex in A ∪ {x}.
The following two theorems supplement results proven for Banach spaces
by J. Lindenstrauss, see [Lin64, Lemma 4.2] and [BL00, Lemma 2.13], and
hence completely answer Problem 1 and Problem 4 raised by N. Aronszajn and
P. Panitchpakdi in [AP56].
Theorem V.2.1. Let X be a complete, almost n-hyperconvex metric space for
n ≥ 3. Then X is n-hyperconvex.
This implies for instance that the metric completion of an n-hyperconvex
metric space is n-hyperconvex as well. Note that there are complete metric
spaces which are almost 2-hyperconvex but not 2-hyperconvex, see [AP56].
Theorem V.3.1. Let X be a complete metric space and let A ⊂ X be an
arbitrarily chosen non-empty subset. Then, the following hold:
(i) X is 4-hyperconvex if and only if X is n-hyperconvex for every n ∈ N.
(ii) A is externally 4-hyperconvex in X if and only if A is externally n-hyperconvex in X for every n ∈ N.
(iii) A is weakly externally 4-hyperconvex in X if and only if A is weakly externally n-hyperconvex in X for every n ∈ N.
4
Observe that this is the best we can hope for, since there are metric spaces
which are 3-hyperconvex but not 4-hyperconvex, e.g. l13 , and there is a subset
A of l∞ (N) which is externally n-hyperconvex for every n ∈ N, but fails to be
hyperconvex, see Example V.3.5.
σ-Convexity. Given a geodesic bicombing σ on a metric space, we can
investigate σ-convex subsets. A subset A of a metric space X, endowed with a
geodesic bicombing σ, is σ-convex if for all x, y ∈ A, it holds that σxy ([0, 1]) ⊂ A.
For sufficiently strong assumptions on the geodesic bicombing, we show that
externally and weakly externally hyperconvex subsets are σ-convex. Moreover,
we prove that σ-convex subsets which are uniformly locally (weakly) externally
hyperconvex are (weakly) externally hyperconvex.
Theorem VI.1.1. Let X be a hyperconvex metric space, let A ⊂ X be any
subset and let σ denote a convex geodesic bicombing on X.
(I) The following are equivalent:
(i) A is externally hyperconvex in X.
(ii) A is σ-convex and uniformly locally externally hyperconvex.
(II) If straight curves in X are unique, the following are equivalent:
(i) A is weakly externally hyperconvex and possesses a consistent, convex geodesic bicombing.
(ii) A is σ-convex and uniformly locally weakly externally hyperconvex.
Chapter V and Chapter VI are a publication in preparation joint with M.
Pavón [MP16].
Hyperconvex Metrics on Median Metric Spaces. Median metric
spaces are another class of metric spaces which where studied intensively. They
are strongly related with CAT(0) cube complexes and B. Bowditch showed that
complete connected median metric spaces of finite rank admit a bi-Lipschitz
equivalent CAT(0) metric [Bow16a]. Adopting his methods, we also construct
a hyperconvex metric on certain median metric spaces.
Theorem VII.3.1. Let (M, ρ) be a proper, geodesic median metric space of
finite rank, then M possesses a bi-Lipschitz equivalent metric d such that (M, d)
is hyperconvex.
Independently, B. Bowditch recently gave another construction for a hyperconvex metric on median metric spaces, see [Bow16b].
Finally, we use the methods from Chapter IV to prove the following localto-global result for median metric spaces.
Theorem VII.4.1. Let (X, d) be a metric space with a reversible, conical geodesic bicombing σ, such that for all x ∈ X and some r > 0, the neighborhood
B(x, r) is a median metric space, then X is a median metric space.
5
CHAPTER I. INTRODUCTION
Geodesic Bicombings. A geodesic bicombing σ : X × X × [0, 1] → X is
called reversible if we have
σyx (t) = σxy (1 − t)
for all x, y ∈ X and t ∈ [0, 1]. We prove that a result of D. Descombes [Des16,
Proposition 1.2] on the existence of reversible geodesic bicombings holds in
general metric spaces.
Proposition A.1.1. Let (X, d) be a complete metric space with a conical geodesic bicombing σ. Then X also admits a reversible, conical geodesic bicombing.
Moreover, we give a first example of a convex geodesic bicombing which
is not consistent. These last results are part of a publication in preparation,
co-authored with G. Basso [BM16].
6
Chapter II
Hyperconvex Metric Spaces
II.1
Basic Properties
First, we fix some notation and then prove some basic facts. Let (X, d) be a
metric space. We denote by
B(x0 , r) := {x ∈ X : d(x, x0 ) ≤ r}
the closed ball of radius r with center in x0 . For any subset A ⊂ X, let
B(A, r) := {x ∈ X : d(x, A) := inf d(x, y) ≤ r}
y∈A
be the closed r-neighborhood of A.
A metric space (X, d) is injective if for every isometric embedding ι : A ,→ Y
of metric spaces and every 1-Lipschitz map f : A → X, there is some 1-Lipschitz
map f¯: Y → X such that f = f¯ ◦ ι. Similarly, a metric space (X, d) is an
absolute 1-Lipschitz retract if for every isometric embedding ι : X ,→ Y of X
into a metric spaces Y , there is a 1-Lipschitz retraction r : Y → X. It is a
classical result due to N. Aronszajn and P. Panitchpakdi that these definitions
coincide with the one of hyperconvex metric spaces [AP56].
Proposition 1.1. Let (X, d) be a metric space. The following statements are
equivalent:
(i) X is hyperconvex.
(ii) X is injective.
(iii) X is an absolute 1-Lipschitz retract.
Proof. First we show (i) ⇒ (ii). Let A ⊂ Y and let f : A → X be a 1-Lipschitz
map. Consider the following set
F := {(Z, g) : A ⊂ Z ⊂ Y, f : Z → X 1-Lipschitz and g|A = f }
with partial order (Z, g) (Z 0 , g 0 ) if and only if Z ⊂ Z 0 and g 0 |Z = g.
7
CHAPTER II. HYPERCONVEX METRIC SPACES
By Zorn’s Lemma, there is a maximal element (Z̄, ḡ) ∈ F. Assume that
there is some y ∈ Y \ Z̄. For z ∈ Z, we define rz := d(z, y) and consider the
collection {B(ḡ(z), rz )}z∈Z of closed balls. We have
d(g(z), g(z 0 )) ≤ d(z, z 0 ) ≤ rz + rz 0
and therefore there is some
x∈
\
B(ḡ(z), rz ).
z∈Z
But then (Z ∪ {y}, f¯) ∈ F with f¯: Z ∪ {y} → X, f¯Z = ḡ and f¯(y) = x
is strictly bigger than (Z̄, ḡ), a contradiction. Hence, we have Z̄ = Y and
ḡ : Y → X is a 1-Lipschitz extension of f .
The implication (ii) ⇒ (iii) is immediate. Note that any 1-Lipschitz extension f¯: Y → X of the identity map id : X → X is a 1-Lipschitz retraction.
Finally, to prove (iii) ⇒ (i), recall that l∞ (X) is hyperconvex and X embeds
via the Kuratowski embedding k : X ,→ l∞ (X), x 7→ dx −dx0 for some fixed x0 ∈
X, where dx denotes the map z 7→ d(x, z). Hence, X is a 1-Lipschitz retract
of the hyperconvex space l∞ (X) and therefore hyperconvex itself. Indeed, fix
a retraction r : l∞ (X) → X and consider a family {B(xi , ri )}i∈I of closed balls
in X with d(xi , xj ) ≤ ri + rj . Then there is some
z∈
\
B(k(xi ), ri ) ⊂ l∞ (X)
i∈I
and thus r(z) ∈
T
i∈I
B(xi , ri ) ∩ X 6= ∅.
We call a non-empty subset of a metric
T space admissible if it can be written
as an intersection of closed balls A = i∈I B(xi , ri ). Furthermore, we denote
by A(X), E(X), W(X), and H(X) the collection of all admissible, externally
hyperconvex, weakly externally hyperconvex, and hyperconvex subsets of X.
We always have E(X) ⊂ W(X) ⊂ H(X). Moreover, it holds that A(X) ⊂ E(X)
if and only if X is hyperconvex.
Lemma 1.2. If A ∈ A(X) and E ∈ E(X) such that A ∩ E 6= ∅, then we have
A ∩ E ∈ E(X). Especially, if X is hyperconvex, we have A(X) ⊂ E(X).
Proof. Since
T A is admissible, there is a collection of balls {B(xi , ri )}0 i∈I0 such
that A = i∈I B(xi , ri ). Now, given a family of closed balls {B(xj , rj )}j∈J
with d(x0j , x0k ) ≤ rj0 + rk0 and d(x0j , A ∩ E) ≤ rj0 , we have d(xi , x0j ) ≤ ri + rj0 and
d(xi , E) ≤ ri , and therefore
A∩E∩
\
j∈J
B(x0j , rj0 ) = E ∩
\
i∈I
B(xi , ri ) ∩
\
B(x0j , rj0 ) 6= ∅
j∈J
since E is externally hyperconvex. If X is hyperconvex, then X ∈ E(X) and
therefore A(X) ⊂ E(X).
8
II.1. INTERSECTION PROPERTY
Recall the following well known facts about admissible and externally hyperconvex subsets; cf. [Sin89, KKM00].
T
Lemma 1.3. Let X be a hyperconvex metric space, A = i∈I B(xi , ri ) ∈ A(X)
and s ≥ 0. Then one has
\
B(A, s) =
B(xi , ri + s) ∈ A(X).
i∈I
T
Proof.
T Clearly, B(A, s) is contained in i∈I B(xi , ri + s). Conversely, for any
z ∈ i∈I B(xi , ri + s) we have d(xi , z) ≤ ri + s. Hence, since X is hyperconvex,
there is some
\
y ∈ B(z, s) ∩
B(xi , ri ) = B(z, s) ∩ A
i∈I
and thus z ∈ B(y, s) ⊂ B(A, s).
Lemma 1.4. Let X be a hyperconvex metric space and A ∈ E(X). Then also
B(A, r) ∈ E(X).
Proof. Let {B(xi , ri )}i∈I be a collection of closed balls with d(xi , xj ) ≤ ri + rj
and d(xi , B(A, r)) ≤ ri . Then we also have
T d(xi , A) ≤ ri + r. Since A is
externally hyperconvex, there is some y ∈ i∈I B(xi , ri + r) ∩ A. Especially, we
have d(xi , y) ≤ ri + r and therefore, since X is hyperconvex, we get
\
\
∅=
6
B(xi , ri ) ∩ B(y, r) ⊂
B(xi , ri ) ∩ B(A, r)
i∈I
i∈I
as desired.
II.2
An Intersection Property for Externally
Hyperconvex Subsets
In this section, we establish the following important intersection property of
externally hyperconvex subsets.
Proposition 2.1. Let (X, d) be a hyperconvex metric space and {Ai }i∈I a family of pairwise intersecting, externally
T hyperconvex subsets such that at least one
of them is bounded. Then we have i∈I Ai 6= ∅.
Observe that externally hyperconvex subsets are closed. The following technical lemma turns out to be the initial step in proving Proposition 2.1.
Lemma 2.2. Let X be a hyperconvex metric space. Let A, A0 ∈ E(X) with
y ∈ A ∩ A0 6= ∅ and x ∈ X with d(x, A), d(x, A0 ) ≤ r. Denote d := d(x, y) and
s := d − r. Then we have
A ∩ A0 ∩ B(x, r) ∩ B(y, s) 6= ∅,
given s ≥ 0. In any case, the intersection A ∩ A0 ∩ B(x, r) is non-empty.
9
CHAPTER II. HYPERCONVEX METRIC SPACES
Proof. For s ≤ 0, we have y ∈ A ∩ A0 ∩ B(x, r). Therefore, let us assume s > 0.
Claim. For each 0 < l ≤ s, there are a ∈ A, a0 ∈ A0 such that d(a, a0 ) ≤ l and
a, a0 ∈ B(x, r) ∩ B(y, s).
We start choosing
a1 ∈ B(y, l) ∩ B(x, d − l) ∩ A
and
a01 ∈ B(y, l) ∩ B(x, d − l) ∩ B(a1 , l) ∩ A0 .
Then, we inductively take
an ∈ B(y, nl) ∩ B(x, d − nl) ∩ B(a0n−1 , l) ∩ A
and
a0n ∈ B(y, nl) ∩ B(x, d − nl) ∩ B(an , l) ∩ A0
as long as n ≤ b sl c =: n0 . Finally, there are
a ∈ B(y, s) ∩ B(x, r) ∩ B(a0n0 , l) ∩ A
and
a0 ∈ B(y, s) ∩ B(x, r) ∩ B(a, l) ∩ A0
as desired.
We now construct recursively two converging sequences (an )n∈N ⊂ A and
0
(an )n∈N ⊂ A0 such that an , a0n ∈ B(x, r) ∩ B(y, s) with
d(an , a0n ) ≤
1
2n+1
and d(an−1 , an ), d(a0n−1 , a0n ) ≤
1
.
2n
First, choose a0 , a00 ∈ B(x, r) ∩ B(y, s) with d(a0 , a00 ) ≤ 21 according to the
claim. Given an−1 , a0n−1 with d(an−1 , a0n−1 ) ≤ 21n , by hyperconvexity, there is
1
1
1
some xn ∈ B(an−1 , 2n+1
) ∩ B(a0n−1 , 2n+1
) ∩ B(x, r − 2n+1
). Now applying the
claim to xn and y, we find
1
an , a0n ∈ B(y, s) ∩ B(xn , 2n+1
) ⊂ B(y, s) ∩ B(x, r)
with d(an , a0n ) ≤
1
.
2n+1
Moreover, we have
d(an−1 , an ) ≤ d(an−1 , xn ) + d(xn , an ) ≤
1
1
1
+
= n.
2n+1 2n+1
2
For m ≥ n, we get
d(an , am ) ≤
m
X
d(ak−1 , ak ) ≤
k=n+1
∞
X
k=n+1
1
1
= n,
2
2k
and similarly for a0n . Hence, the two sequences converge and, since d(an , a0n ) →
0, they have a common limit point a ∈ B(y, s) ∩ B(x, r) ∩ A ∩ A0 .
10
II.2. INTERSECTION PROPERTY
Lemma 2.3. Let X be a hyperconvex metric space and let A0 , A1 , A2 ∈ E(X)
be pairwise intersecting, externally hyperconvex subsets. Then we get
A0 ∩ A1 ∩ A2 6= ∅.
Proof. Choose some point x0 ∈ A1 ∩ A2 and let r := d(x0 , A0 ). By Lemma 2.2,
there is y0 ∈ A0 ∩ A1 ∩ B(x0 , r). Define A00 := A0 ∩ B(y0 , r) ∈ E(X). Using
again the lemma, we have A00 ∩ A2 = A0 ∩ A2 ∩ B(y0 , r) 6= ∅ and therefore, there
is some
z0 ∈ A00 ∩ A2 ∩ B(x0 , r) = A0 ∩ A2 ∩ B(x0 , r) ∩ B(y0 , r).
Then, since A0 is externally hyperconvex, there is some
x̄0 ∈ B(x0 , r) ∩ B(y0 , 2r ) ∩ B(z0 , 2r ) ∩ A0
and using again Lemma 2.2, we find
x1 ∈ A1 ∩ A2 ∩ B(x̄0 , 2r ) ∩ B(x0 , 2r ).
r
Proceeding this way, we get some sequence (xn )P
n ⊂ A1 ∩A2 with d(xn , A0 ) ≤ 2n
m
r
r
and d(xn−1 , xn ) ≤ 2rn . Hence, d(xn , xm ) ≤
k=n+1 2k ≤ 2n and therefore,
(xn )n converges to x ∈ A0 ∩ A1 ∩ A2 6= ∅.
Lemma 2.4. Let X be a hyperconvex metric space. If A0 , A1 ∈ E(X) and
A0 ∩ A1 6= ∅, then it holds A0 ∩ A1 ∈ E(X).
Proof. Let {B(xi , ri )}i∈I be a collection
T of closed balls with d(xi , xj ) ≤ ri + rj
and d(xi , A0 ∩ A1 ) ≤ ri . Define A := i∈I B(xi , ri ). Since, for k = 0, 1, the set
Ak is externally hyperconvex, we have
\
A ∩ Ak =
B(xi , ri ) ∩ Ak 6= ∅,
i∈I
and since admissible sets are externally hyperconvex, we have
\
(A0 ∩ A1 ) ∩
B(xi , ri ) = A0 ∩ A1 ∩ A 6= ∅
i∈I
by Lemma 2.3.
By induction, we therefore get the following proposition.
Proposition 2.5. Let X be a hyperconvex
Tn metric space and A0 , . . . , An ∈ E(X)
with Ai ∩ Aj 6= ∅. Then we have ∅ =
6
k=0 Ak ∈ E(X).
As a consequence of Baillon’s theorem on the intersection of hyperconvex
spaces [Bai88], the following theorem was proven by Espı́nola and Khamsi in
[EK01].
Theorem 2.6. [EK01, Theorem 5.4]. Let {Ai }i∈I be a descending chain of
non-emptyTexternally hyperconvex subsets of a bounded hyperconvex metric space
X. Then i∈I Ai is non-empty and externally hyperconvex in X.
11
CHAPTER II. HYPERCONVEX METRIC SPACES
Similarly to Corollary 8 in [Bai88], we can deduce the following corollary
which implies Proposition 2.1.
Corollary 2.7. Let {Ai }i∈I be a family of pairwise intersecting externally
hyT
perconvex subsets of a bounded hyperconvex metric space X. Then i∈I Ai is
non-empty and externally hyperconvex in X.
Proof. Consider the set
(
F :=
)
J ⊂ I : ∀F ⊂ I finite,
\
Ai 6= ∅ is externally hyperconvex .
i∈J∪F
By Proposition 2.5, clearly ∅ ∈ F. Considering
a chain (Jk )k∈N ∈ F and
T
some finite set F ⊂ I, the sets AJk := i∈Jk ∪F Ai build aSdescending chain
of non-empty
subsets. Define J := k∈N Jk . We have
T externally hyperconvex
T
that A := i∈J∪F Ai = k∈N AJk is non-empty and externally hyperconvex
by Theorem 2.6. Therefore, J ∈ F is an upper bound for (Jk )k∈N . Hence, F
satisfies the hypothesis of Zorn’s Lemma and therefore there is some maximal
element J0 ∈ F. But for i ∈ I, we have J0 ∪ {i} ∈ F and by maximality of J0 ,
we conclude that I = J0 ∈ F.
A first application of Proposition 2.1 yields the following.
Proposition 2.8. Let Y be an externally hyperconvex subset of the metric space
X. Moreover, let A be externally hyperconvex in Y . Then A is also externally
hyperconvex in X.
Proof. Let {B(xi , ri )}i∈I be a collection of closed balls with d(xi , xj ) ≤ ri + rj
and d(xi , A) ≤ ri . Then the sets Ai := B(xi , ri ) ∩ Y are externally hyperconvex
subsets of X and therefore also of Y . We have
Ai ∩ A =
\
B(xi , ri + n1 ) ∩ A 6= ∅
n∈N
and, since Y is externally hyperconvex, we also get
Ai ∩ Aj = B(xi , ri ) ∩ B(xj , rj ) ∩ Y 6= ∅.
Therefore, this is a collection of pairwise intersecting externally hyperconvex
subsets of Y and hence, by Proposition 2.1, it follows that
A∩
\
B(xi , ri ) = A ∩
i∈I
\
i∈I
12
Ai 6= ∅.
II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
II.3
Weakly Externally Hyperconvex Subsets
Let us now have a look at weakly externally hyperconvex subsets . We can
characterize them by the following properties.
Lemma 3.1. Let A be a subset of the hyperconvex metric space X. Then
A is weakly externally hyperconvex in X if and only if for every x ∈ X and
s := d(x, A), the following hold:
(i) The intersection B(x, s) ∩ A is externally hyperconvex in A, and
(ii) for every y ∈ A, there is some a ∈ B(x, s) ∩ A such that
d(x, y) = d(x, a) + d(a, y).
Proof. If A is weakly externally hyperconvex, (i) clearly holds. Moreover, for
y ∈ A, it also follows by weak external hyperconvexity of A that there is some
a ∈ A ∩ B(x, s) ∩ B(y, d(x, y) − s) 6= ∅
and therefore
d(x, y) ≤ d(x, a) + d(a, y) ≤ s + d(x, y) − s = d(x, y).
For
A must be hyperconvex. Indeed, for
T the converse, first observe that
0
x ∈ i∈I B(xi , ri ) with xi ∈ A, let A := B(x, d(x, A)) ∩ A ∈ E(A). Then, we
have d(xi , A0 ) ≤ ri by (ii) and therefore, there is also
\
\
x 0 ∈ A0 ∩
B(xi , ri ) ⊂ A ∩
B(xi , ri )
i∈I
i∈I
by property (i).
Now fix x ∈ X, r ≥ s = d(x, A) and let {B(xi , ri )}i∈I be a family of closed
balls with d(xi , xj ) ≤ ri + rj and d(x, xi ) ≤ r + ri . Then, by (ii), we have
B(x, r) ∩ A = B A (B(x, s) ∩ A, r − s)
and therefore B(x, r) ∩ A is externally hyperconvex in A by (i) and Lemma 1.4.
Moreover, by (ii), for xi ∈ A with d(x, xi ) ≤ r + ri , we have
d(xi , B(x, r) ∩ A) ≤ ri
and therefore
B(x, r) ∩
\
B(xi , ri ) ∩ A 6= ∅
i∈I
since B(x, r) ∩ A is externally hyperconvex in A.
Lemma 3.2. Let X be a metric space, A ∈ W(X) and s ≥ 0. Then there
is an s-constant retraction ρ : B(A, s) → A, i.e. d(ρ(x), ρ(y)) ≤ d(x, y) and
d(x, ρ(x)) ≤ s for all x, y ∈ B(A, s).
13
CHAPTER II. HYPERCONVEX METRIC SPACES
Proof. Consider the partially ordered set
F := {(B, ρ) : B ⊂ B(A, s) and ρ : B → A is an s-constant retraction}.
By Zorn’s Lemma, there is some maximal element (B̃, ρ̃) ∈ F. Assume that
there is some x ∈ B(A, s) \ B̃. For all y ∈ B̃, define ry = d(x, y). Then, we
have
d(x, ρ(y)) ≤ d(x, y) + d(y, ρ(y)) ≤ ry + s
and therefore, since A is weakly externally hyperconvex, there is some
\
z ∈ B(x, s) ∩
B(ρ(y), ry ) ∩ A.
y∈B̃
But then, defining ρ̃(x) := z, we can extend ρ̃ to B̃ ∪ {x} contradicting maximality of (B̃, ρ̃). Hence, we conclude B̃ = B(A, s).
This and the first part of the following result can also be found in [EKL00,
Esp05], where a complete characterization of weakly externally hyperconvex
subsets in terms of retractions is given.
Lemma 3.3. Let A be a weakly externally hyperconvex subset of a hyperconvex
metric space X. Then, for any s ≥ 0, the closed neighborhood B(A, s) is weakly
externally hyperconvex in X. Moreover, if for all x ∈ X we have
B(x, d(x, A)) ∩ A ∈ E(X),
then this also holds for B(A, s), i.e.
B(x, d(x, B(A, s))) ∩ B(A, s) ∈ E(X).
Proof. Let x ∈ X and {xi }i∈I ⊂ B(A, s) with d(xi , xj ) ≤ ri +rj , d(x, xi ) ≤ r+ri
and d(x, B(A, s)) ≤ r. By Lemma 3.2, there is a retraction ρ : B(A, s) → A such
that d(y, ρ(y)) ≤ s. Then, we have d(ρ(xi ), x) ≤ d(ρ(xi ), xi )+d(xi , x) ≤ s+r+ri
and therefore, there is some
\
y ∈ B(x, s + r) ∩
B(ρ(xi ), ri ) ∩ A.
i∈I
As d(x, y) ≤ r + s and d(xi , y) ≤ d(xi , ρ(xi )) + d(ρ(xi ), y) ≤ s + ri , by hyperconvexity of X, there is some
\
\
z ∈ B(x, r) ∩
B(xi , ri ) ∩ B(y, s) ⊂ B(x, r) ∩
B(xi , ri ) ∩ B(A, s)
i∈I
i∈I
as required.
Now, define r := d(x, B(A, s)). We claim that
B(x, r) ∩ B(A, s) = B(x, r) ∩ B(B(x, d(x, A)) ∩ A, s)
and therefore, if B(x, d(x, A)) ∩ A ∈ E(X), we also have B(x, r) ∩ B(A, s) ∈
E(X).
Indeed, if y ∈ B(x, r)∩B(A, s), there is some a ∈ A such that d(y, a) = s and
hence d(x, a) ≤ d(x, y)+d(y, a) ≤ r+s = d(x, A), i.e. a ∈ B(x, d(x, A))∩A.
14
II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
Example 3.4. In general, it is not true that the neighborhood of a hyperconvex
3 of
subset is hyperconvex as well. Consider the isometric embedding ι : R → l∞
the real line given by


t ≤ 0,
(t, t, −t),
ι(t) = (t, t, t),
0 ≤ t ≤ 10,


(20 − t, t, t) t ≥ 10,
3 ). Then the two points x = (−2, 0, 2), y =
and define A := ι(R) ∈ H(l∞
(8, 10, 12) are contained in B(A, 1) since d(x, ι(−1)) = d(y, ι(11)) = 1. Now
look at the intersection B(x, 5) ∩ B(y, 5) = {z = (3, 5, 7)}. But d(z, A) =
d(z, ι(5)) = 2 and therefore B(x, 5) ∩ B(y, 5) ∩ B(A, 1) = ∅, i.e. B(A, 1) is not
hyperconvex, even not geodesic.
Lemma 3.5. Let X be a hyperconvex metric space and let A = I(x, y) be a
metric interval. Then A ∈ W(X) and, for all z ∈ X, we have B(z, d(z, A))∩A ∈
E(X).
Proof. Fix some z ∈ X. First, observe that
B(z, d(z, A)) ∩ A = B(x, rx ) ∩ B(y, ry ) ∩ B(z, rz )
for rx := (y|z)x , ry := (x|z)y and rz := (x|y)z , where
(y|z)x := 12 (d(x, y) + d(x, z) − d(y, z))
denotes the Gromov product. Hence, B(z, d(z, A)) ∩ A ∈ E(X) since it is admissible.
Now, let p ∈ A and t := d(p, z) − rz . Then we have
d(x, p) = d(x, y) − d(y, p) ≤ d(x, y) − d(y, z) + d(p, z)
= rx + ry − ry − rz + rz + t = rx + t
and similarly d(y, p) ≤ ry + t. Therefore, since X is hyperconvex, there is some
p̄ ∈ B(x, rx ) ∩ B(y, ry ) ∩ B(z, rz ) ∩ B(p, t) 6= ∅,
that is p̄ ∈ B(z, d(z, A))∩A with d(p, z) = d(p, p̄)+d(p̄, z). Thus, by Lemma 3.1,
we get A ∈ W(X).
Although we cannot expect a similar result for weakly externally hyperconvex subsets as in Proposition 2.1, it is possible to say more about the intersection
of weakly externally hyperconvex subsets.
Example 3.6. The intersection property as we stated for externally hyperconvex subsets in Proposition 2.1 does not hold for weakly externally hyperconvex
subsets. Indeed, consider the points z := (1, 1), w := (−1, −1) and the half2 : y − y ≥ 2} in l2 . Clearly, B(z, 1), B(w, 1) and H are
space H := {y ∈ l∞
1
2
∞
2 ) and they are pairwise intersecting. However,
all three elements of W(l∞
B(z, 1) ∩ B(w, 1) ∩ H = ∅.
15
CHAPTER II. HYPERCONVEX METRIC SPACES
Lemma 3.7. Let X be a metric space, Y ∈ W(X) and A ∈ E(Y ). Then we
have A ∈ W(X).
Proof. Let x ∈ X, r ≥ d(x, A) and let {B(xi , ri )}i∈I be a collection of closed
balls with xi ∈ A, d(xi , xj ) ≤ ri + rj and d(xi , x) ≤ ri + r. Then the sets
B := B(x, r + ) ∩ Y , Ai := B(xi , ri ) ∩ Y and A are pairwise intersecting,
externally hyperconvex subsets of Y and therefore, by Proposition 2.1, we have
\
\
\
B ∩ A ∩
Ai 6= ∅.
B(x, r) ∩ A ∩
B(xi , ri ) =
>0
i∈I
i∈I
Lemma 3.8. Let X be a metric space and A ∈ E(X), Y ∈ W(X) such that
A ∩ Y 6= ∅. Let {xi }i∈I ⊂ Y be a collection of points with d(x
T i , xj ) ≤ ri + rj
and d(xiT
, A) ≤ ri . Then, for any s > 0, there are a ∈ A ∩ i∈I B(xi , ri ) and
y ∈ Y ∩ i∈I B(xi , ri ) with d(a, y) ≤ s.
T
Proof. Let y0 ∈ A∩Y and d := d(y0 , i∈I B(xi , ri )). Without loss of generality,
we may assume that s ≤ d. Then, since A ∈ E(X) and Y ∈ W(X), there are
a1 ∈
\
B(xi , ri + d − s) ∩ B(y0 , s) ∩ A,
i∈I
y1 ∈
\
B(xi , ri + d − s) ∩ B(a1 , s) ∩ Y.
i∈I
Proceeding this way, we can choose inductively
\
an ∈
B(xi , ri + d − ns) ∩ B(yn−1 , s) ∩ A,
i∈I
yn ∈
\
B(xi , ri + d − ns) ∩ B(an , s) ∩ Y,
i∈I
for n ≤ b ds c =: n0 and finally, there are
a∈
\
B(xi , ri ) ∩ B(yn0 , s) ∩ A,
i∈I
y∈
\
B(xi , ri ) ∩ B(a, s) ∩ Y,
i∈I
as desired.
