Harmonic quasi-isometric maps III :quotients of Hadamard manifolds
HHarmonic quasi-isometric maps III :quotients of Hadamard manifolds
Yves Benoist & Dominique Hulin
Abstract
In a previous paper, we proved that a quasi-isometric map f : X → Y between two pinched Hadamard manifolds X and Y is withinbounded distance from a unique harmonic map.We extend this result to maps f : Γ \ X → Y , where Γ is a convexcocompact discrete group of isometries of X and f is locally quasi-isometric at infinity. The main result in this paper, which is a continuation of [4], [5] and [6], isthe following.
Theorem 1.1
Let X and Y be pinched Hadamard manifolds and Γ ⊂ Is ( X ) be a torsion-free convex cocompact discrete subgroup of the group of isome-tries of X . Assume that the quotient manifold M = Γ \ X is not compact.Let f : M → Y be a map. Assume that f is quasi-isometric or, moregenerally, that f is locally quasi-isometric at infinity (see Definition 1.4).Then, there exists a unique harmonic map h : M → Y within boundeddistance from f , namely such that d ( f, h ) := sup m ∈ M d ( f ( m ) , h ( m )) < ∞ . Let us begin with a short historical background (see [4, Section 1.2] formore references). In the 60’s, Eells and Sampson prove in [11] that anysmooth map f : M → N between compact Riemannian manifolds, where N is assumed to have non positive curvature, is homotopic to a harmonic map h . This harmonic map h actually minimizes the Dirichlet energy (cid:82) M | Dh | among all maps that are homotopic to f . Primary 53C43 ; Secondary 53C24, 53C35, 58E20, 20H10
Key words
Harmonic map, Harmonic measure, Quasi-isometric map, Boundary map,Hadamard manifold, Negative curvature, Convex cocompact subgroup a r X i v : . [ m a t h . DG ] J u l ater on, P.Li and J.Wang conjecture in [18] that it is possible to re-lax the co-compactness assumption in the Eells-Sampson theorem : namely,they conjecture that any quasi-isometric map f : X → Y between non com-pact rank one symmetric spaces is within bounded distance from a uniqueharmonic map. This extends a former conjecture by R. Schoen in [24].The Schoen-Li-Wang conjecture is proved by Markovic and Lemm-Markovic([21], [20], [17]) when X = Y = H n are a real hyperbolic space, and in ourpapers [4], [6] when X and Y are either rank one symmetric spaces or, moregenerally, pinched Hadamard manifolds. See also the recent paper by Si-dler and Wenger [25] and the survey by Guritaud [14]. Our theorem 1.1generalizes these results by allowing some topology in the source manifold. A first concrete example where our theorem applies is the following, whichis illustrated in Figure (a).
Corollary 1.2
Let Γ ⊂ PSl R be a convex cocompact Fuchsian group.Any quasi-isometric map f : Γ \ H → H is within bounded distance froma unique harmonic map h : Γ \ H → H . (b) (c)(a) Examples of harmonic maps from surfaces to H , with prescribed boundary valuesat infinity, such as discussed in this paper. As we will see later on in Paragraph 2.1, proving Theorem 1.1 amounts tosolving a Dirichlet problem at infinity. Recall that a map h ∞ : S k − → S n − ( k, n ≥
2) is quasi-regular if it is locally the boundary value of some quasi-isometric map from H k to H n . The following concrete example is again aspecial case of our theorem, illustrated in Figure (b). Corollary 1.3
Let h ∞ : S k − → S n − be a quasi-regular map. Then, h ∞ extends as a harmonic map h : H k → H n . When k = n , Corollary 1.3 was proved by Pankka and Souto in [22].Figure (c) illustrates Theorem 1.1 in a situation that combines those ofCorollaries 1.2 and 1.3. 2et us now explain the hypotheses and conclusion of Theorem 1.1. Amap f : X → Y between two metric spaces is quasi-isometric if there existsa constant c ≥ c − d ( x, x (cid:48) ) − c ≤ d ( f ( x ) , f ( x (cid:48) )) ≤ c d ( x, x (cid:48) ) + c (1.1)holds for any x, x (cid:48) ∈ X . Such a map needs not be continuous.A smooth map h : X → Y between Riemannian manifolds is harmonicif it is a critical point for the Dirichlet energy (cid:82) | Dh | , namely if it satisfiesthe elliptic PDE Tr D h = 0 . A pinched Hadamard manifold is a complete simply-connected Rieman-nian manifold X with dimension at least 2 whose sectional curvature ispinched between two negative constants, namely − b ≤ K X ≤ − a < . (1.2)Observe that working with pinched Hadamard manifolds provides a nat-ural and elegant framework to deal simultaneously with all the rank onesymmetric spaces.A discrete subgroup Γ ⊂ Is ( X ) is convex cocompact when the convexcore K ⊂ M is compact, see Definition 2.5. We will see in Section 2 (Propo-sition 2.6) that requiring the discrete subgroup Γ ⊂ Is ( X ) to be convex co-compact is equivalent to assuming that the Riemannian quotient M = Γ \ X is Gromov hyperbolic, and that its injectivity radius inj : M → [0 , ∞ [ is aproper function on M . Therefore we can speak of the boundary at infinity ∂ ∞ M of M . Definition 1.4
A map f : M → Y is locally quasi-isometric at infinity ifit satisfies the two conditions : (a) the map f is rough Lipschitz : there exists a constant c ≥ suchthat d ( f ( m ) , f ( p )) ≤ c d ( m, p ) + c for any p, m ∈ M ; (b) each point ξ ∈ ∂ ∞ M in the boundary at infinity of M admits aneighbourhood U ξ in M ∪ ∂ ∞ M such that the restriction f : U ξ ∩ M → Y isquasi-isometric. Note that condition (a) follows from condition (b) when the map f isbounded on compact subsets of M .In Section 3, we give a few counter-examples that emphasise the relevanceof assuming in Theorem 1.1 that Γ is convex cocompact.Note that the assumption that Γ is torsion-free in Theorem 1.1 is onlyused to ensure that the quotient M = Γ \ X is a manifold. See Kapovich [15]for nice examples of non torsion-free groups.3 .3 Structure of the proof Although the rough structure of the proof of Theorem 1.1 is similar to thoseof [4] or [6], we now have to deal with new issues. Two new crucial stepswill be understanding the geometry of the source manifold M = Γ \ X inProposition 2.6, and obtaining uniform estimates for the harmonic measureson M in Corollary 6.20.Let us now give an overview of the proof of existence in Theorem 1.1.We refer to Section 4 for complete proofs of existence and uniqueness.Replacing the quasi-isometric map f by local averages in Lemma 4.1,we may assume that f is smooth. This averaging process even allows us toassume that f has uniformly bounded first and second covariant derivatives,a fact that will be crucial later on to prove the so-called boundary estimatesof Paragraph 4.4.To prove existence, we begin by solving a family of Dirichlet problemson bounded domains of M . Namely, we introduce in Proposition 4.2 anexhausting and increasing family of compact convex domains V R ⊂ M withsmooth boundaries ( R >
R > h R : V R → Y which is solution of the Dirichlet problem “ h R = f on the boundary ∂V R ” (Lemma 4.7).The heart of the proof, in Proposition 4.8, consists in showing that thedistances d ( h R , f ) are uniformly bounded by a constant ¯ ρ that does not de-pend on R . Once we have this bound, we recall in Paragraph 4.3 a standardcompactness argument that ensures that the family of harmonic maps ( h R )converges, when R goes to infinity, to a harmonic map h : M → Y . Thelimit harmonic map h still satisfies d ( h, f ) ≤ ¯ ρ .Let us now briefly explain how we obtain this uniform bound for thedistances ρ R := d ( h R , f ). We assume by contradiction that ρ R is very large.The first step consists in proving that the distance ρ R is achieved at apoint m ∈ M which is far away from the boundary ∂V R . This relies onthe uniform bound for the covariant derivatives Df and D f , and on theequality d ( h R ( p ) , f ( p )) = 0 when p lies in the boundary ∂V R .The second step consists in the so-called interior estimates of Section 7.Since the point m ∈ V R such that ρ R = d ( h R ( m ) , f ( m )) is far away fromthe boundary ∂V R , we may select a large neighbourhood W ( m ) ⊂ M of thepoint m , such that W ( m ) ⊂ V R .Let us describe the neighbourhoods W ( m ). We fix two large constants1 (cid:28) (cid:96) (cid:28) (cid:96) that will not depend on R and need to be properly chosen. If m is far enough from the convex core K ⊂ M , the injectivity radius at the point m will be larger than (cid:96) and we will choose W ( m ) to be the Riemannianball W ( m ) = B ( m, (cid:96) ) ⊂ V R with center m and radius (cid:96) . If m is close tothe convex core, we will chose W ( m ) = V (cid:96) ⊂ V R to be a fixed large compactneighbourhood of the convex core K .4o wrap up the proof in Section 7, we will exploit the subharmonicityof the function q ∈ W ( m ) → d ( f ( m ) , h R ( q )) ∈ [0 , ∞ [. The crucial tool forobtaining a contradiction, and thus proving that ρ R cannot be very large,is uniform estimates for the harmonic measures of the boundaries of theseneighbourhoods W ( m ) relative to the point m that are proved in Sections5 and 6. See Proposition 5.2 and Corollary 6.20. We refer to Paragraph 7.3for a more precise overview of this part the proof. Is ( X ) In this section we characterize convex cocompact subgroups Γ ⊂ Is ( X ) in terms of the geometric properties of the quotient Γ \ X . We first recall a few facts and definitions concerning Gromov hyperbolicmetric spaces, see [12].Let δ >
0. A geodesic metric space X is said to be δ -Gromov hyperbolicwhen every geodesic triangle ∆ in X is δ -thin, namely when each edge of∆ lies in the δ -neighbourhood of the union of the other two edges. Forexample, any pinched Hadamard manifold is Gromov hyperbolic for someconstant δ X that depends only on the upper bound for the curvature.When the Gromov hyperbolic space X is proper, namely its balls arecompact, the boundary at infinity ∂ ∞ X of X may be defined as the set ofequivalence classes of geodesic rays, where two geodesic rays are identifiedwhen they remain within bounded distance from each other. In case X is apinched Hadamard manifold, the boundary at infinity (or visual boundary) ∂ ∞ X naturally identifies with the tangent sphere at any point x ∈ X . Theboundary at infinity provides a compactification X = X ∪ ∂ ∞ X of X .A quasi-isometric map f : X → Y between proper Gromov hyperbolicspaces admits a boundary value ∂ ∞ f : ∂ ∞ X → ∂ ∞ Y , and two quasi-isometric maps f , f : X → Y have the same boundary value if and only if d ( f , f ) < ∞ .Since our quotient M = Γ \ X is Gromov hyperbolic (see Proposition 2.6),we may thus rephrase the conclusion of Theorem 1.1 in case f is assumedto be globally quasi-isometric. Corollary 2.1
Given a quasi-isometric map f : M → Y , there exists aunique harmonic quasi-isometric map h : M → Y which is a solution to theDirichlet problem at infinity with boundary value ∂ ∞ h = ∂ ∞ f : ∂ ∞ M → ∂ ∞ Y . .2 Gromov product In Section 7, we shall use Gromov products in the Gromov hyperbolic man-ifolds M and Y to carry out the proof of Theorem 1.1. Here, we recall thedefinition and two properties of the Gromov product.In a metric space, the Gromov product of the three points x, x , x isdefined as ( x , x ) x = 12 ( d ( x, x ) + d ( x, x ) − d ( x , x )) . Gromov hyperbolicity may be expressed in terms of Gromov products. Also,in a Gromov hyperbolic space, the Gromov product ( x , x ) x is roughlyequal to the distance from x to a minimizing geodesic segment [ x , x ]. Inparticular, the following holds. Lemma 2.2 [12, Chap.2]
Let X be a δ -Gromov hyperbolic space. Then, forany points x, x , x , x ∈ X and any minimizing geodesic segment [ x , x ] ⊂ X , one has ( x , x ) x ≥ min(( x , x ) x , ( x , x ) x ) − δ (2.1)( x , x ) x ≤ d ( x, [ x , x ]) ≤ ( x , x ) x + 2 δ . (2.2)Moreover, the next lemma tells us that Gromov products are quasi-invariantunder quasi-isometric maps. Lemma 2.3 [12, Prop.5.15]
Let X and Y be Gromov hyperbolic spaces, and f : X → Y be a c quasi-isometric map. Then, there exists a constant A > such that, for any three points x, x , x ∈ X : c − ( x , x ) x − A ≤ ( f ( x ) , f ( x )) f ( x ) ≤ c ( x , x ) x + A .
