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Are there non compact smooth manifolds which have the property that every Riemannian metric is geodesically convex?

Note that a manifold for which every Riemannian metric is complete must be compact.

(In particular, it can be proved that every metric has a conformally equivalent one which is bounded, and the claim follows from this).

Reference: Nomizu, Katsumi, and Hideki Ozeki. "The existence of complete Riemannian metrics." Proceedings of the American Mathematical Society 12.6 (1961): 889-891.

(I am trying to understand how much weaker is the notion of geodesic convexity compared with completeness).

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No = there are no non compact smooth manifolds which have the property that every Riemannian metric is geodesically convex.

Note that any non compact $n$-manifold $M$ contains a closed subset homeomorphic to $[0,\infty)\times D^{n-1}$ Indeed choose a bounded metric on the manifold $M$ and a minimizing geodesic $\gamma\colon[0,\ell)\to M$ and then pass to the appropriate closed neighborhood of $\gamma$ [see (*)]

Now it is easy to construct a metric $g$ on $[0,\infty)\times D^{n-1}$ which is not geodesically convex [see (**)]. One can also ensure that there is a pair of points say $x,y\in[0,\infty)\times D^{n-1}$ without a $g$-geodesic between such that intrinsic distance from $x$ to $y$ in $ ([0,\infty)\times D^{n-1},g)$ is much smaller than distance from $x$ to $\partial \left([0,\infty)\times D^{n-1}\right)$. The later condition implies that no matter how you extend this metric to whole $M$, it will fail to be geodesically convex at the pair $(x,y)$.

(*) Given a function $r\colon[0,\ell)\to\mathbb R_{>0}$ consider the subset $W_r$ in the normal bundle to $\gamma$ formed by all the normal vectors $v$ at $\gamma(t)$ such that $|v|<r(t)$. Note that $W_r$ is homeomorphic to $[0,\infty)\times D^{n-1}$ and for small enough functions $r$, the restriction of normal exponential map to $W_r$ is an embedding.

(**) Take the induced metric for $h\colon [0,\infty)\times D^{n-1}\to \mathbb R\times \mathbb R^{n-1}$ where $h(t,x)= (f(t,x),x)$ and $f(x,t)$ is an increasing in $t$ and $s(x)=\lim_{t\to\infty} f(x,t)$ is very small near the center of the disc.

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  • $\begingroup$ pints $\mapsto$ points $\;$ $\endgroup$ – user5810 Aug 3 '15 at 14:24
  • $\begingroup$ @RickyDemer, corrected. $\endgroup$ – Anton Petrunin Aug 3 '15 at 16:13
  • $\begingroup$ @AntonPetrunin, your idea sounds great. Two questions: 1) In finding a suitable neighbourhood of $\gamma$, you mean something analogous to a tubular neighbourhood? (can you please elaborate more on this?) 2) Can you please give some sketch of how to construct a metric on $[0,\infty)\times D^{n-1}$ which is not geodesically convex? (not necessarily satisfying the other properties you mentioned, I will try to fill in these gaps myself) $\endgroup$ – Asaf Shachar Aug 3 '15 at 17:28
  • $\begingroup$ @AsafShachar, see $({*})$ and $({*}{*})$, hope it helps. $\endgroup$ – Anton Petrunin Aug 3 '15 at 23:01

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