Proposition 3.9. Let X be a metric space and let A ∈ E(X), Y ∈ W(X) such
that A ∩ Y 6= ∅. Then we have A ∩ Y ∈ E(Y ), and therefore A ∩ Y ∈ W(X).
Proof. Let {xi }i∈I ⊂ Y with d(xi , xj ) ≤ ri + rj and d(xi , A)
T ≤ ri . We now
construct inductively
two
converging
sequences
(a
)
⊂
n n∈N
i∈I B(xi , ri ) ∩ A
T
1
and (yn )n∈N ⊂ i∈I B(xi , ri ) ∩ Y with d(an , yn ) ≤ 2n+1 as follows:
16
II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
By Lemma 3.8, we may choose
\
a0 ∈
B(xi , ri ) ∩ A,
i∈I
y0 ∈
\
B(xi , ri ) ∩ Y
i∈I
with d(a0 , y0 ) ≤ 12 . Assume now, that we have
an ∈
\
B(xi , ri ) ∩ A,
i∈I
yn ∈
\
B(xi , ri ) ∩ Y
i∈I
1
. Then, applying Lemma 3.8 for the collection of balls
with d(an , yn ) ≤ 2n+1
1
{B(xi , ri )}i∈I ∪ {B(yn , 2n+1
)}, we find
an+1 ∈
\
1
B(xi , ri ) ∩ B(yn , 2n+1
) ∩ A,
i∈I
yn+1 ∈
\
1
)∩Y
B(xi , ri ) ∩ B(yn , 2n+1
i∈I
with d(an+1 , yn+1 ) ≤
1
.
2n+2
Especially, we have d(yn+1 , yn ) ≤
d(an+1 , an ) ≤ d(an+1 , yn ) + d(yn , an ) ≤
1
2n+1
+
1
2n+1
1
2n+1
=
and
1
2n .
Hence the two sequencesTare Cauchy and therefore converge to some common
limit point z ∈ A ∩ Y ∩ i∈I B(xi , ri ) 6= ∅. Finally, we have A ∩ Y ∈ W(X) by
Lemma 3.7.
Looking at the proof carefully, we see that only d(xi , A) ≤ ri is assumed.
Therefore, we may deduce the following corollary:
Corollary 3.10. Let X be a metric space and A ∈ E(X), Y ∈ W(X) such that
A ∩ Y 6= ∅. Then, for all x ∈ Y , we have d(x, A) = d(x, A ∩ Y ).
Corollary 3.11. Let X be a metric space and A ∈ W(Y ) for Y ∈ W(X).
Then we have A ∈ W(X).
Proof. Let x ∈ X with d(x, A) ≤ r and {xi }i∈I ⊂ A with d(xi , xj ) ≤ ri + rj ,
d(xi , x) ≤ ri + r. Then we have
B(x, r) ∩ Y ∈ E(Y ) and B(x, r) ∩ A = (B(x, r) ∩ Y ) ∩ A ∈ E(A)
by Proposition 3.9. Moreover, by applying Corollary 3.10 twice, we have
d(xi , (B(x, r) ∩ Y ) ∩ A) = d(xi , B(x, r) ∩ Y ) = d(xi , B(x, r)) ≤ ri
and hence
T
i∈I
B(xi , ri ) ∩ B(x, r) ∩ A 6= ∅.
17
CHAPTER II. HYPERCONVEX METRIC SPACES
Corollary 3.12. Let X be a hyperconvex metric space and let A ∈ E(X),
Y ∈ W(X). Then there are a ∈ A, y ∈ Y with d(a, y) = d(A, Y ).
Proof. Let s := d(A, Y ). For any n ∈ N, we have B(A, s + 21n ) ∈ E(X) and
Clearly, we have An ∩Am 6= ∅
An := B(A, s+ 21n )∩Y ∈ E(Y ) by Proposition 3.9. T
and therefore, there is some y ∈ Y ∩ B(A, s) = n∈N An by Proposition 2.1.
Now, since A is proximinal, there is some a ∈ A with d(a, y) ≤ s = d(A, Y ) as
required.
The following proposition answers an open question on the intersection of
weakly externally hyperconvex sets stated in [EK01] for proper metric spaces,
i.e. for spaces where all closed balls are compact.
Proposition 3.13. Let X be a proper hyperconvex metric space and let Y and
Y 0 be two weakly externally hyperconvex subsets with non-empty intersection.
Then we have Y ∩ Y 0 ∈ W(X).
Proof. By Corollary 3.11, it is enough to show that Y ∩ Y 0 ∈ W(Y ). Therefore,
let {B(xi , ri )}i∈I be a collection of balls with xi ∈ Y ∩ Y 0 and d(xi , xj ) ≤ ri + rj
and x ∈ Y with d(x, Y ∩ Y 0 ) ≤ r and d(xi , x) ≤ ri + r. Furthermore, let
0
0
s > 0. Since
T d(x, Y ∩ Y ) ≤ r, there is some y0 ∈ Y ∩ Y ∩ B(x, r + s). Define
d := d(y0 , i∈I B(xi , ri )). Then there is some
y00 ∈ B(y0 , s) ∩ Y 0 ∩ B(x, r) ∩
\
B(xi , ri + d).
i∈I
Now, for n ≤ b ds c =: n0 , we can choose inductively
0
yn ∈ B(yn−1
, s) ∩ Y ∩
\
B(xi , ri + d − ns),
i∈I
yn0 ∈ B(yn , s) ∩ Y 0 ∩ B(x, r) ∩
\
B(xi , ri + d − ns).
i∈I
Finally, there are
y ∈ B(yn0 0 , s) ∩ Y ∩
\
B(xi , ri ),
i∈I
y 0 ∈ B(y, s) ∩ Y 0 ∩ B(x, r) ∩
\
B(xi , ri )
i∈I
T
T
and hence d(YT∩ i B(xi , ri ), Y 0 T
∩ i B(xi , ri ) ∩ B(x, r)) ≤ d(y, y 0 ) ≤ s, i.e.
we get d(Y ∩ i∈I B(xi , ri ), Y 0 ∩ i∈I B(xi , ri ) ∩ B(x, r)) = 0, and since X is
proper, both sets are compact and therefore their intersection
\
Y ∩Y0∩
B(xi , ri ) ∩ B(x, r)
i∈I
is non-empty.
18
II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
n are stable
Note that not even linearly convex hyperconvex subsets of l∞
under non-empty intersections, as observed in [Pav16, Example 1.4].
Proposition 3.14. Let X be a metric
S space and let (Yn )n∈N ⊂ W(X) be an
increasing sequence such that Y := n∈N Yn is proper. Then we have Y ∈
W(X).
Proof. Consider a family {B(xi , ri )}i∈I of closed balls with xi ∈ Y such that
d(xi , xj ) ≤ ri +rj . Moreover, let B(x, r) be a closed ball with x ∈ X, d(x, Y ) ≤ r
and d(x, xi ) ≤ r + ri .
There is a decreasing sequence sn ↓ 0 such that d(x, Yn ) ≤ r +
Ssn . Now fix
1
:= m
> 0. Then, for every i ∈ I, there is some yi ∈ B(xi , ) ∩ n∈N Yn . For
n ∈ N, let
In := {i ∈ I : yi ∈ Yn }
and since Yn is weakly externally hyperconvex, there is some
\
zn ∈ Yn ∩ B(x, r + sn ) ∩
B(yi , ri + ).
i∈In
Since Y is proper and (zn )n∈N ⊂ Y ∩ B(x, r + sT
0 ), it follows that there is a
1
convergent subsequence znk → z m ∈ Y ∩B(x, r)∩ i∈I B(xi , ri + m
). Moreover,
T
m
l
since Y is proper, there is a subsequence z → z ∈ Y ∩ B(x, r) ∩ i∈I B(xi , ri ).
This proves that Y is weakly externally hyperconvex in X.
It turns out that a subset of a hyperconvex metric space is already (weakly)
externally hyperconvex if it is (weakly) externally hyperconvex in a uniform
neighborhood. We say that a subset A of a metric space X is proximinal if,
for any x ∈ X, we have B(x, d(x, A)) ∩ A 6= ∅. Note that weakly externally
hyperconvex subsets are proximinal.
Lemma 3.15. Let X be a hyperconvex metric space and A ∈ W(B(A, r)).
Then A is proximinal in B(A, 2r).
Proof. Let x ∈ B(A, 2r) with d(x, A) = r + t. Then for every > 0, there is
some x ∈ B(A, r) with d(x , x) ≤ t + . Since A ∈ W(B(A, r)), there is a
proximinal non-expansive retraction ρ : B(A, r) → A by Proposition 4.1. We
have d(x, ρ(x )) ≤ r + d(x, x ) and hence, there is some
y ∈ B(x, r) ∩
\
B(ρ(x ), d(x, x )).
>0
In particular, we have d(y, A) ≤ d(y, ρ(x )) ≤ t + for all > 0 and therefore
d(y, A) = t ≤ r. Hence, since A is proximinal in B(A, r), there is some z ∈ A
with d(z, y) = t. It follows that
d(x, z) ≤ d(x, y) + d(y, z) ≤ r + t = d(x, A)
as desired.
19
CHAPTER II. HYPERCONVEX METRIC SPACES
Lemma 3.16. Let X be a hyperconvex metric space and A ⊂ X. Assume that
there is some s > 0 such that A ∈ W(B(A, s)). Then we have A ∈ W(X).
Proof. We show that A ∈ W(B(A, r)) implies A ∈ W(B(A, 2r)). It then follows
that A ∈ W(B(A, r)) for any r > 0 and thus A ∈ W(X).
By Corollary 2.1 in [EKL00], it is enough to show that there is a proximinal
non-expansive retraction R : A ∪ {x0 } → A for any x0 ∈ B(x, 2r). For some
given x0 ∈ B(x, 2r) \ B(A, r), let t := d(x0 , A) − r ≤ r. By Lemma 3.15, A is
proximinal in B(A, 2r) and hence, there is some x̄0 ∈ B(A, r) with d(x0 , x̄0 ) = t.
Then, for all x ∈ A, define rx := d(x, x0 ). We have
d(x, x̄0 ) ≤ d(x, x0 ) + d(x0 , x̄0 ) ≤ rx + r.
Hence, since A ∈ W(B(A, r)), there is some
\
y∈
B(x, rx ) ∩ B(x̄0 , r) ∩ A.
x∈A
Especially, we get d(x0 , y) ≤ d(x0 , x̄0 )+d(x̄0 , y) ≤ t+r = d(x0 , A) and therefore,
we can define a proximinal non-expansive retraction R : A ∪ {x0 } → A by
R(x0 ) := y and R(x) := x for x ∈ A.
Lemma 3.17. Let X be a hyperconvex metric space and A ⊂ X. Assume that
there is some s > 0 such that A ∈ E(B(A, s)). Then we have A ∈ E(X).
Proof. By Lemma 3.16, we already have A ∈ W(X). Let us first show that
A ∈ E(B(A, r)) for any r ≥ 0. By assumption, this holds for r = s > 0.
Therefore, it is enough to prove A ∈ E(B(A, r)) ⇒ A ∈ E(B(A, 2r)).
Let {B(xi , ri )}i∈I be a family of closed balls with xi ∈ B(A, 2r), such that
d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri . Define
Ai := B(xi , ri ) ∩ B(A, r) ∈ E(B(A, r)).
Clearly A ∩ Ai = A ∩ B(xi , ri ) 6= ∅. By Lemma 3.3, B(A, r) ∈ W(X) and hence,
by Lemma 3.2, there is a retraction ρ : B(A, 2r) → B(A, r) with
T d(ρ(x), x) ≤ r.
Set yi := ρ(xi ). Since A ∈ E(B(A, r)), there is some z ∈ A ∩ i∈I B(yi , ri ) with
d(xi , z) ≤ ri + r. Therefore, since X is hyperconvex, we get
\
∅=
6 B(z, r) ∩
B(xi , ri ) ⊂ B(A, r).
i∈I
Especially,
Ai ∩ Aj = B(xi , ri ) ∩ B(xj , rj ) ∩ B(A, r) 6= ∅
and hence,
A∩
\
B(xi , ri ) = A ∩
i∈I
\
Ai 6= ∅
i∈I
by Proposition 2.1.
To deduce that A ∈ E(X), let now {B(xi , ri )}i∈I be a family of closed balls in
X with d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri . Define Ai := B(xi , ri ) ∩ A ∈ E(A).
20
II.4. RETRACTS
For fixed i, j ∈ I, we have xi , xj ∈ B(A, r) for some r ≥ 0 and hence, by the
first step, we have
Ai ∩ Aj = A ∩ B(xi , ri ) ∩ B(xj , rj ) 6= ∅.
Therefore, we get
A∩
\
B(xi , ri ) =
i∈I
\
Ai 6= ∅
i∈I
by Proposition 2.1 as before.
To conclude this section, we give some properties for products of hyperconvex metric spaces.
Lemma 3.18. Let X = X 1 ×∞ X 2 be the product of two metric spaces with
d(x, y) = maxλ=1,2 dλ (xλ , y λ ). Moreover, let A = A1 × A2 be a subset of X.
Then, the following properties hold:
(i) B(x, r) = B 1 (x1 , r) × B 2 (x2 , r).
(ii) A ∈ E(X) if and only if Aλ ∈ E(X λ ) for each λ ∈ {1, 2}.
(iii) A ∈ W(X) if and only if Aλ ∈ W(X λ ) for each λ ∈ {1, 2}.
(iv) X is hyperconvex if and only if X λ is hyperconvex for each λ ∈ {1, 2}.
Proof. First, property (i) follows from the fact that d(x, y) ≤ r if and only if
dλ (xλ , y λ ) ≤ r for λ = 1, 2.
For (ii), let first {xi }i∈I ⊂ X be any collection of points with the property
d(xi , xj ) ≤ ri +rj and d(xi , A) ≤ ri . Then we have d(xλi , xλj ) ≤ d(xi , xj ) ≤ ri +rj
λ
and d(xλi , A
if Aλ ∈ TE(X λ ), there is some
T ) ≤ λd(xλi , A) ≤ ri and therefore,
1
2
λ
λ
y ∈ A ∩ i∈I B (xi , ri ) and hence y = (y , y ) ∈ A ∩ i∈I B(xi , ri ).
For the converse, if d(x1i , x1j ) ≤ ri + rj and d(x1i , A1 ) ≤ ri , fix some x2 ∈ A2 .
Then the points xi = (x1i , x2 ) fulfill d(xTi , xj ) ≤ ri + rj and d(xi , A) ≤ ri ,
i.e.
is some y = (y 1 , y 2 ) ∈ A ∩ i∈I B(xi , ri ) and hence y 1 ∈ A1 ∩
T there
1
1
i∈I B (xi , ri ) 6= ∅.
The proof of (iii) is similar and (iv) follows from (ii) by setting Aλ = X λ .
II.4
Retracts
Weakly externally hyperconvex subsets were recognized as the proximinal 1-Lipschitz retracts by Espı́nola in [Esp05].
Proposition 4.1. [Esp05, Theorem 3.6] Let X be a hyperconvex metric space
and let A ⊂ X be non-empty. Then A is a proximinal 1-Lipschitz retract of X
if and only if A is a weakly externally hyperconvex subset of X.
Similarly, we can characterize externally hyperconvex subsets as 1-Lipschitz
retracts with some further properties.
21
CHAPTER II. HYPERCONVEX METRIC SPACES
Proposition 4.2. Let X be a hyperconvex metric space and A ⊂ X a subset.
Then A is externally hyperconvex in X if and only if there is a proximinal
1-Lipschitz retraction ρ : X → A with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all
x, y ∈ X.
Proof. First, assume that there is a proximinal 1-Lipschitz retraction ρ : X →
A, with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all x, y ∈ X.
Let {B(xi , ri )}i∈I be a family of closed balls in X with d(xi , xjT
) ≤ ri +rj and
d(xi , A) ≤ ri . Then, by hyperconvexity of X, there is some z ∈ i∈I B(xi , ri ).
We get d(ρ(z), xi ) ≤ max{d(z, xi ), d(A, xi )} ≤ ri and hence
\
ρ(z) ∈ A ∩
B(xi , ri ) 6= ∅.
i∈I
For the converse, consider the set
F := {(Y, ρ) : Y ⊂ X, ρ : Y → A a good retraction} ,
i.e. ρ : Y → A is a retraction with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all
x, y ∈ X. We endow F with the usual order relation (Y, ρ) 4 (Y 0 , ρ0 ) if and
only if Y ⊂ Y 0 and ρ0 |Y = ρ.
Clearly, F is non-empty, S
since (A, id) ∈ F, and for any chain (Yn , ρn )n∈N ,
the element (Y, ρ) with Y := n∈N Yn and ρ(x) := ρn (x) for x ∈ Yn is an upper
bound. Hence, there is some maximal element (Ȳ , ρ̄) ∈ F.
Assume that there is some x0 ∈ X \ Ȳ . Then for all x ∈ X, let us define
rx := max{d(x0 , x), d(A, x)} and for all y ∈ Ȳ , define sy := d(x0 , y). We have
d(x, x0 ) ≤ d(x0 , x) + d(x0 , x0 ) ≤ rx + rx0 ,
d(x, A) ≤ rx ,
d(ρ̄(y), ρ̄(y 0 )) ≤ d(y, y 0 ) ≤ sy + sy0 ,
d(ρ̄(y), x) ≤ max{d(y, x), d(A, x)} ≤ sy + rx .
Hence, since A is externally hyperconvex, there is some
\
\
z ∈A∩
B(x, rx ) ∩
B(ρ̄(y), sy ).
x∈X
y∈Ȳ
But then (Ȳ ∪ {x0 }, ρ0 ) with ρ0 (y) := ρ̄(y) for y ∈ Ȳ and ρ0 (x0 ) := z is a strictly
bigger element in F, contradicting maximality of (Ȳ , ρ̄).
Therefore, Ȳ = X and ρ̄ : X → A is the desired retraction.
22
Chapter III
Gluing Hyperconvex Metric
Spaces
III.1
Gluing along Weakly Externally Hyperconvex
Subsets
Definition 1.1. Let (Xλ , dλ )λ∈Λ be a family of metric spaces with closed subsets Aλ ⊂ Xλ . Suppose that all Aλ are isometric to some metric space A. For
every λ ∈ Λ, fix an isometry
ϕλ : A → Aλ . We define an equivalence relation
F
on the disjoint union F λ∈Λ Xλ generated by ϕλ (a) ∼ ϕλ0 (a) for a ∈ A. The
resulting space X := ( λ∈Λ Xλ )/ ∼ is called the gluing of the Xλ along A.
X admits a natural metric. For x ∈ Xλ and y ∈ Xλ0 it is given by
(
dλ (x, y),
if λ = λ0 ,
d(x, y) =
inf a∈A {dλ (x, ϕλ (a)) + dλ0 (ϕλ0 (a), y)}, if λ =
6 λ0 .
(1.1)
For more details see for instance [BH99, Lemma I.5.24].
In the following, if there is no ambiguity, indices for dλ are dropped and the
sets Aλ = ϕλ (A) ⊂ Xλ are identified with A.
Balls inside the subset Xλ are denoted by B λ (x, r).
Lemma 1.2. Let X be a hyperconvex metric space obtained by gluing a family
of hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A. Then A is hyperconvex.
Proof. Let {xi }i∈I ⊂ A such that d(x
T i , xj ) ≤ ri + rj . Then for each
T λ, since Xλ
is hyperconvex, there is some yλ ∈ i∈I B(xi , ri )∩Xλ . Moreover, i∈I B(xi , ri )
is path-connected and a path from yλ to yλ0 must intersect A, i.e.
\
B(xi , ri ) ∩ A 6= ∅.
i∈I
23
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
In general, we cannot say more about necessary conditions on A such that
the gluing along A is hyperconvex. For instance, gluing a hyperconvex space
X and any hyperconvex subset A ∈ H(X) along A, the resulting space is
isometric to X and therefore hyperconvex. But there are also plenty of nontrivial examples.
Example 1.3. Let f : R → R be any 1-Lipschitz function. Consider its graph
2 : y = f (x)} and the two sets X = {(x, y) ∈ l2 : y ≤ f (x)},
A = {(x, y) ∈ l∞
1
∞
2 : y ≥ f (x)}. Then l2 = X t X is hyperconvex and occurs
X2 = {(x, y) ∈ l∞
1 A 2
∞
as the gluing of two hyperconvex spaces X1 , X2 .
But if we assume that the gluing set is weakly externally hyperconvex, we
can do better.
Lemma 1.4. Let (X, d) be the metric space obtained by gluing a family of
hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that A is weakly
externally hyperconvex in Xλ for each λ ∈ Λ. For x ∈ Xλ and x0 ∈ Xλ0 with
λ 6= λ0 , there are then points a ∈ B(x, d(x, A)) ∩ A and a0 ∈ B(x0 , d(x0 , A)) ∩ A
such that
d(x, x0 ) = d(x, a) + d(a, a0 ) + d(a0 , x0 ).
Proof. As A is weakly externally hyperconvex in each Xλ , by Lemma II.3.1, for
every y ∈ A there are points a ∈ B(x, d(x, A)) ∩ A and a0 ∈ B(x0 , d(x0 , A)) ∩ A
such that both d(x, y) = d(x, a) + d(a, y) and d(y, x0 ) = d(y, a0 ) + d(a0 , x0 ) hold.
Hence
d(x, x0 ) = d(x, A) + d(B(x, d(x, A)) ∩ A, B(x0 , d(x0 , A)) ∩ A) + d(x0 , A).
But the sets B(x, d(x, A))∩A and B(x0 , d(x0 , A))∩A are externally hyperconvex
in A and therefore, by Corollary II.3.12, there are a ∈ B(x, d(x, A)) ∩ A and
a0 ∈ B(x0 , d(x0 , A)) ∩ A with
d(a, a0 ) = d(B(x, d(x, A)) ∩ A, B(x0 , d(x0 , A)) ∩ A).
Lemma 1.5. Let (X, d) be the metric space obtained by gluing a family of
hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that A is weakly
externally hyperconvex in Xλ for each λ ∈ Λ. Then for λ 6= λ0 , x ∈ Xλ and
r ≥ s := d(x, A), one has
0
B(x, r) ∩ Xλ0 = B λ (B λ (x, s) ∩ A, r − s).
Therefore, if B λ (x, s) ∩ A ∈ E(Xλ0 ), then we also have B(x, r) ∩ Xλ0 ∈ E(Xλ0 ).
Proof. Let x0 ∈ B(x, r) ∩ Xλ0 . By Lemma 1.4, there is some a ∈ B λ (x, s) ∩ A
with d(x, x0 ) = d(x, a) + d(a, x0 ) . We have d(a, x0 ) ≤ r − s and hence
0
x0 ∈ B λ (B λ (x, s) ∩ A, r − s).
24
III.1. WEAKLY EXTERNALLY HYPERCONVEX
Proposition 1.6. Let (X, d) be the metric space obtained by gluing a family of
hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that A is weakly
externally hyperconvex in Xλ for each λ ∈ Λ. If X is hyperconvex, then for all
λ ∈ Λ and all x ∈ X \ Xλ , the set B(x, d(x, A)) ∩ A is externally hyperconvex
in Xλ .
Proof. Set s := d(x, A) and let {xi }i∈I be a collection of point in Xλ and {ri }i∈I
such that d(xi , xj ) ≤ ri +rj and d(xi , B(x, s)∩A) ≤ ri . Then, by hyperconvexity
of X, there is some
\
y ∈ B(x, s) ∩
B(xi , ri ).
i∈I
Since y ∈ B(x, s), we have y ∈ Xλ0 for some λ0 6= λ. Therefore, by Lemma 1.4, for each i ∈ I, there is some yi ∈ B(xi , d(xi , A)) ∩ A with d(y, xi ) =
d(y, yi ) + d(yi , xi ). Define ri0 := ri − d(yi , xi ). We have
d(yi , yj ) ≤ d(yi , y) + d(y, yj ) ≤ ri0 + rj0
and d(x, yi ) ≤ s + ri0 . Hence, since A is weakly externally hyperconvex in Xλ0 ,
there is some
\
\
z∈
B(yi , ri0 ) ∩ B(x, s) ∩ A ⊂
B(xi , ri ) ∩ B(x, s) ∩ A 6= ∅.
i∈I
i∈I
Proposition 1.7. Let (X, d) be the metric space obtained by gluing a family
of hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that, for each
λ ∈ Λ, the set A is weakly externally hyperconvex in Xλ and, for all x ∈ X \Xλ ,
the intersection B(x, d(x, A)) ∩ A is externally hyperconvex in Xλ . Then X is
hyperconvex and Xλ ∈ W(X) for every λ ∈ Λ.
Proof. Let {B(xi , ri )}i∈I be a family of balls in X with d(xi , xj ) ≤ ri + rj . We
divide the proof into two cases.
Case 1. If, for every i, j ∈ I, one has
B(xi , ri ) ∩ B(xj , rj ) ∩ A 6= ∅,
setting Ci := A ∩ B(xi , ri ), we obtain that the family {Ci }i∈I is pairwise intersecting. Moreover, {Ci }i∈I is contained in E(A), sinceTA is weakly externally
hyperconvex.
By Proposition II.2.1, we obtain that i∈I Ci 6= ∅, and hence
T
i∈I B(xi , ri ) 6= ∅.
Case 2. Otherwise, there are i0 , j0 ∈ I with xi0 , xj0 ∈ Xλ0 such that
B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ A = ∅.
Indeed, either there is some i0 ∈ I such that d(xi0 , A) > ri0 and we may take
i0 = j0 , or if
B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ A = ∅
25
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
with d(xi0 , A) ≤ ri0 and d(xj0 , A) ≤ rj0 , we get xi0 , xj0 ∈ Xλ0 by Lemma 1.4.
Observe that, in both cases, we may assume that, for xi ∈ Xλ 6= Xλ0 , we have
d(xi , A) ≤ ri .
Define Aλi 0 := B(xi , ri ) ∩ Xλ0 . The goal is now to show the following claim:
Claim. For every i, j ∈ I, one has Aλi 0 ∩ Aλj 0 6= ∅.
Then, by Lemma 1.5, we have Aλi 0 ∈ E(Xλ0 ) and, by Proposition II.2.1, we
get
\
B(xi , ri ) ∩ Xλ0 =
\
Aλi 0 6= ∅.
To prove the claim, consider first the following two easy cases.
• If xi , xj ∈ Xλ0 , then we are done by hyperconvexity of Xλ0 .
• If xi ∈ Xλ 6= Xλ0 3 xj , we have B(xi , ri )∩B(xj , rj )∩A 6= ∅ by Lemma 1.4
and we are done.
The remaining case is when xi , xj ∈ Xλ 6= Xλ0 . We do this in two steps.
Step I. Set
A0 := B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) = B λ0 (xi0 , ri0 ) ∩ B λ0 (xj0 , rj0 )
and s := d(A, A0 ). By Corollary II.3.12, we have B(A0 , s)∩A 6= ∅. Furthermore,
we get
B(A0 , s) = B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s)
and hence B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ⊂ Xλ0 .
To see this, observe first that by Lemma II.1.3, we have
B(A0 , s) = B λ0 (xi0 , ri0 + s) ∩ B λ0 (xj0 , rj0 + s) ⊂ Xλ0
and therefore
(B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s)) \ B(A0 , s) ⊂ X \ Xλ0 .
Thus, assume that there is some
y ∈ B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ (Xλ \ Xλ0 ) .
Now since
B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ A 6= ∅
and
B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ
is externally hyperconvex in Xλ and thus path-connected, there is some
y 0 ∈ B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ \ Xλ0
26
III.1. WEAKLY EXTERNALLY HYPERCONVEX
with d(y 0 , A) ≤ s. But then by Lemma 1.4 we have
B(xi0 , ri0 ) ∩ B(y 0 , s) ∩ Xλ0 6= ∅,
B(xj0 , rj0 ) ∩ B(y 0 , s) ∩ Xλ0 6= ∅
and therefore
B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ B(y 0 , s) ∩ Xλ0 6= ∅,
i.e. y 0 ∈ B(A0 , s) contradicting y 0 ∈
/ Xλ0 .
Step II. We now show that the family
F := {B(xi0 , ri0 + s) ∩ Xλ , B(xj0 , rj0 + s) ∩ Xλ , B λ (xi , ri ), B λ (xj , rj )}
is pairwise intersecting. We already observed that
(B(xi0 , ri0 + s) ∩ Xλ ) ∩ (B(xj0 , rj0 + s) ∩ Xλ ) 6= ∅.
Further, since xi0 ∈ Xλ0 6= Xλ 3 xi , by Lemma 1.4 one has
(B(xi0 , ri0 + s) ∩ Xλ ) ∩ B λ (xi , ri ) 6= ∅
and similarly for (i0 , i) replaced by (i0 , j) as well as by (j0 , i) and (j0 , j). Finally,
B λ (xi , ri ) ∩ B λ (xj , rj ) ∩ Xλ 6= ∅
by hyperconvexity of Xλ . Hence, we have shown that F is pairwise intersecting.
Since F ⊂ E(Xλ ), it follows by Proposition II.2.1 that
C := B λ (xi0 , ri0 + s) ∩ B λ (xj0 , rj0 + s) ∩ B λ (xi , ri ) ∩ B λ (xi , ri ) 6= ∅.
Since B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ ⊂ A, we have in particular C ⊂ A.
Hence
B λ (xi , ri ) ∩ B λ (xj , rj ) ∩ A ⊃ C ∩ A = C 6= ∅,
and this is the desired result.
