Circumcenter extension maps for non-positively curved spaces
aa r X i v : . [ m a t h . M G ] S e p CIRCUMCENTER EXTENSION MAPS FORNON-POSITIVELY CURVED SPACES
MERLIN INCERTI-MEDICI
Abstract.
We show that every cross ratio preserving homeomorphismbetween boundaries of Hadamard manifolds extends to a continuousmap, called circumcenter extension, provided that the manifolds satisfycertain visibility conditions. We show that this map is a rough isometry,whenever the manifolds admit cocompact group actions by isometriesand we improve the quasi-isometry constants provided by Biswas in thecase of
CAT( − spaces. Finally, we provide a sufficient condition forthis map to be an isometry in the case of Hadamard surfaces. Contents
1. Introduction 12. Preliminaries 72.1. Boundaries at infinity, Gromov products and Busemannfunctions 72.2. Cross ratios 102.3. Metric derivatives 212.4. Convex functions 252.5. Visibility and algebraic visibility 252.6. Jacobi fields 283. Construction of Φ and F Φ F F Introduction
The visual boundary of a geodesically complete
CAT( − space is nat-urally endowed with a cross ratio. In this paper, we show that this crossratio can still be defined on the visual boundary of a geodesically complete CAT(0) space and that it still retains a lot of information about the interiorspace, provided that certain visibility conditions are satisfied.Our core motivation is the following, informal question, which is related toseveral results from geometry and geometric group theory in recent decades.
Question.
Let ( X, d ) and ( Y, d ) be two geodesically complete CAT(0) spacesand f : ∂X → ∂Y a cross ratio-preserving homeomorphism between theirboundaries. Can f be extended to an isometry F : X → Y ?Alternatively, if a group G acts on X and Y by isometries and f is a G -equivariant, cross ratio-preserving homeomorphism, can we construct F to be G -equivariant?This question has seen a series of complete and partial answers for variousspecial cases over the course of the last few decades. The first series of re-sults concern situations where X and Y admit geometric actions, i.e. proper,cocompact actions by isometries, by some group G . Specifically, if X and Y have constant negative curvature, the extension of f to a G -equivariantisometry is used in Thurston’s proof of Mostow rigidity [Thu80]. If X and Y are universal coverings of negatively curved surfaces, the fact that f ex-tends to a G -equivariant isometry is crucial to Otal’s proof of Marked LengthSpectrum Rigidity [Ota90]. In a series of papers, Hamenstädt and Besson-Courtois-Gallot show that f extends to a G -equivariant isometry, if bothspaces are universal coverings of negatively curved manifolds and one ofthem is a locally symmetric space [BCG95, Ham99].A second, more recent series of results mostly drops the assumptions aboutgroup actions. If X and Y are proper, geodesically complete CAT( − spaces, Biswas proved that f can be extended to a rough isometry, i.e. a (1 , C ) -quasi-isometry [Bis15]. One may also consider classes of spaces forwhich the appropriate notions are no longer the visual boundary and isome-tries. Beyrer, Fioravanti and the author proved a similar extension theoremfor CAT(0) cube complexes and their Roller boundaries, equipped with asuitable cross ratio [BFIM18]. Furthermore, Beyrer-Fioravanti proved ad-ditional extension theorems for cubulable hyperbolic groups and for certaingroup actions on nice
CAT(0) cube complexes [BF18a, BF18b].A third collection of results can be found when thinking of the questionabove in more dynamical terms. There are various instances, where the vi-sual boundary is closely related to the geodesic flow of the interior space.For example, if X is the universal covering of a closed, negatively curvedRiemannian manifold, then the cross ratio on its boundary and the MarkedLength Spectrum of the closed manifold determine each other (see for ex-ample [Ham92, Bis15] for how the cross ratio determines the Marked LengthSpectrum and [Ota92, Kim01, Kim04] for the converse). De Simoi-Kaloshin-Wei have proven a Length Spectrum Rigidity result for billiards that featuresufficient symmetries and whose boundaries are sufficiently close to the cir-cle [dSKW17]. We also mention the work of Kaloshin-Sorrentino, who show IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 3 that, if a strictly convex billiard has the same maximal marked length spec-trum as an ellipse, then it is an ellipse [KS18]. It seems to be an openquestion, whether these results can be formulated in a geometric languagesimilar to the one of this paper.Finally, one may take a coarse viewpoint and only require that f coarselypreserves the cross ratio. This is called a quasi-Möbius map. In [CCM19],Charney-Cordes-Murray show that, under a mild stability condition, quasi-Möbius maps between Morse boundaries of finitely generated groups extendto quasi-isometries of groups.While cross ratios have been used on numerous occasions in spaces ofnegative curvature, they have not been studied very much for general non-positively curved spaces. In part, this is due to the fact that visual bound-aries of hyperbolic or CAT( − spaces have several properties that visualboundaries of CAT(0) spaces don’t. In this paper, we show how to workaround these difficulties to define a cross ratio on the visual boundary of aproper, geodesically complete
CAT(0) space. We then show that the cir-cumcenter construction introduced in [Bis15, Bis17b] can be generalized toa large class of non-positively curved manifolds and that this circumcenterextension provides a good framework to study the initially stated question.In order to state our results, we first need to define the cross ratio. Let X be a proper, connected, geodesically complete CAT(0) space. Fix a basepoint o ∈ X . For any admissible quadruple ( ξ , ξ , ξ , ξ ) ∈ ∂X (see section2.2 for the definition of admissibility), we can define the cross ratio cr ( ξ , ξ , ξ , ξ ) := ρ o ( ξ , ξ ) ρ o ( ξ , ξ ) ρ o ( ξ , ξ ) ρ o ( ξ , ξ ) , where ρ o ( ξ, η ) := e − ( ξ | η ) o , with ( ξ | η ) o denoting the Gromov product on ∂X with respect to the base point o . Proposition A.
Let X be a proper, connected, geodesically complete CAT(0) space. The cross ratio cr is well-defined for all admissible quadruples andindependent of the choice of o . It turns out that the boundary, together with the cross ratio, containsa lot of information about the interior space, provided that the boundarysatisfies certain visibility properties. Specifically, we say:(1) The visual boundary ∂X satisfies , if for every quadruple ( ξ , ξ , ξ , ξ ) ∈ ∂X , there exists η ∈ ∂X , such that for all i ∈{ , , , } , ( ξ i | η ) o is finite for some base point o ∈ X .(2) We say that ξ ∈ ∂X is in a rank 1 hinge if there exist η, ζ ∈ ∂X ,such that there is a bi-infinite geodesic from η to ζ and there existbi-infinite rank 1 geodesics from ξ to η and from ξ to ζ .Given two proper, connected, geodesically complete CAT(0) spaces X and Y , we say that a map f : ∂X → ∂Y is Möbius if and only if it preserves thecross ratio.
MERLIN INCERTI-MEDICI
Let
X, Y be Hadamard manifolds, i.e. simply connected, geodesically com-plete Riemannian manifolds such that all sectional curvatures are non-positive.The main result of this paper is a construction that allows us to extendMöbius homeomorphisms that satisfy one mild extra condition to the inte-rior spaces, provided that ∂X and ∂Y satisfy conditions (1) and (2) above.We call this extension the circumcenter extension of f . We prove a very gen-eral Theorem about the circumcenter extension in section 4. This Theoremhas several consequences, as soon as one adds some extra assumption. If X and Y admit a cocompact group action, we obtain Theorem B.
Let
X, Y be Hadamard manifolds such that ∂X, ∂Y satisfy(1) and all points in ∂X and ∂Y satisfy (2). Suppose, the group G actscocompactly by isometries on X and Y . Let f : ∂X → ∂Y be a G -equivariantMöbius homeomorphism, such that f and f − send visible pairs to visiblepairs. Then the circumcenter extension is a G -equivariant (1 , M ) -quasi-isometry F : X → Y for some constant M ≥ . The constant M will be the supremum of a Lipschitz continuous function M : X → [0 , ∞ ) , which essentially measures by how much F fails to be anisometry at a certain point. A better understanding of the function M hasthe potential to significantly improve this result, the main result of the paperand the results stated below.If we drop the assumption about cocompact group actions, we can restrictto more specialised situations and obtain other, sometimes even strongerresults. The first of these results concerns surfaces. Theorem C.
Let
X, Y be Hadamard manifolds whose sectional curvaturesare bounded from below by − b such that ∂X, ∂Y satisfy (1) and all points in ∂X and ∂Y satisfy (2). Assume that X, Y are -dimensional. Let f : ∂X → ∂Y be a Möbius homeomorphism such that f and f − send visible pairsto visible pairs. Then, the circumcenter extension is a homeomorphism F : X → Y , it is locally Lipschitz continuous on a dense subset and differentiablealmost everywhere.Furthermore, to every x ∈ X , we can associate a set K x ⊂ ∂X with thefollowing property: For almost every x , F is differentiable at x and if K x contains at least five points, then DF x is an isometry between tangent spaces. We emphasize that the Theorem above includes the statement that thecircumcenter extension is invertible, a result that we do not obtain in higherdimensions.In [Bis15], Biswas proved that the circumcenter extension provides a (1 , ln(2)) -quasi-isometry, if X and Y are CAT( − spaces. For manifoldswith a lower curvature bound − b , it is a (1 , (1 − b ) ln(2)) -quasi-isometry(cf. [Bis17a]). We can obtain the following constant. IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 5
Theorem D.
Let
X, Y be Hadamard manifolds whose sectional curvaturesare bounded from below by − b and from above by − . Let f : ∂X → ∂Y be aMöbius homeomorphism. Then, the circumcenter extension is a (1 , ln( √ -quasi-isometry.Furthermore, if X and Y are -dimensional, then the circumcenter exten-sion is a (1 , ln( √ )) -quasi-isometry. We end the introduction with a discussion on what kind of spaces satisfythe visibility properties necessary for our results (see section 2 for all def-initions not given here). Many examples can be constructed by using thenotion of visibility points in the boundary. A point ξ ∈ ∂X is called a vis-ibility point if it can be connected with every other point in the boundaryby a bi-infinite geodesic in X . Let ξ ∈ ∂X be a visibility point, η ∈ ∂X and γ a bi-infinite geodesic from ξ to η . It follows that γ is a rank onegeodesic, as the end points of geodesic lines that are not rank one (i.e. thatbound a euclidean halfplane) cannot be visibility points. Since two pointsin the boundary that can be connected by a bi-infinite geodesic always havefinite Gromov product, we conclude that, whenever ∂X contains at least fivevisibility points, both visibility properties introduced above are satisfied.We now present a class of Hadamard manifolds that do not have strictlynegative curvature and do admit five visibility points. Consider a closed,non-positively curved Riemannian manifold M . By the rank rigidity theorem(see Theorem C in [Bal95]), we obtain that its universal covering ˜ M is eithera finite Riemannian product, a higher rank symmetric space, or containsat least one bi-infinite rank one geodesic. By Lemma 1.7 in [BB08], theendpoints of this rank one geodesic in ∂X have Tits distance strictly greaterthan π . Because π ( M ) acts properly and cocompactly, its limit-set, denoted Λ , satisfies Λ = ∂X (see the introduction of [BB08]). By Proposition 1.10in [BB08], this implies that there exists an element g ∈ π ( M ) and a rankone geodesic γ in ˜ M , such that g acts as translation on γ . By Theorem5.4 in [BF09], an axis for some isometry in a proper CAT(0) space is rankone if and only if it is contracting. Therefore, γ is a contracting geodesicline. In particular, both of its endpoints in ∂X are visibility points byProposition 3.6 in [CS15]. Since every orbit of π ( M ) in ∂X is dense (seefor example [Ham09], in particular Lemma 5.1) and isometries send visibilitypoints to visibility points, we conclude that ∂ ˜ M has infinitely many visibilitypoints. Therefore, the universal covering of any closed, non-positively curvedRiemannian manifold M is either a finite Riemannian product, a higherrank symmetric space, or it satisfies the visibility properties (1) and (2).This provides us with a large class of spaces satisfying our assumptions. Inparticular, this includes most graph manifolds.We also mention a non-cocompact example that can be obtained as fol-lows. Consider five copies of the euclidean upper halfplane R × [0 , ∞ ) andglue them together isometrically along their boundary R × { } such thatall five halfplanes intersect at the origin. The space obtained this way is a MERLIN INCERTI-MEDICI
CAT(0) space and its Tits boundary is a circle of circumference π . Usingproperties of the Tits metric and rank one geodesics, one can see that thisspace satisfies visibility properties (1) and (2) as well. While this exampleis only a CAT(0) and not a Riemannian manifold it seems feasible that aHadamard manifold with the same behaviour can be constructed. Both of theexamples above illustrate that there is a large and flexible class of Hadamardmanifolds that satisfy our visibility properties, but do not admit a negativeupper curvature bound, which shows that the circumcenter extension mapindeed can be constructed in a more general setting than previously thought.The remainder of the paper is organised as follows. In section 2, wedevelop all the necessary preliminary theory. Specifically, we give a briefintroduction to asymptotic geometry and generalise several results knownfor
CAT( − spaces to CAT(0) spaces, including the proof of Proposition A;we generalise the theory of metric derivatives as needed and we give a briefprimer on the facts we will need about convex functions and Jacobi fields.We also provide an example of a space whose boundary contains points thathave finite Gromov product but are not visible (this example is the reasonwhy f and f − have to send visible pairs to visible pairs). In section 3, weconstruct the circumcenter extension and define all the notions we will use toprove the results above. In section 4, we prove a result on Hölder continuityof the circumcenter extension and finish the proof of the main result (seeTheorem 4.6). In section 5, we prove Theorems B, C and D. Sections 4 and5 are written so that they can be read independently. Acknowledgments
The author thanks Viktor Schroeder for countlessdiscussions, many helpful questions, ideas and suggestions. The authorthanks Jean-Claude Picaud for several discussions, ideas, for proposing theoriginal project and for his comments on an early draft of the first partof this paper. The author is grateful to Giuliano Basso, Ruth Charney,Matthew Cordes, Peter Feller, Yannick Krifka, Urs Lang, Marc Lischka,Alessandro Sisto, Davide Spriano and Luca Studer for their patient listen-ing, their inputs, suggestions and counter examples concerning all thingsgeometric and to Elia Fioravanti for the same and additionally for teachingthe author to always consult [BH99]. The author thanks Manuela Gehrigand Simone Steinbruechel for their patience, ideas and for pointing out sev-eral dead ends concerning all things analytic. Finally, the author is grate-ful to Kingshook Biswas for his comments, suggestions and for his work in[Bis15, Bis17a, Bis17b, Bis18], which was crucial to the work presented here.Part of this work was carried out at the Simons Semester in spring 2019 sup-ported by the grant 346300 for IMPAN from the Simons Foundation and thematching 2015-2019 Polish MNiSW fund. The author has been supportedby the Swiss National Science Foundation Grant 175567.
IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 7 Preliminaries
Boundaries at infinity, Gromov products and Busemann func-tions.
For a general introduction to spaces of non-positive curvature, werefer to [BGS85] and [BH99]. For more material on asymptotic geometry,we additionally refer to [BS07].Let ( X, d ) be a proper, connected CAT(0) space. (Later, we will specializeto an n -dimensional, connected, geodesically complete Riemannian manifold ( X, g ) , such that all sectional curvatures are non-positive.) Since ( X, d ) is CAT(0) , the functions d ( γ ( t ) , γ ′ ( t )) and d ( x, γ ( t )) are convex and strictlyconvex respectively for all geodesics γ, γ ′ and x ∈ X .A geodesic ray is an isometric embedding γ : [ a, ∞ ) → X . A bi-infinitegeodesic is an isometric embedding defined on R . To make notation easier,all our geodesic rays will start at time a = 0 . A metric space is called geodesically complete if and only if all geodesic segments can be extended togeodesic lines.Assume from now on, that ( X, d ) is a proper, connected, geodesicallycomplete CAT(0) space. Two geodesic rays γ , γ ′ are called asymptotic ifthere exists B > such that for all t ≥ , d ( γ ( t ) , γ ′ ( t )) ≤ B . The boundaryat infinity of X is defined as the space of equivalence classes of geodesic rays ∂X := { γ | γ a geodesic ray } (cid:30) asymptoticity. Given a representative γ of ξ , we sometimes write ξ = γ ( ∞ ) and call ξ the endpoint of γ . A geodesic ray γ : ( −∞ , → X can also be interpretedas a representative of a point ξ at infinity and we write ξ = γ ( −∞ ) . Inparticular, a bi-infinite geodesic γ defines two points γ ( ∞ ) , γ ( −∞ ) in ∂X .Since X is CAT(0) , any two bi-infinite geodesics whose endpoints are ξ, η are parallel. We denote the set of geodesics with γ ( −∞ ) = ξ, γ ( ∞ ) = η by [ ξ, η ] . Whenever there exists a bi-infinite geodesic with endpoints ξ, η we callit a geodesic from ξ to η and we say that ( ξ, η ) is visible . We call an n -tuple ( ξ , . . . , ξ n ) a visible n -tuple , whenever for all i = j, ( ξ i , ξ j ) is visible.We can equip the boundary at infinity with a topology called the visualtopology. It is defined as follows. Fix a base point x ∈ X . For any ξ ∈ ∂X ,denote the unique geodesic ray starting at x , representing ξ by ξ x . (Theexistence of such geodesics is a well-known application of the Theorem ofArzela-Ascoli.) Let ξ ∈ ∂X . For all R > , ǫ > , define U R,ǫ,x ( ξ ) := { η ∈ ∂X | d ( η x ( R ) , ξ x ( R )) < ǫ } . It is easy to see that the collection { U R,ǫ,x ( ξ ) } R,ǫ,ξ forms a basis for a topol-ogy on ∂X , the visual topology. Furthermore, this topology is independentof x (see Part II, Section 8 in [BH99]).The following family of functions is a valuable tool when studying ∂X .Fix a base point x ∈ X and define the Gromov product of two points ξ , MERLIN INCERTI-MEDICI η ∈ ∂X with respect to x to be ( ξ | η ) x := lim t →∞
12 ( d ( x, ξ x ( t )) + d ( x, η x ( t )) − d ( ξ x ( t ) , η x ( t )))= lim t →∞ t − d ( ξ x ( t ) , η x ( t )) . This limit exists, since the function on the right-hand-side is non-decreasingin t , although ( ξ | η ) x may be infinite, e.g. if ξ = η , or if ξ x , η x span a flatsector in X (think of a sector in R ). If there exists ǫ > , such that X is CAT( − ǫ ) , then ( ξ | η ) x is infinite if and only if ξ = η . Note that, if ( ξ | η ) x = ∞ for some base point x , then ( ξ | η ) y = ∞ for all y ∈ X . We define ρ x ( ξ, η ) := e − ( ξ | η ) x . Remark 2.1.
