Asymptotic approximation of harmonic maps to Teichmüller space

Transcription

1 Asymptotic approximation of harmonic maps to Teichmüller space Georgios Daskalopoulos 1 Brown University daskal@math.brown.edu Chikako Mese Johns Hopkins University cmese@math.jhu.edu Last Revision: September 5, 013 Abstract This is the first in the series of papers that studies the behavior of harmonic maps into the metric completion T of Teichmüller space with respect to the Weil-Petersson metric. This current paper studies harmonic maps from a smooth n-dimensional Riemannian domain to a space modeling the normal space to the boundary strata of T. The main result is that the singular set of a harmonic map into two copies of the model space is n 1)- rectifiable. The importance of this result is that harmonic maps into the Teichmüller space at order 1 points are asymptotically approximated by the behavior of harmonic maps into two copies of the model space. This regularity result will be used in the sequels [DM1], [DMW] to study rigidity properties of Teichmüller space. 1 Introduction Consider the smooth Riemannian manifold H, g H ) where H = {ρ, φ) R : ρ > 0, φ R} is the right half-plane and the Riemannian metric g H is given in coordinates ρ, φ) by g H = dρ + ρ 6 dφ. 1 ported by research grant from the Simons Foundation ported by research grant NSF DMS

2 Denote by d H the distance function induced by the metric g H. The Riemannian manifold H, g H ) is incomplete; namely, the boundary {ρ = 0} is at a finite distance from any point P = ρ, φ) H. The metric completion H of H is obtained by adding the boundary {ρ = 0}. More precisely, identify {ρ = 0} as a single point P 0 and define H = H {P 0 }. The distance function d H is extended to H by setting d H P, P 0 ) = ρ for P = ρ, φ) H. The following facts are easy to check cf. [DW]): 1) The surface H, g H ) is geodesically convex. ) The complete metric space H, d) is an NPC space. 3) The space H, d) is not locally compact. Here, recall that NPC means non-positively curved in the sense of Alexandrov. We now let H, d) be the metric space constructed by taking two copies H + and H of H and identifying the point P 0. More precisely, H = H + H / where identifies the P 0 H + with P 0 H. The distance function d on H is defined by setting dp 1, P ) = d H P 1, P ) if P 1, P H ± and dp 1, P ) = ρ 1 + ρ if P 1 = ρ 1, φ 1 ) H + = H + \{P 0 } and P = ρ, φ ) H = H \{P 0 }. The metric space H, d) is an NPC space cf. [BH]). The current paper concerns harmonic maps into the metric space H, d). The motivation for understanding H comes from the study of certain rigidity properties of Teichmüller space which will be explored in the succeed articles. We now explain the relationship between Teichmüller space and H. Let T g,n denote the Teichmüller space of a genus g Riemann surfaces with n punctures and 3g 3 + n > 0 endowed with the Weil-Petersson metric and let T g,n denote its metric completion. It is well known that T g,n is a stratified space where the strata are themselves Teichmüller spaces, possibly of lower genus. The fundamental relation between T g,n and H is that metrically the normal space to the boundary strata of T g,n are products of H. More precisely, at a codimension m stratum, the Weil-Petersson metric is asymptotic to the product of the Weil-Petersson metric on the stratum and H m = H... H cf. [DW], [Wo], [Ya] and [DM3]). Note that in Weil-Petersson geometry, one considers the slightly different metric 4dr + r 6 dθ which is

3 clearly isometric to the one considered here via the change of coordinates ρ = r, φ = θ 8.) One of the difficulties of working with T g,n is that it is not locally compact. Specifically, a geodesic ball centered at a point on T g,n = T g,n \T g,n is not compact. This property is captured by the model space H at the point P 0. The non-compactness means that a tangent space T P0 H of H at P 0 which is isometric to the interval [0, )) does not properly reflect the geometry of H in the neighborhood of P 0. Indeed, the tangent map of a harmonic map at a point in the pre-image of P 0 does not map into T P0 H. Instead, the image of a tangent map can be embedded into a metric space constructed by joining a multiple copies of H at the boundary point P 0 cf. [W]). In particular, the image of a tangent map at an order 1 point is embedded in H. We show precisely how the local image of a harmonic map is reflected by the geometry of H in our follow-up paper [DM1]. This is the main reason for the interest in the behavior of harmonic maps into H, d). The main result of this paper is the following. Theorem 1 If u : Ω, g) H, d) is a harmonic map from an n- dimensional smooth Riemannian domain, then the set u 1 P 0 ) is n 1)- rectifiable. Theorem 1 is the first step in proving rigidity results for the Teichmüller space and the mapping class group. Indeed, as in all problems of geometric errigidity the basic problem is to show that the singular set of a harmonic map u : Ω T g,n sufficiently small so that one can apply Bochner techniques. First, it follows from [DM] that the set of singular points of a harmonic map u : Ω T g,n with order greater than 1 is of Hausdorff co-dimension at least. Second, if x 0 is a point of order 1 such that ux 0 ) maps into a top dimensional stratum of the boundary of T g,n, then u behaves asymptotically like a map into H, d). Thus harmonic maps into H, d) form an asymptotic local model for maps into T g,n near points of order 1. For a definition of the order we refer to the next section.) Harmonic maps Let Ω, g) be a smooth Riemannian domain and H be the NPC space defined in Section 1. For a map v : Ω, g) H, d), let v be the 3

