Harmonic maps from a Riemannian polyhedron

Transcription

1 Harmonic maps from a Riemannian polyhedron Chikako Mese Department of Mathematics Johns Hopkins University cmese@math.jhu.edu Research partially supported by grant NSF DMS Introduction In this paper, we consider study the regularity of harmonic maps from a Riemannian simplicial complex to a non-positively curved metric space. The main theorem asserts that such maps are locally Lipschitz continuous away from the (n 2)-skeleton of the complex. Furthermore, we show that the gradient of energy at a point q in a neighborhood of a point p on the (n 2)-simplex is bounded by Cr α 1 where r is the distance between p and q, α is the order of the map at p and C is a constant dependent on the energy and the distances from p to the (n 3)-skeleton and the boundary of the complex. It is well known in the the classical theory of harmonic maps between smooth Riemannian manifolds that if the target has non-positive sectional curvature, then the harmonic map is smooth. Gromov and Schoen [GS] and Korevaar and Schoen [KS1] studied harmonic maps when the target space is singular. In particular, [KS1] showed that harmonic maps into a metric space of non-positive curvature is Lipschitz continuous. When the domain space is also singular, Hölder continuity was shown in the case when the domain is a Riemannian simplicial complex and the target is a non-positively curved metric space by [Ch] and [EF]. In [DM1] and [M], Lipschitz continuity of harmonic maps away from (n 2)-skeleton of the simplicial complex was shown with the assumption that each simplex is flat, i.e. isometric to Euclidean space. That the Lipschitz continuity is crucial in applications can be seen by [DM1] and [DM2]. We hope that the regularity theory in this paper will lead some further applications of harmonic map theory between singular spaces. 1

2 2 Domain and target spaces 2.1 Polyhedral complexes Let h be a hyperplane in R n. A half space H is one of the connected component of R n h. If the intersection of half spaces H 1,..., H k defined by hyperplanes h 1,..., h k is bounded, we will refer to it as a n-dimensional polytope. For the rest of the paper, all polytopes will be of dimension n. A face of a polytope P is a non-empty intersection of the boundary of P with a linear subspace of R n. A non-empty intersection of a linear subspace of R n with the closure of a face F is itself a face. Such a face that is not equal to F will be called a subface of F. Let P be a set of the closures of polytopes. Let GM be a set of (Euclidean) isometries ϕ : F F where F, F are faces of P, P P respectively. An element ϕ GM is called a gluing map and we say P and P are glued together by ϕ along F and F. Obviously, the faces F and F must be isometric for there to exists a ϕ : F F GM. We require that if ϕ is an element of GM, then the restriction of ϕ to a subface of F is also an element of GM. We allow the possibility the GM glues together a finite number of polytopes. The elements of GM define an equivalence relation on the disjoint union of elements P of P. This disjoint union P modulo this equivalence relation, P/, is referred to as a n-dimensional polyhedron X. A k-face is the equivalence class of k-dimensional faces as defined by GM. For the rest of the paper, we will take the liberty to use k-face to mean the equivalence class of faces or a particular face in the equivalent class whenever the context is clear. We say polytopes P 1,..., P l are incident to a k-face F if F is a face of P i for i = 1,..., l. Any equivalence class containing a face of P i and with F as a subface is also said to be incident to F. Two polytopes are adjacent if they are incident to the same (n 1)-face. We say that a n-dimensional polyhedron X is chainable if given every two polytopes P, P of X, there exists a sequence of polytopes P 1 = P,..., P n = P so that P i is adjacent to P i+1 for i = 1,..., n 1. A boundary of n-dimensional polyhedron X is a union of all faces of the polytopes which is not glued to another face. An interior point of X is a point which is not on a boundary of X. We now define the space which will be the domain space of the harmonic maps. Definition A C m Riemannian n-dimensional polyhedron is a chainable n- dimensional polyhedron X along with g = {g P } so that g P is a Riemannian metric defined on each polytone P of X with the following two properties. (1) The component functions g ij (i, j = 1,..., n) of the metric g P is C m on P with an uniform elliptic bound on P. The uniform elliptic bound means that 2

3 there exists λ, Λ > 0 so that n λ(x i ) 2 i=1 n i,j=1 g ij x i x j i=1 Λ(x i ) 2. Therefore, g P extends to any face F of P as a a C m Riemannian metric g F. (2) If polytopes P and P are glued together along faces F and F, then g F and g F agree (i.e. the gluing map is an isometry of (F, g F ) and (F, g F )). Remark. The 2-dimensional complex considered in the regularity theory of [DM1] is a C m Riemannian polyhedron for any m. There, every polytope P is a Euclidean equilateral triangle and g P is the Euclidean metric. 2.2 Local models We are interested in a local regularity theory for harmonic maps from a n- dimensional polyhedron X. We construct spaces which serve as local models near a given point in X. By a normalized half space, we will mean a half space H so that the hyperplane h that defines H contains the origin 0. We say the normalized half spaces H 1,..., H k are linearly independent if the normals to the hyperplanes h 1,..., h k defining the half spaces are linearly independent. A wedge (or a k-wedge) W is the closure of the intersection of k number of linearly independent normalized half spaces H 1,..., H k. By its construction, every k-wedge W is a n-dimensional cone in R n with 0 as the vertex. A non-empty intersection of W with a linear subspace of R n will also be called a face. The intersection h 1... h k is a face which is a (n k)-dimensional linear subspace of R n. We denote this space by D. We will use the coordinates of R n to label points in W. For simplicity, we always choose the coordinate system (x 1,..., x n ) of R n so that D is given as x n k+1 =... = x n = 0. For a point x W in the interior of W, we let R(x) be the distance of x to W. If x is a point on an interior of a face F, we let R(x) be the distance of x to the closest face that is not F. Given a n-dimensional polyhedron X, let p be a point on a k-face F with l-number of polytopes P 1,..., P l incident to it. For each P i, we associate a (n k)-wedge W i. (Note that since a (n k)-wedge contains a k-dimensional linear subspace D of R n, this is the natural wedge to associate with a polytope incident to a k-dimensional face.) Take the disjoint union of W i s and glue them along the faces according gluing of the faces of P i s defined by GM. Denote this space B and call it the model space at p. In particular, all the W i s are glued together along D. We can view B as a subset of R N for N large so that the wedges W i s are embedded in R N and glued together accordingly. For example, for n = 2, the 2-wedge is the half plane {(x, y) R n : y 0}. 3

4 Consider a model space B for which 3 copies of 2-wedges are glued together along D (which in this case is the set y = 0). Then B can be realized the union of {(x, y, z) R 3 : y 0, z = 0}, {(x, y, z) R 3 : y = 0, z 0} and {(x, y, z) R 3 : y = 0, z 0}. On the other hand, it will be more convenient for us to view each wedge W i as being a subset of R n so that we can use the coordinates (x 1,..., x n ) of R n as mentioned above. The r-ball centered at the origin of B is denoted by B(r). This means that we take an intersection of each wedge W in B with a r-ball B r R n centered at the origin. For the sake of simplicity, we will also refer to this intersection as a wedge (of B(r)). Given a C m Riemannian n-dimensional polyhedron X, two types of points in X will be important in later discussions. These are points of a (n 1)-face and a (n 2)-face. We want to be precise about modelling a neighborhood of such a point. Lemma 1 Let X be a C m Riemannian n-dimensional polyhedron for m 0 and suppose p is a point on a (n 1)-face F. For r sufficiently small, there exists homeomorphism Ψ : B(r) X so that Ψ(0) = p, the restriction of Ψ to any wedge W is a diffeomorphism Ψ : W Ψ(W ) and the pull back metric Ψ g on W is C m and (Ψ g(0)) ij = δ ij. We can also arrange Ψ so that (Ψ g)(x 1,..., x n 1, 0) in = δ in Proof. Let P 1,..., P l be the polytopes incident to F. The model space B of p is then constructed from l number of 1-wedges W 1,..., W n. By definition, a 1-wedge is a half space, so W i s are copies of the set {(x 1,..., x n ) R n : x n 0}. The map Ψ : B(r) X is constructed as follows. Using the fact that g F is a C m Riemannian manifold, let (ξ 1,..., ξ n 1 ) be coordinates of F so that 0 = (0,..., 0) corresponds to the point p and the coordinate representation g ij of the metric g F satisfies g ij ( 0) = δ ij. (If g is C 2, we can simply take the normal coordinates centered at p.) Next, for each i = 1,..., l, let v be an unit vector at point p pointing into a polytope P i so that the angle between v and F is orthogonal with respect to the metric g. Using Euclidean translation in F, extend v to a unit vector field q V q defined along F. Then each point in P i U, where U is a small neighborhood of p, can be represented as tv q for some t 0. (Here, we set 0V q to be the point q.) The map Ψ takes the point (x 1,..., x n ) W i to the point x n V q where q is the point in F with coordinates (ξ 1,..., ξ n 1 ) = (x 1,..., x n 1 ). By construction, the pull back metric Ψ g defined on B(r) is C m on each wedge W i and (Ψ g(0)) ij = δ ij. To obtain the last statment define q v(q) so that v(q) is orthogonal to F for every q F instead of using the Euclidean translation of v(p). q.e.d. Lemma 2 Let X be a C m Riemannian n-dimensional polyhedron for m 0 and assume p is a point on a (n 2)-face F. There exists a model space B at 4

