In this short note we will give a condition for a Riemannian manifold for which every compact hypersurface has an elliptic point (i.e. all principal curvatures are positive). This is a generalization of the well known fact that every compact hypersurface in has an elliptic point.

Theorem 1Every compact hypersurface in a complete simply-connected Riemannian manifold with non-positive curvature has an elliptic point, i.e. all the principal curvatures of at this point have the same sign (which we can assume to be positive by convention).

Although the case where is included in the theorem, we give the proof for the case here for reference, as it is easier.

*Proof in the Euclidean case:* We can assume is in the interior of . Let be the furthest point on from , with distance . For any curve on with , we have

Thus the principal curvatures are .

We now prove the main result.

*Proof of Theorem 1:* By Cartan-Hadamard theorem, we can identify with . We can assume that is in the interior of . Let be a point on whose distance from is the furthest. Let be this distance. We claim that is an elliptic point of .

Let be the geodesic sphere of radius centered at . This is a smooth hypersurface of by Cartan-Hadamard theorem. We need the following two lemmas.

Lemma 2 (Comparison principle)Let and be the second fundamental form of and respectively (we use the convention that and is the outward normal field of ). Then for all ),

*Proof:* Let be a curve on and be a curve on such that , then at ,

On the other hand, similar calculation shows that

So we have

Lemma 3 (Rauch comparison theorem, see also here)

has principal curvatures .

*Proof:* Let be a unit vector at , and let be the Jacobi field along such that and . Then as (since is a variational vector field of a family of geodesic), at

Then differentiating and using the above equation again,

Let . Then . We thus have

As , so for , i.e.

Plugging this into (1), we have at .

By these two lemmas, we see that for all unit vector . This finishes the proof.

Remark 1The assumption to be nonpositively curved can’t be dropped. A counterexample is the equator in the sphere. The simple-connectedness of cannot be dropped also, as the equator of the flat torus is a counterexample. The conclusion may not hold even when is diffeomorphic to : consider an infinite two-dimensional cylinder with one open end and the other end capped with a axially symmetric “spherical cap” (like a hemisphere). The geodesic sphere whose center is the center of this spherical cap is totally geodesic when the radius is large enough, i.e. is on the cylindrical part, thus it can’t have any geodesic point.