We remark that we could define the Gromov product to bethe limit lim n →∞ ( ξ x ( t n ) | η x ( t ′ n )) x for any two sequences t n , t ′ n n →∞ −−−→ ∞ . Dueto monotony, all these limits are equal.We will also use the following notation. In analogy to the Gromov producton the boundary, we define for any triple x, y, z ∈ X the Gromov product of y, z with respect to x by ( y | z ) x := 12 ( d ( x, y ) + d ( x, z ) − d ( y, z )) . In order to understand how the Gromov product depends on the choice ofbase point, we use the Busemann function: Let x , y ∈ X and ξ ∈ ∂X . TheBusemann function is defined by B ( x, y, ξ ) := lim t →∞ d ( y, ξ x ( t )) − d ( x, ξ x ( t )) . Using the triangle-inequality, we see that the function on the right-hand-side is bounded in absolute value and non-increasing, hence this limit existsand is finite. Further, the Busemann function is continuous in x and y .In Appendix A.2 of [DPS12], it is shown that in CAT(0) spaces, for allgeodesic rays γ asymptotic to ξ x , B ( x, y, ξ ) = lim t →∞ d ( y, γ ( t )) − d ( x, γ ( t )) . This independence of the representative of ξ implies that for all x, y, z ∈ X and ξ ∈ ∂X , we have the cocycle equation B ( x, y, ξ ) + B ( y, z, ξ ) = lim t →∞ d ( y, γ ( t )) − d ( x, γ ( t )) + d ( z, γ ( t )) − d ( y, γ ( t ))= B ( x, z, ξ ) . In particular, B ( x, y, ξ ) = − B ( y, x, ξ ) . Finally, the Busemann function is also continuous in ξ with respect to thevisual topology on ∂X Let γ, γ ′ be two asymptotic geodesic rays representing ξ ∈ ∂X . Extendboth of them to bi-infinite geodesic lines. For all T ∈ R , there exists T ′ such IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 9 ξγ (0) γ ′ (0) γ ′ new (0) γ ( T ) γ ′ ( T ′ ) Figure 1.
Any two geodesic rays representing the samepoint ξ at infinity can be reparametrised so that at any giventime, they lie on the same horosphere centered at ξ . In thisfigure, one may reparametrise γ ′ as γ ′ new ( t ) = γ ′ ( t + T ′ − T ) .that γ ( T ) , γ ′ ( T ′ ) lie on the same horosphere of ξ , i.e. B ( γ ′ ( T ′ ) , γ ( T ) , ξ ) = 0 .In fact, T ′ = B ( γ ′ ( T ′ ) , γ ′ (0) , ξ )= B ( γ ′ ( T ′ ) , γ ( T ) , ξ ) + B ( γ ( T ) , γ (0) , ξ ) + B ( γ (0) , γ ′ (0) , ξ )= T + B ( γ (0) , γ ′ (0) , ξ ) . Definition 2.2.
Let x, y ∈ X and ξ ∈ ∂X . For k > , there exists an- up to isometry unique - constellation x, y ∈ H − k , ξ ∈ ∂ H − k , such that d ( x, y ) = d ( x, y ) and B ( x, y, ξ ) = B ( x, y, ξ ) . We call ( x, y, ξ ) a comparisontriangle of ( x, y, ξ ) in H − k . We call the angle ∠ x ( y, ξ ) the comparison angle to ( x, y, ξ ) in curvature − k and denote it by ∠ ( − k ) x ( y, ξ ) .For k = 0 , we analogously find x, y ∈ R , ξ ∈ ∂ R satisfying the sameequations. We use the same terminology and denote the comparison angleto ( x, y, ξ ) in curvature by ∠ (0) x ( y, ξ ) .Note that ∠ ( − k ) x ( y, ξ ) is continuous in x, y and ξ for all k ≥ . TheBusemann function B ( x, y, ξ ) can be described completely in terms of thedistance d ( x, y ) and its comparison triangle with respect to one model spaceof constant curvature. Specifically, ∀ k > e kB ( x,y,ξ ) = cosh( kd ( x, y )) − sinh( kd ( x, y )) cos( ∠ ( − k ) x ( y, ξ )) B ( x, y, ξ ) = − d ( x, y ) cos( ∠ (0) x ( y, ξ )) . The first of these formulas is proven in [Bis17a]. We prove the second onehere, since it is not easy to find in the literature.
Proof.
Let x, y ∈ X, ξ ∈ ∂X and let γ be a geodesic ray in R . Denote x := γ (0) and ξ := [ γ ] . For every t > , there exist exactly two points y ( t ) , y ( t ) ∈ R , such that ( x, y i , γ ( t )) is a comparison triangle to ( x, y, ξ x ( t )) (by definition a triangle with the same side lengths). For every t , choose oneof these two points, denoted y ( t ) , such that y ( t ) varies continuously in t .Since ( x, y ( t ) , γ ( t )) are comparison triangles to ( x, y, ξ x ( t )) , we have d ( y ( t ) , γ ( t )) − d ( x, γ ( t )) = d ( y, ξ x ( t )) − d ( x, ξ x ( t )) t →∞ −−−→ B ( x, y, ξ ) . Since y ( t ) is a bounded curve, it admits a converging subsequence. Theequation above implies that any convergent subsequence of y ( t ) convergesto a point y such that ( x, y, ξ ) is a comparison triangle for ( x, y, ξ ) . Sincethere are exactly two such points and y ( t ) is continuous, we see that y ( t ) converges to one of these points. This implies that ∠ x ( y ( t ) , γ ( t )) t →∞ −−−→ ∠ x ( y, ξ ) = ∠ (0) x ( y, ξ ) . By the law of cosines in Euclidean space, we have d ( y ( t ) , γ ( t )) = d ( x, y ( t )) + d ( x, γ ( t )) − d ( x, y ( t )) d ( x, γ ( t )) cos( ∠ x ( y ( t ) , γ ( t )))= d ( x, y ) + t − d ( x, y ) t cos( ∠ x ( y ( t ) , γ ( t ))) . Therefore, B ( x, y, ξ ) = lim t →∞ ( d ( y, ξ x ( t )) − d ( x, ξ x ( t ))) d ( y, ξ x ( t )) + d ( x, ξ x ( t ))2 t = lim t →∞ d ( x, y ) − d ( x, y ) t cos( ∠ x ( y ( t ) , γ ( t )))2 t = − d ( x, y ) cos( ∠ (0) x ( y, ξ )) . (cid:3) Cross ratios.
Let X be a proper, connected, geodesically complete CAT(0) space. We obtain a family of functions ( ·|· ) x : ∂X × ∂X → [0 , ∞ ] . Definition 2.3.
Let ( ξ , . . . , ξ n ) ∈ ∂X n be an n -tuple. Choose x ∈ X . Wesay ( ξ , . . . , ξ n ) is algebraically visible , if for all i = j , ( ξ i | ξ j ) x < ∞ .As noted in the last section, this definition does not depend on the choice of x . We can reformulate it in terms of the maps ρ x , by requiring ρ x ( ξ i , ξ j ) > for all i = j instead. Note that ρ x is symmetric and non-negative, but it doesnot satisfy the triangle inequality and there may be pairs ξ = η such that ρ x ( ξ, η ) = 0 . Nevertheless, we can use ρ x to define a cross ratio as follows.Define the set of admissible quadruples in ∂X to be the set A := { ( ξ , ξ , ξ , ξ ) ∈ ∂X |∀ i = j = k = i, at least two of the pairs ( ξ i , ξ j ) , ( ξ i , ξ k ) , ( ξ j , ξ k ) are algebraically visible } . In other words, A consists of the quadruples, whose points do not includea chain in which pairs of consecutive points are not algebraically visible. For IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 11 all admissible quadruples, we can define a cross ratio by cr x ( ξ , ξ , ξ , ξ ) := ρ x ( ξ , ξ ) ρ x ( ξ , ξ ) ρ x ( ξ , ξ ) ρ x ( ξ , ξ ) ∈ [0 , ∞ ] . The goal of this section is to prove the following theorem.
Theorem 2.4.
Let X be a proper, connected, geodesically complete CAT(0) space. Then, for all x , y ∈ X , cr x = cr y on all of A . The proof is based on the following
Lemma 2.5.
For all x, y ∈ X , and for all ξ, η ∈ ∂X , ( ξ | η ) x = 12 ( B ( y, x, ξ ) + B ( y, x, η )) + ( ξ | η ) y This formula is well known for
CAT( − spaces (see [Bou95]). However,the case of CAT(0) spaces is hard to find in the literature, which is why weprovide a proof here.
Proof of Lemma 2.5.
Since ( ξ | η ) x = ∞ if and only if ( ξ | η ) y = ∞ , the equa-tion trivially holds in that case. Suppose ( ξ | η ) x < ∞ . We first show theinequality ‘ ≥ ’. Let ǫ > . Since the function ( ξ y ( t ) | η y ( t )) y is monotoneincreasing, we find T ≥ , such that for all t ≥ T , we have ( ξ y ( t ) | η y ( t )) y ≥ ( ξ | η ) y − ǫ . From the properties of Busemann functions in the last section, we knowthat there are T ξ , T η such that B ( ξ x ( T ξ ) , ξ y ( T ) , ξ ) = 0 = B ( η x ( T η ) , η y ( T ) , η ) .Specifically, T ξ = T + B ( y, x, ξ ) T η = T + B ( y, x, η ) . Since B ( ξ x ( T ξ ) , ξ y ( T ) , ξ ) = B ( η x ( T η ) , η y ( T ) , η ) = 0 , there exists S ≥ T such that for all s ≥ S , | d ( ξ y ( T ) , ξ x ( s )) − d ( ξ x ( T ξ ) , ξ x ( s )) | ≤ ǫ | d ( η y ( T ) , η x ( s )) − d ( η x ( T η ) , η x ( s )) | ≤ ǫ . We obtain for all s ≥ S ξ | η ) x ≥ d ( x, ξ x ( s )) + d ( x, η x ( s )) − d ( ξ x ( s ) , η x ( s )) ≥ s + s − d ( ξ x ( s ) , ξ y ( T )) − d ( ξ y ( T ) , η y ( T )) − d ( η y ( T ) , η x ( s )) ≥ s + s − d ( ξ x ( s ) , ξ x ( T ξ )) − d ( η x ( s ) , η x ( T η )) − d ( ξ y ( T ) , η y ( T )) − ǫ = T ξ + T η − d ( ξ y ( T ) , η y ( T )) − ǫ = B ( y, x, ξ ) + B ( y, x, η ) + 2( ξ y ( T ) | η y ( T )) y − ǫ ≥ B ( y, x, ξ ) + B ( y, x, η ) + 2( ξ | η ) y − ǫ. Since ǫ was chosen arbitrarily, we obtain ( ξ | η ) x ≥ ( ξ | η ) y + 12 ( B ( y, x, ξ ) + B ( y, x, η )) The same argument with x and y swapped yields ( ξ | η ) y ≥ ( ξ | η ) x + 12 ( B ( x, y, ξ ) + B ( x, y, η ))= ( ξ | η ) x −
12 ( B ( y, x, ξ ) + B ( y, x, η )) and thus ( ξ | η ) x ≤ ( ξ | η ) y + 12 ( B ( y, x, ξ ) + B ( y, x, η )) , which concludes the proof. (cid:3) Proof of Theorem 2.4.
We know from Lemma 2.5 that for all ξ, η ∈ ∂Xρ y ( ξ, η ) = p e − B ( x,y,ξ ) e − B ( x,y,η ) ρ x ( ξ, η ) Therefore, for all admissible, algebraically visible quadruples ( ξ , ξ , ξ , ξ ) , cr y ( ξ , ξ , ξ , ξ ) = √ e − B ( x,y,ξ ) − B ( x,y,ξ ) − B ( x,y,ξ ) − B ( x,y,ξ ) ρ x ( ξ , ξ ) ρ x ( ξ , ξ ) √ e − B ( x,y,ξ ) − B ( x,y,ξ ) − B ( x,y,ξ ) − B ( x,y,ξ ) ρ x ( ξ , ξ ) ρ x ( ξ , ξ )= cr x ( ξ , ξ , ξ , ξ ) . We are left to check the special cases where ( ξ , ξ , ξ , ξ ) is admissible,but not algebraically visible. If an admissible quadruple is not algebraicallyvisible, there has to be at least one pair in the quadruple that is not al-gebraically visible. If ( ξ , ξ ) or ( ξ , ξ ) is not algebraically visible, then cr x ( ξ , ξ , ξ , ξ ) = 0 = cr y ( ξ , ξ , ξ , ξ ) . If ( ξ , ξ ) or ( ξ , ξ ) is not alge-braically visible, then cr x ( ξ , ξ , ξ , ξ ) = ∞ = cr y ( ξ , ξ , ξ , ξ ) . Otherwise,the equation from above still applies. We conclude that cr x = cr y on all of A . (cid:3) If X is CAT( − , it is a well-known result that ( ξ | η ) x is continuous withrespect to the visual topology. For CAT(0) spaces, this is not true anymore,which is illustrated by the fact that the Gromov product on the boundaryof the euclidean plane obtains exactly the values zero and infinity and theset of pairs for which the Gromov product is infinite is dense. Nevertheless,some continuity properties remain true. We say that a bi-infinite geodesic γ in X is rank 1 if and only if it does not bound an isometrically embeddedhalf plane in X . Note that γ might still have parallel geodesics, howeverthere is a bound on the distance of any such parallel geodesic to γ . Lemma 2.6.
Let ξ n → ξ, η n → η be two converging sequences in ∂X . Thenthe following statements hold.(1) ( ξ | η ) x ≤ lim inf n →∞ ( ξ n | η n ) x .(2) If ( ξ n , η n ) is visible for all n and ( ξ, η ) can be connected by a rank 1geodesic, then lim n →∞ ( ξ n | η n ) x = ( ξ | η ) x . The proof of Lemma 2.6 requires several preliminary results. We beginwith
IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 13
Lemma 2.7.
Let ξ n → ξ , η n → η be two converging sequences in ∂X suchthat ( ξ n , η n ) can be connected by a geodesic line γ n for all n and ( ξ, η ) can beconnected by a geodesic line γ . Then, for all n sufficiently large, there existsa point p n ∈ γ n such that ∠ γ (0) ( ξ, p n ) = π .Proof. By Proposition 9.2 in Part II of [BH99], the function ( ξ, η ) ∠ γ (0) ( ξ, η ) that sends two points in X ∪ ∂X to the angle between the unique geodesicsfrom γ (0) to ξ and η respectively is continuous in the cone topology (whichrestricts to the visual topology on the boundary). This implies that ∠ γ (0) ( ξ, ξ n ) → , ∠ γ (0) ( ξ, η n ) → π. Therefore, we find N such that for all n ≥ N , we have ∠ γ (0) ( ξ, ξ n ) ≤ π , ∠ γ (0) ( ξ, η n ) ≥ π . Let γ n be a geodesic from ξ n to η n . Since the angle function varies contin-uously along γ n , the intermediate value theorem tells us that there has toexist some point p n ∈ γ n , for which ∠ γ (0) ( ξ, p n ) = π . (cid:3) Given a subset A ⊂ X and ǫ > , we denote the ǫ -neighbourhood of A by N ǫ ( A ) := { x ∈ X | d ( x, A ) < ǫ } . Next, we need Lemma 2.8.
Let ξ n → ξ , η n → η be two converging sequences in ∂X , suchthat ( ξ n , η n ) is visible for all n and ( ξ, η ) can be connected by a rank 1 geodesic γ . Let γ n be a geodesic from ξ n to η n . For n sufficiently large, Lemma 2.7allows us to choose p n ∈ γ n , such that ∠ γ (0) ( ξ, p n ) = π . Reparametrize γ n ,such that γ n (0) = p n .Then for all ǫ > , T > , there exists an N such that for all n ≥ N ,there exists a geodesic ˜ γ from ξ to η such that γ n | [ − T,T ] ⊂ N ǫ (˜ γ | [ − T,T ] ) .Proof of Lemma 2.8. We set the convention that, throughout this proof, ˜ γ denotes a bi-infinite geodesic from ξ to η , which is parametrised such that B ( γ (0) , ˜ γ (0) , ξ ) = 0 .Suppose, the statement of the Lemma was not true. Then, we would find ǫ > , T > and subsequences ( ξ n i ) i , ( η n i ) i such that for all ˜ γ , we find t n i ∈ [ − T, T ] such that d (˜ γ ( t n i ) , γ n i ( t n i )) ≥ ǫ .Step 1: There exists N , such that for all n i ≥ N and for all ˜ γ , we have d (˜ γ (0) , γ n i (0)) ≥ ǫ .Suppose, d (˜ γ (0) , γ n i (0)) < ǫ . Since γ is a rank 1 geodesic, the set { ˜ γ (0) | ˜ γ } is bounded. Therefore, we can choose N sufficiently large, such that for all ˜ γ , we have d (˜ γ ( − T ) , ξ n i , ˜ γ (0) ( T )) ≤ ǫ ,d (˜ γ ( T ) , η n i , ˜ γ (0) ( T )) ≤ ǫ . ξ ηξ n i η n i δ n i B ( γ n i (0) , · , η n i ) = 0 B ( γ n i (0) , · , ξ n i ) = 0 Figure 2.
Visualisation of proof: Since the geodesic from ξ n i to η n i stays away from the flat strip from ξ to η , the geodesic δ n i has to move away from that flat strip. The shape ofthe two horospheres illustrates why φ n i is strictly decreasingalong δ n i . As n i → ∞ , any finite segment of δ n i is pushedinto the flat strip from ξ to η , because ξ n i → ξ and η n i → η .By construction and assumption, d ( ξ n i , ˜ γ (0) (0) , γ n i (0)) < ǫ ,d ( η n i , ˜ γ (0) (0) , γ n i (0)) < ǫ . Since distance functions are convex in
CAT(0) spaces, this implies that d ( ξ n i , ˜ γ (0) ( T ) , γ n i ( − T )) < ǫ ,d ( η n i , ˜ γ (0) ( T ) , γ n i ( T )) < ǫ and therefore, d (˜ γ ( t ) , γ n i ( t )) < ǫ for all t ∈ [ − T, T ] . This contradicts our assumption that we have t n i ∈ [ − T, T ] such that for all ˜ γ , d (˜ γ ( t n i ) , γ n i ( t n i )) ≥ ǫ . We conclude that, for all n i ≥ N and for all ˜ γ , d (˜ γ (0) , γ n i (0)) ≥ ǫ .Step 2: Let δ n i be the geodesic from γ (0) to γ n i (0) . Because of the waywe parametrised γ n in the statement of the Lemma, we know that δ n i meets γ | [0 , −∞ ) at a right angle for all sufficiently large n i . By the Theorem ofArzela-Ascoli, δ n i has a converging subsequence in compact-open topology.Passing to a subsequence if necessary, we assume without loss of generalitythat δ n i converges to a geodesic δ . Since δ n i meets γ | [0 , −∞ ) at a right anglefor all n i , we know that the same is true for δ . Therefore, δ cannot be ageodesic ray representing ξ or η .Choose x ∈ γ and denote for all ζ ∈ ∂XB ζ ( x ′ ) := B ( x, x ′ , ζ ) . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 15
Further, we define φ n i ( x ′ ) := B ξ ni ( x ′ ) + B η ni ( x ′ ) − B ξ ni ( γ (0)) − B η ni ( γ (0)) . For x ∈ γ , we have B ( x, x ′ , ξ ) + B ( x, x ′ , η ) ≥ with equality if and onlyif x ′ lies on a geodesic from ξ to η . Using the cocycle equation, we see that φ n i ( γ n i (0)) ≤ and φ n i ( δ (0)) = φ n i ( γ (0)) = 0 . In particular, since Busemann functions are convex, φ n i ( δ n i ( s )) ≤ forall s , where δ n i is defined. Since B ξ ( x ) is continuous in ξ and Lipschitzcontinuous in x , we see that on every compact interval, on which δ n i isdefined for n i large, we have ≥ φ n i ( δ n i ( s )) i →∞ −−−→ B ξ ( δ ( s )) + B η ( δ ( s )) ≥ . Therefore, B ξ ( δ ( s )) + B η ( δ ( s )) ≡ for all s for which δ is defined, whichimplies that δ lies completely in the set of points that are contained ingeodesics from ξ to η . Since ξ and η are connected by a rank 1 geodesic,the geodesic δ can only have infinite length if it represents either ξ or η . Asdiscussed above, this cannot happen given the way we constructed δ . Weconclude that δ has finite length. Therefore, the forward-endpoints of δ n i converge to the forward-endpoint of δ , i.e. γ n i ( t n i ) → p , where p lies on ageodesic connecting ξ with η and – because δ meets γ | [0 , −∞ ) at a right angle– B ( γ (0) , p, ξ ) = 0 . Therefore, p = ˜ γ (0) for some ˜ γ . This is a contradictionto our original assumption that the subsequence γ n i (0) stays away from ˜ γ (0) for all ˜ γ . This completes the proof. (cid:3) We also need another characterisation of the Gromov product. Consider x ∈ X, ξ, η ∈ ∂X and let h, h ′ be the horoballs centered at ξ and η respec-tively such that x ∈ ∂h ∩ ∂h ′ . Denote by h − m the horoball centered at ξ such that for any y ∈ h − m , B ( x, y, ξ ) ≤ − m and analogously for h ′− m . Notethat ξ x ( t ) ∈ ∂h − t and η x ( t ) ∈ ∂h ′− t . Define m x ( ξ, η ) := sup { m ′ | h − m ′ ∩ h ′− m ′ = ∅} . Lemma 2.9.