5 p with the property that for r sufficiently small, there exists a homeomorphism Ψ : B(r) X so that Ψ(0) = p, the restriction of Ψ to any wedge W is a diffeomorphism Ψ : W Ψ(W ) and the pull back metric Ψ g is C m and (Ψ g(0)) ij = δ ij Proof. Let P 1,..., P l be the polytopes incident to F. For each i = 1,..., l, let F 1, F 2 be the (n 1)-faces of P i and also incident to F. (We note that there are exactly two such faces since F is of dimension n 2.) Let v 1, v 2 be vectors at p pointing inward into F 1, F 2 respectively and orthogonal to F with respect to the metric g. Let S be a plane containing v 1, v 2 in P i and α i be the angle between v 1, v 2 with respect to metric g restricted to S. We use polar coordinates (r, θ) with respect to the metric g S to label points of S near p. For each point q F, let S q be the Euclidean translation in P i of S from p to q. The points of S q near q will also be labeled by the polar coordinates (r, θ) inherited from S via the translation. In the model space B at p, we choose the 2-wedge W i that corresponds to a polytope P i to have angle α i ; more precisely, the two hyperplanes defining W i R n will have angle α i. For the 2-wedge, D is a (n 2) plane. Recall we choose coordinates (x 1,..., x n ) in R n so that D is given by x n 1 = x n = 0. For x = (x 1,..., x n 2, 0, 0) D, let Ŝx be defined as the intersection of W i with a 2-plane orthognal (in Euclidean metric) to D at x. Let (R, Θ) be the polar coordinates of S x (with respect to the Euclidean metric). For sufficiently small r, we define Ψ : B(r) X so that Ψ maps S x to S q. More specifically, the point (R, Θ) S x is mapped to (r, θ) S q and where q the point with coordinates (ξ 1,..., ξ n 2 ) = (x 1,..., x n 2 ) in F. By construction, the pull back metric Ψ g defined on B(r) is C m on each wedge W i and (Ψ g(0)) ij = δ ij. q.e.d. 2.3 NPC spaces We now define our target space. Definition A complete metric space (Y, d) is said to be a NPC (non-positively curved) space if the following conditions are satisfied: (i) The space (Y, d) is a length space. That is, for any two points P and Q in Y, there exists a rectifiable curve γ P Q so that the length of γ P Q is equal to d(p, Q) (which we will sometimes denote by d P Q for simplicity). We call such distance realizing curves geodesics. (ii) Let P, Q, R Y. Define Q t to be the point on the geodesic γ QR satisfying d QQt = td QR and d QtR = (1 t)d QR. Then d 2 P Q t (1 t)d 2 P Q + td 2 P R t(1 t)d 2 QR. 5

6 Remark. Simply connected Riemannian manifolds of non-positive sectional curvature, Bruhat-Tits Euclidean buildings associated with actions of p-adic Lie groups and R-trees are examples of NPC spaces. These spaces are also referred to as CAT(0) spaces in literature. We refer to [?] for more details. 2.4 Harmonic maps We will now review the definition of harmonic maps. First, we define the energy of a map. If Ω is a Riemannian domain and Y is a NPC space, then the energy E f of a map f : Ω Y is defined as the weak limit of ɛ-approximate energy density measures which are measures derived from the appropriate average difference quotients. More specifically, define e ɛ : Ω R by e ɛ (x) = { y S(x,ɛ) d 2 (f(x),f(y)) dσ x,ɛ ɛ 2 ɛ n 1 for x Ω ɛ 0 for x Ω Ω ɛ where σ x,ɛ is the induced measure on the ɛ-sphere S(x, ɛ) centered at x and Ω ɛ = {x Ω : d(x, Ω) > ɛ}. Define a family of functionals Eɛ f : C c (Ω) R by setting Eɛ f (ϕ) = ϕe ɛ dµ. We say f has finite energy (or that f W 1,2 (Ω, Y )) if E f := sup ϕ C c(ω),0ϕ1 Ω lim sup Eɛ f (ϕ) <. ɛ 0 It can be shown that if f has finite energy, the measures e ɛ (x)dx converge weakly to a measure which is absolutely continuous with respect to the Lebesgue measure. Therefore, there exists a function e(x), which we call the energy density, so that e ɛ (x)dµ e(x)dµ. In analogy to the case of real valued functions, we write f 2 (x) in place of e(x). In particular, E f = f 2 dµ. Ω For V ΓΩ where ΓΩ is the set of Lipschitz vector fields on Ω, f (V ) 2 is similarly defined. The real valued L 1 function f (V ) 2 generalizes the norm squared on the directional derivative of f. The generalization of the pull-back metric is π f (V, W ) = ΓΩ ΓΩ L 1 (Ω, R) where π f (V, W ) = 1 2 f (V + W ) f (V W ) 2. 6

7 We refer to [KS1] for more details. If X is a C m Riemannian n-dimensional polyhedron, then energy E f of f : X Y is f 2 dµ := f 2 dµ X P X P where the sum is taken over all polytopes of X. (Since we are assuming X is compact, there are only finite number of top dimensional simplices and each of them are finite.) The functions f 2 and f (V ) 2 are defined for almost every point in X. We refer to [EF] for more details. A map f : X Y is said to be harmonic if it is locally energy minimizing. 3 Monotonicity Formula Let B be a local model of some interior point on a C m Riemannian n-dimensional polyhedron. By definition, a wedge W of B is a cone in R n with vertex 0. We will say g is a C m metric defined on B(r) if g defines a C m Riemannian metric on each wedge of B(r) with uniform elliptic bound. We will refer to this structure by (B(r), g). We say (B(r), g) is normalized if g is equal to the Euclidean metric at the origin. For the rest of the paper, all (B(r), g) we consider will be normalized and C m with m 1. We will also need to consider the Euclidean metric on B(r) denoted by δ. This is just the Euclidean metric inherited by the embedding of each wedge W in R n. We will refer to this structure by (B(r), δ) We will denote the distance between x, y B(r) induced by the metric g by x y g. In other words, x y g = inf L g (γ) where inf is taken over all Lipschitz curves γ : [0, 1] B(r) with γ(0) = x, γ(1) = y and 1 L g (γ) = g(γ (t), γ (t))dt. 0 The distance induced by δ will be denoted by x y. For a W 1,2 -map f : B(r) (Y, d), let π f indicate the pull back inner product defined in Section 2.3 of [KS]. We set f x i f x j = π f ( x i, x j ) and f x i 2 = f x i f x i. Additionally, we use the notation E f (Eg f resp.) and f 2 ( f 2 g resp.) to indicate the energy and the energy density function of f with respect to the Euclidean metric δ = (δ ij ) (Riemannian metric g = (g ij ) resp.). More precisely, given a subset S B(r), E f [S] = f 2 dµ and Eg f [S] = f 2 gdµ g, S 7 S

