ON THE LOCAL VERSION OF THE CHERN CONJECTURE: CMC HYPERSURFACES WITH CONSTANT SCALAR CURVATURE IN Sn+1

After nearly 50 years of research the Chern conjecture for isoparametric hypersurfaces in spheres is still an unsolved and important problem and in particular its local version is of great interest, since here one loses the power of Stokes’ Theorem as a method for proving it. Here we present a related result for CMC hypersurfaces in Sn+1 with constant scalar curvature and three distinct principal curvatures.


Introduction
The Chern conjecture for isoparametric hypersurfaces in spheres can be stated as follows. Let M be a closed, minimally immersed hypersurface of the (n + 1)-dimensional sphere S n+1 with constant scalar curvature. Then M is isoparametric.
One obvious generalization is that on non-closed manifolds, i.e., a local version of the conjecture. This has in particular been proposed by Bryant for the case n = 3.
Let M ⊂ S 4 be a minimal hypersurface with constant scalar curvature. Then M is isoparametric.
For more details, a short history and an overview of results we would like to refer to the review article [3] by Scherfner, Weiss and Yau.
Here we will give a result related to the local version. Let n > 3 and M ⊂ S n+1 be a hypersurface with constant mean and scalar curvatures which has three pairwise distinct principal curvatures everywhere, then M is isoparametric.

Preliminaries
Let M be an n-dimensional hypersurface in a unit sphere S n+1 (1). We choose a local orthonormal frame field {e 1 , . . . , e n+1 } in S n+1 (1), so that restricted to M , e 1 , . . . , e n are tangent to M . Let ω 1 , . . . , ω n+1 denote the dual co-frame field in S n+1 (1). We use the following convention for the indices: A, B, C, D range from 1 to n + 1 and i, j, k from 1 to n. The structure equations of S n+1 (1) as a hypersurface of the Euclidean space R n+2 are given by whereR is the Riemannian curvature tensor The contractionsR AC = BRABCB andR = A,BRABAB are the Ricci curvature tensor and the scalar curvature of S n+1 (1), respectively. Next, we restrict all the tensors to M . First of all, since ω n+1 = 0 on M , i ω n+1,i ∧ ω i = dω n+1 = 0. By Cartan's lemma we can write Here h = i,j h ij ω i ω j denotes the second fundamental form of M and the principal curvatures λ i are the eigenvalues of the matrix (h ij ). Furthermore the mean curvature is given by H = 1 n i h ii = 1 n i λ i and K = det(h ij ) = i λ i is the Gauss-Kronecker curvature. We also define Independently of the choice of the e i we have and so on. On M we have where R is the Riemannian curvature tensor on M with components satisfying These structure equations imply the following integrability condition (Gauss equation): For the scalar curvature we have If we consider minimal hypersurfaces, the Ricci curvature and scalar curvature are given by, respectively, It follows from (2.9) that κ is constant if and only if S is constant. The covariant derivative ∇h with components h ijk is given by Then the exterior derivative of (2.8) together with the structure equations yields the following Codazzi equation In addition we have We will use the following result by Otsuki given in [2]. Proof. Let λ, µ und ν be the distinct principal curvatures with corresponding multiplicities r 1 , r 2 and r 3 . From r 1 + r 2 + r 3 = n and the definitions of H and S one has a system of equations with continuous coefficients which the r i solve uniquely. Thus the r i are continuous functions and therefore constant.
Locally we choose the e i such that h is diagonal in every point. For the directional derivatives of the principal curvatures one has Let the principal curvature directions corresponding to the three principal curvatures be called e A , e a and e α . Then (2.12) implies We consider different cases for the multiplicities of the principal curvatures. Without loss of generality, let r 1 ≥ r 2 ≥ r 3 . Case 1: r 1 , r 2 , r 3 > 1. Then Lemma 2.1 implies λ A = µ a = ν α = 0, and with (3.1) it follows that all derivatives of the principal curvatures vanish.
From (2.14) one has and thus Let v a be the column vector of the h aAn for a given a, then this can be expressed in Since the left hand side can only have rank 0 or 1, it follows that z 1 = 0 and therefore h aAn = 0 for all a und A. From (3.1) it follows that On the other hand, z 1 = 0 implies 1 and one has Then (2.15) is of the form From r 1 λ + r 2 µ + ν = nH, r 1 λ 2 + r 2 µ 2 + ν 2 = S, one has Solving for λ yields where w := ± −nr 2 µ 2 + 2nr 2 Hµ + (1 + r 1 )S − n 2 H 2 r 1 (1 + r 1 ) 2 .