To see that Xλ is weakly externally hyperconvex in X, use that for x ∈ X,
r ≥ d(x, Xλ ) and {xi }i∈I ∈ Xλ with d(x, xi ) ≤ r + ri , d(xi , xj ) ≤ ri + rj , we
have B(x, r) ∩ B(xi , ri ) ∩ Xλ 6= ∅ by Lemma 1.4 and therefore the set
{B(x, r) ∩ Xλ } ∪ {B λ (xi , ri )}i∈I
is a family of pairwise intersecting externally hyperconvex subsets of Xλ and
hence
\
B(xi , ri ) ∩ B(x, r) ∩ Xλ 6= ∅
i∈I
by Proposition II.2.1.
27
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
Combining Proposition 1.6 and Proposition 1.7, we get the following.
Theorem 1.8. Let (X, d) be the metric space obtained by gluing a family of
hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that, for each
λ ∈ Λ, A is weakly externally hyperconvex in Xλ . Then X is hyperconvex if and
only if for all λ ∈ Λ and all x ∈ X \ Xλ , the set B(x, d(x, A)) ∩ A is externally
hyperconvex in Xλ . Moreover, if X is hyperconvex, the subspaces Xλ are weakly
externally hyperconvex in X.
As a consequence, we get the following results for gluing along strongly
convex and along externally hyperconvex subsets.
Corollary 1.9. Let (X, d) be the metric space obtained by gluing a collection
(Xλ , dλ )λ∈Λ of hyperconvex metric spaces along some space A, such that A is
closed and strongly convex in Xλ for each λ ∈ Λ. Then X is hyperconvex as
well.
Corollary 1.10. Let (X, d) be the metric space obtained by gluing a family
of hyperconvex metric spaces (Xλ , dλ )λ∈Λ along some set A, such that A is
externally hyperconvex in each Xλ . Then X is hyperconvex. Moreover, A is
externally hyperconvex in X.
In the second case, we clearly have that the gluing set is weakly externally
hyperconvex in each Xλ . Furthermore, it holds that B(x, d(x, A)) ∩ A ∈ E(A)
and therefore, by Proposition II.2.8, we get B(x, d(x, A)) ∩ A ∈ E(Xλ ).
The reason that Corollary 1.9 holds is that a closed, strongly convex subsets
A of a hyperconvex metric space X is gated , that is that for each x ∈ X there
is some x̄ ∈ A such that for all y ∈ A, we have x̄ ∈ I(x, y). Clearly, if such an
x̄ exists, it is unique and we call it the gate of x in A.
Lemma 1.11. Let A be a subset of a hyperconvex metric space X. Then A is
strongly convex and closed if and only if it is gated.
Proof. First assume that A is strongly convex and closed. Fix x ∈ X. Let xn
be a sequence of points in A with d(x, xn ) ≤ d(x, A) + n1 . Note that, for all
x, y, z ∈ X, we have I(x, y) ∩ I(y, z) ∩ I(z, x) 6= ∅, see Proposition V.1.1.
Hence, for n, k ∈ N, we can choose mn,k ∈ I(x, xn ) ∩ I(x, xk ) ∩ I(xn , xk ).
By strong convexity, we get mn,k ∈ A and hence
d(xn , mn,k ) = d(x, xn ) − d(xn , mn,k ) ≤ d(x, A) +
By interchanging xn and xk , we also get d(xk , mn,k ) ≤
d(xn , xk ) = d(xn , mn,k ) + d(mn,k , xk ) ≤
1
1
− d(x, A) = .
n
n
1
k
and therefore
1
1
+ .
n k
Thus (xn )n∈N is a Cauchy sequence and since A is closed, it converges to some
x̄ ∈ A. Moreover, we have d(x, x̄) = d(x, A). We claim that x̄ is a gate for x
in A. Let y ∈ A. Then, there is some z ∈ I(x, x̄) ∩ I(x̄, y) ∩ I(y, x). By strong
28
III.2. GLUING ISOMETRIC COPIES
convexity, we have z ∈ A and therefore d(x, z) ≥ d(x, A) = d(x, x̄). Since
z ∈ I(x, x̄), this implies z = x̄ and hence x̄ ∈ I(x, y) as desired.
On the other hand, if A is gated, for all points x, y ∈ A and z ∈ I(x, y)
we have z = z̄ and hence I(x, y) ⊂ A. Moreover, for all x ∈ X, we have
d(x, A) = d(x, x̄) and therefore x̄ ∈ B(x, d(x, A)) ∩ A, i.e. A is proximinal and
therefore closed.
Therefore, if A is closed and strongly convex in each Xλ , the intersection
B(x, d(x, A)) ∩ A is a point (the gate) and thus externally hyperconvex in X.
Moreover, A is weakly externally hyperconvex in Xλ by Lemma II.3.1.
We can apply Theorem 1.8 to the situation where we glue several hyperconvex metric spaces onto a given space and get the following result:
Theorem 1.12. Let X0 be a hyperconvex metric space and {Xλ }λ∈Λ a family
of hyperconvex metric spaces with subsets Aλ ∈ W(Xλ ), such that, for every
λ ∈ Λ, there is an isometric copy Aλ ∈ W(X0 ) and Aλ ∩ Aλ0 = ∅ for λ 6= λ0 . If
for every xλ ∈ Xλ and every x ∈ X0 , we have B(x
F λ , d(xλ , Aλ )) ∩ Aλ ∈ E(X0 )
and B(x, d(x, Aλ )) ∩ Aλ ∈ E(Xλ ), then X = X0 {Aλ :λ∈Λ} Xλ is hyperconvex.
Proof. First, by Theorem 1.8, we get that Yλ = X0 tAλ Xλ is hyperconvex and
X0 ∈ W(Yλ ).FObserve that X can be obtained by gluing the spaces Yλ along
X0 , i.e. X = X0 Yλ . Therefore, it remains to prove that for λ 6= λ0 and x ∈ Yλ
the intersection B := B(x, d(x, X0 )) ∩ X0 ∈ E(Yλ0 ).
By Corollary II.3.11, we already have B ∈ W(Yλ0 ) and without loss of
generality, we may assume that x ∈
/ X0 . Then we have d(x, X0 ) = d(x, Aλ )
and therefore B = B(x, d(x, Aλ )) ∩ Aλ ∈ E(X0 ), especially B ⊂ Aλ . Hence,
by Corollary II.3.12, we get d(B, Aλ0 ) > 0, i.e. there is some s > 0, such
0
0
that B λ (B, s) ⊂ X0 . Thus B ∈ E(B λ (B, s)) and, by Lemma II.3.17, we get
B ∈ E(Yλ0 ) as desired.
III.2
Gluing Isometric Copies
We now turn our attention to the case where we glue two copies of the same
hyperconvex space. Here we get that the gluing set must be weakly externally
hyperconvex.
Proposition 2.1. Let X be a metric space and A ⊂ X such that X tA X is
hyperconvex. Then the following hold:
(i) A is weakly externally hyperconvex in X.
(ii) For every x ∈ X, the intersection B(x, d(x, A)) ∩ A is externally hyperconvex in X.
Proof. Let us denote the second copy of X by X 0 and for any y ∈ X, let
y 0 denote its corresponding copy in X 0 . Pick x ∈ X and r ≥ 0 such that
d(x, A) ≤ r and let {B(xi , ri )}i∈I be a family of closed balls with xi ∈ A,
29
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
d(x, xi ) ≤ r + ri and d(xi , xj ) ≤ ri + rj . It follows that d(x, x0 ) ≤ 2r and, since
X tA X is hyperconvex, we have
B :=
\
B(xi , ri ) ∩ B(x, r) ∩ B(x0 , r) 6= ∅.
i∈I
By symmetry, there are y, y 0 ∈ B with y ∈ X and y 0 ∈ X 0 . Then, since
intersections of balls are hyperconvex, there is some geodesic [y, y 0 ] ⊂ B, which
must intersect A. Therefore, we get
\
B(xi , ri ) ∩ B(x, r) ∩ A 6= ∅
i∈I
and hence, A is weakly externally hyperconvex.
For (ii), observe that
B(x, d(x, A)) ∩ A = B(x, d(x, A)) ∩ B(x0 , d(x, A))
is admissible in X tA X and therefore externally hyperconvex in X.
Condition (i) is not enough, as the following example shows.
3 . Consider now the gluing
Example 2.2. Let X1 and X2 be two copies of l∞
X := X1 tV X2 where
3
V := {x ∈ l∞
: x1 = x2 and x3 = 0}.
and where the gluing maps are given by the inclusion maps for V . To see that
X is not hyperconvex, consider p1 := (0, 0, 1) in X1 , as well as p2 := (2, 0, 0) and
p02 := (0, −2, 0) both in X2 . Note that B(p1 , 1) ∩ X2 = {(t, t, 0) : t ∈ [−1, 1]}
and hence B(p1 , 1) ∩ B(p2 , 1) = {(1, 1, 0)} ⊂ V , as well as B(p1 , 1) ∩ B(p02 , 1) =
{(−1, −1, 0)} ⊂ V . Moreover, B(p2 , 1) ∩ B(p02 , 1) = {1} × {−1} × [−1, 1] ⊂ X2 .
Hence,
B(p1 , 1) ∩ B(p2 , 1) ∩ B(p02 , 1) = ∅.
As a consequence of Proposition 1.7 and Proposition 2.1, we get a necessary
and sufficient condition for gluings of isometric copies.
Theorem 2.3. Let X be a hyperconvex metric space and let A be a subset.
Then X tA X is hyperconvex if and only if A is weakly externally hyperconvex
in X and for every x ∈ X, the intersection B(x, d(x, A)) ∩ A is externally
hyperconvex in X.
Example 2.4. Let X be a hyperconvex metric space, A ⊂ X and r ≥ 0. If
either A is strongly convex or A = I(x, y) is a metric interval, then B(A, r) is
weakly externally hyperconvex and, for all x ∈ X, the set B(x, d(x, B(A, r))) ∩
B(A, r) is externally hyperconvex in X by Lemma II.3.3 and Lemma II.3.5.
Hence, gluing two copies of X along B(A, r) preserves hyperconvexity.
30
III.3. GLUING HYPERCONVEX LINEAR SPACES
III.3
Gluing Hyperconvex Linear Spaces
We start this section with a classification of externally hyperconvex, strongly
n . We use coordinates
convex and weakly externally hyperconvex subsets of l∞
n .
x = (x1 , . . . , xn ) ∈ l∞
Q
n
A cuboid in l∞ is the product ni=1 Ii of (possibly unbounded) closed, nonempty intervals Ii ⊂ R. Note that these are exactly the sets that can be
described by inequalities of the form σxi ≤ C with σ ∈ {±1} and C ∈ R.
n is externally hyperconvex if and only if it
Proposition 3.1. A subset A of l∞
is a cuboid.
Proof. On the one hand, cuboids are externally hyperconvex by Lemma II.3.18.
On the other hand, if A is externally hyperconvex, it is closed. Hence it is
n with z ∈ I(x , y ) for
enough to show that for any points x, y ∈ A and z ∈ l∞
i
i i
each i ∈ {1, . . . , n}, it follows that z ∈ A. Without loss of generality, we may
assume that xi ≤ zi ≤ yi . Let
max {zi − xi , yi − zi }.
r :=
i∈{1,...,n}
For each i ∈ {1, . . . , n}, define
pi := (z1 , . . . zi−1 , xi − r, zi+1 , . . . , zn ) and ri := zi − xi + r,
q i := (z1 , . . . zi−1 , yi + r, zi+1 , . . . , zn ) and si := yi − zi + r.
Then we have
n
\
B(pi , ri ) ∩
i=1
n
\
B(q i , si ) = {z}.
i=1
Moreover, it holds d(pi , A) ≤ d(pi , x) ≤ ri as well as d(q i , A) ≤ d(q i , y) ≤ si
and therefore, since A is externally hyperconvex,
∅=
6
n
\
i=1
i
B(p , ri ) ∩
n
\
B(q i , si ) ∩ A ⊂ {z}.
i=1
It follows that z ∈ A and this concludes the proof.
Theorem 3.2. Let I 6= ∅ be any index set. Suppose that Q is a non-empty
subset of l∞ (I), given by an arbitrary system of inequalities of the form σxi ≤ C
or σxi + τ xj ≤ C with σ, τ ∈ {±1} and C ∈ R. Then Q is weakly externally
hyperconvex in l∞ (I).
Before giving a proof of Theorem 3.2, we show that a set Q given by a
system of inequalities as in the theorem is proximinal. Recall that l∞ (I) is the
dual space of l1 (I) and observe that Q is a weak*-closed subset of l∞ (I). In
general, the following holds.
31
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
Lemma 3.3. Let V be a normed vector space and V ∗ its dual space. Then
every non-empty weak*-closed subset of V ∗ is proximinal. Especially, if Q is
a non-empty subset of l∞ (I), given by an arbitrary system of inequalities of
the form σxi ≤ C or σxi + τ xj ≤ C with σ, τ ∈ {±1} and C ∈ R, then Q is
proximinal.
Proof. Let A be a non-empty weak*-closed subset of V ∗ , let x be any point in
V ∗ and set r := d(x, A). It follows from the theorem of Banach-Alaoglu that
the set A0 := B(x, r + 1) is weak*-compact. Now, the sets
B x, r + n1 ∩ A n∈N
form a decreasing sequence of non-empty weak*-closed subsets of A0 . By the
closed set criterion for compact sets, it follows that
\
B(x, r) ∩ A =
B(x, r + n1 ) ∩ A 6= ∅.
n∈N
This shows that A is proximinal.
To see that Q is weak*-closed, note that the maps from l∞ (I) to R given
by ϕσei : f 7→ σfi and ϕσei +τ ej : f 7→ σfi + τ fj are continuous in the weak*topology on l∞ (I) since σei and σei + τ ej are both in l1 (I). Since, by assumption,
\
\
Q=
ϕ−1
ϕ−1
σei ((−∞, C(i,σ) ]) ∩
σei +τ ej ((−∞, C(i,j,σ,τ ) ]),
(i,σ)
(i,j,σ,τ )
we deduce that Q is weak*-closed and therefore proximinal.
Now, the proof of Theorem 3.2 is a direct adaptation of the proof of [Lan13,
Proposition 2.4].
Proof of Theorem 3.2. For i ∈ I, denote by Ri the reflection of l∞ (I) that
interchanges xi with −xi .
Let A ∪ {b} be a metric space with ∅ =
6 A and let y ∈ l∞ (I) \ Q. For any
f ∈ Lip1 (A, Q ∪ {y}) satisfying that d(y, Q) ≤ d(a, b) if f (a) = y, we show
that there is an extension f ∈ Lip1 (A ∪ {b}, Q ∪ {y}) such that f (b) ∈ Q. It is
easy to see that this implies that Q ∈ W(l∞ (I)), it is similar to showing that
injectivity implies hyperconvexity.
Let A0 := A \ f −1 ({y}). By proximinality of Q, we have Q ∩ B(y, d(y, Q)) 6=
∅. We can thus assume that
\
0 ∈ Q ∩ B(y, d(y, Q)) ⊂ Q ∩
B(f (a0 ), d(a0 , b)),
(3.1)
a0 ∈f −1 ({y})
so that all constants on the right sides of the inequalities describing Q are
non-negative. First, for a real valued function f ∈ Lip1 (A, R), we combine the
smallest and largest 1-Lipschitz extensions and define f : A ∪ {b} → R by
n
o
n
o
0
0
f (b) := sup 0, sup(f (a) − d(a, b)) + inf 0, inf
(f
(a
)
+
d(a
,
b))
.
0
a ∈A
a∈A
32
III.3. HYPERCONVEX LINEAR SPACES
Note that at most one of the two summands is nonzero since
f (a) − d(a, b) ≤ f (a0 ) + d(a, a0 ) − d(a, b) ≤ f (a0 ) + d(a0 , b).
It is not difficult to check that f is a 1-Lipschitz extension of f and that
R◦f =R◦f
for the reflection R : x 7→ −x of R. Moreover, by (3.1), it follows that
(f (a) − d(a, b)) ≤ 0
sup
a∈f −1 ({y})
and
inf
a∈f −1 ({y})
(f (a) + d(a, b)) ≥ 0.
It follows that f can be defined by taking suprema and infima on A0 instead of
A without changing f (b). We define the extension operator
φ : Lip1 (A, l∞ (I)) → Lip1 (A ∪ {b}, l∞ (I))
such that φ(f ) satisfies φ(f )i = fi for every i. Clearly φ(f ) ∈ Lip1 (A ∪
{b}, l∞ (I)) and
φ(Ri ◦ f ) = Ri ◦ φ(f )
(3.2)
for every i. To see that φ(f )(b) ∈ Q, it thus suffices to show that the components
of φ(f ) satisfy
φ(f )i + φ(f )j ≤ C
(3.3)
whenever fi + fj ≤ C for some pair of possibly equal indices i, j and some
constant C ≥ 0.
Suppose that fi (a) + fj (a) ≤ C for some indices i, j, some constant C ≥ 0
and every a ∈ A0 := A \ f −1 ({y}). Assume that φ(f )i (b) ≥ φ(f )j (b). If
φ(f )j (b) > 0, then
φ(f )i (b) + φ(f )j (b) = sup (fi (a) + fj (a0 ) − d(a, b) − d(a0 , b))
a,a0 ∈A0
≤ sup (fi (a) + fj (a0 ) − d(a, a0 ))
a,a0 ∈A0
≤ sup (fi (a) + fj (a)) ≤ C.
a∈A0
If φ(f )i (b) > 0 ≥ φ(f )j (b), then
φ(f )i (b) + φ(f )j (b) ≤ sup (fi (a) − d(a, b)) + inf
(fj (a0 ) + d(a0 , b))
0
0
a ∈A
a∈A0
≤ sup (fi (a) + fj (a)) ≤ C.
a∈A0
Finally, if φ(f )i (b) ≤ 0, then φ(f )i (b) + φ(f )j (b) ≤ 0 ≤ C.
33
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
Remark 3.4. By Theorem 3.2, and borrowing the notation and terminology
from [Lan13], for any metric space X and for any set A ∈ A (X), note that the
set
P (A) := ∆(X) ∩ H(A)
is isometric to a weakly externally hyperconvex subset of l∞ (X).
Indeed, for any point x0 ∈ X, let us denote by τ : RX → RX the translation
f 7→ f − dx0 , where dx0 : x 7→ d(x0 , x). One has
τ (P (A)) ⊂ τ (E(X)) ⊂ l∞ (X),
see [Lan13, Sections 3 and 4]. Recall that
∆(X) := {f ∈ RX : f (x) + f (y) ≥ d(x, y) for all x, y ∈ X} and
H(A) := {g ∈ RX : g(x) + g(y) = d(x, y) for all {x, y} ∈ A}.
By Theorem 3.2, it follows that τ (P (A)) is in W(l∞ (X)). In particular, it
follows that τ (P (A)) is weakly externally hyperconvex in τ (E(X)) and since τ
is an isometry, P (A) ∈ W(E(X)).
Finally, recall that under suitable assumptions on X, the family of cells
{P (A)}A∈A (X) endows E(X) with a canonical polyhedral structure. Thereby
for any A ∈ A (X), P (A) is a cell of E(X) isometric to a convex polytope in
n .
l∞
In the following, we use the notation In := {1, . . . , n}. To characterize
n , it is
strongly convex and weakly externally hyperconvex linear subspaces of l∞
helpful to recall the characterization for hyperconvex linear subspaces.
n be a linear subspace and
Theorem 3.5. [Pav16, Theorem 2.1] Let ∅ =
6 V ⊂ l∞
let k := dim(V ). Then the following are equivalent:
(i) V is hyperconvex.
(ii) There is a subset J ⊂ In with |J| = kP
such that, for all i ∈ In \ J, there
exist real numbers {ci,j }j∈J such that j∈J |ci,j | ≤ 1 and
(
V =
(x1 , . . . , xn ) ∈
)
n
l∞
: for all i ∈ In \ J , one has xi =
X
ci,j xj .
j∈J
n . Then
Proposition 3.6. Let n ≥ 1 and let V denote any linear subspace of l∞
n
V is strongly convex in l∞ if and only if one of the following three cases occurs:
(i) V = {0},
(ii) V = Re where e denotes a vertex of the hypercube [−1, 1]n , or
n .
(iii) V = l∞
34
III.3. HYPERCONVEX LINEAR SPACES
Proof. Let k = dim V . Since strongly convex subspaces are hyperconvex, by
Theorem 3.5, we may assume that V is of the form
(
)
k
X
n
V = (x1 , . . . , xn ) ∈ l∞ : xl =
cl,i xi for l = k + 1, . . . n
i=1
with
Pk
i=1 |cl,i |
≤ 1. Fix some 1 ≤ i ≤ k and consider
x = (0, . . . 0, 1, 0, . . . 0, ck+1,i , . . . , cn,i ) ∈ V.
We have cl,i 6= 0 since otherwise
y = (0, . . . 0, 21 , 0, . . . 0,
ck+1,i
cl−1,i 1 cl+1,i
cn,i
2 , . . . , 2 , 2, 2 , . . . , 2 )
∈ I(0, x) \ V.
Now take l such that |cl,i | is minimal. Then
|c |
|c |
l,i
l,i
y = (0, . . . 0, |cl,i |, 0, . . . 0, |ck+1,i
| ck+1,i , . . . , |cn,i | cn,i ) ∈ I(0, x) ⊂ V
and therefore cl,i = cl,i · |cl,i |, i.e. |cl,i | = 1. Hence, one has cl,i ∈ {±1} for
P
all l = k + 1, . . . , n and i = 1, . . . , k. Since ki=1 |cl,i | ≤ 1, it thus follows that
either k = 0, which corresponds to (i), k = 1, which corresponds to (ii) or
k = n, which corresponds to (iii).
In the following, relint(S) denotes the relative interior of S and Fi denotes
n : x = 1} of the unit ball [−1, 1]n in ln . Half-spaces
the facet [−1, 1]n ∩ {x ∈ l∞
i
∞
are denoted by Hv := {x ∈ Rn : x · v ≥ 0} (with · denoting the standard scalar
product) and ∂Hv is the boundary of Hv .
n , such that V
Lemma 3.7. Let V denote any hyperconvex linear subspace of l∞
is not contained in any hyperplane of the form ∂Hσei +τ ej . Then the following
hold:
(i) There is (i, σ) ∈ In × {±1} such that V ∩ relint(σFi ) 6= ∅.
n , there is i ∈ I and ν ∈ Rn \ {0} such that V ⊂ ∂H and
(ii) If V 6= l∞
n
ν
∂Hν ∩ (Fi ∪ (−Fi )) = ∅.
Proof. By Theorem 3.5, V can, without loss of generality, be written as


k
k


X
X
V := (x1 , . . . , xk ,
ck+1,j xj , . . . ,
cn,j xj ) : x1 , . . . , xk ∈ R ,


j=1
j=1
where, for every m ∈ {k + 1, . . . , n}, one has
V ⊂ ∂Hν m , where
Pk
j=1 |cm,j |
≤ 1. It follows that
ν m := (cm,1 , . . . , cm,k , 0, . . . , 0, −1, 0, . . . , 0)
35
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
(i.e. −1 in the m-th entry) and
m
kν k1 = 1 +
k
X
|cm,j | ≤ 2 = 2 kν m k∞ .
j=1
Since we assume that V 6⊂ ∂Hσei +τ ej , it follows that, for every m ∈ {k +
1, . . . , n} and every i ∈ {1, . . . , k}, we have |cm,i | < 1. In particular, (i) follows
from the fact that
(1, 0, . . . , 0, ck+1,1 , . . . , cn,1 ) ∈ V ∩ relint(F1 ).
Now, (ii) follows from the fact that V ⊂ ∂Hν , where
ν := (ck+1,1 , . . . , ck+1,k , −1, 0, . . . , 0) ∈ relint(−Fk+1 )
and ∂Hν ∩ (Fk+1 ∪ (−Fk+1 )) = ∅.
We now proceed to the characterization of weakly externally hyperconvex
n .
polyhedra in l∞
n . Then
Theorem 3.8. Let n ≥ 1 and let V denote any linear subspace of l∞
n
V is weakly externally hyperconvex in l∞ if and only if V can be written as the
intersection of hyperplanes of the form ∂Hσei or ∂Hσei +τ ej , where σ, τ ∈ {±1}
n if
and i, j ∈ {1, · · · , n}. Equivalently, V is weakly externally hyperconvex in l∞
and only if
m
V = {0}k × l∞
× Sp1 × · · · × Spq ,
p
j
where Spj denotes a one-dimensional strongly convex
Pq linear subspace of l∞ for
each j ∈ {1, . . . , q}. Note that then n = k + m + j=1 pj and dim(V ) = m + q.
Proof. First, it is easy to see that both representations for V given in the
statement of Theorem 3.8 are equivalent. To see this, note that, if V has the
desired product representation, then using Proposition 3.6, it is easy to express
V as an intersection of hyperplanes as desired. Conversely, if V can be written
as such an intersection, then V is given by a system of linear equations which
one can decompose into a set of minimal linear subsystems. Applying Proposition 3.6, one obtains the desired product representation for V .
n of the form ∂H
Note that the hyperplanes of l∞
σei or ∂Hσei +τ ej are exactly
the ones to which Theorem 3.2 applies. Hence, whenever V can be written
as an intersection of such hyperplanes, it follows that V is weakly externally
hyperconvex. This proves one implication in Theorem 3.8.
n ) and we show that V can be written as
We now assume that V ∈ W(l∞
\
\
V =
∂Hσei ∩
∂Hσei +τ ej .
(i,σ)
(i,σ),(j,τ )
If V ⊂ W , then V ∈ W(W ). Now assume that W = ∂Hσei or W = ∂Hσei +τ ej .
Then there is a canonical bijective linear isometry
n−1
π : W → l∞
, (x1 , . . . , xn ) 7→ (x1 , . . . , xi−1 , xi+1 , . . . xn )
36
III.3. HYPERCONVEX LINEAR SPACES
n−1 ) and we can iterate this process.
with π(V ) ∈ W(l∞
We can therefore, without loss of generality, assume that V is not contained
in any hyperplane of the form ∂Hσei or ∂Hσei +τ ej . It follows by (i) in Lemma 3.7 and hyperconvexity of V that there is (i, σ) ∈ In × {±1} such that
V ∩ relint(σFi ) 6= ∅.
By (ii) in Lemma 3.7, there is a facet τ Fj of [−1, 1]n such that τ Fj ∩ V = ∅.
It follows that the facet τ Fj0 := τ Fj ∩ σFi of σFi satisfies τ Fj0 ∩ V = ∅. But
τ Fj0 can be written as τ Fj0 = ∩i∈{1,2,3} Bi where we pick v (0) ∈ V ∩ relint(σFi )
so that [0, ∞)v (0) ∩ ∂Hτ ej 6= ∅ and with
B1 = B(0, 1),
B2 = B(Rv (0) , kRv (0) k∞ ) and
e ej , kσei + τ ej + Rτ
e ej − (σei + τ ej )k∞ ).
B3 = B(σei + τ ej + Rτ
e big enough, one has B3 ∩ V 6= ∅ since V 6⊂ ∂He and by the
Indeed, for R
j
e chosen even bigger (if needed) and for R big enough, one
choice of v (0) . For R
has σFi = B1 ∩ B2 and τ Fj0 = B1 ∩ B2 ∩ B3 . It follows that the balls pairwise
intersect and
V ∩ B1 ∩ B2 ∩ B3 = V ∩ τ Fj0 = ∅.
n ), which is a contradiction.
This implies that V 6⊂ W(l∞
Let V be a finite dimensional real vector space and let Q be any convex
non-empty subsetSof V . The tangent cone to Q at p ∈ Q is defined to be
the set Tp Q := n∈N (p + n(Q − p)) where Q − p := {q − p : q ∈ Q} and
nQ := {nq : q ∈ Q}.
n . Then the following are
Proposition 3.9. Suppose that Q is a subset of l∞
equivalent:
(i) Q is given by a finite system of inequalities of the form σxi + τ xj ≤ C or
σxi ≤ C with |σ|, |τ | = 1 and C ∈ R,
n ).
(ii) Q ∈ W(l∞
Proof. The fact that (i) implies (ii) is a direct consequence of Theorem 3.2. On
the other hand, the fact that (ii) implies (i) follows by considering the tangent
cones Tp Q for each smooth point p in the boundary of Q. By Theorem VI.1.1,
the set Q is convex and as it is for instance observed at the beginning of the
proof of [BEL05, Theorem 3.1], by virtue of Mazur density theorem, cf. [Hol75],
the set of smooth points of the boundary of Q is dense in ∂Q. It follows that
Q can be written as an intersection of half-spaces which correspond to tangent
cones of Q at smooth points on the boundary. These tangent cones are weakly
externally hyperconvex by Proposition II.3.14 and therefore they are of the
n : σx ≤ C} or {x ∈ ln : σx + τ x ≤ C} by Theorem 3.8. Since
form {x ∈ l∞
i
i
j
∞
there are only finitely many such cones up to translation, Q is an intersection
of finitely many half-spaces and is thus a convex polyhedron of the desired
form.
37
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
We now go back to the study of gluings.
n and X = lm , and let V be a linear subspace of
Lemma 3.10. Let X1 = l∞
2
∞
both X1 and X2 . Consider X = X1 tV X2 . Then, for x ∈ X1 and r ≥ 0, we
have
[
B(x, r) ∩ X2 =
B 2 (y, r − d(x, y)).
y∈B 1 (x,r)∩V
Moreover, B(x, r) ∩ X2 is connected.