8 where f 2 = i f x i f x i and f 2 g = i,j ij f g f x i x j and dµ, dµ g are the volume forms corresponding to δ and g respectively. Additionally, we set and I f (σ) = E f (σ) = E f [B(σ)] and E f g (σ) = E f g [B(σ)] B(σ) d 2 (f, f(0))dσ and Ig f (σ) = B(σ) d 2 (f, f(0))dσ g where dσ and dσ g are the measures on B(t) induced by δ and g respectively. We say a function η defined on B(r) is smooth if the restriction of η to each wedge W of B(r) is smooth up to the boundary of W. The set of smooth function with compact support in B(r) will be denoted by Cc (B(r)). Lemma 3 Let f : (B(r), g) Y be a harmonic map. For any σ (0, r) and η Cc (B(σ)), f 2 g(2 n)η f 2 η g x i + 2 ik η f f g x j dµ g B(σ) x i i x i x j x k i,j,k +O(σ) = 0 (1) where O(σ) cσ and c is a constant dependent on the C 1 -norm of g and the total energy of f. Proof. For t sufficiently small, we define F t : B(r) B(r) by setting F t (x) = (1 + tη(x))x for each x = (x 1,..., x n ) in a wedge W. For f t : B(r) Y definted as f t = f F t, a direct computation (cf. [GS] Section 2) on each wedge W of B(r) gives d dt Eft g [W ] t=0 = f 2 g(2 n)η f 2 g W +remainder i η x i + 2 ik η f g x j x i x i x j i,j,k f dµ g x k Here, the remainder term is given by η g ij f x k f g + f 2 W x k x i x g η j i,j,k i g x i dµ g. x i 8

9 Since we assume the metric g is C 1, there exists a constant c so that gij x k, g x i c, which then implies that the remainder term is bounded by cσ Eg f. Summing over all the wedges W of B(r) we get the right hand side of (1) and this is equals 0 since f 0 = f is harmonic. q.e.d. Lemma 4 If f : (B(r), g) Y satifies (1), then for σ (0, r) 0 = (2 n + O(σ)) f 2 gdµ g + σ f 2 gdσ g 2σ B(σ) B(σ) B(σ) f f r r dσ g (2) where O(σ) cσ and c is a constant dependent only on the C 1 norm of g and the total energy of f. Proof. Let η in (1) approximate the characteristic function of B(σ). q.e.d. Lemma 5 Let f : (B(r), g) Y be a harmonic map. For any Q Y, d 2 (f, Q) 2 f 2 g 0 weakly, i.e. d 2 (f, Q) η dµ g 2 f 2 gη dµ g (3) for any η C c B(r) (B(r)). Proof. The computation for the proof of this inequality can be found in the proof of [EF] Lemma 10.2 which is based on the proof of [GS] Proposition 2.2. q.e.d. B(r) Lemma 6 If f : (B(r), g) Y satisfies (3), then r d2 (f, f(0))dµ g 2Eg f (σ). (4) B(σ) Proof. Let η in (3) approximate the characteristic function of B(σ). q.e.d. Proposition 7 If f : (B(r), g) Y satisfy (2) and (4), then σ e cσ σef g (σ) I f g (σ) is a non-decreasing function. Here, the constant c dependents only on the C 1 - norm of the metric g and the total energy of f. Consequently, α = lim σ 0 e cσ σef g (σ) I f g (σ) exists. The number α is called the order of f at 0. 9

10 Proof. Let { r, θ 1,..., θ n 1 } be the tangent basis corresponding to the polar coordinates (r, θ 1,..., θ n ) on W. (By polar coordinates on W, we mean the restriction of the polar coordinates defined on the Euclidean space; i.e, r gives the radial distance from the origin and θ = (θ 1,..., θ n 1 ) are coordinates on the standard (n 1)-sphere in R n.) We define v(r, θ) = 1 r n 1 ( det g( ) θ i, θ j ). By the assumption that g(0) is the Euclidean inner product, we see that lim v(r, θ) = 1. r 0 Since g is C 1, so is r v(r, θ), and we can apply the mean value theorem to obtain v(σ, θ) 1 = v r (σ 0, θ)σ cσ (5) where σ 0 (0, σ). Define ˆ W (σ) = {x W : x = σ} for 0 < σ < r. Then on each wedge W, we have d d 2 (f, Q)dΣ g dσ ˆ W (σ) = d d 2 (f, Q)σ n 1 v(σ, θ)dθ dσ ˆ W (σ) = ˆ W (σ) + d dr (d2 (f, Q))σ n 1 v(σ, θ)dθ + d 2 (f, Q)σ ˆ W (σ) ˆ W (σ) +c ˆ B + j (σ) +c n 1 dv (σ, θ)dθ dr d dr (d2 (f, Q))σ n 1 v(σ, θ)dθ + n 1 σ d 2 (f, Q)σ n 1 dθ ˆ W (σ) d dr (d2 (f, Q))dΣ g + n 1 σ d 2 (f, Q)dΣ g, ˆ W (σ) This in turn implies ( I f g (σ) ) d dσ I f g (σ) n 1 σ d 2 (f, Q)(n 1)σ n 2 v(σ, θ)dθ ˆ W (σ) d 2 (f, Q)σ n 1 v(σ, θ)dθ ˆ W (σ) d 2 (f, Q)dΣ g ˆ W (σ) + 1 Ig f (σ) B(σ) r (d2 (f, Q))dΣ g + c. (6) 10

11 On the other hand, (2) implies that d dσ (Ef g (σ)) Eg f n (σ) σ E(σ) B(σ) 2 f r dσ g + c (7) for some constant c. Together, (6) and (7) imply ( ) d σe f dσ log g (σ) Ig f (σ) [( ( ) 2 2 Eg f (σ)ig f d 2 (f, Q)dΣ g) f (σ) B(σ) B(σ) r dσ g ( d(f, Q) ) ] 2 r d(f, Q)dΣ g + c (8) B(σ) for some constant c. By using the Schwarz inequality, ( ) 2 d(f, Q) B(σ) r d(f, Q)dΣ g ( B(σ) d 2 (f, Q)dΣ g) ( B(σ) and by the triangle inequality, d(f, Q) r f r, we obtain hence ( ) d σe f dσ log g (σ) Ig f c, (σ) ( ) d dσ log e cσ σef g (σ) Ig f 0. (σ) ( ) ) 2 d(f, Q) dσ g, r which immediately implies the assertions in the Proposition. q.e.d. Remark. The error term of O(σ) in the above proof is equal to 0 when the metric g is the Euclidean metric δ. Corollary 8 If f : (B(r), g) (Y, d) satisfies (2) and (4) and α = lim σ 0 e cσ σef g (σ) I f g (σ) 11

12 then ( σ and ( σ I(σ) e cσ σ 2α+n 1 E(σ) e cσ σ 2α+n 2 are non-decreasing functions. Proof. Proposition 7 implies that αig f (σ) e cσ σeg f (σ). ) ) Furthermore, (4) and (6) imply that 2 Eg f (σ) r d2 (f, f(v))dσ g = B(σ) d dσ (If g (σ)) n 1 O(σn ) Ig f (σ). (9) σ Combining the above two inequalities, we get which implies and hence 2αI f g (σ) σ d dσ (If g (σ)) (n 1 O(σ n ))I f g (σ) 2α + n 1 σ 0 O(σ n 1 ) ( e cσ If g (σ) σ 2α+n 1 d dσ (If g (σ)) I f g (σ) ). Again, by Proposition 7, 2α + n 1 σ O(σ n 1 ) d dσ (If g (σ)) I f g (σ) 1 d σ + dσ (E(σ)) + O(σ 2 ) E(σ) which implies q.e.d. 0 ( e cσ Ef g (σ) σ 2α+n 2 ). 12

