Complete surfaces with negative extrinsic curvature in M2×R

J. Math. Anal. Appl. 423 (2015) 538–546

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Complete surfaces with negative extrinsic curvature in M2 × R ✩ José A. Gálvez ∗ , José L. Teruel Departamento de Geometría y Topología, Universidad de Granada, E-18071 Granada, Spain

a r t i c l e

i n f o

Article history: Received 29 March 2014 Available online 8 October 2014 Submitted by R. Gornet

a b s t r a c t We prove that there exist no complete vertical graphs in M2 × R with extrinsic curvature bounded from above by a negative constant, when M2 is a Riemannian surface with a pole and non-negative curvature. © 2014 Elsevier Inc. All rights reserved.

Keywords: Graphs with negative curvature Product spaces Hyperbolic Monge–Ampère equation

1. Introduction In 1964, Efimov proved that no complete surface can be C 2-immersed in the Euclidean 3-space R3 if its extrinsic curvature Kext is bounded from above by a negative constant (see [3,8]). This generalized Hilbert’s classical result which stated that no complete smooth immersion exists with negative constant extrinsic curvature in R3 . Efimov’s theorem has been tried to extend in several ways (see [4,10,9]). In this sense, not much is known about the behaviour of the extrinsic curvature of a complete surface in product spaces M2 × R (see [1,5]). In this paper we prove that there exist no entire vertical graphs with extrinsic curvature bounded from above by a negative constant in M2 ×R, where M2 is a Riemannian surface with a pole and with non-negative curvature KM . In our arguments we use Heinz’s ideas, who proved this result when R3 is the ambient space (see [7]). As a consequence of a result in [6] we obtain that, if M2 is a Riemannian surface with KM ≥ 0 and with a pole, then every entire vertical graph Σ in M2 × R satisfies inf |Kext | = 0, Σ

where Kext denotes the extrinsic curvature of Σ. ✩ Both authors are partially supported by MICINN–FEDER, Grant No. MTM2013-43970-P and by Junta de Andalucía Grant Nos. FQM325 and P09-FQM-5088. * Corresponding author. E-mail addresses: [email protected] (J.A. Gálvez), [email protected] (J.L. Teruel).

http://dx.doi.org/10.1016/j.jmaa.2014.10.002 0022-247X/© 2014 Elsevier Inc. All rights reserved.

J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

539

Moreover, we construct examples of complete surfaces with constant negative extrinsic curvature in certain product spaces M2 × R. 2. Graphs with negative extrinsic curvature Let M2 be a Riemannian surface and (r, θ) local geodesic polar coordinates around a point p0 ∈ M2 which are well defined for r < R, for a certain R > 0. The induced metric is given by ·,· = dr2 + G(r, θ)dθ2 . We consider M2 × R endowed with the product metric and a vertical graph Σ in M2 × R over the geodesic disk B(p0 , R) centred at p0 and with radius R, given in local coordinates as     ψ(r, θ) = expp0 r cos(θ), r sin(θ) , z(r, θ) , which we identify with (r, θ, z(r, θ)) in order to simplify notation. We denote by {∂r , ∂θ } and {∂r , ∂θ } the partial derivatives with respect to r and θ on M2 and Σ respec∂ ∂ and ∂θ ≡ ∂θ . Then tively, where, for instance, ∂r ≡ ∂r ∂r = ∂r + zr ∂t ∂θ = ∂θ + zθ ∂t . Thus, the induced metric on Σ is given by     ds2 = 1 + zr2 dr2 + 2zr zθ drdθ + G + zθ2 dθ2 and its second fundamental form is given by zrr ∇∂r ∂r , N  =  1 + zr2 +

zθ2 G

1

∇∂θ ∂r , N  =  1 + zr2 +

zθ2 G

1 ∇∂θ ∂θ , N  =  1 + zr2 +

zθ2 G

 