Proof. Since V is proper, for any x ∈ X1 and x0 ∈ X2 , there is some y ∈ V
with d(x, x0 ) = d(x, y) + d(y, x0 ). Since B 1 (x, r) ∩ V is connected, this is also
true for B(x, r) ∩ X2 .
If V is any finite dimensional real vector space and Q is any convex nonempty subset of V , we say that e ∈ Q is an extreme point of Q, and we write
e ∈ ext(Q), if for any c, c0 ∈ Q and t ∈ [0, 1] such that e = (1 − t)c + tc0 , it
follows that t ∈ {0, 1}. Finally, by a convex polytope in V , we mean a bounded
intersection of finitely many closed half-spaces or equivalently the convex hull
of finitely many points. A compact convex set in V is a convex polytope if and
only if it has finitely many extreme points, cf. [Gru02, Section 3.1].
n and X = lm . Moreover, let V be a linear subspace
Lemma 3.11. Let X1 = l∞
2
∞
of both X1 and X2 such that V 6= X1 , X2 and
X = X1 tV X2
is hyperconvex. Then, for x ∈ X1 and r > d(x, V ), the set B(x, r) ∩ X2 is a
cuboid.
Proof. We first show that Q := B(x, r) ∩ X2 is convex. Let p, q ∈ Q. Then
there are p̄, q̄ ∈ V such that
d(x, p̄) + d(p̄, p) = d(x, p)
and d(x, q̄) + d(q̄, q) = d(x, q).
(3.4)
Consider now any point z := p + t(q − p) ∈ p + [0, 1](q − p) ⊂ X2 and let
z̄ := p̄ + t(q̄ − p̄) ∈ V . On the one hand,
d(z, z̄) ≤ (1 − t)d(p, p̄) + td(q, q̄)
and on the other hand,
d(x, z̄) ≤ (1 − t)d(x, p̄) + td(x, q̄).
Summing the above two inequalities, applying the triangle inequality and using
(3.4), we obtain
d(x, z) ≤ (1 − t)d(x, p) + td(x, q) ≤ r.
It follows that z ∈ Q and shows that Q is convex.
Since X is hyperconvex, for every y ∈ X2 with d(x, y) ≤ r + d(y, V ), the set
B(y, d(y, V )) ∩ B(x, r) ∩ X2 = B(y, d(y, V )) ∩ B(x, r)
38
III.3. HYPERCONVEX LINEAR SPACES
is externally hyperconvex in X2 and is therefore a cuboid by Proposition 3.1.
Hence B(x, r) ∩ X2 \ V is locally a cuboid.
To see that Q is a convex polytope, we show that Q has only finitely many
extreme points. Observe that Q has only finitely many extreme points inside
V since Q ∩ V = B 1 (x, r) ∩ V and the right-hand side is a convex polytope
contained in V . Now, note that, since Q is locally a cuboid outside of V , Q
has only finitely many different tangent cones at points outside of V , and this
up to translation (of such cones). It is easy to see that Q cannot have both a
cone and a nontrivial translate of this cone as tangent cones. It follows that
Q can only have finitely many different tangent cones at points outside V . In
particular, Q has only finitely many extreme points and it is thus a convex
polytope. Furthermore, V meets the interior of Q and V has codimension at
least one in X2 , hence V cannot contain any facet of Q. It follows that Q can
be written as the intersection of tangent cones to Q at points in Q \ V . Once
again, since Q is locally a cuboid outside of V , it follows that Q is a cuboid.
n and X = lm . Moreover, let V be a linear subspace
Lemma 3.12. Let X1 = l∞
2
∞
of both X1 and X2 such that V 6= X1 , X2 and
X = X1 tV X2
is hyperconvex. Then V must be weakly externally hyperconvex in X.
Proof. Note that it is enough to show that V is weakly externally hyperconvex
in X1 . First we introduce coordinates:
(
)
k
X
m
V = x ∈ l∞ : for every l ∈ {k + 1, . . . , n}, one has xl =
cl,i xi .
i=1
P
As V must be hyperconvex by Lemma 1.2, we may assume that ki=1 |cl,i | ≤ 1
for all l by Theorem 3.5. For x ∈ X1 and r > d(x, V ), we can write
A := B(x, r) ∩ X2 =
m
Y
1
[a−1
i , ai ]
i=1
by Lemma 3.11. Let ν be a vertex of [−1, 1]m , that is ν ∈ {−1, 1}m , and
aν = (aν11 , . . . , aνmm ) the corresponding vertex of A. By Lemma 3.10, for every
aν there must be some āν ∈ V such that d(aν , āν ) = r − d(x, āν ). Since aν is a
vertex of the cuboid A, āν must lie on the diagonal aν + Rν or more precisely
āν = aν − tν ν
(3.5)
for tν = d(aν , āν ) ≥ 0.
Hence, for every l ∈ {k + 1, . . . , m}, one has
āνl =
k
X
i=1
cl,i āνi i =
k
X
cl,i aνi i − tν
i=1
39
k
X
i=1
cl,i νi .
(3.6)
CHAPTER III. GLUING HYPERCONVEX METRIC SPACES
Fix now some l ∈ {k + 1, . . . , m}. Then, for every ν ∈ {−1, 1}m , consider
∈ {−1, 1}m with νl+ = 1, νl− = −1 and νi+ = νi− = νi for i 6= l. Clearly,
ν coincides with either ν + or ν − . From (3.5), we get
ν −, ν +
+
−
ν−
ν+
āνl − āνl = (al l − tν + νl+ ) − (al l − tν − νl− ) = a1l − a−1
l − tν + − tν − .
Observe that, for ν + , ν − , on the right-hand
P side of (3.6) all parameters except
tν + , tν − are equal. Hence, setting c := ki=1 cl,i νi ∈ [−1, 1], we get
+
−
āνl − āνl = c(tν + − tν − ) ≤ |tν + − tν − |.
Let us assume that tν + ≥ tν − (the other case is analogous). We then get
−1
a+1
l − al − tν + − tν − ≤ tν + − tν − ,
−1
or equivalently 2tν + ≥ a+1
l − al . But this inequality must be an equality since
+
+
−
+
B 2 (āν , tν + ) ⊂ A. Moreover, we have aν , aν ∈ B 2 (āν , tν + ). We conclude
that āν can always be chosen such that
d(aν , āν ) = tν =
1 +1
.
al − a−1
l
2
−1
Especially, we have āνl = 12 a+1
. Hence, all vertices of A and therefore,
l + al
by convexity, all points a ∈ A are contained in a ball B 2 (ā, t) with ā ∈ B 1 (x, r)∩
V and maximal radius
t = r − d(x, ā) =
1 +1
al − a−1
.
l
2
Note that, for y ∈ B 1 (x, r) ∩ V , the radius r − d(x, y) is maximal if and only if
d(x, y) = d(x, V ) and therefore ā ∈ B(x, d(x, V )) ∩ V .
We now use Lemma II.3.1 to conclude that V is weakly externally hyperconvex in X1 . First, we have
\
B(x, d(x, V )) ∩ V =
B(x, d(x, V ) + n1 ) ∩ X2 ∈ E(X2 )
n∈N
using Lemma 3.11 and therefore B(x, d(x, V )) ∩ V ∈ E(V ). Furthermore, for
y ∈ V , assume that r = d(x, y) > d(x, V ). Then, by the above, there is some
a ∈ V ∩ B(x, d(x, V )) such that
r = d(x, y) ≤ d(x, a) + d(a, y) ≤ d(x, V ) + r − d(x, V ) = r,
i.e. d(x, y) = d(x, a) + d(a, y).
n and X = lm . Moreover, let V be a linear
Theorem 3.13. Let X1 = l∞
2
∞
subspace of both X1 and X2 such that V 6= X1 , X2 . Then X = X1 tV X2 is
k ×X 0 , X = lk ×X 0 ,
hyperconvex if and only if there is some k such that X1 = l∞
2
∞
1
2
k × V 0 , and V 0 is strongly convex in both X 0 = ln−k and X 0 = lm−k .
V = l∞
∞
∞
1
2
40
III.3. HYPERCONVEX LINEAR SPACES
Proof. If X is hyperconvex, by Lemma 3.12, the subspace V is weakly externally
hyperconvex in X1 and in X2 , we can thus apply Theorem 3.8 and write
µ
k1
k1
µ0
µ1
V = l∞
× S0 × S1 × . . . × Sp ⊂ l∞
× l∞
× l∞
× . . . × l∞p = X1 ,
µi
µi
where for all i ∈ {0, . . . , p}, the set Si ( l∞
is strongly convex in l∞
and for
i 6= 0, Si is one dimensional (for i = 0 we also might have S0 = {0}) and
similarly
ν
k2
k2
ν0
ν1
V = l∞
× T0 × T1 × . . . × Tq ⊂ l∞
× l∞
× l∞
× . . . × l∞q = X2 .
Let {e1 , . . . , en } and {f1 , . . . , fm } be the standard basis for X1 and X2 ,
respectively. First, observe that, if Rei ⊂ V , then there is some fj with Rei =
Rfj ⊂ V . Indeed, if Rei ⊂ V for every r ≥ 0, we have
\
[−r, r]ei =
B(re, r) ∈ E(X).
e∈{−1,1}i−1 ×{0}×{−1,1}n−i ⊂X1
Note that, as [−r, r]ei ∈ E(X2 ), it is a cuboid in X2 and therefore we get
[−r, r]ei = [−r, r]fj .
k
Hence, we have k = k1 = k2 and can split off the maximal component l∞
k
0
k
0
from X1 , X2 and V , that is, we can write V = l∞ × V ⊂ l∞ × Xi = Xi with
µ
µ0
µ1
V 0 = S0 × S1 × . . . × Sp ⊂ l∞
× l∞
× . . . × l∞p = X10
as well as
ν
ν0
ν1
V 0 = T0 × T1 × . . . × Tq ⊂ l∞
× l∞
× . . . × l∞q = X20 .
Assume now that there are at least two non-trivial factors in the first deµi
µi
composition, i.e. p ≥ 1. Since, for every factor l∞
, there is some xi ∈ l∞
with
0
d(xi , Si ) = 1, there is some x = (x0 , x1 , . . . , xp−1 , 0) ∈ X1 with
B(x, d(x, V 0 )) ∩ V 0 = B(x, 1) ∩ V 0 = {0}p × Sp ∩ B Sp (0, 1) ,
where B Sp (0, 1) denotes the unit ball inside Sp .
But {0} × (Sp ∩ B Sp (0, 1)) ⊂ X20 is not externally hyperconvex in X20 since
it is not a cuboid. It follows that B(x, d(x, V 0 )) ∩ V 0 ∈
/ E(X20 ) which is a
contradiction to Proposition 1.6. Therefore, V must consist of only one strongly
convex factor.
k × (X 0 t 0 X 0 ). The second factor
For the other direction, note that X = l∞
1 V
2
0
is hyperconvex by Corollary 1.9 since V is strongly convex. Hence also the
product is hyperconvex by Lemma II.3.18.
Note that Theorem 3.13 further strengthens the weak optimality of the
assumption in Theorem 1.8 that the gluing set is weakly externally hyperconvex.
41
Chapter IV
Local to Global
IV.1 The Cartan-Hadamard Theorem for Metric
Spaces with Local Geodesic Bicombings
Let us first fix some notation. In a metric space X, we denote by
U (x, r) := {y ∈ X : d(x, y) < r}
the open ball of radius r around x ∈ X and by
B(x, r) := {y ∈ X : d(x, y) ≤ r}
the closed one.
Let X be a metric space and γ : [0, 1] → X a continuous curve. The length
of γ is given by
( n
)
X
L(γ) := sup
d(γ(tk−1 ), γ(tk )) : 0 = t0 < . . . < tn = 1 .
k=1
Then
¯ y) := inf {L(γ) : γ : [0, 1] → X, γ(0) = x, γ(1) = y}
d(x,
¯ we
defines a metric on X, called the induced length metric. If we have d = d,
say that (X, d) is a length space.
For a metric space X, let G(X) := {c : [0, 1] → X} be the set of all geodesics
in X, i.e. continuous maps c : [0, 1] → X with
d(c(s), c(t)) = |s − t| · d(c(0), c(1))
for all s, t ∈ [0, 1]. We equip G(X) with the metric
D(c, c0 ) := sup d(c(t), c0 (t)).
t∈[0,1]
Let c ∈ G(X) and 0 ≤ a ≤ b ≤ 1, then we denote by c[a,b] the reparametrized
geodesic given by c[a,b] : [0, 1] → X with
c[a,b] (t) := c((1 − t)a + tb).
43
CHAPTER IV. LOCAL TO GLOBAL
For a metric space X, a geodesic bicombing is a map σ : X × X × [0, 1] → X
that selects, for every pair of points (x, y) ∈ X × X, a geodesic σxy := σ(x, y, ·)
from x to y. If this selection satisfies the inequality
d(σxy (t), σx0 y0 (t)) ≤ (1 − t)d(x, x0 ) + td(y, y 0 )
for all x, y, x0 y 0 ∈ X and all t ∈ [0, 1], we call it a conical geodesic bicombing.
Note that this does not imply that the map t 7→ d(σxy (t), σx0 y0 (t)) is convex. If
this is the case, we say that the geodesic bicombing is convex. A sufficient (but
not necessary) condition for a conical geodesic bicombing to be convex is that
the selection of geodesics is consistent. This is
σσxy (s1 )σxy (s2 ) (t) = σxy ((1 − t)s1 + ts2 )
for x, y ∈ X, 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1].
Finally, we call a geodesic bicombing reversible if σyx (t) = σxy (1 − t) for all
x, y ∈ X and t ∈ [0, 1].
Definition 1.1. A local geodesic bicombing on a metric space X is a local
selection of geodesics, i.e. a map σ : U × [0, 1] → X, (y, z, t) 7→ σyz (t), with the
following properties:
(i) For all x ∈ X, there is some rx > 0 such that, for all y, z ∈ U (x, rx ), there
is a geodesic σyz : [0, 1] → U (x, rx ) from y to z, and
U = {(y, z) ∈ X × X : y, z ∈ U (x, rx ) for some x}.
(ii) The selection is consistent with taking subsegments of geodesics, i.e.
σσyz (s1 )σyz (s2 ) (t) = σyz ((1 − t)s1 + ts2 )
for (y, z) ∈ U , 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1].
We call a local geodesic bicombing σ convex if it is locally convex, i.e. for
y, z, y 0 , z 0 ∈ U (x, rx ), it holds that
t 7→ d(σyz (t), σy0 z 0 (t))
is a convex function. Furthermore, σ is reversible if
σzy (t) = σyz (1 − t)
for all (y, z) ∈ U and t ∈ [0, 1].
Remark. Observe that, by consistency, a local geodesic bicombing is convex if
and only if
d(σyz (t), σy0 z 0 (t)) ≤ (1 − t)d(y, y 0 ) + td(z, z 0 )
for all y, z, y 0 , z 0 ∈ U (x, rx ) and t ∈ [0, 1].
A (local) geodesic c : [0, 1] → X is consistent with the local geodesic bicombing σ if
c[s1 ,s2 ] (t) = σc(s1 )c(s2 ) (t)
for all 0 ≤ s1 ≤ s2 ≤ 1 with (c(s1 ), c(s2 )) ∈ U .
44
IV.1. CARTAN-HADAMARD THEOREM
The goal of this section is now to prove the following generalization of the
Cartan-Hadamard Theorem for metric spaces with a local geodesic bicombing.
Theorem 1.2. Let X be a complete, simply-connected metric space with a
convex local geodesic bicombing σ. Then the induced length metric on X admits
a unique consistent, convex geodesic bicombing σ̃ which is consistent with σ. As
a consequence, X is contractible. Moreover, if the local geodesic bicombing σ is
reversible, then σ̃ is reversible as well.
To prove this, we roughly follow the structure of Chapter II.4 in [BH99].
Adapting the methods of S. Alexander and R. Bishop [AB90], we can prove the
following key lemma.
Lemma 1.3. Let X be a complete metric space with a convex local geodesic
bicombing σ and let c be a local geodesic from x to y which is consistent with σ.
Then there is some > 0 such that, for all x̄, ȳ ∈ X with d(x, x̄), d(y, ȳ) < ,
there is a unique local geodesic c̄ from x̄ to ȳ with D(c, c̄) < which is consistent
with σ. Moreover, we have
L(c̄) ≤ L(c) + d(x, x̄) + d(y, ȳ)
and if c̃ is another consistent geodesic from x̃ to ỹ with D(c, c̃) < , then
t 7→ d(c̃(t), c̄(t))
is convex.
Proof. Let > 0 be such that σ U (c(t),2)×U (c(t),2)×[0,1] is a convex geodesic
bicombing for all t ∈ [0, 1]. Now, let P(A) be the following statement:
P(A): For all a, b ∈ [0, 1] with 0 ≤ b − a ≤ A and for all p, q ∈ X with
d(c(a), p), d(c(b), q) < , there is a unique local geodesic c̄pq : [0, 1] → X
from p to q with D(c[a,b] , c̄pq ) < which is consistent with σ. Moreover,
for all such local geodesics the map t 7→ d(c̄pq (t), c̄p0 q0 (t)) is convex.
By our assumption, P( l(c)
) holds. Therefore, let us show P(A) ⇒ P( 3A
2 ).
2
1
Given a, b ∈ [0, 1] with 0 ≤ b − a ≤ 3A
2 , we define p0 := c( 3 a + 3 b) and
1
2
q0 := c( 3 a + 3 b). Then, by P(A), there are consistent local geodesics c1 from
p to q0 and c01 from p0 to q. Inductively, we set pn := cn ( 12 ) and qn := c0n ( 12 ),
where cn is a consistent local geodesic from p to qn−1 and c0n from pn−1 to q.
Observe that, by convexity of the cn , c0n , we have d(pn−1 , pn ), d(qn−1 , qn ) < 2n
and hence the sequences (pn )n∈N and (qn )n∈N converge to some p∞ and q∞ ,
respectively, and we have d(p∞ , p0 ), d(q∞ , q0 ) < . Furthermore, by convexity,
the cn , c0n converge to the consistent local geodesics c∞ from p to q∞ and c0∞
from p∞ to q, which coincide between p∞ = c∞ ( 12 ) and q∞ = c0∞ ( 12 ). Hence,
they define a new local geodesic cpq from p to q which is consistent with σ and
p∞ = cpq ( 13 ), q∞ = cpq ( 32 ).
45
CHAPTER IV. LOCAL TO GLOBAL
Now, given two local geodesics cpq and cp0 q0 with D(c[a,b] , cpq ) < and
D(c[a,b] , cp0 q0 ) < , set s := d(p, p0 ), t := d(q, q 0 ), s0 := d(cpq ( 31 ), cp0 q0 ( 13 )) and
0
s0 +t
0
t0 := d(cpq ( 23 ), cp0 q0 ( 23 )). Then we have s0 ≤ s+t
2 , t ≤ 2 and therefore, we get
s0 ≤
s
2
+
s0
4
+
t
4
⇒ s0 ≤
2s+t
3 ,
0 0
and similarly, t0 ≤ s+2t
3 follows. Hence, we get convexity of t 7→ d(cpq (t), cp q (t))
and therefore also uniqueness follows.
It remains to prove that L(c̄) ≤ L(c) + d(x, x̄) + d(y, ȳ). Let c̃ be the unique
consistent local geodesic from x to ȳ with D(c, c̃) < . For t small enough, we
have
tL(c̃) = d(c̃(0), c̃(t)) = d(c(0), c̃(t))
≤ d(c(0), c(t)) + d(c(t), c̃(t)) ≤ tL(c) + td(c(1), c̃(1)),
i.e. L(c̃) ≤ L(c) + d(y, ȳ) and similarly L(c̄) ≤ L(c̃) + d(x, x̄).
Definition 1.4. Let X be some metric space with a local geodesic bicombing
σ. For some fixed x0 ∈ X, we define
X̃x0 := {c : [0, 1] → X local geodesic with c(0) = x0 , consistent with σ}.
We equip X̃x0 with the metric
D(c, c0 ) = sup d(c(t), c0 (t))
t∈[0,1]
and define the map
exp : X̃x0 → X, c 7→ c(1).
If X is complete, this map has the following properties.
Lemma 1.5. Let X be a complete metric space with a convex local geodesic
bicombing σ. Then the following holds:
(i) The map exp : X̃x0 → X is locally an isometry. Hence, σ naturally induces
a convex local geodesic bicombing σ̃ on X̃x0 .
(ii) X̃x0 is contractible.
(iii) For each x̃ ∈ X̃x0 , there is a unique local geodesic from x̃0 to x̃ which is
consistent with σ̃, where x̃0 is the constant path x̃0 (t) = x0 .
(iv) X̃x0 is complete.
Proof. By Lemma 1.3, for every c ∈ X̃x0 , there is some > 0 such that the map
exp U (c,) : U (c, ) → U (c(1), )
is an isometry. Hence, σ naturally induces a convex local geodesic bicombing σ̃
on X̃x0 .
46
IV.1. CARTAN-HADAMARD THEOREM
Consider the map r : X̃x0 × [0, 1] → X̃x0 , (c, s) 7→ (rs (c) : t 7→ c(st)). This
defines a deformation retraction of X̃x0 onto x̃0 .
A continuous path c̃ : [0, 1] → X̃x0 is a local geodesic in X̃x0 which is consistent with σ̃ if and only if exp ◦c̃ is a local geodesic in X which is consistent with
σ. Therefore, for any c ∈ X̃x0 , the map s 7→ rs (c) is the unique local geodesic
from x̃0 to c.
Finally, if (cn )n∈N is a Cauchy sequence in X̃, by completeness of X, for
every t ∈ [0, 1], the sequences (cn (t))n∈N converge in X, to c(t) say. Locally,
i.e. inside U (c(t), rc(t) ), the subsegment c|[t−,t+] is the limit of the consistent
geodesics (cn |[t−,t+] )n∈N and hence c is consistent with σ by the convexity of
the local geodesic bicombing.
The following criterion will ensure that exp is a covering map.
Lemma 1.6. Let p : X̃ → X be a map of length spaces such that
(i) X is connected,
(ii) p is a local homeomorphism,
(iii) for all rectifiable curves c̃ : [0, 1] → X̃, we have L(c̃) ≤ L(p ◦ c̃),
(iv) X has a convex local bicombing σ, and
(v) X̃ is complete.
Then p is a covering map.
Proof. The proof of Proposition I.3.28 in [BH99] also works in our setting. In
the second step, take U = U (x, rx ) and define the maps sx̃ : U (x, rx ) → X̃ by
sx̃ (y) = σ̃xy (1), where σ̃xy is the unique lift of σxy with σ̃xy (0) = x̃.
Remark. For a local isometry p, conditions (ii) and (iii) are satisfied.
Corollary 1.7. Let (X, d) be a complete, connected metric space with a convex
local geodesic bicombing σ. Then exp : X̃x0 → X is a universal covering map.
Proof. Consider the induced length metrics d¯ and D̄ on X and X̃x0 . Since
(X, d) locally is a length space, the metrics d and D locally coincide with d¯
and D̄, respectively. Hence p still is a local isometry with respect to the length
metrics and σ is a convex local bicombing. Thus Lemma 1.6 applies.
Proof of Theorem 1.2. First, we show that for all x, y ∈ X, there is a unique
consistent local geodesic from x to y. Since X is simply-connected, the covering
map exp : X̃x → X is a homeomorphism which is a local isometry and, by Lemma 1.5, there is a unique consistent local geodesic σ̃xy from x to y.
Next, we prove that σ̃xy is a geodesic. To do so, it is enough to show that,
for every curve γ : [0, 1] → X and every t ∈ [0, 1], we have
L(σ̃γ(0)γ(t) ) ≤ L(γ|[0,t] ).
47
CHAPTER IV. LOCAL TO GLOBAL
Let
n
o
A := s ∈ [0, 1] : ∀t ∈ [0, s] we have L(σ̃γ(0)γ(t) ) ≤ L(γ|[0,t] ) .
Clearly, A is non-empty and closed. To prove that A is open, consider s ∈ A.
For δ > 0 small enough, by Lemma 1.3, we have
L(σ̃γ(0)γ(s+δ) ) ≤ L(σ̃γ(0)γ(s) ) + d(γ(s), γ(s + δ))
≤ L(γ|[0,s] ) + L(γ|[s,s+δ] ) = L(γ|[0,s+δ] ).
Hence, A = [0, 1] as desired.
Finally, we show that t 7→ d(σ̃xy (t), σ̃x̄ȳ (t)) is convex. By Lemma 1.3, there
is a sequence 0 = t1 < . . . < tn = 1 and k > 0 such that
• the balls U (σ̃xx̄ (t1 ), 1 ), . . . , U (σ̃xx̄ (tn ), n ) cover σ̃xx̄ ,
• the balls U (σ̃yȳ (t1 ), 1 ), . . . , U (σ̃yȳ (tn ), n ) cover σ̃yȳ , and
• for all p, p̄ ∈ U (σ̃xx̄ (tk ), k ) and for all q, q̄ ∈ U (σ̃yȳ (tk ), k ) the map
t 7→ d(σ̃pq (t), σ̃p̄q̄ (t))
is convex.
Consider now a sequence 0 = s0 < s1 < . . . < sn = 1 with
σ̃xx̄ (sk ) ∈ U (σ̃xx̄ (tk ), k ) ∩ U (σ̃xx̄ (tk+1 ), k+1 ),
σ̃yȳ (sk ) ∈ U (σ̃yȳ (tk ), k ) ∩ U (σ̃yȳ (tk+1 ), k+1 ),
for k = 1, . . . , n − 1. Then we get
d(σ̃xy (t), σ̃x̄ȳ (t))
n
X
≤
d(σ̃σ̃xx̄ (sk−1 )σ̃yȳ (sk−1 ) (t), σ̃σ̃xx̄ (sk )σ̃yȳ (sk ) (t))
k=1
≤ (1 − t)
n
X
!
d(σ̃xx̄ (sk−1 ), σ̃xx̄ (sk ))
k=1
+t
n
X
!
d(σ̃yȳ (sk−1 ), σ̃yȳ (sk ))
k=1
= (1 − t)d(x, x̄) + td(y, ȳ).
Hence, σ̃ is a convex geodesic bicombing on X.
∗ (t) := σ̃ (1 − t) also defines a convex geodesic
If σ is reversible, then σ̃xy
yx
bicombing on X which is consistent with σ. Therefore, by uniqueness, σ̃ ∗ and
σ̃ coincide, i.e. σ̃ is reversible.
48
IV.2. LOCALLY HYPERCONVEX METRIC SPACES
IV.2
Locally Hyperconvex Metric Spaces
Hyperconvex metric spaces are an important example of metric spaces with a
convex geodesic bicombing. In fact, D. Descombes and U. Lang showed that
every proper, hyperconvex metric space of finite combinatorial dimension admits a unique convex geodesic bicombing, which is consistent and reversible,
see [DL15]. Such spaces occur, for instance, as injective hulls of hyperbolic
groups [Lan13, Theorem 1.4] and therefore, every hyperbolic group acts properly and cocompactly by isometries on a space with a reversible, consistent,
convex geodesic bicombing [DL15, Theorem 1.3].
Recall that every hyperconvex metric space is complete, geodesic and contractible. Now, knowing that under the above conditions hyperconvex metric
spaces possess a consistent, convex geodesic bicombing, we deduce the following
local-to-global theorem for hyperconvex metric spaces.
Theorem 2.1. Let X be a complete, locally compact, simply-connected, locally
hyperconvex length space with locally finite combinatorial dimension. Then X
is a hyperconvex metric space.
A metric space X is locally hyperconvex if, for every x ∈ X, there is some
rx > 0 such that B(x, rx ) is hyperconvex. If we can choose rx = r for all x ∈ X,
we call X uniformly locally hyperconvex.
Lemma 2.2. Let X be a metric space with the property that every closed ball
B(x, r) is hyperconvex, then X is itself hyperconvex.
Proof. Let {B(xi , ri )}i∈I be a family of closed balls with d(xi , xj ) ≤ ri + rj .
Fix some i0 ∈ I and set Ai := B(xi , ri ) ∩ B(xi0 , ri0 ). Since, for r big enough,
we have xi , xj ∈ B(xi0 , r), we get that the Ai ’s are externally hyperconvex in
Ai0 and Ai ∩ Aj 6= ∅ for all i, j ∈ I. Hence, it follows
\
B(xi , ri ) =
i∈I
\
Ai 6= ∅
i∈I
by Proposition II.2.1.
Proposition 2.3. Let X be a uniformly locally hyperconvex metric space with
a reversible, conical geodesic bicombing σ. Then X is hyperconvex.
Proof. Consider the following property:
P(R): For every family
T {B(xi , ri )}i∈I with d(xi , xj ) ≤ ri + rj and ri ≤ R, there
is some x ∈ i∈I B(xi , ri ).
Since X is uniformly locally hyperconvex, this clearly holds for some R0 > 0.
Next, we show P(R) ⇒ P(2R) and therefore P(R) holds for any R ≥ 0.
Let {B(xi , ri )}i∈I be a family of closed balls with d(xi , xj ) ≤ ri + rj and
ri ≤ 2R. For i, j ∈ I, define yij := σxi xj ( 21 ). By convexity of σ, we have
d(yij , yik ) = d(σxi xj ( 21 ), σxi xk ( 12 )) ≤ 12 d(xj , xk ) ≤
49
rj
2
+
rk
2 .
CHAPTER IV. LOCAL TO GLOBAL
Hence, for every i ∈ I, there is some
\
r
zi ∈
B(yij , 2j ).
j∈I
Now, observe that
d(zi , zj ) ≤ d(zi , yij ) + d(yij , zj ) ≤
rj
2
+
ri
2
=
ri
2
+
rj
2
and therefore, we find
x∈
\
B(zi , r2i ) ⊂
i∈I
\
B(xi , ri ).
i∈I
Since all balls with center in B(x, r) and radius larger than 2r contain
B(x, r), P(R) for R = 2r implies that B(x, r) is hyperconvex. Hence, by Lemma 2.2, the metric space X is hyperconvex.
