Welcome to MathOverflow

Question

3

Given a subset $S$ of a Hadamard manifold $M$. Is there a curvature criterion (for $\partial S$) to decide whether $S$ is convex.

I am looking for a ganeralization of the following statement:

A connected subset of $\mathbb{R}^2$ bounded by a smooth curve $\gamma$ with $||\dot{\gamma}||=1$ is convex, if $\gamma$ "always turns in the same direction" - more formally if $t\mapsto det(\dot{\gamma}(t),\ddot{\gamma}(t))$ never changes sign.

edited Jan 10 2011 at 14:31

HenrikRüping

community wiki

2

I believe that the appropriate generalization is a positive semidefinite second fundamental form. – Deane Yang Jan 10 2011 at 14:33

@Henrik: Can you specify what you consider a convex subset of a Riemannian manifold? – Dror Atariah Jan 10 2011 at 14:48

convex for me means geodesically convex. is there another reasonable meaning? – Deane Yang Jan 10 2011 at 15:02

2

In Hadamard manifolds there is a unique geodesic between any two points so convexity can only mean geodesic convexity. – Igor Belegradek Jan 10 2011 at 15:08

1

Thank you. After some googling I found the paper "Locally convex hypersurfaces of negatively curved spaces" by S. Alexander (ams.org/journals/proc/1977-064-02/…), which adresses precisely this question. – HenrikRüping Jan 10 2011 at 15:48

show 1 more comment

HenrikRüping · Accepted Answer · 2011-01-13 18:39:10Z UTC

1

Thank you. After some googling I found the paper "Locally convex hypersurfaces of negatively curved spaces" by S. Alexander (ams.org/journals/proc/1977-064-02/…), which adresses precisely this question.

answered Jan 13 2011 at 18:39

HenrikRüping

community wiki

Convex subsets of Hadamard manifolds

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.