We begin with definitions that are classical in the hyperbolic space H k . SeeBowditch [8] for a reference dealing with pinched Hadamard manifolds. Definition 2.4
Let X be a pinched Hadamard manifold and Γ ⊂ Is ( X ) bea discrete subgroup of the group of isometries of X .The limit set Λ Γ ⊂ ∂ ∞ X of the group Γ is the closed subset of ∂ ∞ X defined as Λ Γ = Γ x (cid:114) Γ x , where x is any point in X and the closure Γ x ofits orbit is taken in the compactification X = X ∪ ∂ ∞ X .The domain of discontinuity of Γ is Ω Γ = ∂ ∞ X (cid:114) Λ Γ . It is an open subsetof the boundary at infinity ∂ ∞ X . The group Γ acts properly discontinuouslyon X ∪ Ω Γ ⊂ X . Recall that a subset C ⊂ M of a Riemannian manifold is geodesically convexif, given two points m , m ∈ C , any minimizing geodesic segment [ m , m ]joining these two points lies in C . 6 efinition 2.5 Let X be a pinched Hadamard manifold, Γ ⊂ Is ( X ) be aninfinite discrete subgroup of the isometry group of X and M = Γ \ X .The convex hull convΛ Γ ⊂ X of Λ Γ is the smallest closed geodesicallyconvex subset C of X such that C (cid:114) C = Λ Γ (where the closure C is againtaken in the compactification X ). The convex core of M is the quotient K := Γ \ (convΛ Γ ) ⊂ M .The group Γ is said to be convex cocompact if the convex core K ⊂ M iscompact. This is equivalent to requiring the quotient Γ \ ( X ∪ Ω Γ ) ⊂ Γ \ X tobe compact. See [8, Th.6.1]. The hypothesis in Theorem 1.1 that Γ is convex cocompact will be usedin this paper through the following characterization in terms of geometricproperties of the Riemannian quotient M . Proposition 2.6
Let X be a pinched Hadamard manifold and Γ ⊂ Is ( X ) be an infinite torsion-free discrete subgroup. Then, the group Γ is convexcocompact if and only if M = Γ \ X satisfies the following conditions : • the quotient M is Gromov hyperbolic ; • the injectivity radius inj : M → [0 , ∞ [ is a proper function. Proposition 2.6 is proved in Paragraphs 2.4, 2.5 and 2.6 below. M = Γ \ X In this paragraph, X is our pinched Hadamard manifold, Γ ⊂ Is ( X ) is atorsion-free convex cocompact subgroup and M = Γ \ X . We want to provethat there exists a constant δ > T ⊂ M in thequotient is δ -thin. To do this, we start with quadrilaterals. We first recall aresult by Reshetnyak that compares quadrilaterals in a Hadamard manifoldwith quadrilaterals in the model hyperbolic plane H ( − a ) with constantcurvature − a . Lemma 2.7
Reshetnyak comparison lemma [23]
Let X be a Hadamard manifold satisfying the pinching assumption (1.2) .For every quadrilateral [ x, y, z, t ] in X , there exists a convex quadrilateral [¯ x, ¯ y, ¯ z, ¯ t ] in H ( − a ) and a map j : [¯ x, ¯ y, ¯ z, ¯ t ] → X that is -Lipschitz, thatsends respectively the vertices ¯ x, ¯ y, ¯ z, ¯ t on x, y, z, t , and whose restriction toeach one of the four edges of [¯ x, ¯ y, ¯ z, ¯ t ] is isometric. We now deduce from the Reshetnyak Lemma a standard property of quadri-laterals in X , that will be used again in the proof of Proposition 6.16. TheGromov hyperbolicity of X ensures that any edge of a quadrilateral in X lies in the 2 δ X -neighbourhood of the union of the three other edges. Onecan say more under an angle condition. Namely :7 emma 2.8 Thin quadrilaterals in X Let X be a pinched Hadamard manifold. Let ε > . There exists an angle < α ε < π/ and a distance ˜ δ ε such that for any quadrilateral [ x, x , y , y ] in the Hadamard manifold X with d ( x , y ) ≥ ε , and whose angles at bothvertices x and y satisfy ∠ x ≥ π/ − α ε and ∠ y ≥ π/ − α ε , the ˜ δ ε -neighbourhood of the edge [ x, y ] contains the union of the other three edges,namely : [ x, x ] ∪ [ x , y ] ∪ [ y , y ] ⊂ V ˜ δ ε ([ x, y ]) . Proof
When X is the hyperbolic plane, the easy proof is left to the reader.The general case follows from this special case and the Reshetnyak com-parison lemma 2.7. Indeed, the comparison quadrilateral also satisfies thedistance and angles conditions at the points ¯ x and ¯ y . (cid:3) We now turn to quadrilaterals in our quotient space M = Γ \ X . In thewhole paper, unless otherwise specified, all triangles and quadrilaterals in M will be assumed to be minimizing, namely their edges will be minimizinggeodesic segments. Lemma 2.9
Thin quadrilaterals in M Assume that X is a pinched Hadamard manifold, that Γ ⊂ Is ( X ) is atorsion-free convex cocompact subgroup, and let M = Γ \ X . Let V ⊂ M be a compact convex subset of M whose lift ˜ V ⊂ X is convex.(1) There exists δ such that, for any quadrilateral [ p, p , q , q ] in M withboth p , q ∈ V , any edge of this quadrilateral lies in the δ -neighbourhood ofthe union of the three other edges.(2) Let ε > . There exists δ ε such that if we assume moreover that p , q ∈ V are respectively the projections of the points p and q on the convexset V , and that d ( p , q ) ≥ ε , then : [ p, p ] ∪ [ p , q ] ∪ [ q , q ] ⊂ V δ ε ([ p, q ]) . Proof
Lift successively the minimizing geodesic segments ( p p ), ( pq ) and( qq ) to geodesic segments ( x x ), ( xy ) and ( yy ) in X , so that the geodesicsegment ( x y ) projects to a curve c from p to q that lies in V .Since both the curve c and the geodesic segment lie in the compact set V ,the conclusion follows from Lemma 2.8, with δ = 2 δ X + d V and δ ε = ˜ δ ε + d V ,where d V denotes the diameter of V . (cid:3) Corollary 2.10
Gromov hyperbolicity of the quotient Γ \ X Assume that X is a pinched Hadamard manifold and that Γ ⊂ Is ( X ) is atorsion-free convex cocompact subgroup. Then, the quotient M = Γ \ X isGromov hyperbolic. roof Let V ⊂ M be a compact convex neighbourhood of the convex corewith smooth boundary, whose lift in X is convex. Such a neighbourhoodwill be constructed in Proposition 4.2. Let ε = inj( V ) / V ) =inf m ∈ V inj( m ) denotes the injectivity radius on V ).Let T = [ p, q, r ] be a triangle in M . In case where at least one of thevertices of T lies in V , Lemma 2.9 applied to a quadrilateral with two equalvertices proves that T is δ -thin.We now turn to the case where none of p, q, r lie in V and denote by p , q , r their projections on V .Assume first that d ( p , q ) ≤ ε and d ( q , r ) ≤ ε . Then, the triangle[ p , q , r ] is homotopically trivial. Writing p = exp p u with u a normalvector to V at point p (and similarly for q and r ) we construct an homotopybetween [ p, q, r ] and a constant map, so that the triangle [ p, q, r ] lifts to atriangle [ x, y, z ] in X . Since X is δ X -Gromov hyperbolic, [ x, y, z ] is δ X -thin,hence [ p, q, r ] is δ X -thin too.Assume now that d ( p , q ) ≥ ε and d ( q , r ) ≥ ε . The first part of Lemma2.9, applied to the quadrilateral [ p, p , r , r ], yields that [ p, r ] lies in the δ -neighbourhood of [ p, p ] ∪ [ p , r ] ∪ [ r , r ], so that [ p, r ] lies in the ( δ + d V )-neighbourhood of [ p, p ] ∪ [ r , r ] (recall that d V is the diameter of V ). Thesecond part of Lemma 2.9 now applied to both quadrilaterals [ p, p , q , q ]and [ q, q , r , r ] yields that [ p, r ] lies in the ( δ + δ ε + d V )-neighbourhood of[ p, q ] ∪ [ q, r ].Assume finally that d ( p , q ) ≤ ε and d ( q , r ) ≥ ε . It follows fromthe first part of Lemma 2.9 that [ p, r ] lies in the ( δ + d V )-neighbourhoodof [ p, p ] ∪ [ r , r ] and that [ p, p ] lies in the ( δ + d V )-neighbourhood of[ p, q ] ∪ [ q , q ], while the second part of this Lemma ensures that [ q, q ] ∪ [ r , r ]lies in the δ ε -neighbourhood of [ q, r ].Hence, every triangle in M is (2( δ + d V ) + δ ε + δ X )-thin, so that M isGromov hyperbolic. (cid:3) M = Γ \ X Our aim in this paragraph is the following.
Proposition 2.11
Properness of the injectivity radius
Assume that X is a pinched Hadamard manifold and that Γ ⊂ Is ( X ) is atorsion-free convex cocompact subgroup. Let M = Γ \ X .Then, the injectivity radius inj : M → [0 , ∞ [ is a proper function. Proof
For p ∈ M , we denote by i ( p ) the injectivity radius at the point p ,and let j ( p ) = inf { d (˜ p, γ ˜ p ) | γ ∈ Γ ∗ } , where ˜ p ∈ X is a lift of p and Γ ∗ is theset of non trivial elements in Γ. Since M has non positive curvature, thereare no conjugate points in M hence j ( p ) = 2 i ( p ).9e proceed by contradiction and assume that there exists a sequence( p n ) of points in M going to infinity, and such that the injectivity radii i ( p n ) remain bounded. Let q n denote the projection of the point p n on theconvex core K ⊂ M ( n ∈ N ). Since Γ is convex cocompact, there exists acompact L ⊂ X such that each q n lifts in X to a point ˜ q n ∈ L . Then, thegeodesic segment [ q n , p n ] lifts as [˜ q n , ˜ p n ]. By hypothesis, there exists I > γ n ∈ Γ ∗ such that d (˜ p n , γ n ˜ p n ) ≤ I . Since the projection X → conv(Λ Γ ) commutes to the action of Γ and is 1-Lipschitz, it followsthat d (˜ q n , γ n ˜ q n ) ≤ I . By compactness of L , and since Γ is discrete, onemay thus assume that the sequence (˜ q n ) converges to a point ˜ q ∈ L andthat the bounded sequence ( γ n ) is constant, equal to γ ∈ Γ ∗ . The boundaryat infinity ∂ ∞ X being compact, we may also assume that the sequence ofgeodesic rays ([˜ q n , ˜ p n [) converges to a geodesic ray [˜ q, ξ [ where ξ ∈ ∂ ∞ X .By construction, the geodesic ray [˜ q, ξ [ is within bounded distance I fromits image [ γ ˜ q, γξ [, hence ξ ∈ ∂ ∞ X is a fixed point of γ . The group Γ beingdiscrete and torsion-free, it has no elliptic elements, so that ξ ∈ Λ Γ . Thus,the whole geodesic ray [˜ q, ξ [ lies in conv(Λ Γ ), a contradiction to the fact thatthe sequence ( p n ) goes to infinity in M . (cid:3) In this paragraph, we complete the proof of Proposition 2.6 by proving thefollowing.
Proposition 2.12
Let X be a pinched Hadamard manifold, and Γ ⊂ Is ( X ) be a torsion-free discrete subgroup. Assume that the quotient manifold M =Γ \ X is Gromov hyperbolic, and that the injectivity radius inj : M → [0 , ∞ [ is a proper function.Then, the group Γ is convex cocompact. We want to prove that the convex core K := Γ \ (convΛ Γ ) is a compact subsetof M , where convΛ Γ denotes the convex hull of the limit set of Γ. We willrather work with the join of the radial limit set Λ r Γ . Definition 2.13
A geodesic ray c : [0 , ∞ [ → M is said to be recurrent whenit is not a proper map, that is if there exists a sequence t n → + ∞ and acompact set L c ⊂ M (that might depend on c ) with c ( t n ) ∈ L c .The radial limit set Λ r Γ ⊂ Λ Γ of Γ is the set of endpoints ξ ∈ ∂ ∞ X ofgeodesic rays in X that project to a recurrent geodesic ray in M . Since there exist closed geodesics in the negatively curved manifold M , theradial limit set Λ r Γ is not empty, and it is a Γ-invariant subset of ∂ ∞ X .Hence, the limit set Λ Γ being a minimal Γ-invariant closed subset of ∂ ∞ X ,it follows that Λ r Γ is dense in Λ Γ . 10 efinition 2.14 The join of a closed subset Q ⊂ ∂ ∞ X is defined as join Q = (cid:91) ξ ,ξ ∈ Q ] ξ , ξ [ ⊂ X where ] ξ , ξ [ denotes the geodesic line with endpoints ξ and ξ . The join is thus the first step towards the construction of the convex hull.One has join Q ⊂ conv Q . The following result by Bowditch [8] investigateshow much we miss in the convex hull by considering only the join. Proposition 2.15 [8, Lemma 2.2.1 and Proposition 2.5.4]
Let X be a pinched Hadamard manifold. There exists a real number λ > that depends only on the pinching constants such that, for any closed subset Q ⊂ ∂ ∞ X , the convex hull of Q lies in the λ -neighbourhood of the join of Q . Let us now proceed with our proof. The projection K rj := Γ \ (join Λ r Γ ) ⊂ M of the join of the radial limit set is the union of all the recurrent geodesicson M , namely of all geodesics that are recurrent both in the future and inthe past. We begin with the following. Proposition 2.16
Under the assumptions of Proposition 2.12, the join ofthe radial limit set of Γ projects in M to a bounded set K rj . Proof
Proposition 2.16 is an immediate consequence of the definition ofthe join and of Lemma 2.18 below. (cid:3)
For r >
0, define M r = { m ∈ M | inj( m ) ≤ r } . Under our hypotheses, M r is a compact subset of M . Let δ > M is δ -Gromov hyperbolic. Lemma 2.17
Any geodesic segment in M whose image lies outside the com-pact subset M δ ⊂ M is minimizing. Proof
We proceed by contradiction and assume that the geodesic segment c : [0 , l ] → M (cid:114) M δ is minimizing, but ceases to be minimizing on any largerinterval. Since M has negative curvature, there are no conjugate points, sothat there exists another minimizing geodesic ¯ c : [0 , l ] → M with the sameendpoints c and c l as c .The manifold M being δ -hyperbolic, the geodesic segments c and ¯ c arewithin distance δ from each other. We may thus choose a subdivision( x p ) ≤ p ≤ P of c ([0 , l ]) with x = c , x P = c l , and such that d ( x p , x p +1 ) ≤ δ when 0 ≤ p < P , and another sequence ( y p ) ≤ p ≤ P with y p ∈ ¯ c ([0 , l ]) and d ( x p , y p ) ≤ δ when 0 ≤ p ≤ P . In particular, d ( y p , y p +1 ) ≤ δ , so that thelength of any of the quadrilaterals q p = [ x p , x p +1 , y p +1 , y p ] is at most 6 δ .Since the injectivity radius at the point x p is larger than 3 δ , it follows thateach quadrilateral q p is homotopically trivial, so that c ([0 , l ]) and ¯ c ([0 , l ])themselves are homotopic. But X being a Hadamard manifold then yields c = ¯ c , which is a contradiction. (cid:3) emma 2.18 Let d δ denote the diameter of M δ . Under the assumption ofProposition 2.12, any recurrent geodesic in M lies in a fixed compact subsetof M . More precisely, such a geodesic lies in the d δ -neighbourhood of M δ . Proof
Let c : R → M be a geodesic that lifts to a geodesic ˜ c : R → X withboth endpoints in Λ r Γ .We first claim that both geodesic rays c | [0 , ∞ [ and c | ] −∞ , must keepvisiting M δ . If this were not the case, Lemma 2.17 would ensure that oneof these geodesic rays would be eventually minimizing, thus contradictingthe assumptions that both endpoints of c lie in the radial limit set.Now assume by contradiction that the image of c does not lie in the d δ -neighbourhood of M δ . Then we can find an interval [ a, b ] such that c ( a ) , c ( b ) ∈ M δ but with c ( t ) / ∈ M δ for every t ∈ ] a, b [, and such that c ([ a, b ]) exits the d δ -neighbourhood of M δ . Hence this geodesic segment c ([ a, b ]) has length at least 2 d δ . By Lemma 2.17 it would be minimizing, acontradiction to the fact that d ( c ( a ) , c ( b )) ≤ d δ . (cid:3) Proof of Proposition 2.12
We noticed earlier that the radial limit setΛ r Γ ⊂ Λ Γ is dense in the limit set of Γ. Thus, the join of the full limit setlies in the closure of the join of the radial limit set. Hence Proposition 2.16ensures that join Λ Γ projects to a bounded subset of M .Now Proposition 2.15 ensures that convΛ Γ also projects to a boundedsubset of M or, in other words, that the group Γ is convex cocompact. (cid:3) Before going into the proof of Theorem 1.1, we proceed with afew examples and counter-examples.