Since compact, locally hyperconvex metric spaces are always uniformly locally hyperconvex, we conclude the following.
Corollary 2.4. Let X be a compact, locally hyperconvex metric space with a
reversible, conical geodesic bicombing σ. Then X is hyperconvex.
Corollary 2.5. Let X be a proper, locally hyperconvex metric space with a
reversible, conical geodesic bicombing σ. Then X is hyperconvex.
Proof. Let {B(xi , ri )}i∈I be a family of balls with d(xi , xj ) ≤ ri + rj . Fix some
i0 ∈ I and define In := {i ∈ I : d(xi , xi0 ) ≤ n} for n ∈ N. Since B(xi0 , n) is
compact, by the previous corollary, there is some
\
yn ∈
B(xi , ri ).
i∈In
Especially, (yn )n∈N ⊂ B(xi0 , ri0 ) and hence, there is some converging subsequence
\
ynk → y ∈
B(xi , ri ).
i∈I
Remark. In [Lan13], U. Lang proves that every hyperconvex metric space admits
a reversible, conical geodesic bicombing (Proposition 3.8). Therefore, we get
the following equivalence statement: A metric space is hyperconvex if and only
if it is uniformly locally hyperconvex and admits a reversible, conical geodesic
bicombing.
If a hyperconvex metric space X is proper, it also admits a convex geodesic
bicombing [DL15, Theorem 1.1] and if X has finite combinatorial dimension,
this convex geodesic bicombing is consistent, reversible and unique [DL15, Theorem 1.2]. The combinatorial dimension of a metric space X was introduced by
50
IV.3. ABSOLUTE 1-LIPSCHITZ NEIGHBORHOOD RETRACTS
A. Dress [Dre84] and is given by the supremum of the dimensions of the polyhedral complexes E(Y ) that arise as the tight span of finite subsets Y ⊂ X.
For instance, every finite dimensional hyperconvex metric space also has finite
combinatorial dimension as E(Y ) embeds isometrically into X.
Proposition 2.6. Every proper, hyperconvex metric space with finite combinatorial dimension admits a unique reversible, convex geodesic bicombing.
Recall that, by the Hopf-Rinow Theorem, any complete, locally compact
length space is proper.
Corollary 2.7. Let X be a locally compact, locally hyperconvex metric space
with locally finite combinatorial dimension. Then X admits a reversible, convex
local geodesic bicombing.
Proof. For every x ∈ X, there is some rx > 0 such that B(x, 3rx ) is compact, hyperconvex and has finite combinatorial dimension. This also holds for
B(x, rx ) and therefore, there is a reversible, convex geodesic bicombing σ x on
B(x, rx ), i.e.
σ x : B(x, rx ) × B(x, rx ) × [0, 1] → B(x, rx ).
We will check that for B(x, rx ) and B(y, ry ) with B(x, rx ) ∩ B(y, ry ) 6= ∅ the
two geodesic bicombings σ x , σ y coincide on the intersection. Assume without
loss of generality that rx ≥ ry and hence B(x, rx ), B(y, ry ) ⊂ B(x, 3rx ). Then
the convex geodesic bicombing τ on B(x, 3rx ) restricts to both B(x, rx ) and
B(y, ry ) since, for p, q ∈ B(z, rz ), we have
d(z, τpq (t)) ≤ (1 − t)d(z, p) + td(z, q) ≤ rz .
Hence, by uniqueness, the geodesic bicombings σ x , σ y are both restrictions of τ
and thus coincide on B(x, rx ) ∩ B(y, ry ).
Therefore σ, defined by
σ|B(x,rx )×B(x,rx ) := σ x |B(x,rx )×B(x,rx ) ,
is a reversible, convex local geodesic bicombing on X.
Proof of Theorem 2.1. Let X be a complete, locally compact, simply-connected, locally hyperconvex length space with locally finite combinatorial dimension. By Corollary 2.7, X has a reversible, convex local geodesic bicombing,
which induces a reversible, convex geodesic bicombing by Theorem 1.2. Hence,
we can apply Corollary 2.5 and deduce that X is hyperconvex.
IV.3
Absolute 1-Lipschitz Neighborhood Retracts
It is well known that hyperconvex metric spaces are the same as absolute 1-Lipschitz retracts. For Lipschitz retracts, the weaker notion of absolute Lipschitz
uniform neighborhood retracts is common; e.g. see [HJ14]. Absolute 1-Lipschitz
uniform neighborhood retracts are locally hyperconvex but the converse is not
true as we can see in Example 3.3. In fact, it turns out that the following holds.
51
CHAPTER IV. LOCAL TO GLOBAL
Theorem 3.1. Let X be a locally compact absolute 1-Lipschitz uniform neighborhood retract with locally finite combinatorial dimension. Then X is an absolute 1-Lipschitz retract.
A metric space X is an absolute 1-Lipschitz neighborhood retract if, for every
metric space Y with X ⊂ Y , there is some neighborhood U of X in Y and a
1-Lipschitz retraction ρ : U → X. Furthermore, if we can take U = U (X, ) for
some > 0, we call X an absolute 1-Lipschitz uniform neighborhood retract. In
this case, can be chosen independent of Y ; see [HJ14, Proposition 7.78].
Lemma 3.2. Let X be an absolute 1-Lipschitz (uniform) neighborhood retract.
Then X is (uniformly) locally hyperconvex.
Proof. Consider X ⊂ l∞ (X). Since X is an absolute 1-Lipschitz neighborhood
retract, there is some neighborhood U of X and a 1-Lipschitz retraction ρ : U →
X. For x ∈ X, there is some rx > 0 such that B(x, rx ) ⊂ U . Let now
{B(xi , ri )}i∈I be a family of closed balls with xi ∈ B(x, rx ) ∩ X and d(xi , xj ) ≤
ri + rj . Then, since l∞ (X) is hyperconvex, there is some
y ∈ B(x, rx ) ∩
\
B(xi , ri ) ⊂ U.
i∈I
Hence, we have
ρ(y) ∈ B(x, rx ) ∩
\
B(xi , ri ) ∩ X 6= ∅
i∈I
and therefore B(x, rx ) ∩ X is hyperconvex.
If X is an absolute 1-Lipschitz uniform neighborhood retract, we have that
U = U (X, ) for some > 0 and therefore, we can choose rx = 2 for all
x ∈ X.
The converse is not true, as the following example shows.
Example 3.3. Consider the unit sphere S 1 endowed with the inner metric.
Since, for every x ∈ S 1 and ∈ (0, π2 ], the ball B(x, ) is isometric to the
interval [−, ], the unit sphere S 1 is uniformly locally hyperconvex.
But S 1 is not an absolute 1-Lipschitz neighborhood retract. Fix some inclusion S 1 ⊂ l∞ (S 1 ). We choose three points x, y, z ∈ S 1 with
r := d(x, y) = d(x, z) = d(y, z) =
2π
3 .
Let U be a neighborhood of S 1 in l∞ (S 1 ). As U is open, there is some ∈ (0, 2r )
such that B(x, ) ⊂ U . By hyperconvexity of l∞ (S 1 ), there is some
p ∈ B(x, ) ∩ B(y, r − ) ∩ B(z, r − ) ⊂ U.
But since
B(x, ) ∩ B(y, r − ) ∩ B(z, r − ) ∩ S 1 = ∅,
there is no 1-Lipschitz retraction ρ : S 1 ∪ {p} → S 1 .
52
IV.3. ABSOLUTE 1-LIPSCHITZ NEIGHBORHOOD RETRACTS
In fact, the notion of an absolute 1-Lipschitz uniform neighborhood retract
is quite restrictive.
Lemma 3.4. Let X be an absolute 1-Lipschitz uniform neighborhood retract.
Then, X is
(i) complete,
(ii) geodesic, especially a length space, and
(iii) simply-connected.
Proof. Fix some inclusion X ⊂ l∞ (X) and r = 2 > 0 such that there is a
1-Lipschitz retraction ρ : U (X, ) → X.
First, if (xn )n∈N is a Cauchy sequence in X it converges to some x ∈ U (X, ).
It follows that x = ρ(x) ∈ X.
Next, assume that there is a geodesic in X between points at distance less
than d. By Lemma 3.2, this is clearly true for d = r. Consider two points
x, y ∈ X with d(x, y) ≤ d + r. Now, since l∞ (X) is geodesic, there is some
z ∈ l∞ (X) with d(x, y) = d(x, z)+d(z, y), d(x, z) ≤ r and d(z, y) ≤ d. But then
we have ρ(z) ∈ X with d(x, y) = d(x, ρ(z)) + d(ρ(z), y) and, by our hypothesis,
there are geodesics from x to ρ(z) and from ρ(z) to y, which combine to a
geodesic from x to y.
Finally, since X is locally simply-connected, every curve is homotopic to a
curve of finite length and hence it is enough to consider loops of finite length.
We show that every such loop in X is contractible.
Let γ be a loop in X of length L(γ) = 2πR with R > r. We denote by
2 := {x ∈ R3 : |x| = R} the sphere of radius R endowed with the inner metric
SR
2 : 0 ≤ x ≤ R sin( r )} be the region bounded by the
and let A := {x ∈ SR
3
R
2 : x = 0} and c0 := {x ∈ S 2 : x = R sin( r )}. Let
two circles c := {x ∈ SR
3
3
R
R
f : c → X be a parametrization of γ by arclength and let f¯: A → l∞ (X) be a
1-Lipschitz extension. Then γ 0 := ρ ◦ f¯(c0 ) is a loop of length L(γ 0 ) ≤ L(c0 ) =
2πR0 with R0 := R cos( Rr ), which is homotopic to γ. Since cos( Rr0 ) ≤ cos( Rr ),
we find inductively a loop γn with L(γn ) ≤ 2πR cos( Rr )n , which is homotopic
to γ.
If L(γ) = 2πR with R ≤ r, we can use the same argument with A replaced
by the upper halfsphere of radius R to show that γ is contractible.
We conclude that an absolute 1-Lipschitz uniform neighborhood retract is
a complete, simply-connected, locally hyperconvex length space and therefore
Theorem 3.1 follows directly from Theorem 2.1.
53
Chapter V
Finite Hyperconvexity
V.1
Basic Properties
We denote the collection of all n-hyperconvex, externally n-hyperconvex and
weakly externally n-hyperconvex subsets of X by Hn (X), En (X) and Wn (X),
respectively. It holds that En (X) ⊂ Wn (X) ⊂ Hn (X) and Hn+1 (X) ⊂ Hn (X).
Remark. Obviously, every metric space is 1-hyperconvex. A subset A of X is
(weakly) externally 1-hyperconvex if and only if it is proximinal , i.e. for every
x ∈ X, there is some a ∈ A with d(x, a) = d(x, A). Recall that proximinal
subsets are closed. Furthermore, a metric space X is 2-hyperconvex if and only
if it is metrically convex , i.e. for every x0 , x1 ∈ X and every t ∈ [0, 1], there
is some xt ∈ X with d(x0 , xt ) = td(x0 , x1 ) and d(xt , x1 ) = (1 − t)d(x0 , x1 ),
compare [BL00, Definition 1.3].
A metric space (X, d) is called modular if, for all x, y, z ∈ X, the median
set
M (x, y, z) := I(x, y) ∩ I(y, z) ∩ I(z, x)
is non-empty, where I(x, y) := {z ∈ X : d(x, y) = d(x, z) + d(z, y)} denotes the
metric interval between x and y.
Proposition 1.1. A metric space X is 3-hyperconvex if and only if it is metrically convex and modular.
Proof. First, assume that X is 3-hyperconvex. Recall the Gromov product
(y|z)x := 12 (d(x, y) + d(x, z) − d(y, z)) for x, y, z ∈ X. Let x1 , x2 , x3 ∈ X and
define ri := (xj |xk )i for {i, j, k} = {1, 2, 3}. Then we have
M (x1 , x2 , x3 ) =
3
\
B(xi , ri ) 6= ∅.
i=1
Conversely, let X be metrically convex and modular. Let
T3 x1 , x2 , x3 ∈ X and
r1 , r2 , r3 ∈ R with d(xi , xj ) ≤ ri + rj . If M (x1 , x2 , x3 ) ∩ i=1 B(xi , ri ) = ∅, we
might assume without loss of generality that r3 < (x1 |x2 )x3 . We may then take
m ∈ M (x1 , x2 , x3 ) and define rm := min{r1 − d(x1 , m), r2T− d(x2 , m)}. Since X
is metrically convex, we get ∅ =
6 B(m, rm ) ∩ B(x3 , r3 ) ⊂ 3i=1 B(xi , ri ).
55
CHAPTER V. FINITE HYPERCONVEXITY
V.2
Almost n-Hyperconvex Metric Spaces
We now turn our attention to almost n-hyperconvex metric spaces and eventually prove the following.
Theorem 2.1. Let X be a complete, almost n-hyperconvex metric space for
n ≥ 3. Then X is n-hyperconvex.
A subset A of a metric space X is called almost externally n-hyperconvex in
X if, for every family {B(xi , ri )}ni=1 of closed balls with d(xi , xj ) ≤ ri + rj and
d(xi , A) ≤ ri , we have
n
\
B(xi , ri + ) ∩ A 6= ∅
i=1
for every > 0. We denote the set of all closed, almost externally n-hyperconvex
subsets of X by Ẽn (X).
Similarly, we say that a subset A of a metric space X is almost weakly
externally n-hyperconvex in X if it is almost externally n-hyperconvex in A∪{x}
for every x ∈ X. We denote all such subsets which are closed by W̃n (X).
N. Aronszajn and P. Panitchpakdi already showed that every complete,
almost (n+1)-hyperconvex metric space is n-hyperconvex [AP56, Theorem 3.4].
This result easily extends to almost (weakly) externally hyperconvex subsets.
Lemma 2.2. Let X be a complete metric space. Then we have
(i) A ∈ Ẽn+1 (X) ⇒ A ∈ En (X),
(ii) A ∈ W̃n+1 (X) ⇒ A ∈ Wn (X), and
(iii) if X is almost (n + 1)-hyperconvex, then X is n-hyperconvex.
Proof. We will prove (i). The other statements then follow since
• A ∈ Wn (X) ⇔ A ∈ En (A ∪ {x}) for all x ∈ X, and
• A is n-hyperconvex ⇔ A ∈ En (A).
Let {B(xi , ri )}ni=1 be a family
with d(xi , xj ) ≤ ri + rj and
Tnof closed balls
1
d(xi , A) ≤ ri . Starting with y1 ∈ i=1 B(xi , ri + 2 )∩A, we construct inductively
a sequence (yk )k by choosing
yk+1 ∈ B(yk , 21k +
1
)∩
2k+1
n
\
B(xi , ri +
1
)
2k+1
∩ A.
i=1
1
Since d(yk , yk+1 ) ≤ 21k + 2k+1
, this sequence is Cauchy and converges to some
y ∈ A with
d(xi , y) = lim d(xi , yk ) ≤ lim ri + 21k = ri .
k→∞
That is y ∈
Tn
i=1 B(xi , ri )
k→∞
∩ A 6= ∅.
56
V.2. ALMOST n-HYPERCONVEX METRIC SPACES
Lemma 2.3. Let X be a complete metric space, A ∈ Ẽn (X) and k ≤ n − 2.
Then, for every family {B(xi , ri )}ki=1 with d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri ,
T
we have A0 := A ∩ ki=1 B(xi , ri ) ∈ Ẽn−k (X).
Proof. Let {B(xi , ri )}ni=k+1 be a family of closed balls with d(xT
i , xj ) ≤ ri + rj
and d(xi , A0 ) ≤ ri . Fix some > 0. Then, there is some y ∈ A ∩ ni=1 B(xi , ri +
2 ) and by Lemma 2.2, we get
∅=
6 A∩
k
\
B(xi , ri ) ∩ B(y,
2)
0
⊂A ∩
i=1
n
\
B(xi , ri + ).
i=k+1
Corollary 2.4. Let X be a complete, almost n-hyperconvex metric space and
let k ≤ n − 2. Then for every family {B(xi , ri )}ki=1 with d(xi , xj ) ≤ ri + rj , we
T
have B := ki=1 B(xi , ri ) ∈ Ẽn−k (X).
By investigating the proof of Lemma II.2.2, we see that the requirements
there can be weakened as follows.
Lemma 2.5. Let X be a complete, almost 3-hyperconvex metric space. Let
A, A0 ∈ Ẽ2 (X) with y ∈ A ∩ A0 6= ∅ and x ∈ X with d(x, A), d(x, A0 ) ≤ r.
Denote d := d(x, y) and s := d − r. Then, for every > 0 and every δ > 0, we
have
A ∩ A0 ∩ B(x, r + δ) ∩ B(y, s + ) 6= ∅,
given s ≥ 0. In any case, the intersection A ∩ A0 ∩ B(x, r + δ) is non-empty.
Proof. We will split the proof into three steps.
Step I. For all > 0 and δ > 0, there are a ∈ Cδ := B(x, r + δ) ∩ A and
a0 ∈ Cδ0 := B(x, r + δ) ∩ A0 such that d(a, a0 ) ≤ and d(y, a) ≤ s + .
Let 0 ≤ ˜ ≤ 2 , n0 :=
s
˜
and 0 < δ ≤
˜
n0 +1 .
We start by choosing
a1 ∈ B(y, ˜ + 2δ ) ∩ B(x, d − ˜ + 2δ ) ∩ A,
a2 ∈ B(a1 , ˜ +
δ
2
+ 4δ ) ∩ B(x, d − 2˜
+
δ
2
+ 4δ ) ∩ A0 .
Then, as long as n ≤ n0 , we can inductively pick
an ∈ B(an−1 , ˜ + ∆n ) ∩ B(x, d − n˜
+ ∆n ) ∩ A
if n is odd, and
an ∈ B(an−1 , ˜ + ∆n ) ∩ B(x, d − n˜
+ ∆ n ) ∩ A0
if n is even, where ∆n :=
Pn
δ
i=1 2i
≤ δ. Observe that we have d(y, an ) ≤ n(˜
+δ).
57
CHAPTER V. FINITE HYPERCONVEXITY
Finally, assuming without loss of generality an0 ∈ A0 (otherwise interchange
A and A0 ), there are
a ∈ B(an0 , ˜ + ∆n0 +1 ) ∩ B(x, r + ∆n0 +1 ) ∩ A ⊂ Cδ
and
a0 ∈ B(a, ˜ + ∆n0 +2 ) ∩ B(x, r + ∆n0 +2 ) ∩ A0 ⊂ Cδ0 .
We have d(a, a0 ) ≤ ˜+δ ≤ and d(a, y) ≤ (n0 +1)(˜
+δ) ≤ s+˜
+(n0 +1)δ ≤ s+.
Step II. For all > 0, δ > 0 and all n ≥ 1, there are an ∈ A and a0n ∈ A0 with
• d(an , a0n ) ≤
2n ,
• d(an−1 , an ), d(a0n−1 , a0n ) ≤
2+δ
2n ,
P
• d(x, an ), d(x, a0n ) ≤ r + ∆n , where ∆n := ni=1 2δi , and
P
• d(y, an ), d(y, a0n ) ≤ s + En , where En := ni=1 2i .
We start by choosing
a1 ∈ A ∩ B(x, r + 2δ ),
a01 ∈ A0 ∩ B(x, r + 2δ ),
with d(a1 , a01 ) ≤ 2 and d(y, a1 ) ≤ s + 2 according to Step I.
We then continue inductively as follows. First, since X is almost 3-hyperconvex, we can pick some
xn ∈ B(an−1 , 2n +
We denote rn :=
2n
δ
)
2n+2
+
∩ B(a0n−1 , 2n +
δ
2n+2
δ
)
2n+2
∩ B(x, r + ∆n−1 −
2n
+
δ
).
2n+2
and sn := s + En−1 . Observe that
d(xn , y) ≤ d(xn , an−1 ) + d(an−1 , y) ≤ rn + sn .
Now, by Step I, there are
δ
),
2n+1
δ
a0n ∈ A0 ∩ B(xn , rn + 2n+1
),
d(y, an ) ≤ sn + 2n = s + En .
an ∈ A ∩ B(xn , rn +
with d(an , a0n ) ≤
2n
and
Moreover, we have
d(an−1 , an ) ≤ d(an−1 , xn ) + d(xn , an ) ≤ 2rn +
δ
2n+1
=
2+δ
2n ,
d(x, an ) ≤ d(x, xn ) + d(xn , an )
≤ r + ∆n−1 −
2n
+
δ
2n+2
+ rn +
δ
2n+1
= r + ∆n ,
as desired.
Step III. Observe that the sequences (an )n∈N , (a0n )n∈N are Cauchy and since
the distance d(an , a0n ) → 0, they have a common limit point
a ∈ A ∩ A0 ∩ B(y, s + ) ∩ B(x, r + δ) 6= ∅.
This concludes the proof.
58
V.2. ALMOST n-HYPERCONVEX METRIC SPACES
Lemma 2.6. Let X be a complete, almost 3-hyperconvex metric space and let
A0 ∈ Ẽ3 (X) and A1 , A2 ∈ Ẽ2 (X) be pairwise intersecting subsets. Then
A0 ∩ A1 ∩ A2 6= ∅.
Proof. Choose a point x0 ∈ A1 ∩ A2 and set r0 := d(x0 , A0 ). By Lemma 2.5,
there is
r0
y0 ∈ A0 ∩ A1 ∩ B(x0 , r0 + 12
).
Let A00 := A0 ∩ B(y0 , 76 r) ∈ Ẽ2 (X). Using again Lemma 2.5, we have
13
A00 ∩ A2 = A0 ∩ A2 ∩ B(y0 , 12
r0 +
r0
12 )
6= ∅
and therefore, there is some
z0 ∈ A00 ∩ A2 ∩ B(x0 , 13
12 r0 +
r0
12 )
= A0 ∩ A2 ∩ B(x0 , 67 r0 ) ∩ B(y0 , 76 r0 ).
Then, since A0 is almost externally 3-hyperconvex, there is some
x̄0 ∈ B(x0 , r0 +
r0
12 )
7
∩ B(y0 , 12
r0 +
r0
12 )
7
∩ B(z0 , 12
r0 +
r0
12 )
∩ A0
and using again Lemma 2.5, we find
x1 ∈ A1 ∩ A2 ∩ B(x̄0 , 23 r0 +
r0
12 )
5
∩ B(x0 , 12
r0 +
r0
12 ).
Note that d(x1 , A0 ) ≤ d(x1 , x̄0 ) ≤ 43 r0 =: r1 and d(x0 , x1 ) ≤ 21 r0 . Proceeding
n
this way, we get some
sequence (xn )n∈N ⊂ A1 ∩A2 with d(xn , A0 ) ≤ 43 r0 and
n
d(xn , xn+1 ) ≤ 21 34 r0 . Hence, (xn )n∈N is a Cauchy sequence and therefore
converges to some x ∈ A0 ∩ A1 ∩ A2 6= ∅ since A0 is closed.
We are now able to give a proof of Theorem 2.1. In fact, we will show the
following more general result.
Proposition 2.7. Let X be a complete metric space. Then, for every n ≥ 3,
we have
(i) A ∈ Ẽn (X) ⇒ A ∈ En (X),
(ii) A ∈ W̃n (X) ⇒ A ∈ Wn (X), and
(iii) if X is an almost n-hyperconvex metric space, then X is n-hyperconvex.
Proof. First note that (ii) and (iii) directly follow from (i) by the same argument
as in Lemma 2.2. To prove (i), we now consider two cases.
Case 1. Let A ∈ Ẽ3 (X) and let {B(xi , ri )}3i=1 be a family of closed balls with
d(xi , xj ) ≤ ri + rj . By Lemma 2.2, A is externally 2-hyperconvex and hence,
59
CHAPTER V. FINITE HYPERCONVEXITY
there is some y0 ∈ A ∩ B(x1 , r1 ) ∩ B(x2 , r2 ). Set s := d(y0 , x3 ) − r3 . By Lemma 2.3, we have Ai := A ∩ B(xi , ri ) ∈ Ẽ2 (X) and d(x3 , Ai ) ≤ r3 for i = 1, 2,
since A ∩ B(xi , ri ) ∩ B(x3 , r3 ) 6= ∅. By Lemma 2.5, we therefore find inductively
y1 ∈ A1 ∩ A2 ∩ B(x3 , r3 + 2 ) ∩ B(y0 , s + ),
y2 ∈ A1 ∩ A2 ∩ B(x3 , r3 + 4 ) ∩ B(y1 , 2 + 2 ),
..
.
yn ∈ A1 ∩ A2 ∩ B(x3 , r3 +
2n )
∩ B(yn−1 , 2n−1
+
).
2n−1
This is a Cauchy sequence which converges to some
y ∈ A1 ∩ A2 ∩ B(x3 , r3 ) = A ∩
3
\
B(xi , ri ) 6= ∅.
i=1
Case 2. Assume now that n ≥ 4 and let {B(xi , ri )}ni=1 be aTfamily of closed
balls with d(xi , xj ) ≤ ri + rj , d(xi , A) ≤ ri . Then B := A ∩ nj=3 B(xj , rj ) ∈
Ẽ2 (X) by Lemma 2.3. Moreover, for i = 1, 2, we haveTAi := A ∩ B(xi , ri ) ∈
Ẽ3 (X) by Lemma 2.3 and Ai ∩ B = A ∩ B(xi , ri ) ∩ nj=3 B(xj , rj ) 6= ∅ by
Lemma 2.2. Furthermore, observe that A1 , A2 , B ⊂ A and A is almost nhyperconvex. Hence, by Lemma 2.6, we get
A∩
n
\
B(xi , ri ) = A1 ∩ A2 ∩ B 6= ∅,
i=1
as desired.
This proves Theorem 2.1 and as a consequence, we get the following corollary.
Corollary 2.8. Let X be an n-hyperconvex metric space for n ≥ 3. Then its
metric completion is n-hyperconvex as well.
According to the previous propositions and the following lemma, in a complete, 4-hyperconvex metric space X, there is no difference between Ẽn (X) and
En (X) for n ≥ 2.
Lemma 2.9. Let X be a complete, 4-hyperconvex metric space. If A ∈ Ẽ2 (X)
then we have A ∈ E2 (X).
Proof. Let {B(xi , ri )}2i=1 with d(x1 , x2 ) ≤ r1 + r2 and d(xi , A) ≤ ri . Then we
have B(xi , ri ) ∈ Ẽ3 (X) by Lemma 2.3 and A ∩ B(xi , ri ) 6= ∅ by Lemma 2.2.
Therefore, we get B(x1 , r1 ) ∩ B(x2 , r2 ) ∩ A 6= ∅ by Lemma 2.6.
60
V.3. 4-HYPERCONVEX METRIC SPACES
V.3
4-Hyperconvex Metric Spaces
Let us now turn to 4-hyperconvex metric spaces. We will prove the following
theorem step by step.
Theorem 3.1. Let X be a complete metric space and let A ⊂ X be an arbitrarily chosen non-empty subset. Then, the following hold:
(i) X is 4-hyperconvex if and only if X is n-hyperconvex for every n ∈ N.
(ii) A is externally 4-hyperconvex in X if and only if A is externally n-hyperconvex in X for every n ∈ N.
(iii) A is weakly externally 4-hyperconvex in X if and only if A is weakly externally n-hyperconvex in X for every n ∈ N.
Lemma 3.2. Let X be a complete, 4-hyperconvex metric space. If A0 ∈ E3 (X),
A1 ∈ E2 (X) and A0 ∩ A1 6= ∅, then
A0 ∩ A1 ∈ E2 (X).
Proof. Let B(x1 , r1 ), B(x2 , r2 ) be closed balls such that d(x1 , x2 ) ≤ r1 + r2 and
d(xi , A0 ∩ A1 ) ≤ ri . Define A2 := B(x1 , r1 ) ∩ B(x2 , r2 ) ∈ E2 (X). Since, for
k = 0, 1, the sets Ak are externally 2-hyperconvex, we have
A2 ∩ Ak = B(x1 , r1 ) ∩ B(x2 , r2 ) ∩ Ak 6= ∅.
Therefore, we get
A0 ∩ A1 ∩ B(x1 , r1 ) ∩ B(x2 , r2 ) = A0 ∩ A1 ∩ A2 6= ∅
by Lemma 2.6.
Proposition 3.3. Let X be a complete, 4-hyperconvex metric space. Then
(i) X is n-hyperconvex for every n ∈ N,
(ii) if A ∈ E2 (X), we have A ∈ En (X) for every n ∈ N, and
(iii) T
if A1 , . . . , Am ∈ E2 (X) with Ai ∩ Aj 6= ∅ for all i, j ∈ {1, . . . , m}, then
m
i=1 Ai ∈ En (X) for every n ∈ N.
Proof. In order to show (i), we will prove by induction that the following claim
is true.
n
Claim. For
Tn every family of closed balls {B(xi , ri )}i=1 with d(xi , xj ) ≤ ri + rj ,
we have i=1 B(xi , ri ) ∈ E2 (X).
This clearly holds for n = 2. For n ≥ 2, consider {B(xi , ri )}n+1
i=1 with
d(xi , xj ) ≤ ri + rj . Observe that B(x1 , r1 ), B(x2 , r2 ) ∈ E3 (X) and, by the
61
CHAPTER V. FINITE HYPERCONVEXITY
T
induction hypothesis, A := n+1
i=3 B(xi , ri ) ∈ E2 (X) and B(xi , ri ) ∩ A 6= ∅ for
i = 1, 2. Hence, by Lemma 2.6, we get
n+1
\
B(xi , ri ) = B(x1 , r1 ) ∩ B(x2 , r2 ) ∩ A 6= ∅.
i=1
Again by the induction hypothesis, we have A0 :=
therefore, we conclude that
n+1
\
Tn
i=1 B(xi , ri )
∈ E2 (X) and
B(xi , ri ) = A0 ∩ B(xn+1 , rn+1 ) ∈ E2 (X)
i=1
by Lemma 3.2.