zθ Gr − + zrθ 2G



 zr Gr zθ Gθ − + zθθ , 2 2

where ∇ is the Levi-Civita connection on M2 × R and N=

−1 1 + zr2 +

  zθ ∂ z ∂ + − ∂ r r θ t G zθ2 G

is the upwards pointing unit normal vector of the graph. By using the above formulae, a straightforward computation gives √ ( G)r d

 zr2 +

       2 √ zθ zθ2 zθ z2 − √ dzr = 2 G 1 + zr2 + θ Kext (dr ∧ dθ), dθ + d zr d √ G G G G

where Kext denotes the extrinsic curvature of Σ.

(1)

J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

540

Now we define the following auxiliary function which will be useful for our purposes   √  zθ2 , G 1+ f (r) = G

r>0

Br

where Br denotes the ball centred at the origin of R2 and of radius r, with r < R. Here, we identify R2 with the tangent plane Tp0 M2 of M2 at p0 , in the usual way.  We note that throughout this section we will work with functions of type Br h(ρ, θ), where h(ρ, θ) is well defined in Br \ {(0, 0)}. However all these functions h(ρ, θ) can be continuously extended to the origin as √ √ it happens for the previous f (r). This is due to the facts that limp→p0 G(p) = 0, limp→p0 ( G)ρ (p) = 1 (see [2]) and the functions | √zθG | and |zρ | are bounded in a neighbourhood of the origin. Lemma 1. Under these conditions, we have 2  12   √ zθ2 2 |Br | ≤ f (r) ≤ |Br | G 1 + zρ + , G Br

where |Br | denotes the area of the geodesic disk B(p0 , r) in M2 . Proof. Since has

zθ2 G

≥ 0, the first inequality is clear. Moreover, by using the Cauchy–Schwarz inequality, one

 f (r) ≤

√

 12   G

Br

√

2  12 2  12    √ z2 z2 G 1 + zρ2 + θ = |Br | G 1 + zρ2 + θ . G G

Br

2

Br

Lemma 2. Let us denote by KM the Gauss curvature of M2 . Then in the previous conditions we have 2π

d dr

0

    2π √ √ zθ2 zθ2 2 2 √ dθ = ( G)r zr dθ + zρ + KM G G G 0

Br

2  √  zθ2 2 −2 G 1 + zρ + Kext . G Br

Proof. From (1) we can write

2

2  √  z2 G 1 + zρ2 + θ Kext G

Br

        √ zθ zθ2 zθ 2 − √ dzρ ( G)ρ d zρ + dθ + d zρ d √ = G G G Br





√  ( G)ρ − 1 d

= Br

+

zρ2 +

  zθ2 dθ G

        zθ z2 zθ − √ dzρ . d zρ2 + θ dθ + d zρ d √ G G G

Br

(2)

J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

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Using the Green theorem and integrating by parts one gets 

 √ ( G)ρ − 1 d

Br



 d

=

   zθ2 2 dθ zρ + G

      √    z2 z 2  √ zρ2 + θ d ( G)ρ − 1 dθ ( G)ρ − 1 zρ2 + θ dθ − G G

Br

Br

      √  z2 z2  √ zρ2 + θ ( G)ρρ ( G)r − 1 zr2 + θ dθ − G G

2π = 0

Br

   2π √ 2π 2π √ √   zθ2 zθ2 2 2 2 zr + = ( G)r zr dθ − zr dθ + dθ + KM G, ( G)r − 1 G G 0

0

0

(3)

Br

√

rr where we have used that KM = −( √G) . G On the other hand, using again the Green theorem one obtains

        zθ z2 zθ − √ dzρ d zρ2 + θ dθ + d zρ d √ G G G

Br

2π zr2 +

= 0

0

2π zr2 +

= 0

   2π  zθ zθ2 zθ − √ dzr dθ dθ + zr d √ G G G  zr zθθ zθ2 zr zθ Gθ zθ zrθ √ − √ dθ. + √ − G G 2G G G