We begin with a family of straightforward applications of Theorem 1.1.Let A ( r ) = { z ∈ C | /r < | z | < r } and D = { z ∈ C | | z | < } be anannulus ( r >
1) or the disk equipped with their complete hyperbolic metrics.
Example 3.1
For every α ∈ R , there exists a unique harmonic quasi-isometric map h α : A ( r ) → D with boundary value g α : ∂ ∞ ( A ( r )) → ∂ ∞ D defined as g α ( re iθ ) = e iθ g α ( e iθ /r ) = e i ( θ + α ) . Note that the case where the parameter α is equal to 0 is rather easy. Indeed,the uniqueness of the harmonic map with boundary value g ensures that h has a symmetry of revolution and thus reads as h ( te iθ ) = u ( t ) e iθ , and thecondition that h is harmonic reduces to a second order ordinary differentialequation on u whose solutions can be expressed in terms of elliptic integrals.12e now give an illustration of Theorem 1.1 where the restriction of theboundary map to each connected component of the boundary at infinity ∂ ∞ M is not injective. Example 3.2
The map g : ∂ ∞ ( A ( r )) → ∂ ∞ D defined as g ( re iθ ) = e iθ g α ( e iθ /r ) = e − iθ is the boundary value of a harmonic map h : A ( r ) → D . In the next two paragraphs, we provide counter-examples to emphasize theimportance of assuming that Γ is convex cocompact.We first prove non-existence for hyperbolic surfaces with a cusp.
Proposition 3.3
Let S be a non compact hyperbolic surface of finite topo-logical type admitting at least one cusp, and Y be a pinched Hadamard man-ifold. Then, there exists no harmonic quasi-isometric map h : S → Y . Note that, such a surface S being quasi-isometric to the wedge of a finitenumber of hyperbolic disks and rays, there always exist quasi-isometric maps f : S → H n for any n ≥ Lemma 3.4
Let τ > and Σ be the quotient of { z ∈ C | Im z ≥ } , equippedwith the hyperbolic metric | dz | (Im z ) under the map z → z + τ .Let Y be a pinched Hadamard manifold. Then, there is no harmonicquasi-isometric map h : Σ → Y . To prove lemma 3.4, we will use the following result, which will also becrucial for the proof of uniqueness in Theorem 1.1.
Lemma 3.5 [6, Lemma 5.16]
Let M , Y be Riemannian manifolds, andassume that Y has non positive curvature. Let h , h : M → Y be harmonicmaps. Then, the distance function m ∈ M → d ( h ( m ) , h ( m )) ∈ R issubharmonic. Proof of Lemma 3.4
Assuming that there exists a harmonic quasi-isometricmap h : Σ → Y , we will prove that h takes its values in a bounded domainof Y . This is a contradiction since Σ is unbounded.Let H t = { Im z = e t } /< τ > ⊂ Σ denote the horocyle at distance t ≥ H . Pick a point z ∈ H and let m = h ( z ) ∈ Y . Since h is harmonic,Lemma 3.5 ensures that the function ϕ : z ∈ Σ → d ( h ( z ) , m ) ∈ [0 , ∞ [ issubharmonic. Since h is quasi-isometric, there exists a constant k > ϕ ( z ) ≤ k ( t + 1) for any z ∈ H t , with t ≥ T > η T ( z ) = k + k ( T + 1) e − T Im z . By construction, ϕ ≤ η T on H ∪ H T , hence the maximum principle ensuresthat ϕ ( z ) ≤ η T ( z ) holds for any z ∈ H t with 0 ≤ t ≤ T . In other words, d ( h ( z ) , m ) ≤ k + k ( T + 1) e − T Im z if T ≥ Im z . Letting T → ∞ provesthat h takes its values in the ball B ( m , k ), a contradiction since h is quasi-isometric. (cid:3) We now give another counter-example where the surface has no cusp, thatis when Γ has no parabolic element.Let Σ now denote a compact hyperbolic surface whose boundary is theunion of two totally geodesic curves of the same length. We consider a noncompact hyperbolic surface S obtained by gluing together infinitely manycopies of Σ along their boundaries, in such a way that S admits a naturalaction τ of Z “by translation”. The quotient S/τ is then a compact Riemannsurface without boundary. Observe that S , being quasi-isometric to R , isGromov hyperbolic and that the injectivity radius of S is bounded belowbut is not a proper function on S .Our aim is to prove the following. Proposition 3.6
There exist quasi-isometric functions ϕ : S → R that arenot within bounded distance from any harmonic function.Thus, there exist quasi-isometric maps f : S → H that are not withinbounded distance from any harmonic map. 𝜏 Sc -1 1 c c c Σ The surface S We will first describe all quasi-isometric harmonic functions on S . By ap-plying the afore mentionned theorem by Eeels and Sampson [11] to maps S/τ → R / Z between these compact manifolds, we construct a harmonicfunction η : S → R that satisfies the relation η ( τ m ) = η ( m ) + 1 for anypoint m ∈ S . We may think of the function η as a projection from S onto R . It is a quasi-isometric map. Lemma 3.7
Let η : S → R be a quasi-isometric harmonic function. Then,there exist constants α ∈ R ∗ and β ∈ R such that η = αη + β . roof We denote by ( c n ) n ∈ Z the memories in S of the boundary of Σ, with c n = τ n c ( n ∈ Z ). All these curves have the same finite length.By adding a suitable constant to η , we may assume that η vanishes atsome point m ∈ c . Let m n = τ n ( m ) ∈ c n so that η ( m n ) = n for n ∈ Z .Let t n = η ( m n ) ( n ∈ Z ). Replacing the quasi-isometric function η by − η if necessary, we may assume that t ± n → ±∞ when n → + ∞ . Better, thereexist a constant k > N ∈ N such that | η − n | ≤ k and | η − t n | ≤ k on c n for n ∈ Z ,where n/k ≤ εt εn ≤ kn for ε = ± n ≥ N .For any n ≥
1, let ( α n , β n ) ∈ R be the solution of the linear system − nα n + β n = t − n and nα n + β n = t n . The sequence ( α n ) is bounded since | α n | ≤ k when n ≥ N . The harmonicfunction η − ( α n η + β n ) vanishes at both points m n ∈ c n and m − n ∈ c − n ,so that | η − ( α n η + β n ) | ≤ k ( k + 1) on c n ∪ c − n if n ≥ N . By the maximumprinciple, it follows that | η − ( α n η + β n ) | ≤ k ( k +1) on the compact subset of S cut out by c n ∪ c − n . By applying this estimate at the point m , we obtainthat the sequence ( β n ) is also bounded. By going to a subsequence, we maythus assume that α n → α and β n → β when n → ∞ , so that the limitharmonic function η − ( αη + β ) : S → R is bounded. Since S is a nilpotentcover of a compact Riemannian manifold, a theorem by Lyons-Sullivan [19,Th.1] ensures that this bounded harmonic function is constant. (cid:3) Proof of Proposition 3.6
Let ϕ : S → R be a quasi-isometric functionsuch that– ϕ ( m n ) = n when n = ± p – ϕ ( m n ) = 2 n when n = ± p +1 .This function ϕ is quasi-isometric, but is not within bounded distance fromany function αη + β .We now embed isometrically the real line in the hyperbolic plane as ageodesic γ : R → H and define f = γ ◦ ϕ : S → H . Then f is a quasi-isometric map. Assume by contradiction that there exists a harmonic map h : S → H within bounded distance from f . Let h = σ ◦ h : S → H ,where σ : H → H is the symmetry with respect to geodesic γ . By Lemma3.5, the bounded function t ∈ S → d ( h ( t ) , h ( t )) ∈ [0 , ∞ [ is subharmonic.Since S is a Z -cover of a compact Riemannian manifold, another theorem byLyons-Sullivan [19, Th.4] ensures that this bounded subharmonic function isconstant. Therefore, the harmonic map h takes its values in a curve whichis equidistant to γ , hence in γ . This means that h reads as h = γ ◦ η with η : S → R harmonic within bounded distance from ϕ , a contradiction. (cid:3) Proof of Theorem 1.1
In this section, we prove Theorem 1.1, taking Proposition 4.8below for granted.Recall that both X and Y are pinched Hadamard manifolds, that Γ ⊂ Is ( X ) is a torsion-free convex cocompact discrete subgroup of the group ofisometries of X which is not cocompact, and that we let M = Γ \ X .Let f : M → Y be a quasi-isometric map or, more generally, a map thatsatisfies the hypotheses in Theorem 1.1. We want to prove that there existsa unique harmonic map h : M → Y within bounded distance from f . f We first observe that we can assume that the initial map f : M → Y issmooth, with bounded covariant derivatives. Lemma 4.1
Let f : M → Y be a rough Lipschitz map, namely that satisfies d ( f ( m ) , f ( p )) ≤ c d ( m, p ) + c for some constant c , and any pair of points m, p ∈ M . Then, there existsa C ∞ map F : M → Y with uniformly bounded first and second covariantderivatives, and such that d ( f, F ) < ∞ . Proof
Lift f : M → Y to ˜ f : X → Y . Since ˜ f is still rough-Lipschitz,the construction in [4, Section 3.2] provides a smooth map ˜ F : X → Y withbounded first and second covariant derivatives, and within bounded distancefrom ˜ f . This construction being Γ-invariant, the map ˜ F goes to the quotientand yields the smooth map F : M → Y we were looking for. (cid:3) Our goal in this paragraph is to construct the family ( V R ) of compact convexneighbourhoods with smooth boundaries of the convex core, on which wesolve in Lemma 4.7 the bounded Dirichlet problems with boundary value f . Proposition 4.2
There exists a compact convex set V ⊂ M with smoothboundary which is a neighbourhood of the convex core K ⊂ M . For any R > , the R -neighbourhood V R of V is also a convex subset of M withsmooth boundary. The proof will rely on Proposition 4.4, which is due to Greene-Wu.
Definition 4.3
Strictly convex functions
Let ϕ : M → R be a continuous function defined on a Riemannian manifold M . We say that the function ϕ is strictly convex if, for every compact subset ⊂ M , there exists a constant α > such that, for any unit speed geodesic t → c t ∈ L , the function t → ϕ ( c t ) − α t is convex.When ϕ is C , this definition means that D ϕ > on M . Proposition 4.4 [13, Th.2]
Let M be a (possibly non complete) Rieman-nian manifold and ϕ : M → R be a strictly convex function. Then, thereexists a sequence ( ϕ n ) of smooth strictly convex functions on M that con-verges uniformly to ϕ on compact subsets of M . The main tool used in [13] to prove Proposition 4.4 is a smoothing procedurecalled Riemannian convolution.
