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Is it possible to recover a unique vector field given a curve of which we assume that it is the auto-parallel curve, which is constructed in a unit-length vector field? Probably not without further information, anything could happen far away from the curve, right? But is there some type of "analytic continuation" of vector fields at least locally?

By auto-parallel I mean \begin{equation} \nabla_X X = 0 \end{equation} with $X$ being a vector field and one could say that the auto-parallel curve at a specific point is the given information. For now let's also consider this connection to be the Levi-Civita Connection and the metric $g$ is given as well.

The question could be boiled down to. If a specific curve with the tangent $\dot\gamma(s)=X(s)$ is given, is there any sensible rule with construct the unit vector field in at least the close proximity?

Michael Paris
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  • Could you please clarify: 1. What do you mean by an "auto-parallel curve"? (A curve whose parallel curves are isometric...?) 2. What data do you assume are given? (A plane curve? A parametrization? A (smooth, unit?) vector field along the curve?) 3. What are you looking for? (A unit vector field extending a given unit vector field along a curve?) Your laying out the question in short, declarative sentences would be appreciated. :) – Andrew D. Hwang Jul 10 '17 at 12:58
  • I specified it a bit more, hope I was able to express what my question is. – Michael Paris Jul 10 '17 at 13:22
  • So you have a unit-speed geodesic and you want to extend its tangent vector field to a vector field $X$ on the manifold? Do you want the extension to satisfy $\nabla_X X = 0$ everywhere, or only along the curve? – Anthony Carapetis Jul 11 '17 at 14:04
  • Yes! Exactly! The geodesic condition has to be fulfilled only along the starting curve. – Michael Paris Jul 11 '17 at 14:13
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    Then there are many, many such extensions - think about the analogous problem of prescribing the value of a function the $x$-axis and asking for an extension to the plane. For a "sensible rule", I would probably extend by parallel transport in directions orthogonal to the curve. When the curve is embedded, this extension is smooth on a small enough normal tube. – Anthony Carapetis Jul 12 '17 at 03:52
  • Thanks. I will look into it. – Michael Paris Jul 12 '17 at 09:04

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