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).


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.

This entry was posted in Geometry. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s