Lemma 4.5
Strict convexity of ϕ C Let C be a non empty closed convex subset of the pinched Hadamard manifold X . Then the function ϕ C = d ( ., C ) , square of the distance function to C ,is strictly convex on the complement M = X (cid:114) C of the convex set C . Proof of Lemma 4.5
We only need to prove that, for every ε >
0, thereexists an α > t → c t with d ( c t , C ) ≥ ε , one has2 ϕ C ( c ( s + t ) / ) − ϕ C ( c t ) − ϕ C ( c s ) ≤ − α ( t − s ) . Denote by π the projection on the closed convex set C . Applying the Reshet-nyak comparison lemma 2.7 to the quadrilateral [ c t , π ( c t ) , π ( c s ) , c s ], we arereduced to the well-known case where C is a geodesic segment in the hyper-bolic plane H . (cid:3) Remark 4.6
The function ψ C = d ( ., C ) , distance function to the convexsubset C of X , is convex. Moreover, there exists α > such that ∆ ψ C ≥ α outside the -neighbourhood of C . This follows from the same arguments asabove. Proof of Proposition 4.2
Applying Lemma 4.5 to the convex hull C :=conv(Λ Γ ) of the limit set, we obtain that the function ϕ C is strictly convexon X (cid:114) C . Hence, the function ϕ K = d ( ., K ) is strictly convex on M (cid:114) K .We may thus apply Proposition 4.4 on the manifold M = M (cid:114) K to thefunction ϕ K , and obtain a smooth strictly convex function ϕ n on M suchthat, for every m with 1 / ≤ d ( m, K ) ≤
2, one has | ϕ n ( m ) − ϕ ( m ) | ≤ / V as the set V = K ∪ ϕ − n ([0 , V is a convex neighbourhood of the convex core K . Since ϕ n does not reacha minimum on the boundary ∂V = ϕ − n (1), the differential of this convexfunction does not vanish on ∂V , so that V has smooth boundary. (cid:3) .3 Existence To prove Theorem 1.1, it follows from Paragraph 4.1 that we may assumethat the map f : M → Y we are starting with is not only rough Lipschitz butsatisfies, as well as Condition (b) of Definition 1.4, the stronger condition :(a (cid:48) ) There exists a constant c > such that f : M → Y is smooth with (cid:107) Df (cid:107) ≤ c and (cid:107) D f (cid:107) ≤ c . (4.1)Recall that V is the compact convex neighbourhood, with smooth boundary,of the convex core K ⊂ M that we constructed in Proposition 4.2. Lemma 4.7
For
R > , let V R ⊂ M denote the R -neighbourhood of V .Then, there exists a unique harmonic map h R : V R → Y solution of theDirichlet problem h R = f on the boundary ∂V R . Proof
This is a consequence of a theorem by R. Schoen [10, (12.11)], since V R ⊂ M is a compact manifold with smooth boundary (Lemma 4.2), Y is aHadamard manifold and f : M → Y is a smooth map. (cid:3) The crucial step in the proof of existence in Theorem 1.1 consists in thefollowing uniform estimate.
Proposition 4.8
There exists a constant ¯ ρ > such that d ( f, h R ) ≤ ¯ ρ forany large radius R > . Since the map f is c -Lipschitz, we infer that the smooth harmonic maps h R are locally uniformly bounded. The following statement due to Cheng, andwhich is local in nature, provides a uniform bound for their differentials aswell. Proposition 4.9
Cheng Lemma [9]
Let Z be a k -dimensional completeRiemannian manifold with sectional curvature − b ≤ K Z ≤ , and Y be aHadamard manifold. Let z ∈ Z , r > and let h : B ( z, r ) ⊂ Z → Y be aharmonic C ∞ map such that the image h ( B ( z, r )) lies in a ball of radius r (cid:48) .Then one has the bound (cid:107) D z h (cid:107) ≤ k brr r (cid:48) . Corollary 4.10
There exists a constant κ ≥ that depends only on M and Y , and with the following property. Let R > and assume that d ( f, h R ) ≤ ρ for some constant ρ ≥ c . Let p ∈ M such that B ( p, ⊂ V R . Then, (cid:107) D p h R (cid:107) ≤ κ ρ . roof The map f being c -Lipschitz, it follows from d ( f, h R ) ≤ ρ that h R ( B ( p, ⊂ B ( h R ( p ) , ρ + c ) ⊂ B ( h R ( p ) , ρ ) . Thus, the Cheng Lemma 4.10 applies and yields that (cid:107) D p h R (cid:107) ≤ κρ , with κ = 2 k (1 + b ). (cid:3) Proof of Theorem 1.1
Applying the Ascoli-Arzela’s theorem, it followsfrom Propositions 4.8 and 4.10 that we can find a increasing sequence of radii R n → ∞ such that the sequence of harmonic maps ( h R n ) converges locallyuniformly towards a continuous map h : M → Y which is within boundeddistance ¯ ρ from f . The Schauder estimates then provide a uniform boundfor the C ,α -norms of the ( h R n ), hence we may assume that this sequenceconverges in the C norm, so that the limit map h is smooth and harmonic.We refer to [6, Section 3.3] for more details. (cid:3) In this paragraph, we make the first step towards the proof of Proposition4.8 by proving the so-called boundary estimates.The boundary estimates state that, if the distance d ( h R , f ) is very large,this distance is reached at a point which is far away from the boundary ∂V R of the domain where h R is defined. More specifically, we have the following. Proposition 4.11
There exists a constant B such that, for any R ≥ andany point m ∈ V R , then d ( f ( m ) , h R ( m )) ≤ B d ( m, ∂V R ) . The proof of this proposition is similar to that of [6, Proposition 3.7], butwe must first construct a strictly subharmonic function Ψ : M → [0 , ∞ [ thatcoincides, outside the 1-neighbourhood V of V , with the distance function ψ V = d ( ., V ). Lemma 4.12
There exists a constant α > such that the function ψ V = d ( ., V ) satisfies the inequality ∆ m ψ V ≥ α at any point m ∈ M (cid:114) V . Proof
This follows from Remark 4.6 applied to the convex set ˜ V ⊂ X ,which is the lift of V ⊂ M . (cid:3) Corollary 4.13
There exists a continuous function
Ψ : M → [0 , ∞ [ whichis uniformly strictly subharmonic, namely with ∆Ψ ≥ ε (weakly) on M forsome ε > , and such that Ψ = ψ V outside V . Proof If ν denotes the outgoing unit normal to V , we have ψ V = 1 and ∂ψ V ∂ν = 1 on ∂V . Let η : V → R be the solution of the Dirichlet problem∆ η = 1 on V , and η = 0 on ∂V .19ote that there exists a constant c η > − c η ≤ η ≤ V , and0 ≤ ∂η∂ν ≤ c η on ∂V . Let ψ = 1 + ηc η : V → [0 , ∞ [, and define a functionΨ : M → [0 , ∞ [ by lettingΨ = ψ on V and Ψ = ψ V outside V .The function Ψ is positive and continuous on the whole M . Moreover, since ∂ψ V ∂ν = 1 ≥ ∂ψ ∂ν on ∂V , it follows that ∆Ψ ≥ inf( α , /c η ) weakly on M . (cid:3) Proof of Proposition 4.11
Let m ∈ V R . Choose a point y ∈ Y such that f ( m ) lies on the geodesic segment [ h R ( m ) , y ], and such that d ( y, f ( V R )) ≥ B > v : p ∈ V R → d ( y, h R ( p )) − d ( y, f ( p )) − ( R − Ψ( p )) B/ ∈ R . The choice of y ensures that d ( h R ( m ) , f ( m )) = d ( y, h R ( m )) − d ( y, f ( m )). Ifthe point m lies in V R (cid:114) V , we have R − Ψ( m ) = d ( m, ∂V R ) while, if m ∈ V ,the inequality R − Ψ( m ) ≤ R ≤ R − ≤ d ( m, ∂V R ) holds. Therefore,we only have to prove that v ( m ) ≤
0. Since the function v vanishes on theboundary ∂V R , we will be done if we prove that, for a suitable choice of theconstant B , the function v is subharmonic on V R .Since h R is a harmonic map, the function p → d ( y, h R ( p )) is subharmonic(Lemma 3.5). Since f is smooth with uniformly bounded first and secondorder covariant derivatives, and we chosed y ∈ Y with d ( y, f ( V R )) ≥
1, itfollows that there exists a constant β such that the absolute value of theLaplacian of the function p ∈ V R → d ( y, f ( p )) ∈ R is bounded by β (see [6,(2.3)]). We infer from Corollary 4.13 that∆ v ≥ − β + εB/ > B is large enough, hence the result. (cid:3) Before going into the main part of the proof of Proposition 4.8, we settle thematter of uniqueness. This is where the non compactness of M is needed. Proof of uniqueness in Theorem 1.1
We rely on arguments in [5, Section 5]. See also [18, Lemma 2.2]. Let h , h : M → Y be two harmonic maps within bounded distance from f .We assume by contradiction that δ := d ( h , h ) > m ∈ M → d ( h ( m ) , h ( m )) ∈ [0 , ∞ [ achieves its maximum, hence isconstant. As in [5, Corollary 5.19], it follows that both maps h and h take their values in the same geodesic of Y . Hence each end of M is quasi-isometric to a geodesic ray. This is a contradiction, since Γ being convexcocompact implies that the injectivity radius is a proper function on M .20ssume now that there exists a sequence of points ( m i ) i ∈ N in M , thatgoes to infinity, and such that d ( h ( m i ) , h ( m i )) → δ . We may also assumethat the sequence ( m i ) converges to a point ξ ∈ ∂ ∞ M .Since the injectivity radius is a proper function on M , it follows fromHypothesis (b) on f that there exist a sequence of radii r i → ∞ and aconstant c ξ such that r i < inj( m i ) and the restriction of f to each ball B ( m i , r i ) is a quasi-isometric map for some constant c ξ .Applying [5, Lemma 5.16] ensures that, going if necessary to a subse-quence, there also exist two limit C pointed Hadamard manifolds ( X ∞ , x ∞ )and ( Y ∞ , y ∞ ) with C Riemannian metrics such that the maps h , h : B ( m i , r i ) → Y respectively converge to quasi-isometric harmonic maps h , ∞ , h , ∞ : X ∞ → Y ∞ with d ( h , ∞ ( x ) , h , ∞ ( x )) = δ for every x ∈ X ∞ .Applying again [5, Corollary 5.19], we infer that both maps h , ∞ , h , ∞ take their values in the same geodesic of Y . This is a contradiction, sinceboth maps h i, ∞ are quasi-isometric. (cid:3) To prove Proposition 4.8 in Section 7, we will need uniformbounds for the harmonic measures on specific domains of M .In this section, we deal with the lower bounds. Assume that M is a Riemannian manifold, and let W ⊂ M be a relativelycompact domain with smooth boundary. For any continuous function u : ∂W → R , there exists a unique continuous function η u : W → R which issmooth on W , and is solution to the Dirichlet problem∆ η u = 0 on W and η u = u on ∂W .This gives rise to a family of Borel probability measures σ m,W supportedon ∂W , indexed by the points m ∈ W and such that, for any continuousfunction u ∈ C ( ∂W ) : η u ( m ) = (cid:90) ∂W u ( z ) dσ m,W ( z ) . (5.1)The measure σ m,W is the harmonic measure of W relative to the point m .In our previous paper [5], we worked in pinched Hadamard manifolds,and obtained the following uniform upper and lower bounds for the harmonicmeasures on balls relative to their center. Theorem 5.1 [5]
Let < a ≤ b and k ≥ . There exist positive constants C, s depending only on a, b and k , and with the following property. et X be a k -dimensional pinched Hadamard manifold, whose sectionalcurvature satisfies − b ≤ K X ≤ − a .Then for any point x ∈ X , any radius R > and angle θ ∈ [0 , π/ , theharmonic measure σ x,R of the ball B ( x, R ) relative to the center x satisfies C θ s ≤ σ x,R ( C θx ) ≤ C θ /s (5.2) where C θx ⊂ X denotes any cone with vertex x and angle θ . Since the measure σ x,R is supported on the sphere S ( x, R ), the expression σ x,R ( C θx ) means σ x,R ( C θx ∩ S ( x, R )).To prove Theorem 1.1, we will need similar estimates in the quotientmanifold M . The lower bound is provided in the next paragraph. Theupper bound will require more work and will be carried out in Section 6. M The description of the positive harmonic functions on the quotient M = Γ \ X of a pinched Hadamard manifold by a convex cocompact group is due toAnderson-Schoen in [3, Corollary 8.2].Our goal in this paragraph is to obtain the following lower bound for themass of a ball of fixed radius λ >
0, with respect to the harmonic measureson suitable domains of M . Proposition 5.2
Lower bound for harmonic measures on M = Γ \ X Let λ > and L ≥ . Then, there exists a constant s ( λ, L ) > with thefollowing property. For any pair of points m, q in M with < d ( m, q ) ≤ L ,there exists a bounded domain with smooth boundary W m,q with m ∈ W m,q and q ∈ ∂W m,q , and whose harmonic measure relative to the point m satisfiesthe inequality σ m,W m,q (cid:0) B ( q, λ ) (cid:1) ≥ s ( λ, L ) . (5.3)This statement will derive from a compactness argument, together with thefollowing continuity lemma. Lemma 5.3
