The Curvature of Minimal Surfaces in Singular Spaces

Transcription

1 The Curvature of Minimal Surfaces in Singular Spaces Chikako Mese November 27, Introduction Let be an unit disk in R 2 and (M, g) a smooth Riemannian manifold. If an immersed surface u : M is minimal, i.e. stationary with respect to the area functional det(τ ij )dx where (τ ij )=u g is the pull back metric on, then b 11 + b 22 = 0 with b ij the components of the vector valued second fundamental form. The Gauss equation then gives, K Σ = K M + b 11 b 22 b 2 12 K M where K Σ is the Gauss curvature of the surface and K M is the sectional curvature of the tangent plane to the surface in the manifold. This shows that the curvature of a minimal surface is less than or equal to that of the ambient space. In this paper, we will show that this fundamental curvature property of minimal surfaces also holds in certain singular spaces. When a smooth surface Σ has a conformal metric with conformal factor λ, it is well known that the Gaussian curvature K Σ is given by the formula K Σ = 1 log λ. 2λ Hence, the condition that the curvature be bounded from above by κ reduces to the inequality log λ 2κλ. Our main theorem states that this same type of inequality holds when we replace the smooth Riemannian manifold with a complete metric space of curvature bounded from above by κ. We will call a map from a surface a minimal surface if it is conformal and locally energy minimizing. Recall that these conditions on a map are equivalent to minimality in the smooth setting. Because our target space can be quite singular, we can only 1

2 expect the following weak inequality: Main Theorem Let λ be a conformal factor of the pull back metric under a minimal surface u : X where (X, d) is a complete metric space of curvature bounded from above by κ. Then for all non-negative ϕ Cc (), log λ ϕ 2κ ϕλ. The complete metric spaces considered are length spaces, i.e. any two points can be joined by a distance realizing curve. Furthermore, we impose a curvature bound from above; here, the curvature bound is defined in terms of comparing geodesic triangles with comparison triangles in a constant curvature surface (see Section 3 for the precise definition.) Sometimes called Alexandrov spaces with curvature bounded from above, they were studied by A.. Alexandrov [A] in the 1950 s and advanced by him and the Russian school of mathematicians. They include smooth Riemannian manifolds with an upper bound on the sectional curvature but allow singularities of a very general type. In fact, no restriction is made on the singularities. If we consider a class of Riemannian manifolds with upper bound κ on sectional curvature and a lower bound on the injectivity radius, the completion by Gromov-Hausdorff metric turns out to be these metric spaces of curvature bounded from above by! κ. The motivation of this paper is twofold. One, we wish to extend the study of minimal surfaces in Riemannian manifolds to spaces with singularities. Second, we wish to introduce analytical tools in the study of singular spaces. Recently, there has been much interest in the study of harmonic map theory for spaces with singularities. A general existence and regularity theories of harmonic maps into Riemannian simplicial complexes of non-positive curvature was developed in [GS] to solve certain rigidity problems. This theory was further generalized for maps into complete metric spaces with non-positive curvature by [KS] and independently by [J]. The case of curvature bounded from above by some constant is treated in [S]. Some examples of the application of the harmonic maps into certain singular spaces are Wolf s [W1],[W2],[W3] investigations of Teichmüller spaces and the actions of fundamental groups of closed surfaces and Hardt and Lin s [HL] study of neumatic liquid crystals. In light of the these successful studies of the harmonic map theory in singular spaces, it is natural to consider an extension of minimal surface theory to these settings. This paper is organized as follows. In Section 2, we outline Sobolev space theory for metric space targets due to [KS]. In Section 3, we recall the notion of metric spaces with curvature bounds and derive some inequalities for the distance functions that will be important in Section 4. Furthermore, we define area for maps into these spaces which allows us to consider the Plateau Problem. We note that Nikolaev [N2] is the first to consider the Plateau Problem; there he takes a defintion of area that is different from ours. Our definition is a natural extension of the definition of area for maps into smooth 2

3 Riemannian manifolds and allows us to use a classical approach in the solution of the Plateau Problem. In Section 4, we will prove an inequality satisfied by the energy density function e(u) of an energy minimizing map by a careful consideration of the curvature bound of the target. If u : M N is an energy minimizing ma! p between smooth Riemannian manifolds, then the Bochner s formula gives 1 2 e(u) = du 2 α,β < R N (u e α, u e β )u e α, u e β > + i Ric M (u θ i, u θ i ) where e 1,..., e n is an orthonormal basis for T M and θ 1,..., θ k is an orthonormal basis for T N. In particular, if M is flat and N has sectional curvature bounded from above by κ, then e(u) 2κe(u) 2. We will see that the energy density function of energy minimizing maps into metric spaces of curvature bounded from above by κ weakly satisfies the above inequality. This inequality will be the starting point of the proof of the main theorem. Section 5 is devoted to the proof of the main theorem. In Section 6, we make a geometric interpretation of the analytical result of Section 5; namely, we consider the natural distance function induced by the metric λ(dx 2 + dy 2 ) which defines a metric space of curvature bounded from above by κ. In the case when the map minimizes area, Professor Nikolaev has pointed out that this result follows from the works of Reshetnayk [R1],[R2]. We thank him for communicating this observation. Acknowledgment. This is a part of the author s doctoral dissertation at Stanford University. She would like to express deep gratitude to advisor Richard M. Schoen for his support and encouragement. She also thanks the refees for their comments. 2 Sobolev Space Theory for Maps to Metric Spaces Let Ω be a compact domain in R n and (X, d) any complete metric space. In [KS], Korevaar and Schoen develop the space W 1,2 (Ω, X). Here we define this space and collect some of their results. A Borel measurable map u : Ω X is said to be in L 2 (Ω, X) if for P X, d 2 (u(x), P )dx <. Ω Note that by the triangle inequality, this definition is independent of P chosen. u L 2 (Ω, X), we can construct an ɛ approximate energy function e ɛ : Ω ɛ R, e ɛ (x) = n B ɛ (x) 1 B ɛ(x) 3 d 2 (u(x), u(y)) dσ. ɛ 2 For