4 energy density as defined in [KS1]. The energy of v in a subset Ω is E v [Ω ] = v dµ <. Ω The map v is said to be harmonic if it is energy minimizing in every bounded subset Ω of Ω with respect to all finite energy maps with the same trace cf. [KS1]). Let B R 0) = {x = x 1,..., x n ) R : x < R}, g a smooth Riemannian metric on B R 0) and u : B R 0), g) H be a harmonic map. In a neighborhood of x B R 0) {u ρ 0}, u maps into a smooth Riemannian manifold either H + or H ), and we can write u = u ρ, u φ ) in terms of coordinates ρ, φ). The local Lipschitz continuity of u cf. [KS1] Theorem.4.6) implies that, for every r 0, R), there exists a constant C dependent only on r and the total energy of u such that u ρ C and u 3 ρ u φ C in B r 0) {u ρ 0}. 1) Furthermore, in B r 0) {u ρ 0}, u satisfies the harmonic map equations u ρ u ρ = 3u 6 ρ u φ and u 6 ρ u φ = 6u 5 ρ u ρ u φ. ) Moreover, u satisfies the harmonic map equations ) weakly in B r 0). The harmonic map u : B R 0), g) H has the following important monotonicity formula. Define E u σ) := u dµ and I u σ) := d u, u0))dσ. B σ0) B σ0) There exists a constant c > 0 depending only on the C norm of the metric on g with c = 0 when g is the standard Euclidean metric) such that σ Ord u 0, σ) := e σ cσ Eu σ) I u is non-decreasing 3) σ) for any x B r 0) and σ > 0 sufficiently small. As a non-increasing limit of continuous functions, Ord u 0) := lim σ 0 Ord u 0, σ) 4

5 is an upper semicontinuous function and Ord u 0) 1. The value Ord u 0) is called the order of u at 0. See Section 1. of [GS] with [KS1] and [KS] to justify various technical steps.) A point 0 is called an order 1 point if Ord u 0) = 1 and a higher order point if Ord u 0) > 1. The homogeneous cordinates ρ, Φ) of H ± are defined by setting Φ = ρ 3 φ. It can be easily seen that the metric g H is invariant under the scaling ρ λρ, Φ λφ. Here, we remark that the scaling is also equivalently given by ρ λρ, φ λ φ. 4) Thus, the distance function of H is homogeneous of degree 1 under this scaling. More precisely, for P given by ρ, Φ) in homogeneous coordinates if P P 0 and λ 0, ), we denote by λp the point given by λρ, λφ) and λp 0 = P 0. Then dλp, λq) = λdp, Q). Furthermore, if γs) is an arclength parameterized geodesic, then s λ 1 γλs) is also an arclength parameterized geodesic. Lemma In homogeneous coordinates ρ, Φ), the metric is given by or equivalently g H = 1 + 9Φ ρ 3ρ 1 Φ 3ρ 1 Φ 1 g H = 1 + 9Φ ρ )dρ 3ρ 1 ΦdρdΦ 3ρ 1 ΦdΦdρ + dφ = dρ + 3Φρ 1 dρ dφ) Proof. Let ˆρ = ρ and write g H = aˆρˆρ dˆρ + aˆρφ dˆρdφ + a Φˆρ dφdˆρ + a ΦΦ dφ 5

6 and note that and Thus, ˆρ = ˆρ ρ ˆρ + φ ˆρ φ = ρ 3Φˆρ 4 φ Φ = ρ Φ ρ + φ Φ φ = ˆρ 3 φ. aˆρˆρ = g H ˆρ, ˆρ ) = 1 + 9Φ ρ, a ΦΦ = 1 and q.e.d. aˆρφ = 3ρ 1 Φ = a Φˆρ. Convention 3 For the rest of the paper, we apply the change of variables ρ, φ) ρ, φ) to obtain new coordinates for H. Furthermore, 0, φ) for any φ R is identified with the point P 0. Thus, for ρ, φ) H, ρ > 0 implies H +, ρ < 0 implies H and ρ = 0 implies ρ, φ) = P 0. Lemma 4 For P 1 = ρ 1, φ 1 ), P = ρ, φ ) H, ρ 1 ρ dp 1, P ) ρ 1 ρ + Φ 1 Φ where Φ 1 = ρ 3 1 φ 1 and Φ = ρ 3 φ. Proof. Let γ = γ ρ, γ φ ) : [0, 1] H be a geodesic from P 1 to P. Then ρ 1 ρ 1 0 γ ρ ds 1 0 γ gh ds = dp 1, P ). Next, note that since ρ, φ) ρ, φ φ 1 ) is an isometry, we can assume without the loss of generality that φ 1 = 0. Along the line Φ = ρ 3 φ = 0, the metric g H is given in coordinates ρ, Φ) by g H = Thus, setting αt) = 1 t)ρ 1 + tρ, 0), we have 1 0 α ρ gh dt = ρ 1 ρ 6

7 Furthermore, setting βt) = ρ, tφ ), we have β Φ gh = Φ. The join of the curves αt) with 0 t 1 and βt) with 0 t 1 connects P 1 to P. Thus, we obtain dp 1, P ) 1 1 α ρ gh dt + β Φ gh dt 0 0 = ρ 1 ρ + Φ. q.e.d. Definition 5 A smooth Riemannian metric g on B R 0) R n is said to be normalized if the standard Euclidean coordinates x 1,..., x n ) are normal coordinates of g. The metric g s for s 0, R] on B 1 0) is defined by g s x) = gsx). Given R 0, 1], a normalized metric g on B R 0) and a harmonic map u : B R 0), g) H, d), the homogeneous coordinates can be used to define blow up maps of u at 0. More precisely, we write u = u ρ, u Φ ) in coordinates ρ, Φ). For σ 0, R], define a harmonic map which will be referred to as a blow-up map) by setting u σ = u σρ, u σφ ) : B 1 0), g σ ) H, d) 5) u σρ x) = µ 1 σ)u ρ σx) and u σφ x) = µ 1 σ)u Φ σx) or equivalently, writing u σ = u σρ, u σφ ), u σρ x) = µ 1 σ)u ρ σx) and u σφ x) = µ σ)u φ σx) where µσ) = I u σ). 6) σn 1 7