(4)

We observe that 2π 0

  2π  zr zθθ ∂ zr zθ z zθ Gθ zθ zrθ √ − r √ √ dθ = dθ = 0. + √ ∂θ G 2G G G G

(5)

0

Moreover, it is easy to check 2π 2 0

zθ zrθ d √ dθ = dr G

2π 0

z2 √θ dθ + G

2π √ z2 ( G)r θ dθ. G

(6)

0

Using formulae (4)–(6) we have         zθ z2 zθ − √ dzρ d zρ2 + θ dθ + d zρ d √ G G G

Br

2π zr2 dθ +

= 0

2π  0

2π 2 √  z2 d z √θ dθ. 1 − ( G)r θ dθ − G dr G

Finally, by summing up (3) and (7), we obtain (2). 2

0

(7)

J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

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Recall that a point p0 in a manifold M2 is said to be a pole if the exponential map is a diffeomorphism from the tangent plane at p0 onto M2 . In such a case, M2 is complete. Remark 2.1. Let M2 be a Riemannian surface as above with a pole at p0 . Let us observe that, in these √ conditions, if the Gauss curvature KM of M2 is non-negative, then ( G)r is a non-negative function. Indeed, since √ √ ( G)rr = −KM G ≤ 0, √ one has that G is concave as a function of r. Fixed an arbitrary θ0 , if there existed a point r0 > 0 such √ √ √ that ( G)r (r0 , θ0 ) < 0 then, from the concavity of G(·, θ0 ) we would have that G(r, θ0 ) = 0 for a certain r > r0 . Now, we can establish the main theorem of this section. Theorem 1. Let M2 be a Riemannian surface with a pole. If the Gaussian curvature of M2 is non-negative then there is no entire vertical graph Σ in M2 × R with extrinsic curvature bounded from above by a negative constant. Proof. We will assume that there exists a positive constant α such that Kext ≤ −α and we will show that this leads to a contradiction. Differentiating f twice and using Lemma 2 one has d f (r) = dr 

 2π√  2π √ 2π 2 zθ2 d z √θ dθ dθ = ( G)r dθ + G 1+ G dr G 0

0

0

   2π √ √   zθ2 2 2 zρ + KM G = ( G)r 1 + zr dθ + G 0

Br

2  √  zθ2 2 G 1 + zρ + Kext . −2 G Br

Since KM ≥ 0, it is clear from Remark 2.1 that the following inequality holds    2π √ √   zθ2 2 2 zρ + KM G ≥ 0. ( G)r 1 + zr dθ + G 0

Br

Thus, using that Kext ≤ −α for a certain positive constant α f  (r) ≥ −2

2 2  √   √  z2 z2 2α f (r)2 G 1 + zρ2 + θ Kext ≥ 2α G 1 + zρ2 + θ ≥ G G |Br |

Br

Br

by Lemma 1. From the last inequality  2α 4α 4α d  d   2 f (ρ) = 2f  (ρ)f  (ρ) ≥ 2f  (ρ) f (ρ)2 ≥ f  (ρ) f (ρ)2 = f (ρ)3 . dρ |Bρ | |Br | 3|Br | dρ

J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

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By integrating both sides of the above inequality one gets r

d   2 f (ρ) ≥ dρ

r





 4α d  f (ρ)3 , 3|Br | dρ

where 0 <  < r. Thus, f  (r)2 ≥

  4α  4α  f (r)3 − f ()3 + f  ()2 ≥ f (r)3 − f ()3 . 3|Br | 3|Br |

It is clear from the definition of f that f () → 0 when  → 0, hence f  (r)2 ≥

4α f (r)3 3|Br |

and so 1 3 3 d 1 1 f (r)− 2 = f (r)− 2 f  (r) ≥ f (r)− 2 − dr 2 2

3 4α f (r) 2 = 3|Br |

α . 3|Br |

Let R1 and R2 be real numbers such that 0 < R1 < R2 . By integrating we have on one hand R2 1 1 1 d f (r)− 2 dr = −f (R2 )− 2 + f (R1 )− 2 . − dr R1

