# A characterization of quadric constant mean curvature hypersurfaces of spheres

###### Abstract.

Let be an immersion of a complete -dimensional oriented manifold. For any , let us denote by the function given by and by , the function given by , where is a Gauss map. We will prove that if has constant mean curvature, and, for some and some real number , we have that , then, is either a totally umbilical sphere or a Clifford hypersurface. As an application, we will use this result to prove that the weak stability index of any compact constant mean curvature hypersurface in which is neither totally umbilical nor a Clifford hypersurface and has constant scalar curvature is greater than or equal to .

###### Key words and phrases:

constant mean curvature, Clifford hypersurface, stability operator, first eigenvalue###### 2000 Mathematics Subject Classification:

Primary 53C42, Secondary 53A10## 1. Introduction

Let be an immersion of a complete -dimensional oriented manifold. For every we will denote by the tangent space of at . Sometimes, specially when we are dealing with local aspects of , we will identify with the set , and the space with the linear subspace of . Let us denote by , a normal unit vector field along , i.e., for every , is perpendicular to the vector and to the vector space . The shape operator , is given by where and is any smooth curve in such that and . It can be shown that the linear map is symmetric, therefore it has real eigenvalues . These eigenvalues are known as the principal curvatures of at . The mean curvature of at is the average of the principal curvatures,

and the norm square of the shape operator is defined by the equation

### 1.1. Examples: Totally umbilical spheres and Clifford hypersurfaces

In this section we will describe two families of examples that are related with the main result of this paper.

###### Example 1.

Let be a fixed unit vector and a real number with . Let us define

Clearly, is a hypersurface of . In this case the map given by

is a normal unit vector field along . Therefore, for every the shape operator is the map , where is the identity map, and

for all . It is not difficult to show that these examples are the only totally umbilical complete hypersurfaces of . In this case

are both constant on .

###### Example 2.

Given any integer and any real number , let us define and

It is not difficult to see that for any one gets

Therefore, the map given by

defines a normal unit vector field along , i.e. it is a Gauss map on . Notice that the vectors in of the form define a dimensional space. A direct computation, using the expression for , gives us that if , then,

Therefore is an eigenvalue of with multiplicity . In the same way we can show that is an eigenvalue of with multiplicity . Therefore, the principal curvatures of are given by

and we also have that

are both constant. Hypersurfaces that, up to a rigid motion, are equal to for some and , are called Clifford hypersurfaces.

### 1.2. Two families of geometric functions on hypersurfaces in spheres

Given a fixed vector , let us define the functions and by and , where is a Gauss map. When we consider all possible we obtain the families

These two families are very useful in the study of the spectrum of important elliptic operators defined on like the Laplacian an the stability operator. For example, in [10] and [11], Solomon computed the whole spectrum for the Laplace operator of every minimal isoparametric hypersurface of degree 3 in spheres using these two families of functions. For the totally umbilical spheres we have that if , then dim and dim. Indeed, it is not difficult to prove that if for some compact hypersurface in , we have that either or , then for some unit vector , [8, Lemma 3.1].

If we take , and we consider the example we observe that if is a vector perpendicular to the vector , then

We also have this kind of relation between the function and the function in the Clifford hypersurfaces; more precisely, if we consider the example and we take then we have that

Also, if we take , then, we have that

In this paper we will prove that these two examples are the only hypersurfaces with constant mean curvature in where the relation , for some non-zero vector , is possible. More precisely, we will prove the following result.

###### Theorem 3.

Let be an immersion with constant mean curvature of a complete -dimensional oriented manifold. If for some non-zero vector and some real number , we have that , then, is either a totally umbilical sphere or a Clifford hypersurface.

Recall that constant mean curvature hypersurfaces in are characterized as critical points of the area functional restricted to variations that preserve a certain volume function. As is well-known, the Jacobi operator of this variational problem is given by , with associated quadratic form given by

and acting on the space

Precisely, the restriction means that the variation associated to is volume preserving.

In contrast to the case of minimal hypersurfaces, in the case of hypersurfaces with constant mean curvature one can consider two different eigenvalue problems: the usual Dirichlet problem, associated with the quadratic form acting on the whole space of smooth functions on , and the so called twisted Dirichlet problem, associated with the same quadratic form , but restricted to the subspace of smooth functions satisfying the additional condition . Similarly, there are two different notions of stability and index, the strong stability and strong index, denoted by and associated to the usual Dirichlet problem, and the weak stability and weak index, denoted by and associated to the twisted Dirichlet problem. Specifically, the strong index of the hypersurface is characterized as

and is called strongly stable if and only if . On the other hand, the weak stability index of is characterized by

and is called weakly stable if and only if . From a geometrical point of view, the weak index is more natural than the strong index. However, from an analytical point of view, the strong index is more natural and easier to use (for further details, see [1]).

As an application of our Theorem 3, we will prove that the weak stability index of a compact constant mean curvature hypersurface in with constant scalar curvature must be greater than or equal to whenever is neither a totally umbilical sphere nor a Clifford hypersurface (see Theorem 9). This result complements the one obtained in [2] where the authors showed that the weak index of a compact constant mean curvature hypersurface in which is not totally umbilical and has constant scalar curvature is greater than or equal to , with equality if and only if is a Clifford hypersurface with radius . At this respect, it is worth pointing out that the weak stability index of the Clifford hypersurfaces depends on , reaching its minimum value when , and converging to as converges either to or (see [2, Section 3] for further details).