Let W ⊂ R k be a bounded domain with C boundary, and ( g n ) n ∈ N be a sequence of Riemannian metrics on W that converges, in the C ( W ) sense, to a Riemannian metric g ∞ on W .Let u ∈ C ( ∂W ) be a continuous function on the boundary. For each n ∈ N ∪ {∞} , denote by η n ∈ C ( W ) ∩ C ( W ) the solution of the Dirichletproblem with fixed boundary value u for the metric g n , namely such that ∆ n η n = 0 ( η n ) | ∂W = u . Here ∆ n denotes the Laplace operator corresponding to the metric g n .Then, the sequence ( η n ) n ∈ N converges uniformly on W to η ∞ . roof Introduce the solution ηηη : W → R of the boundary value problem∆ ∞ ηηη = 1 ηηη | ∂W = 0 . Let ε >
0. If n is large enough, we have∆ n ( η ∞ − η n + εηηη ) ≥ n ( η ∞ − η n − εηηη ) ≤ η ∞ − η n ± εηηη vahishing on theboundary ∂W , the maximum principle ensures that | η ∞ − η n | ≤ ε sup | ηηη | . (cid:3) Each domain W m,q ⊂ M will either be a ball, or the image under asuitable diffeomorphism of a fixed domain W ⊂ R k . This diffeomorphismwill be defined using the normal exponential map along a geodesic segmentcontaining [ m, q ]. We first observe the following, where inj( M ) > M . Lemma 5.4 (1) Let [ m, q ] ⊂ M be a minimizing geodesic segment. Then,the extended geodesic segment I mq = [ m q , q ] defined by the conditions [ m, q ] ⊂ [ m q , q ] and d ( m, m q ) = inj( M ) / is still injective.(2) For any compact subset Z ⊂ M , there exists < r ≤ such that,when [ m, q ] is a minimizing geodesic segment with m ∈ Z and d ( m, q ) ≤ L ,the normal exponential map ν mq along I mq is a diffeomorphism from thebundle of normal vectors to I mq with norm at most r onto its image. Proof (1) derives easily from the definition of the injectivity radius.(2) follows since the L -neighbourhood of Z is also compact. (cid:3) Let α = inj( M ) / L and introduce the segment J = [ − α, ⊂ R . Wechoose W to be a convex domain of revolution W ⊂ J × B (0 , r ) ⊂ R × R k − whose boundary ∂W is smooth and contains both points ( − α,
0) and (1 , Proof of Proposition 5.2
We may assume that λ ≤ inj( M ).If d ( m, q ) ≤ λ/
2, then choose W m,q to be the ball with center m andradius d ( m, q ), so that W m,q ⊂ B ( q, λ ).We now assume that d ( m, q ) > λ/
2. Since the injectivity radius is aproper function on M , the set Z = { m ∈ M | inj( m ) ≤ L + λ } is a compact subset of M .Assume first that the point m does not belong to the compact set Z .Then, the ball B ( m, L + λ ) is isometric to a ball with radius L + λ in the23adamard manifold X while B ( q, λ ) ⊂ B ( m, L + λ ). Choosing W m,q = B ( m, d ( m, q )), the required estimate follows easily from the lower bound inTheorem 5.1.Assume now that m ∈ Z , and that λ/ ≤ d ( m, q ) ≤ L . Pick a minimizinggeodesic segment [ m, q ]. Identify R with the geodesic line containing [ m, q ]through the constant speed parameterization c : R → M defined by c (0) = m and c (1) = q , so that c ( J ) ⊂ I mq . Introduce W m,q = ν m,q ( W ). Byconstruction, W m,q is a bounded domain of M with smooth boundary, suchthat m ∈ W m,q and q ∈ ∂W m,q .Let us prove that the harmonic measures σ m,W m,q (cid:0) B ( q, λ ) (cid:1) are uniformlybounded below. We proceed by contradiction and assume that there existtwo sequences of points m n ∈ Z , and q n ∈ M with λ/ ≤ d ( m n , q n ) ≤ L ,and such that σ m n ,W mn,qn (cid:0) B ( q n , λ ) (cid:1) → n → ∞ .Since Z is compact, we may assume that m n → m ∞ ∈ Z , that q n → q ∞ with m ∞ (cid:54) = q ∞ , and that the sequence of minimizing geodesic segments([ m n , q n ]) converges to a minimizing geodesic segment [ m ∞ , q ∞ ]. Denotingby g n ( n ∈ N ∪ {∞} ) the Riemannian metrics on W obtained by pull-back of the Riemannian metric of M under the map ν m n ,q n , we may evenassume that g n → g ∞ in the C sense on W . Hence Lemma 5.3 yields σ m ∞ ,W m ∞ ,q ∞ (cid:0) B ( q ∞ , λ ) (cid:1) = 0, a contradiction to the maximum principle. (cid:3) α m qW m,q W Construction of the domain W m,q The main goal of this section is to obtain, in Corollary 6.20,the upper bound for the harmonic measures on M needed forthe proof of Proposition 4.8.One of the major technical tools for estimating or constructing harmonicfunctions on Hadamard manifolds is the so-called Anderson-Schoen barriers.Given two opposite geodesic rays in a pinched Hadamard manifold, thecorresponding Anderson-Schoen barrier is a positive superharmonic functionthat decreases exponentially along one of the geodesics rays, and is greaterthan 1 on a cone centered around the other geodesic ray [3].Our first step is to obtain an analogous to the Anderson-Schoen barrierfunctions for the quotient manifold M = Γ \ X in Proposition 6.18. This willrely on the work by Ancona [1], Anderson [2] and Anderson-Schoen [3].24n the whole section, X will denote our pinched Hadamard manifoldsatisfying (1.2). The torsion-free convex cocompact subgroup Γ of Is ( X )will only come into the picture starting from Paragraph 6.4. X We first recall some fundamental, and by now classical, results concerningharmonic measures on Hadamard manifolds.In the sequel
T, α, c ◦ , c r ≥ T, α, c ◦ depend only on the pinched Hadamard manifold X , and c r also depends onthe distance r > Lemma 6.1 [26]
For a positive harmonic function η : B ( x, ⊂ X → ]0 , ∞ [ defined on a ball with radius , one has | D x (log η ) | ≤ c . Another fundamental tool is the Green function G : X × X → ]0 , ∞ ]. ThisGreen function is continuous on X × X , and is uniquely defined by theconditions ∆ y G ( x, y ) = − δ x lim y →∞ G ( x, y ) = 0for every x ∈ X . One can prove that G is symmetric i.e. G ( x, y ) = G ( y, x )for x, y in X . Moreover, the Green function satisfies the following estimates. Proposition 6.2
1. If d ( x, y ) ≥ , one has c − ◦ d ( x, y ) − c ◦ ≤ log(1 /G ( x, y )) ≤ c ◦ d ( x, y ) + c ◦ . (6.1)
2. Let x, y, z ∈ X such that d ( x, y ) ≥ and d ( y, z ) ≥ . One has c − ◦ G ( x, y ) G ( y, z ) ≤ G ( x, z ) . (6.2) If d ( x, z ) ≥ and d ( y, [ x, z ]) ≤ r , one has G ( x, z ) ≤ c r G ( x, y ) G ( y, z ) . (6.3)The first assertion is (2.4) in Anderson-Schoen [3]. The second assertionfollows from (6.1), using Harnack inequality and the maximum principle,while the third one is Ancona’s inequality [1, Theorem 5].Let us now turn to the bounded harmonic functions on X . Andersonproved in [2] that the Dirichlet problem at infinity on X has a unique solutionfor any continuous boundary value. Hence there exists, for every point25 ∈ X , a unique Borel measure ω x on ∂ ∞ X such that, for any continuousfunction u ∈ C ( ∂ ∞ X ), the function η u : x ∈ X → (cid:90) ∂ ∞ X u ( ξ ) dω x ( ξ ) ∈ R is harmonic on X and extends continuously to X with boundary value atinfinity equal to u . The measure ω x is the harmonic measure on ∂ ∞ X atthe point x . The Harnack inequality of Lemma 6.1 ensures that two suchharmonic measures ω x and ω y are absolutely continuous with respect toeach other, and that their Radon-Nikodym derivatives dω y dω x are uniformlybounded when d ( x, y ) ≤
1. We will also need a control on these Radon-Nikodym derivatives for d ( x, y ) ≥
1, that will be given in Lemma 6.3.In [3], Anderson-Schoen study the positive harmonic functions on X , andprovide an identification of the Martin boundary of X with the boundaryat infinity ∂ ∞ X . More precisely, they obtain the following results.Fix a base point o ∈ X and introduce the normalized Green function atthe point o ∈ X with pole at z ∈ X , which is defined by k ( o, x, z ) = G ( z, x ) G ( z, o ) . Letting the point z ∈ X converge to ξ ∈ ∂ ∞ X , the limit k ( o, x, ξ ) = lim z → ξ k ( o, x, z )exists and x → k ( o, x, ξ ) is now a positive harmonic function on the whole X that extends continuously to the zero function on ∂ ∞ X (cid:114) { ξ } , and suchthat k ( o, o, ξ ) = 1.In [3, Theorem 6.5] Anderson-Schoen prove that these positive harmonicfunctions are the minimal ones. Using the Choquet representation theorem,they provide the following Martin representation formula. For any positiveharmonic function η on X , there exists a unique finite positive Borel measure µ η on ∂ ∞ X such that, for every x ∈ X , η ( x ) = (cid:90) ∂ ∞ X k ( o, x, ξ ) dµ η ( ξ ) . The minimal harmonic functions relate to the harmonic measures at infinity.
Lemma 6.3
1. Let o, x ∈ X . The following holds for ω o -a.e. ξ ∈ ∂ ∞ X : k ( o, x, ξ ) = dω x dω o ( ξ ) . (6.4)
2. For o, x ∈ X and ξ ∈ ∂ ∞ X with d ( o, x ) ≥ , we have k ( o, x, ξ ) G ( o, x ) ≤ c ◦ . f moreover d ( x, [ o, ξ [) ≤ r , then we also have c − r ≤ k ( o, x, ξ ) G ( o, x ) . Proof
1. is proved in [3, § z ∈ [ o, ξ [ and letting z → ξ . (cid:3) Lemma 6.3 asserts that, if d ( o, x ) ≥ ξ ∈ ∂ ∞ X which is inthe shadow of the ball B ( x, r ) seen from o , all the densities dω x dω o ( ξ ) are closeto ( G ( o, x )) − hence do not depend too much on ξ . Is ( X ) on ∂ ∞ X We now investigate, using Lemma 6.3, the action of Is ( X ) on the harmonicmeasures at inifinity.Introduce the function defined, for any pair of points x, y ∈ X , by d G ( x, y ) = log + (1 /G ( x, y ))where log + ( t ) = sup(log t,
0) denotes the positive part of the logarithm.Proposition 6.2 tells us that, at large scale, the function d G behaves roughlylike a distance that would be quasi-isometric to the Riemannian distance d . In particular, (6.2) ensures that there exists a constant c ◦ such that thefollowing weak triangle inequality holds for every x, y, z ∈ X : d G ( x, z ) ≤ d G ( x, y ) + d G ( y, z ) + c ◦ . (6.5)Although d G is not exactly a distance on X , we will thus nevertheless agreeto think of d G as of the Green distance. We would like to mention thatBlachre-Hassinski-Mathieu [7] already used such a Green distance in thesimilar context of random walks on hyperbolic groups.From now on, we choose a base point o ∈ X . We associate to d G ananalog to the Busemann functions by letting, for ξ ∈ ∂ ∞ X and x ∈ X : β ξ ( o, x ) = − log k ( o, x, ξ ) = lim z → ξ d G ( x, z ) − d G ( o, z ) . These Busemann functions relate to the harmonic measures at infinity, since(6.4) reads as dω x dω o ( ξ ) = e − β ξ ( o,x ) . (6.6)Define the length of g ∈ Is ( X ) as | g | o = d G ( o, go ). We want to compare β ξ ( o, g − o ) with | g | o . Notation 6.4
When g ∈ Is ( X ) does not fix the point o , we introduce theendpoints ξ + ( g ) , ξ − ( g ) ∈ ∂ ∞ X of the geodesic rays with origin o that containrespectively the points go and g − o . 27 emma 6.5 (1) For g ∈ Is ( X ) and ξ ∈ ∂ ∞ X , one has | g | o = | g − | o and | β ξ ( o, g − o )) | ≤ | g | o + c ◦ . (2) For every ε > there exists a constant a ε ≥ such that, when g ∈ Is ( X ) and ξ ∈ ∂ ∞ X satisfy ∠ o ( ξ, ξ − ( g )) ≥ ε , then | g | o ≤ β ξ ( o, g − o ) + a ε . Proof (1) We observe that, since the Green function G is symmetric andinvariant under isometries, we have d G ( y, x ) = d G ( x, y ) = d G ( gx, gy ) forevery g ∈ Is ( X ) and x, y ∈ X . The equality | g | o = | g − | o follows.Using the weak triangle inequality (6.5) for d G yields β ξ ( o, g − o ) = lim z → ξ d G ( g − o, z ) − d G ( o, z ) ≤ d G ( g − o, o ) + c ◦ = | g | o + c ◦ . The lower bound follows by observing that the invariance of d G under isome-tries ensures that β ξ ( o, g − o ) = β gξ ( go, o ) = − β gξ ( o, go ) ≥ −| g | o − c ◦ . (2) We may suppose that | g | o > | g | o = − log G ( o, g − o ). Since K X ≤ − a ,the condition ∠ o ( ξ, ξ − ( g )) ≥ ε ensures thatthe distance of the point o to the geodesicray [ g − ( o ) , ξ [ is bounded above by a con-stant r ε that depends only on ε . Lemma 6.3(2) yields | g | o ≤ β ξ ( o, g − o ) + log c r ε . (cid:3) The following corollary provides useful es-timates for action of isometries on the har-monic measures on ∂ ∞ X . o ξ (g) +- ξ (g) g o -1 g o ξ Corollary 6.6
For A ⊂ ∂ ∞ X a measurable set and g ∈ Is ( X ) , one has e − c ◦ e −| g | o ω o ( A ) ≤ ω o ( gA ) ≤ e c ◦ e | g | o ω o ( A ) . If we assume that ∠ o ( ξ, ξ − ( g )) ≥ ε for every ξ ∈ A , one has ω o ( gA ) ≤ e a ε e −| g | o ω o ( A ) where a ε is the constant in Lemma 6.5. Proof
Immediate consequence of Lemma 6.5 and Equality (6.6). (cid:3) .3 Harmonic measures of cones in X We now recall estimates for the harmonic measures at infinity, that are dueto Anderson-Schoen.