For (ii), we also do induction on n. Let {B(xi , ri )}n+1
a collection of
i=1 be T
balls with d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri . We have A0 := ni=1 B(xi , ri ),
A1 := B(xn+1 , rn+1 ) ∈ E3 (X) by (i) and A0 ∩A 6= ∅ by the induction hypothesis.
Hence, we get
n+1
\
B(xi , ri ) ∩ A = A0 ∩ A1 ∩ A 6= ∅
i=1
by Lemma 2.6.
Finally, statement (iii) is a direct consequence of (ii), Lemma 2.6 and Lemma 3.2.
Proof of Theorem 3.1. Statement (i) was shown in Proposition 3.3(i). Thus,
we will start proving (ii). Let A ∈ E4 (X) and let {B(xi , ri )}ni=1 be a collection
of balls with d(xi , xj ) ≤ ri + rj and d(xi , A) ≤ ri . Define Ai := B(xi , ri ) ∩ A ∈
E3 (A). We have
Ai ∩ Aj = B(xi , ri ) ∩ B(xj , rj ) ∩ A 6= ∅
and therefore, we get
n
\
B(xi , ri ) ∩ A =
i=1
n
\
Ai 6= ∅
i=1
by Proposition 3.3(iii). This is A ∈ En (X).
Finally, let A ∈ W4 (X) and let {B(xi , ri )}n−1
i=1 be a collection of balls in A
with d(xi , xj ) ≤ ri + rj and let x ∈ X, r ≥ 0 with d(x, A) ≤ r, d(x, xi ) ≤ r + ri .
Then, we have A0 := A ∩ B(x, r) ∈ E3 (A) and therefore A0 ∈ En−1 (A) by
Proposition 3.3(ii). Moreover, B(xi , ri ) ∩ A0 6= ∅ and therefore
n−1
\
B(xi , ri ) ∩ B(x, r) ∩ A =
i=1
n−1
\
B(xi , ri ) ∩ A0 6= ∅.
i=1
This proves A ∈ Wn (X) and hence establishes (iii).
Proposition 3.4. Let X be a complete, 4-hyperconvex metric space and let
A ∈ W2 (X). Then we have A ∈ Wn (X) for every n ∈ N.
62
V.3. 4-HYPERCONVEX METRIC SPACES
Proof. It is enough to show that A ∈ Wn (X) ⇒ A ∈ Wn+1 (X) for n ≥ 2.
For x ∈ X, let {x0 , . . . , xn } ⊂ A ∪ {x} be a set of points with d(xi , xj ) ≤
ri + rj and d(xi , A) ≤ ri . Without loss of generality, we may assume that
x0 ∈ A. Define
n
\
B :=
B(xi , ri ) ∈ E2 (X).
i=1
Since A ∈ Wn (X), there is some y ∈ A ∩ B 6= ∅.
Fix > 0 and let d := d(x0 , y), s := d − r0 and n0 := s˜ . We start by
choosing
a1 ∈ B(y, ) ∩ B(x0 , d − ) ∩ A,
a2 ∈ B(a1 , ) ∩ B(x0 , d − 2) ∩ B.
Then, as long as n ≤ n0 , we can inductively pick
an ∈ B(an−1 , ) ∩ B(x0 , d − n) ∩ A
if n is odd, and
an ∈ B(an−1 , ) ∩ B(x0 , d − n) ∩ B
if n is even. Finally, assuming without loss of generality an0 ∈ B, there is some
a ∈ B(an0 , ) ∩ B(x0 , r) ∩ A.
Especially, we have
a∈A∩
n
\
B(xi , ri + ) 6= ∅
i=0
and thus A ∈ W̃n+1 (X). By Proposition 2.7, we conclude that A ∈ Wn+1 (X),
as desired.
Example 3.5. Consider c0 ⊂ l∞ (N), the subspace of all null sequences. As it is
mentioned in [Lin64], c0 is finitely hyperconvex but not hyperconvex. Moreover,
c0 is externally n-hyperconvex in l∞ (N) for every n ∈ N.
Indeed, we will show that c0 ∈ E2 (l∞ (N)) and then use Proposition 3.3(ii).
Let x := (xn )n∈N , y := (yn )n∈N ∈ l∞ (N) with d∞ (x, y) ≤ r + s, d∞ (x, c0 ) ≤ r
and d∞ (y, c0 ) ≤ s. For every n ∈ N, choose some
zn ∈ B(xn , r) ∩ B(yn , s) with |zn | = inf{|ζ| : ζ ∈ B(xn , r) ∩ B(yn , s)}.
Define z := (zn )n∈N . Clearly, we have d(xn , zn ) ≤ r and d(yn , zn ) ≤ s. Moreover, limn→∞ zn = 0, since lim supn→∞ |xn | ≤ r and lim supn→∞ |yn | ≤ s.
Hence
z ∈ B(x, r) ∩ B(y, s) ∩ c0 6= ∅.
63
Chapter VI
Convexity of Weakly
Externally Hyperconvex
Subsets
VI.1
Main Results
In this chapter, we relate weak notions of convexity of subsets of a metric space
to the notions of (weak) external hyperconvexity. The first challenge at this
point consists in defining convexity in general hyperconvex metric spaces. For
doing this, we use again the notion of geodesic bicombings. Given a metric space
X with a geodesic bicombing σ, we say that a subset A of X is σ-convex if, for
every x, y ∈ A, we have σxy ([0, 1]) ⊂ A. If X is a normed vector space with the
linear geodesic bicombing, this notion coincides with the ordinary convexity.
We will show that, under the appropriate assumptions the metric space X
and on the geodesic bicombing σ, (weakly) externally hyperconvex subsets are
σ-convex. We then combine these convexity results with local-to-global properties of (weakly) externally hyperconvex subsets with a geodesic bicombing.
A subset A of a metric space X is uniformly locally (weakly) externally
hyperconvex if there is some r > 0 such that, for every x ∈ A, the set A∩B(x, r)
is (weakly) externally hyperconvex in B(x, r).
Theorem 1.1. Let X be a hyperconvex metric space, let A ⊂ X be any subset
and let σ denote a convex geodesic bicombing on X.
(I) The following are equivalent:
(i) A is externally hyperconvex in X.
(ii) A is σ-convex and uniformly locally externally hyperconvex.
(II) If straight curves in X are unique, the following are equivalent:
(i) A is weakly externally hyperconvex and possesses a consistent, convex geodesic bicombing.
(ii) A is σ-convex and uniformly locally weakly externally hyperconvex.
65
CHAPTER VI. CONVEXITY
Note that, if X is a proper hyperconvex metric space with finite combinatorial dimension, then a convex geodesic bicombing σ exists, the extra assumption in (II) holds and a weakly externally hyperconvex subset always possesses
a consistent, convex geodesic bicombing.
Corollary 1.2. Let X be a proper, hyperconvex metric space with finite combinatorial dimension and let σ be the unique consistent, convex geodesic bicombing
on X. Then A ⊂ X is weakly externally hyperconvex in X if and only if A is
σ-convex and locally weakly externally hyperconvex in X.
Theorem 1.1 shows that there are substantial differences between hyperconvex subsets and externally hyperconvex or weakly externally hyperconvex
subsets of a metric space since hyperconvex subsets are not convex in general.
As an application, consider now the simple case of the n-dimensional normed
n := (Rn , k · k ), where k · k
space l∞
∞
∞ denotes the maximum norm. If A is a
n
hyperconvex subset of l∞ , then A does not need to be convex. However, as
n , it follows by Proposisoon as A is weakly externally 2-hyperconvex in l∞
tion V.3.4 that A is weakly externally hyperconvex and thus A must be convex
by Corollary 1.2.
This leads to the following Helly type result. A family of subsets of a given
set X is a Helly family of order k if every finite subfamily such that every k-fold
intersection is non-empty has non-empty total intersection. For instance, for a
hyperconvex metric space X, the set E(X) of externally hyperconvex subsets is
a Helly family of order 2.
n ) of weakly externally hyperconvex subsets
Theorem 1.3. The collection W(l∞
n
of l∞ is a Helly family of order n + 1. Moreover, the order is optimal.
VI.2
σ-Convexity
The goal of this section is to prove σ-convexity for (weakly) externally hyperconvex subsets and therefore establish the implications (i) ⇒ (ii) in Theorem 1.1.
Afterwards, we conclude this section with the proof of Theorem 1.3.
A curve c : [0, 1] → X is straight if, for every x ∈ X, the function t 7→
d(x, c(t)) is convex. Note that, on the one hand, straight curves are geodesics
and on the other hand, the geodesics of a convex geodesic bicombing are straight
curves. Furthermore, we say that a metric space X has unique straight curves if
for all x, y ∈ X, there is at most one straight curve c : [0, 1] → X with c(0) = x
and c(1) = y. For instance, normed vector spaces and metric spaces with finite
combinatorial dimension have unique straight curves, compare [DL15].
Proposition 2.1. Suppose that X is a metric space with a geodesic bicombing
σ such that the geodesics σxy are straight curves. Moreover, let E ∈ E2 (X).
Then E is σ-convex.
Proof. Assume, by contradiction, that E is not σ-convex. Then there are x, y ∈
E such that
σxy ([0, 1]) * E.
66
VI.2. σ-CONVEXITY
Hence, there are 0 ≤ t1 < t2 ≤ 1 such that σxy ([t1 , t2 ]) ∩ E = {x̄, ȳ} with
x̄ = σxy (t1 ) and ȳ = σxy (t2 ). Let
R := max d(σxy (t), E) > 0.
t∈[t1 ,t2 ]
Define s := min{t0 ∈ [t1 , t2 ] : d(σxy (t0 ), E) = R} and z := σxy (s). In particular,
R
we have d(E, z) = R. For any ∈ (0, 2d(x,y)
), we set z − := σxy (s − ) and
+
−
−
z := σxy (s + ). We then have R := d(z , E) < R and R+ := d(z + , E) ≤ R.
Moreover, by the choice of , we get
R− = d(z − , E) ≥ d(z, E) − d(z, z − ) = R − d(x, y) >
and similarly R+ >
R
2.
R
2
In particular, we have
d(z − , z + ) = 2d(x, y) <
R
2
+
R
2
< R− + R+
and therefore, by external 2-hyperconvexity of E, we can pick
e ∈ E ∩ B(z − , R− ) ∩ B(z + , R+ ).
Now, since the curve t 7→ σxy (t) is straight in X, it follows that
d(E, σxy (s)) ≤ d(e, σxy (s)) ≤ 21 d(e, z − ) + d(e, z + ) ≤ 12 R− + R+ < R,
which is a contradiction to the definition of s. Hence, E is σ-convex.
Proposition 2.2. Suppose that X is a metric space with unique straight curves
and let σ X be a convex geodesic bicombing on X. Moreover, we assume that
W ∈ W3 (X) and W possesses a consistent geodesic bicombing σ W . Then it
follows that σ W = σ X |W ×W ×[0,1] . In particular, W is σ X -convex.
Proof. We claim that the geodesics of σ W are straight curves in X. Consider
z ∈ X and let s := d(z, W ). For w, w0 ∈ W , let w̄, w̄0 ∈ E := B(z, s) ∩ W such
that
s + d(w̄, w) = d(z, w̄) + d(w̄, w) = d(z, w),
(2.1)
s + d(w̄0 , w0 ) = d(z, w̄0 ) + d(w̄0 , w0 ) = d(z, w0 ).
We have
W
W
W
W
d(z, σww
0 (t)) ≤ d(z, σw̄ w̄ 0 (t)) + d(σw̄ w̄ 0 (t), σww 0 (t)).
(2.2)
To bound the first term on the right-hand side of (2.2), note that E ∈ E2 (W ) and
W (t)) = s.
thus E is in particular σ W -convex by Proposition 2.1. Hence, d(z, σw̄
w̄0
On the other hand, to bound the second term on the right-hand side of (2.2),
note that, by convexity of σ W , we get
W
W
0
0
d(σw̄
w̄0 (t), σww0 (t)) ≤ (1 − t)d(w̄, w) + td(w̄ , w ).
Putting those two estimates together and using (2.1), we obtain that
W
0
d(z, σww
0 (t)) ≤ (1 − t)d(z, w) + td(z, w ).
It follows that the consistent geodesic bicombing σ W consists of straight curves
in X. Since straight curves in X are unique, σ W must coincide with the restriction σ X |W ×W ×[0,1] . Therefore, W is σ X -convex.
67
CHAPTER VI. CONVEXITY
Remark 2.3. In [Lan13], U. Lang proves that the injective hull of certain discretely geodesic metric spaces (including hyperbolic groups) has the structure of
a locally finite polyhedral complex of finite combinatorial dimension. The cells
are weakly externally hyperconvex subsets of this complex by Remark III.3.4
and from Proposition 2.2 it follows now that the unique consistent, convex
geodesic bicombing on this complex is linear inside the cells.
Lemma 2.4. Let X be a metric space with a conical geodesic bicombing σ and
let A ⊂ X be σ-convex. Then the following holds:
(i) For all x, y ∈ X and t ∈ [0, 1], we have
d(σxy (t), A) ≤ (1 − t)d(x, A) + td(y, A).
(ii) If σ is consistent, then, for all x, y ∈ X, the function t 7→ d(σxy (t), A) is
convex.
Proof. We prove (ii). For s1 , s2 ∈ [0, 1] and > 0, there are p, q ∈ A with
d(σxy (s1 ), p) ≤ d(σxy (s1 ), A) + and
d(σxy (s2 ), q) ≤ d(σxy (s2 ), A) + .
We get
d(σxy ((1 − t)s1 + ts2 ), A) ≤ d(σσxy (s1 )σxy (s2 ) (t), σpq (t))
≤ (1 − t)d(σxy (s1 ), A) + td(σxy (s2 ), A) + for all > 0. Finally, let ↓ 0.
Corollary 2.5. Suppose that X is a proper metric space of finite combinatorial
dimension which admits a consistent geodesic bicombing σ. Then, for all x, y ∈
X and W ∈ W3 (X), the function t 7→ d(σxy (t), W ) is convex.
Proof of Theorem 1.3. By Proposition 2.2, weakly externally hyperconvex subn are convex in the usual sense. By the classical Helly theorem, it follows
sets of l∞
n ) is a Helly family of order n + 1. It thus remains to be proved that
that W(l∞
n )
n+1 is optimal. In other words, we need to find A := {A1 , . . . , An+1 } ⊂ W(l∞
such that
T
(a) one has n+1
i=1 Ai = ∅ and
T
(b) for any j ∈ {1, . . . , n + 1}, one has Bj := A∈A\{Aj } A 6= ∅.
68
VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
We can define the following weakly externally hyperconvex half-spaces
A1 := {x ∈ Rn : x1
≥0
},
n
≥ x1
},
n
A3 := {x ∈ R : x3
..
.
≥ x2
},
An := {x ∈ Rn : xn
≥ xn−1 },
A2 := {x ∈ R : x2
n
An+1 := {x ∈ R : −1
≥ xn
T
Note, if x ∈ ni=1 Ai , then x1 , . . . , xn ≥ 0 and thus
Moreover, note that
}.
Tn+1
i=1
Ai = ∅ as desired.
(−1, −1, −1, . . . , −1, −1, −1) ∈ B1 ,
( 0, −1, −1, . . . , −1, −1, −1) ∈ B2 ,
( 0, 0, −1, . . . , −1, −1, −1) ∈ B3 ,
..
.
( 0, 0, 0, . . . , 0, 0, −1) ∈ Bn ,
( 0, 0, 0, . . . , 0, 0, 0) ∈ Bn+1 .
This concludes the proof.
VI.3
Locally Weakly Externally Hyperconvex
Subsets
In Section IV.2, we showed that a uniformly locally hyperconvex metric space
with a geodesic bicombing is hyperconvex. We will now extend this result to the
classes of weakly externally hyperconvex and externally hyperconvex subsets.
Let X be a metric space. A subset A ⊂ X is called locally (weakly) externally
hyperconvex in X if, for all x ∈ A, there is some rx > 0 such that A ∩ B(x, rx )
is (weakly) externally hyperconvex in B(x, rx ).
If we can choose rx = r > 0 for all x ∈ A, we call A uniformly locally
(weakly) externally hyperconvex in X.
Lemma 3.1. Let X be a hyperconvex metric space. Then A ⊂ X is locally
(weakly) externally hyperconvex in X if and only if, for every x ∈ A, there is
some rx > 0 such that A ∩ B(x, rx ) is (weakly) externally hyperconvex in X.
Proof. If A ∩ B(x, rx ) is (weakly) externally hyperconvex in B(x, rx ), then
A ∩ B(x, r2x ) = (A ∩ B(x, rx )) ∩ B(x, r2x )
is (weakly) externally hyperconvex in
B(A ∩ B(x, r2x ), r2x ) ⊂ B(x, rx )
and therefore, by Lemma II.3.16 and Lemma II.3.17, B(x, r2x ) is (weakly) externally hyperconvex in X. The converse is obvious.
69
CHAPTER VI. CONVEXITY
Lemma 3.2. Let X be a metric space and A ⊂ X. Then A ∈ E(X) if and only
if A ∩ B(z, R) ∈ E(B(z, R)) for all z ∈ X, R ≥ d(z, A).
Proof. If A ∈ E(X), then A ∩ B(z, R) ∈ E(B(z, R)) by Lemma II.1.2.
For the other direction, we first observe that A is hyperconvex by Lemma IV.2.2. Let {xi }i∈I ⊂ X with d(xi , A) ≤ ri and d(xi , xj ) ≤ ri + rj . We
define Bi := A ∩ B(xi , ri ). Since, for fixed i, j ∈ I and some s > 0, there are
z ∈ X and R > 0 such that B(xi , ri + s), B(xj , rj + s) ⊂ B(z, R), we have
Bi ∩ Bj 6= ∅ and Bi ∈ E(A) by Lemma II.3.17. Hence, we can conclude
\
\
A∩
B(xi , ri ) =
Bi 6= ∅
i∈I
i∈I
by Proposition II.2.1.
Lemma 3.3. Let X be a metric space and A ⊂ X. Then A ∈ W(X) if and
only if A ∩ B(z, R) ∈ W(B(z, R)) for all z ∈ X, R ≥ d(z, A).
Proof. If A ∈ W(X), then we obviously have A ∩ B(z, R) ∈ W(B(z, R)) by
Proposition II.3.9.
For the other direction, we first observe that A is hyperconvex by Lemma IV.2.2. Let x ∈ X, {xi }i∈I ⊂ A with d(x, A) ≤ r, d(x, xi ) ≤ r + ri and
d(xi , xj ) ≤ ri + rj . Define B̄ := A ∩ B(x, r) and Bi := A ∩ B(xi , ri ). Since,
for fixed i, j ∈ I and some s > 0, there are z ∈ X and R > 0 such that
B(x, r + s), B(xi , ri + s), B(xj , rj + s) ⊂ B(z, R), we have B̄ ∩ Bi ∩ Bj 6= ∅ and
B̄, Bi ∈ E(A) by Lemma II.3.17. Hence, we can conclude
\
\
A ∩ B(x, r) ∩
B(xi , ri ) = B̄ ∩
Bi 6= ∅
i∈I
i∈I
by Proposition II.2.1.
Lemma 3.4. Let X be a hyperconvex metric space and assume that we have
A ∩ B(z, R) ∈ W(B(z, R)) for all z ∈ B(A,R). Then A must be proximinal
in
B A, 45 R . Especially, for all z̄ ∈ B A, 54 R , we have A ∩ B z̄, 54 R 6= ∅.
Proof. Let z̄ ∈ B(A, 45 R) and d(z̄, A) = R + t with t ∈ 0, R4 . Then, for
every ∈ 0, R4 , there is some z ∈ B(A, R) with d(z, z ) ≤ t + . Especially,
we have d(z , z R ) ≤ R, i.e. z ∈ B(z R , R), and, by our assumption, there is
4
4
some proximinal non-expansive retraction ρ : B(z R , R) → A ∩ B(z R , R). Since
4
4
d(ρ(z ), z̄) ≤ R + t + and d(ρ(z ), ρ(z0 )) ≤ d(z , z0 ) ≤ 2t + + 0 , there is some
\
y ∈ B(z̄, R) ∩
B (ρ(z ), t + ) .
>0
Finally, we have d(y, A) = t ≤ R and hence, there is some ȳ ∈ A with d(ȳ, y) = t
and thus
d(ȳ, z̄) ≤ d(ȳ, y) + d(y, z̄) ≤ t + R = d(z̄, A),
i.e. A is proximinal in B A, 54 R .
70
VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
We are now able to prove our local-to-global result for (weakly) externally
hyperconvex subsets of a metric space with a geodesic bicombing.
Proposition 3.5. Let X be a hyperconvex metric space with a geodesic bicombing σ. If A ⊂ X is σ-convex and uniformly locally (weakly) externally
hyperconvex, then A is (weakly) externally hyperconvex in X.
Proof. The two cases are similar but we prove them separately to be precise.
Case 1. Let A be σ-convex and uniformly locally externally hyperconvex.
Let P(R) be the property given by the following expression:
P(R): ∀z ∈ B(A, R) : A ∩ B(z, R) ∈ E(B(z, R)).
We want to show that this is true for every R > 0. Then, by Lemma 3.2,
the set A is externally hyperconvex in X.
Since A is uniformly locally externally hyperconvex, P(R) clearly holds for
some small R > 0. Hence,
we need to show that P(R) ⇒ P( 45 R).
For z ∈ B A, 54 R , let
X 0 := B z, 45 R , A0 := A ∩ X 0 and B 0 (x, r) := B(x, r) ∩ X 0 for x ∈ X 0 .
By Lemma II.3.17, it is enough to show that
A0 ∈ E B 0 A0 , R4
.
(3.1)
First, note that A0 6= ∅ by Lemma 3.4.
We establish now the following property for all R0 > 0:
P0 (R0 ): ∀{xi }i∈I ⊂ B 0T
(A0 , R4 ) with d(xi , xj ) ≤ ri + rj , d(xi , A0 ) ≤ ri and ri ≤ R0 ,
we have A0 ∩ i∈I B 0 (xi , ri ) 6= ∅.
Observe that P0 ( 25 R) implies (3.1) since balls with center in B(z, 54 R) and
radius bigger than 25 R contain B(z, 54 R).
Step I. P0 ( R2 ) holds.
We have d(xi , z) ≤ 54 R, d(xi , xj ) ≤ R and hence, there is some
\
z̄ ∈ B(z, R) ∩
B(xi , R2 ).
i∈I
Then we have d(z̄, A) ≤ d(z̄, xi )+d(xi , A) ≤ R and A∩B(z̄, R) ∈ E(B(z̄, R)) by
P(R). Clearly, z, xi ∈ B(z̄, R) and therefore, if we have d(xi , A ∩ B(z̄, R)) ≤ ri
and d(z, A ∩ B(z̄, R)) ≤ 45 R, there is some
\
y ∈ A ∩ B(z̄, R) ∩ B z, 54 R ∩
B(xi , ri ) 6= ∅
i∈I
and thus y ∈ A0 ∩
T
i∈I
B 0 (xi , ri ) 6= ∅.
71
CHAPTER VI. CONVEXITY
Indeed, since, for every ∈ 0, R4 , there is some
a ∈ A0 = A ∩ B z, 45 R
with d(a , xi ) ≤ min ri , R4 + . It follows
d(a , z̄) ≤ d(a , xi ) + d(xi , z̄) ≤
R
4
++
R
2
≤ R.
and therefore a ∈ A ∩ B(z̄, R). Hence, we get
d(xi , A ∩ B(z̄, R)) ≤ d(xi , a ) ≤ ri + for all > 0, i.e.
d(xi , A ∩ B(z̄, R)) ≤ ri , and
d(z, A ∩ B(z̄, R)) ≤ d(z, a ) ≤ 45 R,
as desired.
Step II. We establish P0 (R0 ) ⇒ P0 (2R0 ).
Let yij := σxi xj
1
2
and zi := σxi z
1
2
. We get
d(zi , A) ≤ d(zi , xi ) + d(xi , A) ≤ 85 R +
d(yij , zi ) ≤
d(z, zi ) =
1
2 d(xj , z) ≤
1
2 d(z, xi ) ≤
R
4
≤ R,
5
8 R ≤ R,
5
8 R ≤ R,
i.e. yij , z ∈ B(zi , R), and A ∩ B(zi , R) ∈ E(B(zi , R)) by P(R). Note that there
is some a ∈ A0 = A ∩ B(z, 54 R) with d(a, xi ) ≤ d(xi , A0 ) + R8 ≤ 83 R and thus
d(zi , a) ≤ d(zi , xi ) + d(xi , a) ≤ 58 R + 38 R = R.
Hence, it follows a ∈ Ai := A ∩ B(zi , R) ∩ B(z, 54 R) 6= ∅ and therefore we get
Ai ∈ E(B(zi , R)). Moreover, we have
d(yij , yik ) ≤ 12 d(xj , xk ) ≤
d(yij , A0 ) ≤
1
2
rj
2
+
rk
2 ,
d(xi , A0 ) + d(xj , A0 ) ≤ min{ r2i +
rj R
2, 4}
by Lemma 2.4. For ∈ 0, R8 , let a ∈ A0 with d(yij , a ) ≤ d(yij , A0 ) + . Then
we have d(a , zi ) ≤ d(a , yij ) + d(yij , zi ) ≤ R4 + + 85 R ≤ R, i.e. a ∈ Ai , and
therefore
r
d(yij , Ai ) = d(yij , A0 ) ≤ min{ r2i + 2j , R4 }.
T
r
Hence, there are x̄i ∈ B(Ai , min{ r2i , R4 }) ∩ B(z, 54 R) ∩ j∈I B(yij , 2j ) with
d(x̄i , x̄j ) ≤ d(x̄i , yij ) + d(yij , x̄j ) ≤
ri
2
+
rj
2,
d(x̄i , A0 ) ≤ d(x̄i , Ai ) ≤ min{ r2i , R4 }.
T
T
Therefore, by P0 (R0 ), we get y ∈ A0 ∩ i∈I B(x̄i , r2i ) ⊂ A0 ∩ i∈I B(xi , ri ) =
6 ∅
as desired.
72
VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS
Case 2. Let A be σ-convex and uniformly locally weakly externally hyperconvex.
Let P(R) be the property given by the following expression:
P(R): ∀z ∈ B(A, R) : A ∩ B(z, R) ∈ W(B(z, R)).
We want to show that this is true for every R > 0. Then, by Lemma 3.3,
the set A is weakly externally hyperconvex in X.
Since A is uniformly locally externally hyperconvex, P(R) clearly holds for
some small R > 0. Hence,
we need to show that P(R) ⇒ P( 45 R).
5
For z ∈ B A, 4 R , let
X 0 := B z, 45 R , A0 := A ∩ X 0 and B 0 (x, r) := B(x, r) ∩ X 0 for x ∈ X 0 .
By Lemma II.3.16, it is enough to show that
A0 ∈ W B 0 A0 , R4
.
(3.2)
First, note that A0 6= ∅ by Lemma 3.4.
We establish now the following property for all R0 > 0:
P0 (R0 ): ∀x ∈ B 0 A0 , R4 , {xi }i∈I ⊂ A0 with d(x, A0 ) ≤ r, d(x, xi ) ≤ r + ri ,
d(xi , xj ) ≤ ri + rj and r,Tri ≤ R0 ,
we have A0 ∩ B 0 (x, r) ∩ i∈I B 0 (xi , ri ) 6= ∅.
Observe that P0 ( 25 R) implies (3.2) since balls with center in B(z, 54 R) and
radius bigger than 25 R contain B(z, 54 R).
Step I. P0 ( R2 ) holds.
We have d(x, z), d(xi , z) ≤ 54 R, d(x, xi ), d(xi , xj ) ≤ R and hence, there is some
z̄ ∈ B(z, R) ∩ B(x, R2 ) ∩
\
B(xi , R2 ).
i∈I
For ∈ (0, R4 ], let a ∈ A0 with d(x, a ) ≤ d(x, A0 ) + ≤
z, x, xi ∈ B(z̄, R),
d(z̄, A) ≤ d(z̄, x) + d(x, a ) ≤ R
R
2.
Then we have
and
a ∈ A ∩ B(z̄, R) ∩ B(z, 54 R) 6= ∅.
Hence A ∩ B(z̄, R) ∩ B(z, 45 R) ∈ W(B(z̄, R)) by P(R) and Proposition II.3.9,
and d(x, A ∩ B(z̄, R) ∩ B(z, 45 R)) ≤ r. Therefore, there is some
y ∈ A ∩ B(z̄, R) ∩ B(z, 54 R) ∩ B(x, r) ∩
\
i∈I
and thus y ∈ A0 ∩ B 0 (x, r) ∩
T
i∈I
B 0 (xi , ri ) 6= ∅.
73
B(xi , ri ) 6= ∅
CHAPTER VI. CONVEXITY
Step II. We establish P0 (R0 ) ⇒ P0 (2R0 ).
