Minimal surfaces of finite total curvature in H × R - IMPA

Minimal surfaces of finite total curvature inH × RLaurent Hauswirth and Harold RosenbergMay 20, 2007We dedicate this work to Renato Tribuzy, on his 60’th birthday.1 IntroductionWe consider complete minimal surfaces Σ in H × R, H the hyperbolic plane.Let C(Σ) denote the total curvature of Σ, C(Σ) = ∫ KdA, K the intrinsicΣcurvature of Σ. We shall prove that C(Σ) is an integer multiple of 2π, whenit is finite. We give examples of such Σ with total curvature −2πm, m anynon-negative integer.In R 3 , complete minimal surfaces of finite total curvature have total curvaturean integer multiple of −4π. This results from the Gauss map of thesurface, that extends meromorphically to the conformal compactification.In H × R, we have no conformal Gauss map. We have the holomorphicquadratic differential of the harmonic height function: projection on the Rfactor of H × R.Now we describe a simply connected example. Let Γ be an ideal polygonof H with m+2 vertices at infinity, 2m+2 sides, A 1 , B 1 , A 2 , B 2 , ..., A m+1 , B m+1 .Let D be the convex hull of Γ.In [3], the authors find necessary and sufficient conditions on the ”lengths”of the A i and B j which ensure the existence of a minimal graph u : D −→ R,taking the values +∞ on each A i and −∞ on each B j .They prove the graph of such a u is complete and of total curvature −2πm.The Γ obtained from the m+2 roots of unity satisfies the ”length” conditions.Thus this gives examples of total curvature −2πm for each integer m ≥ 1.1

For m = 0, take Σ = γ × R, γ a complete geodesic of H. It wouldbe interesting to construct non-simply connected examples of finite totalcurvature. For example an annulus of total curvature −4π.2 PreliminariesWe consider X : Σ → H × R a minimal surface conformally embedded inH × R, H the hyperbolic plane. We denote by X = (F, h) the immersionwhere F : Σ −→ H is the vertical projection to Σ = Σ×(0), and h : Σ −→ Rthe horizontal projection. We consider local conformal parameters z = x+iyon Σ. The metric induced by the immersion is of the form ds 2 = λ 2 (z)|dz| 2 .If H is isometrically embedded in L 3 the Minkowski space, the meancurvature vector is (see B.Lawson [8], page 8)2 −→ H = (△X) T X(H×R) = ((△F) T F H , △h) = 0Then F is a harmonic map and h is a real harmonic function. In the following,we will use the unit disk model for H. We will note (D, σ 2 (u)|du| 2 ) thedisk with the hyperbolic metric σ 2 (u)|du| 2 . We will denote |v| 2 σ = σ2 |v| 2 ,〈v 1 , v 2 〉 σ = σ 2 〈v 1 , v 2 〉 where |v| and 〈v 1 , v 2 〉 stands for the standard norm andinner product in R 2 . The harmonic map equation in the complex coordinateu = u 1 + iu 2 of D (see [12], page 8) isF z¯z + 2(log σ ◦ F) u F z F¯z = 0 (1)where 2(log σ ◦ F) u = 2 ¯F(1 − |F | 2 ) −1 . In the theory of harmonic maps thereare two global objects to consider. One is the holomorphic quadratic Hopfdifferential associated to F:Q(F) = (σ ◦ F) 2 F z ¯Fz (dz) 2 := φ(z)(dz) 2 (2)The function φ depends on z, whereas Q(F) does not. An other object is thecomplex coefficient of dilatation (see Alhfors [1]) of a quasi-conformal map,which does not depend on z, a conformal parameter on Σ:a = F¯zF z2

