We study hypersurfaces either in the De Sitter space S n+1 1 ⊂ R n+2 1 or in the anti De Sitter space H n+1 1 ⊂ R n+2 2 whose position vector ψ satisfies the condition L k ψ = Aψ + b, where L k is the linearized operator of the (k+1)-th mean curvature of the hypersurface, for a fixed k = 0, . . . , n − 1, A is an (n + 2) × (n + 2) constant matrix and b is a constant vector in the corresponding pseudo-Euclidean space. For every k, we prove that when A is selfadjoint and b = 0, the only hypersurfaces satisfying that condition are hypersurfaces with zero (k + 1)-th mean curvature and constant k-th mean curvature, open pieces of standard pseudo-Riemannian products in S n+1When H k is constant and b is a non-zero constant vector, we show that the hypersurface is totally umbilical, and then we also obtain a classification result (see Theorem 2). (2010): 53C50, 53B25, 53B30 and hectorfabian.ramirez@um.es c ⊂ R n+2 q satisfies the condition L k ψ = Aψ + b, for some self-adjoint constant matrix A ∈ R (n+2)×(n+2) and some non-zero constant vector b. Since H k is assumed to be constant on M , from Lemma 9 we know that H k+1 is also constant on M . The case H k+1 = 0 cannot occur, because in that case we have b = 0 (see Example 1).
Mathematics Subject ClassificationsLet us assume that H k+1 is a non-zero constant. From (35) we obtain that b ⊤ = 0 and that the function b, ψ is constant on M . Now we use (34) to deduce that b, N = cH k H k+1 b, ψ = constant. Since b = ε b, N N + c b, ψ ψ, taking covariant derivative in this equation we have −ε b, N SX + c b, ψ X = 0, for any tangent vector field X. If b, N = 0, then M is totally umbilical (but not totally geodesic). Otherwise, b = c b, ψ ψ and then b, ψ = 0, but this implies b = 0. That concludes the proof of Theorem 2.