Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
I believe that the appropriate generalization is a positive semidefinite second fundamental form. –  Deane Yang Jan 10 '11 at 14:33
    
@Henrik: Can you specify what you consider a convex subset of a Riemannian manifold? –  Dror Atariah Jan 10 '11 at 14:48
    
convex for me means geodesically convex. is there another reasonable meaning? –  Deane Yang Jan 10 '11 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 '11 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/S0002-9939-1977-0448262-6/…), which adresses precisely this question. –  HenrikRüping Jan 10 '11 at 15:48
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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