8 The choice of the scaling constant µσ) implies that I uσ 1) = d u σ, P 0 )dσ = 1. 7) B 1 0) By the monotonicity property stated above, E uσ 1) Ord u 0) for σ > 0 sufficiently small. Thus, by [KS1] Theorem.4.6, {u σ } has a uniform modulus of continuity. In turn, this implies that given a sequence u σi with σ i 0, there exists a subsequence converging locally uniformly in the pullback sense to a map u : B 1 0) Y, d ) into an NPC space cf. [KS1] Proposition 3.7). In particular, du σi ), u σi )) d u ), u )) uniformly on compact sets. Following [GrSc], we can show that u is a homogeneous map of degree α = Ord u 0), i.e. du x), u 0)) = x α du x x, u0)) and the curve t u tx) is a geodesic in H for each x B 1 0). By [W], see also [DM] Section 4), u is piecewise linear, i.e. the distance function du, u 0)) is a linear function in each connected component of {x B 1 0) : u x) u 0)}. In particular, if 0 is an order 1 point, then the pullback distance function of u is equal to that of a linear function fx) = Ax 1 8) where A is a constant. Given a harmonic map u : Ω, g) H, d) from a smooth Riemannian domain and x Ω, define Ord u x) and a blow-up map u σ at x for σ > 0 sufficiently small by identifying x = 0 by normal coordinates. The set of higher order points hs the following property. S 0 u) = {x Ω : Ord u x) > 1} Theorem 6 [DM] Theorem ) If u : Ω, g) H, d) is a harmonic map from a smooth Riemannian domain, then the set of higher order points is of Hausdorff codimension at least ; i.e. dim H S 0 u)) n. Our main result Theorem 1) claims that the set u 1 P 0 ) is n 1)- rectifiable. Thus, by the virture of Theorem 6, we only need to consider the set of order 1 points. For order 1 points, we have the following. 8

9 Lemma 7 If u : Ω, g) H, d) is a harmonic map from a smooth Riemannian domain and x Ω such that ux) = P 0 and Ord u x) = 1, then there exists a sequence σ i 0, a rotation R : R n R n, a sequence of homogeneous degree 1 maps l σi : B 1 0) H defined by Ax 1, φ + σ i ) x 1 > 0 l σi x) = P 0 x 1 = 0 Ax 1, φ σ i ) x 1 < 0 for a constant A > 0 and sequences {φ + σ i }, {φ σ i } such that, for any r 0, 1), lim du σi R, l σi ) = 0 i B r0) where u σi are the blow up maps u at x. Proof. As mentioned above, after a rotation of the domain if necessary, we may assume that the sequence of blow up maps u σi converges locally uniformly in the pullback sense to a linear function fx) = Ax 1 as in 8). The assertion follows from applying [DM] Lemma 14 to B 1 ± 0) = {x B 10) : x 1 ) 0}. q.e.d. 9) 3 Essential Regularity A harmonic map between two C Riemannian manifolds is C whenever one assumes an NPC condition on the target. On the other hand, one of the difficulties in the application of harmonic maps into singular targets is that these maps lack the regularity of harmonic maps between smooth spaces. For a harmonic map into a generic NPC space, one can reasonably only expect local Lipschitz regularity. However, stronger regularity results can be obtained by assuming additional structure on the target space. This is the idea exploited in [GrSc]; indeed, if the target is an m-dimensional Euclidean building, then a harmonic map is regular in a neighborhood of an order 1 point. Moreover, a neighborhood of such a point is mapped into a flat, i.e. a closed convex subspace isometric to an m-dimensional Euclidean space. For the space H, the role of a flat is played by the following set: H [φ 0 ] = {ρ, φ) H : φ φ 0 } where φ 0 > 0. 9

10 Since s s, φ 0 ) and s s, φ 0 ) are geodesics, H [φ 0 ] is a closed convex set. The goal of this section is to show that H [φ 0 ] satisfies a property which is similar to being essentially regular in the sense of [GrSc] Section 5. More precisely, harmonic maps into H [φ 0 ] are regular in a sense given by Theorem 10 below. We remark that so far we are unable to prove that H [φ 0 ] is essentially regular in the strict sense of [GrSc]; in particular, we do not make any assertion about the regularity of v φ in this paper. However, the weaker notion that we prove in Theorem 10 is sufficient for our applications. For a harmonic map v : B R 0) H, define v Υ := v ρ 3 v5 ρv φ and v Φ := v 3 ρv φ. The harmonic map equations ) imply that in B R 0)\v 1 P 0 ), we have and v Υ = 45 v9 ρv φ v φ 30v 3 ρv φ v ρ 1v 4 ρv φ v ρ v φ 10) v Φ = 9v 7 ρv φ v φ + 6v ρ v φ v ρ. 11) Lemma 8 Let R [ 1, 1), E 0 > 0, A 0 > 0 and a normalized metric g on B R 0) cf. Definition 5) be given. Then there exists C > 0 such that if φ 0 > 0, s 0, 1],, ϑ 0, 1] and v : B ϑr 0), g s ) H [ φ 0 ] is a ϑ harmonic map with then and E v ϑ in E 0 and v ρ L B 15ϑR 0)) A 0 ϑ, vρ v 3 φ L B 15ϑR 0)) Cφ 0, v Υ L B 15ϑR 0)) Cϑ 1 φ 0. Proof. Throughout the proof, C will denote a generic constant dependent on only E 0, A 0 and g. The assumption on the bounds of v ρ and v φ imply that v Φ L B ϑr 0)) = v 3 ρv φ L B ϑr 0)) Cϑφ 0. 10

11 Furthermore, 1) and 11) imply v Φ L B ϑr 0)) Cϑ 1 φ 0. Next, note that v Υ and v Φ satisfy the equalities 10) and 11) weakly in B R 0). Thus, by elliptic regularity, for any α 0, 1) ) ϑ v Φ L B 15ϑR 0)) C ϑ v Φ L B ϑr 0)) + v Φ L B ϑr 0)) Cϑφ 0. 1) Since v Φ = v 3 ρv φ ) = v 3 ρ v φ + 3v ρv φ v ρ, we have v 3 ρ v φ 3 v ρv φ v ρ + v Φ. The assumption on the bounds of v ρ and v φ and 1) imply that first term on the right hand side above is bounded in B 15ϑR 0) by Cφ 0. By 1), the second term is also bounded by Cφ 0. Thus, we obtain first estimate of the lemma. Combining the first estimate, the assumption on the bounds of v ρ and v φ, 1) and 10), we obtain the second estimate of the lemma. q.e.d. Definition 9 We say that a map l = l ρ, l φ ) : B 1 0) H is an almost affine map if l ρ x) = a x + b for a R n and b R, i.e. l ρ is an affine function. Theorem 10 Let R [ 1, 1), E 0 > 0, A 0 > 0 and a normalized metric g on B R 0) cf. Definition 5) be given. There exist C 1 and α > 0 with the following property: For φ 0 > 0, σ 0, 1] ϑ 0, 1], if v = v ρ, v φ ) : B ϑr 0), g σ ) H [ φ 0 ϑ ] is a harmonic map with then E v ϑ n E 0 and v ρ L B 15ϑR 0)) ϑa 0, B rϑ 0) dv, ˆl) Cr 1+α B ϑr 0) dv, L) + Crϑφ 0, r 0, R ] where ˆl = ˆl ρ, ˆl φ ) : B 1 0) H is the almost affine map given by ˆlρ x) = v ρ 0) + v ρ 0) x, ˆlφ x) = v φ x) and L : B 1 0) H is any almost affine map. 11