4 Here Ω ɛ is the set of points in Ω with distance from the boundary more than ɛ and B ɛ (x) is a ball of radius ɛ centered at x. Letting e ɛ (x) = 0 for Ω Ω ɛ, we have that e ɛ (x) L 1 (Ω) and by integrating against continuous functions with compact support, these functions define linear functionals E ɛ : C c (Ω) R. We say u L 2 (Ω, X) has finite energy (or that u W 1,2 (Ω, X)) if E u sup f C c(ω),0 f 1 lim sup E ɛ (f) <. ɛ 0 It can be shown that if u has finite energy, the measures e ɛ (x)dx converge in the weak* topology to a measure which is absolutely continuous with respect to the Lebesgue measure. Hence, there exists a function e(x), which we call the energy density, so that e ɛ (x)dx e(x)dx. In analogy to the case of real valued functions, we write u 2 (x) in place of e(x). In particular, E u = u 2 dx. Ω Similarly, the directional energy measures u (Z) 2 dx for Z Γ Ω can also be defined as the weak* limit of measures Z e ɛ dx, where Furthermore, for Z T Ω, Z e ɛ (x) = d2 (u(x), u(x + ɛz)) ɛ 2. a.e. x Ω. Finally, we have d(u(x), u(x + ɛω) u (Z) (x) = lim, ɛ 0 ɛ u 2 = u (Z) 2 dσ(z). S n 1 This definition of Sobolev space W 1,2 (Ω, X) is consistent with the usual definition when X is a Riemannian manifold. The following theorems allow us to use variational methods in the setting where the target space of maps is a complete metric space. Note that u k u in L 2 (Ω, X) will mean d(u k, u) converges to 0 in L 2 (Ω), i.e. lim d 2 (u k, u) = 0. k Theorem 2.1 ([KS] Theorem 1.6.1) If {u k } W 1,2 (Ω, X) is a sequence with uniformly bounded W 1,2 (Ω, X) norms and u k u in L 2 (Ω, X), then u W 1,2 (Ω, X) and E(u) = lim inf k 4 E(u k).

5 The following is a generalization of the W 1,2 trace theory. Theorem 2.2 ([KS] Theorem ) Any u W 1,2 (Ω, X) has a well-defined trace map (denoted tr(u)), with tr(u) L 2 ( Ω, X). If {u k } W 1,2 (Ω, X) is a sequence with uniformly bounded energies and u k u in L 2 (Ω, X), then tr(u k ) converges to tr(u) in L 2 ( Ω, X). We also have the following Rellich type precompactness theorem. Theorem 2.3 ([KS] Theorem 1.13) Let (X, d) be locally compact. If {u k } W 1,2 (Ω, X) satisfy d 2 (u k (x), Q)dx + E(u k ) C, Ω where Q is a fixed point in X, then a subsequence of {u k } converges in L 2 (Ω, X) to a finite energy map u. Using these theorems, one can solve the following. The irichlet Problem Let (X, d) be a complete locally compact metric space. ψ W 1,2 (Ω, X). efine Let W 1,2 ψ = {u W 1,2 (Ω, X) : tr(u) = tr(ψ)}. Let E ψ = inf{e(v) : v W 1,2 1,2 ψ }. There exists u Wψ such that E(u) = E ψ. If we assume an upper curvature bound on the target (see efinition 3.1 in the next section), we get nice regularity properties of the solution. In fact, [KS] shows that the solution is Lipschitz when X is non-positively curved and [S1] shows that the same holds in the case when curvature is bounded from above by some constant provided that the boundary data lies in a small geodesic ball. In both cases, the map is Hölder continuous to the boundary. 3 Metric Spaces of Curvature Bounded from Above In this section, we will recall the definition of curvature bounds in a metric space, give some technical propositions and define the notion of area for maps into these singular spaces. 5

6 3.1 The efinition efinition 3.1 A complete metric space (X, d) is said to have curvature bounded from above by κ if the following conditions hold: (i) (X, d) is a length space; that is, if P, Q X there exists a distance realizing curve connecting P and Q. (We call such distance realizing curves geodesics.) (ii) Let S κ be a surface of constant curvature κ. For any three points P, Q, R X (with d P Q + d QR + d RS < π κ if κ > 0) and choices of geodesics γ P Q (of length r), γ QR (of length p) and γ P R (of length q) connecting the respective points, call a triangle ( P Q R) in S κ with vertices P, Q, R and opposite side lengths p, q, r a comparison triangle in S κ. For any 0 < λ < 1 write Q λ for the point on γ QR so that d(q, Q λ ) = λp and d(q λ, R) = (1 λ)p and define Q λ S κ analogously to Q λ, then d(p, Q λ ) d Sκ ( P, Q λ ). Remark: These spaces are sometimes defined in terms of an angle excess (see [ABN] for example). The upper angle between geodesics are defined as follows: if γ and σ are geodesics having a common point P with R γ, Q σ and r = d(p, Q), q = d(p, R), we let αγσ(r, κ q) be the angle at P of the comparison triangle ( P Q R) of S κ. The upper angle between γ and σ is α(γ, σ) = lim sup αγσ(r, κ q). r,q 0 This definition is independent of κ. (X, d) is said to be a metric space of curvature bounded from above by κ if for every triangle (P QR) in X (with d P Q + d QR + d RP < π κ ), α + β + γ α κ + β κ + γ κ where α, β, γ are the upper angles of (P, Q, R) and α κ, β κ, γ κ are angles of the comparison triangle in S κ. This definition is equivalent to the above definition of a curvature bound. These spaces are referred to as CAT (κ) spaces in literature. If κ = 1, then S κ is a standard unit sphere S 2. Note that if κ > 0, we can make X into a CAT (1) space by rescaling the distance function. If κ = 1, then S κ is the hyperbolic plane H 2. Again, note that if κ < 0, then by rescaling the distance function, we can make X into a CAT ( 1) space. 3.2 Technical Propositions This important result is given in [R1] and will be basis of the propositions that follow. 6

7 Theorem 3.2 (Reshetnyak) Let (X, d) be a metric space of curvature bounded above by κ and Γ be a closed rectifiable curve in X (of length less than or equal to π κ if κ > 0). Then there exists a convex domain V in S κ and a map ϕ : V X such that ϕ( V ) = Γ, the lengths of the corresponding arcs coincide, and d Sκ (x, y) d(ϕ(x), ϕ(y)), for x, y V. Let X be a CAT (κ) space and P, Q, R, S X. If d P S, d QR < π κ, then there is a unique geodesic between P and S (Q and R, resp.). We denote by P t (resp. Q t ) the point on this geodesic such that d P Pt = td P S (resp. d QQt = td QR ). Let (X, d) be a CAT (1) space. Given ordered sequence {P, Q, R, S} X with d P Q + d QR +d RS +d SP < π. Theorem 3.2 asserts that there is an ordered sequence { P, Q, R, S} S 2 such that the quadrilateral associated with it (i.e. the four ordered points, the geodesics between consecutive points and its interior) is convex and d(p, Q) = d S 2( P, Q), d(q, R) = d S 2( Q, R), d(r, S) = d S 2( R, S), d(s, P ) = d S 2( S, P ), d(p t, Q s ) d S 2( P t, Q s ). We will call { P, Q, R, S} a spherical subembedding for {P, Q, R, S}. (X, d) is a CAT ( 1) space, we can define a hyperbolic subembedding Similarly, when { ˆP, ˆQ, ˆR, Ŝ} H2. In the propositions below, 0 n ( ) denotes terms that are nth order in the specified variables. Proposition 3.3 Let (X, d) be a metric space of curvature bounded from above by κ with κ = 1 or 1. Then for {P, Q, R, S} X (with d P Q + d QR + d RS + d SP < π if κ = 1), the following inequalities hold: For κ = 1, cos d P Qt + cos d RQ1 t cos d P Q cos td QR + sin d P Q sin td QR cos α + cos d RS cos td QR + sin d RS sin td QR cos β (1) cos d PtQ t sin(1 t)d P S sin d P S (cos d P Q cos td QR + sin d P Q sin td QR cos α) + sin td P S sin d P S (cos d RS cos td QR + sin d RS sin td QR cos β) (2) where α = P Q R and β = S R Q and { P, Q, R, S} is a spherical subembedding for {P, Q, R, S}. 7