On the other hand, since KM ≥ 0, from the volume comparison theorem we have |Br | ≤ πr2 , and one has from the above inequality

R2 R2 R2  d α α 1 − 12 f (r) dr = − dr ≥ dr dr 3|Br | 3 |Br |

R1

R1

≥

R1

α 3

R2

1 √ dr = r π

  R2 α log . 3π R1

R1

So, by Lemma 1

|BR1 |− 2 ≥ f (R1 )− 2 ≥ f (R1 )− 2 − f (R2 )− 2 ≥ 1

1

1

1

Hence

that is,

  R2 3π − 12 |BR1 | , ≥ log α R1

  α R2 . log 3π R1

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J.A. Gálvez, J.L. Teruel / J. Math. Anal. Appl. 423 (2015) 538–546

 R2 ≤ R1 exp

1 3π α |BR1 |



for every 0 < R1 < R2 , with R2 an arbitrary real value. 2 We observe that if M2 is compact then, for any entire vertical graph Σ in M2 × R there exist points in Σ with non-negative extrinsic curvature. This happens at the points where the height function achieves a local maximum or a local minimum. Finally, as a consequence of Theorem 1 and [6, Theorem 1] we have Corollary 1. Let M2 be a Riemannian surface with a pole and non-negative Gauss curvature. If Σ is an entire vertical graph in M2 × R with extrinsic curvature Kext then inf |Kext | = 0. Σ

3. Complete surfaces with constant negative extrinsic curvature In this section we will give an example of a complete surface with strictly negative extrinsic curvature in a certain product manifold. Consider M2 as the surface R2 with the induced metric ·,· = dρ2 + G(ρ)dθ2 which is well defined in R2 \ {(0, 0)}, where (ρ, θ) are the usual polar coordinates of R2 . The function G(ρ) will be defined later. Let Σ be a rotational surface in M2 × R parametrized by     Ψ (r, θ) = k(r) cos(θ), k(r) sin(θ), z(r) ≡ k(r), θ, z(r) where k(r) > 0 and z(r) are smooth functions, and r denotes the arc length parameter of the curve β(r) = Ψ (r, 0). Thus k (r)2 + z  (r)2 = 1.

(8)

If we denote by {∂r , ∂θ } and {∂ρ , ∂θ } the respective partial derivatives of Σ and M2 we have     ∂r (r, θ) = k (r)∂ρ k(r), θ + z  (r)∂t k(r), θ   ∂θ (r, θ) = ∂θ k(r), θ . So from (8), the induced metric in our surface is given by   ds2 = dr2 + G k(r) dθ2 , and its second fundamental form is   II (r, θ) = −k (r)z  (r) + k (r)z  (r) dr2 +



 z  (r)Gρ (k(r)) dθ2 . 2

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Hence, the principal curvatures of Σ can be written as λ1 = −k (r)z  (r) + k (r)z  (r) λ2 = z  (r)

Gρ (k(r)) , 2G(k(r))

and the extrinsic curvature of the surface is Kext = −z  (r)

 Gρ (k(r))   k (r)z  (r) − k (r)z  (r) . 2G(k(r))

(9)

By differentiating (8) we have k (r)k (r) + z  (r)z  (r) = 0, and so (9) becomes the following ODE for k k (r) = −

2G(k(r)) Kext . Gρ (k(r))

(10)

4

Now we take, for instance, the function G(ρ) = ρ4 eρ and impose that the extrinsic curvature of Σ is constant −1. Under these conditions, (10) becomes k (r) =

k(r) . 2(1 + k(r)4 )

(11)