Let X be a proper, geodesically complete CAT(0) space. Forall x ∈ X, ξ, η ∈ ∂X , we have ( ξ | η ) x = m x ( ξ, η ) . Furthermore, if ( ξ, η ) is visible, then every geodesic γ from ξ to η containsa point in ∂h − m x ( ξ,η ) ∩ ∂h ′− m x ( ξ,η ) and every p ∈ h − m x ( ξ,η ) ∩ h ′− m x ( ξ,η ) lieson a geodesic from ξ to η . Before we prove this lemma, we introduce a convenient notation. Giventwo real numbers a, b and δ > , we write a ≍ δ b , whenever | a − b | ≤ δ . ξ ηxp ( ξ | η ) x Figure 3.
The horospheres in the picture are the smallestones centered at ξ and η that have non-empty intersection,provided that we shrink them at equal speed, starting withthe horospheres containing x . Proof of Lemma 2.9.
Denote m := m x ( ξ, η ) . We first show that ξ | η ) x ≤ m . Suppose ( ξ | η ) x < ∞ . Let m ′ > m and ǫ > . There exists t ≥ , suchthat ξ | η ) x ≤ t − d ( ξ x ( t ) , η x ( t )) + ǫ. Let γ t be the geodesic from ξ x ( t ) to η x ( t ) . Denote the unique intersectionpoint of γ t with ∂h − m ′ by p ( t ) and the unique intersection point of γ t with ∂h ′− m ′ by q ( t ) . Since m ′ > m , we know that there is a segment of γ t thatlies outside of h − m ′ ∪ h ′− m ′ . For t > t , we compute ξ | η ) x ≤ t − d ( ξ x ( t ) , p ( t )) − d ( p ( t ) , q ( t )) − d ( q ( t ) , η x ( t )) + ǫ ≤ t − d ( ξ x ( t ) , p ( t )) − d ( q ( t ) , η x ( t )) + ǫ ≤ B ( p ( t ) , x, ξ ) + B ( q ( t ) , x, η ) + ǫ = 2 m ′ + ǫ, as p ( t ) ∈ ∂h − m ′ , q ( t ) ∈ ∂h − m ′ . Since this computation applies for all ǫ > and m ′ > m , we conclude that ( ξ | η ) x ≤ m whenever ( ξ | η ) x < ∞ . If ( ξ | η ) x = ∞ , we do the same computation as above, except that we drop ǫ and instead find for every C > a time t , such that C ≤ t − d ( ξ x ( t ) , p ( t )) − d ( p ( t ) , q ( t )) − d ( q ( t ) , η x ( t )) t →∞ −−−→ m ′ .Now, let m ′ < m , i.e. h − m ′ ∩ h ′− m ′ = ∅ and define γ t as above. Choose p ∈ ∂h − m ′ ∩ h ′− m ′ . Let ǫ > . Then, for t sufficiently large, m ′ ≤ B ( p, x, ξ ) + B ( p, x, η ) ≤ t − d ( ξ x ( t ) , p ) − d ( p, η x ( t )) + ǫ ≤ t − d ( ξ x ( t ) , η x ( t )) + ǫ ≤ ξ | η ) x + ǫ. IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 17
Therefore, ( ξ | η ) x ≥ m , which concludes the proof of the first statement.For the second statement of the Lemma, suppose ( ξ, η ) is visible and againdenote m := m x ( ξ, η ) . Let γ be a geodesic from ξ to η . Denote the uniqueintersection point of γ with ∂h − m by p .We claim that p ∈ ∂h ′− m . Suppose not. Since m = sup { m ′ | h − m ′ ∩ h ′− m ′ = ∅} , this implies that B ( x, p, η ) > − m . Therefore, there exists ǫ > suchthat B ( x, p, η ) > − m + 2 ǫ . Since m = m x ( ξ, η ) , we find q ∈ h − m + ǫ ∩ h ′− m + ǫ .We compute B ( x, p, ξ ) + B ( x, p, η ) > − m + 2 ǫ ≥ B ( x, q, ξ ) + B ( x, q, η )= B ( x, p, ξ ) + B ( x, p, η ) + B ( p, q, ξ ) + B ( p, q, η ) . This implies that B ( p, q, ξ ) + B ( p, q, η ) < , which is a contradiction to thefact that p ∈ γ . We conclude that p ∈ ∂h ′− m .Note that h − m ∩ h ′− m = ∂h − m ∩ ∂h − m . Otherwise, we would find a point p such that B ( x, p, ξ ) = B ( x, p, η ) = − m − ǫ with ǫ > , which contradicts theassumption that m = m x ( ξ, η ) . We now show that every q ∈ ∂h − m ∩ ∂h ′− m lies on a geodesic from ξ to η as well. Let p be as above. We compute B ( p, q, ξ ) + B ( p, q, η ) = B ( p, x, ξ ) + B ( x, q, ξ ) + B ( p, x, η ) + B ( x, q, η )= m + ( − m ) + m + ( − m ) = 0 . Since p lies on a geodesic from ξ to η , the sum B ( p, q, ξ ) + B ( p, q, η ) equalszero if and only if q also lies on a geodesic from ξ to η . This proves the secondpart of the Lemma. (cid:3) Proof of Lemma 2.6.
We first prove (1). Let ǫ > . There exists T ≥ , suchthat for all t ≥ T , ( ξ x ( t ) | η x ( t )) x ≥ ( ξ | η ) x − ǫ . Since ξ n → ξ and η n → η , wefind N such that for all n ≥ N, ξ n ∈ U T, ǫ ,x ( ξ ) and η n ∈ U T, ǫ ,x . Thus, ( ξ n | η n ) x ≥ ( ξ n,x ( T ) | η n,x ( T )) x = T − d ( ξ n,x ( T ) , η n,x ( T )) ≥ T −
12 ( d ( ξ x ( T ) , η x ( T )) + 2 ǫ ) ≥ ( ξ | η ) x − ǫ. Since ǫ was chosen to be any positive number, we conclude that lim inf n →∞ ( ξ n | η n ) x ≥ ( ξ | η ) x .To prove (2), we start by using Lemma 2.9 to describe the Gromov productas follows. Let ( ξ, η ) be visible and let h, h ′ be as in the definition of m x ( ξ, η ) .Denote the unique point where ξ x intersects ∂h − m x ( ξ,η ) by p and the uniquepoint where η x intersects ∂h ′− m x ( ξ,η ) by q . Since ( ξ, η ) is visible, we knowfrom Lemma 2.9 that h − m x ( ξ,η ) ∩ h ′− m x ( ξ,η ) is non-empty and contains only points that are contained in a geodesic from ξ to η . Let r ∈ h − m x ( ξ,η ) ∩ h ′− m x ( ξ,η ) . The Gromov product is equal to m x ( ξ, η ) which is the same asthe distance d ( x, p ) . Note that the following equations hold by construction: d ( x, p ) = d ( x, q ) B ( p, r, ξ ) = B ( q, r, η ) = 0 . Let ǫ > . Let γ n be a bi-infinite geodesic from ξ n to η n . By Lemma2.8, there exists (after reparametrisation) a subsequence γ n i converging to ageodesic γ from ξ to η . Choose r from above such that r ∈ γ . We find triples ( p n , q n , r n ) as in the construction above, where we choose r n ∈ γ n . By (1),we know that lim inf n →∞ d ( x, p n ) ≥ d ( x, p ) .Suppose, lim inf n →∞ d ( x, p n ) ≥ d ( x, p ) + ǫ . By choice of γ , we know that γ n i converges to γ in compact-open topology. In particular, for n i sufficientlylarge, r ∈ N ǫ ( γ n i ) and we find r ′ n i ∈ γ n i , such that d ( r, r ′ n i ) < ǫ . Therefore,for n i sufficiently large, B ( r, r n i , ξ n i ) ≍ ǫ B ( r ′ n i , r n i , ξ n i ) .Furthermore, since ξ n → ξ and η n → η , we can choose n i sufficiently largesuch that d ( ξ x ( d ( x, p )) , ξ n i ,x ( d ( x, p ))) < ǫ and d ( η x ( d ( x, p )) , η n i ,x ( d ( x, p ))) < ǫ . Together with our assumption on the convergence behaviour of d ( x, p n ) ,we obtain that for all n i sufficiently large B ( p n i , p, ξ n i ) ≥ B ( p n i , ξ n i ,x ( d ( x, p )) , η n i ) − ǫ ≥ d ( x, p n i ) − d ( x, p ) − ǫ ≥ ǫ ,B ( q n i , q, η n i ) ≥ B ( q n i , η n i ,x ( d ( x, p )) , η n i ) − ǫ ≥ d ( x, q n i ) − d ( x, q ) − ǫ ≥ ǫ , and thus, B ( p n i , r, ξ n i ) ≥ B ( p, r, ξ n i ) + 3 ǫ ,B ( q n i , r, η n i ) ≥ B ( q, r, η n i ) + 3 ǫ . Finally, since Busemann functions B ( x, y, ξ ) are continuous in ξ , we canchoose n i sufficiently large such that B ( p, r, ξ n i ) ≥ B ( p, r, ξ ) − ǫ ,B ( q, r, η n i ) ≥ B ( q, r, η ) − ǫ . Altogether, this implies that there exists N ∈ N such that for all n ≥ N , B ( p n i , r n i , ξ n i ) ≥ B ( p n i , r, ξ n i ) + B ( r ′ n i , r n i , ξ n i ) − ǫ ≥ B ( p, r, ξ n i ) + B ( r ′ n i , r n i , ξ n i ) + ǫ ≥ B ( p, r, ξ ) + B ( r ′ n i , r n i , ξ n i ) + ǫ B ( r ′ n i , r n i , ξ n i ) + ǫ IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 19 B ( q n i , r n i , η n i ) ≥ B ( q n i , r, η n i ) + B ( r ′ n i , r n i , η n i ) − ǫ ≥ B ( q, r, η n i ) + B ( r ′ n i , r n i , η n i ) + ǫ ≥ B ( q, r, η ) + B ( r ′ n i , r n i , η n i ) + ǫ B ( r ′ n i , r n i , η n i ) + ǫ . However, since r n i , r ′ n i both lie on the geodesic γ n i from ξ n i to η n i , wehave B ( p n i , r n i , ξ n i ) + B ( q n i , r n i , η n i )) ≤ and B ( r ′ n i , r n i , ξ n i ) + B ( r ′ n i , r n i , η n i )) = 0 , which is a contradiction to the inequalities above. We conclude that lim inf n →∞ d ( x, p n ) = d ( x, p ) . Since this argument applies to any subsequence of ( ξ n , η n ) as well,we conclude that lim n →∞ d ( x, p n ) exists and equals d ( x, p ) . This concludes theproof. (cid:3) We prove one more Lemma that characterizes rank 1 geodesics in termsof a local visibility property.
Lemma 2.10.
Let γ be a rank 1 geodesic from ξ to η . Then, there exists anopen neighbourhood U × V of ( ξ, η ) , such that for all ( ξ ′ , η ′ ) ∈ U × V , ( ξ ′ , η ′ ) is a visible pair.In particular, a pair ( ξ, η ) can be connected by a rank 1 geodesic if andonly if there exists a neighbourhood U of ξ , such that for all ξ ′ ∈ U , ( ξ ′ , η ) is visible.Proof. The proof uses a similar idea as the proof of Lemma 2.8. Since γ is arank 1 geodesic, there exists a constant C > , such that every geodesic γ ′ from ξ to η is parallel to γ and has Hausdorff distance d Haus ( γ, γ ′ ) ≤ C .Suppose the Lemma was not true. Then, there exist sequences ξ i → ξ , η i → η , such that for all i , ( ξ i , η i ) is not visible. Denote x := γ (0) , γ − i thegeodesic ray starting at x representing ξ i , and γ + i the geodesic ray startingat x representing η i . Let γ T,i be the unique geodesic from γ − i ( T ) to γ + i ( T ) .Note that, if we fix i , the paths γ T,i vary continuously in T in the sense that d Haus ( γ T + ǫ,i , γ T,i ) ≤ ǫ . Denote the points at infinity obtained by extending γ T,i by ξ T,i and η T,i respectively.Since γ T,i varies continuously in T , there exists a time T i such that d ( x, γ T,i ) = 2 C and there exists a unique point x i ∈ γ T i ,i satisfying d ( x, x i ) =2 C . Note that T i i →∞ −−−→ ∞ , as ξ i → ξ and η i → η . We reparametrise γ T i ,i such that it is an arc-length geodesic with γ T i ,i (0) = x i . Since X is assumedto be proper, the Arzela-Ascoli theorem implies the existence of a subse-quence γ T ni ,n i that converges to a bi-infinite geodesic line ˜ γ from ˜ ξ to ˜ η with ξ ηγ x η i γ + i γ + i ( T i ) ξ i γ − i γ − i ( T i ) γ T i ,i x i ˜ γ ˜ x ˜ η ˜ ξ Figure 4.
The times T i are chosen such that d ( x, x i ) = 2 C .This provides us with a subsequence of the geodesics γ T i ,i thatconverges to ˜ γ . The angles of the triangles ( x, x i , γ ± i ( T i )) tellus that ˜ x cannot lie in the flat strip from ξ to η (indicated bythe dotted lines). However, the endpoints of ˜ γ turn out to be ξ and η , which leads to a contradiction. x n i converging to a point ˜ x ∈ ˜ γ . Without loss of generality, we denote thesesubsequences by γ T i ,i and x i .We claim that ∠ x (˜ x, ξ ) = ∠ x (˜ x, η ) = π . To prove this, we denote α i := ∠ x ( x i , γ − i ( T i )) , α ′ i := ∠ x ( x i , γ + i ( T i )) , β i := ∠ x i ( x, γ − i ( T i )) , β ′ i := ∠ x i ( x, γ + i ( T i )) . Since x i minimizes the distance d ( x, γ T i ,i ( t )) , we concludethat β i , β ′ i ≥ π . Since the sum of angles of a triangle in a CAT(0) space is atmost π , this implies that α i , α ′ i ≤ π . However, since ξ i → ξ and η i → η , wehave that lim i →∞ α i + α ′ i ≥ π (the limit exists, since x i → ˜ x ). We concludethat lim i →∞ α i = lim i →∞ α ′ i = π , which means that ∠ x (˜ x, ξ ) = ∠ x (˜ x, η ) = π . Combined with the fact that d ( x, ˜ x ) = 2 C and any geodesic parallel to γ is contained in the C -neighbourhood of γ , this implies that ˜ x does not lie inthe flat strip spanned by all geodesic lines from ξ to η .We now claim that ˜ ξ = ξ and ˜ η = η , contradicting the fact that ˜ x does notlie on any geodesic from ξ to η . We show this by proving that d Haus ( γ, ˜ γ ) < ∞ . Fix R > . By the convergences established above, there exists I such that for all i ≥ I , we have T i ≥ R , and for all | t | ≤ R , we have d ( γ T i ,i ( t ) , ˜ γ ( t )) ≤ C , and d ( γ ( t ) , γ ± i ( | t | )) ≤ C . We estimate for all | t | ≤ R , d ( γ ( t ) , ˜ γ ( t )) ≤ d ( γ ( t ) , γ ± i ( | t | ))+ d ( γ ± i ( | t | ) , γ T i ,i ( t ))+ d ( γ T i ,i ( t ) , ˜ γ ( t )) ≤ C, where we used the fact that γ ± i ( T i ) ∈ γ T i ,i and convexity of distance functionsto estimate d ( γ ± i ( | t | ) , γ T i ,i ( t )) ≤ max( d ( x, x i ) , d ( γ ± i ( T i ) , γ T i ,i ( T i ))) = 2 C .This implies that ˜ γ and γ have bounded Hausdorff distance and, therefore,they are parallel. In particular, ξ = ˜ ξ and η = ˜ η and ˜ x lies on a geodesic IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 21 from ξ to η . However, ˜ x was constructed so that it cannot lie on such ageodesic. This is a contradiction and proves the Lemma. (cid:3) Corollary 2.11.
Let x ∈ X, ξ, γ a rank 1 geodesic from ξ to η . Then, ( · |· ) x : ∂X × ∂X → [0 , ∞ ] is continuous at ( ξ, η ) . We now define one of the properties necessary to make the circumcenterextension construction work.
Definition 2.12.
Let X be a proper, connected, geodesically complete CAT(0) space, ξ ∈ ∂X . We say that ξ is in a rank 1 hinge if there exist η, ζ ∈ ∂X , such that ( η, ζ ) is algebraically visible and the pairs ( ξ, η ) , ( ξ, ζ ) both can be connected by a rank 1 geodesic.2.3. Metric derivatives.
In order to extend cross ratio preserving mapsto maps of the interior, we need to generalize the notion of metric deriva-tives, which has been developed for general metric spaces (see [Bis15, Bis17a,Bis18]). In this subsection, we show how this tool can be extended toboundaries of
CAT(0) spaces that ‘have sufficiently many algebraically vis-ible pairs’. Since the underlying theory is more general, we will state thedefinitions and results in a more general form and then return to
CAT(0) -spaces and boundaries.Let Z be a topological space, ρ and ρ ′ two non-negative, symmetric maps ρ , ρ ′ : Z × Z → [0 , ∞ ] such that for all z ∈ Z, ρ ( z, z ) = ρ ′ ( z, z ) = 0 . By analogyto the previous section, we call an n -tuple ( x , . . . , x n ) ∈ Z n algebraicallyvisible with respect to ρ if and only if for all i = j , ρ ( x i , x j ) > . We say thata quadruple ( x , x , x , x ) ∈ Z is admissible with respect to ρ if it containsno triple ( x i , x j , x k ) with i = j = k = i such that ρ ( x i , x j ) = ρ ( x j , x k ) = 0 .Denote the set of quadruples admissible with respect to ρ by A ρ . We willnot indicate the ρ , whenever it is clear from context. Using the same formulaas before, ρ and ρ ′ both define a cross ratio cr ρ and cr ρ ′ on the set A ρ and A ρ ′ respectively. We say that ρ and ρ ′ are Möbius equivalent if A ρ = A ρ ′ and cr ρ = cr ρ ′ . We write ρ M ∼ ρ ′ . Note that A ρ = A ρ ′ if and only if ρ and ρ ′ define the same algebraically visible pairs. Definition 2.13.