8 For κ = 1, cosh d P Qt + cosh d RQ1 t cosh d P Q cosh td QR sinh d P Q sinh td QR cos α + cosh d RS cosh td QR sinh d RS sinh td QR cos β cosh d PtQ t sinh(1 t)d P S sinh d P S (cosh d P Q cosh td QR sinh d P Q sinh td QR cos α) + sinh td P S sinh d P S (cosh d RS cosh td QR sinh d RS sinh td QR cos β) where α = ˆP ˆQ ˆR and β = Ŝ ˆR ˆQ and { ˆP, {P, Q, R, S}. ˆQ, ˆR, Ŝ} is a hyperbolic subembedding for Proof: We will only prove the case when κ = 1 by comparing the distance function d to the distance function d S 2 of the sphere. The proof of the case when κ = 1 follows analogously by considering the distance function of the hyperbolic plane instead of the sphere. Let {X, Y, Z} S 2 such that d XY + d Y Z + d ZX < π. We let Y t be the point on the geodesic between Y and Z such that d YtZ = td Y Z and θ = S2Y XZ. We have the following equalities: Hence we have, and cos d Y Z = cos d XY cos d XZ + sin d XY sin d XZ cos θ cos d XYt = sin(1 t)d Y Z sin d Y Z cos d XY + sin td Y Z sin d Y Z cos d XZ. cos d P Qt + cos d R Q1 t = cos d P Q cos td Q R + sin d P Q sin td Q R cos α + cos d R S cos td Q R + sin d R S sin td Q R cos β, cos d Pt = sin(1 t)d P S (cos d Qt sin d P Q cos td Q R + sin d P Q sin td Q R cos α) P S + sin td P S (cos d sin d R S cos td Q R + sin d R S sin td Q R cos β). (3) P S The result follows immediately from Theorem

9 Proposition 3.4 Let (X, d) be a metric space of curvature bounded from above by κ with κ = 1 or 1 and {P, Q, R, S} X (with d P Q + d QR + d RS + d SP < π if κ = 1). Furthermore, let α, β as in Proposition 3.3. Then and for κ = 1, we have d P Q cos α + d RS cos β = d QR d P S (d P Q, d RS ) (4) d P Q cos α + d RS cos β d QR d P S 1 2 d P Qd RS d 2 QR +0 2 (d P Q, d RS )0 3 (d QR ) (5) and for κ = 1, we have d P Q cos α + d RS cos β d QR d P S d P Qd RS d 2 QR +0 2 (d P Q, d RS )0 3 (d QR ). (6) Proof: Once again, we will only prove the case when κ = 1 since the case κ = 1 follows analogously. Again, let { P, Q, R, S} be the spherical subembedding of {P, Q, R, S}. Assume w.l.o.g., Q = (1, 0, 0), R = (cos θ, sin θ, 0) S 2 R 3. Note for any X, Y S 2 R 3, we have that cos d S 2(X, Y ) = X Y. (Here denotes the usual dot product in R 3.) In particular, we see that d Q R = θ. Let γ (resp. σ) be a unit speed parameterization of a geodesic on S 2 emanating from Q (resp. R) such that, for t > 0, S 2γ(t) Q R = α (resp. S 2σ(t) R Q = β). If ϕ(t) = (cos t, sin t, 0), then γ and σ must satisfy: Hence, we have that and that γ (0) ϕ (0) = cos α, γ (0) = 1 σ (0) ϕ (θ) = cos(π β), σ (0) = 1. γ(0) = (1, 0, 0) γ (0) = (0, cos α, sin α) γ (0) = ( 1, 0, 0) σ(0) = (cos θ, cos θ, 0) σ (0) = (cos β sin θ, cos β cos θ, sin β) σ (0) = ( cos θ, sin θ, 0). 9

10 Using the Taylor series expansion, γ(t) σ(s) = cos θ + t cos α sin θ + s cos β sin θ t2 + s 2 cos θ 2 ts cos α cos β cos θ + ts sin α sin β (t, s). Let f(t, s) = γ(t) σ(s). Then again using Taylor series expansion, d S 2(γ(t), σ(s)) = arccos f(0, 0) + t t (arccos f) (0,0) + s s (arccos f) (0,0) This shows + t2 2 2 t (arccos f) 2 (0,0) + s2 2 2 s (arccos f) 2 (0,0) +ts 2 t s (arccos f) (0,0) (t, s) = θ t cos α s cos β t 2 cos θ cos 2 α cos θ + s2 2 sin θ 2 ts sin α sin β sin θ (t, s). cos θ cos 2 β cos θ sin θ d P Q cos α + d R S cos β = d Q R d P S (d P Q, d R S). Hence equation 4 follows the above equality. By Cauchy-Schwartz inequality, and thus we obtain ts sin α sin β t2 sin 2 α 2 + s2 sin 2 β, 2 t 2 2 (cos θ cos2 α cos θ) + s2 2 (cos θ cos2 β cos θ) ts sin α sin β = t2 2 (cos θ cos2 α cos θ) + s2 2 (cos θ cos2 β cos θ) ts sin α sin β cos θ ts(1 cos θ) sin α sin β ts(1 cos θ) sin α sin β ts(1 cos θ). 10