12 Proof. Throughout the proof, let C be a generic constant that depends on E 0, A 0 and g. Since v Υ ˆl ρ = v ρ ˆl ρ 3 v5 ρv φ, we have v Υ ˆl ρ L B Rϑ 0)) v ρ ˆl ) ρ L B Rϑ 0)) Cϑ 5 φ0 = Cϑφ 0. Thus, elliptic regularity and the second estimate of Lemma 8 imply that ϑ 1+α [ v Υ ˆl ρ )] C α B 7ϑR 8 0)) Cϑ v Υ ˆl ρ ) L B 15ϑR Cϑφ 0 + C v ρ ˆl ρ L B 15ϑR 0)). Since v ρ 0) = ˆl ρ 0), we have B rϑ 0) ϑ 0)) + v Υ ˆl ρ L B 15ϑR 0))) v ρ ˆl ρ ) Crϑ) ϑ α α φ 0 + ϑ 1 α v ρ ˆl ρ L B 15ϑR 0)) Furthermore, since Cr φ α 0 + ϑ 1 v ρ ˆl ρ L B 15ϑR 0)) v Υ ˆl ρ ) = v ρ ˆl ρ ) 3vρv 5 φ v φ 15 v4 ρvφ v ρ, the first estimate of Lemma 8 and 1) imply v ρ ˆl ρ L B 15ϑR 0)) v Υ ˆl ρ L B 15ϑR 0)) ) ). 13) Cφ 0. 14) Combined with the mean value inequality, we therefore conclude that for r 0, R ] v ρ ˆl ρ rϑ B rϑ 0) rϑ B rϑ 0) B rϑ 0) rϑ Cr α v ρ ˆl ρ ) v Υ ˆl ρ ) + Cφ 0 ) by 14)) φ 0 + ϑ 1 v ρ ˆl ρ L B 15ϑR 0)) Cr 1+α v ρ ˆl ρ + Cr 1+α ϑφ 0 + Crϑφ 0 B Rϑ 0) ) ) + Cφ 0 by 13)) Cr 1+α v ρ ˆl ρ + Crϑφ 0. 15) B Rϑ 0) 1

13 We claim v ρ ˆl ρ C v ρ L ρ + ϑφ 0. ) B ϑr 0) B ϑr 0) Assuming ) is true, the above two inequalities imply v ρ ˆl ρ Cr 1+α v ρ L ρ + Crϑφ 0, r 0, R B ϑr 0) B ϑr 0) ]. Since v φ x) = ˆl φ x), this estimate and Lemma 4 imply the lemma. We are left to prove ). First, since ˆl ρ and L ρ are affine functions and ˆl ρ 0) = v ρ 0), we have for any s 0, R ] ˆl ρ L ρ B ϑr 0) B ϑr 0) = s 1 R ) ˆl ρ L ρ ) ˆl ρ 0) L ρ 0)) + ˆl ρ 0) L ρ 0) B ϑs 0) ) ˆl ρ L ρ ) ˆl ρ 0) L ρ 0)) + ˆl ρ 0) L ρ 0) = s 1 R ) ˆl ρ L ρ ) v ρ 0) L ρ 0)) + v ρ 0) L ρ 0) B ϑs 0) s 1 R ˆl ρ L ρ + s 1 R v ρ L ρ. B ϑs 0) B ϑr 0) Apply the triangle inequality and 15) to estimate the first term above by s 1 R ˆl ρ L ρ B ϑs 0) s 1 R v ρ L ρ + v ρ ˆl ρ B ϑs 0) B ϑs 0) s 1 R v ρ L ρ + CRs α v ρ ˆl ρ + CRϑφ 0 B ϑr 0) B ϑr 0) Combining the above two inequalities and noting R 1, we obtain B ϑr 0) ˆl ρ L ρ 3s 1 B ϑr 0) Triangle inequality along with 17) imply v ρ ˆl ρ B ϑr 0) v ρ L ρ + ˆl ρ L ρ B ϑr 0) B ϑr 0) ) v ρ L ρ +Cs α v ρ ˆl ρ +Cϑφ 0. 17) B ϑr 0) 1 + 3s 1 ) v ρ L ρ + Cs α v ρ ˆl ρ + Cϑφ 0. B ϑr 0) B ϑr 0) 13

14 Choose s R such that Csα 1 to obtain ). q.e.d. 4 First order approximation The goal of this section is to prove Theorem 15 below; i.e. we prove that if a harmonic map u into H is close to a homogeneous degree 1 map, then u ρ is approximately a linear function to first order. The main step of the proof of Theorem 15 is contained in Lemma 14. We start with a preliminary lemma. Lemma 11 Given ɛ 0 0, 1), D 0 0, ɛ 0 ), θ 0, 1] and i = 0, 1,,..., if Q = ρ, φ) satisfies dq, H [ θ i ɛ 0 ) 3 θ i D 0 i ]) θi D 0 i, then ρ θ i ɛ 0 or Q H [ θ i ) 3 ɛ 0 θ i D 0 i ]). Proof. On the contrary, pose there exists a point Q = ρ, φ) with ρ θ i ɛ 0, Q / H [ θ i ) 3 ɛ 0 θ i D 0 i ]) and dq, H [ θ i ɛ 0 Let γ = γ ρ, γ φ ) : [0, 1] H be a geodesic with γ0) = Q and γ1) H [ θ i ɛ 0 ) 3 θ i D 0 i ]) θi D 0 i. ) 3 θ i D 0 i 3 ] ) where γ1) is the point in H [ θ i 3 ɛ 0 θ i D 0 ] closest to Q. We first claim i 3 min γ ρt) θi ɛ 0 t [0,1]. 18) Indeed, assume on the contrary that γ ρ t 0 ) < θi ɛ 0 for some t 0 0, 1]. Then since γ ρ 0) θ i ɛ 0, we obtain θ i ɛ 0 < γ ρ t 0 ) γ ρ 0) 14

