If $M$ is a smooth closed manifold together with a non-vanishing (maybe unit) vector field $X$. In what condition can we construct a Riemannian metric on $M$ s.t $X$ be the geodesic field of on $TM$?
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asked Sep 6, 2016 at 12:31
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2 Answers
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This is not always possible. For example there exists smooth vector fields on the three sphere without any closed orbit, this is due to Kuperberg. However any geodesic vector field on a closed manifold must carry a closed orbit, a theorem due to Lyusternik and Fet.
The keyword that you want to search for is geodesible flow. I think you can spend your time in a worse way than looking at the website of A. Rechtman, and references in her papers.
answered Sep 6, 2016 at 12:49
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The magic word is "geodesible", and the question in full generality is open. For surfaces, it has been answered by H. Gluck:
Gluck, Herman. "Dynamical behavior of geodesic fields." Global theory of dynamical systems. Springer Berlin Heidelberg, 1980. 190-215.
Obstructions have been found in some cases:
Johnson, D.L. & Naveira, A.M. Geom Dedicata (1981) 11: 347. doi:10.1007/BF00149358
answered Sep 6, 2016 at 13:01