Since we consider conformal immersions, we have|F x | 2 σ + (h x ) 2 = |F y | 2 σ + (h y ) 2〈F x , F y 〉 σ+ h x .h y = 0hence (h z ) 2 (dz) 2 = −Q(F) (see [10]).Then the zeroes of Q are double and we can define η as the holomorphic oneform η = ±2i √ Q. The sign is chosen so that:∫h = Re η (3)When X is a conformal immersion then the unit normal vector n in H×Rhas third coordinate:〈n, ∂ ∂t 〉 = n 3 = |g|2 − 1|g| 2 + 1whereg 2 := − F zF¯z= − 1 aThen we define the function ω on Σ (which has poles where Σ is horizontal)by n 3 = tanh ω. By identification we have(4)ω = 1 2 ln |F z||F¯z |(5)Using the equations above (2),(4) we can express the differential dF independentlyof z by:1dF = F¯z d¯z + F z dz =2σ ◦ F g−1 η − 1 gη (6)2σ ◦ FThe metric ds 2 = λ| dw| 2 is given [10] in a local coordinate z by:ds 2 = (|F z | σ + |F¯z | σ ) 2 |dz| 2 (7)Thus combining equations 6 and 7, we derive the metric in terms of gand η byds 2 = 1 4 (|g|−1 + |g|) 2 |η| 2 = 4 cosh 2 ω|Q| (8)3

We remark that the zeroes of Q correspond to the poles of ω so that theimmersion is well defined. Moreover the zeroes of Q are points of Σ, wherethe tangent plane is horizontal.It is a well known fact (see [12] page 9) that harmonic mappings satisfythe Böchner formula:△ 0 log |F z||F¯z | = −2K HJ(F) (9)where J(F) = σ 2 ( |F z | 2 − |F¯z | 2) is the Jacobian of F with |F z | 2 = F z F z .Hence taking into account (2), (4), (5) and (9):△ 0 ω = 2 sinh(2ω)|Q| (10)where △ 0 denote the laplacian in the euclidean metric |dz| 2 . From this wededuce△ Σ ω = n 3where △ Σ is the Laplacian in the metric ds 2 .The Gauss curvature is given by:K Σ = K(X x , X y ) + K ext = − tanh 2 ω −|∇ω| 24 cosh 4 ω|Q|This formula follows from the fact that the sectional curvature of the tangentplane to Σ at a point z is −n 2 3 and the second fundamental form isω xII =cosh ω (dx)2 −ω xcosh ω (dy)2 + 2 ω ycosh ω (dxdy)The total curvature is defined by∫C(Σ) = K Σ dA3 Minimal surfaces of finite total curvatureTheorem 3.1. Let X be a complete minimal immersion of Σ in H × R withfinite total curvature. Thena) Σ is conformally M − {p 1 ...., p n }, a Riemann surface punctured in afinite number of points.4Σ

) Q is holomorphic on M and extends meromorphically to each puncture.If we parameterize each puncture p i by the exterior of a disk of radius R 0 ,and if Q(z) = z 2m i(dz) 2 at p i then m i ≥ −1.c) The third coordinate of the unit normal vector n 3 → 0 uniformly ateach puncture.d)The total curvature is a multiple of 2π:∫(−KdA) = 2π(2 − 2g − 2k −n∑m i )Proof. The proof of this theorem uses arguments of harmonic diffeomorphismstheory as can be found in the work of Han, Tam, Treibergs andWan[4] ,[13], [5] and Minsky [11].The conformal type is an application of Huber’s theorem ([7]). Σ isconformally a compact Riemann surface minus a finite number of points (theends).We consider M(r 0 ) = M − ∪ i D(p i , r 0 ); the surface minus a finite numberof disks removed around the punctures p i . Around each puncture weconsider a conformal parametrization of the punctured disk D ∗ (p i , r 0 ). Weparametrize these ends by the exterior of the disk of radius R 0 in C. In thisparameter we express the metric as ds 2 = λ 2 |dz| 2 with λ 2 = 4 cosh 2 ω|φ| in aconformal parameter z. Then −Kλ 2 = 2∆ 0 lnλ where ∆ 0 = 4∂ 2 z¯z.Let us define u = ln cosh 2 ω, a subharmonic function by Böchner’s formula:∆ 0 u = 8 sinh 2 ω|φ| + 2|∇ω|2cosh 2 ω ≥ 0The function u is globally defined, since ω is globally defined on Σ.Step 1: We prove that the holomorphic quadratic differential Q has afinite number of zeroes on M.Since the zeroes are isolated, Q has a finite number of zeroes on thecompact part M(r). Then we assume that there is a disk D ∗ (p i , r) whichcontains an infinite number of zeroes of Q, {z i }. We parametrize conformallythis disk on the exterior of the disk of radius R 0 . In this parameter Q(z) =φ(z)(dz) 2 and if ∆ 0 is the laplacian in the flat metric |dz| 2 at the puncture:i=1∆ 0 ln |φ| = ∑ 2πδ zi5