We say that ( Z, ρ ) satisfies the assumption, ifthe following holds:(4v) For every quadruple ( z, x, x ′ , y ′ ) ∈ Z , there exists w ∈ Z , such that w is algebraically visible with z , x , x ′ , y ′ . Remark 2.14.
For any n ∈ N + , we can define the assumption ( n v) byreplacing quadruples by n -tuples. Note that ( n v) implies ( k v) for all k ≤ n and whenever Z satisfies ( n v), it has to contain at least n + 1 points, asotherwise we could choose an n -tuple that contains all points in Z to createa contradiction to ( n v).Further note that, if ( Z, ρ ) satisfies (4v) and ρ M ∼ ρ ′ , then ( Z, ρ ′ ) satisfies(4v) as well. We say that a point z in ( Z, ρ ) is approximable , if there exists a sequence z n ∈ Z , such that z n n →∞ −−−→ z and ( z, z n ) is algebraically visible for all n .Note that, if ρ M ∼ ρ ′ , then a point is approximable in ( Z, ρ ) if and only if itis approximable in ( Z, ρ ′ ) . Definition/Proposition 2.15 (cf. [Bis15]) . Suppose ρ M ∼ ρ ′ . Additionally,assume that ( Z, ρ ) (and thus ( Z, ρ ′ ) ) satisfies (4v)). Let z ∈ Z and choose x, y ∈ Z such that ( x, y, z ) is an algebraically visible triple with respect to ρ (and thus ρ ′ ). Then, the expression R z ( x, y ) := ρ ( z, x ) ρ ( z, y ) ρ ′ ( x, y ) ρ ( x, y ) ρ ′ ( z, x ) ρ ′ ( z, y ) , is independent of the choice of x , y ; it is continuous, whenever ρ and ρ ′ are continuous and, if ρ and ρ ′ are continuous, the following equality holdsfor every point z ∈ Z that is approximable with respect to ρ : R z ( x, y ) = lim z ′ → z,ρ ( z,z ′ ) =0 ρ ( z, z ′ ) ρ ′ ( z, z ′ ) . This equation motivates to define the derivative of ρ by ρ ′ at z by ∂ρ∂ρ ′ ( z ) := ρ ( z, x ) ρ ( z, y ) ρ ′ ( x, y ) ρ ( x, y ) ρ ′ ( z, x ) ρ ′ ( z, y ) Proof.
We start by showing that every z ∈ Z admits x , y ∈ Z , such that ( x, y, z ) is an algebraically visible triple. Let z ∈ Z . We can extend z toa quadruple ( z, a, b, c ) ∈ Z . By (4v), there exists a point x ∈ Z that isalgebraically visible to z , a , b , c . In particular, z = x . By extending thepair ( z, x ) to a quadruple and using (4v) again, we obtain y ∈ Z , that isalgebraically visible to both z and x . We conclude that ( x, y, z ) is an alge-braically visible triple.Next, we show independence of x and y for all possible choices of x, y .Let x ′ , y ′ ∈ Z be another pair such that ( x ′ , y ′ , z ) is an algebraically visibletriple. We proceed in two steps.Step 1: Suppose, one of the pairs ( x, x ′ ) , ( x, y ′ ) , ( y, x ′ ) , ( y, y ′ ) is alge-braically visible. Let’s assume that ( x, x ′ ) is. We want to show that R z ( x, y ) = ρ ( z, x ) ρ ( z, y ) ρ ′ ( x, y ) ρ ( x, y ) ρ ′ ( z, x ) ρ ′ ( z, y ) = ρ ( z, x ′ ) ρ ( z, y ′ ) ρ ′ ( x ′ , y ′ ) ρ ( x ′ , y ′ ) ρ ′ ( z, x ′ ) ρ ′ ( z, y ′ ) = R z ( x ′ , y ′ ) . This is true if and only if ρ ( z, x ) ρ ( z, y ) ρ ( x ′ , y ′ ) ρ ( x, x ′ ) ρ ( x, y ) ρ ( z, x ′ ) ρ ( z, y ′ ) ρ ( x, x ′ ) = ρ ′ ( z, x ) ρ ′ ( z, y ) ρ ′ ( x ′ , y ′ ) ρ ′ ( x, x ′ ) ρ ′ ( x, y ) ρ ′ ( z, x ′ ) ρ ′ ( z, y ′ ) ρ ′ ( x, x ′ ) , which is the same as cr ρ ( z, y, x ′ , x ) cr ρ ( z, x, y ′ , x ′ ) = cr ρ ′ ( z, y, x ′ , x ) cr ρ ′ ( z, x, y ′ , x ′ ) . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 23
This last equation is true, since all appearing quadruples are admissibleby assumption and ρ M ∼ ρ ′ . The cases, where ( x, y ′ ) , ( y, x ′ ) or ( y, y ′ ) is alge-braically visible are analogous.Step 2: Suppose, all the pairs above are not algebraically visible. Byassumption (4v), there exists a point w ∈ Z , which is algebraically visiblewith z , x , x ′ and y ′ . By Step 1, we obtain that R z ( x, y ) = R z ( w, y ′ ) = R z ( x ′ , y ′ ) . Therefore, R z ( x, y ) = R z ( x ′ , y ′ ) for any two algebraically visible triples ( x, y, z ) , ( x ′ , y ′ , z ) .In order to prove continuity, note that, if ρ is continuous, algebraic visibil-ity with respect to ρ is an open condition and analogously for ρ ′ . Therefore,for any z ∈ Z , we find an open neighbourhood U and a pair ( x, y ) , such thatfor all z ′ ∈ U , ( x, y, z ′ ) is an algebraically visible triple with respect to ρ and ρ ′ . Thus, for all z ′ ∈ U , ∂ρ∂ρ ′ ( z ′ ) = R z ′ ( x, y ) , which is continuous in z ′ bycontinuity of ρ and ρ ′ .Finally, if z is approximable in ( Z, ρ ) , we find a sequence of points z n thatare algebraically visible with z and converging to z . By continuity of ρ and ρ ′ , we find a point y ∈ Z , such that ( z n , y, z ) is an algebraically visible triplefor all sufficiently large n . Using the continuity of ρ and ρ ′ again, we obtain ∂ρ∂ρ ′ ( z ) = lim n →∞ R z ( z n , y )= lim n →∞ ρ ( z, z n ) ρ ( z, y ) ρ ′ ( z n , y ) ρ ( z n , y ) ρ ′ ( z, z n ) ρ ′ ( z, y )= ρ ( z, y ) ρ ′ ( z, y ) ρ ( z, y ) ρ ′ ( z, y ) lim n →∞ ρ ( z, z n ) ρ ′ ( z, z n ) = lim n →∞ ρ ( z, z n ) ρ ′ ( z, z n ) . This implies that lim z ′ → z ρ ( z,z ′ ) ρ ′ ( z,z ′ ) exists and the desired equality, whichcompletes the proof. (cid:3) We require a few properties of these derivatives. If ρ, ρ ′ are metrics, theseproperties are shown in [Bis15] and the proof is the same as here. Lemma 2.16 (cf. [Bis15]) . Let ρ M ∼ ρ ′ M ∼ ρ ′′ , z, z ′ ∈ Z . Then(Chain rule) ∂ρ∂ρ ′ ( z ) ∂ρ ′ ∂ρ ′′ ( z ) = ∂ρ∂ρ ′′ ( z ) (Geometric mean value theorem) ρ ( z, z ′ ) = ∂ρ∂ρ ′ ( z ) ∂ρ∂ρ ′ ( z ′ ) ρ ′ ( z, z ′ ) Proof.
For the Chain rule, choose x, y ∈ Z such that ( z, x, y ) is an alge-braically visible triple with respect to ρ, ρ ′ , ρ ′′ . Then ∂ρ∂ρ ′ ( z ) ∂ρ ′ ∂ρ ′′ ( z ) = ρ ( z, x ) ρ ( z, y ) ρ ′ ( x, y ) ρ ′ ( z, x ) ρ ′ ( z, y ) ρ ′′ ( x, y ) ρ ( x, y ) ρ ′ ( z, x ) ρ ′ ( z, y ) ρ ′ ( x, y ) ρ ′′ ( z, x ) ρ ′′ ( z, y ) = ∂ρ∂ρ ′′ ( z ) . For the Geometric mean value theorem, if ρ ( z, z ′ ) = 0 , the equation followsfrom A ρ = A ρ ′ . If ρ ( z, z ′ ) = 0 , we can choose x ∈ Z such that ( z, z ′ , x ) is analgebraically visible triple. Then ∂ρ∂ρ ′ ( z ) ∂ρ∂ρ ′ ( z ′ ) = ρ ( z, z ′ ) ρ ( z, x ) ρ ′ ( z ′ , x ) ρ ( z ′ , z ) ρ ( z ′ , x ) ρ ′ ( z, x ) ρ ( z ′ , x ) ρ ′ ( z, z ′ ) ρ ′ ( z, x ) ρ ( z, x ) ρ ′ ( z ′ , z ) ρ ′ ( z ′ , x ) = ρ ( z, z ′ ) ρ ′ ( z, z ′ ) (cid:3) Remark 2.17.
If at least one point z ∈ Z is approximable in ( Z, ρ ) andboth ρ and ρ ′ are continuous, then it is easy to see from the characterizationof the derivative at the approximable point z by ∂ρ∂ρ ′ ( z ) = lim z ′ → z ρ ( z,z ′ ) ρ ′ ( z,z ′ ) thatthe Geometric mean value theorem uniquely determines the derivative of ρ by ρ ′ . Remark 2.18 (cf. [Bis15]) . Using Lemma 2.5 and the additivity of Buse-mann functions, it is easy to see that on boundaries of
CAT(0) spaces, ∂ρ x ∂ρ y ( ξ ) = e B ( x,y,ξ ) . Lemma 2.19 (cf. [Bis15]) . Let ρ M ∼ ρ ′ . Additionally, assume that Z iscompact, for all z, z ′ ∈ Z , ρ ( z, z ′ ) ≤ , ρ ′ ( z, z ′ ) ≤ and that for every z ∈ Z there exist ¯ z, ¯ z ′ ∈ Z such that ρ ( z, ¯ z ) = 1 and ρ ′ ( z, ¯ z ′ ) = 1 . Then, max z ∈ Z (cid:26) ∂ρ∂ρ ′ ( z ) (cid:27) min z ∈ Z (cid:26) ∂ρ∂ρ ′ ( z ) (cid:27) = 1 Note that, if Z = ∂X and ρ = ρ x , ρ ′ = ρ x ′ , then the assumptions ofLemma 2.19 are satisfied, so this Lemma applies in the context that we willbe considering. Proof.
Let z ∈ Z such that ∂ρ∂ρ ′ ( z ) is maximal and z ′ ∈ Z such that ∂ρ∂ρ ′ ( z ′ ) isminimal. Denote the obtained maximum and minimum by µ and λ respec-tively. Let ¯ z ′ ∈ Z be such that ρ ′ ( z, ¯ z ′ ) = 1 . then ≥ ρ ( z, ¯ z ′ ) = ∂ρ∂ρ ′ ( z ) ∂ρ∂ρ ′ (¯ z ′ ) ρ ′ ( z, ¯ z ′ ) ≥ µλ. On the other hand, let ¯ z ∈ Z be such that ρ ( z ′ , ¯ z ) = 1 . Then ≥ ρ ′ ( z ′ , ¯ z ) = ∂ρ ′ ∂ρ ( z ′ ) ∂ρ ′ ∂ρ (¯ z ) ρ ( z ′ , ¯ z ) ≥ λ µ . We conclude that µ · λ = 1 . (cid:3) IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 25
Convex functions.
We need some basic results about convex func-tions. A function f : I → R defined on an interval I ⊂ R is called convex iffor all a, b ∈ I and t ∈ [0 , , we have f ((1 − t ) a + tb ) ≤ (1 − t ) f ( a ) + tf ( b ) .A function is called strictly convex if this inequality is a strict inequality forall t ∈ (0 , .In a geodesic metric space X , a function f : X → R is called convex if forany geodesic γ on X and any a, b on the domain of γ , we have ∀ t ∈ [0 ,
1] : f ( γ ((1 − t ) a + tb )) ≤ (1 − t ) f ( γ ( a )) + tf ( γ ( b )) . Remark 2.20.
There is a sufficient, but generally not necessary analyticcondition for (strict) convexity. If f : I → R is a C -function, then f is con-vex if and only if ∂ f∂t ≥ everywhere. Furthermore, if ∂ f∂t > everywhere,then f is strictly convex. However, the converse is not necessarily true, as isillustrated by the example t t at the point zero.We recall the following standard result about convex functions. Lemma 2.21.
Let f z : X → R be a family of convex functions on a con-nected, geodesic CAT(0) space X parametrized by z ∈ Z . Define F ( x ) :=sup z ∈ Z { f z ( x ) } . Then F : X → R is convex. Visibility and algebraic visibility.
Let
X, Y be proper, connected,geodesically complete
CAT(0) spaces. A map f : ∂X → ∂Y , is called Möbius if and only if it sends algebraically visible pairs to algebraically visible pairsand preserves the cross ratio, i.e. ∀ ( ξ , ξ , ξ , ξ ) ∈ A : cr X ( ξ , ξ , ξ , ξ ) = cr Y ( f ( ξ ) , f ( ξ ) , f ( ξ ) , f ( ξ )) . In order to construct our extension map, we require that f is not onlyMöbius but also that f and f − both preserve visible pairs. It is temptingto try and show that Möbius maps always preserve visible pairs by arguingthat a pair ( ξ, η ) in ∂X is visible if and only if it is algebraically visible. Itis known that visible pairs are always algebraically visible. However, whilethe converse is true if X admits a cocompact group action by isometries, itis not true in general, as the following example – provided by Jean-ClaudePicaud and Viktor Schroeder – illustrates.Consider the manifold R with coordinates ( x, y ) and equip it with theRiemannian metric dx + f ( x ) dy , where f : R → R is a C -function, suchthat f ( x ) > for all x and lim x →∞ f ( x ) = 1 . The curvature of this metricat ( x, y ) is given by − f ′′ ( x ) f ( x ) . Hence, if f is strictly convex, this space hasnegative curvature everywhere. We equip the tangent space of R with thestandard basis e , e everywhere. We denote the inner product with respectto the Riemannian metric above by h· , ·i f .This Riemannian manifold is the universal covering of a surface of revolu-tion R × S with coordinates ( x, ϑ ) and Riemannian metric dx + f ( x ) dϑ .By abuse of notation, we call the projection of the vector fields e , e onto the surface of revolution by e , e as well. It is a classical result that a path γ on asurface of revolution is a geodesic in the Riemannian sense if and only if thefunction h γ ′ ( t ) , e ( γ ( t )) i f is constant. (This is called Clairaut’s constant,cf. [dC15].) We observe from this fact that a geodesic γ ( t ) = ( x ( t ) , y ( t )) with x ′ (0) > , will have monotone increasing x ( t ) for all t ≥ if and onlyif its Clairaut constant h γ ′ ( t ) , e i f ≤ lim t →∞ h e , e i f = lim t →∞ f ( t ) = 1 .Else, the geodesic γ will eventually change its x -direction and have decreas-ing x ( t ) . This argumentation carries over to the universal covering, wherewe conclude that a geodesic ray γ represents a point in the boundary with x ( t ) t →∞ −−−→ ∞ if and only if |h γ ′ (0) , e i f | ≤ . The Clairaut constant alsoimplies that no two geodesic rays with x ( t ) t →∞ −−−→ ∞ can be connected by abi-infinite geodesic. Thus, any pair of geodesics with x ′ (0) > and Clairautconstant at most one is a non-visible pair.We focus our attention on the borderline case where the absolute value ofthe Clairaut constant equals one, i.e. |h γ ′ ( t ) , e i| ≡ . Fixing ( x , y ) ∈ R ,there are exactly two geodesic rays starting at ( x , y ) whose Clairaut con-stant in absolute value equals . We will show that, depending on the choiceof the function f , this pair of points in the boundary may be algebraicallyvisible or not algebraically visible.We start with some general arguments that will allow us to reverse en-gineer the function f and y , assuming that we know the x -coordinate of ageodesic with Clairaut constant one. Suppose, we have a geodesic γ with aknown x -coordinate. We know that the following two equations hold: ± ≡ h γ ′ ( t ) , e i f = f ( x ( t )) y ′ ( t ) . ≡ h γ ′ ( t ) , γ ′ ( t ) i f = x ′ ( t ) + f ( x ( t )) y ′ ( t ) This implies that y ′ ( t ) = ± f ( x ( t )) ,f ( x ( t )) = ± − x ′ ( t ) . We now use these equations in two concrete cases.
Example 2.22.
Restrict to t ≥ and suppose, x ( t ) = ln( t ) . By the equa-tions above, using the fact that we also require f ( x ) > , we obtain f ( x ( t )) = r t t − t √ t − ,y ′ ( t ) = ± (cid:18) − t (cid:19) . In particular, we obtain f ( x ) = e x √ e x − , IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 27 which is a strictly convex function for x > with lim x →∞ f ( x ) = 1 , as directcomputation shows.For every starting point p , we obtain two geodesic rays starting at thatpoint that are described by the equations above. Choose some p and de-note the two geodesics starting there by γ + , γ − . We claim that their Gro-mov product is finite. Since R ∞ t dt < ∞ , we obtain that there existssome constant C such that y + ( t ) > t − C and y − ( t ) < − t + C for all t .Let δ be the path connecting γ + ( t ) , γ − ( t ) . Since the euclidean inner prod-uct satisfies h· , ·i Eucl ≤ h· , ·i f , we see that the euclidean distance satisfies t − C ≤ d Eucl ( γ + ( t ) , γ − ( t )) ≤ d f ( γ + ( t ) , γ − ( t )) . Therefore, the Gromovproduct satisfies ([ γ + ] | [ γ − ]) p ≤ C < ∞ . In particular, here we have anexample of a non-visible pair that is algebraically visible. Example 2.23.
