# Convex subsets of Hadamard manifolds

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.

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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
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
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