Then with −Kλ 2 − 1∆ 2 0 ln |φ| = 1∆ 2 0u we have on the annulus C(R) ={R 0 ≤ |z| ≤ R}:∫(−KdA) − mπ = ∫1 ∆ C(R) 2 C(R) 0u ≥ 0Then m has to be finite and ∫ C(r) ∆ 0u ≤ C 0Step 2: An upper bound.∫ ∫ ∫∂u 2π∫∆ 0 u =∂n = ∂u 2π∂R Rdθ −C(R)∂C(R)0= R ddRNow let I(r) := ∫ 2π0u(r, θ)dθ. Then∫ 2π0ddR I(R) ≤ C 1R0∂u∂R R 0dθu(R, θ)dθ − R 0∫ 2π0∂u∂r dθ ≤ 2C 0I(R) − I(R 0 ) ≤ 2C 1 ln R R 0Then for R >> R 0 large we have, with a > 0, b > 0:I(r) ≤ a ln r + bStep 3: Since φ has a finite number of zeroes, we prove at each puncturecosh 2 ω|φ| ≤ β|z| α |φ|.To prove φ extends meromorphically to the punctures, we will use a theoremof Osserman [9] (recall that the metric is complete). For R > R 0 , φ is withoutzeroes and for |z| = R large enough, u is subharmonic, henceu(z) ≤ 4 uπR∫B(z,R/2)2≤ 4uπ|z|∫B(0,3|z|/2)−B(0,|z|/2)2≤ 4 ∫ 3|z|/2I(r)rdr ≤ 4 ∫ 3|z|/2(a ln r + b)rdrπ|z| 2 π|z| 2|z|/2≤ α ln |z| + β6|z|/2

Thenand2 lnλ = u + ln |φ| ≤ α ln |z| + β + ln |φ|λ 2 = cosh 2 ω|φ| ≤ e β |z| α |φ|Thus the function φ extends meromorphically to the puncture by Osserman[9].Step 4: We now prove that the function φ(z) ≡ z 2m , with m ≥ −1 ateach puncture.If m ≤ −2, then we can conformally parametrize the end on the punctureddisk by w = 1/z. Then Q(w) = ψ(w)(dw) 2 with φ(1/w) = w 4 ψ(w), whereψ(w) has a pole of order 2m + 4. If 2m + 4 ≤ 0, the following integral isfinite:∫|φ|dz < ∞D(p i ,r)Now, by the finite total curvature hypothesis we will show the area of theend is finite:∫HenceD∫K Σ dA =D∫2∆ 0 ln λ =∫=DD∫Area(D) =∫8 sinh 2 ω|φ| +∫8 cosh 2 ω|φ| −DDDcosh 2 ω|φ| < ∞.2|∇ω| 2cosh 2 ω∫8|φ| +D2|∇ω| 2cosh 2 ωBut a complete end of Σ has infinite area by the monotonicity formula (see[2]).c) Now we prove that n 3 → 0 uniformly at each puncture. We adaptestimates on positive solutions of sinh-Gordon equations by Minsky [11],Wan [13] and Han [5] to our context.7