Restrict to t > , choose α ∈ (0 , ) and suppose, x ( t ) = − α t − α . We obtain f ( x ( t )) = 1 √ − t − α ,y ′ ( t ) = ± (1 − t − α ) ,y ( t ) = ± ( t − − α t − α ) + C. In particular, f ( x ) = 1 q − (1 − α ) − α − α x − α − α . Abbreviating σ := − α and τ := (1 − α ) − α − α , we rewrite f ( x ) = 1 q − τ x − α − α y ( t ) = ± ( t − σt − α ) + C. Again, a computation shows that f ′′ > and f ( x ) x →∞ −−−→ . Again, weobtain two geodesics γ + , γ − starting at the same starting point p , describedby these equations. We claim that their Gromov product is infinite. Forthis, it is sufficient to show that d ( γ + ( t ) , γ − ( t )) ≤ t − ψ ( t ) for some function ψ t →∞ −−−→ ∞ . Since d Eucl ( γ + ( t ) , γ − ( t )) ≤ t − σt − α ) + C ′ is the euclideanlength of the euclidean geodesic between γ + ( t ) , γ − ( t ) and since σ > , we obtain that d f ( γ + ( t ) , γ − ( t )) ≤ f ( x ( t ))( t − σt − α ) + f ( x ( t )) C ′ ≤ √ − t − α ( t − t − α ) + C ′ √ − t − α = 2 t p − t − α + C ′ √ − t − α ≤ t (1 − t − α ) + C ′ √ − t − α = 2 t − ψ ( t ) , where ψ ( t ) = t − α − C ′ √ − t − α t →∞ −−−→ ∞ . We conclude that ([ γ + ] | [ γ − ]) p = ∞ .These examples illustrate why we will assume not only that f is Möbiusbut also that it preserves visible pairs in the coming sections.2.6. Jacobi fields.
We now move fully into the realm of Riemannian man-ifolds. We refer to [dC15] for all necessary background informations. Let X be an n -dimensional, connected, geodesically complete Riemannian manifoldsuch that all sectional curvatures are non-positive. Let γ be a geodesic in X . A vector field J defined along γ is called a Jacobi-field if and only if itsatisfies the following second-order ordinary differential equation: D dt J ( t ) + R ( J ( t ) , γ ′ ( t )) γ ′ ( t ) . where R denotes the Riemannian curvature tensor and Ddt the covariantderivative along γ with respect to the Levi-Civita connection. Any Jacobifield along γ is uniquely determined by the initial conditions J (0) , DJdt (0) .The space of Jacobi fields along γ forms a real n -dimensional vector space.On complete manifolds, Jacobi fields are uniquely characterised as thevector fields arising from smooth one-parameter families of geodesics γ s with γ = γ . The Jacobi field corresponding to ( γ s ) s is given by J ( t ) = dds | s =0 γ s ( t ) . A Jacobi field is called perpendicular , if J ( t ) ⊥ γ ′ ( t ) for all t . AJacobi field is called stable if sup t ≥ {k J ( t ) k } < ∞ . A Jacobi field is called parallel if k J ( t ) k is constant along all of γ .We now define a subset of X that consists of all the points that have‘asymptotic features of flatness’. Specifically, F X := { x ∈ X |∃ γ geodesic ray, starting at x and ∃ J perpendicular, parallel Jacobi field along γ } . We first note that, whenever x ∈ F X , we find a geodesic γ as in thedefinition of F X and every point on γ is contained in F X . The followingresult that goes back to Eberlein connects the complement of F X with aconvexity property of horospheres. IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 29
Proposition 2.24 (Lemma 3.1 in [HIH77]) . Let X be a Hadamard manifold, Ξ the radial field in the direction of ξ ∈ ∂X and B a Busemann functioncentered at ξ . Then Ξ = − grad( B ) , Ξ is C and ∇ v Ξ =
DJdt (0) for all v ∈ T x X , where J is the unique stable Jacobi field along the geodesic ray ξ x such that J (0) = v . The covariant derivative ∇ v Ξ can be thought of as a second derivative ofthe Busemann function B , because for all v, w ∈ T x X , ddt | t =0 dds | s =0 B ( x, γ ( t, s ) , ξ ) = h∇ v Ξ , w i , where γ ( t, s ) = exp x ( tv + sw ) .3. Construction of Φ and F For the rest of this paper, let X , Y be n -dimensional, connected, simplyconnected, geodesically complete Riemannian manifolds such that their sec-tional curvatures are bounded by − b ≤ curv ≤ . Further, assume that ∂X and ∂Y satisfy (4v) and that all points in ∂X and ∂Y are in a rank 1hinge. We denote the unit tangent bundle of X by T X . Further, we havethe canonical projection π X : T X → X . If the manifold X is clear fromthe context, we simply write π . For all x ∈ X, ξ ∈ ∂X we denote the unittangent vector in T x X that ‘points to ξ ’, i.e. whose induced geodesic rayrepresents ξ , by −→ xξ . This provides us with a homeomorphism between ∂X and T x X equipped with the standard topology. Analogously, for any twopoints x, x ′ ∈ X , we denote the tangent vector of the arc-length geodesicfrom x to x ′ at x by −→ xx ′ .Let f : ∂X → ∂Y be a Möbius homeomorphism such that f and f − bothpreserve visible pairs. Our goal is to extend f to a map F : X → Y . Theconstruction presented in this section is a generalisation of a construction byBiswas for CAT( − spaces. Its most similar presentation to the one belowcan be found in [Bis17b].3.1. Constructing Φ . We start by constructing a map between the tangentbundles of X and Y . However, it turns out that this map can only be definedafter identifying certain vectors in the tangent bundle.Let v ∈ T X . The geodesic flow on X provides us with a unique bi-infinitegeodesic γ such that γ ′ (0) = v . Denote the two endpoints of γ at infinityby v −∞ := γ ( −∞ ) and v ∞ := γ ( ∞ ) . Let v, w ∈ T X and denote theirprojection in X by x and x ′ respectively. We say that v ∼ w , if k v k = k w k , v ∞ = w ∞ , v −∞ = w −∞ and B ( x, x ′ , v ∞ ) . Note that this is equivalent tothe convex hull of the geodesics induced by v and w being a flat strip (seeTheorem 2.13 in Part II of [BH99]) and the foot points of v and w beingon the same horosphere with respect to either endpoint of the strip. Thisdefines an equivalence relation on T X and we denote the quotient by
T X .Denote the quotient of the unit tangent bundle by the same equivalence vv ′ Φ( v )Φ( v ′ ) v ∞ v ′∞ f ( v ′∞ ) f ( v ′−∞ ) f ( v ∞ ) Figure 5.
The vector v is sent to the vector Φ( v ) . Thederivative ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) determines, which horosphere Φ( v ) needs to be placed on. If f ( v ′∞ ) , f ( v ′−∞ ) have several con-necting bi-infinite geodesics, the choice of Φ( v ′ ) is no longerunique and we obtain a non-trivial equivalence class.relation by T X . The equivalence class of a vector v will be denoted by [ v ] .Since v ∼ w ⇔ − v ∼ − w , we define − [ v ] := [ − v ] .We construct a map Φ : T X → T Y which will be a geodesic conjugacyin the sense of Lemma 3.3. Let v ∈ T x X be a unit-vector. As above, weobtain two points v ∞ , v −∞ at infinity. Since f preserves visible pairs, thereexists at least one geodesic from f ( v −∞ ) to f ( v ∞ ) . Choose one such geodesicand denote it by γ . The image of [ v ] under Φ will be the equivalence classof a unit-vector on the geodesic γ pointing towards f ( v ∞ ) . All that is left isto choose the foot point on γ . Lemma 3.1 (cf. [Bis15]) . There exists a unique y ∈ γ , such that ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) =1 . Furthermore, if γ ′ is another geodesic from f ( v −∞ ) to f ( v ∞ ) and y ′ theunique point on γ ′ such that ∂f ∗ ρ x ∂ρ ′ y ( f ( v ∞ )) = 1 , then −−−−→ yf ( v ∞ ) ∼ −−−−−→ y ′ f ( v ∞ ) .Finally, if v ∼ v ′ and π ( v ′ ) =: x ′ , then for all y ∈ Y , ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) = ∂f ∗ ρ x ′ ∂ρ y ( f ( v ∞ )) . We define
Φ([ v ]) to be the equivalence class of the unit vector at thisunique point y that points to f ( v ∞ ) (see Figure 5). By Lemma 3.1, Φ iswell-defined. Whenever we use an equivalence class [ v ] as an input for Φ , wesimply write Φ( v ) . Proof.
Let γ be a bi-infinite geodesic from f ( v −∞ ) to f ( v ∞ ) and y , y ′ ∈ γ .By the Chain Rule for metric derivatives, we have IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 31 ∂f ∗ ρ x ∂ρ y ′ ( f ( v ∞ )) = ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) ∂ρ y ∂ρ y ′ ( f ( v ∞ ))= ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) e B ( y,y ′ ,f ( v ∞ )) . Since for any geodesic representative γ of ξ , B ( γ ( t ) , γ ( t ′ ) , ξ ) = t − t ′ , andsince the metric derivative is always a positive number by the way it is intro-duced in Definition 2.15, any point y ′ ∈ γ provides us with a unique number t − t ′ and a unique point y ∈ γ such that ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) = 1 . This impliesexistence and uniqueness.For the second statement, let γ ′ be another geodesic from f ( v −∞ ) to f ( v ∞ ) and y ′ the unique point on γ ′ such that ∂f ∗ ρ x ∂ρ y ′ ( f ( v ∞ )) = 1 . Using the ChainRule and Remark 2.18, we get B ( y ′ , y, f ( v ∞ )) = ln (cid:18) ∂ρ y ′ ∂ρ y ( f ( v ∞ )) (cid:19) = ln (cid:18) ∂ρ y ′ ∂f ∗ ρ x ( f ( v ∞ )) · ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) (cid:19) = ln(1) = 0 . Since γ and γ ′ have the same endpoints, it follows that the unit vectorsat y and y ′ respectively, pointing at f ( v ∞ ) are equivalent.To prove the last statement, let v ∼ v ′ , π ( v ′ ) =: x ′ and y ∈ Y . For thesame reasons as above, we have ∂f ∗ ρ x ∂ρ y ( f ( v ∞ )) = ∂ρ x ∂ρ x ′ ( v ∞ ) · ∂f ∗ ρ x ′ ∂ρ y ( f ( v ∞ ))= e B ( x,x ′ ,v ∞ ) ∂f ∗ ρ x ′ ∂ρ y ( f ( v ∞ ))= ∂f ∗ ρ x ′ ∂ρ y ( f ( v ∞ )) . (cid:3) Lemma 3.2.
For all v ∈ T x X , we have Φ( − v ) = − Φ( v ) .Proof. By the Geometric mean value theorem, for all y on a geodesic γ from f ( v −∞ ) to f ( v ∞ ) , ∂f ∗ ρ x ρ y ( f ( v ∞ )) ∂f ∗ ρ x ρ y ( f ( v −∞ )) = ρ y ( f ( v ∞ ) , f ( v −∞ )) ρ x ( v ∞ , v −∞ ) = 11 = 1 . This implies that ∂f ∗ ρ x ρ y ( f ( v ∞ )) = 1 if and only if ∂f ∗ ρ x ρ y ( f ( v −∞ )) = 1 . (cid:3) Throughout the following, we will want to consider Busemann functionsthat are evaluated on a point in the set π ([ v ]) . We denote π ◦ Φ( v ) to be thefoot point of a chosen representative of Φ( v ) . Lemma 3.3 (cf. [Bis15]) . For all x, x ′ ∈ X, ξ ∈ ∂X , B ( π ◦ Φ( −→ xξ ) , π ◦ Φ( −→ x ′ ξ ) , f ( ξ )) = B ( x, x ′ , ξ ) . By Lemma 3.1, the left-hand-side does not depend on the choice of repre-sentative and is thus well-defined.Proof. B ( π ◦ Φ( −→ xξ ) , π ◦ Φ( −→ x ′ ξ ) , f ( ξ )) = ln ∂ρ π ◦ Φ( −→ xξ ) ∂ρ π ◦ Φ( −→ x ′ ξ ) ( f ( ξ )) ! = ln ∂ρ π ◦ Φ( −→ xξ ) ∂f ∗ ρ x ( f ( ξ )) · ∂f ∗ ρ x ∂f ∗ ρ x ′ ( f ( ξ )) · ∂f ∗ ρ x ′ ∂ρ π ◦ Φ( −→ x ′ ξ ) ( f ( ξ )) ! = ln (cid:18) ∂ρ x ∂ρ x ′ ( ξ ) (cid:19) = B ( x, x ′ , ξ ) . (cid:3) The map Φ is natural in the following sense. Lemma 3.4.
Given two Möbius bijections f : ∂X → ∂Y, g : ∂Y → ∂Z thatare homeomorphisms and preserve visible pairs, we have Φ g ◦ Φ f = Φ g ◦ f . Furthermore, Φ Id = Id .Proof. Let u ∈ T X with π ( u ) = x . Choose v ∈ Φ f ( u ) , w ∈ Φ g ( v ) , w ′ ∈ Φ g ◦ f ( u ) and denote y := π ( v ) , z := π ( w ) , z ′ := π ( w ′ ) . By construction of Φ , w ′∞ = g ( f ( u ∞ )) = w ∞ , w ′−∞ = g ( f ( u −∞ )) = w −∞ and ∂ ( g ◦ f ) ∗ ρ x ∂ρ z ( w ′∞ ) = 1 .By the Chain Rule, ∂g ∗ ρ y ∂ρ z ( w ∞ ) · ∂f ∗ ρ x ∂ρ y ( v ∞ ) = ∂g ∗ f ∗ ρ x ∂ρ z ( w ∞ ) = 1 and therefore, w is in the equivalence class of Φ g ◦ f ( v ) . The identity Φ Id = Id is immediate. (cid:3) Remark 3.5.
We did not say that Φ is part of a functor because there areopen questions regarding a potential category of boundaries to use. Specifi-cally, for an object to be part of a ‘boundary category’ on which the construc-tion above makes sense, this object needs to admit a ‘filling’ by a Hadamardmanifold. This is sometimes called the inverse problem for Möbius geometry.The only case the author is aware of, where the inverse problem is solved, isthe case when the boundary is a circle (see [Buy19]). IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 33
Lemma 3.4 implies in particular that Φ is invertible and its inverse is themap induced by f − . In [Bis15], Biswas shows that, if X and Y are both CAT( − spaces, the map Φ is a homeomorphism. Since X, Y can containflat strips under our assumptions, his proof does not generalize directly. Wewill present a way around this in the next section. Nevertheless, we raise thefollowing
Question.
Is the map
Φ : T X → T Y a homeomorphism?3.2. Constructing F . Let x ∈ X . Consider the unit-tangent sphere T x X at x . Every point ξ ∈ ∂X can be represented by a unit vector −→ xξ ∈ T x X .Applying the map Φ to all −→ xξ , we obtain a collection of equivalence classesin T Y . Note that we may not be able to choose representatives of theseequivalence classes, such that all representatives share the same foot point.We want F ( x ) ∈ Y to be ’in the middle’ of the family Φ( T x X ) . For all x ∈ X , y ∈ Y , ξ ∈ ∂X , we define u x,y ( ξ ) := B ( π ◦ Φ( −→ xξ ) , y, f ( ξ )) . By Lemma 3.1, the expression above is independent of the choice of π ◦ Φ( −→ xξ ) . We start by showing important properties of u x,y ( ξ ) . Lemma 3.6 (cf. [Bis15]) . For all x ∈ X , y ∈ Y , ξ ∈ ∂X , ∂f ∗ ρ x ∂ρ y ( f ( ξ )) = e u x,y ( ξ ) . Proof.
We have ∂f ∗ ρ x ∂ρ y ( f ( ξ )) = ∂f ∗ ρ x ∂ρ π ◦ Φ( −→ xξ ) ( f ( ξ )) ∂ρ π ◦ Φ( −→ xξ ) ∂ρ y ( f ( ξ ))= 1 · e B ( π ◦ Φ( −→ xξ ) ,y,f ( ξ )) = e u x,y ( ξ ) , where we used the definition of Φ and Remark 2.18 in the second step. (cid:3) Lemma 3.7.
The map u x,y ( ξ ) is continuous in x, y and ξ .Proof. Clearly, u is continuous in y . To show continuity in x , denote theextension of f − by Ψ : T X → T Y . By Lemma 3.4, Ψ = Φ − . By Lemma3.3, we have u x,y ( ξ ) = B ( π ◦ Φ( −→ xξ ) , π ◦ Φ(Ψ( −−−→ yf ( ξ ))) , f ( ξ )) = B ( x, π ◦ Ψ( −−−→ yf ( ξ )) , ξ ) , which is continuous in x .To prove continuity in ξ , we note that this is equivalent to continuity of ∂f ∗ ρ x ∂ρ y ( f ( ξ )) in ξ by Lemma 3.6. Since f is continuous by assumption, we areleft to prove continuity of specific metric derivatives. By definition, ∂f ∗ ρ x ∂ρ y ( f ( ξ )) = ρ x ( ξ, η ) ρ x ( ξ, ζ ) ρ y ( f ( η ) , f ( ζ )) ρ x ( η, ζ ) ρ y ( f ( ξ ) , f ( η )) ρ y ( f ( ξ ) , f ( ζ )) for any η, ζ ∈ ∂X such that ( ξ, η, ζ ) is an algebraically visible triple. Sinceevery point in ∂X is in a rank 1 hinge, we can additionally choose η and ζ , such that ( ξ, η ) and ( ξ, ζ ) are connected by a rank 1 geodesic. Since f preserves visible pairs, Lemma 2.10, implies that ( f ( ξ ) , f ( η )) and ( f ( ξ ) , f ( ζ )) can be connected by a rank 1 geodesic. Corollary 2.11 then implies that theexpression above is continuous in ξ . This proves continuity of u in ξ . (cid:3) Since ∂X is compact, continuity implies that the supremum-norm k u x,y k ∞ < ∞ . By Lemma 2.21, we know that k u x,y k ∞ is convex in y . Furthermore,the function y
7→ k u x,y k ∞ is proper, since for any diverging sequence y n , wehave sup ξ ∈ ∂X { u x,y n ( ξ ) } = sup ξ ∈ ∂X { u x,y ( ξ ) + B ( y , y n , f ( ξ )) } n →∞ −−−→ ∞ , because for every n , we can choose ξ such that f ( ξ ) is the endpoint of thegeodesic from y to y n which yields sup ξ ∈ ∂X { B ( y , y n , f ( ξ )) } = d ( y , y n ) →∞ , while k u x,y k ∞ < ∞ .Since y
7→ k u x,y k ∞ is proper and convex, the function M ( x ) := min y ∈ Y {k u x,y k ∞ } is well-defined. In addition, we define M x := { y ∈ Y |k u x,y k ∞ = M ( x ) } the set of points where the minimum is obtained. Finally, we define for any x ∈ X, y ∈ Y K x,y := { ξ ∈ ∂X | u x,y ( ξ ) = k u x,y k ∞ } the set of points in the boundary where the u x,y obtains its supremum.Analogously, for every y ∈ Y , we obtain sets M y ⊂ X and K y,x ⊂ ∂Y byworking with f − and Φ − . Lemma 2.19 and Lemma 3.6 together imply that ∀ y ∈ M x : M ( x ) = max ξ ∈ ∂X { u x,y ( ξ ) } = − min ξ ∈ ∂X { u x,y ( ξ ) } . In particular, we conclude that K x,y is non-empty for all x ∈ X, y ∈ M x .We would like to define F ( x ) to be the unique point in M x . However, if Y is not a CAT( − space, it is absolutely not clear that M x consists onlyof one point. As we will see in a moment, issues arise whenever Φ( −→ xξ ) isan equivalence class that contains more than one vector. We solve this bydefining an equivalence relation on Y : We define ∼ to be the equivalencerelation generated by demanding that y ∼ y ′ , whenever there exists x ∈ X such that y, y ′ ∈ M x . Denote Y := Y (cid:30) ∼ . We define F : X → Y to be themap that sends x [ M x ] . We call F the circumcenter extension of f .We can characterise elements of M x as follows. Lemma 3.8 (cf. [Bis17b]) . Let x ∈ X, y ∈ Y . The following are equivalent:(1) y ∈ M x (2) For all w ∈ T y Y , there exists ξ ∈ K x,y such that h w, −−−→ yf ( ξ ) i ≤ . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 35 (3) The convex hull of the set {−−−→ yf ( ξ ) | ξ ∈ K x,y } in T y Y contains the zerovector.Proof of Lemma 3.8. (1) ⇒ (2): Suppose not. Then, we find x ∈ X, y ∈ M x and w ∈ T y Y such that for all ξ ∈ K x,y , h w, −−−→ yf ( ξ ) i > . Let γ be thegeodesic passing through y at time zero with tangent vector w . Since theinner product is continuous and K x,y is compact, we find ǫ, ǫ ′ > and aneighbourhood N of K x,y , such that for all ξ ′ ∈ N, h w, −−−→ yf ( ξ ′ ) i > ǫ and forall ξ ′ ∈ X (cid:31) N , u x,y ( ξ ′ ) < M ( x ) − ǫ ′ . Using the fact that the gradient ofthe map y B ( y ′ , y, η ) is equal to −−→ yη , we obtain for all ξ ′ ∈ N and t sufficiently small u x,γ ( t ) ( ξ ′ ) = B ( π ◦ Φ( −→ xξ ) , γ ( t ) , f ( ξ ))= B ( π ◦ Φ( −→ xξ ) , y, f ( ξ )) + B ( y, γ ( t ) , f ( ξ )) ≤ M ( x ) + t ( −h−−−→ yf ( ξ ) , w i ) + o ( t ) < M ( x ) For ξ ′ ∈ ∂X (cid:31) N , we have u x,γ ( t ) ( ξ ′ ) = u x,y ( ξ ′ ) + B ( y, γ ( t ) , f ( ξ ′ ))
The following is an important property of the function M . Lemma 3.9 (cf. [Bis18]) . The map M : X → R is -Lipschitz continuous.Furthermore, the maps x
7→ k u x,y k ∞ for fixed y and y
7→ k u x,y k ∞ for fixed x are both -Lipschitz.Proof of Lemma 3.9. Let x, x ′ ∈ X , y ∈ M x , y ′ ∈ M x ′ , ξ ∈ K x ′ ,y . UsingLemma 3.3, we compute M ( x ′ ) = k u x ′ ,y ′ k ∞ ≤ k u x ′ ,y k ∞ = B ( π ◦ Φ( −→ x ′ ξ ) , y, f ( ξ ))= B ( x ′ , x, ξ ) + B ( π ◦ Φ( −→ xξ ) , y, f ( ξ )) ≤ d ( x, x ′ ) + k u x,y k ∞ . We conclude that M ( x ′ ) ≤ d ( x, x ′ )+ M ( x ) . Since the argument is symmetricin x, x ′ , we conclude that M is -Lipschitz continuous. This estimate alsoproves the -Lipschitz continuity of the map x
7→ k u x,y k ∞ . For the last map,the proof is analogous with ξ ∈ K x,y ′ . (cid:3) It turns out that F has several nice properties. Lemma 3.10.