15 t0 0 dγ ρ t0 dt dt dγ 0 dt dt dq, H [ θ i ɛ 0 ) 3 θ i D 0 i ]) θi D 0 i. This contradicts the assumption that D 0 0, ɛ 0 ) and proves 18). In turn, 18) implies that θ i ) 3 ɛ 0 γ 3 ρt), t [0, 1]. 19) Therefore θ i ) 3 ɛ 0 φ θ i ɛ 0 which in turn implies ) 3 θ i D 0 i θ i ) 3 ɛ 1 0 dγ φ 0 dt t) dt 1 γρ 6 dγ φ 0 dt t) dt 1 dγ ρ 0 dt t) + γρ 6 dγ φ dt t) = lengthγ) = dq, πq)) θi D 0 i dt φ θ i ) 3 ɛ 0 θ i D 0 i. q.e.d. Lemma 1 Given A > 0, ɛ 0 > 0, D 0 0, ɛ 0 ), a normalized metric g on B 1 0) and i = 1,,..., if il : B θ i0) H [ θ i ɛ 0 ) 3 θ i D 0 i ] and u : B θ i0) H 15

16 satisfies and u ρ Ax 1 < θ i ɛ 0 0) B θ i0) du, i l) < θi D 0 B θ i0) i, 1) then there exists σ 0 > 0 such that for all σ 0, σ 0 ) V ol gσ x B θ i0) : ux) / H [ θ i ) 3 ɛ 0 θ i D 0 i ] < 4ɛ θin 0 A. Proof. By and assumption 1) and the fact that i lx) H [ ], ɛ 3 0 i 3 we have for x B θ i0) D 0 dux), H [ θ i ɛ 0 Thus, Lemma 11 implies that x B θ i0) : ux) / H [ Furthermore, assumption 0) implies ) 3 θ i D 0 i ]) du, i l) < θi D 0 B θ i0) i. θ i ) 3 ɛ 0 θ i D 0 i ] {x B θ i0) : u ρ x) θ i ɛ 0 }. u ρ x) θ i ɛ 0 Ax 1 Ax 1 u ρ x) + u ρ x) < θ i ɛ 0 in B θ i0). Hence {x B θ i0) : u ρ x) θ i ɛ 0 } {x B θ i0) : Ax 1 < θ i ɛ 0 }. Let σ 0 > 0 be such that for σ 0, σ 0 ), q.e.d. V ol gσ {x B θ i0) : Ax 1 < θ i ɛ 0 } θ in 4ɛ 0 A.

17 Lemma 13 If u 0, u 1 : B θ i0), g) H are harmonic functions, then d u 0, u 1 ) is a weakly subharmonic function; i.e. d u 0, u 1 ) ηdx 0 B θ i0) for any η C c B θ i0)) with η 0. Proof. Apply [KS1] Lemma.4. and use the minimizing property of u 0 and u 1 to obtain d u 0, u 1 ) ηdx Qη, η) B θ i0) where Qη, η) consists of integrable terms which are quadratic in η and η. Replace η by tη, divide by t and let t 0 to obtain the desired inequality. q.e.d. Lemma 14 Inductive step) Given E 0 > 0, A > 0, δ 0 > 0 and a normalized metric g on B 1 0), there exist σ 0 > 0, θ 0, 1 8 ), ɛ 0 > 0 and D 0 > 0 satisfying D 0 θ 1 < δ 0 ) with the following property: For any harmonic map u : B 1 0), g σ ) H with σ 0, σ 0 ], u0) = P 0, E u E 0, 3) an almost affine map cf. Definition 9) il : B θ i0) H [ and a constant i δ > 0 such that θ i ɛ 0 du, i l) < θ i D 0 B θ i 0) i B θ i 0) ) 3 θ i D 0 i ] i+1 u ρ Ax 1 < i δ < θ i k=0 4) D 0 k, 17

18 there exists an almost affine map i+1l : B θ i+10) H [ and a constant i+1 δ > 0 such that du, i+1 l) < θ i+1 D 0 B θ i+1 0) i+1 B θ i+1 0) θ i+1 ɛ 0 u ρ Ax 1 < i+1 δ := i δθ + θ i+1 D 0 ) 3 θ i+1 D 0 i+1 ] i+ < θi+1 i+ k=0 5) D 0 k. Proof. Let σ 0 > 0 be chosen smaller than the constant σ 0 that appears in Lemma 1 and such that if denote V ol to be the volume with respect to the standard Euclidean metric 15 V ols) V ol g σ S) 17 V ols) for any σ 0, σ 0 ) and any smooth subset S of B 1 0). Let and the constant c 0 1 be such that B 15r 0) A 0 = A 6) f c 0 r) i 1 B r0) fdσ gσ 7) for any r [0, 1], any σ 0, 1] and any weakly subharmonic function f : B 1 0) R. For R = 1, E 0 > 0, A 0 > 0 and the metric g given in the statement of the theorem, let C 1 and α > 0 be as in Theorem 10. 8) Let θ 0, min{ 1 8, σ 0, 1 A }) sufficiently small such that Cθ < 1, 9) Cθ α < 1 5, 30) and Cθ 3 < ) 18

