3
$\begingroup$

Let $M$ be a Riemannian manifold of dimension $n$. Let $N\subset M$ be a subset with smooth boundary $\Sigma=\partial N$. If one assume the second fundamental form $II$ with respect to inner normal direction of $\Sigma$ is nonnegative. (here we use the convention that the second fundamental form of round sphere in $\mathbb R^n$ is positive respective to inner normal.) It is easy to show that for any point $x\in \Sigma$ there exists a neighborhood $U$ around $x$ in $M$ such that $U\cap N$ is convex in the sense that $p, q\in U$ then any short geodesic connecting $p$ and $q$ remains in $U\cap N$.

Is it true that for any two point $p, q\in N$ any short geodesic(realize the distance) connecting them remains in $N$?

$\endgroup$
6
  • $\begingroup$ @Wille, I've correct the typos. Thanks! $\endgroup$
    – Ralph
    Commented Feb 18, 2013 at 19:47
  • 4
    $\begingroup$ And if I interpret it correctly, the answer is no. Consider the sphere $\mathbb{S}^n$ and the equatorial $\mathbb{S}^{n-1}$. Let the closure of the northern hemisphere be $C$. Its boundary, the equatorial sphere, is totally geodesic. But two antipodal points on the equatorial sphere have a short geodesic that goes through the south pole. More generally you get counterexamples if you take a cylinder $[0,1]\times \mathbb{S}^{n-1}$ with spherical caps. $\endgroup$ Commented Feb 18, 2013 at 19:52
  • $\begingroup$ @Willie, That's a good example. How about we strength the condition to positivity of second fundamental form? $\endgroup$
    – Ralph
    Commented Feb 18, 2013 at 19:57
  • $\begingroup$ Did you want to write: It is easy to show that for any point $x\in \Sigma$ there exists a neighborhood $U$ around $x$ in $M$ such that $U\cap N$ is convex in the sense that $p, q\in U\cap N$ then any short geodesic connecting $p$ and $q$ remains in $U\cap N$. $\endgroup$ Commented Feb 18, 2013 at 19:59
  • $\begingroup$ @Peter, you are right. I meant $U\cap N$. $\endgroup$
    – Ralph
    Commented Feb 18, 2013 at 20:05

1 Answer 1

7
$\begingroup$

From the comment of Willie Wong , the answer is NO.

Yet simpler example is $\mathbb{S}^1\times\mathbb{R}$, where you have locally convex discs which are not globally convex.

If you want "yes" as an answer, you have to assume bit more. For example simply connectedness plus sectional curvature $\le 0$ (It was proved by S. Alexander).

An other related statement: if an immersed hypersurface is locally convex plus curvature is nonnegative then it bounds an immersed locally convex ball. (this was conjectured by S. Alexander and proved by Gromov independently).

$\endgroup$
6
  • $\begingroup$ Can you clarify "curvature is nonnegative"? Do you mean curvature of the hypersurface or the curvature of the background? And do you mean nonpositive? $\endgroup$ Commented Feb 18, 2013 at 20:17
  • $\begingroup$ @Anton, What are the papers of S. Alexander and Gromov you mentioned? $\endgroup$
    – Ralph
    Commented Feb 18, 2013 at 20:35
  • $\begingroup$ @Ralph: the refs are: Alexander, S. Locally convex hypersurfaces of negatively curved spaces. Proc. Amer. Math. Soc. 64 (1977), no. 2, 321–325. and Gromov, M. Sign and geometric meaning of curvature. Rend. Sem. Mat. Fis. Milano 61 (1991), 9–123 (1994). $\endgroup$ Commented Feb 18, 2013 at 21:15
  • $\begingroup$ @Willie Wong: the curvature of ambient space is nonnegative. $\endgroup$ Commented Feb 18, 2013 at 21:16
  • $\begingroup$ @Anton: thanks. And also for the references! $\endgroup$ Commented Feb 20, 2013 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.