The map F is continuous with respect to the quotient topology.Proof of Lemma 3.10. Since X is first countable, it is enough to show that F is sequentially continuous. Let y ∈ Y such that [ y ] ∈ Im( F ) . Let [ y ] ∈ U ⊂ Y open, i.e. [ y ] ⊂ U := P − ( U ) , U open, where P denotes the projection Y → Y . Let x ∈ X such that y ∈ M x and thus, F ( x ) = [ y ] . We have k u x,y k ∞ = M ( x ) . Let x n → x and let y n ∈ M x n .We first show that ( y n ) n is bounded. Suppose not. Since the map y u x,y k ∞ is proper, we conclude that there is a subsequence, also denoted ( y n ) n such that k u x,y n k ∞ → ∞ . On the other hand, since M is -Lipschitz, k u x n ,y n k ∞ = M ( x n ) → M ( x ) = k u x,y k ∞ . In addition, since x
7→ k u x,y ′ k ∞ is -Lipschitz for all y ′ , we conclude that M ( x ) ≥ M ( x n ) − d ( x, x n ) ≥ k u x,y n k ∞ − d ( x, x n ) → ∞ . This is a contradiction, hence ( y n ) n is bounded.By properness of Y , any subsequence of ( y n ) n has a converging subse-quence ( y n i ) i that converges to some y ′ ∈ Y . We claim that y ′ ∼ y . Since k u x n ,y n k ∞ is -Lipschitz continuous in both variables, we have k u x,y k ∞ = M ( x ) n →∞ ←−−− M ( x n ) = k u x n ,y n k ∞ n →∞ −−−→ k u x,y ′ k ∞ . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 37
Therefore, y ′ ∈ M x . Suppose now, that ( y n ) n admits a subsequence ( y n i ) n such that for all n i , y n i / ∈ U . No subsequence of ( y n i ) i can converge toan element in M x , a contradiction to our argument above. Therefore, forall large n , [ y n ] ∈ U and [ y n ] → [ y ] . We conclude that F is sequentiallycontinuous. (cid:3) We are now ready to prove that the equivalence relation on Y affects onlyspecific parts of Y . Proposition 3.11.
The union E Y := S x ∈ X : | M x |≥ M x satisfies E Y ⊂ F Y .In particular, the projection P : Y → Y is a homeomorphism on Y (cid:31) F Y . In order to prove this, we need to do some preparation which will be offurther use in later sections.
Definition 3.12.
Let x ∈ X, ξ ∈ ∂X . We define a x : ∂X → ∂X to be themap that sends ξ ∈ ∂X to the forward endpoint of the geodesic ray inducedby the vector −−→ xξ . We call a x the antipodal map with respect to x .By definition, a x = exp x ◦ ( − Id ) ◦ exp − x . Since the visual topology co-incides with the standard topology on the unit tangent sphere T x X , weimmediately see that a x is a homeomorphism. Proposition 3.13 (cf. [Bis18]) . Fix x ∈ X and y ∈ M x . Let ξ ∈ ∂X . If ∂f ∗ ρ x ∂ρ y ( f ( ξ )) is minimal among all ξ , then there exists a bi-infinite geodesic γ ∈ [ f ( a x ( ξ )) , f ( ξ )] such that y lies on γ . In particular, f ( a x ( ξ )) = a y ( f ( ξ )) .Proof. A point y lies on a geodesic from f ( a x ( ξ )) to f ( ξ ) if and only if ρ y ( f ( a x ( ξ )) , f ( ξ )) = 1 . Combining Lemma 2.19 and Lemma 3.6, we knowthat the minimal value obtained by ∂f ∗ ρ x ∂ρ x ( f ( ξ )) is equal to e − M ( x ) . Wecompute ρ y ( f ( a x ( ξ )) f ( ξ )) = ∂ρ y ∂f ∗ ρ x ( f ( a x ( ξ ))) ∂ρ y ∂f ∗ ρ x ( f ( ξ )) f ∗ ρ x ( f ( a x ( ξ )) , f ( ξ )) = ∂ρ y ∂f ∗ ρ x ( f ( a x ( ξ ))) e M ( x ) ≥ e − M ( x ) e M ( x ) = 1 , where we used the fact that x ∈ ( a x ( ξ ) , ξ ) by construction. This concludesthe proof. (cid:3) Corollary 3.14 (cf. [Bis18]) . If x ∈ X, y ∈ M x , ξ ∈ ∂X , then ∂f ∗ ρ x ∂ρ y ( f ( ξ )) is maximal if and only if ∂f ∗ ρ x ∂ρ y ( f ( a x ( ξ ))) is minimal.Proof. If ∂f ∗ ρ x ∂ρ y ( f ( a x ( ξ ))) is minimal, then y lies on a geodesic from f ( a x ( ξ )) to f ( ξ ) . Then, ∂f ∗ ρ x ∂ρ y ( f ( ξ )) = ∂f ∗ ρ x ∂ρ y ( f ( a x ( ξ ))) − = e M ( x ) by the Geometricmean value theorem.On the other hand, if ∂f ∗ ρ x ∂ρ y ( f ( ξ )) is maximal, then ∂f − ∗ ρ y ∂ρ x ( ξ ) is minimal bythe Chain rule. By Proposition 3.13, this implies that a x ( ξ ) = f − ( a y ( f ( ξ ))) . The argument above implies that ∂f − ∗ ρ y ∂ρ x ( a x ( ξ )) is maximal and therefore, ∂f ∗ ρ x ∂ρ y ( f ( a x ( ξ ))) is minimal. (cid:3) Corollary 3.15.
For all x ∈ X, y ∈ M x , the set K x,y contains at least threepoints.Proof. By Lemma 3.8, K x,y contains at least two points, as any non-trivialconvex combination requires at least two vectors. Suppose it consisted ofexactly two points ξ, η . Then α −−−→ yf ( ξ ) + α −−−→ yf ( η ) for α , α > . Sincethis is a sum of unit vectors, we conclude that a y ( f ( ξ )) = f ( η ) . As ξ ∈ K x,y ,Corollary 3.14 implies that u x,y ( a x ( ξ )) is minimal. By Proposition 3.13, f ( a x ( ξ )) = a y ( f ( ξ )) = f ( η ) . Thus, u x,y ( η ) is both maximal and minimal.Since min ξ ∈ ∂X { u x,y ( ξ ) } = − max ξ ∈ ∂X { u x,y ( ξ ) } , we obtain that u x,y ( η ) =0 and u x,y ≡ . Therefore, K x,y = ∂X , which contains infinitely manypoints. (cid:3) The following result provides us with more information about M x , whichmay be of general interest in further study of this construction. Lemma 3.16.
Let x ∈ X . The set M x is convex and contained in anintersection of at least three horospheres in Y . Furthermore, diam( M x ) ≤ M ( x ) < ∞ . In particular, M x is compact and has codimension at least twoin Y .Proof of Lemma 3.16. If M x consists of exactly one point, this is trivial.Suppose, M x contains at least two points. Let y = y ′ ∈ M x and denote thegeodesic from y to y ′ by γ . Since k u x,γ k ∞ is convex, greater or equal to M ( x ) and equal to M ( x ) at both endpoints, we conclude that k u x,γ k ∞ ≡ M ( x ) .Therefore, M x is convex.Let p be any point on γ strictly between y and y ′ . Since u x,p is continu-ous, we find at least one ξ ∈ K x,p . Since u x,γ ( ξ ) is convex (for any ξ ∈ ∂X ),we obtain that it is either constant or increasing in one direction. If it wasincreasing, then k u x,γ k ∞ > | u x,p ( ξ ) | = M ( x ) for some γ ( t ) near, but notequal, to p . This contradicts the fact that k u x,γ k ∞ ≡ M ( x ) . Therefore, u x,γ ( ξ ) ≡ M ( x ) along γ .By Proposition 3.13, we conclude that every y ∈ M x lies on a geodesicfrom f ( a x ( ξ )) to f ( ξ ) . Therefore, γ is contained in a horosphere centeredat f ( ξ ) intersected with a flat strip from f ( a x ( ξ )) to f ( ξ ) . Furthermore,we see that for every point p on γ that is not an end point, any element ξ ∈ K x,p realises the supremum k u x,γ ( t ) k ∞ at every point on the geodesic γ . In particular, if we extend γ to its maximal length such that it is stillcontained in M x , the points in ∂X that obtain u x,γ ( ξ ) = M ( x ) are the samealong the entire geodesic, except for some extremal points that appear onlyat the endpoints of the extended geodesic.Choose y on γ not an endpoint. By Lemma 3.8 and Corollary 3.15, thereexist k ≥ , ξ , . . . , ξ k ∈ K x,y and α , . . . , α k > such that P ki =1 α i = 1 IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 39 and P ki =1 α i −−−→ yf ( ξ i ) = 0 . In particular, γ is contained in the intersectionof horospheres centered at f ( ξ ) , . . . f ( ξ k ) . Suppose, M x is not containedin the intersection of these horospheres. Then we find y ∈ M x such that B ( y, y , ξ i ) = 0 for some i . Without loss of generality, B ( y, y , ξ ) = 0 .Since M ( x ) ≥ u x,y ( ξ i ) = u x,y ( ξ i ) + B ( y , y, ξ i ) = M ( x ) + B ( y , y, ξ i ) , weconclude that B ( y , y, ξ i ) ≤ for all i and B ( y , y, ξ ) < . Let δ be thegeodesic from y to y . By convexity, B ( y , δ ( t ) , ξ ) is decreasing for small,positive t . Therefore, > ddt | t =0 B ( y , δ ( t ) , f ( ξ )) = h δ ′ (0) , −−−−→ yf ( ξ ) i . On the other hand, h δ ′ (0) , k X i =1 α i −−−→ yf ( ξ i ) i = α h δ ′ (0) , −−−−→ yf ( ξ ) i + k X i =2 α i h δ ′ (0) , −−−→ yf ( ξ i ) i . Since α i > for all k , we conclude that ddt | t =0 B ( y , δ ( t ) , f ( ξ i )) = h δ ′ (0) , −−−→ yf ( ξ i ) i > for some i ≥ . In particular, u x,δ ( t ) ( ξ i ) = u x,y ( ξ i ) + B ( y , δ ( t ) , f ( ξ i )) >u x,y ( ξ i ) = M ( x ) for t > sufficiently small. Since y, y ∈ M x and M x isconvex, we have found an element δ ( t ) in M x for which k u x,δ ( t ) k ∞ is notminimal, a contradiction. Therefore, there can be no point y outside of theintersection of the horospheres centered at the points f ( ξ ) , . . . , f ( ξ k ) . Since,among any three distinct horospheres with non-empty intersection, at leasttwo of them intersect transversely and horospheres have codimension one,we conclude that M x has codimension at least two. This proves the Lemmaexcept for the bound on the diameter.To estimate the diameter, consider y, y ′ ∈ M x , let δ be the geodesic from y to y ′ and let ξ ∈ ∂X such that f ( ξ ) is the forward end-point of the geodesicray induced by δ . Then u x,y ( ξ ) = u x,y ′ ( ξ ) + B ( y ′ , y, f ( ξ )) = u x,y ′ ( ξ ) + d ( y, y ′ ) . Since y, y ′ ∈ M x , the expressions u x,y ( ξ ) , u x,y ′ ( ξ ) are both bounded inabsolute value by M ( x ) . The equation above shows that, whenever d ( y, y ′ ) > M ( x ) , i.e. the length of δ greater than M ( x ) , this bound is violated byat least one of the two terms. We obtain that any two points in M x areconnected by a geodesic of length at most M ( x ) . This provides the boundon the diameter. (cid:3) Proof of Proposition 3.11.
By the proof of Lemma 3.16, if M x contains atleast two points, any geodesic in M x is contained within a flat strip. There-fore, E Y is contained in the union of all flat strips in Y . Since every bi-infinitegeodesic in a flat strip admits a perpendicular, parallel Jacobi field, everyflat strip in Y is contained in F Y . Therefore, E Y ⊂ F Y . (cid:3) Based on Lemma 3.16, it makes sense to define the set K x := { ξ ∈ ∂X |∀ y ∈ M x : u x,y ( ξ ) = M ( x ) } . The set K x is non-empty, compact and, by the proof of Lemma 3.16, con-tains at least three points.In [Bis17a], the map F is constructed as the limit of a sequence of circum-centers. There is another geometric interpretation of M x and M ( x ) , whichwe present here. Any vector v ∈ T X defines a horoball in X , namely the set HB ( v ) := { x ∈ X | B ( π ( v ) , x, v ∞ ) ≤ } . Consider the horoballs HB (Φ( v )) for all v ∈ T x X . Define Φ( v ) t to be thevector obtained by applying the geodesic flow on Y to the vector Φ( v ) (thegeodesic flow sends equivalence classes in T y Y to equivalence classes). Since Φ( − v ) = − Φ( v ) , we know that the intersection T v ∈ T x X HB (Φ( v )) is thesmallest non-empty intersection in the sense that T v ∈ T x X HB (Φ( v ) t ) = ∅ for all t > . If this intersection is empty, there is a minimal t , such that T v ∈ T x X HB (Φ( v ) − t ) is non-empty. This minimal t equals M ( x ) and theintersection of the horoballs Φ( v ) − t equals M x (see figure 6 for the situationwhere M x consists of one point).4. Hölder and Lipschitz continuity of F Recall that, in section 2.6, we defined F X to be the set of all points in X that admit a geodesic ray γ starting at x and a perpendicular, parallel Jacobifield along γ . The goal of this section is to prove that F is locally Höldercontinuous on F − ( Y (cid:31) F Y ) and to provide a sufficient condition for F tobe locally Lipschitz continuous. To do so, we will use geometric propertiesarising from bounds on the second derivative of the Busemann function.We first introduce some notation. Given a function g : Y → R that is twicecontinuously differentiable, we can consider its Hessian, i.e. the bilinear forminduced by its second differential. Since Busemann functions on CAT(0) manifolds are twice continuously differentiable, we can consider the Hessianof the Busemann function y B ( y ′ , y, η ) , which we denote by H y B η ( y ) .Since a change of y ′ changes the function y B ( y ′ , y, η ) by a constantindependent of y , H y B η ( y ) is independent of y ′ . Since Busemann functionsare convex in their second variable, H y B η ( y ) is semi-positive definite.Let γ be the geodesic ray from y to η . Since ddt | t =0 B ( y ′ , γ ( t ) , η ) = − ,we see that H y B η ( y )( γ ′ (0) , w ) = 0 for all w ∈ T y Y . Therefore, we areinterested in the restriction of the Hessian to the orthogonal complement of γ ′ (0) = −→ yη , which we denote by −→ yη ⊥ . Let w ∈ T y Y , η ∈ ∂Y . We write w ⊥ η for the orthogonal projection of w onto −→ yη ⊥ . Lemma 4.1.
Let y ∈ Y (cid:31) F Y . Then there exists an open neighbourhood U ⊂ Y (cid:31) F Y of y and a constant ǫ > , such that for all y ∈ U , w ∈ T y Y , IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 41 Φ( v )Φ( v )Φ( v ) M x Figure 6.
If we flow the images Φ( v i ) backwards in Y , weincrease the drawn horoballs until they all intersect (whichhappens for the first time in the case of the dashed horoballs).The candidates for F ( x ) are all the points in the mutual inter-section of the dashed horoballs when going over all v ∈ T x X . we have H y B η ( w, w ) ≥ ǫ k w ⊥ η k . Proof.
Since y ∈ Y (cid:31) F Y , we know that for all η ∈ ∂Y and all w ∈ −→ y η ⊥ with k w k = 1 , the unique stable Jacobi field J w along the geodesic ray η y satisfies ddt | t =0 k J w ( t ) k < . Since ddt | t =0 k J w ( t ) k depends continuously on y , η and w and since ∂Y and −→ yη ⊥ ∩ T y Y are compact for all y ∈ Y , we find some constant ǫ > andan open neighbourhood U of y , such that U ⊂ Y (cid:31) F Y and for all y ∈ U , all η ∈ ∂Y , all w ∈ −→ yη ⊥ with k w k = 1 and all stable Jacobi fields J w along thegeodesic from y to η , we have ddt | t =0 k J w ( t ) k ≤ − ǫ. Using Proposition 3.1 in [HIH77] (see section 2.6), we have for all y ∈ U, η ∈ ∂Y, w ∈ −→ yη ⊥ with k w k = 1 : H y B η ( y )( w, w ) = h∇ w ( −−→ yη ) , w i = − ddt | t =0 k J w ( t ) k ≥ ǫ. Since H y B η ( y )( −→ yη, w ) = 0 for all w ∈ T y Y and since the Hessian is bilinear,we obtain the estimate stated in the Lemma. (cid:3) We need one more piece of notation before stating the results on localHölder and local Lipschitz continuity. We define the sets D X := F − ( Y (cid:31) F Y ) L X := { x ∈ D X |∃ U open neighbourhood of x, ∃ ǫ > ∀ x ′ ∈ U, ∀ w ∈ T F ( x ′ ) Y, ∃ ξ ∈ K x : h w, −−−−−−→ F ( x ′ ) f ( ξ ) i > ǫ } . Remark 4.2.