19 Define Finally, we let ) A ɛ 0 := 13 θ < 1. 3) c 0 D 0 := min { ɛ C, A 4, δ } 0θ By the definition of D 0, inequality ) is satisfied and also 33) 4D 0 A. 34) By the definition of D 0, the assumption 0) of Lemma 1 is satisfied. Indeed, 4), the definition 3) of ɛ 0, the inequality θ 1 A and the definition of D 0 imply D 0 ɛ 0 and, combining this with 4), we obtain u ρ Ax 1 < θ i D 0 < θ i ɛ 0. B θ i 0) Additionally, assumption 1) of Lemma 1 is contained in 4). Thus, we conclude θ i ) 3 V ol gσ x B θ i0) : ɛ 0 θ i D 0 i u φ x) < 4ɛ θin 0 A, and there exists R [ 1, 7 8 ] with the property that θ i ) 3 V ol gσ x B ɛ 0 θ i D 0 θ i R0) : i u φ x) < θin 1) 4 ɛ 0 A. Let π : H H [ θ i ) 3 ɛ 0 θ i D 0 i ] be the closest point projection map into a closed convex set) and 35) v : B θ i R0), g σ ) H [ θ i ) 3 ɛ 0 θ i D 0 i ] be the harmonic map with prescribed boundary condition v Bθ i R 0) = π uθi BR 0). 36) 19

20 ) By the definition of π and the fact that i lx) H [ θ i 3 ɛ 0 θ i D 0 ] by i assumption, we conclude dux), vx)) du, i lx)), x B θ i R0). 37) ) Since u θi x) π ux) implies θ i 3 ɛ 0 θ i D 0 u i φ x), inequality 35) implies that V ol gθ i {x B θ i R0) : ux) vx)} < θ i 1 4 ɛ 0 A. 38) Furthermore, since u, v : B θ i R0), g σ ) H are harmonic maps, Lemma 13 asserts that d u, v) is a subharmonic function. 39) We therefore conclude B 15θ i R d u, v) 0) c 0 θ i R) n 1 d u, v)dσ gθ i by 7)) B θ i R 0) In other words, 4 ɛ 0 c 0 A 4 ɛ 0 c 0 A d u, v) by 38)) B θ i R 0) d u, i l) by 37)) B θ i R 0) 4 ɛ 0 c 0 A θi D 0 i by 4)) = θ i+ D 0 i+9 by 3)). 40) B 15θ i R 0) Combining 4) and 41), we obtain B θ i 0) dv, i l) B θ i 0) du, v) θ i+1 D 0. 41) i+4 du, v) + B θ i 0) du, i l) θ i D 0. 4) i 1 We will now check that we can apply Theorem 10. First, note that the Lipschitz continuity of u implies E u θ i ) θ in E 0. Since a projection into a convex set in an NPC space is distance non-increasing, we obtain E v 0

21 θ in E 0. Next, Lemma 4, 41), 4) and 34) imply that in B 15θ i R 0), we have v ρ v ρ u ρ + u ρ Ax 1 + Ax 1 θ i+1 D 0 i+4 +θi D 0 +θ i A θ i A = θ i A 0. Thus, with il = L, i+1 l = ˆl, ϑ = θ i, r = θ and R = 1 in Theorem 10, we have by the choice of the constants in 8) that B θ i+1 0) dv, i+1 l) Cθ 1+α B θ i 0) dv, i l) + Cθ i+1 ɛ0 ) ) 3 D 0 i. 43) Using 4) to estimate the first term on the right hand side of 43), we obtain dv, i+1 l) Cθ i+1 θ α D 0 B θ i+1 0) i 1 + D Cθi+1 0 ɛ 6 0 i 8 Combined with 41), we obtain Cθ i+1 θ α D 0 i 1 + θi+1 D 0 i+5 by 33)) < θ i+1 D 0 i+4 by 30)). du, i+1 l) du, v) + B θ i+1 0) B θ i+1 0) B θ i+1 0) dv, i+1 l) < θ i+1 D 0. 44) i+3 This implies the first inequality of 5). To see that i+1 l maps into ) H [ θ i+1 3 ɛ 0 θ i+1 D 0 ], note that i+1 i+1 l φ = v φ by definition cf. Theorem 10). Since θ 0, 1 8 ), i+1 l φ x) = v φ x) θ i ɛ 0 ) 3 θ i D 0 i θ i+1 ɛ 0 ) 3 θ i+1 D 0 i+1 We now proceed with proof of the second inequality of 5). Since i+1l ρ x) and Ax 1 are both affine functions and u0) = P 0, we have for 1.

22 every x B θ i0) ˆl ρ θx) Aθx 1 = 1 θ)ˆl ρ 0) + θˆl ρ x) Ax 1 ) 1 θ) ˆl ρ 0) + θ ˆl ρ x) Ax 1 1 θ) ˆl ρ 0) + θ ˆl ρ x) u ρ x) + θ u ρ x) Ax 1 1 θ)dˆl0), u0)) + θdˆlx), ux)) + θ u ρ x) Ax 1 < 1 θ)θ i+1 D 0 i+3 + θi+ D 0 i+3 + iδθ by 44) and 4)). 45) which implies that for x B θ i+10) i+1 l ρ x) Ax 1 = ˆl ρ θx) Aθx 1 < i δθ + θ i+1 D 0 i+. This is the second inequality of 5). q.e.d. Theorem 15 Let E 0 > 0, A > 0, δ 0 > 0 and a normalized metric g on B 1 0) be given. There exists σ 0 > 0 and D 0 > 0 such that if u : B 1 0), g σ ) H is a harmonic map that satisfies σ 0, σ 0 ], u0) = P 0, E u E 0, and then du, l) < D 0 where lx) = Ax 1, 0), 46) B 1 0) s 1 u ρ Ax 1 < δ 0. B s0) Proof. For E 0, A and δ 0 given in the statement of the theorem, let σ 0 > 0, θ 0, 1 8 ), ɛ 0 > 0 and D 0 > 0 be as in Lemma 14. By letting 0l = l and 0 δ = D 0, Lemma 4 implies du, 0 l) < D 0 B 1 0) B 1 0) u ρ Ax 1 < 0 δ < D 0. Apply Lemma 14 inductively to conclude that for all i = 0, 1,,..., i u ρ Ax 1 < i δ < θ i D 0 B θ i0) k < θi D 0. 47) k=0

