Cones in the Euclidean space with vanishing scalar curvature

Given a hypersurface M on a unit sphere of the Euclidean space, we define the cone based on M as the set of half-lines issuing from the origin and passing through M . By assuming that the scalar curvature of the cone vanishes, we obtain conditions under which bounded domains of such cone are stable or unstable.


INTRODUCTION
A natural generalization of minimal hypersurfaces in Euclidean spaces was introduced in (Reilly 1973).Reilly considered the elementary symmetric functions S r , r = 0, 1, . . ., n, of the principal curvatures k 1 , . . ., k n of an orientable hypersurface x : M n → R n+1 given by Here, k i 1 , . . ., k i n are the eigenvalues of A = −dg, where g : M n → S n (1) is the Gauss map of the hypersurface.Reilly showed that orientable hypersurfaces with S r+1 = 0 are critical points of the functional A r = M S r dM for variations of M with compact support.Thus, such hypersurfaces generalize the fact that minimal hypersurfaces are critical points of the area functional A 0 = M S 0 dM for compactly supported variations.

JOÃO LUCAS M. BARBOSA and MANFREDO DO CARMO
A breakthrough in the study of these hypersurfaces occurred in the last five years of last century: in (Hounie and Leite 1995) and (Hounie and Leite 1999) conditions for the linearization of the partial differential equation S r+1 = 0 to be an elliptic equation were found.This linearization involves a second order differential operator L r (see the definition of L r in Section 2) and the Hounie-Leite conditions read as follows: L r is elliptic ⇐⇒ rank(A) > r + 1 ⇐⇒ S r+2 = 0 everywhere.
In this paper, we will be interested in the case S 2 = 0.For this situation, since rank(A) cannot be two, the ellipticity condition is equivalent to rank (A)≥ 3.
In (Alencar et al. 2003) a general notion of stability was introduced for bounded domains of hypersurfaces of Euclidean spaces with S r+1 = 0.In the case we are interested, namely S 2 = 0, it can be shown that if we assume that L 1 is elliptic, an orientation can be chosen so that a bounded domain D ⊂ M is stable if In what follows, we denote by B r (0) the ball of radius r centered at the origin 0 of R n+1 .Let M n−1 be a smooth hypersurface of the sphere S n (1).A cone C(M) in R n+1 is the union of half-lines starting at 0 and passing through the points of M. It is clear that C(M) ∩ S n (1) = M.It is easy to show that C(M) − {0} is a smooth n-dimensional hypersurface of R n+1 .The manifold C(M) is referred to as the cone based on M n−1 .The part of the cone contained in the closure of the ring B 1 (0) \ B ε (0), 0 < ε < 1, is called a truncated cone and is denoted by C(M) ε .
In this note we present the following two theorems which provide a nice description of the stability of truncated cones in R n+1 based on compact, orientable hypersurfaces of S n (1), with S 2 = 0 and S 3 = 0 everywhere.Theorem 1.Let M n−1 , n ≥ 4, be an orientable, compact, hypersurface of S n (1) with S 2 = 0 and S 3 = 0 everywhere.Then, if n ≤ 7, there exists an ε > 0 so that the truncated cone C(M) ε is not stable.
Although Theorems 1 and 2 are interesting on their own right, a further motivation to prove these theorems is that, for the minimal case, they provide the geometric basis to prove the generalized Bernstein theorem, namely, that a complete minimal graph y = f (x 1 , . . ., x n−1 ) in R n , n ≤ 8, is a linear function (See (Simons 1968), Theorems 6.1.1,6.1.2,6.2.1, 6.2.2).
For elliptic graphs in R n with vanishing scalar curvature the question appears in a natural way.Of course, since we want to consider graphs with S 2 = 0 and S 3 never zero, we must start with n ≥ 4, and the solution cannot be a hyperplane.Thus the question is whether there exists an elliptic graph in R n , n ≥ 4, with vanishing scalar curvature.Cienc (2004) 76 (4) CONES WITH VANISHING SCALAR CURVATURE 633

PRELIMINARIES
For notational reasons, it will be convenient to denote the hypersurface of the Introduction by x : M → R n+1 .We first need to consider the Newton Transformations P r , that are inductively given by (1) and then define the differential operator L r by L r f = trace{P r Hessf } . (2) It turns out that L r is self-adjoint and that L r f = div(P r gradf ).
The second variation formula for the variational problem of the functional A 1 is, up to a positive constant, given by for test functions f of compact support in M.
Consider now a compact orientable (n − 1)-dimensional manifold M immersed as a hypersurface of the unit sphere S n (1) of the Euclidean space R n+1 .The cone C(M) based on M is described by Of course, the geometry of C(M) is closely related to the one of M and it is simple to compute the second fundamental form Ā of C(M) in terms of the second fundamental form A of M. In fact, one finds From this relation on the second fundamental forms it follows that Proposition 1.If Sr represents the elementary symmetric function of order r of C(M) and Pr its Newton transformations, then: Proof.The proof is direct except for the last item.But this can be done using finite induction and the definition of Pr . Let Proposition 2. With the above notation we have: Proof.The proof of this Lemma follows the same lines used to find the expression of the Laplacian in polar coordinates and using the previous proposition.