13 4 Convergence in the pull back sense Given a map u : B(r) (Y, d), we recall the following construction of [KS2]. First, we let Ω 0 = B(r), u 0 = u and d 0 : Ω Ω R + be the pseudodistance function d 0 (x, y) = d(u 0 (x), u 0 (y)). Next, we inductively define Ω i+1 = Ω i Ω i [0, 1] and identify Ω i as a subset of Ω i+1 by the inclusion map x (x, x, 0). Extend u i : Ω (Y, d) to u i+1 : Ω (Y, d) by and let Then and u i+1 (x, y, λ) = (1 λ)u i (x) + λu i (y) d i+1 (x, y) = d(u i+1 (x), u i+1 (y)). d i+1 ((x, x, 0), (y, y, 0)) = d i (x, y) d i+1 ((x, y, λ), (x, y, µ)) = λ µ d i (x, y) d i+1 (z, (x, y, λ)) (1 λ)d i+1 (z, (x, x, 0)) +λd i+1 (z, (y, y, 0)) λ(1 λ)d i+1 ((x, x, 0), (y, y, 0)). (10) Let Ω = Ω i and define u : Ω (Y, d) by setting u = u i on Ω i. With d (x, y) := d(u (x), u (y)), define (Y, d ) as the completion of the quotient space from (Ω, d ) and let π : Ω Y be the natural projection map. Equation (10) implies that the metric space (Y, d ) is a NPC space. The unique extension of u to Y is an isometry U : (Y, d ) C(u(B(r))) Y to the closed convex hull of the image of u. Furthermore, if ι : B(r) = Ω 0 Ω is the inclusion map, then u = U π ι. (cf. [KS2]) Definition Let v k : B(r) (Y k, d k ) be a sequence of maps to NPC spaces. We say v k converge to v in the pullback sense if there exists a pseudodistance function d : Ω Ω R + with the following property. Let (Y, d ) be the completed quotient space from (Ω, d ) and π : Ω Y the natural projection map. Furthermore, let v k = u in the above paragraph and let d k, : Ω Ω R + the corresponding pullback distance function of v k, (which equals u above). Then d k, converges pointwise to d and v = π ι. Remark. If we let v = u with u as in the paragraph preceeding the definition above and d,i (d, resp.) the corresponding pullback distance function of v,i = u i (v, = u resp.), then d = d,. Definition Suppose v k converge to v in the pullback sense. Let d k,i (d,i resp.) be the corresponding pullback distance function to v k,i : Ω i (Y k, d k ) (v,i : Ω i (Y, d ) resp.). We say that the convergence is locally uniform if the convergence of d k,i to the limit d,i is uniform on each compact subset of Ω i Ω i. 13

14 Proposition 9 Let v k : B(r) (Y k, d k ) be a sequence of maps to NPC spaces for which there is uniform modulus of continuity control, i.e. assume for each x B(r) and R > 0 there is a positive function ω(x, R) which is monotone in R, satisfying lim ω(x, R) = 0, R 0 and so that for each k Z max d(v k(x), v k (y)) ω(x, R). y B(x,R) Then there is a NPC space (Y, d ) and a subsequence v k of the v k which converges locally uniformly in the pullback sense to a limit map v : B(r) (Y, d ), and v satisfies the same modulus of continuity estimates. Here, (Y, d ) is the completed quotient of (Ω, d, ) where d, = lim k d k,. Proof. The proposition follows from the argument of the proof of Lemma 3.1 and Proposition 3.7 in [KS2] since the fact that B(r) is not a Riemannian domain plays no consequence in the argument. q.e.d. 5 The tangent map Let f : (B(r), g) (Y, d) be a map so that the function is non-decreasing and let σ e cσ σef g (σ) I f g (σ) α = lim σ 0 e cσ σef g (σ) I f g (σ). For λ sufficiently small so that λr < 2, define the λ-blow up map f λ : (B(2), g λ ) (Y, d λ ) by setting: g λ (x) = g(λx) µ λ = (λ 1 n I(λ)) 1/2 d λ (p, q) = µ 1 λ d(p, q) f λ (x) = f(λx). The mean value inequality (cf. (5)) implies that g ij (x) δ ij c x and g ij (x) δ ij cσ. (11) 14

15 Hence, (g λ ) ij (x) δ ij cλ x and (g λ ) ij δ ij cλ x. (12) If we have uniform modulus of continuity control for the sequence f λ, then by Proposition 9, there exists a sequence λ k 0 and a NPC space (Y, d ) so that f λk converges locally uniformly in the pullback sense to a limit map f : B(1) (Y, d ). Definition The map f is called a tangent map of f at 0. Lemma 10 The map f is not identically constant. Proof. This follows immediately from the argument in the proof of Proposition 3.3 [GS]. q.e.d. and By change of variables f λ 2 g λ dµ gλ B(σ) B(σ) d 2 λ(f λ, f λ (0))dΣ gλ = µ 2 λ λ2 n f 2 gdµ g B(λσ) = µ 2 λ λ1 n d 2 (f, f(0))dσ g. B(λσ) Thus, the definition of µ λ implies d 2 λ(f λ, f λ (0))dΣ gλ = 1 and B(1) lim λ 0 Ordf λ (0, 1, 0) = α. Consequently, by choosing λ sufficiently small, we have f λ 2 dµ gλ 2α. (13) B(1) We want to prove a uniform modulus of continuity of the blow-up maps to show that harmonic maps have tangent maps. We first need: Lemma 11 Let f : (B(1), g) (Y, d) be a harmonic map into an NPC space. For each r, there exist constants C and γ depending only on r and the energy of f so that d(f(x), f(y)) C x y γ g for all x, y B(r). 15

16 Proof. In this setting, this follows from the argument in [Ch]. Also see [EF] where a more general domain for the harmonic map is considered. q.e.d. Proposition 12 A harmonic map f : (B(r), g) (Y, d) has a tangent map f which satisfies d(f (x), f (y)) C x y γ for all x, y B(r) where γ is as in Lemma 11 and C is a constant dependent on the C 1 norm of g. Proof. Equation (13) and Lemma 11 implies d λ (f λ (x), f λ (y)) C x y γ g λ, x, y B(r), where C and γ are independent of λ. Furthermore, (12) implies x y gλ c x y. Therefore, we are done by letting C = cc. q.e.d. Given a W 1,2 map u, recall (cf. structure is defined by By Lemma of [KS1], and thus, Thus, i,j [KS1]) that its pull back inner product u x i u x j = 1 4 u ( i + j ) u ( i j ) u ( i + j ) u ( i j ) u ( i + j ) u ( i ) u ( j ) [( u ( i ) + u ( j ) ) 2 u ( i ) 2 u ( j ) 2 = u ( i ) u ( j ) 1 2 ( u ( i ) 2 + u ( j ) 2 ), u u 1 ( u u + u u ). x i x j 2 x i x i x j x j cλ u x i u x j i,j Ncλ 2 ( cλ u u + u u ) 2 x i x i x j x j u u + x i i x i j u u x j x j = Ncλ i u u. x i x i 16

