## Existence of elliptic point

In this short note we will give a condition for a Riemannian manifold ${M}$ 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 ${\mathbb{R}^n}$ has an elliptic point.

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

Although the case where ${M=\mathbb{R}^n}$ is included in the theorem, we give the proof for the ${\mathbb{R}^n}$ case here for reference, as it is easier.

Proof in the Euclidean case: We can assume ${0}$ is in the interior of ${\Sigma}$. Let ${p}$ be the furthest point on ${\Sigma}$ from ${0}$, with distance ${r_0}$. For any curve ${\gamma}$ on ${\Sigma}$ with ${\gamma'(0)=v\in T_p\Sigma}$, we have

$\displaystyle 0\geq \langle \gamma,\gamma\rangle''= 2 \langle \gamma'',\gamma\rangle + 2|\gamma'|^2= 2 \langle \gamma'',r_0n\rangle + 2|v|^2=-2 r_0 h(v,v)+2|v|^2.$

Thus the principal curvatures are ${\geq \frac{1}{r_0}>0}$. $\Box$

We now prove the main result.
Proof of Theorem 1: By Cartan-Hadamard theorem, we can identify ${M}$ with ${\mathbb{R}^n}$. We can assume that ${0}$ is in the interior of ${\Sigma}$. Let ${p}$ be a point on ${\Sigma }$ whose distance ${r}$ from ${0}$ is the furthest. Let ${r_0}$ be this distance. We claim that ${p}$ is an elliptic point of ${\Sigma}$.

Let ${S_{r_0}}$ be the geodesic sphere of radius ${r_0}$ centered at ${0}$. This is a smooth hypersurface of ${M}$ by Cartan-Hadamard theorem. We need the following two lemmas.

 Lemma 2 (Comparison principle) Let ${h^\Sigma}$ and ${h^S}$ be the second fundamental form of ${\Sigma}$ and ${S_{r_0}}$ respectively (we use the convention that ${h^\Sigma (u,v)= \langle \nabla _u n, v\rangle}$ and ${n}$ is the outward normal field of ${\Sigma}$ ). Then for all ${v\in T_p\Sigma(=T_p S_{r_0}}$), $\displaystyle h ^\Sigma(v,v)\geq h^S(v,v).$

Proof: Let ${\gamma }$ be a curve on ${\Sigma}$ and ${\beta }$ be a curve on ${S_{r_0}}$ such that ${\gamma'(0)=\beta'(0)=v}$, then at ${p}$,

$\displaystyle \begin{array}{rcl} 0\geq (r\circ \gamma)''(0)=\langle \partial_r ,\gamma'\rangle' &=& \langle \nabla _{\gamma'}\partial_r , \gamma'\rangle+ \langle \partial_r , \nabla_{\gamma'}\gamma'\rangle\\ &=&\langle \nabla_v \partial_r, v\rangle+ \langle \partial_r , \nabla_{\gamma'}\gamma'\rangle . \end{array}$

On the other hand, similar calculation shows that

$\displaystyle 0= (r\circ \beta)''(0)=\langle \nabla_v\partial_r, v\rangle + \langle \partial_r ,\nabla_{\beta'}\beta'\rangle.$

So we have

$\displaystyle -h^S(v,v)=\langle \partial_r ,\nabla_{\beta'}\beta'\rangle \geq \langle \partial_r ,\nabla_{\gamma'}\gamma'\rangle= -h^\Sigma(v.v).$

$\Box$

 Lemma 3 (Rauch comparison theorem, see also here) ${S_r}$ has principal curvatures ${\geq \frac{1}{r}}$.

Proof: Let ${X}$ be a unit vector at ${p=\exp_0(r_0 v)\in S_{r_0}}$, and let ${Y(r)}$ be the Jacobi field along ${\exp_0(r v)}$ such that ${Y(0)=0 }$ and ${Y(r_0)=X}$. Then as ${[Y, \partial_r]=0}$ (since ${Y}$ is a variational vector field of a family of geodesic), at ${p}$

$\displaystyle h^S(X, X) = \langle \nabla _Y \partial _r , Y\rangle = \langle \nabla_r Y, Y\rangle = \frac{1}{2}\partial_r\langle Y,Y\rangle =|Y(r_0)||Y|'(r_0).\ \ \ \ \ (1)$

Then differentiating and using the above equation again,

$\displaystyle \begin{array}{rcl} |Y|''&=& (\langle Y'', Y\rangle +\langle Y',Y'\rangle )|Y|^{-1}-|Y|^{-3}\langle Y',Y\rangle^2\\ &=& (\langle - R(Y,\partial_r)\partial_r , Y\rangle +\langle Y',Y'\rangle )|Y|^{-1}-|Y|^{-3}\langle Y',Y\rangle^2\\ &\geq& |Y|^{-3}(\langle Y',Y'\rangle |Y|^{2}-\langle Y',Y\rangle^2)\\ &\geq&0. \end{array}$

Let ${s(r)=r}$. Then ${s''=0}$. We thus have

$\displaystyle (s|Y|'-s'|Y|)'=s |Y|''\geq0.$

As ${(s|Y|'-s'|Y|)(0)=0}$, so ${s|Y|'\geq s'|Y|}$ for ${r>0}$, i.e.

$\displaystyle \frac{|Y|'}{|Y|}\geq \frac{1}{r}.$

Plugging this into (1), we have ${h^S(X, X)\geq \frac{1}{r}}$ at ${p}$.$\Box$

By these two lemmas, we see that ${h^\Sigma(v,v)\geq \frac{1}{r_0}>0}$ for all unit vector ${v\in T_p\Sigma}$. This finishes the proof. $\Box$

 Remark 1 The assumption ${M}$ to be nonpositively curved can’t be dropped. A counterexample is the equator in the sphere. The simple-connectedness of ${M}$ cannot be dropped also, as the equator of the flat torus is a counterexample. The conclusion may not hold even when ${M}$ is diffeomorphic to ${\mathbb{R}^n}$: 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 ${\Sigma}$ whose center is the center of this spherical cap is totally geodesic when the radius is large enough, i.e. ${\Sigma}$ is on the cylindrical part, thus it can’t have any geodesic point.