Definition 6.7
Let o ∈ X , ξ ∈ ∂ ∞ X and θ ∈ [0 , π ] . The closed cone C θoξ ⊂ X with vertex o , axis [ o, ξ [ and angle θ is the union of all the geodesicrays [ o, ζ [ whose angle with [ o, ξ [ is at most θ . The trace of the cone C θoξ onthe sphere at infinity ∂ ∞ X will be denoted by S θoξ . By analogy with the case where X has constant curvature, we will take theliberty of calling a cone D oξ := C π/ oξ ⊂ X with angle θ = π/ o , and its boundary H oξ a hyperplane, also with vertex o . We will denote by S oξ the trace of the half-space D oξ on the boundary atinfinity, and we will call it a half-sphere at infinity seen from the point o .Although half-spaces in the pinched Hadamard manifold X may not beconvex, the following lemma tells us that they are not far from being so. Lemma 6.8 [8, Prop.2.5.4]
There exists a constant λ , that depends only onthe pinching constants of X , such that the convex hull of a half-space D ⊂ X lies within its λ -neighbourhood : Hull ( D ) ⊂ V λ ( D ) . Proof
This statement follows from Proposition 2.15, due to Bowditch.Observe indeed that D is included in the join of its half-sphere S at infinity,and that this join lies in the δ X -neighbourhood of D . (cid:3) The following uniform bounds for harmonic measures of cones in pinchedHadamard manifolds are due to Anderson-Schoen.
Lemma 6.9
There exists a constant c ◦ ≥ such that one has, for any point o ∈ X and any ξ ∈ ∂ ∞ X , ω o ( S π/ oξ ) ≥ /c ◦ . A more precise statement is given by Kifer-Ledrappier in [16, Theorem 4.1].
Notation 6.10
We now fix a base point o ∈ X . When ξ ∈ ∂ ∞ X , we denoteby t ∈ R → x tξ ∈ X the unit speed geodesic with origin o that converges to ξ ∈ ∂ ∞ X in the future. We will let D tξ stand for the half-space D x tξ ξ withvertex x tξ and axis [ x tξ , ξ [, and will denote accordingly by H tξ and S tξ thecorresponding hyperplane and half-sphere at infinity. Proposition 6.11
There exist a distance
T > and two constants α > and c ◦ ≥ such that the following holds for every ξ ∈ ∂ ∞ X : ω x − tξ ( S oξ ) ≤ c ◦ e − αt for every t ≥ ω y ( S oξ ) ≥ /c ◦ for every y ∈ D Tξ . roof The first assertion is [3, Corollary 4.2].Thanks to the upper bound on the cur-vature K X of X , there exists a distance T > X such thatthe half-space D Tξ is seen from the point o under an angle at most π/
4, namely suchthat D Tξ ⊂ C π/ oξ .If now y ∈ D Tξ and ζ y ∈ ∂ ∞ X denotes theendpoint of the geodesic ray such that y ∈ [ o, ζ y [, it follows that C π/ yζ y ⊂ D oξ . Lemma6.9 yields ω y ( S oξ ) ≥ ω y ( S π/ yζ y ) ≥ /c ◦ . (cid:3) o ξ x T y ζ y ξ ξ y ζ y π /4 T0 ξ -t x ξ ξ M In this paragraph, we introduce embedded half-spaces in the quotient M =Γ \ X (Proposition 6.16), that will be needed in the sequel of this paper.Recall that we fixed a base point o ∈ X . We keep Notation 6.10. Notation 6.12
We now introduce the projection m ∈ M = Γ \ X of ourbase point o ∈ X . When ξ ∈ ∂ ∞ X , we will denote by t ∈ R → m tξ ∈ M the(perhaps non minimizing) geodesic obtained by projection of the geodesic t ∈ R → x tξ ∈ X . Definition 6.13
A closed embedded half-space with vertex m tξ is the pro-jection in M of a closed half-space D tξ ⊂ X that embeds in M , namely thatsatisfies D tξ ∩ γ D tξ = ∅ for every non trivial element γ ∈ Γ , where D tξ ⊂ X denotes the closure of D tξ in the compactification of X . We first remark that there exist many embedded half-spaces in M . Choosea relatively compact subset Ω ⊂ Ω Γ of the domain of discontinuity. Lemma 6.14
There exists t > such that, for every ξ ∈ Ω and every t ≥ t , the half-space D tξ ⊂ X embeds in M . Proof
We proceed by contradiction, and assume that there exist sequences t n → + ∞ , ξ n ∈ Ω , z n ∈ D t n ξ n and γ n ∈ Γ ∗ such that γ n z n ∈ D t n ξ n forevery n ∈ N . By compactness of Ω , we may assume that the sequence ( ξ n )converges to some ξ ∞ ∈ Ω . Since t n → ∞ and ξ n → ξ ∞ , both sequences( z n ) and ( γ n z n ) converge to ξ ∞ in X . Since the action of Γ on X ∪ Ω Γ isproperly discontinuous, and Γ is torsion-free, it follows that γ n is trivial for n large, a contradiction. (cid:3) otation 6.15 For ξ ∈ Ω and t ≥ t , we will denote by the Roman letters D tξ ⊂ M the closed embedded half-space with vertex m tξ and by H tξ itsboundary in M , respectively obtained as the projections of D tξ and H tξ .Our goal in the remaining of this section is to prove that an embedded half-space in the quotient M = Γ \ X is not far from being geodesically convex : Proposition 6.16
There exists τ ≥ such that, for every ξ ∈ Ω and t ≥ t , and for every pair of points p , p ∈ D t + τ ξ , there exists only oneminimizing geodesic segment [ p , p ] ⊂ M , and it lies in D tξ . Proposition 6.16 relies on an analogous property for the half-space D t + t ξ inthe pinched Hadamard manifold X that we proved in Lemma 6.8. We firststate an elementary property of obtuses triangles in X . Lemma 6.17
There exists a constant ˆ δ that depends only on X such thatif [ y, x, z ] ⊂ X is a triangle with an angle at least π/ at the vertex x , then d ( y, z ) ≥ d ( x, y ) + d ( y, z ) − ˆ δ . Proof
Same as for Lemma 2.8. (cid:3)
Proof of Proposition 6.16
Let τ > p , p ∈ D t + τ ξ . Consider aminimizing geodesic segment [ p , p ] ⊂ M . Lift the points p , p ∈ D t + τ ξ as y , y ∈ D t + τ ξ . Then, there exists γ ∈ Γ such that [ p , p ] lifts as a geodesicsegment [ y , γy ] ⊂ X .Assume first that γ = e . If we choose τ ≥ λ X , the λ X -neighbourhood of thehalf-space D t + τ ξ lies in D tξ . Hence, it fol-lows from Lemma 6.8 that [ y , y ] ⊂ D tξ sothat [ p , p ] ⊂ D tξ .Proceed now by contradiction and assumethat γ ∈ Γ ∗ is non trivial. Observe that,since the boundary of D tξ is a union ofgeodesic rays emanating from the point x tξ , the geodesic segment ] x tξ , γx tξ [ staysout of both half-spaces D tξ and γ D tξ . ξ x t y y γγ ξ x t+ τγ ξ x t+ τγ ξ x t ξ t Let ε = inf { d ( y, γy ) | y ∈ X, γ ∈ Γ ∗ } >
0, and α ε be the correspondingangle in Lemma 2.8. As in the proof of Proposition 6.11, choose τ largeenough so that any half-space D t + τ ξ is seen from the point x tξ under anangle less than α ε . We may then apply Lemma 2.8 to the quadrilateral[ y , x tξ , γx tξ , γy ] ⊂ X to obtain d ( y , γy ) ≥ d ( y , x tξ ) + d ( x tξ , γx tξ ) + d ( γx tξ , γy ) − δ ε ≥ d ( y , x tξ ) + d ( y , x tξ ) − δ ε . y i , x t + τ ξ , x tξ ] ( i = 1 ,
2) yields d ( y i , x tξ ) ≥ d ( y i , x t + τ ξ ) + τ − ˆ δ . This is a contradiction if τ > (ˆ δ + ˆ δ ε ), since the triangle inequality yields d ( y , γy ) ≥ d ( y , y ) + 2 τ − δ + ˆ δ ε ) . (cid:3) M To construct an analogous to the Anderson-Schoen barrier functions for thequotient manifold M = Γ \ X , we work in the Hadamard manifold X .Recall that we fixed a base point o ∈ X and that Ω ⊂ Ω Γ is relativelycompact. We keep the notation of Proposition 6.11 and Lemma 6.14. Proposition 6.18
There exists C ≥ such that, for every ξ ∈ Ω , theharmonic measure at infinity of the saturation of S tξ under Γ satisfies ω o (cid:0) (cid:91) γ ∈ Γ γ S tξ (cid:1) ≤ C e − αt for every t ≥ t ω y (cid:0) (cid:91) γ ∈ Γ γ S tξ (cid:1) ≥ /C for every y ∈ D t + Tξ . We will need the following fact relative to the domain of discontinuity of Γ.