SKETCH OF PROOF OF THEOREM 1
First of all let us observe that since S 2 ≡ 0 then (S 1 ) 2 = |A| 2 ≥ 0. Hence, at a point where S 1 = 0 we would have that all the entries of the matrix A are zero and so S 3 = 0 what is forbidden by our hypothesis.Therefore, we will have (S 1 ) 2 > 0 everywhere.
According to Proposition 1, our hypotheses then imply that, for the cone C(M), we have S2 ≡ 0 and S1 and S3 never zero.
It was proved in (Hounie and Leite 1999) that, for a hypersurface of R n+1 with Sr ≡ 0, 2 ≤ r < n, the operator Lr−1 is elliptic if and only if Sr+1 is never zero.Then we conclude that L 1 and L1 are elliptic.
To prove the theorem, we are going to show the existence of a truncated cone C(M) for which the second variation formula attains negative values.Hence, from now on we are going to work on a truncated cone, with test functions f that have a support contained in the interior of the truncated cone.As we did before, for each test function f : C(M) → R and each fixed t we define ft : M → R by ft (m) = f (m, t).From Proposition 2 we have that The volume element of C(M) is easily seen to be An Acad Bras Cienc ( 2004) 76 (4) CONES WITH VANISHING SCALAR CURVATURE

635
Hence, using (3), ( 5) and the expression of the volume, the second variation formula on f becomes Since S 1 > 0, then t n−4 S 1 dt ∧ dM is a volume element in C(M), in particular in C(M) .We will represent it by dS.In fact, dS is a product of two measures.The first one on the real line: dξ = t n−4 dt; the second, on M, given by dµ = S 1 dM.So, dS = dξ ∧ dµ.We can then rewrite the second variation formula on f as: Define, now, the following two operators: Observe that we are considering the space C ∞ (M) with the inner product: and C ∞ [ , 1] with the inner product: Since L 1 is elliptic and M is compact then L 1 , and so L 1 , is strongly elliptic.The same is true for the operator Using orthonormal bases of eigenfunctions for theses operators one deduces the following Lemma: Lemma 1.For any test function f we have There exists a test function f such that I (f ) < 0 if and only if λ 1 + δ 1 < 0.
The operator L 2 is well known.In fact it has been used in (Simmons 1968) to prove his celebrated theorem.The following lemma contains all the information we need about this operator: An Acad Bras Cienc (2004) 76 (4) Lemma 2. The operator L 2 has eigenvalues where 1 ≤ k < ∞.
We will also need the following lemma whose proof uses Lemmas (3.7) and (4.1) in (Alencar et al. 1993).
Lemma 3. Let M n−1 be a compact, orientable, immersed hypersurface of S n (1) with S 2 ≡ 0 e S 3 never zero.Suppose n ≥ 4. The first eigenvalue of the operator L 1 in M satisfy: Finally, we observe that the lemma below completes the proof of Theorem 1.
Lemma 4. Let M n−1 be a compact, orientable, immersed hypersurface of S n (1) with S 2 ≡ 0, S 3 never zero and n ≥ 4. If n ≤ 7 then there exists > 0 such that the truncated cone CM is not stable.
Proof of the Lemma: From Lemmas 2 and 3 we have It is trivial to verify that the sum of the first two terms of the right hand side of this inequality is a quadratic polynomial, with positive second order term, whose roots are approximately 2.2 and 7.8 .Hence it is strictly negative for values of n ∈ {4, 5, 6, 7}, in fact, it is less than or equal to −1.Hence, Choosing sufficiently small we can guarantee that the right hand side is negative.Now, by Lemma 1, we see that CM is not stable.This proves Lemma 4 and completes the proof of the Theorem 1.

EXISTENCE OF STABLE CONES
In this section we sketch the proof o Theorem 2.
The following example has been considered by various people in different contexts (see e.g.(Chern 1968) and (Alencar et al. 2002) When ξ 1 describes S r (1) ⊂ R r+1 and ξ 2 describes S s (1) ⊂ R s+1 , by taking positive numbers a 1 and a 2 with a 2 1 + a 2 2 = 1, we have that x = a 1 ξ 1 + a 2 ξ 2 describes a submanifold M of dimension p = r + s of the sphere S p+1 (1) ⊂ R p+2 .The manifold M is diffeomorphic to S r (1) × S s (1) and so is compact and orientable.It can be shown that a 1 and a 2 can be chosen so that S 2 = 0 and S 3 = 0. We will show that, in this case, the truncated cone C(M) ε is stable as a hypersurface of R r+s+1 when r + s + 1 ≥ 8. 1 S 1 L 1 + 3 S 3 S 1 will be given by where the last equality comes from a long but straightforward computation.Therefore, using Lemma 2, the above value for λ 1 , and the fact that, in our case, n = p + 1, we obtain For n ≥ 8, the sum of the first two terms becomes > 1/4.Thus, for any choice of ε, λ 1 + δ 1 > 0. Together with Lemma 1, this completes the proof of Theorem 2.