11 Since, 1 cos θ = θ (θ) and 1 sin θ = 1 θ ( (θ)), we obtain Hence we obtain, d S 2(γ(t), σ(s)) ts(1 cos θ) θ t cos α s cos β sin θ θ t cos α s cos β tsθ 2 + ts03 (θ). d P Q cos α + d R S cos β d Q R d P S 1 2 d P Qd R Sd Q R Now inequality 5 follows immediately (d P Q, d R S)0 3 (d Q R). Proposition 3.5 Let (X, d) be a metric space of curvature bounded from above by κ with κ = 1 or 1 and {P, Q, R, S} X (with d P Q + d QR + d RS + d SP < π if κ = 1). Let d 0 = d QR, d 1 = d P S, and l t = d PtQt where, P t (resp. Q t ) is the point on the unique geodesic between P and S (resp. Q and R) such that d P Pt = td P S (resp. d QQt = td QR ). Then for κ = 1, we have and for κ = 1, we have l 2 t + l 2 1 t l l tl 0 l 1 d (t)0 2 (l 0, l 1 ) 0 4 (l 0, l 1 ) + t0 2 (l 0, l 1 )0 3 (d 0, d 1 ) + t0 3 (l 0, l 1 ) (7) l 2 t + l 2 1 t l l 2 1 2tl 0 l 1 d (t)0 2 (l 0, l 1 ) 0 4 (l 0, l 1 ) + t0 2 (l 0, l 1 )0 3 (d 0, d 1 ) + t0 3 (l 0, l 1 ). Proof: Again, we will only prove the case when κ = 1. Inequality 7 follows immediately from Theorem 3.2 if we can prove the same inequality in the sphere. Hence, let { P, Q, R, S} be four points in the sphere with d P Q+d Q R+d R S +d S P < π and let d 0 = d Q R, d 1 = d P S, and l t = d Pt where, P Qt t (resp. Qt ) is the point on the unique geodesic between P and S (resp. Q and R) such that d P = td Pt P S (resp. d Q = td Qt Q R). From equation 3, cos l t + cos l 1 t = ( sin td 1 sin d 1 cos(1 t)d 0 + sin(1 t)d 1 sin d 1 cos td 0 )(cos l 0 + cos l 1 ) +( sin td 1 sin(1 t)d 0 + sin(1 t)d 1 sin td 0 )(sin l 0 cos α + sin l 1 cos β). sin td 1 sin d 1 11

12 By expanding sin d 0 and sin d 1 using Taylor series and then using inequality 5, we have sin l 0 cos α + sin l 1 cos β = l 0 cos α + l 1 cos β (l 0, l 1 ) Also note that Hence d 0 d 1 l 0l 1 d (l 0, l 1 )0 3 (d 0 ) (l 0, l 1 ). cos l t + cos l 1 t = 2 l2 t 2 l2 1 t (l 0, l 1 ). 2 l2 t 2 l2 1 t (l 0, l 1 ) = (2 l2 0 2 l (l 0, l 1 ))( sin td 1 sin d 1 cos(1 t)d 0 + sin(1 t)d 1 sin d 1 cos td 0 ) +(d 0 d 1 l 0l 1 d We now use, and to obtain (l 0, l 1 )0 3 (d 0 ) (l 0, l 1 ))( sin td 1 sin d 1 sin(1 t)d 0 + sin(1 t)d 1 sin d 1 sin td 0 ). sin td 1 cos(1 t)d 0 + sin(1 t)d 1 cos td 0 = (t) sin d 1 sin d 1 sin td 1 sin(1 t)d 0 + sin(1 t)d 1 sin td 0 = 2t (t) sin d 1 sin d l2 t 2 + l2 1 t (l 0, l 1 ) = 2( sin td 1 sin d 1 cos(1 t)d 0 + sin(1 t)d 1 sin d 1 cos td 0 ) Hence, + l l (t)0 2 (l 0, l 1 ) (l 0, l 1 ) (d 0 d 1 )( sin td 1 sin(1 t)d 0 + sin(1 t)d 1 sin td 0 ) sin d 1 sin d 1 +tl 0 l 1 d (t)0 2 (l 0, l 1 ) + t0 2 (l 0, l 1 )0 3 (d 0, d 1 ) + t0 3 (l 0, l 1 ). 4 + l 2 t + l 2 1 t = 2 sin d 1 [2(sin td 1 cos(1 t)d 0 + sin(1 t)d 1 cos td 0 ) +(d 0 d 1 )(sin td 1 sin(1 t)d 0 + sin(1 t)d 1 sin td 0 )] +l l tl 0 l 1 d (t)0 2 (l 0, l 1 ) +0 4 (l 0, l 1 ) + t0 2 (l 0, l 1 )0 3 (d 0, d 1 ) + t0 3 (l 0, l 1 ). 12

13 If we let F (x) = 2[sin td 1 cos(1 t)(x + d 1 ) + sin(1 t)d 1 cos t(x + d 1 )] +x(sin td 1 sin(1 t)(x + d 1 ) + sin(1 t)d 1 sin t(x + d 1 )), then the lemma will follow from the following claim with x = d 0 d 1. Claim: There exists σ > 0 such that for x σ, then F (x) 2 sin d 1. Proof of claim: It is easy to check that F (0) = 2 sin d 1, F (0) = 0. Furthermore, F (0) = 2(1 t) 2 sin td 1 cos(1 t)d 1 2t 2 sin(1 t)d 1 cos td 1 +2(1 t) sin td 1 cos(1 t)d 1 + 2t sin(1 t)d 1 cos td 1 = 2t(1 t) sin td 1 cos(1 t)d 1 + sin(1 t)d 1 cos td 1 = 2t(1 t) sin d 1 0. Since F is a C function, the claim follows. 3.3 The Pull-back Inner Product and the Area We make sense of the notion of area for maps u W 1,2 (, X) when X has an upper curvature bound. We do this by defining an inner product structure on which generalizes the pull-back metric for a smooth map between smooth Riemannian manifolds. The proof of the existence of such an inner product structure is an easy generalization of the proof in [KS] for maps into NPC space using the following technical lemma. Lemma 3.6 Let (X, d) be a CAT (1) space. Let P, Q, R, S X. Then where σ = max{d 2 P Q, d 2 QR, d 2 RS, d 2 P S}. We use inequality 3 to obtain d 2 P R + d 2 QS d 2 P Q + d 2 QR + d 2 RS + d 2 P S + o(σ), cos d P R + cos d QS 1 2 (cos d P Q cos d QR + cos d P S cos d RS + cos d P Q cos d P S + cos d QR cos d RS ) + sin d P Q sin d QR cos Q + sin d P S sin d RS cos S + sin d P Q sin d P S cos P + sin d QR sin d RS cos R). 13

14 Expanding terms, we obtain, d 2 P R + d 2 QR d 2 P Q + d 2 QR + d 2 RS + d 2 P S d P Q d QR cos Q d P S d RS cos S d P Q d P S cos P d QR d RS cos R + o(σ). Now consider ordered points { P, Q, R, S} R 2 such that, Q, the quadrilateral associated with it is convex and d P Q = P Q, d QR = Q R, d RS = R S, d P S = P S. Letting P = S P Q, Q = P Q R, R = Q R S, S = R S P in R 2, we have P + Q + R + S = 4π. Furthermore, P P, Q Q, R R, and S S. By the Gauss-Bonnet Theorem, P + Q + R + S 4π = da. Hence, Q Q = 0 2 (d P Q, d QR, d RS, d P S ). We can rewrite equation 3.6: Q d 2 P R + d 2 QR o(σ) d 2 P Q + d 2 QR + d 2 RS + d 2 P S d P Q d QR cos Q dp S d RS cos S d P Q d P S cos P dqr d RS cos R. Letting A, B, C, be the oriented vectors pointing to the consecutive vertices of the Euclidean quadrilateral Q, i.e. we have that, A = Q P, C = S R, B = R Q B = P S, P Q Q R cos Q P S R S cos S P Q P S cos P Q R R S cos R = A B + B C + C + A = (A + C) (B + C) = A + C 2. Here, we have used the fact that A + B + C + = 0. Hence, d 2 P R + d 2 QR d 2 P Q + d 2 QR + d 2 RS + d 2 P S + o(σ). As a result of the above, the directional energy functions satisfy a parallelogram law: 14