Now, let k(r) be a solution of the previous ODE with initial conditions k(0) = k0 > 0, k (0) = 0, where k0 is a positive value which will be determined later. It is well known that the solution k(r) to the previous problem is uniquely determined in a neighbourhood of the origin. Let us see in fact that k(r) is well defined for every r ∈ R. First, we observe that, from the uniqueness to the previous problem, k(r) = k(−r) since k(−r) is also a solution to (11) satisfying the same initial conditions. Thus, in order to study the behaviour of k(r), we will only work for r ≥ 0. From (11) we have d d   2 k (r) = dr dr



   1 arctan k(r)2 2

and so k (r)2 = c0 +

  1 arctan k(r)2 . 2

(12)

Hence there exists a constant c1 such that |k (r)| ≤ c1 , that is, k(r) is globally defined as we required. Since k(0) = k0 > 0 and k (0) = 0, if we take k0 > 0 small enough, then we will have c0 ∈ ( π4 − 1, 0). Thus, from (12) one gets |k (r)| < 1 for every r ∈ R. Therefore, from (8), z(r) is a well defined smooth function in R. Let us see that k(r) ≥ k(0) = k0 > 0 for every r ∈ R. In such a case, Σ is a regular surface due to the fact that the generatrix curve β(r) = Ψ (r, 0) does not touch the rotation axis. For that purpose, let us first observe that k(r) > 0, for every r > 0.

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Let us assume that there exists a first point  > 0 such that k() = 0. Since k(r) is a positive function in the interval [0, ), from (11) we obtain that k(r) is a convex function in that interval. Moreover, k (0) = 0, thus k() ≥ k(0) = k0 > 0, and this is a contradiction. On the other hand, since from (11) the function k(r) is a global convex function in R and k (0) = 0, then k(r) ≥ k(0) = k0 for every r ∈ R and lim k(r) = +∞.

r→±∞

In particular, this shows that Σ is properly immersed in M2 × R. One can easily check that our metric ·,· is not well defined at the origin. Since k(r) ≥ k0 > 0 for every r ∈ R, we can perturb the metric ·,· for instance for 0 ≤ ρ < k20 , and have a complete metric in M2 well defined at the origin and such that Σ preserves the same induced metric. As Σ is properly immersed in M2 × R, then Σ is a complete surface with negative constant extrinsic curvature as we claimed. References [1] J.A. Aledo, J.M. Espinar, J.A. Gálvez, Complete surfaces of constant curvature in M2 × R and S2 × R, Calc. Var. Partial Differential Equations 29 (2007) 347–363. [2] M.P. Do Carmo, Riemannian Geometry, Birkhäuser Boston Inc., Boston, MA, 1992. [3] N.V. Efimov, Generation of singularities on surfaces of negative curvature, Mat. Sb. (N.S.) 64 (1964) 286–320. [4] N.V. Efimov, Differential homeomorphism tests of certain mappings with an application in surface theory, Mat. Sb. (N.S.) 76 (1968) 499–512. [5] J.M. Espinar, J.A. Gálvez, H. Rosenberg, Complete surfaces with positive extrinsic curvature in product spaces, Comment. Math. Helv. 84 (2009) 351–386. [6] J.A. Gálvez, V. Lozano, Existence of barriers for surfaces with prescribed curvatures in M2 × R, J. Differential Equations 255 (2013) 1828–1838. [7] E. Heinz, Über Flächen mit eineindeutiger Projektion auf eine Ebene, deren Krümmungen durch Ungleichungen eingenschränkt sind, Math. Ann. 129 (1955) 451–454. [8] T. Klotz-Milnor, Efimov’s theorem about complete immersed surfaces of negative curvature, Adv. Math. 8 (1972) 474–543. [9] J.M. Schlenker, Surfaces à courbure extrinsèque négative dans l’espace hyperbolique, Ann. Sci. Ec. Norm. Super. 34 (2001) 79–130. [10] B. Smyth, F. Xavier, Efimov’s theorem in dimension greater than two, Invent. Math. 90 (1987) 443–450.