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Suppose $(M, g)$ is an open complete nonnegatively curved Riemannian manifold with $d$ its distance.

A totally convex set $C\subset M$ has the property that for any two point $x, y \in C$ any geodesic (not only the minimal ones) joining them must lie in $C$.

If $C$ is totally convex, the fattened set $C^a=\{x \in M \mid d(x,C)\leq a\}$ is still totally convex? Is it at least for small $a$?

This question is somehow related to question "Examples on small cut radius of totally convex set in non-negatively curved manifold", Examples on small cut radius of totally convex set in non-negatively curved manifold

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up vote 6 down vote accepted

No. Consider a rotation-symmetric metric on $\mathbb R^2$ resembling a small spherical cap extented by a flat cone. A sufficiently short geodesic segment at the origin is totally convex in your sense. But a small neighborhood of such a segment is not convex due to positive curvature.

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thanks, perfect! – Carlo Mantegazza Jan 24 '13 at 16:57

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