15 Lemma 3.7 Let Ω R n and let X be a CAT (1) space. If u W 1,2 (Ω, X), then for any Z, W Γ(T Ω), the parallelogram identity holds for a.e. x Ω. u (Z + W ) 2 + u (Z W ) 2 = 2 u (Z) u (W ) 2 Proof: Use Lemma 3.6 with P = u(x), Q = u(x + ɛz), R = u(x + ɛw ), S = u(x + ɛ(z + W )). ivide by ɛ 2 and let ɛ 0 to obtain u (Z + W ) 2 + u (Z W ) 2 2 u (Z) u (W ) 2. for a.e. x Ω. Repeat using Z + W and Z W in place of Z and W to get the opposite inequality. For Z, W Γ(T Ω), we define π(z, W ) = 1 4 u (Z + W ) u (Z W ) 2. Proposition 3.8 The operator π defined above, π : Γ(T Ω) Γ(T Ω) L 1 (Ω, R) is continuous, symmetric, bilinear, non-negative and tensorial. Proof: The proof is the same as the one given in [KS] (Theorem 2.3.2). efinition 3.9 π as above is the pull back inner product under the map u. We can now define the area functional A : W 1,2 (, X) R by A(u) = = det π u dx 1 dx 2 (π u ) 11 (π u ) 22 (π u ) 2 12dx 1 dx 2 where (π) ij = π u ( i, j ). Thus, we can formulate: The Plateau Problem Let be a disk and Γ be a closed Jordan curve in X and let C Γ = {u W 1,2 (, X) : u parametrizes Γ monotonically}. 15

16 There exists u C Γ so that A(u) = inf{a(v) v C Γ }. Moreover, u is weakly conformal, i.e. π 11 = π 22 and π 12 = 0 = π 21, and Lipschitz in the interior of. We can solve the Plateau Problem for a locally compact CAT (κ) space. Since the arguments are essentially the same as the classical approach (see for example [M]), we omit the proof here. Because conformal energy minimizing maps into smooth Riemannian manifolds are minimal, it is natural to define: efinition 3.10 Let X be a complete metric space of curvature bounded from above by κ. We say u : X is a minimal surface if u is a weakly conformal energy minimizing map. λ = π 11 = π 22 is called the conformal factor of the pull back metric under u. 4 The Energy ensity Inequality Before we can prove our main inequality, we will need to prove another inequality which is of interest in itself. As mentioned in the introduction, this can be seen as a generalization of the Bochner s inequality for harmonic maps between smooth Riemannanian manifolds. [KS] proves the weak subharmonicity for the energy density of a harmonic map when the target is an NPC space. We generalize their result by proving the following inequality when the target is a space of curvature bounded from above. Theorem 4.1 Let u : X be an energy minimizing map into a CAT (κ) space (i.e. a metric space of curvature bounded from above by κ). Then for any η Cc 2 () with η 0, u 2 η 2κ η u 4. (8) If u is minimal (i.e. also weakly conformal) with conformal factor λ, then λ η 2κ ηλ 2. Proof: The case κ = 0 is the result of [KS]. We will first prove the above for the case when κ = 1. Note that the result for κ > 0 follows immediately from rescaling the target distance function. The case κ = 1 is proven analogously. In the proof below, we prove equation 8 for η Cc 2 () with 0 η 1. By rescaling 2 η, we see that equation 8 holds for any non-negative Cc 2 function η. For two given points x, y, define η = min{η(x), η(y)}. Let u 0 and u 1 be energy minimizing maps such that sup d(u 0, u 1 ) < π x supp(η) 2. 16

17 We let L 0, L 1 be the Lipschitz constants of u 0 and u 1 in supp(η) and let L = max{l 0, L 1 }. If x y π, then d(u 2L 0(x), u 0 (y)), d(u 1 (x), u 1 (y)) < π. Let u 2 η W 1,2 (, X) be defined by taking the geometric interpolation of u 0 and u 1. In other words, let u η (x) be the point on the (unique) geodesic between u 0 (x) and u 1 (x) such that d(u 0 (x), u η (x)) d(u 0 (x), u 1 (x)) = η(x) and d(u η (x), u 1 (x)) d(u 0 (x), u 1 (x)) = 1 η(x). If η = η(y), we consider the ordered sequence {u η (y), u η (x), u 1 η (x), u 1 η (y)} and let t = η(x) η(y) and apply Proposition 3.3. If η 1 2η(y) = η(x), we interchange the roles of x and y and apply Proposition 3.3. Using the shorthand notation, we deduce in both cases, l t = d(u t (x), u t (y)) l η = d(u η (x), u η (y)) l 1 η = d(u 1 η (x), u 1 η (y)) d x = d(u 0 (x), u 1 (x)) d y = d(u 0 (y), u 1 (y) η = η(x) η(y), cos l η + cos l 1 η [cos l η + cos l 1 η ] cos( ηd x ) +[sin l η cos α + sin l 1 η cos β] sin( ηd x ) (9) where α = P Q R and β = S R Q and { P, Q, R, S} is a spherical subembedding for {u η (y), u η (x), u 1 η (x), u 1 η (y)}. By expanding the above, we obtain, l 2 η + l 2 1 η (l η, l 1 η ) l 2 η + l 2 1 η 2(l η cos α + l 1 η cos β) ηd x +0 2 ( ηd x ) (l η, l 1 η ) Applying Proposition 3.5 to {u 0 (y), u 0 (x), u 1 (x), u 1 (y)} and t = η, we obtain, lη 2 + l1 η 2 l0 2 + l η l 0 l 1 d 2 x (η )0 2 (l 0, l 1 ) +0 4 (l 0, l 1 ) + η 0 2 (l 0, l 1 )0 3 (d x, d y ) + η 0 3 (l 0, l 1 ) Applying Proposition 3.4 to {u η (y), u η (x), u 1 η (x), u 1 η (y)}, we obtain, 17