Lemma 6.19
Let L ⊂ Ω Γ be any compact subset of the domain of discon-tinuity. Then, there is only a finite number of elements γ ∈ Γ ∗ with γo (cid:54) = o and ξ − ( γ ) ∈ L . Proof
Let ( γ n ) be a sequence of pairwise distinct elements of Γ such that ξ − ( γ n ) → ζ ∈ ∂ ∞ X . Since Γ is discrete, d ( γ − n o, o ) → ∞ so that γ − n o → ζ hence ζ ∈ Λ Γ . The result follows readily. (cid:3) Proof of Proposition 6.18
The lower bound is an immediate consequenceof Proposition 6.11.Let now ε > ∠ o ( ξ, Λ Γ ) ≥ ε for every ξ ∈ Ω .Lemma 6.19 ensures that the set Γ ε = { γ ∈ Γ ∗ | ∠ o ( ξ − ( γ ) , Λ Γ ) ≥ ε } is finite.Assuming that t ≥ t , we now seek an upper bound for ω o ( (cid:83) γ ∈ Γ γ S tξ ).We first observe that Corollary 6.6 and Lemma 6.14 ensure that e − c ◦ ω o ( S tξ ) (cid:0)(cid:88) γ ∈ Γ e −| γ | o (cid:1) ≤ (cid:88) γ ∈ Γ ω o ( γ S tξ ) = ω o ( (cid:91) γ ∈ Γ γ S tξ ) , so that the series (cid:80) γ ∈ Γ e −| γ | o converges. When γ ∈ Γ (cid:114) Γ ε is non trivial,one has ∠ o ( ξ, ξ − ( γ )) ≥ ε for every ξ ∈ Ω . Hence Corollary 6.6 again yields (cid:88) γ ∈ Γ ω o ( γ S tξ ) ≤ (cid:16) e c ◦ (cid:88) Γ ε e | γ | o + e a ε (cid:88) Γ (cid:114) Γ ε e −| γ | o (cid:17) ω o ( S tξ ) , (cid:3) Recall that V is the compact convex subset of M that we introduced inProposition 4.2. From now on, we will assume that the base point o ∈ X isso chosen that its projection m ∈ M belongs to V . Corollary 6.20
Upper bound for harmonic measures on M There exists a constant C such that the following holds. For every compactdomain with smooth boundary W whose interior contains V , every ξ ∈ Ω ,every t ≥ and every point m ∈ V : σ m,W ( ∂W ∩ D tξ ) ≤ C e − αt . (6.7) Proof
It suffices to prove the assertion when t is large. Recall that T hasbeen defined in Proposition 6.11. For ξ ∈ Ω and t ≥ t + T , introduce thepositive harmonic function defined by˜ η tξ : y ∈ X → ω y (cid:0) (cid:91) γ ∈ Γ γ S t − Tξ (cid:1) ∈ [0 , . The function ˜ η tξ is Γ-invariant and thus goes to the quotient to a harmonicfunction η tξ : M → [0 , t − T ≥ t )ensures that η tξ satisfies η tξ ( p ) ≥ /C for every p ∈ D tξ η tξ ( m ) ≤ C e αT e − αt . Applying the Harnack inequality (Lemma 6.1) to ˜ η tξ yields η tξ ( m ) ≤ e c d V C e αT e − αt for every m ∈ V , (6.8)where d V denotes the diameter of the compact convex set V .The function C η tξ is everywhere positive, and is greater or equal to 1on ∂W ∩ D tξ . Thus, the maximum principle ensures that C η tξ ( p ) ≥ σ p,W ( ∂W ∩ D tξ )holds for every p ∈ W , and the claim now follows from (6.8). (cid:3) Note that the constants t , τ , C and C that we introduced in theprevious paragraphs depend only on the group Γ, on the base point o , onΩ and on the compact subset V ⊂ M .33 .6 Gromov products and embedded half-spaces We end this chapter with Proposition 6.21, that relates half-spaces corre-sponding to the same point at infinity with level sets of Gromov products.Recall that we choose a base point o ∈ X whose projection m ∈ M lies in the compact subset V ⊂ M and that t and τ ≥ Proposition 6.21
There exists a constant g such that, for every ξ ∈ Ω ,every t ≥ t + τ , every point p ∈ D t + τ ξ and every point m ∈ V : { q ∈ M | ( p, q ) m ≥ t + g } ⊂ D tξ . Proof
Let p ∈ D t + τ ξ and q / ∈ D tξ . Consider a minimizing geodesic segment[ p, q ] ⊂ M and introduce two points p (cid:48) and q (cid:48) where [ p, q ] intersects thehyperplanes H t + τ ξ and H tξ .Since t ≥ t + τ , it follows from Proposition 6.16 that any, hence theonly, minimizing quadrilateral [ p (cid:48) , m t + τ ξ , m tξ , q (cid:48) ] lies in the embedded half-plane D t ξ and is thus isometric to a quadrilateral in X .This quadrilateral [ p (cid:48) , m t + τ ξ , m tξ , q (cid:48) ] isright-angled at both vertices m tξ and m t + τ ξ , and d ( m tξ , m t + τ ξ ) = τ ≥ m tξ is within distance ˜ δ ofthe edge [ p (cid:48) , q (cid:48) ]. Since [ p (cid:48) , q (cid:48) ] ⊂ [ p, q ]and d ( m, m tξ ) ≤ t + d V , it follows fromLemma 2.2 that( p, q ) m ≤ d ( m, [ p, q ]) ≤ d ( m, [ p (cid:48) , q (cid:48) ]) ≤ t +( d V +˜ δ ) . Dp' q'm τ D ξ t ξ m t+ τ ξ m t ξ m V The claim follows for g > d V + ˜ δ . (cid:3) In this final section, we wrap up the proof of Proposition 4.8,that gives a uniform bound for the distances d ( f, h R ).We split the proof into two parts. In the first part, where we assume thatthe point m ∈ V R where the distance d ( f, h R ) is reached lies far away fromthe convex core, the proof reduces to the proof of the main theorem of [6].In the second part, where we assume that the point m lies in a fixedneighbourhood of the convex core, we must deal with the topology of thequotient M = Γ \ X . 34 .1 Harmonic quasi-isometric maps H : X → Y In [6], we proved that a quasi-isometric map F : X → Y between two pinchedHadamard manifolds X and Y is within bounded distance from a uniqueharmonic map. As in the present paper, this harmonic map was obtained asthe limit of a family of solutions of Dirichlet problems on bounded domainswith boundary value F , where a uniform bound for the distances between F and the solutions of these Dirichlet problems ensured the convergence ofthe family.In the following technical statement, which is local in nature, we gathersome information obtained in [6] that was used to obtain this uniform bound.The first part of our proof of Proposition 4.8 will derive easily from thisstatement, see Proposition 7.3. Fact 7.1
Let c ≥ . There exist (cid:96) > and ρ > with the followingproperty.Let F, H : B ( x, (cid:96) ) → Y be two smooth maps defined on a ball B ( x, (cid:96) ) ⊂ X with radius (cid:96) , and such that the distance ρ := sup z ∈ B ( x,(cid:96) ) d ( F ( z ) , H ( z )) is reached at the center x of the ball, namely ρ = d ( F ( x ) , H ( x )) . Assumethat H is harmonic and that the map F satisfies c − d ( z, z (cid:48) ) − c ≤ d ( F ( z ) , F ( z (cid:48) )) ≤ c d ( z, z (cid:48) ) for any z, z (cid:48) ∈ B ( x, (cid:96) ) , (7.1) Then ρ ≤ ρ . Proof
This fact is proven in [6, Section 4]. Alternatively, one may followparagraphs 7.3 through 7.5 below, replacing the domain V (cid:96) with the ball B ( m, (cid:96) ) and using the uniform estimates (5.2) for the harmonic measuresof balls in the Hadamard manifold X instead of the new estimates (5.3) and(6.7). (cid:3) In this paragraph, we introduce a finite family of embedded half-spacesin M , whose union is a neighbourhood of infinity, and that will be usedthroughout the proof of Proposition 4.8. Then, we prove Proposition 4.8 incase the distance d ( f, h R ) is reached far away from the convex core.Recall that, after smoothing, the map f : M → Y is assumed to be c -Lipschitz (4.1) and that each point ξ ∈ ∂ ∞ M admits a neighbourhood towhich f restricts as a quasi-isometric map.From now on, Ω ⊂ Ω Γ is a fixed compact neighbourhood of a funda-mental domain for the action of Γ on Ω Γ .35 emma 7.2 (1) There exists a finite number of embedded half-spaces D t i ζ i ( ≤ i ≤ N ) with ζ i ∈ Ω and t i ≥ t such that, taking perhaps a largerconstant c in (4.1) : • each restriction f : D t i ζ i → Y is a quasi-isometric map with constant c • ∪ Ni =1 D t i + τ ζ i is a neighbourhood of infinity in M .(2) We may also assume that V has been choosen large enough so that • ∂V ⊂ ∪ Ni =1 D t i + τ ζ i (7.2) • for m / ∈ V , one has inj( m ) ≥ (cid:96) (7.3) • for m / ∈ V , the restriction of f to B ( m, (cid:96) ) is c quasi-isometric. (7.4) Proof (1) We proved in Lemma 6.14 that, for every ξ ∈ Ω and any t ≥ t , the half-space D tξ embeds in M . Hence, the claim follows from thehypothesis on f , since the boundary at infinity ∂ ∞ M is compact.(2) The injectivity radius inj : M → [0 , ∞ [ is a proper function, and thecomplement in M of ∪ Ni =1 { m ∈ M | B ( m, (cid:96) ) ⊂ D t i ζ i } is bounded. Hence, it suffices to replace the convex set V of Proposition 4.2by its R -neighbourhood for some large R to ensure these conditions. (cid:3) We want a uniform upper bound for the distance d ( f, h R ) when R islarge. We may thus assume that R ≥ (cid:96) + 1 . (7.5)In the next proposition, we obtain such a bound in case the distance d ( f, h R )is reached at some point m which is far away from the convex core, that is if m / ∈ V . The case where m ∈ V will be carried out in the next paragraphs. Proposition 7.3
Suppose that the distance d ( f, h R ) = d ( f ( m ) , h R ( m )) isreached at some point m ∈ V R (cid:114) V . Then d ( f, h R ) ≤ ρ + B(cid:96) , where B is the constant in Proposition 4.11, and ρ is defined in Fact 7.1. Proof
Assume first that d ( m, ∂V R ) ≤ (cid:96) . It follows from Proposition 4.11that d ( f, h R ) = d ( f ( m ) , h R ( m )) ≤ B d ( m, ∂V R ) ≤ B(cid:96) . Assume now that d ( m, ∂V R ) > (cid:96) , so that the ball B ( m, (cid:96) ) lies in V R . Since m / ∈ V , Condition (7.3) ensures that this ball B ( m, (cid:96) ) is isometric to aball with radius (cid:96) in the Hadamard manifold X . Fact 7.1 applies to therestrictions F = f | B ( m,(cid:96) ) and H = ( h R ) | B ( m,(cid:96) ) , so that ρ ≤ ρ . The resultfollows. (cid:3) .3 Part two : estimate close to the core, an overview To complete the proof of Proposition 4.8, we assume from now on that thedistance ρ = d ( f, h R ) is reached at some point m that belongs to the fixedcompact set V . We pick a large (cid:96) (namely (cid:96) will have to satisfy Conditions(7.6), (7.11) and (7.13)), and we will mainly work on the compact convexset with smooth boundary V (cid:96) which is the (cid:96) -neighbourhood of V . Note that V (cid:96) does not depend on R .We will assume that (cid:96) ≥ d V + 1 , (7.6)where d V denotes the diameter of the domain V . Since we want an upperbound for the distance ρ = d ( f, h R ) when R is large, we may also assumefrom now on that ρ ≥ c (7.7) R ≥ (cid:96) + 2 . (7.8)For any point p in the 1-neighbourhood of V (cid:96) , namely for p ∈ V (cid:96) +1 , Condition(7.8) ensures that B ( p, ⊂ V R , so that Corollary 4.10 yields (cid:107) D p h R (cid:107) ≤ κρ . (7.9)We introduce y = f ( m ) ∈ Y , which is the image under f of the point m ∈ V where the distance d ( f, h R ) is reached. For any point p ∈ ∂V (cid:96) , we shall studythe three following Gromov products relative to this point y : g ( p ) = ( f ( p ) , h R ( m )) y , g ( p ) = ( f ( p ) , h R ( p )) y , g ( p ) = ( h R ( p ) , h R ( m )) y . V y=f(m) f(p)h (p) R (m) R h g (p) pm V g (p) ∂ R V If ρ is large, we shall prove that on a suitable subset U (cid:96),R of the boundary ∂V (cid:96) , both g and g are large (Lemma 7.8 and Corollary 7.10) while themeasure of U (cid:96),R is large enough (Lemma 7.5) to ensure that g cannot bethat large on the whole U (cid:96),R (Lemma 7.6). This will yield a contradictionthanks to Inequality (2.1) satisfied by Gromov products.37he arguments we develop here are similar to those of [6, Section 4], andthat led to Fact 7.1. In the setting of our previous paper [6], they relied onthe uniform upper and lower bounds for harmonic measures of balls in theHadamard manifold relative to their center. In our new context, they relyon the uniform upper and lower bounds for harmonic measures obtained inProposition 5.2 and Corollary 6.20. U (cid:96),R ⊂ ∂V (cid:96) We introduce the domain U (cid:96),R ⊂ ∂V (cid:96) that will play a central role in theproof of Proposition 4.8. Definition 7.4
Let U (cid:96),R be the set of those points p ∈ ∂V (cid:96) where the dis-tance d ( y, h R ( p )) is close to ρ = d ( y, h R ( m )) , namely U (cid:96),R = { p ∈ ∂V (cid:96) | d ( y, h R ( p )) ≥ ρ − (cid:96) c } . In the next lemma, we give a lower bound for the “size” of the domain U (cid:96),R ⊂ ∂V (cid:96) , which is uniform with respect to R and to the choice of (cid:96) . Lemma 7.5
The harmonic measure σ m,V (cid:96) of the set U (cid:96),R ⊂ ∂V (cid:96) relative tothe point m ∈ V satisfies σ m,V (cid:96) ( U (cid:96),R ) ≥ c . Proof
We first observe that, for any point q ∈ V (cid:96) , one has d ( y, h R ( q )) ≤ ρ + 2 c (cid:96) . (7.10)Indeed, since by (4.1) the function f is c -Lipschitz, the triangle inequalityyields d ( f ( m ) , h R ( q )) ≤ d ( f ( m ) , f ( q )) + d ( f ( q ) , h R ( q )) ≤ c ( d V + (cid:96) ) + ρ where as usual d V denotes the diameter of V . Thus (7.10) follows, since weassumed in (7.6) that (cid:96) ≥ d V .Now, Lemma 3.5 asserts that the function u : q → d ( y, h R ( q )) − ρ issubharmonic on V R . Moreover (7.10) ensures that u is bounded above by2 c (cid:96) on V (cid:96) . Since u ( m ) = 0, this yields0 ≤ (cid:90) ∂V (cid:96) u ( p ) dσ m,V (cid:96) ( p ) ≤ σ m,V (cid:96) ( U (cid:96),R ) (2 c (cid:96) ) − (1 − σ m,V (cid:96) ( U (cid:96),R )) (cid:96) c , thus 1 ≤ σ m,V (cid:96) ( U (cid:96),R ) (1 + 4 c ) hence the claim, since we assumed c ≥ (cid:3) .5 Upper bound for the Gromov product g ( p ) on U (cid:96),R In this paragraph we prove that, if (cid:96) is large enough, the Gromov products g ( p ) = ( f ( p ) , h R ( m )) y cannot be uniformly large on the whole U (cid:96),R .We first prove that the image f ( U (cid:96),R ) ⊂ Y , seen from the point y = f ( m ),is relatively spread out. Lemma 7.6
There exists a distance ¯ (cid:96) and a constant ¯ g > such that, if (cid:96) ≥ ¯ (cid:96) , (7.11) then there exist two points p , p ∈ U (cid:96),R ⊂ ∂V (cid:96) with ( f ( p ) , f ( p )) y ≤ ¯ g . Corollary 7.7 If (cid:96) ≥ ¯ (cid:96) , then there exists a point p ∈ U (cid:96),R such that g ( p ) ≤ ¯ g + 2 δ Y . Proof
Follows from Lemma 7.6 and Inequality (2.1). (cid:3)
Proof of Lemma 7.6
We first construct the points p , p ∈ U (cid:96),R .We proved in Lemma 7.5 that the harmonic measure σ m,V (cid:96) of U (cid:96),R isbounded below by 1 / (5 c ). It thus follows from (7.2) that, for any choice of (cid:96) , there exists an index 1 ≤ j ≤ N with σ m,V (cid:96) ( U (cid:96),R ∩ D t j + τ ζ j ) ≥ / (5 N c ).Fix t such that C e − αt < / (5 N c ). Assume moreover that t ≥ t + τ .Let ¯ (cid:96) > ∂V (cid:96) ⊂ ∪ ξ ∈ Ω D t + τ ξ for each (cid:96) ≥ ¯ (cid:96) . Pick apoint p ∈ U (cid:96),R ∩ D t j + τ ζ j and choose ξ ∈ Ω such that p ∈ D t + τ ξ . Proposition6.21, that applies since t ≥ t + τ , and Corollary 6.20 ensure that σ m,V (cid:96) ( { p ∈ U (cid:96),R | ( p , p ) m ≥ t + g } ) ≤ σ m,V (cid:96) ( D tξ ) ≤ C e − αt < / (5 N c ) . Hence, there exists p ∈ U (cid:96),R ∩ D t j + τ ζ j such that ( p , p ) m ≤ t + g .We now turn our attention to the two images f ( p ) , f ( p ). If f : M → Y were assumed to be quasi-isometric with constant c , we would infer imme-diately from Lemma 2.3 that ( f ( p ) , f ( p )) y ≤ ¯ g , with ¯ g = c ( t + g ) + A .But under the hypotheses of Theorem 1.1, where we only assume thatthe restriction of f to the half-space D t j ζ j is a c quasi-isometric map, wehave to make a slight adjustment to this elementary proof. For simplicityof notation, let us denote by m j = m t j + τ ζ j . Introduce the compact set W = V ∪ { m , · · · , m N } , and let d W be its diameter. Observe that( p , p ) m j ≤ ( p , p ) m + d W ≤ t + g + d W .