17 which implies (1 Ncλ) i u u (δ ij cλ) u u x i x i x i,j i x j i,j i,j ij u g u x i x j (δ ij + cλ) u x i u x j (1 + Ncλ) i u u. x i x i Let f : (B(r), g) (Y, d) be a harmonic map and let the sequence of blowup maps f λk : B(1) (Y, d λk ) converge to a tangent map f. Let h λ : B(1) (Y, d λ ) a map which is harmonic with respect to the Euclidean domain metric and constrained to the boundary condition h λ B(1) = f λ B(1). The above discussion implies (1 Ncλ) n i=1 f λ x i f λ x i n (g λ ) ij f λ f n λ f λ (1 + Ncλ) f λ x i x j x i x i i,j=1 i=1 and (1 Ncλ) n i=1 h λ h λ x i x i n (g λ ) ij h λ h λ x i x j i,j=1 (1 + Ncλ) n i=1 h λ x i h λ x i which in turn implies and In particular, we have that (1 Ncλ) δ E f λ g λ E f λ (1 + Ncλ) δ E f λ (14) (1 Ncλ) δ E h λ g λ E h λ (1 + Ncλ) δ E h λ. (15) δ E h λ δ E f λ 1 1 Ncλ g λ E f λ 2α 1 Ncλ. (16) Thus, Proposition 9 and Lemma 11 imply that there exists a subsequence of λ k (which we will still denote λ k by an abuse of notation) and a NPC space (Ȳ, d ) so that h λk converge locally uniformly in the pullback sense to h : B(1) (Ȳ, d ). Set h k := h λk, f k := f λk and g k = g λk. Furthermore, let d k (x, y) = d λk (f k (x), f k (y)) and d k (x, y) = d λk (h k (x), h k (y)). Then in any compactly contained subset of B(1) B(1), d k, d k converge uniformly to (the restriction to B(1) = Ω 0 of) d, d. 17

18 Lemma 13 The pseudodistance functions d and d above are equal. Consequently, the tangent maps f and h are the same map. Proof. By the repeated use of the triangle inequality, d k (x, y) d k (x, y) d λk (f k (x), h k (x)) + d λk (f k (y), h k (y)). Therefore, for any r < 1, the Lebesgue dominated convergence theorem and the Poincaré inequality (cf. [EF]) implies d (x, y) d (x, y) 2 dµ(x)dµ(y) B(r) B(r) = lim d k (x, y) d k (x, y) 2 dµ(x)dµ(y) k 0 B(r) B(r) = 4vol(B(r)) lim d 2 λ k (f k (x), h k (x))dµ(x) (17) Equations (14) and (15) imply δ E f λ k 0 B(r) 4C lim d 2 λ k 0 k (f k (x), h k (x)) dµ(x). (18) B(r) 1 1 Ncλ g λ E f λ Therefore, if we let w = 1 2 f λ h λ, 1 1 Ncλ 2 δ E w δ E f λ + δ E h λ 1 2 = 2 δ E h λ + O(λ) 1 2 g λ E h λ B(r) B(r) 1 + Ncλ 1 Ncλ d 2 λ(f λ, h λ ) dµ d 2 λ(f λ, h λ ) dµ by equation (2.2iv) of [KS1]. Since E h λ E w, this in turn implies d 2 λ(f λ, h λ ) dµ 0 B(r) δ E h λ. as λ 0. This, combined with equation (18) and continuity of d and d, shows that d (x, y) = d (x, y) which in turn implies that (Y, d ) = (Ȳ, d ) and h = f. q.e.d. 6 Harmonic maps from a flat domain For each wedge W R n of B(r) modeling a point p of a k-face, recall that coordinates (x 1,..., x n ) of R n is arranged to that D(r) is given by x n k+1 =... = x n = 0. We show that the gradient of harmonic map h : (B(r), δ) (Y, d) for points in D(r) in the direction parallel to D(r) is bounded. 18

19 Lemma 14 Let h : (B(1), δ) (Y, d) be a harmonic map. Let V be a unit vector parallel to D(r) and let g(x) = h(x + ɛv ) for 0 < ɛ << 1. Then 0 η d(h, g)dµ (19) B(1) for η C c (B(1 ɛ)). Proof. Define a map h η : B(1 ɛ) R by setting h η (x) = (1 η(x))h(x) + η(x)g(x). Here, (1 τ)p + τq for P, Q Y denotes the point on the unique geodesic between P and Q at a distance τd(p, Q) from P and (1 τ)d(p, Q) from Q. Since spt(η) B(1 ɛ), we see that and h η B(1 ɛ) = h B(1 ɛ) h 1 η B(1 ɛ) = g B(1 ɛ). By following the proofs of Lemma and of [KS1], we see that h η, h 1 η W 1,2 (B(r)), and on each half ball B + j (r), j = 1,..., N, we have h η 2 + h 1 η 2 B + j (1 ɛ) B + j (1 ɛ) h 2 + g 2 B + j (1 ɛ) B + j (1 ɛ) 2 η d 2 (h, g) + Q(η, η) B + j (1 ɛ) B + j (1 ɛ) where Q(η, η) consists of integrable terms which are quadratic in η and η. Taking the sum of all B + j (r) and noting that h 2 h η 2 and we have B(1 ɛ) B(1 ɛ) g 2 B(1 ɛ) B(1 ɛ) h 1 η 2, 0 2 η d 2 (f, g) + Q(η, η) B(1 ɛ) B(1 ɛ) by the harmonicity of h. By replacing η by tη, dividing by t and letting t 0, we obtain (19). q.e.d. 19

20 For a point x in B(1) belonging to a wedge W, recall that R(x) is defined in the following way: if x is in the interior of W, R(x) is the distance of x to W. If x is a point on a boundary piece b, we let R(x) be the distance of x to the closest boundary piece that is not b Lemma 15 Let h : (B(1), δ) Y be a harmonic map and V be a unit vector parallel to D(r). For x B(1), where C = Vol( B(1)). h (V ) 2 (x) n CR n (x) Eh x(r(x)) Proof. Let η approximate the characteristic function of B x (σ) in (19) to obtain r d2 (h, g)dσ 0. Let for 0 < σ < R(x). Then J (σ) = This implies that B x(σ) B x(σ) J(σ) = B x(σ) d 2 (h, g)dσ r d2 (h, g)dσ + n 1 σ J(σ) n 1 σ J(σ). ( ) J(σ) σ n 1 0, and hence J(σ) J(R) σn 1 R n 1 for 0 < σ < R < R(x). Taking the limit of the left-hand side as σ 0, we obtain CR n 1 d 2 (h(x), g(x)) d 2 (h, g)dσ. We now integrate R over (0, R(x)) to obtain, d 2 n (h(x), g(x)) CR n (x) Divide by ɛ 2 and let ɛ 0 to get h (V ) 2 n (x) CR n (x) q.e.d. B x(r(x)) B x(r) B x(r(x)) h (V ) 2 dµ d 2 (h, g)dµ. n CR n (x) Eh x(r(x)). 20

21 Lemma 16 Let h : (B(1), δ) Y be a harmonic map. Let r (0, 1). Let x, y be a pair of points in a wedge of B(r) so that the distance to D(r) is equal to ρ (0, 1 r). Then d(h(x), h(y)) c x y ρn/2 for some constant c which depends only on n and E h. If x, y are point on D(r), then d(h(x), h(y)) L x y for some constant L depending only on n, E h and r. Proof. Let γ : [0, 1] Y be a constant speed parameterization of the line between x and y. If the distance between x or y and D(r) is equal to ρ, the distance between γ(t) and D(r) is equal to ρ. Thus, by Lemma 15, 1 d(h(x), h(y)) h (γ ne (t)) dt h x y Cρn 0 0 q.e.d. If x, y D(r), then R(x) = 1 r. Thus, 1 d(h(x), h(y)) h (γ ne (t)) dt h x y C(1 r) n Proof. First, suppose B is a model space of a point p on a (n 1)-face. In this case, wedges are half spaces x n 0 and D is a hyperplane x n = 0. If x = (x 1,..., x n ) is a point in a wedges so that x < r and x n < δ/2, let ˆx = (x 1,..., x n 1, 0) be the projection of x to D. Thus, R(x) = x n. Therefore, B x (R(x)) Bˆx (2R(x)). Monotonicity then implies E h x(r(x)) R n (x) Thus, Lemma 15 implies Ehˆx (2R(x)) R n (x) 2n Ehˆx (2R(x)) (2R(x)) n h (V ) 2 2n E h (2δ) n and the result follows immediately. q.e.d. 2n Ehˆx (δ) (2δ) n 2n E h (2δ) n. 7 At a point on a (n-1)-face Let X be a C 1 Rimannian n-dimensional polyhedral complex. In this section, we assume that B the model space of a point p on a (n 1)-face with the pull back metric as given in Lemma 1. Thus, B(r) comes with a metric (which will 21