18 (l η cos α + l 1 η cos β) (1 2η )(d y d x ) (l η, l 1 η )0 2 (d x, d y ) Thus, the Cauchy-Schwartz inequality gives, Hence, inequality 9 implies (l η cos α + l 1 η cos β) ηd x 1 2 (1 2η ) η(d 2 y d 2 x) + η0 2 (l η, l 1 η )0 3 (d y, d x ) l 2 η + l 2 1 η (l η, l 1 η ) l l η l 0 l 1 d 2 x + (1 2η ) η(d 2 y d 2 x) +0 2 (η )0 2 (l 0, l 1 ) (l 0, l 1 ) + (η + η)0 2 (l 0, l 1 )0 3 (d x, d y ) +η 0 3 (l 0, l 1 ) ( ηd x ) (l 0, l 1 ) Let Z Γ(T Ω) be a vector field. By taking y = x + ɛz, dividing by ɛ 2, and letting ɛ 0, we deduce that for a.e. x Ω, (u η ) (Z) 2 + (u 1 η ) (Z) 2 (u 0 ) (Z) 2 + (u 1 ) (Z) 2 + η( (u 0 ) (Z) 2 + (u 1 ) (Z) 2 )d 2 (u 0, u 1 ) (1 2η)η (Z)d 2 (u 0, u 1 ) (Z) (η)0 2 ( (u 0 ) (Z), (u 1 ) (Z) ) +(η + η (Z))0 2 ( (u 0 ) (Z), (u 1 ) (Z) )0 3 (d(u 0, u 1 )) (η (Z)d x ) In the above, substitute η by tη, divide by t and let t 0 to obtain, (u η ) (Z) 2 + (u 1 η ) (Z) 2 (u 0 ) (Z) 2 + (u 1 ) (Z) 2 + η( (u 0 ) (Z) 2 + (u 1 ) (Z) 2 )d 2 (u 0, u 1 ) η (Z)d 2 (u 0, u 1 ) (Z) + (η + η (Z))0 2 ( (u 0 ) (Z), (u 1 ) (Z) )0 3 (d(u 0, u 1 )) (10) Adding the above equation with Z = x to the above with Z = y, we obtain for a.e. x Ω, u η 2 + u 1 η 2 u u η( u u 1 2 )d 2 (u 0, u 1 ) η d 2 (u 0, u 1 ) + (η + η)0 2 ( u 0, u 1 )0 3 (d(u 0, u 1 )) If u 0 and u 1 are energy minimizers, then integrating over gives, 0 d 2 (u 0, u 1 ) η + η( u u 1 2 )d 2 (u 0, u 1 ) + η + η C 0 2 ( u 0, u 1 )0 3 (d(u 0, u 1 )) supp(η) 18

19 Let u be an energy minimizing map and u ω (x) = u(x + δw ), with W 1. dividing by δ 2 and letting δ 0, we obtain, 2 η( u 2 u (W ) 2 u (W ) 2 η. Then, Adding the above equation with W = x to the above with W = y, we obtain 2 η u 4 u 2 η. Now if u is conformal, then inequality 10 implies λ η + λ 1 η λ 0 + λ 1 + η(λ 0 + λ 1 )d 2 (u 0, u 1 ) η (Z)d 2 (u 0, u 1 ) (Z) + (η + η (Z))0 2 (λ 0, λ 1 )0 3 (d(u 0, u 1 )). Substituting Z = η η if η 0 and following the same procedure as above we obtain 2κ ηλ 2 λ η. 5 The Curvature Inequality In this section, we prove our main result. As mentioned in the introduction, if X is a smooth Riemannian manifold of sectional curvature bounded from above by κ, then the inequality in the following theorem implies that the curvature of the surface is also bounded from above by κ. Theorem 5.1 Let u : X be a minimal surface (i.e. a weakly conformal energy minimizing map) with conformal factor λ where (X, d) is a metric space of curvature bounded from above by κ. Then for all non-negative ϕ Cc (), log λ ϕ 2κ ϕλ. (11) Proof: We will prove this for the case of κ = 1. The result κ arbitrary is obtained in the same manner as below. Before we proceed with the proof of Theorem 5.1, we need the following preliminary lemmas: Lemma 5.2 Let λ be a conformal factor of a minimal surface u : X where X is a CAT (1) space. Then λ H 1 loc(). 19

20 Proof. Let K. Since λ is bounded locally, we let Λ be such that λ Λ in K. Choose g C (), non-negative such that g Λ 2. Then, by Theorem 8 we have (λ + g) φ = φ( λ 2 + g) 0 for any φ Cc (K). Hence, and λ + g is weakly subharmonic in K and is a non-negative function. Thus λ + g H 1 (K) and λ Hloc(). 1 Lemma 5.3 Let λ be a conformal factor of a minimal surface u : X where X is a CAT (1) space. Then for any harmonic function h : R, ϕ(λe h ) 2 ϕλ 2 e h. Proof. Let w(z) : be a conformal change of coordinates. Then v = u w is harmonic. Let λ = v 2 be the conformal factor for the pull-back metric on under the map v. By Theorem 4.1, λ z ϕdz = 2 ϕ λ 2 dz for all non-negative ϕ Cc (). Note that Hence, z = dw dz w, dw = dw dz, dz λ = (u w) = u dw dw = λ. dz dz dw dz ( wϕ)λ dw dz dw dz 1 dw 2 ϕ dw dz 2 λ 2 dw dz 1 dw, and we get the desired result by choosing w such that dw dz = eh. Hence, let w = e ψ where ψ is an analytic function such that Reψ = h. Lemma 5.4 Let λ be a conformal factor of a minimal surface map u : X where X is a CAT (1) space. Assume λ λ 0 > 0. Then for any harmonic function h : R, ϕ log λ ϕ(2λ + (log λ + h) 2 ) Proof. Since λ is bounded away from zero and locally bounded above, we can assume that log λ Hloc(). 1 Let h be any harmonic function. By Lemma 5.3 and by the fact that C functions are dense in H 1, for any non-negative ψ Hloc(), 1 ψ(λe h ) 2 ψλ 2 e h. 20

21 Let ϕ C c () be a non-negative function, then ϕ log λ = ϕ(log λ + log e h ) = ϕ log λe h = ϕ log λe h ϕ = λe h (λeh ) = ( ϕ 2 = 2 λe ) h (λeh ) ϕ (λeh ) 2 (λe h ) 2 ϕ λe h λ2 e h ϕ (log λ + h) 2 ϕλ ϕ (log λ + h) 2. Now we proceed with the proof of Theorem 5.1 in the special case that λ λ 0 > 0. Let δ > 0 be given. Since log λ L 2, by the Lebesque Point Lemma, 1 F = {x lim log λ(y) log λ(x) 2 dµ(y) = 0} σ 0 πσ 2 y B σ(x) is of full measure in. For x F, let σ x be such that and y B 5σx (x) 0 < σ x < 1 dist(x, ) 5 log λ(y) log λ(x) 2 dy 25δπσ 2 x. Note that {B σx (x)} x F is a collection of closed balls such that x F B σx (x) is of full measure in. By the Five Times Covering Lemma, we can choose a disjoint subcollection {B σxi (x i )} i=1 such that B σx (x) B σ5xi (x i ). x F i=1 Let ϕ Cc ɛ such that be a non-negative function. Since log λ ϕ, ϕλ L 1 (), there exists log λ ϕ < δ, A A ϕλ < δ 21