By Lemma 3.8, any x ∈ X and any w ∈ T F ( x ) Y admits ξ ∈ K x such that h w, −−−−−−→ F ( x ) f ( ξ ) i ≥ . However, in dimension three and higher, it isvery unclear, if strict inequality can be obtained in general and if it can beobtained uniformly in an open neighbourhood of x . Proposition 4.3.
The map F is locally -Hölder continuous on D X andlocally Lipschitz continuous on L X . The statement on Hölder continuity generalises a result by Biswas in[Bis17a]. The proof is, however, different.
Proof.
Let x ∈ D X . By Lemma 4.1, we find an open neighbourhood U of x and ǫ > , such that for all x ∈ U , the Hessian H y B f ( ξ ) ( F ( x )) is posi-tive definite on the subspace −−−−−−→ F ( x ) f ( ξ ) ⊥ and its positive Eigenvalues are atleast ǫ . Let x, x ′ ∈ U and let f ( ξ ) be the point represented by the geo-desic ray obtained by extending the geodesic from F ( x ′ ) to F ( x ) . Note that −−−−−−−→ F ( x ) F ( x ′ ) = −−−−−−−−→ F ( x ) f ( ξ ) . There are two cases.Case 1: If ξ ∈ K x , then d ( F ( x ) , F ( x ′ )) = B ( F ( x ) , F ( x ′ ) , f ( ξ ))= B ( F ( x ) , π ◦ Φ( −→ xξ ) , f ( ξ )) + B ( x, x ′ , ξ ) + B ( π ◦ Φ( −−→ x ′ ξ ) , F ( x ′ ) , f ( ξ )) ≤ − M ( x ) + d ( x, x ′ ) + M ( x ′ ) ≤ d ( x, x ′ ) , where we use that M is -Lipschitz continuous by Lemma 3.9.Case 2: Suppose, ξ / ∈ K x . By continuity of the Riemannian metric, thereexists δ > , such that for all ξ ∈ K x , h−−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) i ≤ − δ .By Lemma 3.8, we find ξ ∈ K x such that h−−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) i ≥ . In IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 43 particular, this ξ satisfies k−−−−−−−→ F ( x ) F ( x ′ ) ⊥ f ( ξ ) k ≥ δ . By Taylor approximation,we know that B ( F ( x ) , F ( x ′ ) , f ( ξ )) = − D −−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) E d ( F ( x ) , F ( x ′ ))+ H x B f ( ξ ) (cid:16) −−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−−→ F ( x ) F ( x ′ ) (cid:17) d ( F ( x ) , F ( x ′ )) + o (cid:0) d ( F ( x ) , F ( x ′ )) (cid:1) . Let < λ < . For d ( F ( x ) , F ( x ′ )) sufficiently small (‘sufficiently small’depending on λ ), this implies B ( F ( x ) , F ( x ′ ) , f ( ξ )) ≥ λǫ (cid:13)(cid:13)(cid:13) −−−−−−−→ F ( x ) F ( x ′ ) ⊥ f ( ξ ) (cid:13)(cid:13)(cid:13) d ( F ( x ) , F ( x ′ )) ≥ λǫδ d ( F ( x ) , F ( x ′ )) Let U λ ⊂ U be an open neighbourhood of x , such that for all x, x ′ ∈ U , d ( F ( x ) , F ( x ′ )) is sufficiently small in the sense above. On the other hand, B ( F ( x ) , F ( x ′ ) , f ( ξ )) = B ( F ( x ) , π ◦ Φ( −→ xξ ) , f ( ξ )) + B ( x, x ′ , ξ ) + B ( π ◦ Φ( −→ x ′ ξ ) , F ( x ′ ) , f ( ξ )) ≤ − M ( x ) + d ( x, x ′ ) + M ( x ′ ) ≤ d ( x, x ′ ) . We conclude that for all x, x ′ ∈ U λ ⊂ U , d ( F ( x ) , F ( x ′ )) ≤ λǫδ d ( x, x ′ ) . Combining both cases, we conclude that F is locally -Hölder continuous.The proof of local Lipschitz continuity follows the same line of com-putation. Let x ∈ L X . We find an open neighbourhood U of x and ǫ > , such that for all x ∈ U, w ∈ T F ( x ) Y , there is a ξ ∈ K x such that −h w, −−−−−−→ F ( x ) f ( ξ ) i > ǫ . Additionally, we choose U sufficiently small such thatfor all x ∈ U , the positive Eigenvalues of the Hessian H y B f ( ξ ) ( F ( x )) areat least ǫ ′ > . Let x, x ′ ∈ U and let f ( ξ ) be the point represented by thegeodesic ray obtained by extending the geodesic from F ( x ′ ) to F ( x ) . Wehave the same cases as before.Case 1: If ξ ∈ K x , then d ( F ( x ) , F ( x ′ )) ≤ d ( x, x ′ ) by the same argumentas above.Case 2: If ξ / ∈ K x , then – as before – we find δ > , such that for all ξ ∈ K x , h−−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−→ F ( x ) f ξ ) i ≤ − δ . By construction of U , we find ξ ∈ K x ,such that ǫ < h−−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) i ≤ − δ and k−−−−−−−→ F ( x ) F ( x ′ ) ⊥ f ( ξ ) k ≥ δ . Therefore, we have B ( F ( x ) , F ( x ′ ) , f ( ξ )) = −h−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) i d ( F ( x ) , F ( x ′ ))+ H y B f ( ξ ) ( F ( x )) (cid:16) −−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−−→ F ( x ) F ( x ′ ) (cid:17) d ( F ( x ) , F ( x ′ )) + o ( d ( F ( x ) , F ( x ′ )) ) ≥ ǫd ( F ( x ) , F ( x ′ )) + ǫ ′ δ d ( F ( x ) , F ( x ′ )) + o ( d ( F ( x ) , F ( x ′ )) ) For d ( F ( x ) , F ( x ′ )) sufficiently small, we obtain ǫd ( F ( x ) , F ( x ′ )) ≤ B ( F ( x ) , F ( x ′ ) , f ( ξ )) ≤ − M ( x ) + d ( x, x ′ ) + M ( x ′ ) ≤ d ( x, x ′ ) . Let x ∈ V ⊂ U with V open such that for all x, x ′ ∈ V , d ( F ( x ) , F ( x ′ )) issufficiently small in the sense of the inequality above. We conclude that, forall x, x ′ ∈ V , d ( F ( x ) , F ( x ′ )) ≤ ǫ d ( x, x ′ ) . Therefore, F is locally Lipschitzcontinuous near all x ∈ L X . (cid:3) Corollary 4.4.
The map F : L X → Y is differentiable almost everywhere,i.e. there exists a Lebesgue zero set in L X , such that F is differentiableoutside of this zero set. This is an immediate application of Rademacher’s theorem, exploiting thefact that manifolds are second countable.
Remark 4.5.
It is important to note that it is a-priori not clear, whether D X = ∅ . One of the most crucial obstacles to proving that D X is non-emptyis the lack of injectivity results for the map F . If F was locally injective,some assumptions about F Y being small would carry over to F − ( F Y ) – e.g. F Y having codimension one. If we additionally understood more about thetopology of Y , even more general conditions about F Y being small – e.g. F Y being nowhere dense – would translate into statements about F − ( F Y ) being small.As we will see in the next section, there are results of this type for certainspecial cases, but at the time of writing, little is known about injectivity inthe general case.Summarising the last two sections, we have proven the following theorem. Theorem 4.6.
Let
X, Y be Hadamard manifolds whose sectional curvaturesare bounded from below by − b such that ∂X, ∂Y satisfy (4v) and all pointsin ∂X and ∂Y are in a rank 1 hinge. Let f : ∂X → ∂Y be a Möbius homeo-morphism, such that f and f − send visible pairs to visible pairs. Then, thereexists an equivalence relation ∼ on Y , such that the projection P : Y → Y (cid:30) ∼ restricted to Y (cid:31) F Y is a homeomorphism onto its image and there exists acontinuous map F : X → Y (cid:30) ∼ , which is locally -Hölder continuous on X (cid:31) F − ( F Y ) . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 45 Applications
We now turn to several special cases, in which we can show additionalproperties of the map F . The proofs below are all based either on gettingmore out of the continuity proof in section 4, or on a better understandingof the function M .5.1. Surfaces.
The goal of this section is to prove the following result.
Theorem 5.1.
Let
X, Y be -dimensional Hadamard manifolds whose sec-tional curvature is bounded from below by − b , such that ∂X and ∂Y sat-isfy (4v) and all points in ∂X and ∂Y are in a rank 1 hinge. Suppose, f : ∂X → ∂Y is a Möbius homeomorphism such that f and f − preservevisible pairs. Then the circumcenter extension of f is a homeomorphism F : X → Y . In addition, it is locally Lipschitz continuous on a dense subsetand differentiable almost everywhere.Furthermore, if F and M are differentiable at x and K x contains at leastfive points, then DF x : T x X → T F ( x ) Y is an isometry of tangent spacesequipped with their respective Riemannian metric. In particular, if K x hasat least five points for almost every x , then F is a metric isometry. We start by showing that F is a map between X and Y in this instance.Let x ∈ X . By Lemma 3.16, the set M x is contained in an intersection of atleast three distinct horospheres. Since two horospheres centered at ξ , ξ canonly intersect non-transversally in points that lie on a geodesic line connect-ing ξ with ξ , we conclude that at least two of these horospheres intersecttransversally. Consequently, codim( M x ) ≥ . Since Y is -dimensional, thisimplies that M x has dimension . Since M x is convex, this implies that M x is a single point. We conclude that F : X → Y is well-defined on all of X .Next, we show that F is invertible. Proposition 5.2.
Let F denote the circumcenter extension of f and G thecircumcenter extension of f − . Then G = F − . The proof relies on an elementary result about -dimensional vector spaces.Let V be a -dimensional, real vector space with an inner product. The setof all unit vectors in V with respect to this inner product is homeomorphicto the -dimensional circle. After choosing an orientation on the circle, wecan speak of the order of a set of points on the unit-sphere in V . We havethe following result. Lemma 5.3.
Let V be a -dimensional, real vector space with an innerproduct. Let v , v , v be unit-vectors with respect to this inner product. Then,the following are equivalent:(1) The zero vector is contained in the convex hull of { v , v , v } .(2) After reordering the indices, the vectors {± v , ± v , ± v } are orderedas ( v , − v , v , − v , v , − v ) . Proof of Proposition 5.2.
By Lemma 3.8, we know that F ( x ) is characterisedas the unique point y ∈ Y such that ∈ T y Y is contained in the convex hullof the set {−−−→ yf ( ξ ) | ξ ∈ K x,y } . By Carathéodory’s theorem on convex hullsand since Y is -dimensional, we know that the zero vector can be expressedby a convex combination of at most three vectors of the form −−−→ yf ( ξ ) with ξ ∈ K x,y . Combining this with Corollary 3.15, we find ξ , ξ , ξ ∈ K x suchthat the convex hull of {−−−−−−→ F ( x ) f ( ξ i ) | i ∈ { , , }} contains ∈ T y Y . ByLemma 5.3, this means that, after rearranging the indices, the following sixpoints have the following ordering in ∂Y : ( f ( ξ ) , a F ( x ) ( f ( ξ )) , f ( ξ ) , a F ( x ) ( f ( ξ )) , f ( ξ ) , a F ( x ) ( f ( ξ ))) . Since f is a homeomorphism, we conclude that, after changing the orien-tation of ∂X if necessary, we have the following ordering on ∂X : ( ξ , a x ( ξ ) , ξ , a x ( ξ ) , ξ , a x ( ξ )) . Using Lemma 5.3 again, we conclude that the convex hull of the vectors {−−→ xξ i | i ∈ { , , }} contains ∈ T x X . By Lemma 2.19, Corollary 3.14 andthe Chain rule for metric derivatives, we know that K F ( x ) ,x = a x ( K x,F ( x ) ) ,and therefore, the zero vector in T x X is contained in the convex hull of theset {−→ xξ | ξ ∈ K F ( x ) ,x } . By Lemma 3.8, this implies that G ( F ( x )) = x .We conclude that G ◦ F = Id X . By symmetry, the same argument alsoproves that F ◦ G = Id Y . Therefore, G = F − . (cid:3) Next, we show that F is differentiable almost everywhere. We do this byshowing that the pointwise Lipschitz constant of F is finite for all x ∈ X . Proposition 5.4.
The map F is locally Lipschitz continuous on a densesubset of X . Furthermore, the pointwise Lipschitz constant Lip x ( F ) :=lim sup x ′ → x d ( F ( x ) ,F ( x ′ )) d ( x,x ′ ) is finite for all x ∈ X .Proof. Let x ∈ X . We need to distinguish two cases.Case 1: Suppose, M ( x ) = 0 . Let x ∈ X and ξ ∈ ∂X such that f ( ξ ) isthe endpoint of the geodesic ray obtained by extending the geodesic segmentfrom F ( x ) to F ( x ) . As in the proof of Proposition 4.3, we have d ( F ( x ) , F ( x )) = B ( F ( x ) , F ( x ) , f ( ξ )) ≤ M ( x ) + B ( x , x, ξ )) + M ( x ) ≤ d ( x , x ) . If x lies in the interior of the set { x ∈ X | M ( x ) = 0 } , then we findan open neighbourhood U of x , such that the estimate above becomes d ( F ( x ) , F ( x ′ )) ≤ d ( x, x ′ ) for all x, x ′ ∈ U .Case 2: Suppose, M ( x ) > . The proof has three steps. IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 47
Step 1: We show that there exists ǫ > and an open neighbourhood U of x , such that for all x ∈ U, ξ ∈ K x and ξ ′ ∈ ∂X such that h−−−−−−→ F ( x ) f ( ξ ) , −−−−−−→ F ( x ) f ( ξ ′ ) i ≤− ǫ , we have ξ ′ / ∈ K x .Since M ( x ) = 0 , we know that for all ξ ∈ K x , a x ( ξ ) / ∈ K x . In fact, u x,F ( x ) ( a x ( ξ )) = − M ( x ) < . Suppose, the statement of Step 1 was nottrue. Then we find a sequence x n → x and sequences ξ n , ξ ′ n ∈ K x n such that h−−−−−−−−→ F ( x n ) f ( ξ n ) , −−−−−−−−→ F ( x n ) f ( ξ ′ n ) i < − n . Since ∂X is compact, we can assumewithout loss of generality that ξ n → ξ and ξ ′ n → ξ ′ (choosing subsequences ifnecessary). Since u x,F ( x ) ( ξ ) is Lipschitz continuous in its two index variablesand continuous in ξ , we have u x n ,F ( x n ) ( ξ n ) → u x,F ( x ) ( ξ ) . On the other hand, ξ n ∈ K x n and therefore, u x n ,F ( x n ) ( ξ n ) = M ( x n ) → M ( x ) . We conclude that ξ, ξ ′ ∈ K x . However, h−−−−−−−−→ F ( x n ) f ( ξ n ) , −−−−−−−−→ F ( x n ) f ( ξ ′ n ) i ≤ − n → − , which implies that −−−−−−→ F ( x ) f ( ξ ′ ) = −−−−−−−→ F ( x ) f ( ξ ) . Therefore, ξ ′ = a x ( ξ ) and bothof them are contained in K x by the argument above. This is a contradiction.We thus find U and ǫ > as described in the statement of Step 1.Step 2: Let U be the open neighbourhood from Step 1. We show thatthere exists δ > such that for all x ∈ U, w ∈ T F ( x ) Y there exists ξ ∈ K x such that h w, −−−−−−→ F ( x ) f ( ξ ) i ≥ δ .We first introduce the following notation. Given a vector w ∈ T y Y and α > , we define S α ( w ) := { w ′ ∈ T y Y | ∠ ( w, w ′ ) ≤ α } . This is a sector in T y Y , whose middle line is generated by the vector w . Notethat the angle-width of the sector S α ( w ) is α .Suppose the statement of Step 2 is not true. We find sequences x n ∈ U and w n ∈ T F ( x n ) Y such that for all ξ ∈ K x n , h w n , −−−−−−−→ F ( x n ) f ( ξ ) i < n . Equivalently,the angle between these two vectors satisfies ∠ ( w n , −−−−−−−→ F ( x n ) f ( ξ )) > π − α n with α n → . Define α := π − cos − ( − ǫ ) ∈ (0 , π ) , where ǫ is the numberfound in Step 1. Choose n so that α n < α . We conclude that the sector S π − α ( w n ) = { w ′ ∈ T F ( x n ) Y | ∠ ( w ′ , w n ) ≤ π − α } does not contain any elements of the form −−−−−−−→ F ( x n ) f ( ξ ) with ξ ∈ K x .By Step 1, we know that for all ξ ∈ K x n , the sector S α ( −−−−−−−−→ F ( x n ) f ( ξ )) = n w ′ ∈ T F ( x n ) Y | ∠ ( w ′ , −−−−−−−−→ F ( x n ) f ( ξ )) ≤ α o does not contain any elements of the form −−−−−−−→ F ( x n ) f ( ξ ′ ) with ξ ′ ∈ K x n .Since T F ( x n ) Y is -dimensional, Lemma 3.8 implies that there exists ξ ∈ K x n such that π > ∠ ( w n , −−−−−−−→ F ( x n ) f ( ξ )) > π − α . We conclude that, for this ξ , π < ∠ ( w n , −−−−−−−−→ F ( x n ) f ( ξ )) < π α . This implies that the two sectors S α ( −−−−−−−−→ F ( x n ) f ( ξ )) and S π − α ( w n ) inter-sect. Thus, their union is a sector S β ( u ) . Since −−−−−−−−→ F ( x n ) f ( ξ ) / ∈ S π − α ( w n ) ,the angle-width of this union is strictly greater than π − α ) + α = π . Sinceboth S α ( −−−−−−−−→ F ( x n ) f ( ξ )) and S π − α do not contain any vector of the form −−−−−−−→ F ( x n ) f ( ξ ′ ) with ξ ′ ∈ K x n , we conclude that the set {−−−−−−−→ F ( x n ) f ( ξ ′ ) | ξ ′ ∈ K x n } is contained in the complement of S β ( u ) , which is a sector with angle-widthstrictly less than π . Therefore, the convex hull of {−−−−−−−→ F ( x n ) f ( ξ ′ ) | ξ ′ ∈ K x n } can-not contain the zero-vector of T F ( x n ) Y , which is a contradiction to Lemma3.8. We conclude that for all x ∈ U, w ∈ T F ( x ) Y , we find ξ ∈ K x , suchthat ∠ ( w, −−−−−−→ F ( x ) f ( ξ )) ≤ π − α . Applying cosine to this inequality, we find δ > , such that for all x ∈ U , w ∈ T F ( x ) Y , we find ξ ∈ K x such that h w, −−−−−−→ F ( x ) f ( ξ ) i ≥ δ . This proves Step 2.Step 3: We show Lipschitz-continuity on U . Let x, x ′ ∈ U . By Step 2, wefind ξ ∈ K x such that − cos( ∠ F ( x ) ( F ( x ′ ) , f ( ξ ))) = h−−−−−−−−→ F ( x ) F ( x ′ ) , −−−−−−→ F ( x ) f ( ξ ) i ≥ δ . Since Y is non-positively curved, we know that ∠ (0) F ( x ) ( F ( x ′ ) , f ( ξ )) ≥ ∠ F ( x ) ( F ( x ′ ) , f ( ξ )) and, therefore, − cos( ∠ (0) F ( x ) ( F ( x ′ ) , f ( ξ ))) ≥ − cos( ∠ F ( x ) ( F ( x ′ ) , f ( ξ ))) ≥ δ. We have δd ( F ( x ) , F ( x ′ )) ≤ − cos( ∠ (0) F ( x ) ( F ( x ′ ) , f ( ξ ))) d ( F ( x ) , F ( x ′ ))= B ( F ( x ) , F ( x ′ ) , f ( ξ )) ≤ B ( F ( x ) , π ◦ Φ( −→ xξ ) , f ( ξ )) + B ( x, x ′ , ξ ) + B ( π ◦ Φ( −→ x ′ ξ ) , F ( x ′ ) , f ( ξ )) ≤ − M ( x ) + d ( x, x ′ ) + M ( x ′ ) ≤ d ( x, x ′ ) . We conclude that for all x, x ′ ∈ U , d ( F ( x ) , F ( x ′ )) ≤ δ d ( x, x ′ ) . Therefore, F is Lipschitz continuous on U . IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 49
Combining the two cases, we conclude that F is locally Lipschitz-continuouson X (cid:31) ∂ { x | M ( x ) = 0 } . Since the set { x | M ( x ) = 0 } is closed, the comple-ment of its topological boundary is dense in X . This completes the proof. (cid:3) Using Stepanov’s theorem, we conclude that F is differentiable almosteverywhere. Applying Stepanov’s theorem (or Rademacher’s) to M , we con-clude that M is differentiable almost everywhere as well. We conclude thatfor almost every x , both F and M are differentiable at x . We are left toprove the sufficient condition for isometry from Theorem 5.1. Lemma 5.5.
