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Question: Is every open convex subset $C$ of a Riemannian manifold $M$, necessarily contractible?

Here by a "convex subset" I mean a set $C$ having the property that between each pair of points in $C$ there is a unique geodesic contained in $C$. (For example an open hemisphere is convex inside the sphere)

I have tried to construct a deformation retract along the unique geodesic connecting every point in $C$ to a fixed point in $C$. But is this map always continuous? If the answer is not positive in general, I'm also interested in the special case of complete manifolds with non-positive curvature.

Added. For the case of complete manifolds with non-positive curvature one can argue as follows: Let's $q\in C$ be the fixed point and $p\in C$. Now if $\gamma(t) = \mathrm{exp}(tv) (t\in[0,1])$ is the unique geodesic in $C$ connecting $q$ to $p$, there is a neighborhood $U$ of $v$ in $T_q M$ s.t. $\exp([0,1]*U)$ is contained in $C$. Now since $\exp$ has no critical points in non-positive curvature, $\exp(U)$ contains an open neighborhood of $q$ and we can use inverse mapping theorem.

Thanks!

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Fix $q \in C$. The set $D\subseteq T_q M$ of all tangent vectors $v$ such that $[0,1]\ni t \mapsto \exp_q(tv)$ is the unique geodesic in $C$ connecting $q$ with $p$ is open and star shaped, hence contractible.

Then $\exp_q: D \to C$ is a continuous map between manifods with the same dimension. It is also one-to-one, by your convexity hypothesis. By Browuer invariance of domain it is necessarily an homeomorphism.

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  • $\begingroup$ I believe your argument as soon as $q$ has no conjugate points along those unique geodesics, because then $\mathrm{exp}$ is a local diffeo. But I am not sure how to exclude conjugate points - once you pass a conjugate point, your geodesic is no longer a shortest curve. But without geodesic completeness, how can you see that there must therefor be another, shorter geodesic, leading to a contradiction? $\endgroup$
    – Sebastian Goette
    Commented Nov 12, 2015 at 12:26
  • $\begingroup$ I am not assiming that the geodesic is minimizing. The only thing left to check is the statement that $D$ is open (which is trivial if there are no critical points). $\endgroup$
    – Raziel
    Commented Nov 12, 2015 at 12:37
  • $\begingroup$ Exactly, and that is why I was worrying about conjugate points, which are exactly the critical points of the exponential map. $\endgroup$
    – Sebastian Goette
    Commented Nov 12, 2015 at 13:58
  • $\begingroup$ In fact I don't see why $D$ should be open. I don't have counter-examples either. I will think about it. $\endgroup$
    – Raziel
    Commented Nov 12, 2015 at 14:12
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Yes. The magic words are star-shaped, Exponential map, and Gauss' Lemma.

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    $\begingroup$ Could you please write a complete answer? (Note also that my definition of convexity differs a bit from others. I don't assume that the convex set contains the shortest path. So for example a circle minus one point is convex by this definition.) $\endgroup$
    – Mostafa - Free Palestine
    Commented Nov 11, 2015 at 19:36
  • $\begingroup$ But a circle minus a point is convex inside the circle. $\endgroup$
    – Deane Yang
    Commented Nov 11, 2015 at 21:24
  • $\begingroup$ @DeaneYang Yes, circle minus one point is convex inside the circle and also inside itself. In fact this notion is independent of embedding. $\endgroup$
    – Mostafa - Free Palestine
    Commented Nov 11, 2015 at 21:30
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    $\begingroup$ That's a good point. So if you use the convex subset itself and forget the original Riemannian manifold, then the geodesic joining a pair of points is in fact the shortest path between those two points. Then doesn't that lead to a solution? $\endgroup$
    – Deane Yang
    Commented Nov 11, 2015 at 21:39
  • $\begingroup$ @DeaneYang Could you briefly explain why the unique geodesic is the shortest path? Do we need something like the curve-shortening flow? $\endgroup$
    – Fan Zheng
    Commented Nov 12, 2015 at 1:18

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