At each puncture we can choose R 0 such that φ(z) is without zeroes on|z| ≥ R 0 /2. Since φ(z) is without zeroes, the minimal surface is transverse tohorizontal sections H×{c} and we parametrize locally simply connected subdomainsof the end by w = i 2 (x 3 + ix ∗ 3 ) = ∫ √ φdz so that |dw| 2 = |φ(z)||dz| 2is a flat metric. If we consider z ∈ C R0 , then on the disk D(z, |z|/2), we havethe conformal coordinate w = √ φ(z)dz, with the flat metric |dw| 2 = |φ||dz| 2 .In this metric, under the hypothesis that m ≥ −1, the disk D(z, |z|/2) containsa ball of radius at least c ln |z|, where c is independant of z.The function ω satisfies the sinh-Gordon equation∆ |φ| ω = 2 sinh 2ωwhere ∆ |φ| is the laplacian in the flat metric |dw| 2 . For |z| ≥ R 0 , we can finda disk of radius at least r = c ln |z| around z in the |dw| 2 metric. When z islarge, the radius r diverges to +∞.Then for R 0 large enough we can find a disk with radius 1 in the |dw| 2metric around any point z with |z| ≥ R 0 . On this disk D |φ| (z, 1), we considerthe hyperbolic metric given by (w is w − z in the following step):dσ 2 = µ 2 |dw| 2 :=4(1 − |w| 2 ) 2 |dw|2Then µ take infinite values on ∂D |φ| (z, 1) and since the curvature of thismetric is K = −1, the function ω 2 = ln µ satisfies the equation∆ |φ| ω 2 = e 2ω 2≥ e 2ω 2− e −2ω 2= 2 sinh 2ω 2Now we apply a maximun principle to bound ω above as in Wan [13].The same holds with (˜ω = −ω):Let η = ω − ω 2 . Then∆η = e 2ω − e −2ω − e 2ω 2= e 2ω 2(e 2η − e −4ω 2e −2η − 1)which can be written in the metric d˜σ 2 = e 2ω 2|dw| 2 , as∆˜σ η = e 2η − e −4ω 2e −2η − 1Since ω 2 goes to +∞ on the boundary of the disk, the function η isbounded above and attains its max at an interior point p 0 , η(p 0 ) = ¯η and∆¯η ≤ 0. At this point we have8

Hencee 2η − e −4ω 2e −2η ≤ 1e 2η ≤ 1 + √ 1 + a 22where a = e −2ω 2(p 0 ) ≤ sup 1 ≤ 1 . Then at any point of the diskµ 2 4ω ≤ ω 2 + 1 2 ln 1 + √ 1 + 1/42The same estimate holds with ˜ω = −ω. Then at z (i.e. w = 0) we have|ω(z)| ≤ ln4 + 1 2 ln 1 + √ 1 + 1/4:= K 02uniformly on R ≥ R 0 . Using this estimate we can apply a maximun principleas in Minsky [11]. For |z| large, we can find a disk D |φ| (z, r) with r large too.We consider the functionF(x, y) = K 0cosh r cosh √ 2x cosh √ 2yThen F ≥ K 0 ≥ ω on ∂D |φ| (z, r). Since ∆F = 4F, we apply the maximunprinciple to have ω ≤ F. If p 0 is a point where ω(p 0 ) ≥ F(p 0 ) is a minimunof F − ω, then 0 ≤ ω(p 0 ) ≤ sinh ω(p 0 ) and∆(F − ω) = 4F − 2 sinh 2ω ≤ 4(F(p 0 ) − ω(p 0 )) ≤ 0Hence ω ≤ F on the disk. We have |ω| ≤ F by considering the same argumentwith F + ω. Hence|ω(z)| ≤ K 0cosh rAnd |ω| → 0 uniformly at the puncture i.e. the tangent plane becomevertical.d) Now we compute the total curvature. We apply Gauss-Bonnet on thecompact piece M(r) = M − ∑ 1≤i≤k D(p i, r) and we obtain∫ ∫K Σ dA + k g = 2π(2 − 2g − k)M(r)∂M(r)9