22 whenever m(a) < ɛ. On the other hand, since i=1 m(b σxi (x i )) π, there exists N such that m( B 5σxi (x i )) < ɛ. i=n+1 Set A = i=n+1 B 5σxi (x i ). Let {χ i } N i=1 be a partition of unity subordinate to {B 5σxi (x i )} N i=1. Then ϕ log λ = + log λ ϕ A A δ + A N log λ ( ϕχ i ) i=1 N = δ + log λ (ϕχ i ) i=1 A N δ (ϕχ i )(2λ + log λ + h i 2 ) i=1 A N δ 2 ϕλ ϕ log λ + h i 2 A i=1 B 5σxi N 2δ 2 ϕλ ϕ log λ + h i 2. i=1 B 5σxi where {h i } is any collection of harmonic functions in. For each i, we choose h i to be a linear function, bounded uniformly away from 0, such that h i = log λ(x i ). Thus, N i=1 = B 5σxi log λ + h i 2 N log λ log λ(x i ) i=1 B 5σxi (x i ) N 25δπ σx 2 i. i=1 But since {B σxi (x i )} is a disjoint set, N i=1 σ 2 x i 1 and thus, Since the choice of δ was arbitrary, ϕ log λ 2δ 25δ ϕ π 2 ϕλ ϕ log λ 2 22 ϕλ.

23 Finally, since the choice of ϕ Cc () was also arbitrary, we have the desired result. The general case can be handled using the following lemma. Lemma 5.5 Let f n : R be a decreasing sequence of functions converging to a nonnegative function f such that f n M for all n = 1, 2,... If for all non-negative ϕ C c for all non-negative ϕ C c log f n ϕ 2 ϕf n, (), we also have that (). log f ϕ 2 ϕf, Proof: We will show: lim n lim ϕf n = n log f n ϕ = ϕf log f ϕ. The first equality follows immediately from the Lebesgue Convergence Theorem. To prove the second equality, let g C () such that g 2M. Then (log f n + g) ϕ ϕ( 2f n + g) 0. Hence log f n + g is subharmonic. By the mean value inequality (and assuming w.l.o.g. that f(0) 0), < log f(0) + g(0) log f n (0) + g(0) 1 log f n + g π log M + g. In particular, log f n in uniformly bounded. Let F n = log M log f n, and F = log M log f. Then F n is an increasing sequence of non-negative functions. Hence by the Monotone Convergence Theorem, lim F n = F. n 23

24 In other words, we have that lim log f n = log f. n In particular, log f L 1 (). For non-negative ϕ Cc (), we have that log f n ϕ log f ϕ a.e. and log f n ϕ ϕ log f n. Hence by the ominated Convergence Theorem, we have the desired result as we take δ 0. Now consider the space X endowed with the distance function d δ defined by d 2 δ((p, z), (Q, w)) = d 2 (P, Q) + δ z w 2, for P, Q X and z, w. It can be easily checked that (X, d δ ) is a CAT (1) space and u δ : (X, d δ ) defined by u δ (z) = (u(z), z). is a u δ is a minimal surface if u is. We let λ δ be the conformal factor of the pull back metric. Note that λ δ is a decreasing sequence of functions converging to λ and λ δ δ. By the special case above, we have that ϕ log λ δ 2 ϕλ δ, for all δ. Hence by Lemma 5.5, we get the desired result when we take δ 0. 6 Surfaces with Conformal Factor λ As mentioned in the introduction, when X is a smooth Riemannian manifold of sectional curvature bounded from above by κ, the inequality of Theorem 5.1 implies that the curvature of the minimal surface is also bounded from above by κ. In this section, we will see that this interpretation of Theorem 5.1 also makes sense in the setting where X is a metric space of curvature bounded from above by κ; we show that the conformal factor λ induces a metric space on which has upper curvature bound of κ. Theorem 6.1 Let (X, d) be a complete metric space of curvature bounded from above by κ and let u : X be a minimal surface (i.e. a weakly conformal energy minimizing map) with conformal factor λ. Let γ : [0, 1] be a piecewise C 1 curve and let l(γ) = 1 0 λ(γ(t)) γ (t) dt. For x, y, we define the distance between x and y as d λ (x, y) = inf{l(γ) : γ piecewise C 1 and γ(0) = x, γ(1) = y}. Then (, d λ ) is a metric space with curvature bounded from above by κ (locally if k > 0). The metric topology is equivalent to the surface topology. 24

25 Remark: The fact that λ Hloc() 1 follows from the inequality of Theorem 5.1. Hence the definition of l(γ) makes sense. The statement that a space has curvature bounded from above by κ locally means that each point is contained in a neighborhood which has an upper curvature bound of κ. Proof: The fact that d λ defines a length space and the statement about the equivalence of the topologies follow from the work of Reshetnyak [R3] and the weak inequality of Theorem 5.1. (Reshetnyak considers a metric λ(dx 2 + dy 2 ) where log λ is a difference of two subharmonic functions.) We need to show the curvature bound. It is sufficient to consider the cases κ = 1, κ = 0 and κ = 1. The general case then follows by simply scaling the distance function d of X so that the curvature is either κ = 1, 0 or 1. We let λ σ, (log λ) σ be symmetric mollifications (i.e. mollification by a symmetric mollifier) of λ, log λ and let λ σ = e (log λ)σ. Also let σ = {z : z < 1 σ}. By applying Theorem 5.1 with ϕ the mollifier, we have log λ σ = (log λ σ ) 2κλ σ = 2κ ( ) λσ λ σ λ σ (12) for every z σ. By Jensen s inequality, λ σ λ σ. Hence, for κ = 1 or κ = 0, 1 2λ σ log λσ κ. Thus for κ = 1 and κ = 0, ( σ, λ σ (dx 2 + dy 2 )) is a smooth Riemannian surface with curvature bounded from above by 1 and 0, respectively. Furthermore, since λ is subharmonic, λ σ λ. This implies that d σ d λ where d σ is the distance function induced by λ σ (dx 2 + dy 2 ). Combining this with the fact that λ σ λ in H 1, it is easy to check that d σ d λ. From the above discussion, the curvature bound for the case κ = 1 and κ = 0 follows easily: Let x, y, z and let y t (resp. y σ t ) be the point on the geodesic from y to z with respect to the distance function d λ (resp. d σ ) so that d λ (y, y t ) = td λ (y, z) (resp. d σ (y, y σ t ) = td σ (y, z)). Claim: For y t and y σ t defined above, we have d λ (y t, y σ t ) 0 as σ 0. Proof: Assume κ = 1. Consider the geodesic triangle (y t, y, z) with respect to d σ. By using the curvature bound of ( σ, d σ ) we have cosh d σ (y t, y σ t ) sinh(1 t)dσ (y, z) sinh d σ (y, z) cosh d σ (y t, y) + sinh tdσ (y, z) sinh d σ (y, z) cosh dσ (y t, z). 25