22 be given by g = (g ij ) by an abuse of notation) defined on each wedge. Recall that g ij (0) = δ ij and g in (x 1,..., x n 1, 0) = δ in. Also recall that wedges are half spaces x n 0 and D is a hyperplane given by the equation x n = 0 and B is constructed by gluing together the wedges along D. We will assume that p is a interior point, so that B has more than one wedge. Note that we continue to use the notation of B(r) to denote the set of point of B which is distance < r from the origin 0. We first prove some properties of harmonic maps from the model space equipped with the Euclidean metric. Lemma 17 Let h : (B(1), δ) Y be a harmonic map. For every β, r (0, 1), there exists B only dependent on β, r and the total energy of the map h so that for every x, y B(r). d(h(x), h(y)) B x y β Proof. By Lemma 16, h is Lipschitz when restricted to D(t 0 ), t 0 = r+1 2. Thus Hölder regularity of h restricted to a B + j (r) with any Hölder exponent β (0, 1) follows from the boundary regularity result of Serbinowski [Se] where the Hölder constant B is only dependent on the choice of β, r and the total energy of the map h. q.e.d. Lemma 18 Let h k be as defined in Section 5 (see the paragraph preceeding Lemma 13) and fix R (0, 1). Set D ɛ (r) to be the ɛ-neighborhood of D(r) in B(r), i.e. it is the union of the ɛ-neighborhoods of D(r) in the wedges W of B(r) or equivalently D ɛ (r) = {x = (x 1,..., x n ) W : x n ɛ} B(r) where the union is over the wedges W. Then any r (0, R), there exists constants C, δ > 0, k 0 sufficiently large and ɛ 0 > 0 sufficiently small (depending only on R) so that E h k [D ɛ (r)] Cɛ δ, k > k 0, ɛ < ɛ 0 Proof. Let B x (r) be a ball of radius r centered at x. We will use the notation, Ex(r) g = g 2 dµ and Ix(r) g = B x(r) B x(r) d 2 (g, g(x))dσ for any map g : B(r) (Y, d). Let r 0 := 1 R. By Lemma 10 and continuity, there exists c 1 > 0 so that I f x (r 0 ) 2c 1, x D(R). 22

23 Thus, by the local uniform convergence, there exists k 0 so that I h k x (r 0 ) c 1, x D(R), k > k 0. By (16), we may assume we have chosen k sufficiently large so that λ k0 (0, 1 Nc ) and hence Thus, E h k x (r 0 ) E h k (1) E f k 2α (1), x D(R), k > k 0. 1 Ncλ k0 By Proposition 7, and, by Lemma 17, E h k x r 0 E h k x (r 0 ) 2r 0 α I h =: c k 2, x D(R), k > k 0. x (r 0 ) (1 Ncλ k0 )c 1 ɛe h k x (ɛ) I h k x (ɛ) ec/2 c 2 =: c 3, x D(R), k > k 0, (ɛ) c 3I h k x (ɛ) c 3B 2 ɛ 2β+n 1 = c 3 B 2 ɛ 2β+n 2, x D(R), k > k 0. ɛ ɛ Here, we have choosen β (1/2, 1). Since D ɛ (r) can be covered by ( 4r ɛ number of (3ɛ)-balls centered at points in D(r), E h k [D ɛ (r)] c 3 B 2 (3ɛ) 2β+n 2 ( ) n 1 4r ɛ = c 3 B 2 (4r) n 1 3 2β n 2 ɛ 2β 1 =: c 4 ɛ 2β 1. ) n 1 The result follows from the fact that the choice of β implies 2β 1 > 0. q.e.d. Let f k, h k, f = h be as in Section 5. Lemma 19 For r (0, 1), and lim k 0 Eh k (r) = E h (r) (20) lim k Ef k g k (r) = E f (r), (21) and the directional energies of f k, and h k converge to that of f. 23

24 Proof. By the regularity result of harmonic maps from smooth domains ([KS1] Theorem 2.4.6), h k is uniformly Lipschitz in B( 1+r 2 ) D ɛ ( 1+r 2 2 ) for r (0, 1). Thus by [KS1] Theorem 3.11, the energy densities of h k converge to those of f in any smooth domain Ω B( 1+r 2 ) D ɛ ( 1+r 2 2 ), j = 1,..., N. Let Ω 1,..., Ω L be such domains and so that B(r) D ɛ (r) Ω 1... Ω L. Since h k 2 dµ converge to f 2 dµ, we have that for any Ω Ω l. If for l 0 = 1,..., L, then By Lemma 18, Ω l 0 E h k [Ω ] E f [Ω ] = Ω l0 (Ω 1... Ω l0 1 D ɛ (r)) lim [B(r) D k Eh ɛ (r)] = lim k for any ɛ sufficiently small. Thus, = L E h k [Ω l 0 ] l 0=1 L E f [Ω l 0 ] l 0=1 E h k [D ɛ (r)] Cɛ δ = E f [B(r) D ɛ (r)] lim k Eh k (r) Cɛ δ lim E h [B(r) D ɛ (r)] k 0 By lower semicontinuity of energy, = E f [B(r) D ɛ (r)] E f (r). E f (r) lim inf k 0 Eh k (r) lim E h k (r) E f (r) + Cɛ δ. k 0 Since ɛ > 0 can be made arbitrarily small, this proves (20). To prove (21), we see that E f (r) lim inf k (1 Ncλ k) E f k (r) lim inf k lim inf k Ef k g k (r) Eh k g k (r) lim (1 + Ncλ k) E h k (r) k = E h (r) = E f (r). 24

25 Since there is no loss of total energy, we see that the directional energies converge by using the lower semicontinuity. q.e.d. Lemma 20 Let f : (B(r), g) Y be a harmonic map and f : B(1) Y its tangent map. For every β, r (0, 1), there exists B so that d (f (x), f (y)) B x y β for all x, y B(r) and B is only dependent on the choice of β, r and the total energy of f. Proof. This follows from the uniform convergence, as well as the convergence of the energy, of h k to f and Lemma 17. q.e.d. Lemma 21 Let f : (B(r), g) Y be a harmonic map. Its tangent map f : B(1) Y is homogenous of order α where α is the order of f at 0, i.e. d (f (x), f (0)) = x α d (f ( x x ), f (0)) and the image of τ f (τx), 0 τ 1, is a geodesic. Proof. Using (1) and (3) with f replaced by f k, noting that the remainder in (1) is 0 because the domain is Euclidean, and using the convergence of f k and its directional energies to f and its directional energies, we have and 2E f (σ) ( E f (σ) ) = 2 B(σ) B(σ) 2 f r dσ d(f, f (0)) r d(f, f (0))dΣ. Thus, we obtain (8) with f replaced by f ; in other words, ( σe f ) (σ) I f (σ) [( ) ( ) 2 2 d 2 (f, f (0))dΣ f E f (σ)i f (σ) B(σ) B(σ) r dσ (d(f, f (0)) r ) ] 2 d(f, f (0))dΣ g 0. (22) B(σ) 25