## 2. Preliminaries and auxiliary results

Let us start this section by computing the gradient of the functions and . For any fixed vector in , let us define the tangent vector field by

where, as in the previous section, is a Gauss map. Clearly, is a tangent vector field on because and for every . More precisely, is the orthogonal projection of the vector on .

###### Proposition 4.

If is a smooth hypersurface of and denotes its shape operator with respect to the unit normal vector field then, the gradient of the functions and are given by:

###### Proof.

For any vector , let be a curve such that and . Notice that

Since the equality above holds true for every and , then, . For the function , we have

Therefore, . ∎

We also have the following expressions for the Laplacian of the functions and .

###### Proposition 5.

If is a smooth hypersurface of with constant mean curvature , and denotes the shape operator with respect to the unit normal vector field then, the Laplacian of the functions and are given by:

###### Proof.

For any vector , we have

where denotes here the intrinsic derivative on . Let be an orthonormal basis of . Then, the Laplacian of at the point is given by

On the other hand, using Codazzi equation we also have that

Therefore

since the mean curvature is constant. ∎

The following two lemmas will be used in the proof of our main theorem. The first one is an elementary geometric lemma whose proof is left to the reader.

###### Lemma 6.

Let be a smooth hypersurface of and let be a regular curve such that

where is a smooth function and is a normal vector field along , i.e. is orthogonal to . If is the arc-length parameter for , then satisfies that is a normal vector field along , i.e. is a geodesic in .

The other one is an algebraic lemma.

###### Lemma 7.

If are polynomials of degree 1, , with the property that whenever , then, the polynomials

are linearly independent. Moreover, an equation of the form

with and real numbers, can not hold true unless all the ’s and are zero.

###### Proof.

By the condition on the numbers we have that at every polynomial , except the polynomial , vanishes. Therefore, if there exists constants such that

then, taking we get that for every . Therefore, the polynomials ’s are linearly independent. On the other hand, notice that the second equation in the lemma can be written as

where is a polynomial of degree . Since the expression on the left of the last equation is a polynomial of degree , we obtain that the constant on the right hand side must be zero. Then the second part of the lemma follows by the independence of the polynomials ’s. ∎

## 3. Proof of Theorem 3

We are now ready to give our main argument and prove Theorem 3. Since most of the arguments are local and the thesis of the theorem is on and not on , we will identify with and with . By multiplying the equation by an appropriated constant we may assume that . We will also assume that is not constant, otherwise for some , which implies, using the completeness of , that .

Notice that, since is not constant, then . Taking the gradient in both sides of the expression we obtain that

(1) |

at every point .

Step 1: The integral curves of in are Euclidean circles. Let us take a point such that does not vanish. Let be the integral curve of the vector field such that . Since

then,

Here

(3) |

is a normal vector field along and

Therefore if is the arc-length parameter for the curve with , and is the inverse of the function , we have, by Lemma 6, that is a geodesic in . Moreover, from (3) and (3) we also get that

(4) |

If we differentiate (4), we get that the function moves along a circle because it satisfies the equation

More precisely, if we define , then

and

(5) |

Step 2: The intersection is non-empty. Let us compute and in order to obtain an explicit expression for . From the definition of we have that and

(6) |

Notice that

at every point . Therefore

From this last expression we obtain that , at every , and

(7) |

Let us define , and . By (7) we have that , because . With this notation, we obtain that , and

where we have used (4) to derive the last equation. Now, using these equations jointly with (5) we get that

Notice that with . Therefore for some we have

so that

Notice that when moves from to , we have that never reaches the values , therefore by (7) and all these belong to the integral curve of the vector field . In particular, and . This argument shows that

is not empty. Observe that if we were assuming that were compact instead of complete, the fact that is not empty would have followed from the fact that the function must reach its maximum value and a minimum value on , and the fact that necessarily these values must be , since must vanish at its critical points. From now on we will assume that the that we were considering before is an element in , i.e, we will assume that , and therefore and .

Step 3: The intersection as a hypersurface of and as a hypersurface of . Clearly the set is an -dimensional manifold because is a regular value of the function on . Moreover, for every we have that is a constant vector, and therefore is a totally geodesic hypersurface of . Notice that for every we have that and . Therefore we can take vectors in , all of them orthogonal to , such that . Since the vectors ’s are perpendicular to , they form a basis for . On the other hand, notice that is also a hypersurface of the unit -dimensional sphere , and that for every , gives a unit vector field normal to in (see Figure 1).

Taking into account that is totally geodesic in and that is totally geodesic in , it follows from the fact that is both normal to in and normal to in that, for every , are the principal curvatures of as a hypersurface of with respect to

Step 4: Computation of the principal curvatures of along the integral curves of . Under the assumption that , we obtain from (6) that

Therefore, from (4) and (5) we get the following expression for ,

(8) |

By differentiating two times this equation, and using the equation (4), we obtain the following expression,

(9) |

Recall that, if , then

(10) |

Observe that if is a smooth curve in such that and , then by (8), we have that the curve

is a curve on such that . A direct computation shows that

(11) |

where

The computation above shows us that the vectors ’s are also elements in for every . Actually, it follows directly from (11) that if then ; hence by a continuity argument, since the equation has finitely many solutions on , we conclude that for every .

Recall that, by (10) and (1), is a principal curvature at the point , for every , with associated principal direction in the direction of . Let us compute now the other principal curvatures of at the point . Since for every , then the expression (9) holds true when replacing by and then we have that

Differentiating this equation with respect to at and using (11), we get that