26 As σ 0, d σ (y, z) d λ (y, z), d σ (y t, y) d λ (y t, y) = td λ (y, z) and d σ (y t, z) d λ (y t, z) = (1 t)d λ (y, z). Hence the right hand side of the above inequality converges to 1 and since d λ (y t, y σ t ) d σ (y t, y σ t ), this proves the claim. The case κ = 0 is proved analogously.. For κ = 1, we want to show that cosh d λ (x, y t ) sinh(1 t)d λ(x, y) sinh d λ (x, y) cosh d λ (x, y) + sinh(1 t)d λ(x, z) sinh d λ (x, z) cosh d λ (x, z) which is equivalent to showing the triangle comparison property of efinition 3.1. Since ( σ, d σ ) has curvature bounded from above by 1, we have cosh d σ (x, y σ t ) sinh(1 t)dσ (x, y) sinh d σ (x, y) cosh d σ (x, y) + sinh(1 t)dσ (x, z) sinh d σ (x, z) cosh d σ (x, z) and the desired inequality follows by taking σ 0 and using the claim. The proof for κ = 0 follows analogously. Now we treat the case κ = 1. First we show (, d λ ) has curvature bounded from above by 2. Let r (z 0 ) = {z : z z 0 < r}. Since ( σ, d σ ) is a smooth Riemannian surface, by the isoperimetric inequality of [Hu], ( r(z 0 ) ) 2 λσ ds ( ) 4π ( log λ σ ) + dxdy λ σ dxdy ( r(z 0 ) r(z 0 ) ) 4π 2 λ σ dxdy λ σ dxdy r(z 0 ) r(z 0 ) where ( log λ σ ) + = max{ log λ σ, 0}. By taking σ 0, we have ( ) 2 ( 2 λds 4π λdxdy 2 λdxdy). r(z 0 ) r(z 0 ) r(z 0 ) [R2] says that if a surface with a metric λ(dx 2 + dy 2 ) has an isoperimetric inequality for disks r (z 0 ) of the form L 2 4πA κa 2 where L is the length of r (z 0 ) and A is area of r (z 0 ), then the surface has an upper curvature bound of κ. This implies (, d λ ) has an upper curvature bound of 2. Let us call k the best curvature bound of (, d λ ) if for every geodesic triangle T with diam(t ) < π 2 and angles α, β, γ, α + β + γ α k + β k + γ k where α k, β k, γ k are angles of a comparison triangle in S k. By the above, we know that k 2. We wish to show k 1. Suppose not, i.e. 1 < k 2. We need the following claim 26

27 to obtain a contradiction. Claim: Suppose 1 < k < k. There exists a constant C k,k > 1 such that for any geodesic triangles T k S k and T k S k with same side lengths and the sum of side lengths less than π 2, we have area(t k ) C k,k area(t k ). Furthermore, C k,k 1 as k k. Proof: Let a, b, c be the side lengths of geodesic triangle T S k. From spherical geometry, as a 0, b 0 or c 0, area(t ) approaches the area of a Euclidean triangle with side lengths a, b, c. Thus if T k,i S k and T k,i S k are geodesic triangles with side lengths a i, b i, c i and if they form a maximizing sequence of the ratio area(t k,i ) area(t k,i) then we can extract a subsequence so that a i a > 0, b i b > 0, c i c > 0. Thus, C k,k = area( T k ) area( T k ) where T k S k, T k S k are geodesic triangles with side lengths a, b, c. The last assertion is obvious. Let T be any geodesic triangle and let α, β, γ be the angles of T. (see [R3], Theorem 8.1.7), we have α + β + γ 1 log λdxdy + π 2 T λdxdy + π T = area(t ) + π By Gauss-Bonnet Let T k be a the comparison triangle in S k. Clearly, area(t ) area(t k ). By claim, for 1 < k < k and comparison triangle T k in S k, area(t k ) C k,k area(t k ). We note that C k,k is independently of T chosen. We choose k sufficiently close to k so that C k,k < k. Applying Gauss-Bonnet on T k S k, we obtain, area(t k ) k area(t k ) α k + β k + γ k π 27

28 where α k, β k, γ k are angles of T k. Thus, α + β + γ α k + β k + γ k. Since T can be chosen arbitrarily, this implies that the best curvature bound for (, d λ ) is k. This contradiction implies that the best curvature bound is not greater than 1. References [A] A.. Alexandrov. A Theorem for Triangles in a Metric Space and Some Applications of it. Tr. Mat. Inst. Steklova, 38 (1951) [ABN] A.. Alexandrov, V.N. Berestovskii, and I.G. Nikoleav. Generalized Riemann Spaces. Russian Mathematical Surveys, 41 (1986), [GT]. Gilbarg and N.S. Trudinger. Elliptic Partial ifferential Equations of Second Order. Springer-Verlag. Berlin, [GS] M. Gromov and R. Schoen. Harmonic Maps into Singular Spaces and p-adic Superrigidity for Lattices in Groups of Rank One. IHES Publications Math ematique, 76 (1992), pp [HL] R. Hardt and F-H Lin. Harmonic Maps into Round Cones and Singularities of Neumatic Liquid Crystals. Math Z., 213 (1993) No. 4, [J] J. Jost. Convex Functionals and Generalized Harmonic Maps into Spaces of NOnpositive Curvature. Comment. Math. Helv. 70 no. 4 (1995) [HK] W.K. Hayman and P.B. Kennedy. Subharmonic Functions, vol.1. Academic Press. London, [Hu] A. Huber, On the Isoperimetric Inequality on Surfaces of Variable Gaussian Curvature. Annals of Math, 60 no.2 (1954) [KS] N. Korevaar and R. Schoen. Sobolev Spaces and Harmonic Maps for Metric Space Targets. Communication in Analysis and Geometry, 1 (1993), [KS2] N. Korevaar and R. Schoen. Global Existence Theorems for Harmonic Maps to Non-locally Compact Spaces. Preprint. [M] [N1] C.B. Morrey. The Problem of Plateau on a Riemannian Manifold. Annals of Mathematics, 49 (1948), I.G. Nikolaev. Synthetic Methods in Riemann Geometry. Lecture Notes. University of Illinois,

29 [N2] I.G. Nikolaev. Solution of the Plateau Problem of Curvature At Most K. Siberain Mathematical Journal, 20 (1979), [R1] Y.G. Reshetnyak. Nonexpanding Maps in a Space of Curvature No Greater than K. Siberian Mathematical Journal, 9 (1968), [R2] [R3] [S1] [S2] Y.G. Reshetnyak. On the Isoperimetric Property of Two imensional Manifolds of Curvature Not Greater than K Vestn. Leningr. Univ. 16 (1961), No.19, (Review in English Zbl. 106, 365.) Y.G. Reshetnyak. Encyclopaedia of Mathematical Sciences, vol 70. Geometry IV. Springer Verlag. Berlin, T. Serbinowski. Harmonic Maps into Metric Spaces with Curvature Bounded Above Ph.. thesis. University of Utah, T. Serbinowski. Boundary Regularity of Harmonic Maps to Nonpositively Curved Metric Spaces. Communications in Analysis and Geometry, 2 no.1 (1994), mese@math.usc.edu University of Southern California 29