26 On the other hand, Lemma 19 implies that, σe f (σ) I f (σ) σe f k g = lim k (σ) k I f k g k (σ) σµ 2 λ = lim k λ 2 n k Eg f (λ k σ) k µ 2 λ k λ 1 n k Ig f (λ k σ) σλ k Eg f (λ k σ) = lim k Ig f (λ k σ) = α. Thus, Hence, and 0 = = ( σe f (σ) I f (σ) ( ) d 2 (f, f (0))dΣ B(σ) B(σ) ) ( B(σ) f r ) 2 dσ ( d(f, f (0)) r d(f, f (0))dΣ g ) 2. r d(f, f (0)) = f r a.e. 2 f 2 dµ = B(σ) B(σ) r (d2 (f, f (0)))dΣ. We can now follow the proof of Proposition 3.1 [GS] to show the homogeneity of f. q.e.d. Theorem 22 Let f : (B(r), g) (Y, d) be a harmonic map. Then the order α of f at 0 is 1. Proof. Since f is homogeneous of degree α, d (f (τx), f (0)) = τx α d (f ( x x ), f (0)). On the other hand, for any β (0, 1) and τ small, there exists a constant B so that d (f (τx), f (0)) B τx β by Lemma 20. Thus, d (f ( x x ), f (0)) τx β α B. 26

27 If α < 1, choose β so that β > α and take the limit as τ 0 to obtain d (f ( x x ), f (0)) = 0. Since the choice of x B(1) is arbitrary, this contradicts Lemma 10. q.e.d. Using the fact that the order at a point on D is 1, we can prove Lipschitz continuity in B(1). Corollary 23 Let f : (B(2), g) Y be a harmonic map. Then f is Lipschitz in B(1) with Lipschtiz constant dependent on g and the total energy E f of f. Proof. Let x B(1) D(1), δ = dist(x, D(1)) and ˆx D(1) so that δ = x ˆx g. For σ << δ (hence B x (σ) W for some wedge W ), the subharmonicity of the function d 2 (f, f(x)) and the DiGiorgi-Nash-Moser theory implies there exists a constant C 0 depending only on the metric g so that for any y B x (σ/2), d 2 (f(y), f(x)) C 0 σ n 1 Also, by Corollary 8, I f x (σ) σ n+1 B x(σ) C If x (δ) δ n+1 d 2 (f, f(x))dσ C 0 σ n 1 If x (σ). C Ef x(δ) δ n, where C depends only on g. Here, we have used the fact that the order of f at x is 1 which is implies by the regularity result of [KS1] Theorem for Riemannian domains. Thus, d 2 (f(y), f(x)) C Ef x(δ) δ n σ2 where C depends only on g. If δ 1 2, then d 2 (f(y), f(x)) 2 n CE f σ 2. Assume δ < 1 2. By the fact that B δ(x) B 2δ (ˆx), E f x(δ) δ n 2n E fˆx (2δ) (2δ) n. Since the order of f at ˆx 1 by Theorem 23, Corollary 8 implies that E fˆx (2δ) CEfˆx (1) (2δ) n 1 n CE f. 27

28 Thus, again we have d 2 (f(y), f(x)) 2 n CE f σ 2. This immediately implies the gradient estimate f 2 (x) 2 n CE f. and the Lipschitz continuity of f. q.e.d. The k-skeleton of X, denoted by X (k), is the union of all simplices of dimension at most k. We now state the regularity of harmonic maps away from the (n 2)-skeleton of X. Theorem 24 Let X be a C 1 Riemannian n-dimensional polyhedron, (Y, d) a NPC space and f : X Y a harmonic map. For p X X (n 2), let δ = dist(p, X (n 2) ). If p is an interior point, then f is Lipschitz continuous in B p (δ/2) with Lipschitz constant dependent on the C 1 norm of the metric, the total energy of f and δ. Proof. After rescaling the metric g of X by a factor of δ/2, B p (2) is contained in X X (n 2). Thus, from Corollary 23, f is Lipschitz continuous in B p (1) with Lipschitz constant dependent on the C 1 norm of the (rescaled) metric and the energy of f in B p (2). Therefore, the Lipschitz constant of f in B p (δ/2) depends on the C 1 norm of g, total energy of f and δ. q.e.d. Assuming stronger regularity conditions on the domain and target spaces, we can prove a stronger regularity result for a harmonic map. We say X is a C (C ω resp.) Riemannian n-complex if the component functions for g P is C (real analytic) for any polytope P in X. First, we prove a preliminary lemma where we still assume the metric on the domain is C m for m 1 and the target is a smooth Riemannian manifold instead of a NPC space. Lemma 25 Let f : (B(r), g) N be a harmonic map from a m-dimensional C to a Riemannian manifold N with non-sectional Riemannian curvature. Assume that f(b(r)) is contained in a coordinate chart U of N and let f 1,..., f m be the coordinate functions of f. Let W 1,..., W J be the wedges of B(r). For any α = 1,..., m, lim ɛ 0 k=1 J η(x Wk 1,..., x n 1, ɛ) f α x n (x1,..., x n 1, ɛ)dσ g = 0 (23) where dσ g is the measure induced on D ɛ (r) = {(x 1,..., x n ) W k : x n = ɛ} by g on W k and η Cc (B(r)). 28

29 Proof. Let ϕ α be a Lipschitz function defined in B(r) with compact support for α = 1,..., m. For sufficiently small t, define f t : B(r) U by setting f t = f + tϕ = (f 1 + tϕ 1,..., f m + tϕ m ). Then using the notation fi α to denote f α x, we have i E(f t ) = g ij (x)h αβ (f(x) + tϕ(x))(fi α (x) + tϕ α i (x))(f β j (x) + tϕβ j (x))dµ. Because f = f 0 is energy minimizing, we obtain 0 = d dt E(f t) t=0 = (2g ij h αβ (f)fi α ϕ β j + gij h αβ,γ (f)fi α f β j ϕγ ) gdx = 2 ( gg ij h αβ (f)fi α ϕ β ) j dx 2 ( gg ij fi α ) j h αβ ϕ β dx 2 g ij fi α f γ j h αβ,γϕ β gdx + g ij h αβ,γ fi α f β j ϕc γdx = 2 ( gg ij h αβ (f)fi α ϕ β ) j gdx 2 ( gg ij fi α ) j h αβ ϕ β dx + g ij (h αβ,c + h γβ,α h αγ,β )fi α f γ j ϕβ gdx. Now set η α (x) = h αβ (f(x))ϕ β (x). Then 0 = ( gg ij fi α η α ) j dx ( g fi α gij Γ α βγf β i f γ j g)ηα dx. Since f is a smooth harmonic map in the interior of each wedge, we have the pointwise equality, g f α + g ij (x)γ α βγ(f(x))f β i f γ j = 0. Therefore, by the monotone convergence theorem and the fact that f is Lipschitz continuous in B(r), we conclude 0 = d ( gg ij dx j fi α ) η α dµ = lim ɛ 0 J k=1 W k D ɛ(r) For j = 1,..., n 1, Since η α has compact support, d ( g ij x j fi α ) η α dµ = 0 W k D ɛ(r) d dx j ( gg ij f α i η α ) dµ 29

30 since η α has compact support. Hence, d ( 0 = lim gg ɛ 0 W in k D ɛ(r) dx n fi α ) η α dµ ( = lim gg ɛ 0 W in f α ) i η α dσ. k D ɛ(r) By the Lipschitz continuity of f, fi α L for some constant L in the support of η α. Therefore, using the fact that g in (x 1,..., x n 1, ɛ) δ in, we obtain (23) by the dominate convergence theorem. q.e.d. Lemma 26 Let f : (B(r), g) N be a harmonic map to a C Riemannian manifold N with non-positive sectional curvature. Let W 1,..., W J be the wedges of B(r) and assume that f(b(r)) is contained in a coordinate chart U of N and let f 1,..., f m be the coordinate functions of f. Then for r < r, f α C 1,β (B(r ) W j ). Proof. Let B r R n be a ball of radius r centered at the origin. Fix α 0 {1,..., m} and j 0 {1,..., J} and, for r > 0 sufficiently small, define ψ : B r R by setting { f ψ(ˆx, x n α 0 Wj0 ) = (ˆx, xn ) when x n 0 f α0 Wj0 (ˆx, x n ) + 2 J j=1 f α0 Wj (ˆx, x n ) when x n < 0 where we simplify notation by using (ˆx, x n ) = (x 1,..., x (n 1), x n ). By the (Lipschitz) continuity of f, f α0 j (ˆx, 0) = f α0 Wj0 (ˆx, 0) for all j = 1,..., J. Therefore, f α0 Wj0 (ˆx, 0) + 2 J f α0 Wj (ˆx, 0) j=1 = f α0 Wj0 (ˆx, 0) + 2f α0 Wj0 (ˆx, 0) = f α0 Wj0 (ˆx, 0). This implies that ψ is Lipschitz continuous in B r. Additionally, for fixed ɛ > 0, ψ α0 (ˆx, ɛ) = f xn n Wj0 (ˆx, ɛ) 2 J j=1 J j=1 fn α0 Wj (ˆx, ɛ). By Lemma 25, lim ɛ 0 J B r {x n =ɛ} j=1 ϕ(ˆx, ɛ)f α0 n Wj (ˆx, ɛ) = 0 30

