If $(M,g)$ is a smooth Riemannian manifold and $c : [a,b] \to M$ is a smooth embedded simple curve on $M$, it is always possible to choose locally a Riemannian metric $g_0$ on $M$ for which $c$ is a geodesic for $g_0$. As I understand, this can be done by pulling back a tubular neighborhood of $c$ to a disc bundle along $c$. There is always a flat Riemannian metric for which the $0$-section ($c$ itself) is a smooth geodesic. My question is: is it possible to locally realize $c$ as a geodesic to a metric such that the tube is positively curved? For instance, one of constant sectional curvature? I mean, can $g_0$ be positively curved? If it helps, $M$ can be assumed to be a surface.
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2$\begingroup$ some manifolds don't admit metrics with positive curvature, for example it's the case when their fundamental group is infinite. Maybe you meant the tube should be positively curved? $\endgroup$– alesiaCommented Oct 25, 2020 at 17:44
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$\begingroup$ @alesia, thats right, I am sorry, I mean that the tub is positively curved $\endgroup$– L.F. CavenaghiCommented Oct 25, 2020 at 17:46
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3$\begingroup$ It appears to me that your question does not involve the metric $g$ at all. Any connected embedded curve in a smooth $n$-manifold has a tubular neighborhood diffeomorphic to $T = I \times D$, where $I$ is an interval or a circle and $D$ is the unit disk $D^{n-1}$. You can map $T$ into the unit sphere $S^n$ such that the zero section maps into a geodesic and and T$ onto a tubular neighborhood of that geodesic. $\endgroup$– Deane YangCommented Oct 25, 2020 at 17:47
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$\begingroup$ @DeaneYang that is exactly what I was trying to show! Thank you very much, in this case the tube is positively curved, right? $\endgroup$– L.F. CavenaghiCommented Oct 25, 2020 at 17:48
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3$\begingroup$ Since it's just the standard metric on the sphere pulled back to the tubular neighborhood of the curve in $M$, it has constant sectional curvature $1$. $\endgroup$– Deane YangCommented Oct 25, 2020 at 18:13
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It appears to me that your question does not involve the metric $g$ at all. Any connected embedded curve in a smooth $n$-manifold has a tubular neighborhood diffeomorphic to $T = I\times D$, where $I$ is an interval or a circle and $D$ is the unit $(n-1)$-dimensional disk. You can map $T$ into the unit sphere such that $I\times \{0\}$ maps into a geodesic and and $T$ onto a tubular neighborhood of that geodesic. If $g_0$ is the pullback of the standard metric on the sphere, then it has positive sectional curvature $1$.
answered Jan 28, 2023 at 20:33