Here k g is the geodesic curvature of ∂M(r) = Γ 1 ∪...Γ k on the surface M(r).Now consider a puncture p i parameterized on R ≥ R 0 . We consider w = x+iya parametrization of the punctured disk (with w = ∫ √ φdz). In the w-plane,if φ(z) = z 2m there are 2m+2 horizontal asymptotic directions i.e. directionswith Im(w) = 0 (diverging curves at zero level) which define some angularsector in R ≥ R 0 . Now for C 1 >> 0 large, we consider the ”polygon” Γ(C 1 )which is the union of segments of curves Re(w) = ±C 1 and Im(w) = ±C 1 ,alternatively. At each change of direction the exterior angle is π/2. Thesecurves, with Γ i = {R = R 0 } bound an annulus Ω(r, C 1 , p i ) and∫ ∫ ∫K Σ dA + k g − k g = −(2m + 2)πΩ Γ(C 1 ) Γ iNow we let C 1 → 0. If we prove ∫ k Γ(C 1 ) g → 0, we will establish that∫K Σ dA = 2π(2 − 2g − 2k − ∑ m i )M iwhere φ(z) = z 2m iat each p i .Now we prove ∫ Γ(C 1 ) k g → 0. This fact comes from the exponential decreasingproperty of the function ω. First we prove∫Im(w)=C 1k g ds → 0The curve Im(w) = C 1 is a horizontal curve at level C 1 , parameterized byRe(w) = x. In Hauswirth [6], we find an expression of the curvature of thecurve as function of ω. In the w variable (recall that the Hopf differential isQ = 1 4 (dw)2 ):k g (x) = −ω ycosh ωNow we need a gradient estimative of ω. Schauder’s estimate gives ( withthe exponential decreasing property of ω proved above):|u| 2,α ≤ C(| sinhω| 0,α + |ω| 0 ) ≤ Ce −R .√On the curve x + iC 1 , we have |∇ω| ≤ Ce −|C1| e − x 2 C −21 +1 and∫∫ +∞|k g |ds = |ω y |dx ≤ C|C 1 |e −|C 1|Im(w)=C 1−∞10

which is converging to zero as |C 1 | → +∞. Now we prove∫Re(w)=C 1k g ds → 0.Now suppose the curve is not horizontal. If γ(y) = (F(C 1 , y), y) where Fis the harmonic map into H, we have γ ′ (y) = (F y , 1) and |γ ′ (y)| 2 = 1 +sinh 2 ω (conformal coordinate). Then the curvature of the curve in H × R ishorizontal. This curvature can be expressed as a function of ω byk g (y) =ω xsinh ωNow one can argue as above to prove the result.References[1] L. Ahlfors. On quasi-conformal mapping; McGraw-Hill,New York.[2] T. H. Colding and W. P. Minicozzi II, Minimal surfaces, Courant LectureNotes in Mathematics, 4. New York University, Courant Instituteof Mathematical Sciences, New York ,1999.[3] P. Collin and H. Rosenberg, Construction of harmonic diffeomorphismsand minimal graphs, arxiv.org/math.DG/0701547..[4] Z-C Han, L-F Tam, A Treibergs and T Wan, Harmonic maps from thecomplex plane into surfaces with nonpositive curvature , Communicationsin Analysis and Geometry, Vol 3,1(1995),85-114.[5] Z-C Han, Remarks on the geometric behavior of harmonic maps betweensurfaces; Elliptic and Parabolic Methods in Geometry, 1996 A.K Peters,Ltd.[6] L. Hauswirth, Generalized Riemann examples in three-dimensional manifolds,Pacific Journal of Math.,224(2006), no. 1,91-117.[7] A. Huber, On subharmonic functions and differential geometry in thelarge. Comment. Math. Helv. 32 (1957), 13-72.11

[8] B. Lawson. Lectures on minimal submanifolds; T1, Mathematics LectureSeries; 009, Publish or Perish.[9] R. Osserman, A survey of minimal surfaces, Dover, 1986.[10] R. Sa Earp and Eric Toubiana. Screw motion surfaces in H × R and§ 2 × R; Illinois Journal of Mathematics, 2005.[11] Y. Minsky, Harmonic maps, length, and energy in Teichmüller space,Journal of Diff. Geom. 35(1992), 151-217.[12] R. Schoen and S. T. Yau. Lectures on harmonic maps; Conference Proceedingsand lecture Notes in Geometry and Topology, II. InternationalPress, Cambridge, MA, 1997.[13] Tom Y.H. Wan, Constant mean curvature surface, harmonic maps anduniversal Teichmüller space, J. Differential Geometry, 35 (1992),643-657.12