31 for any ϕ Cc (B r ), we see that lim ϕ(ˆx, ɛ) ψ (ˆx, ɛ) ɛ 0 xn B r {x n = ɛ} = lim ɛ 0 B r {x n = ɛ} ϕ(ˆx, ɛ)fn α0 Wj0 (ˆx, ɛ) 2 lim = lim ϕ(ˆx, ɛ)fn ɛ 0 B α0 Wj0 (ˆx, ɛ) r {x n = ɛ} = lim ɛ 0 B r {x n =ɛ} = lim ɛ 0 B r {x n =ɛ} ϕ(ˆx, ɛ)fn α0 Wj0 (ˆx, ɛ) + lim ϕ(ˆx, ɛ) ψ (ˆx, ɛ) + lim xn ɛ 0 J ɛ 0 B r {x n = ɛ} j=1 ɛ 0 B r {x n =ɛ} B r {x n =ɛ} ϕ(ˆx, ɛ)f α0 n Wj (ˆx, ɛ) (ϕ(ˆx, ɛ) ϕ(ˆx, ɛ))f α0 n Wj0 (ˆx, ɛ) (ϕ(ˆx, ɛ) ϕ(ˆx, ɛ))fn α0 Wj0 (ˆx, ɛ). The last term is equal to zero by the dominated convergence theorem since ϕ(ˆx, ɛ) ϕ(ˆx, ɛ) converges uniformly to 0 as ɛ 0 and fn α0 Wj0 is bounded. Thus, lim ϕ ψ ɛ 0 B r {x n = ɛ} x n = lim ϕ ψ ɛ 0 B r {x n =ɛ} x n. (24) Also, since f j is a smooth harmonic map in the interior of a n-simplex, f α + g ik (x)γ α βγ(f(x))f β i f γ k = 0. (25) Let Γ α j (ˆx, x n ) = g ik (x)γ α βγ(f j (ˆx, x n ))f β i W j (ˆx, x n )f γ k W j (ˆx, x n ) and F : B r R be defined by { Γ α 0 F (ˆx, x n j ) = 0 (ˆx, x n ) when x n 0 Γ α0 j 0 (ˆx, x n ) + 2 J j=1 Γα0 j (ˆx, xn ) when x n < 0 The equality (25) implies that ψ(ˆx, x n ) = F (ˆx, x n ) for x n > 0 and x n < 0. Thus, integration by parts implies ϕ ψ = ϕ ψ + ϕ ψ B r {x n >ɛ} B r {x n >ɛ} B r {x n =ɛ} x n = ϕf ϕ ψ x n and ϕ ψ = B r {x n < ɛ} = B r {x n >ɛ} B r {x n < ɛ} B r {x n < ɛ} B r {x n =ɛ} ϕ ψ ϕf + B r {x n = ɛ} B r {x n = ɛ} ϕ ψ x n ϕ ψ x n 31

32 for any ϕ Cc (B r ). Summing the above two equalities, letting ɛ 0 and noting (24), we see that ϕ ψ = ϕf. B r B r Since F is bounded, the standard elliptic regularity theorem implies that ψ C 1,β (B r ) and ψ W 2,2 (B r ) for r < r. Since ψ(ˆx, x n ) = f α0 Wj0 (ˆx, x n ) for x n 0 and α 0 {1,..., m}, j 0 {1,..., J} are arbitrary choices, this shows f α Wj C 1,β (B r {x n 0}) and f α Wj W 2,2 (B r {x n > 0}) for any α = 1,..., n and j = 1,..., J. q.e.d. Corollary 27 Let f : (B(r), g) N be a harmonic map to a Riemannian manifold N with non-positive sectional curvature. Assume that f(b(r)) is contained in a coordinate chart U of N and let f 1,..., f m be the coordinate functions of f. Let W 1,..., W J be the wedges of B(r). For any α = 1,..., m, J j=1 f α Wj x n (x1,..., x n 1, 0) = 0 for all α = 1,..., m and all (x 1,..., x n 1, 0) V E. Proof. This pointwise equality follows immediately from Theorem 25 and the regularity result of Theorem 26. q.e.d. Theorem 28 Let f : (B(r), g) N be a harmonic map from to a m-dimensional Riemannian manifold N with non-positive sectional curvature. Assume that f(b(r)) is contained in a coordinate chart U of N and let f 1,..., f m be the coordinate functions of f. Let W 1,..., W J be the wedges of B(r). If g is C (resp. C ω ) and N is a C (resp. C ω ), then f is C (resp. C ω ). Proof. Let B r + = {x = (x 1,..., x n ) R n : x < r, x n 0} and define ϕ αk : B r + R by setting ϕ αk = f α Wk. The set {ϕ αk } satisfies the system of equations g ϕ αk + g ij (x)γ α ϕ βk ϕ γk βγ x i x j in B + and boundary conditions = 0 for α = 1,..., m and k = 1,..., J ϕ αk (x 1,..., x n 1, 0) = ϕ αl (x 1,..., x n 1, 0) for α = 1,..., J and k, l = 1,..., J and n k=1 ϕ αk x n (x1,..., x n 1, 0) = 0 for α = 1,..., m 32

33 It is verify that this system is elliptic and coersive (cf. [Mo] (see also [KiSt] or [KNS]). The result follows from the regularity theorem of [Mo] Theorem q.e.d. Corollary 29 Let X be a C (resp. C ω ) Riemannian n-complex, Y a C (resp. C ω ) Riemannian manifold of non-positive sectional curvature, and f : X Y a harmonic map. Then f is C (resp. C ω ) in X X (n 2). Proof. Follows immediately from Theorem 28. q.e.d. 8 At a point of a (n 2)-face. Let X be a C 1 Rimannian n-dimensional polyhedral complex. In this section, we assume that B the model space of a point p on a (n 1)-face with the pull back metric as given in Lemma 2. Assume that p is an interior point of X. Lemmas 30 and 31 below is a restatement of Lemmas 18 and 19 corresponding to the current situation. Lemma 30 Let h k, f k, h = f be as in Section 5. Let D ɛ (r) be the ɛ-neighborhood of D(r). Fix R (0, 1). For any r (0, R), there exists C and δ > 0, k 0 sufficiently large and ɛ 0 > 0 sufficiently small so that E h k [D ɛ (r)] Cɛ δ, k > k 0, ɛ < ɛ 0. Proof. As in the proof of Lemma 19, there exists a constant c 3 so that Thus, by Lemma 11, E h k x We can cover D ɛ (r) be ɛe h k x (ɛ) I h k x (ɛ) c 3. (ɛ) c 3E h k x (ɛ) c 3C 2 ɛ 2γ+n 1 =: c 4 ɛ 2γ+n 2. ɛ ɛ c 5 ɛ n 2 E h k [D ɛ (r)] number of (2ɛ)-balls. Thus, c t ɛ n 2 c 4ɛ 2γ+n 2 =: c 6 ɛ 2γ. The lower semicontinuity of energy implies that q.e.d. E f [D ɛ (r)] c 6 ɛ 2γ. 33