Let yi := σxxi ( 21 ), yij := σxi xj ( 12 ), z̄ := σxz ( 12 ) and zi := σxi z ( 12 ). Moreover, let
ā ∈ A0 such that d(x, ā ) ≤ d(x, A0 ) + ≤ 38 R. We get
d(z̄, ā ) ≤ d(z̄, x) + d(x, ā ) ≤ 58 R +
R
4
≤ R,
d(z̄, A) ≤ d(z̄, ā) ≤ R,
d(zi , A) ≤ d(zi , xi ) ≤ 58 R ≤ R,
d(yij , zi ) ≤ 12 d(xj , z) ≤ 58 R ≤ R,
d(yi , zi ) ≤ 21 d(x, z) ≤ 85 R ≤ R,
d(z, zi ) = 12 d(z, xi ) ≤ 58 R ≤ R,
i.e. yi , yij , z ∈ B(zi , R), and A ∩ B(zi , R) ∈ W(B(zi , R)) by P(R).
Define Ai := A ∩ B(zi , R) ∩ B(z, 45 R) and Ā := A ∩ B(z̄, R) ∩ B(z, 45 R).
We have yij ∈ Ai 6= ∅ and ā ∈ Ā 6= ∅ and therefore, Ai ∈ W(B(zi , R)) and
Ā ∈ W(B(z̄, R)) by Proposition II.3.9. Furthermore, we get
d(x, Ā) ≤ d(x, ā ) ≤ r + , i.e. d(x, Ā) ≤ r,
d(yi , Ā) ≤ 21 d(x, Ā) ≤
d(x, yi ) =
d(yi , yj ) ≤
d(yi , yij ) ≤
d(yij , yik ) ≤
r
2
r
2
(Lemma 2.4),
1
+ r2i ,
2 d(x, xi ) ≤
rj
ri
1
2 d(xi , xj ) ≤ 2 + 2 ,
rj
1
r
2 d(x, xj ) ≤ 2 + 2 ,
rj
rk
1
2 d(xj , xk ) ≤ 2 + 2 .
Therefore, there areTsome points x̄ ∈ B(Ā, 2r ) ∩ B(x, 2r ) ∩
r
x̄i ∈ Ai ∩ B(yi , 2r ) ∩ j∈I B(yij , 2j ) with
d(x̄, x̄i ) ≤ d(x̄, yi ) + d(yi , x̄i ) ≤
r
2
+
ri
2,
ri
2
+
d(x̄i , x̄j ) ≤ d(x̄i , yij ) + d(yij , x̄j ) ≤
T
i∈I
B(yi , r2i ) and
rj
2.
Hence, by P0 (R0 ), we get
y ∈ A0 ∩ B(x̄, 2r ) ∩
\
B(x̄i , r2i ) ⊂ A0 ∩ B 0 (x, r) ∩
i∈I
\
B 0 (xi , ri ) 6= ∅
i∈I
as desired.
Proof of Theorem 1.1. Note that the first equivalence directly follows from Proposition 2.1 and Proposition 3.5.
For the second equivalence, note first that, by the uniqueness of straight
lines, σ is a consistent geodesic bicombing and thus, if A is a σ-convex subset of
X, it possesses a consistent geodesic bicombing. The implications then follow
from Proposition 2.2 and Proposition 3.5.
Together with Theorem 1.1 and Theorem 1.2 in [DL16], we deduce Corollary 1.2.
74
Chapter VII
Hyperconvex Metrics on
Median Metric Spaces
In [Bow16a], B. Bowditch constructs a bi-Lipschitz equivalent CAT(0)-metric
on a complete, connected median metric space of finite rank. In this chapter,
we give the construction of a further bi-Lipschitz equivalent metric which turns
out to be hyperconvex in several cases. We start with some general facts on
median metric spaces, then turn our attention to cube complexes and finally
construct the announced metric.
To conclude, we give a local-to-global result for median metric spaces with
a conical geodesic bicombing.
VII.1
Median Metric Spaces
Let (M, ρ) be a metric space. For a, b ∈ M , we define the metric interval
Iρ (a, b) between a and b by Iρ (a, b) := {x ∈ M : ρ(a, x) + ρ(x, b) = ρ(a, b)}. We
then call (M, ρ) a median metric space if, for all a, b, c ∈ M , the intersection
Iρ (a, b) ∩ Iρ (b, c) ∩ Iρ (c, a)
consists of exactly one point of M .
Definition 1.1. Let M be a set and µ : M 3 → M a ternary operator. (M, µ)
is a median algebra if it satisfies the following for all a, b, c, d, e ∈ M :
(M1) µ(a, b, c) = µ(b, c, a) = µ(b, a, c),
(M2) µ(a, a, b) = a,
(M3) µ(a, b, µ(c, d, e)) = µ(µ(a, b, c), µ(a, b, d), e).
Let a, b ∈ M . We say that x ∈ M lies between a and b if x = µ(a, b, x) and we
define
Iµ (a, b) := {x ∈ M : µ(a, b, x) = x},
the set of all points between a and b.
75
CHAPTER VII. MEDIAN METRIC SPACES
Proposition 1.2. Let (M, ρ) be a median metric space. For a, b, c ∈ M , we
write µ(a, b, c) for the unique point in Iρ (a, b) ∩ Iρ (b, c) ∩ Iρ (c, a). This defines
an operator µ : M 3 → M . Then (M, µ) is a median algebra and we have
Iρ (a, b) = Iµ (a, b).
Proof. By symmetry of the metric, we clearly have (M1). Property (M2) follows
since µ(a, a, b) ⊂ Iρ (a, a) = {a}. The proof of (M3) is quite technical and can
be found in [Sho54], see also the comment on this in [Bow16a]. Finally, observe
that µ(a, b, x) = x ⇔ x ∈ Iρ (a, b) since x ∈ Iρ (a, x) ∩ Iρ (b, x), and therefore
Iρ (a, b) = Iµ (a, b).
For any two points a, b in a median algebra (M, µ), there is a natural projection map π : M → Iµ (a, b), given by π(x) := µ(a, b, x). Indeed, we have
µ(a, b, π(x)) = µ(a, b, µ(a, b, x))
= µ(µ(a, b, a), µ(a, b, b), x) = µ(a, b, x) = π(x),
i.e. π(x) ∈ Iµ (a, b).
Lemma 1.3. Let (M, µ) be a median algebra and π : M → Iµ (a, b) the projection map given by π(x) = µ(a, b, x). Then we have
π(Iµ (c, d)) = Iµ (π(c), π(d)),
for all c, d ∈ M .
Proof. First, let x ∈ π(Iµ (c, d)). Then there is some y ∈ Iµ (c, d) with x = π(y)
and we get
x = µ(a, b, y) = µ(a, b, µ(c, d, y))
= µ(µ(a, b, c), µ(a, b, d), y) = µ(π(c), π(d), y),
i.e. x ∈ Iµ (π(c), π(d)).
Conversely, if x ∈ Iµ (π(c), π(d)), we have
x = µ(µ(a, b, c), µ(a, b, d), x) = µ(a, b, µ(c, d, x)) = π(µ(c, d, x)),
i.e. x ∈ π(Iµ (c, d)).
An n-cube in a median algebra M is a subalgebra which is isomorphic to
{−1, 1}n . The rank of the median algebra M is the maximal n ∈ N such that
there is an n-cube in M . If there is some n-cube for every n ∈ N, we say that
M is of infinite rank. A face of a cube is a convex subset and an edge is a face
consisting of exactly two points. The set {a, b} is a diagonal of the cube Q if
{a, b} ⊂ Q and Q ⊂ Iµ (a, b).
76
VII.2. CUBE COMPLEXES
VII.2
Cube Complexes
Q
A real n-cube is the product I n = [0, 1]n . A subspace of the form F = ni=1 Fi
with Fi ∈ {{0}, {1}, [0, 1]} is called a face of I n .
F
Let X = λ∈Λ Iλnλ be the disjoint union of cubes, ∼ an equivalence relation
on X and pλ : Iλnλ → X/ ∼ the projection maps. Assume that the following
holds:
(i) The map pλ : Iλnλ → X/ ∼ is injective for every λ ∈ Λ.
n
n
(ii) If pλ (Iλnλ ) ∩ pλ (Iλ0λ ) 6= ∅, then there are faces Fλ ⊂ Iλnλ , Fλ0 ⊂ Iλ0λ and
0
0
n0λ
an isometry hλ,λ0 : Fλ → Fλ0 such that, for all x ∈ Iλnλ , x0 ∈ Iλ , we have
pλ (x) = pλ0 (x0 ) if and only if hλ,λ0 (x) = x0 .
We call a subset C ⊂ X/ ∼ a cube if it is the image pλ (F ) of some face F ⊂ Iλnλ .
The collection C := {pλ (F ) : F face of Iλnλ , λ ∈ Λ} of all cubes in X/ ∼ is then
a cube complex and |C| := X/ ∼ its geometric realization.
Beside the standard Euclidean metric d2 , we can also choose other metrics
on the cubes I n . We equip each cube C with a length metric dC respecting the
topology of I n ⊂ Rn such that the maps hλ,λ0 still are isometries, for instance
v
u n
uX
p
dp (x, y) = kx − ykp = t
|xi − yi |p ,
for p ≥ 1, or
i=1
d∞ (x, y) = kx − yk∞ = max{|x1 − y1 |, . . . , |xn − yn |}.
An (m-)string Σ between two points x, y ∈ |C| is a tuple (x0 , . . . , xm ) of points
in C such that x = x0 , y = xm and, for any two consecutive points xi , xi+1 , there
is some cube Ci ∈ C containing xi , xi+1 . For each m-string Σ = (x0 , . . . , xm ),
we then get its length
m−1
X
l(Σ) :=
dCi (xi , xi+1 ).
i=0
Using the length of strings, we define the distance between any two points
x, y ∈ C by taking the infimum over the length of all strings joining them.
d(x, y) := inf{l(Σ) : Σ is a string from x to y}.
If all metrics on the cubes are induced by the same lp -norm, we emphasize this
by writing dp .
First of all, this is a pseudo-metric on the cube complex C. But, by the proof
of Theorem I.7.13 in [BH99], this gives a complete length space. For Euclidean
cubes, Bridson’s Theorem [Bri91] applies and tells us that finite dimensional
cube complexes with metric d2 even are geodesic. But this also holds for all
other metrics dp .
77
CHAPTER VII. MEDIAN METRIC SPACES
Theorem 2.1. Let C be a finite dimensional cube complex and let dp be the
metric on C induced by the lp -norm for p ∈ [1, ∞]. Then (|C|, dp ) is a complete,
geodesic space.
We will show the following slightly more general result and, since both properties (2.1) and (2.2) are satisfied by all lp -norms, we deduce that Theorem 2.1
holds.
Theorem 2.2. Let C be a cube complex with finitely many isometry types of
cubes where the metrics on the cubes [0, 1]n are induced by norms on Rn satisfying
kyk ≤ kxk for all x, y ∈ Rn with |y l | ≤ |xl | for l = 1, . . . , n.
(2.1)
|xl | ≤ kxk for all x ∈ Rn and l = 1, . . . , n.
(2.2)
Then (|C|, d) is a complete, geodesic space.
Our strategy is based on finding taut strings that approximate distances.
Since geodesics are not unique, we must demand further properties for taut
strings than in the proof of Theorem I.7.19 in [BH99].
In the following, let C be a cube complex endowed with the metric d defined
as above. Given some string Σ = (x0 , . . . , xm ), let Ci be the largest cube
containing xi , xi+1 . Furthermore, let Fi = Ci ∩ Ci−1 be the common face of
the two consecutive
inductively coordinates (x1i , . . . , xni ) on
Qn cubes. Introduce
each cube Ci = l=1 Il ⊂ [0, 1]n with Il = {0} or [0, 1] such that the following
properties hold:
(i) xl0 ≤ xl1 for all l = 1 . . . , n,
(ii) (0, . . . , 0) ∈ Fi , and
(iii) on Fi , the coordinates of Ci−1 and Ci coincide (modulo 1).
An m-string Σ = (x0 , . . . , xm ) in C is taut if it satisfies the following three
conditions:
(i) There is no cube containing {xi−1 , xi , xi+1 } for i = 1, . . . , m − 1.
(ii) For i = 1, . . . , m − 1, if xi−1 , xi ∈ Ci−1 and xi , xi+1 ∈ Ci , then we have
xi ∈ ILi (xi−1 , xi+1 )
in Li = Ci−1 ∪ Ci , where ILi denotes the metric interval with respect to
the induced length metric on Li .
(iii) For each l = 1, . . . , n and i = 0, . . . , m − 1, we have xli ≤ xli+1 in Ci .
Remark. In the Euclidean case, property (ii) already implies that the coordinates xli are monotone, but, for a general metric d, we have to demand (iii)
additionally to ensure that taut strings which cross many cubes are long.
78
VII.2. CUBE COMPLEXES
Lemma 2.3. Let C be a finite dimensional cube complex with dimension bounded by n. If Σ = (x0 , . . . , xn+2 ) is a taut (n + 2)-string, then l(Σ) ≥ 1.
Proof. By property (iii), we get that xli ≤ xli+1 in Ci for all i = 0, . . . , n + 1
and l = 1, . . . , n. For i = 1, . . . , n + 1, we then make the following observations.
i
Since xi ∈ Fi but xi+1 ∈
/ Fi , there is some li such that xlii = 0 and xli+1
6= 0
li
li
(in Ci ). This leads to chains 0 = xi < xi+1 ≤ . . . for i = 1, . . . , n + 1. Note
that here the coordinate xlki coincides in Ck−1 and Ck if xlki 6= 1 in Ck−1 . Since
li ∈ {1, . . . , n}, by the pigeonhole principle, there are i1 < i2 with li1 = li2 . But
li
li
li
this implies that the chain 0 = xi11 < xi11+1 ≤ . . . ends with some xi01 = 1 (in
Ci0 −1 ), i0 < i2 and therefore
0
l(Σ) ≥
i
X
0
Ci−1
d
(xi−1 , xi ) ≥
i=i1 +1
i
X
li
li
li
li
1
|xi 1 − xi−1
| = xi01 − xi11 = 1
i=i1 +1
li
li
1
by (2.2) and since xi 1 ≥ xi−1
.
Corollary 2.4. Let C be a finite dimensional cube complex. Then there is
some constant α such that, for every taut m-string Σ = (x0 , . . . , xm ), we have
l(Σ) ≥ αm − 1.
Lemma 2.5. For some fixed m, let Σ = (x0 , . . . , xm ) be an m-string from x to
y of minimum length among all m-strings joining these points. Then there is
some taut m̄-string Σ̄ = (x̄0 , . . . , x̄m̄ ) from x to y with l(Σ̄) = l(Σ) and m̄ ≤ m.
Proof. We proceed by induction on m. A 1-string clearly is taut. Therefore,
let us assume m ≥ 2. If there are three consecutive points xi−1 , xi , xi+1 all
contained in some cube C, we can simply cancel xi and get an (m − 1)-string
Σ̄ = (x0 , . . . , xi−1 , xi+1 , . . . , xm ) with the same length by the triangle inequality
and minimality. Hence, we may assume that property (i) of taut strings holds.
If property (ii) does not hold, we can easily find some x̄i ∈ Fi with
d(xi−1 , x̄i ) + d(x̄i , xi+1 ) < d(xi−1 , xi ) + d(xi , xi+1 ),
contradicting minimality of Σ.
Thus, it remains to achieve that x̄li ≤ x̄li+1 in Ci . By induction, we may
assume that this is true for i ≤ m − 2. For each l with xlm−1 > xlm (in Cm−1 ),
consider the maximal chain 0 ≤ xli0 ≤ . . . ≤ xlm−1 , where the coordinates
xlk are the same in Ck−1 and Ck for k = i0 + 1, . . . , m − 2 and xli0 , xlm−1 are
taken in Ci0 , Cm−2 , respectively. If xli0 ≤ xlm , define x̄li := min{xli , xlm } for
i = i0 , . . . , m − 1. Otherwise, we have xli0 > xlm ≥ 0 and therefore i0 = 0, i.e.
xi0 = x. If so, define x̄li = xl for i = 1, . . . , m − 1 and reflect the corresponding
coordinate axes at 12 . In either case, we have |x̄li − x̄li+1 | ≤ |xli − xli+1 | and hence
Σ̄ is not longer than Σ. Observe that, if xlm = 0, it may happen that property
(i) does not hold anymore for Σ̄ but then we can proceed as described at the
beginning of the proof.
79
CHAPTER VII. MEDIAN METRIC SPACES
Lemma 2.6. Let L be a finite cube complex and x, y ∈ |L| two points that can
be joined by an m-string for some fixed m. Then there is a shortest m-string
from x to y.
Proof. Let S ⊂ |L|m+1 be the set of all m-strings joining x and y. Note that S
is a closed subset and hence it is compact. Therefore, since the length l(Σ) is
a continuous function, there is some m-string Σ̄ ∈ S which attains the minimal
length.
Lemma 2.7. Let C be a finite dimensional cube complex and x, y ∈ |C| two
points that can be joined by an m-string for some fixed m. Then there is a
shortest m-string joining them.
Proof. Since the dimension of the cubes is bounded, there are only finitely many
types of cubes in C. Hence, there are also only finitely many isometry types
of subcomplexes consisting of m cubes. Now every m-string is contained in a
subcomplex with at most m cubes and, for every type of finite subcomplex, there
is some smallest m-string joining x and y by the previous lemma. Therefore we
can find a smallest m-string between x and y in |C|.
Proposition 2.8. Let C be a finite dimensional cube complex. Then, for any
two points x and y in |C|, we have
d(x, y) = inf{l(Σ) : Σ is a taut string from x to y}.
Proof. By Lemma 2.7 and Lemma 2.5, for every m-string Σ, there is some taut
n-string Σ̄ with the same endpoints and l(Σ̄) ≤ l(Σ).
Proof of Theorem 2.2. Let x, y be two points in |C| with distance d := d(x, y).
Corollary 2.4 then tells us that there is some m0 such that we can approximate
d∞ (x, y) only using m-stings with m ≤ m0 , i.e.
d(x, y) = inf{l(Σ) : Σ is an m-string from x to y, m ≤ m0 }.
But then, by Lemma 2.7, P
there is some shortest string Σ = (x0 . . . . , xm )
joining x and y, i.e. d(x, y) = ni=1 d(xi−1 , xi ). Connecting the xi ’s by straight
lines, we finally get a geodesic from x to y.
We will now prove that, if (|C|, d∞ ) is a hyperconvex metric space, then
(|C|, d2 ) is CAT(0). For euclidean cube complexes, Gromov [Gro87] gives a
combinatorial criterion for the CAT(0) property. We use a version stated in
[Che00].
Theorem 2.9. A cube complex C is CAT(0) if and only if it is simply-connected
and the following holds: Given three (n + 2)-cubes intersecting in an n-cube and
pairwise sharing an (n + 1)-face, then there is some (n + 3)-cube containing all
of them.
80
VII.2. CUBE COMPLEXES
Lemma 2.10. Let C be a cube complex endowed with the maximum metric d∞
such that (|C|, d∞ ) is geodesic. Assume that there are cubes C, C0 ∈ C with
C ⊂ C0 , a point p ∈ C0 and r ∈ [0, 21 ] with d∞ (p, ∂C) ≥ r and d∞ (p, C) ≤ r.
Then, for q ∈ B(p, r), there is some cube D containing q and C.
Proof. Let Σ = (p0 , p1 , . . . , pk ) be a taut string from p to q and C0 , C1 , . . . , Ck−1
maximal cubes with pi , pi+1 ∈ Ci . Observe that, for i 6= 0, k, we have pi ∈ ∂Ci .
We denote the dimensions by n := dim C, ni := dim Ci and identify C with
the cube [0, 1]n × {0}ni −n ⊂ [0, 1]ni = Ci if C ⊂ Ci . By our assumption, we
have p = (x1 , . . . , xn0 )C0 with r ≤ xl ≤ 1 − r for l = 1, . . . , n and xl ≤ r for
l = n + 1, . . . , n0 .
Claim. For i = 0, . . . , k − 1, we have C ⊂ Ci and pi = (x1 , . . . , xni )Ci with
r − d∞ (p, pi ) ≤ xl ≤ 1 − r + d∞ (p, pi )
for l = 1, . . . , n and xl ≤ r + d∞ (p, pi ) for l = n + 1, . . . ni .
For i = 0, this clearly is true. Assume that the claim holds for some i ≤ k−2.
Then for pi+1 = (y1 , . . . , yni )Ci , we can easily check the inequalities using the
properties |xl − yl | ≤ d∞ (pi , pi+1 ) and d∞ (p, pi+1 ) = d∞ (p, pi ) + d∞ (pi , pi+1 ).
Since pi+1 6= q, we have d∞ (p, pi+1 ) = d∞ (p, q) − d∞ (q, pi+1 ) < r and therefore
yl 6= 0, 1 for l = 1, . . . , n and yl 6= 1 for l = n + 1, . . . , ni . Thus, the subcube
Di+1 of Ci ∩ Ci+1 with pi+1 ∈ Int Di+1 also contains C.
To deduce the lemma, note that, by the claim, Ck−1 contains both q = pk
and C.
Theorem 2.11. Let C be a cubical complex such that (|C|, d∞ ) is geodesic,
simply connected and locally 3-hyperconvex. Then (|C|, d2 ) is CAT(0).
Proof. We must check the link condition. Let A, B, C be three (k + 2)-cubes
such that A ∩ B ∩ C is a k-cube and A ∩ B, A ∩ C and B ∩ C are (k + 1)-cubes.
We need to show that A, B and C are contained in some (k + 3)-cube. Assume
that they are not.
Note that then there is also no cube containing C and A ∩ B. Indeed, if
they are contained in such a cube D, then A ∩ C and A ∩ B are two different
(k + 1)-faces of the same (k + 2)-face of D, which therefore coincides with A.
Similarly B ⊂ D.
The goal is now to find three closed balls in C that intersect pairwise but
have no common intersection point. For doing this, we introduce coordinates
on our cubes A, B and C such that
A ∩ B = {(x1 , . . . , xk+2 )A : xk+2 = 0}
= {(y1 , . . . , yk+2 )B : yk+2 = 0}
A ∩ C = {(x1 , . . . , xk+2 )A : xk+1 = 0}
= {(z1 , . . . , zk+2 )C : zk+2 = 0}
B ∩ C = {(y1 , . . . , yk+2 )B : yk+1 = 0}
= {(z1 , . . . , zk+2 )C : zk+1 = 0}
81
CHAPTER VII. MEDIAN METRIC SPACES
Choose ∈ (0, 1) and define
x = ( 12 , . . . , 21 , 8 , 0)A = ( 12 , . . . , 12 , 8 , 0)B ,
y = ( 21 , . . . , 21 , 3
8 , 8 )C ,
z = ( 21 , . . . , 21 , 8 , 3
8 )C .
It is easy to check that d∞ (x, y), d∞ (x, z) ≤ 4 + 8 and d∞ (y, z) ≤ 8 + 8 . Hence,
since (|C|, d∞ ) is geodesic, the balls B(x, 4 ), B(y, 8 ), B(z, 8 ) intersect pairwise.
Assume now that there is some
p ∈ B(x, 4 ) ∩ B(y, 8 ) ∩ B(z, 8 ).
Then, since p ∈ B(y, 8 ) ∩ B(z, 8 ), we have d(p, ∂C) ≥ 2
8 and d(p, C) ≤
Hence, by Lemma 2.10, there is some cube D containing p and C. Thus,
8.
p = (u1 , . . . , uk , 4 , 4 , uk+3 , . . . , un )D
with ui ∈ [ 21 − 8 , 21 + 8 ] for i = 1, . . . , k and ui ∈ [0, 8 ] for i = k + 3, . . . , n.
Especially, p satisfies the assumption of Lemma 2.10 for r = 4 and since there
is no cube containing x and C, we have d∞ (p, x) > 4 , which contradicts the
assumption p ∈ B(x, 4 ).
Since hyperconvex metric spaces are geodesic and simply connected, we
conclude the following.
Corollary 2.12. If the cube complex (|C|, d∞ ) is a hyperconvex metric space,
then (|C|, d2 ) is CAT(0).
But the converse is false in general as the following example already mentioned in [MT91] shows.
Example 2.13. For every n ∈ N, fix some cube Cn = [0, 1]n . We glue them
0
0
by identifying Cn = [0, 1]n × {0}n −n ⊂ [0, 1]n = Cn0 for n ≤ n0 . Then the
resulting cube complex is not complete and hence not hyperconvex. For instance
xn = ( 21 , 41 , . . . , 21n ) ∈ Cn defines a divergent Cauchy sequence.
Together with the work of V. Chepoi, J.H. Mai–Y. Tang and M.L.J. van
de Vel [Che00, MT83, vdV98], we finally get the following characterization of
finite CAT(0) cube complexes.
Theorem 2.14. For finite cube complexes, the following are equivalent:
1. (|C|, d2 ) is CAT(0),
2. (|C|, d1 ) is median,
3. (|C|, d∞ ) is hyperconvex.
82
VII.3. HYPERCONVEX METRICS ON MEDIAN METRIC SPACES
VII.3
Hyperconvex Metrics on Median Metric
Spaces
In this section, we now adapt B. Bowditch’s construction in [Bow16a] in such
a way that we eventually find a hyperconvex metric.
Theorem 3.1. Let (M, ρ) be a proper geodesic median metric space of finite
rank, then M possesses a bi-Lipschitz equivalent metric d such that (M, d) is
hyperconvex.
Let (M, ρ) be a median metric space with finite rank n. For a cube Q in M
with edge lengths t1 , t2 , . . . , tm , we define
m
ω̃(Q) := max ti .
i=1
Remark. If {a, b} is the diagonal of two cubes Q, Q0 ⊂ M with Q0 ⊂ Q, then
we have
ω̃(Q0 ) ≥ ω̃(Q).
Given a, b ∈ M , denote by Q(a, b) the maximal cube in M with diagonal
{a, b} and
ω̃(a, b) := ω̃(Q(a, b)).
Remark. If M is of finite rank n, then, for a, b ∈ M , we have
ρ(a, b)
≤ ω̃(a, b) ≤ ρ(a, b).
n
(3.1)
Remark. Let π : M → I(c, d) be the projection map onto the interval I(c, d).
Then
ω̃(π(a), π(b)) ≤ ω̃(a, b).
Definition 3.2. For a monotone sequence a = a0 , a1 , . . . , ap , that is a sequence
with ai ∈ I(ai−1 , ai+1 ) and therefore
ρ(a0 , ap ) =
p
X
ρ(ai−1 , ai ),
i=1
let
ω̃(a) :=
p
X
ω̃(ai−1 , ai ).
i=1
For a, b ∈ M , we define the following metric on M :
d(a, b) := inf ω̃(a),
a
where the infimum is taken over all monotone sequences a from a to b.
Remark. From equation (3.1), we immediately get
ρ(a, b)
≤ d(a, b) ≤ ρ(a, b).
n
83
CHAPTER VII. MEDIAN METRIC SPACES
Lemma 3.3. The function d : M × M → R is a metric on M .
Proof. It only remains to verify the triangle inequality. Let a, b, c ∈ M and
set x := µ(a, b, c). Then for monotone sequences a = a0 , a1 , . . . , ap = c and
c = b0 , b1 . . . . , bq = b, look at the projections onto I(a, b), which we denote by
a = a00 , a01 , . . . a0p = x and x = b00 , b01 , . . . , b0q = b. Observe that
a = a00 , a01 , . . . a0p , b00 , b01 , . . . , b0q = b
is a monotone sequence from a to b and we get
X
X
ω̃(ai−1 , ai ) + inf
ω̃(bi−1 , bi )
d(a, c) + d(c, b) = inf
a
≥ inf
≥ inf
X
a
c
b
i
X
ω̃(a0i−1 , a0i )
i
+ inf
i
X
b
ω̃(b0i−1 , b0i )
i
ω̃(ci−1 , ci ) = d(a, b),
i
where c is in the set of monotone sequences from a to b.
Let (Π, ρ) be a finite median metric space. Then Π corresponds to the vertex
set of a CAT(0) cube complex Υ(Π) and the metric ρ naturally extends to a
median metric ρΥ on Υ(Π) by the d1 -metric on each cell. From Theorem 2.14,
it follows that dΥ defined as above is a hyperconvex metric on Υ(Π).
Lemma 3.4. [Bow16a, Lemma 7.5] If Π is a finite subalgebra of a geodesic median metric space (M, ρ), then there is a median monomorphism f : Υ(Π) → M
extending the inclusion Π ,→ M such that f : (Υ(Π), ρΥ ) → (M, ρ) is an isometric embedding.
Lemma 3.5. Let f : Υ(Π) → M be the median monomorphism from Lemma 3.4 above. If M has finite rank, then the map f : (Υ(Π), dΥ ) → (M, d) is
1-Lipschitz.
Proof. Let a, b ∈ Υ(Π) and fix some > 0. Then, choose a monotone sequence
a = a0 , a1 , . . . , ap in Υ(Π) from a to b such that
p
X
ω̃(ai−1 , ai ) ≤ dΥ (a, b) + .
i=1
Denote by Qi the maximal cube in Υ(Π) with diagonal {ai−1 , ai }. Then
f (Qi ) is a cube in M with diagonal {f (ai−1 ), f (ai )}. We get
d(f (ai−1 ), f (ai )) ≤ ω̃(f (Qi )) = ω̃Υ (Qi ) = ω̃Υ (ai−1 , ai )
and therefore
d(f (a), f (b)) ≤
p
X
d(f (ai−1 ), f (ai )) ≤
i=1
p
X
ω̃Υ (ai−1 , ai ) ≤ dΥ (a, b) + .
i=1
Finally, by taking → 0, we have d(f (a), f (b)) ≤ dΥ (a, b).