39e are now ready to use the fact that the restriction of f to D t j ζ j is aquasi-isometric map. Indeed, the three points p , p , m j belong to the half-space D t j + τ ζ j , whose convex hull is included in D t j ζ j (Proposition 6.16). Thisconvex hull being isometric to a convex subset of X , Lemma 2.3 yields( f ( p ) , f ( p )) f ( m j ) ≤ c ( t + g + d W ) + A .
To prove our claim, replace the origin f ( m j ) with y = f ( m ) in this Gromovproduct, observing that, since f is c -Lipschitz, we have d ( f ( m j ) , f ( m )) ≤ c d W . (cid:3) g ( p ) on U (cid:96),R ⊂ ∂V (cid:96) Our second estimate for the Gromov products is the only one that relies onthe left-hand side of Condition (1.1).
Lemma 7.8
There exists a constant ¯ g such that g ( p ) = ( f ( p ) , h R ( p )) y ≥ (cid:96) c − ¯ g holds for every p ∈ U (cid:96),R . Proof
In case the map f is c quasi-isometric on the whole M , the proof isstraightforward. Indeed, using the bound d ( f ( p ) , h R ( p )) ≤ ρ , the definitionof U (cid:96),R and observing that, since m ∈ V and p ∈ ∂V (cid:96) , one has d ( m, p ) ≥ (cid:96) ,we obtain2( f ( p ) , h R ( p )) y = d ( f ( p ) , f ( m )) + d ( h R ( p ) , f ( m )) − d ( f ( p ) , h R ( p )) ≥ ( (cid:96)c − c ) + ( ρ − (cid:96) c ) − ρ = (cid:96) c − c . When the map f is not supposed to be globally quasi-isometric, weproceed as in the proof of Lemma 7.6 and introduce the index 1 ≤ j ≤ N such that p ∈ D j , so that d ( f ( p ) , f ( m j )) ≥ (1 /c ) d ( p, m j ) − c . The same computation as above gives2( f ( p ) , h R ( p )) f ( m j ) ≥ (cid:96) c − c − c d W . Using again the fact that f is c -Lipschitz on M to change the base point,we obtain the result. (cid:3) .7 Lower bound for the Gromov product g ( p ) on ∂V (cid:96) This last estimate relies on the uniform lower bound for the harmonic mea-sures of a family of suitable subdomains of M proven in Paragraph 5.2. Lemma 7.9
There exists a constant ρ ( (cid:96) ) > that depends only on (cid:96) (andnot on R ) and such that if ρ > ρ ( (cid:96) ) (7.12) then d ( y, h R ( q )) ≥ ρ/ holds for every point q ∈ V (cid:96) . Proof
Assume that there exists a point q ∈ V (cid:96) where d ( y, h R ( q )) < ρ/
2. Be-cause of the bound (7.9) for the covariant derivative of h R on V (cid:96) +1 , it followsthat one has d ( y, h R ( z )) ≤ ρ/ z in the ball B ( q, / κ ).Let λ = 1 / κ and L = 2 (cid:96) , so that (7.6) yields 0 < d ( m, q ) ≤ L . Withthe notation of Proposition 5.2, we introduce the constant s ( (cid:96) ) = s ( λ, L ).We proceed as in the proof of Lemma 7.5. Consider the subharmonicfunction u : z → d ( y, h R ( z )) − ρ on the domain W m,q we introduced inProposition 5.2. This function u vanishes at the point m . We just provedthat u ≤ − ρ/ B ( q, λ ), while thanks to (7.6) : u ( z ) ≤ d ( f ( m ) , f ( z )) + d ( f ( z ) , h R ( z )) − ρ ≤ c (cid:96) for any point z ∈ V (cid:96) +1 , and in particular on the boundary ∂W m,q . We thusinfer that 0 ≤ − ρ σ m,W m,q ( B ( q, λ )) + 8 c (cid:96) , hence ρ ≤ c(cid:96)/s ( (cid:96) ). This provesour claim, with ρ ( (cid:96) ) = 8 c(cid:96)/s ( (cid:96) ). (cid:3) Corollary 7.10
There exists a constant ¯ g ( (cid:96) ) , that depends on (cid:96) but not on R , such that if ρ > ρ ( (cid:96) ) g ( p ) = ( h R ( m ) , h R ( p )) y ≥ ρ − ¯ g ( (cid:96) ) log ρ holds for any point p ∈ ∂V (cid:96) . Proof
Let p ∈ ∂V (cid:96) and let [ m, p ] be a minimizing geodesic segment from m to p . By Assumption (7.6), its length is at most d V + (cid:96) ≤ (cid:96) . We inferfrom the bound (7.9) for the covariant derivative of h R that the length ofthe curve h R ([ m, p ]) ⊂ Y is at most 2 (cid:96)κρ .Since this curve stays away from the large ball B ( y, ρ/ y . Indeed, select a subdivision ( z i ) ≤ i ≤ n − of h R ([ m, p ]),with 1 ≤ d ( z i , z i +1 ) ≤
2. One thus has n ≤ log ρ + ν for some constant ν >
0. Since d ( z i , z i +1 ) ≤
2, Lemma 7.9 gives ( z i , z i +1 ) y ≥ ρ/ − ≤ i ≤ n − . Then, the triangle inequality for Gromov products (Lemma2.2) yields ( z i , z i +1) ) y ≥ ρ/ − − δ Y when 0 ≤ i ≤ n − . Iterating theprocess yields ( h R ( m ) , h R ( p )) y ≥ ρ − − n δ Y as claimed. (cid:3) .8 Proof of Proposition 4.8 We now prove that, if ρ is too large, the estimates for the three Gromovproducts g , g and g that we obtained in the previous sections lead to acontradiction, thus completing the proof of Proposition 4.8 and of our maintheorem 1.1.Let us first stress the fact that both constants ¯ g and ¯ g do not dependon R , ρ nor (cid:96) , while ¯ g depends on (cid:96) but not on R nor ρ .We begin by choosing a radius (cid:96) large enough to satisfy (7.6) and (7.11),as well as (cid:96) c − ¯ g − δ Y > ¯ g , (7.13)where ¯ g and ¯ g are defined in Lemmas 7.6 and 7.8.We let R ≥ (cid:96) + (cid:96) + 2, so that Conditions (7.5) and (7.8) are satisfied.Assume by contradiction that the distance ρ := ρ R = d ( f, h R ) is very large,namely that ρ satisfies (7.7), (7.12) and ρ − ¯ g ( (cid:96) ) log ρ ≥ (cid:96) c − ¯ g . (7.14)We proved in Lemma 7.8 and Corollary 7.10 that, under these assumptions :( f ( p ) , h R ( p )) y = g ( p ) ≥ (cid:96) c − ¯ g when p ∈ U (cid:96),R ( h R ( m ) , h R ( p )) y = g ( p ) ≥ ρ − ¯ g ( (cid:96) ) log ρ when p ∈ ∂V (cid:96) .Thus (2.1) and (7.14) yield, for any point p ∈ U (cid:96),R , the lower bound( f ( p ) , h R ( m )) y = g ( p ) ≥ min( g ( p ) , g ( p )) − δ Y ≥ (cid:96) c − ¯ g − δ Y > ¯ g + 2 δ Y thanks to our choice of (cid:96) in (7.13). This is a contradiction to Corollary 7.7.This ends the proof of Proposition 4.8 in case m / ∈ V .The case where m ∈ V has already been dealt with in Proposition 7.3. (cid:3) eferences [1] A. Ancona. Negatively curved manifolds, elliptic operators, and the Martinboundary. Ann. of Math. , 125:495–536, 1987.[2] M. Anderson. The Dirichlet problem at infinity for manifolds of negativecurvature.
J. Differential Geom. , 18:701–721 (1984), 1983.[3] M. Anderson and R. Schoen. Positive harmonic functions on complete mani-folds of negative curvature.
Ann. of Math. , 121:429–461, 1985.[4] Y. Benoist and D. Hulin. Harmonic quasi-isometric maps between rank onesymmetric spaces.
Ann. of Math. , 185:895–917, 2017.[5] Y. Benoist and D. Hulin. Harmonic measures on negatively curved manifolds.
Annales Inst. Fourier , 69:2951–2971, 2019.[6] Y. Benoist and D. Hulin. Harmonic quasi-isometric maps II: negatively curvedmanifolds.
JEMS , 2020. To appear.[7] S. Blach`ere, P. Ha¨ıssinsky, and P. Mathieu. Harmonic measures versus qua-siconformal measures for hyperbolic groups.
Ann. Sci. ´Ec. Norm. Sup´er. (4) ,44, 2011.[8] B. H. Bowditch. Geometrical finiteness with variable negative curvature.
DukeMath. J. , 77:229–274, 1995.[9] S. Cheng. Liouville theorem for harmonic maps. In
Geometry of the Laplaceoperator , pages 147–151. Amer. Math. Soc., 1980.[10] J. Eells and L. Lemaire. A report on harmonic maps.
Bull. London Math.Soc. , 10:1–68, 1978.[11] J. Eells and J Sampson. Harmonic mappings of Riemannian manifolds.
Amer.J. Math. , 86:109–160, 1964.[12] E. Ghys and P. de la Harpe.
Sur les groupes hyperboliques d’apr`es MikhaelGromov . Progress in Mathematics. Birkh¨auser, 1990.[13] R. Greene and H. Wu. C ∞ convex functions and manifolds of positive curva-ture. Acta Math. , 137:209–245, 1976.[14] F. Guritaud. Applications harmoniques en courbure ngative, d’aprs Benoist,Hulin, Markovic... hal-02410011, 2019.[15] M. Kapovich. A note on Selberg’s Lemma and negatively curved Hadamardmanifolds. arXiv:1808.01602, 2018.[16] Y. Kifer and F. Ledrappier. Hausdorff dimension of harmonic measures onnegatively curved manifolds.
Trans. Amer. Math. Soc. , 318(2):685–704, 1990.[17] M. Lemm and V. Markovic. Heat flows on hyperbolic spaces.
Journal Diff.Geom. , 108:495–529, 2018.[18] P. Li and J. Wang. Harmonic rough isometries into Hadamard space.
AsianJ. Math. , 2:419–442, 1998.[19] T. Lyons and D. Sullivan. Function theory, random paths and covering spaces.
J. Differential Geom. , 19:299–323, 1984.
20] V. Markovic. Harmonic maps between 3-dimensional hyperbolic spaces.
In-vent. Math. , 199:921–951, 2015.[21] V. Markovic. Harmonic maps and the Schoen conjecture.
J. Amer. Math.Soc. , 30:799–817, 2017.[22] H. Pankka and J. Souto. Harmonic extensions of quasiregular maps. ArXiv:1711.08287, 2017.[23] J. Reshetnyak. Non-expansive maps in spaces of curvature no greater than K.
Sibirsk. Mat. Z. , 9:918–927, 1968.[24] R. Schoen. The role of harmonic mappings in rigidity and deformation prob-lems. In
Complex geometry , pages 179–200. Dekker, 1993.[25] H. Sidler and S. Wenger. Harmonic quasi-isometric maps into Gromov hyper-bolic CAT(0)-spaces. ArXiv: 1804.06286, 2018.[26] S.T. Yau. Harmonic functions on complete Riemannian manifolds.
Comm.Pure Appl. Math. , 28:201–228, 1975.Y. Benoist & D. Hulin, CNRS & Universit Paris-SaclayLaboratoire de mathmatiques dOrsay, 91405, Orsay, [email protected] & [email protected], 28:201–228, 1975.Y. Benoist & D. Hulin, CNRS & Universit Paris-SaclayLaboratoire de mathmatiques dOrsay, 91405, Orsay, [email protected] & [email protected]