Research Article, Res Rep Math Vol: 1 Issue: 1
A Simpler Proof of the Characterization of Quadric CMC Hypersurfaces in Sn+1
1Department of Mathematics, Universidade Federal do Piaui, Teresina, Brazil
2Department of Mathematics, Universidade Federal de Campina Grande, Campina Grande, Paraiba, Brazil
*Corresponding Author : Aquino CP
Department of Mathematics, Universidade Federal do Piaui, Teresina, Brazil
Tel: (86) 3215-5525
E-mail: cicero.aquino@ufpi.edu.br
Received: August 10, 2017 Accepted: September 01, 2017 Published: September 07, 2017
Citation: Aquino CP, De Lima HF (2017) A Simpler Proof of the Characterization of Quadric CMC Hypersurfaces in Sn+1. Res Rep Math 1:1.
Abstract
In this short article, we present a new and simpler proof of a characterization of the quadric constant mean curvature hypersurfaces of the Euclidean sphere Sn+1, originally due to Alias, Brasil and Perdomo.
Keywords: Euclidean sphere; Constant mean curvature hypersurfaces; Support functions; Totally umbilical hypersurfaces; Clifford torus
Introduction
In 2008, Alias, Brasil and Perdomo studied complete hypersurfaces immersed in the unit Euclidean sphere , whose height and angle functions with respect to a fixed nonzero vector of the Euclidean space are linearly related. Let us recall that, for a fixed arbitrary vector the height and the angle functions naturally attached to a hypersurface endowed with an orientation ν are defined, respectively, by and In this setting, they showed the following characterization result concerning the quadric constant mean curvature hypersurfaces of Sn+1 [1,2]:
Theorem 1
Let be a complete hypersurface immersed in Sn+1 with constant mean curvature. la=λ fa for some non-zero vector and some real number λ, then Σn is either a totally umbilical hypersurface or a Clifford torus , for some k = 0; 1;..; n and some k=0,1,…,n and ρ>0.
Later on, working with a different approach of that used in [2], the first and second authors characterized the totally umbilical and the hyperbolic cylinders of the hyperbolic space Hn+1as the only complete hypersurfaces with constant mean curvature and whose support functions with respect to a fixed nonzero vector a of the Lorentz- Minkowski space are linearly related (see Theorem 4:1 of [3,4], for the case that a is either space like or time like, and Theorem 4:2 of [5], for the case that a is a nonzero null vector). In this short article, our purpose is just to use a similar approach of that in [4,5] in order to present a new and more simple proof of Theorem 1 (cf. Section 3). For this, in Section 2 we recall some preliminaries facts concerning hypersurfaces immersed in Sn+1.
Preliminaries
Let be an orientable hypersurface immersed in the Euclidean sphere. We will denote by A the Weingarten operator of Σn with respect to a globally defined unit normal vector ν.
In order to set up the notation, let us represent by ∇0 , ∇and ∇ the Levi-Civita connections of , Sn+1 and Σn respectively. Then the Gauss and Weingarten formulas for Σn in Sn+1 are given, respectively, by
and
for all tangent vector fields
In what follows, we will work with the first three symmetric elementary functions of the principal curvatures λ1,… λn of ψ, namely:
where i, j, k ∈ {1,…, n}.
As before, for a fixed arbitrary vector a let us consider the height and the angle functions naturally attached to which are defined, respectively, by and . A direct computation allows us to conclude that the gradient of such functions are given by and , where aΤ is the orthogonal projection of a onto the tangent bundle Τ Σ , that is,
Taking into account that and using Gauss and Weingarten formulas, we obtain for all . We use this previous identity jointly with Codazzi equation to deduce that
For all that Thus according to [6] (see also [3]), it follows from the last two identities that
(2.1)
(2.2)
where H = (1/n) S1 is the mean curvature function of Σn
For what follows, it is convenient to consider the so-called Newton transformation
P1= S1-A
where I is the identity operator. Naturally associated with the Newton transformation P1, we have the Cheng-Yau’s square operator [7], which is the second order linear di erential operator given by
(2.4)
Here stands for the self-adjoint linear operator metrically equivalent to the hessian of h, and it is given by
For all X, Y∈ x (Σ) .
Based on Reilly’s seminal paper [8-10], Rosenberg [6] showed the following idenfitities related to the action of on the functions la and fa:
(2.5)
And
(2.6)
To close this section, we quote a suitable Simons-type formula which can be found in [1] or [11].
(2.7)
Proof of Theorem
Now, we are in position to proceed with our alternative proof of Theorem 1.1. If λ=0 then la= λfa=0, that is
for all and, consequently, Σn is a totally umbilical sphere of Sn+1.
So, let us assume that λ≠0. We have Δla=λΔfa and using the fact that H is constant, from (2.1) and (2.2) we conclude that
Or equivalently,
Hence, we get that
(3.1)
By (3.1), we obtain
Thus,
(3.2)
We define a function
by
Suppose that h(p0)≠0 for some p0 ∈ Σn Since h is smooth, there exists a neighbourhood u of p0 in Σn in which h(p)≠0 all p∈ u From (3.2) we conclude that la=0 in u and, hence fa=0 in u. since λ≠0. we arrive at a contradiction because in Σn we have
Therefore, h = 0 on Σn, that is,
(3.3)
Consequently, S2 is constant on Σn . Repeating the previous argument for the operator L1 and using the fact that S2 is constant, we also obtain that
(3.4)
We observe that the above equation shows that S3 is also constant on Σn . We also note that this argument shows, in fact, that Sr is a constant function on Σn for all 2≤r≤n. From (2.7) we get
More precisely,
(3.5)
We observe that if H=0, then S1=0 and, consequently, |A|2=−2s2 From (3.3), we have 2S2 =−n and |A|2 = n. Therefore, since
We have that and, hence, from Theorem 4 of [9], we conclude that Σn must be a Clifford torus
, for some k=0,1,…,n and some ρ>0.
Now, suppose that H≠0 By equation (3.4) we get
that is,
(3.7)
From equation (3.5) we have
(3.8)
Furthermore, from a straightforward computation we can verify that
(3.9)
Hence, if S2 = 0 we obtain of (3.9) that consequently, |∇A|2 =0 and, since Σn is complete, it follows once more from Theorem 4 of [9] that Σn must be a Clifford torus.
If S2≠0 then implies
(3.10)
We note that (3.10) and (3.9) imply and, hence, repeating the previous argument we also get that Σn is a Clifford torus. Therefore, we conclude the proof of Theorem 1.
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