84
VII.3. HYPERCONVEX METRICS ON MEDIAN METRIC SPACES
Lemma 3.6. Let (M, ρ) be a geodesic median metric space of finite rank and
A ⊂ M a finite subset. Then, for every δ > 0, there is a compact, hyperconvex
metric space (Υ, dΥ ) with A ⊂ Υ and a 1-Lipschitz map f : (Υ, dΥ ) → (M, d)
extending the inclusion of A into M such that, for all a, b ∈ A, it holds
d(a, b) ≤ dΥ (a, b) ≤ d(a, b) + δ.
Proof.PFor each pair a, b ∈ A, choose a monotone sequence a = a0 , a1 , . . . , ap = b
with pi=1 ω̃(ai−1 , ai ) ≤ d(a, b) + δ. LetSQi be the maximal cube in M with
diagonal {ai−1 , ai }. We define Aa,b := pi=1 Qi to be the union of all these
cubes.
S
Finally, let A0 := a,b∈A Aa,b . Observe that this is a finite union of finite sets
and hence finite. Take a finite subalgebra Π of M containing A0 and Υ = Υ(Π)
as above. We have
dΥ (a, b) ≤
p
X
ω̃Υ (ai−1 , ai ) ≤
i=1
p
X
ω̃(Qi ) =
i=1
p
X
ω̃(ai−1 , ai ) ≤ d(a, b) + δ,
i=1
as desired.
Proposition 3.7. Let (M, ρ) be a compact, geodesic median metric space of
finite rank. Then (M, d) is hyperconvex.
Proof. Since every compact metric space is complete, it is enough to show that
(M, d) is almost hyperconvex. Let {B(xi , ri )}i∈I be a collection of closed balls
in (M, d) with d(xi , xj ) ≤ ri + rj . For δ > 0, let A be a finite subset of M
such that, for all x ∈ M , there is some a ∈ A with ρ(x, a) < δ. Let Υ be as in
Lemma 3.6.
For each xi , choose ai ∈ A with ρ(xi , ai ) < δ. Then
dΥ (ai , aj ) ≤ d(ai , aj ) + δ ≤ d(xi , xj ) + 3δ ≤ ri + rj + 3δ.
Therefore, since Υ is hyperconvex, there is some
y∈
\
B(ai , ri + 2δ).
i∈I
For the image of y in M , we get
d(xi , f (y)) ≤ d(ai , f (y)) + δ ≤ dΥ (ai , y) + δ ≤ ri + 3δ.
We conclude that, for each > 0, we have
T
i∈I
B(xi , ri + ) 6= ∅.
Proof of Theorem 3.1. By Proposition 3.7, every ball B(x, r) is hyperconvex
and hence, by Lemma IV.2.2, (M, d) is hyperconvex as well.
85
CHAPTER VII. MEDIAN METRIC SPACES
VII.4
Locally Median Metric Spaces
Theorem 4.1. Let (M, ρ) be a metric space with a reversible, conical geodesic
bicombing σ : M × M × [0, 1] → M such that, for all x ∈ M and some r > 0,
the neighborhood B(x, r) is a median metric space, then M is a median metric
space.
Proof. Let x1 , x2 , x3 ∈ M with ρ(xi , xj ) ≤ 2r. For {i, j, k} = {1, 2, 3}, define
yi := σxj xk ( 12 ). Then
ρ(yi , yj ) ≤ 21 ρ(xi , xj ) ≤ r and ρ(xi , yj ) = 12 ρ(xi , xk ) ≤ r.
Therefore, by our assumption, there are zi := µ(xi , yj , yk ). Observe that
ρ(xi , xj ) = ρ(xi , yk ) + ρ(yk , xj )
= ρ(xi , zi ) + ρ(zi , yk ) + ρ(yk , zj ) + ρ(zj , xj )
≥ ρ(xi , zi ) + ρ(zi , zj ) + ρ(zj , xj ) ≥ ρ(xi , xj ),
i.e. I(zi , zj ) ⊂ I(xi , xj ). Moreover, we have
ρ(xi , zi ) = (yk | yj )xi =
≥
=
1
2 (ρ(xi , yk ) + ρ(xi , yj ) − ρ(yk , yj ))
1
4 (ρ(xi , xj ) + ρ(xi , xk ) − ρ(xj , xk ))
1
2 (xj | xk )xi
and therefore
ρ(zi , zj ) = ρ(xi , xj ) − ρ(xi , zi ) − ρ(xj , zj )
≤ ρ(xi , xj ) −
1
2
(xj | xk )xi + (xi | xk )xj
= 21 ρ(xi , xj ) ≤ r.
Hence, there is some m := µ(z1 , z2 , z3 ). But since m ∈ I(zi , zj ) ⊂ I(xi , xj ) it
follows that m ∈ M (x1 , x2 , x3 ).
To prove uniqueness, assume that there are m1 , m2 ∈ M (x1 , x2 , x3 ). Let
qi := σm1 xi ( 21 ) and p := σm1 m2 ( 12 ). Then, we clearly have m1 ∈ M (q1 , q2 , q3 )
and d(qi , qj ) = 21 d(xi , xj ) ≤ r. Moreover, we have
ρ(qi , qj ) ≤ ρ(qi , p) + ρ(p, qj ) ≤ 12 ρ(xi , m2 ) + 12 ρ(xj , m2 ) = 21 ρ(xi , xj ) = ρ(qi , qj ),
i.e. p ∈ M (q1 , q2 , q3 ). By uniqueness of µ(q1 , q2 , q3 ), we therefore have p = m1
and thus ρ(m1 , m2 ) = 2ρ(m1 , p) = 0.
86
Appendix A
Some Results on Geodesic
Bicombings
A.1
Reversible Geodesic Bicombings
Recall that a geodesic bicombing σ : X × X × [0, 1] → X is reversible if, for all
x, y ∈ X and for all t ∈ [0, 1], we have σyx (t) = σxy (1 − t). In this section, we
show that it is no loss of generality to assume that a conical geodesic bicombing
is reversible as the following holds.
Proposition 1.1. Let (X, d) be a complete metric space with a conical geodesic
bicombing σ. Then X also admits a reversible, conical geodesic bicombing.
This generalizes the result for proper metric spaces established in [Des16,
Proposition 1.2]. To prove Proposition 1.1, we need the following midpoint
construction.
Lemma 1.2. Let (X, d) be a complete metric space. If σ : X × X × [0, 1] → X
is a conical geodesic bicombing, then there is a midpoint map m : X × X → X
with the following properties: For all points x, y, x̄, ȳ ∈ X, we have
(i) m(x, y) = m(y, x),
(ii) d(x, m(x, y)) = d(y, m(x, y)) = 12 d(x, y),
(iii) d(m(x, y), m(x̄, ȳ)) ≤ 12 d(x, x̄) + 12 d(y, ȳ).
Proof. Let x, y ∈ X. Set x0 := x, y0 := y and define recursively
xn+1 := σ(xn , yn , 12 ),
yn+1 := σ(yn , xn , 21 ).
We have
d(xn+1 , yn+1 ) = d(σ(xn , yn , 12 ), yn+1 )
≤ 21 d(xn , yn+1 ) + 21 d(yn , yn+1 ) = 12 d(xn , yn )
87
APPENDIX A. SOME RESULTS ON GEODESIC BICOMBINGS
and therefore d(xn , yn ) ≤ 21n d(x, y), d(xn , xn−1 ) ≤ 21n d(x, y). Hence, the sequences (xn )n≥0 , (yn )n≥0 are Cauchy and converge to some common limit point
m(x, y).
By the construction, we clearly have property (i). To prove (ii), we claim
that d(x, xn ), d(x, yn ), d(y, xn ), d(y, yn ) ≤ 12 d(x, y) for all n ≥ 1. This follows
by induction since
d(x, xn+1 ) ≤ 12 d(x, xn ) + 12 d(x, yn ) ≤ 12 d(x, y)
and similar for all other distances. By taking limits, we get
d(x, m(x, y)), d(y, m(x, y)) ≤ 12 d(x, y)
and equality follows since
d(x, y) ≤ d(x, m(x, y)) + d(m(x, y), y) ≤ 21 d(x, y) + 12 d(x, y) = d(x, y).
It remains to show (iii). If we repeat the construction for x̄, ȳ ∈ X, we get
some sequences (x̄n )n≥0 , (ȳn )n≥0 with limit point m(x̄, ȳ). We now prove by
induction that d(xn , x̄n ), d(yn , ȳn ) ≤ 12 d(x, x̄) + 12 d(y, ȳ) for all n ≥ 1. Indeed,
we have
d(xn+1 , x̄n+1 ) = d(σ(xn , yn , 12 ), σ(x̄n , ȳn , 12 ))
≤ 21 d(xn , x̄n ) + 21 d(yn , ȳn ) ≤ 12 d(x, x̄) + 21 d(y, ȳ),
and similarly d(yn+1 , ȳn+1 ) ≤ 21 d(x, x̄) + 12 d(y, ȳ). Hence, statement (iii) follows
by taking the limit n → ∞.
Proof of Proposition 1.1. We define a new bicombing τ : X × X × [0, 1] → X
by
τ (x, y, t) := m(σ(x, y, t), σ(y, x, 1 − t)).
(1.1)
For two points x, y ∈ X, this defines a geodesic from x to y since, for s, t ∈ [0, 1],
we have
d(τ (x, y, t), τ (x, y, s))
= d(m(σ(x, y, t), σ(y, x, 1 − t)), m(σ(x, y, s), σ(y, x, 1 − s))
≤ 21 d(σ(x, y, t), σ(x, y, s)) + 21 d(σ(y, x, 1 − t), σ(y, x, 1 − s))
= |s − t|d(x, y).
Moreover, the conical inequality holds, as we have
d(τ (x, y, t), τ (x̄, ȳ, t))
= d(m(σ(x, y, t), σ(y, x, 1 − t)), m(σ(x̄, ȳ, t), σ(ȳ, x̄, 1 − t)))
≤ 21 d(σ(x, y, t), σ(x̄, ȳ, t)) + 21 d(σ(y, x, 1 − t), σ(ȳ, x̄, 1 − t))
≤ (1 − t)d(x, x̄) + td(y, ȳ)
for all x, y, x̄, ȳ ∈ X and t ∈ [0, 1].
88
A.2. A NON-CONSISTENT CONVEX GEODESIC BICOMBING
A.2
A Non-Consistent Convex Geodesic Bicombing
The goal of this section is to construct a convex geodesic bicombing that is not
consistent. To this end, we consider the following norm on R2 :
n
o
√
k(x, y)k := max |x|, 22 k(x, y)k2 ,
p
where k(x, y)k2 = x2 + y 2 is the Euclidean norm. Observe that k(x, y)k = |x|
if and only if |y| ≤ |x|. Now define
1
X := (x, y) ∈ R2 : −3 ≤ x ≤ 3, 0 ≤ y ≤ 32
max{0, 1 − x2 }
and equip X with the metric d induced by k · k, see Figure A.1.
δ
σpq
p
-3
-2
-1
0
q
1
2
3
δ .
Figure A.1: The metric space X with a geodesic σpq
The space X naturally splits into three pieces, namely X = X− ∪ X0 ∪ X+
with
X− := [−3, −1] × {0},
X0 := (x, y) ∈ R2 : −1 < x < 1, 0 ≤ y ≤
1
32 (1
− x2 ) ,
X+ := [1, 3] × {0}.
1
Definition 2.1. For δ ∈ [0, 64
], we define a geodesic bicombing σ δ : D(X) → X
δ to be the geodesic from p to q which is linear
as follows. Generally, we take σpq
inside X0 , but if both endpoints lie on the antennas X− , X+ , we slightly modify
it. In more details, σ δ is defined as follows:
For p := (px , py ), q := (qx , qy ) ∈ X with px ≤ qx let
δ
σpq
(t) := (xpq (t), ypq (t)) ,
with
xpq (t) := px + t(qx − px ),

δ max{qx − px − 4, 0} max{0, (1 − xpq (t)2 )},




qy


max{0, qx +1 (xpq (t) + 1)},
y
ypq (t) := max{0, pxp−1
(xpq (t) − 1)},


py + t(qy − py ),




0,
and
δ
δ
σqp
(t) := σpq
(1 − t).
89
for p ∈ X− , q ∈ X+ ,
for p ∈ X− , q ∈ X0 ,
for p ∈ X0 , q ∈ X+ ,
for p, q ∈ X0 ,
otherwise.
APPENDIX A. SOME RESULTS ON GEODESIC BICOMBINGS
1
Proposition 2.2. For δ ∈ (0, 64
], the map σ δ is a reversible, convex geodesic
bicombing which is not consistent.
Remark. Observe that for δ = 0, the geodesic bicombing σ δ coincides with
the piece-wise linear bicombing, which is the unique consistent conical geodesic
bicombing on X by Theorem 1.4 in [BM16]. Hence, we have a family of nonconsistent, convex geodesic bicombings σ δ converging to the unique consistent,
convex geodesic bicombing σ 0 .
Alternatively, we can modify the geodesics leading from X− to X+ .
Definition 2.3. Define σ̃ δ : D(X) → X by
δ
δ
σ̃pq
(t) := σpq
(t),
except for p ∈ X+ , q ∈ X− , let
δ
σ̃pq
:= (xpq (t), 0).
1
Proposition 2.4. For δ ∈ (0, 64
], the map σ̃ δ is a convex geodesic bicombing
which neither is reversible nor consistent.
Let us first show that we defined geodesic bicombings.
1
Lemma 2.5. For δ ∈ [0, 64
], the maps σ δ and σ̃ δ are geodesic bicombings.
Proof. The linear case is clear. For the piecewise linear case, observe that, if
p ∈ X− , q ∈ X0 (and similarly in all other cases), we have that the slope m of
δ satisfies
σpq
m=
qy
≤
qx + 1
1
32 (1
− qx2 )
=
1 + qx
1
32 (1
− qx ) ≤
1
16
≤1
and therefore
δ
δ
d(σpq
(s), σpq
(t)) = |xpq (s) − xpq (t)| = |s − t||qx − px | = |s − t|d(p, q).
Finally, let p ∈ X− , q ∈ X+ . For x, x0 ∈ [−1, 1] we have
|δ(qx − px − 4)(1 − x2 ) − δ(qx − px − 4)(1 − x02 )|
≤ δ|qx − px − 4| · |x + x0 | · |x − x0 | ≤
1
16 |x
− x0 |
δ (s), σ δ (t)) = |x (s) − x (t)| as before.
and hence d(σpq
pq
pq
pq
To prove Proposition 2.2 and Proposition 2.4, we will use the following
characterization of convexity; see Lemma 3.5 in [LM10].
Lemma 2.6. A continuous function f : [0, 1] → R is convex if and only if for
every t ∈ (0, 1), there is some τ0 > 0 such that, for all τ ∈ [0, τ0 ], we have
2f (t) ≤ f (t − τ ) + f (t + τ ).
90
A.2. A NON-CONSISTENT CONVEX GEODESIC BICOMBING
Proof of Proposition 2.2. It is immediate that the geodesic bicombing σ δ is nonconsistent and reversible. Hence, it remains to prove convexity.
δ (t), σ δ (t)) is
Given p, q, p0 , q 0 ∈ X, we need to show that f (t) := d(σpq
p0 q 0
convex on [0, 1]. Clearly t 7→ |xpq (t) − xp0 q0 (t)| is convex for t ∈ [0, 1]. Hence, in
δ (t), σ δ (t)) = |x (t) − x 0 0 (t)|, we have
the situation when d(σpq
pq
pq
p0 q 0
δ
2d(σpq
(t), σpδ0 q0 (t)) = 2|xpq (t) − xp0 q0 (t)|
≤ |xpq (t − τ ) − xp0 q0 (t − τ )| + |xpq (t + τ ) − xp0 q0 (t + τ )|
δ
δ
≤ d(σpq
(t − τ ), σpδ0 q0 (t − τ )) + d(σpq
(t + τ ), σpδ0 q0 (t + τ )).
Therefore, it remains to check
δ
2kσpq
(t) − σpδ0 q0 (t)k2
δ
δ
≤ kσpq
(t − τ ) − σpδ0 q0 (t − τ )k2 + kσpq
(t + τ ) − σpδ0 q0 (t + τ )k2
√
2
δ (t), σ δ (t)) =
δ
δ
for τ > 0 small, whenever d(σpq
p0 q 0
2 kσpq (t) − σp0 q 0 (t)k2 , that is,
|xpq (t) − xp0 q0 (t)| ≤ |ypq (t) − yp0 q0 (t)|.
To this end, observe that, for (x, 0) ∈ X− ∪ X+ and (x0 , y 0 ) ∈ X, it holds
that d((x, 0), (x0 , y 0 )) = |x − x0 | and therefore, if xpq (t) ∈
/ (−1, 1), we always
δ (t), σ δ (t)) = |x (t) − x 0 0 (t)|. Hence, we only need to consider
have d(σpq
pq
pq
p0 q 0
points that satisfy xpq (t), xp0 q0 (t) ∈ (−1, 1).
δ and σ δ
First, if both geodesics σpq
p0 q 0 are (piece-wise) linear, then locally
they are linear geodesics inside a normed vector space and hence
d(σpq (t), σp0 q0 (t)) = kσpq (t) − σp0 q0 (t)k
is locally convex, thus convex.
δ is not linear, i.e. p ∈ X , q ∈ X , l := d(p, q) ≥
Let us now assume that σpq
−
+
δ
4. We look at the different options for σp0 q0 separately. But before doing so, let
us first fix some notation. We define p0 := σpq (t), p± := σpq (t±τ ), p∗ = (x∗ , y∗ )
(∗ ∈ {0, +, −}), D := δ(l − 4), := τ l and accordingly for σpδ0 q0 . We then get
y0 = D(1 − x20 ), x± = x0 ± and y± = D(1 − (x0 ± )2 ).
In each case, we need to consider the situation where x0 , x00 ∈ (−1, 1) and
|x0 − x00 | ≤ |y0 − y00 |.
Case 1. p0 ∈ X∓ , q 0 ∈ X± and l0 := d(p0 , q 0 ) ∈ [4, l].
0
0
0
0
0
0
0 2
As above, we have y00 = D0 (1 − x02
0 ), x± = x0 ± , y± = D (1 − (x0 ± ) ) and
l0
with λ := l we get 0 = λ.
We claim that
2kp0 − p00 k2 ≤ kp− − p0− k2 + kp+ − p0+ k2
for > 0 (i.e. τ > 0) small enough.
91
APPENDIX A. SOME RESULTS ON GEODESIC BICOMBINGS
First, note that
kp− − p0− k22
= kp0 − p00 k22 − 2(x0 − x00 )(1 − λ) + (1 − λ)2 2 + 2(y0 − y00 )a + a2 2 ,
kp+ − p0+ k22
= kp0 − p00 k22 + 2(x0 − x00 )(1 − λ) + (1 − λ)2 2 + 2(y0 − y00 )b + b2 2 ,
for
a := 2(x0 D − λx00 D0 ) − (D − λ2 D0 ),
b := −2(x0 D − λx00 D0 ) − (D − λ2 D0 ),
with a + b = −2(D − λ2 D0 ), a − b = 4(x0 D − λx00 D0 ) and either ab = (D −
λ2 D0 )2 2 or ab < 0 for small. In the following, we assume ab < 0. The other
case is similar. Moreover, we have
kp− − p0− k22 · kp+ − p0+ k22 = kp0 − p00 k42
+ 4ab(y − y 0 )2 − 4(x0 − x00 )2 (1 − λ)2 + 4(x0 − x00 )(1 − λ)(y0 − y00 )(a − b)
+ 2(1 − λ)2 + a2 + b2 − 4(y0 − y00 )(D − λ2 D0 ) · kp0 − p00 k22 2 + O(3 )
and, with
√
u+t=
√
u+
t
√
2 u
+ O(t2 ) and u = kp0 − p00 k42 , it follows
q
2 kp− − p0− k22 · kp+ − p0+ k22
0 2
≥ 2kp0 − p0 k2 + 2(1 − λ)2 + a2 + b2 + 4ab − 4(y0 − y00 )(D − λ2 D0 )
4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2
+ O(3 ).
+
(x0 − x00 )2 + (y0 − y00 )2
We therefore get
kp− − p0− k2 + kp+ − p0+ k2
2
q
kp− − p0− k22 + kp+ − p0+ k22 + 2 kp− − p0− k22 · kp+ − p0+ k22
0 2
≥ 4kp0 − p0 k2 + 4(1 − λ)2 + 2(a + b)2 − 8(y0 − y00 )(D − λ2 D0 )
4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2
+ O(3 )
+
(x0 − x00 )2 + (y0 − y00 )2
=
=
4kp0 − p00 k22 + C2 + O(3 )
≥
4kp0 − p00 k22
for > 0 small enough, provided that
4(1 − λ)2 − 8(y0 − y00 )(D − λ2 D0 )
C=
+
16(x0 − x00 )(y0 − y00 )(1 − λ)(x0 D − λx00 D0 ) − 4(x0 − x00 )2 (1 − λ)2
> 0.
(x0 − x00 )2 + (y0 − y00 )2
92
A.2. A NON-CONSISTENT CONVEX GEODESIC BICOMBING
Observe that a + b = O().
First, assuming y0 > y00 , we have
2
y0 − y00 = D(1 − x20 ) − D0 (1 − x20 ) = (D − D0 )(1 − x20 ) + D0 (x02
0 − x0 )
≤ δ(l − l0 ) + δ(l0 − 4)(x00 + x0 )(x00 − x0 ) ≤ δ(l − l0 ) + 4δ(y0 − y00 )
and therefore
|y0 − y00 | ≤
δ
(l − l0 ).
1 − 4δ
Moreover,
|D − λ2 D0 |l2 = δ(l3 − 4l − l03 + 4l0 )
= δ(l − l0 )(l2 + ll0 + l02 − 4(l + l0 )) ≤ 60δ(l − l0 ),
|x0 D − λx00 D0 |l ≤ |x0 |(D − λD0 )l + |x0 − x00 |D0 l0
0
0
≤ δ(l − l )(l + l − 4) + 12δ|y0 −
y00 |
≤
12δ 2
8δ +
1 − 4δ
(l − l0 ).
Hence, we finally get
Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2
− 8(y0 − y00 )(D − λ2 D0 )l2 (x0 − x00 )2 + (y0 − y00 )2
+ 16(x0 − x00 )(y0 − y00 )(l − l0 )(x0 D − λx00 D0 )l
960δ 2
192δ 2
≥
4−
− 128δ −
(l − l0 )2 (y0 − y00 )2
1 − 4δ
1 − 4δ
4 − 144δ − 640δ 2
=
(l − l0 )2 (y0 − y00 )2 > 0
1 − 4δ
for δ <
1
40 .
This is especially true for δ ≤
1
64 .
Case 2. σp0 q0 piece-wise linear with p0 ∈
/ X0 or q 0 ∈
/ X0 .
Let m be the slope of σp0 q0 at p00 . If p0 ∈ X− and q 0 ∈ X0 , then we have
1
02
0
1
−
q
x
qy
32
1
1
1
(1 − qx0 ) ≤ 32
(4 − l0 ) ≤ 32
(l − l0 ),
m= 0
≤
= 32
qx + 1
1 + qx
1
and, similarly, we also get |m| ≤ 32
(l − l0 ) in all other cases and especially
|m| ≤ 1.
0
Moreover, we have l ∈ [4, 6], l0 ∈ [0, 4] and, for 0 = τ l0 , λ = ss , we get
0 = y 0 ± m0 and 0 = λ.
x0± = x00 ± 0 , y±
0
We can proceed as before with
a = λm + 2Dx0 − D,
b = −λm − 2Dx0 − D,
93
APPENDIX A. SOME RESULTS ON GEODESIC BICOMBINGS
and we finally get the constant
4(1 − λ)2 − 8(y0 − y00 )D
C=
+
8(x0 − x00 )(y0 − y00 )(1 − λ)(λm + 2Dx0 ) − 4(x0 − x00 )2 (1 − λ)2
.
(x0 − x00 )2 + (y0 − y00 )2
With
D = δ(l − 4) ≤ δ(l − l0 ),
y0 − y00 ≤ D(1 − x20 ) ≤ D ≤ δ(l − l0 ),
it follows
Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2
− 8(y0 − y00 )Dl2 (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(l − l0 )(ml0 + 2Dx0 l)
≥
4 − 576δ 2 − 1 − 96δ (l − l0 )2 (y0 − y00 )2
=
3 − 96δ − 576δ 2 (l − l0 )2 (y0 − y00 )2 > 0
for δ < 0.026.
Case 3. σp0 q0 linear with p0 , q 0 ∈ X0 .
Let m again denote the slope of σp0 q0 . We distinguish two subcases.
(a) If |m| ≤ 1, we have l ∈ [4, 6], l0 ∈ [0, 2] and
|ml0 | =
|qy0 − p0y | 0
l = |qy0 − p0y | ≤
|qx0 − p0x |
1
32 .
0
0 = y 0 ± m0 and
Moreover, for 0 = τ l0 , λ = ss , we get x0± = x00 ± 0 , y±
0
0 = λ as before and again, we get the constant
4(1 − λ)2 − 8(y0 − y00 )D
C=
+
8(x0 − x00 )(y0 − y00 )(1 − λ)(λm + 2Dx0 ) − 4(x0 − x00 )2 (1 − λ)2
.
(x0 − x00 )2 + (y0 − y00 )2
Now, we estimate
Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2
− 8(y0 − y00 )Dl2 (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(l − l0 )(ml0 + 2Dx0 l)
≥
4 − 576δ 2 − 81 − 96δ (l − l0 )2 (y0 − y00 )2
2
31
−
96δ
−
576δ
(l − l0 )2 (y0 − y00 )2 > 0
=
8
for δ < 0.033.
94
A.2. A NON-CONSISTENT CONVEX GEODESIC BICOMBING
(b) If |m| > 1, we have l ∈ [4, 6] and
l0 =
√
2
2
q
(qx − px )2 + (qy − py )2 ≤ |qy − py | ≤
1
32 .
Furthermore, let 0x = x+ − x0 and 0y = y+ − y0 . Then we have 0y = m0x
and
2
2
2(τ l0 )2 = 0x + (m0x )2 = (1 + m2 )0x .
Thus, we get
0x = λx for λx :=
√
l0 √ 2
l 1+m2 ,
0y = λy for λy := mλx =
√
l0 √ 2m
l 1+m2 ,
x0± = x00 ± 0x , and
0
y±
= y00 ± 0y .
We proceed again as before and get the constant
4(1 − λx )2 − 8(y0 − y00 )D
C=
+
8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 ) − 4(x0 − x00 )2 (1 − λx )2
,
(x0 − x00 )2 + (y0 − y00 )2
with
D = δ(l − 4) ≤ δ(l − l0 ) ≤ 6δ(1 − λx ),
y0 − y00 ≤ D(1 − x20 ) ≤ D ≤ 6δ(1 − λx ),
√
√
l0
2
2
1
λy = q
≤
≤ (1 − λx ).
l
128
64
1
+1
m2
Now, we estimate
Ckp0 − p00 k22 =
4(1 − λx )2 (y0 − y00 )2
− 8(y0 − y00 )D (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 )
1
2
≥
4 − 576δ −
− 96δ (1 − λx )2 (y0 − y00 )2
64
2
255
=
(1 − λx )2 (y0 − y00 )2 > 0
64 − 96δ − 576δ
for δ < 0.034. Hence, this is again true for δ ≤
1
64 .
Observe that, for m → +∞, we get λx = 0 and λy =
holds.
95
√
0
2 ll and the same
APPENDIX A. SOME RESULTS ON GEODESIC BICOMBINGS
Proof of Proposition 2.4. Clearly, the geodesic bicombing σ̃ δ is non-consistent
and non-reversible.
For convexity, the same arguments as in the proof of Proposition 2.2 apply.
The only new case is p0 ∈ X+ and q 0 ∈ X− . With the notions from above with
0
x0± = x00 ∓ 0 for 0 = τ l0 and λ = ll we obtain the constant
4(1 + λ)2 − 8y0 D
C=
+
16(x0 − x00 )y0 (1 + λ)x0 D − 4(x0 − x00 )2 (1 + λ)2
.
(x0 − x00 )2 + y02
With the inequalities D = δ(l − 4) ≤ 2δ and |y0 | ≤
Ckp0 − p00 k22 =
1
32 ,
we get
4(1 + λ)2 y02
− 8y0 D (x0 − x00 )2 + y02
+ 16(x0 − x00 )y0 (1 + λ)x0 D
for δ <
4
33 ,
hence for all δ ≤
≥
(4 − δ − 32δ) (1 + λ)2 y02 > 0
≥
(4 − 33δ) (1 + λ)2 y02 > 0
1
64 .
96
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INDEX
Index
absolute 1-Lipschitz retract, 7
absolute 1-Lipschitz uniform
neighborhood retract, 52
admissible, 8
rank, 76
median metric space, 75
metric interval, 2, 15, 55, 75
metric space
proper, 18
metrically convex, 55
modular, 55
combinatorial dimension, 50
cube complex, 77
curve
length, 43
straight, 66
n-hyperconvex, 4, 55
almost, 4
almost externally, 56
almost weakly externally, 56
externally, 4, 55
weakly externally, 4, 55
extreme point, 38
gated, 28
geodesic, 43
consistent, 44
geodesic bicombing, 3
conical, 44
consistent, 3, 44
convex, 3, 44
reversible, 3, 6, 44, 87
gluing, 23
proximinal, 19, 55
s-constant retraction, 13
σ-convex, 5, 65
strongly convex, 2
tangent cone, 37
Helly family, 66
hyperconvex, 1, 8
externally, 2, 8
locally, 49
locally externally, 65, 69
locally weakly externally, 65, 69
weakly externally, 2, 8, 13
injective metric space, 7
length space, 43
local geodesic bicombing, 44
convex, 44
median algebra, 75
101

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