23 For s 0, 1], let i be a nonnegative integer such that s θ i+1, θ i ]. Then u ρ Ax 1 u ρ Ax 1 D 0 θ i < D 0 θ 1 s. B s0) B θ i0) The assertion of the lemma now follows from ) of Lemma 14. q.e.d. With future application in mind cf. [DM1]), we state a following extention of Theorem 15 below. First, we note that φ is a Jacobi field generated by the one-parameter family of unit speed geodesics ρ γ c ρ) = ρ, c) and g H is g H = 1 0, 0 φ We now consider a family of Riemannian metrics G defined on R with the standard coordinates denoted by ϱ, ϕ) such that ) 1 0 g = 0 J ϱ, ϕ) G satisfies the following conditions: A1) Let H, g denote the Riemanian manifold R, g). For any c R, the curve ϱ γ c = ϱ, c) is a unit speed geodesic with respect to g. In particular, the subspace H, g [ φ 0 ϑ ] := {ϱ, ϕ) H, g : ϕ φ 0 ϑ } of H, g is totally geodesic. The norm of the Jacobi field Moreover, J ϱ, ϕ) = ϕ g. ϕ is inf J ϱ, ϕ) ϱ ) where the infimum is taken over {ϱ, ϕ) R : ϱ ϱ 0 }. A) Let R [ 1, 1), E 0 > 0, A 0 > 0 and a normalized metric g on B R 0) cf. Definition 5) be given. There exist C 1 and α > 0 such that for any g G, the assertion of Theorem 10 is valid with H, g replacing H and H, g [ φ 0 ] replacing H ϑ [ φ 0 ]. ϑ 3

24 Theorem Let E 0 > 0, A > 0, δ 0 > 0 and a normalized metric g on B 1 0) be given. There exists c > 0, σ 0 > 0 and D 0 > 0 with the following property. If σ 0, σ 0 ], g G and u : B 1 0), g σ ) H, g is a map satisfiying u0) = P 0, E u E 0, and du, l) < D 0 where lx) = Ax 1, 0) B 1 0) then For any σ 0, σ 0 ] and a harmonic map w : B 1 0), g σ ) H, there exists a constant C > 0 such that d u, w) C d u, w)dσ gσ + cσ 4, 49) B 15σR 0) B σr 0) s 1 u ρ Ax 1 < δ 0. B s0) Proof. The ingredients of the proof of the crucial Induction step cf. Lemma 14) are Theorem 10, Lemma 1 and Lemma 13. Assumptions A) replaces Theorem 10. To complete the proof of this lemma, we need to check that we can replace Lemma 1 and Lemma 13. First, we claim that 48) implies the assertion of Lemma 11 with ) H, g replacing H and H, g [ θ i 3 ɛ 0 θ i D 0 ] replacing H i [ θ i ɛ 0 Indeed, given a curve t [0, 1] γt) = γ ϱ t), γ ϕ t)) with γ ϱ t) θi ɛ 0, we have by 48) that θ i ) 3 ɛ 0 J γ ϱ t), γ ϕ t)) ) 3 θ i D 0 i ]. Replacing 19) in the proof of Lemma 11 with the above inequality, we can arrive at the conclusion of Lemma 11. This immediately implies Lemma 1. Next, we claim that inequality 49) replaces the harmonicity of u in the proof of Lemma 14. Indeed, applying inequality 49) with σ = θ i, we obtain B 15θ i R d u θi, w) = θ i+ D 0 0) i+9 + Oθi )θ 4i 4

25 instead of inequality 40) in the proof of Lemma 14. By 3) and 33), θ 4 = C 18 c 1 0 A 1 D0. Thus, by assuming that θ in addition to satisfying 9), 30) and 31)) is chosen such that { } A 1 1 θ < min C 18 c 1, 0 37, and we obtain by Therefore, Oθ i )θ 7 1, Oθ i )θ i θ i 7 Oθ i )θ 5 θ i 7 Oθ i )D 0 < D 0 i+9. B 15θ i R d u, w) < θ D 0 0) i+8. This in turn implies 41), and we can continue with the rest of the proof of Lemma 14. q.e.d. 5 Proof of the main theorem In this section, we prove Theorem 1. Lemma 17 Let g be a normalized metric defined on B R 0) cf. Definitio 5) and u : B R 0), g) H, d) be a harmonic map with Ord u 0) = 1 and u0) = P 0. Furthermore, let A > 0 be determined by the blow up sequence given in Lemma 7. For δ 0 > 0, there exist σ > 0 and a homogeneous degree 1 map l σ : B 1 0) H satisfying l σ 0) = P 0 and such that dl σ, P 0 ) = A B 1 0) s 1 u σρ l σρ < δ 0, s 0, 1) B s0) where u σ is a blow up map of u at 0 as defined in 5). 5

26 Proof. Apply Lemma 7 to assert that, after rotating the domain if necessary, there exists a subsequence σ i 0 such that where lim du σ i, l σi ) = 0 i Ax 1, φ + σ i ) x 1 > 0 l σi x) = P 0 x 1 = 0 Ax 1, φ σ i ) x 1 < 0. For each i, define an isometry ι σi : H H by setting ρ, φ φ + σ i ) if P = ρ, φ) with ρ > 0 ι σi P ) = P 0 if P = P 0 ρ, φ φ σ i ) if P = ρ, φ) with ρ < 0. 50) In particular, we then have l := ι σi l σi = Ax 1, 0) and lim dι σ i u σi, l) = 0. i Furthermore, by the normalization 7) and the fact that Ord u x 0 ) = 1, we have that lim σ i 0 Euσ i 1) = 1. For E 0 =, A > 0 contained in 50), δ 0 > 0 and g given in the statement of the lemma, let σ 0 > 0 and D 0 > 0 be as in Theorem 15. Fix σ i 0, σ 0 ] sufficiently small such that and let Since ι σ u σ satisfies dι σi u σi, l) < D 0 and E uσ i 1) < σ = σ i, and u = ι σi u σi = ι σ u σ. u0) = P 0, E u, the lemma follows immediately from Theorem 15. q.e.d. Lemma 18 If g is a normalized metric defined on B R 0) cf. Definition 5) and u : B R 0), g) H, d) is a harmonic map with Ord u 0) = 1 and u0) = P 0, then I u I u r) := lim 0. r 0 rn+1 6