Let x ∈ X such that F and M are differentiable at x . For all v ∈ T x X and ξ ∈ K x , we have h∇ M x , v i = h v, −→ xξ i − h DF x ( v ) , −−−−−−→ F ( x ) f ( ξ ) i . Proof.
Let v ∈ T x X and let x ′ vary along the geodesic that starts at x andgoes in direction v . For ξ ∈ K x , we compute M ( x ′ ) − M ( x ) ≥ B ( π ◦ Φ( −→ x ′ ξ ) , F ( x ′ ) , f ( ξ )) − B ( π ◦ Φ( −→ xξ ) , F ( x ) , f ( ξ ))= B ( x ′ , x, ξ ) + B ( π ◦ Φ( −→ xξ ) , F ( x ) , f ( ξ )) + B ( F ( x ) , F ( x ′ ) , f ( ξ )) − B ( π ◦ Φ( −→ xξ ) , F ( x ) , f ( ξ ))= h v, −→ xξ i t − h DF x ( v ) , −−−−−−→ F ( x ) f ( ξ ) i t + O ( t ) . We obtain that, for all v ∈ T x X and for all ξ ∈ K x , h∇ M x , v i ≥ h v, −→ xξ i − h DF x ( v ) , −−−−−−→ F ( x ) f ( ξ ) i . Replacing v by − v yields the opposite inequality, which implies equality. (cid:3) Lemma 5.6.
Let x ∈ X such that F and M are differentiable at x . Suppose K x contains at least five points. Then DF x is an isometry between tangentspaces.Proof. By definition of adjoint maps, Lemma 5.5 implies that for every ξ ∈ K x , DF ∗ x ( −−−−−−→ F ( x ) f ( ξ )) = −→ xξ − ∇ M x . Furthermore, since F is invertible by Proposition 5.2, the map DF ∗ x is in-vertible. Therefore, the map DF ∗ x + ∇ M x : T F ( x ) Y → T x X is an invertibleaffine map that sends a subset of the unit circle T F ( x ) Y – namely the set {−−−−−−→ F ( x ) f ( ξ ) | ξ ∈ K x } – to a subset of the unit circle T x X .We are given an invertible, affine map x Ax + b between -dimensionalvector spaces with an inner product. Since affine maps send ellipses to ellipsesand thus circles to ellipses, there are three possibilities what the image ofthe unit circle under this map may look like.(1) The image of the unit circle is an ellipse with non-vanishing eccen-tricity. It can intersect the unit circle in the target space in at mostfour points. (2) The image of the unit circle is a circle, but not the unit circle of thetarget space. It can intersect the unit circle in the target space in atmost two points.(3) The image of the unit circle is equal to the unit circle in the targetspace. Then the affine map is of the form x Ax and A is norm-preserving. Since an inner product can be expressed purely in termsof its induced norm, A is orthogonal.Since every point in K x corresponds to a unit vector in T F ( x ) Y which issent to a unit vector by DF ∗ x + ∇ M x , we see that, if K x contains at leastfive points, the map DF ∗ x + ∇ M x has to be the last of the options above.This implies that M x = 0 and DF ∗ x is orthogonal. Thus, DF x is orthogonal,i.e. an isometry of tangent spaces equipped with the Riemannian metric. (cid:3) If K x contains at least five points for almost every x , then the Lemmaabove implies that for almost every x , F is differentiable and DF has op-erator norm at most . It is a standard result that such a map is -Lipschitz. Since F − equals the circumcenter extension of f − and, there-fore, K F ( x ) = f ( a x ( K x )) , we conclude that F − is -Lipschitz as well. Thisimplies that F is a metric isometry and concludes the proof of Theorem 5.1. Remark 5.7. If X and Y are higher-dimensional and we have a situationwhere we can show that F is differentiable, then Lemma 5.5 implies that K x is contained in the intersection of an ( n − -dimensional ellipsoid withthe ( n − -dimensional unit sphere, or, if DF x is not invertible, in theintersection of a ‘full’ ellipsoid of dimension at most n − with the ( n − -dimensional unit sphere. In either case, this tells us that DF x is an isometryof tangent spaces, whenever K x is not distributed in a rather specific way.This criterion may be worth further investigation. However, differentiabilityof F remains an issue in higher dimensions for now.5.2. Rough isometries for CAT(-1) spaces.
The goal of this section isto prove the following result.
Theorem 5.8.
Let
X, Y be Hadamard manifolds whose sectional curvature isbounded from below by − b and suppose that X, Y are also
CAT( − spaces.Let f : ∂X → ∂Y be a Möbius homeomorphism such that f and f − preservevisible pairs. Then the circumcenter extension of f is a (cid:0) , ln(2) (cid:1) -quasi-isometry.Furthermore, if Y is -dimensional, then the circumcenter extension is a (cid:16) , ln (cid:16) √ (cid:17)(cid:17) -quasi-isometry.Proof. Since Y is CAT( − , it contains no flat strips and M x consists ofexactly one point for every x ∈ X . Therefore, the circumcenter extension isa map from X to Y . By Theorem 1.1 in [Bis17a], we know that F is coarselysurjective. Let x, x ′ ∈ X and ξ ∈ ∂X such that f ( ξ ) is represented by the IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 51 geodesic ray obtained by extending the geodesic from F ( x ′ ) to F ( x ) . Wecompute d ( F ( x ) , F ( x ′ )) = B ( F ( x ) , F ( x ′ ) , f ( ξ ))= B ( F ( x ) , π ◦ Φ( −→ xξ ) , f ( ξ )) + B ( x, x ′ , ξ ) + B ( π ◦ Φ( −→ x ′ ξ ) , F ( x ′ ) , f ( ξ )) ≤ M ( x ) + d ( x, x ′ ) + M ( x ′ ) . Putting ξ ′ ∈ ∂X to be represented by the geodesic ray obtained by ex-tending the geodesic from x ′ to x , we obtain d ( F ( x ) , F ( x ′ )) ≥ B ( F ( x ) , F ( x ′ ) , f ( ξ ′ ))= B ( F ( x ) , π ◦ Φ( −→ xξ ) , f ( ξ ′ )) + B ( x, x ′ , ξ ′ ) + B ( π ◦ Φ( −→ x ′ ξ ) , F ( x ′ ) , f ( ξ ′ )) ≥ − M ( x ) + d ( x, x ′ ) − M ( x ′ ) . We conclude that, if M is bounded on X , then F is a (1 , k M k ∞ ) -quasi-isometry. We are left to prove that M is bounded.Let x ∈ X and ξ, ξ ′ ∈ K x,F ( x ) . Since ∂f ∗ ρ x ∂ρ F ( x ) ( ξ ) = e u x,F ( x ) ( ξ ) and M ( x ) = u x,F ( x ) ( ξ ) = u x,F ( x ) ( ξ ′ ) , we have ρ x ( ξ, ξ ′ ) = e M ( x ) ρ F ( x ) ( f ( ξ ) , f ( ξ ′ )) and therefore, ( f ( ξ ) | f ( ξ ′ )) F ( x ) − M ( x ) = ( ξ | ξ ′ ) x . Since Gromov products are non-negative, this implies that M ( x ) ≤ ( f ( ξ ) | f ( ξ ′ )) F ( x ) for all ξ, ξ ′ ∈ K x,F ( x ) .Let Y be of dimension at least three and let ξ ∈ K x,F ( x ) . By Lemma 3.8,we know that there exists ξ ′ ∈ K x,F ( x ) such that h−−−−−−→ F ( x ) f ( ξ ) , −−−−−−→ F ( x ) f ( ξ ) i ≤ ,i.e. the angle between f ( ξ ) and f ( ξ ′ ) at F ( x ) is at least π . Since Y is CAT( − , we have that ( f ( ξ ) | f ( ξ ′ )) F ( x ) ≤ ( η | η ′ ) z , where η, η ′ ∈ ∂ H such that their representing geodesic rays starting at z ∈ H depart at an angle of π . We are left to compute the Gromov productof two specific geodesics in H .If Y is -dimensional, since K x,F ( x ) contains at least three points by Corol-lary 3.15, we conclude that there are ξ, ξ ′ ∈ K x,F ( x ) such that ∠ F ( x ) (cid:16) −−−−−−→ F ( x ) f ( ξ ) , −−−−−−→ F ( x ) f ( ξ ′ ) (cid:17) ≥ π . We are left to compute the Gromov product of two geodesic rays in H that start at the same point and depart at an angle of π .The Theorem now follows from the following formula, which is a standardcomputation. Lemma 5.9.
Let γ, ˜ γ be geodesic rays in H that start at the same point o and depart at an angle α . Then ( γ | ˜ γ ) o = − ln (cid:16) sin (cid:16) α (cid:17)(cid:17) . Let α = π . Since sin (cid:0) π (cid:1) = √ , we obtain ( γ π | ˜ γ π ) i = ln (cid:16) √ (cid:17) . In the -dimensional case, we put α = π . Since sin (cid:0) π (cid:1) = √ , we obtain ( γ π | ˜ γ π ) i = ln (cid:18) √ (cid:19) ≈ . . This proves the Theorem. (cid:3)
Adding a cocompact action.
In this section, we prove the followingresult.
Theorem 5.10.
Let
X, Y be Hadamard manifolds whose sectional curvatureis bounded from below by − b , such that ∂X and ∂Y satisfy (4v) and allpoints in ∂X and ∂Y are in a rank 1 hinge. Suppose, there is a group G which acts cocompactly by isometries on X and Y . Let f : ∂X → ∂Y be a G -equivariant Möbius homeomorphism such that f and f − preserve visiblepairs. Then, the function M : X → R is bounded and there is a G -equivariant (1 , k M k ∞ ) -quasi-isometry F : X → Y .Proof. Since f is G -equivariant and G acts by isometries, we have for all g ∈ G, x ∈ X, y ∈ Y, ξ ∈ ∂X . ∂f ∗ ρ x ∂ρ y ( f ( ξ )) = ∂f ∗ ρ gx ∂ρ gy ( gf ( ξ )) . This implies that, u gx,gy ( gξ ) = u x,y ( ξ ) , k u gx,gy k ∞ = k u x,y k ∞ and there-fore, M gx = g ( M x ) . In particular, thinking of the circumcenter extensionas a map F : X → Y (cid:30) ∼ , the action of G sends equivalence classes of ∼ toequivalence classes and F is G -equivariant. If M x consists of more than onepoint, we can choose y ∈ M x and use its G -orbit to define ˜ F ( gx ) := gy inorder to get a map ˜ F : X → Y . By abuse of notation, we call this map F as well. Note that this map may not be continuous anymore. Since F is G -equivariant and G acts cocompactly on Y , F is coarsely surjective.The same argument as in the proof of Theorem 5.8 shows that F is a (1 , k M k ∞ ) -quasi-isometry, if M is bounded. Since M ( gx ) = k u gx,F ( gx ) ( · ) k ∞ = k u x,F ( x ) ( g − · ) k ∞ = M ( x ) , we obtain that it is sufficient to bound M on a compact fundamental domainof the G -action on X . Since M is Lipschitz continuous, M is bounded onany compact set. We conclude that M is bounded and F is a G -equivariant (1 , k M k ∞ ) -quasi-isometry. (cid:3) References [Bal95] Werner Ballmann.
Lectures on spaces of nonpositive curvature , volume 25 of
DMV Seminar . Birkhäuser Verlag, Basel, 1995. With an appendix by MishaBrin.
IRCUMCENTER EXTENSION MAPS FOR NON-POSITIVELY CURVED SPACES 53 [BB08] Werner Ballmann and Sergei Buyalo. Periodic rank one geodesics in Hadamardspaces. In
Geometric and probabilistic structures in dynamics , volume 469 of
Contemp. Math. , pages 19–27. Amer. Math. Soc., Providence, RI, 2008.[BCG95] G. Besson, G. Courtois, and S. Gallot. Entropies et rigidités des espaces lo-calement symétriques de courbure strictement négative.
Geom. Funct. Anal. ,5(5):731–799, 1995.[BF09] Mladen Bestvina and Koji Fujiwara. A characterization of higher rank sym-metric spaces via bounded cohomology.
Geom. Funct. Anal. , 19(1):11–40, 2009.[BF18a] Jonas Beyrer and Elia Fioravanti. Cross ratios and cubulations of hyperbolicgroups, 2018.[BF18b] Jonas Beyrer and Elia Fioravanti. Cross ratios on
CAT(0) cube complexes andmarked length-spectrum rigidity, 2018.[BFIM18] Jonas Beyrer, Elia Fioravanti, and Merlin Incerti-Medici.
CAT(0) cube com-plexes are determined by their boundary cross ratio, 2018.[BGS85] Werner Ballmann, Mikhael Gromov, and Viktor Schroeder.
Manifolds of non-positive curvature , volume 61 of
Progress in Mathematics . Birkhäuser Boston,Inc., Boston, MA, 1985.[BH99] Martin R. Bridson and André Haefliger.
Metric spaces of non-positive curva-ture , volume 319 of
Grundlehren der Mathematischen Wissenschaften [Funda-mental Principles of Mathematical Sciences] . Springer-Verlag, Berlin, 1999.[Bis15] Kingshook Biswas. On Moebius and conformal maps between boundaries of
CAT( − spaces. Ann. Inst. Fourier (Grenoble) , 65(3):1387–1422, 2015.[Bis17a] Kingshook Biswas. Circumcenter extension of moebius maps to cat(-1) spaces,2017.[Bis17b] Kingshook Biswas. Hyperbolic p-barycenters, circumcenters and moebiusmaps, 2017.[Bis18] Kingshook Biswas. Moebius rigidity for compact deformations of negativelycurved manifolds, 2018.[Bou95] Marc Bourdon. Structure conforme au bord et flot géodésique d’un
CAT( − -espace. Enseign. Math. (2) , 41(1-2):63–102, 1995.[BS07] Sergei Buyalo and Viktor Schroeder.
Elements of asymptotic geometry . EMSMonographs in Mathematics. European Mathematical Society (EMS), Zürich,2007.[Buy19] Sergei Buyalo. Inverse problem for möbius geometry on the circle, 2019.[CCM19] Ruth Charney, Matthew Cordes, and Devin Murray. Quasi-Mobius homeomor-phisms of Morse boundaries.
Bull. Lond. Math. Soc. , 51(3):501–515, 2019.[CS15] Ruth Charney and Harold Sultan. Contracting boundaries of
CAT(0) spaces.
J. Topol. , 8(1):93–117, 2015.[dC15] Manfredo P. do Carmo.
Geometria Riemanniana . Projeto Euclides. [EuclidProject]. Instituto de Matemática Pura e Aplicada (IMPA), Rio de Janeiro,2015. Fifth edition, second printing of [ MR0651516].[DPS12] Françoise Dal’bo, Marc Peigné, and Andrea Sambusetti. On the horoboundaryand the geometry of rays of negatively curved manifolds.
Pacific J. Math. ,259(1):55–100, 2012.[dSKW17] Jacopo de Simoi, Vadim Kaloshin, and Qiaoling Wei. Dynamical spectral rigid-ity among Z -symmetric strictly convex domains close to a circle. Ann. of Math.(2) , 186(1):277–314, 2017. Appendix B coauthored with H. Hezari.[Ham92] Ursula Hamenstädt. Time-preserving conjugacies of geodesic flows.
ErgodicTheory Dynam. Systems , 12(1):67–74, 1992.[Ham99] U. Hamenstädt. Cocycles, symplectic structures and intersection.
Geom.Funct. Anal. , 9(1):90–140, 1999. [Ham09] Ursula Hamenstädt. Rank-one isometries of proper
CAT(0) -spaces. In
Discretegroups and geometric structures , volume 501 of
Contemp. Math. , pages 43–59.Amer. Math. Soc., Providence, RI, 2009.[HIH77] Ernst Heintze and Hans-Christoph Im Hof. Geometry of horospheres.
J. Dif-ferential Geom. , 12(4):481–491 (1978), 1977.[Kim01] Inkang Kim. Marked length rigidity of rank one symmetric spaces and theirproduct.
Topology , 40(6):1295–1323, 2001.[Kim04] Inkang Kim. Rigidity on symmetric spaces.
Topology , 43(2):393–405, 2004.[KS18] Vadim Kaloshin and Alfonso Sorrentino. On the local Birkhoff conjecture forconvex billiards.
Ann. of Math. (2) , 188(1):315–380, 2018.[Ota90] Jean-Pierre Otal. Le spectre marqué des longueurs des surfaces à courburenégative.
Ann. of Math. (2) , 131(1):151–162, 1990.[Ota92] Jean-Pierre Otal. Sur la géometrie symplectique de l’espace des géodésiquesd’une variété à courbure négative.
Rev. Mat. Iberoamericana , 8(3):441–456,1992.[Thu80] William P. Thurston. Geometry and topology of three-manifolds, 1980.
Institut für Mathematik, Universität Zürich, Switzerland
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