27 Proof. The fact that the limit as r 0 of the ratio Iu r) exists r n+1 follows from [GrSc] also see [DM4]). Let A be as in Lemma 17. By choosing δ 0 0, A ) in Lemma 17 and applying the triangle inequality we obtain Therefore, As < l σρ u σρ l σρ < u σρ. B s0) B s0) B s0) 0 A lim 1 s 0 s u σρ. B s0) The assertion now follows from the fact that q.e.d. I u r) r n+1 = Iu σ) σ n 1 σ Iuσ σr) σr) n+1. Remark 19 As shown in [DM5], Lemma 18 implies a strong uniqueness statement of tangent maps of u. Let g be a normalized metric defined on B R 0) cf. Definition 5) and u : B R 0), g) H, d) be a harmonic map with u0) = P 0 and Ord u 0) = 1. By virtue of Lemma 18, there exists a constant λ > 0 such that λs µs) λ 1 s where µ is defined in 6). Thus, we will consider blowup maps of u at x 0 normalized by 1 t instead of µ 1 t). These maps u t x) := 1 utx) 51) t will be referred to as the re-normalized blow up maps. We now prove the following uniqueness statement. Theorem 0 If g is a normalized metric defined on B R 0) cf. Definition 5) and u : B R 0), g) H, d) is a harmonic map with Ord u 0) = 1 and u0) = P 0. Then there exists a rotation R : R n R n and a constant A such that for all x = x 1,..., x n ) B 1 0). lim u t ρ Rx) Ax 1 = 0 t 0 B 1 0) 7

28 Proof. By Lemma 17, given δ 0 > 0, there exist σ > 0 and a homogeneous degree 1 map l σ : B 1 0) H such that u σρ l σρ < δ 0 s, s 0, 1) B s0) where we can assume after applying a rotation of the domain if necessary) that Ax 1, φ + 0 ) x1 > 0 l σ x) = P 0 x 1 = 0 Ax 1, φ 0 ) x1 < 0. In particular, with we have u s σx) = 1 s u σsx) and l s σx) = 1 s l σsx) u s σρx) Ax 1 < δ 0, s 0, 1). B 1 0) The lemma now follows immediately since σ is fixed. q.e.d. Definition 1 For a harmonic map u : Ω, g) H from a Riemannian domain, let S 1 u) := {x B 1 0) : ux) = P 0 and Ord u x) = 1}. For x S 1 u), identify x = 0 via normal coordinates and let A x and R x be the constant A and the rotation R of Theorem 0. Let l x : B 1 0) H, l x = l R where lx) = A x x 1, 0). The l x is called the unique tangent map of u at x. Theorem For a harmonic map u : Ω, g) Y from a smooth Riemannian domain and x S 1 u), let l x be the unique tangent map of u at x. Then x l x is continuous on S 1 u) with the topology of uniform convergence on compact sets. Proof. By the definition of the re-normalized blow up maps in 51) and of the unique tangent map in Definition 1, lim du t x, l x ) = 0, r 0, 1). 5) t 0 B r0) 8

29 Fix x 0 S 1 u). By applying a rotation if necessary, we can assume without the loss of generality that l x0 x) = A x0 x 1. For a given r 0, 1) and D 0 > 0, 5) implies that we can fix t > 0 sufficiently small such that du t x 0, l x0 ) < D 0 B r0). The Lipschitz continuity of u implies that for x S 1 u) sufficiently close to x 0 depending on t), du t x, u t x 0 ) < D 0 B r0). Thus, triangle inequality and the above inequality imply du t x, l x0 ) < D 0 53) B r0) for x S 1 u) sufficiently close to x 0. Given δ 0 > 0, we choose D 0 > 0 as in Theorem 15 to conclude which immediately implies Letting s 0 above, we obtain Since l xφ = 0 = l x0 φ, this implies q.e.d. u t xρ l x0 ρ < δ 0 s B sr0) u t x) s ρ l x0 ρ < δ 0. B r0) l xρ l x0 ρ < δ 0. B r0) dl x, l x0 ) < δ 0. B r0) Proof of Theorem 1. The set S 0 u) of higher order points of Hausdorff codimension at least by Theorem 6. The set S 1 u) is C 1 by Theorem. q.e.d. 9

30 References [DM1] [DMW] [DM] [DM3] [DM4] [DM5] [DMV] [DW] [GrSc] G. Daskalopoulos and C. Mese. Harmonic maps into the model space for the Weil-Petersson Metric. Preprint. G. Daskalopoulos, C. Mese and R. Wentworth. Rigidities of Teichmüller space. In preparation. G. Daskalopoulos and C. Mese. High order singularites of maps into NPC Spaces. Preprint. G. Daskalopoulos and C. Mese. C 1 -Estimates for the Weil- Petersson metric. In preparation. G. Daskalopoulos and C. Mese. On the singular set of harmonic Maps into DM-Complexes. Preprint. G. Daskalopoulos and C. Mese. Monotonicity properties of harmonic maps into NPC spaces. J. Fixed Point Theory Appl ) 543. G. Daskalopoulos, C. Mese and A. Vdovina. Superrigidity of hyperbolic buildings. Geometric and Functional Analysis 1 011) G. Daskalopoulos and R. Wentworth. Classification of Weil- Petersson isometries. Amer. J. Math., 15 4): , 003. M. Gromov and R. Schoen. Harmonic maps into singular spaces and p-adic errigidity for lattices in groups of rank one. IHES Publ. Math ) 5-46 [BH] M. Bridson and A. Haefliger. Metric spaces of Non- Positive Curvature, Grundlehren der Mathematischen Wissenschaften, Vol 319, Springer-Verlag, Berlin [KS1] [KS] N. Korevaar and R. Schoen. Sobolev spaces and harmonic maps for metric space targets. Communications in Analysis and Geometry ), N. Korevaar and R. Schoen. Global existence theorem for harmonic maps to non-locally compact spaces. Communications in Analysis and Geometry ),

31 [W] [Wo] [Ya] R. Wentworth. Regularity of harmonic maps from surfaces to the Weil-Petersson completion of Teichmüller space. Unpublished Manuscript. S. Wolpert. Geometry of the Weil-Petersson completion of Teichmüller space. Surveys in Differential Geometry, VIII: Papers in Honor of Calabi, Lawson, Siu and Uhlenbeck, editor S. T. Yau, International Press, 003. S. Yamada. Weil-Petersson completion of Teichmüller spaces and mapping class group actions. Preprint,