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According to answers to this question every metrics on $S^3$ admits a simple closed geodesic. Given a knot (or link) $K$, it's also quite simple to build a metric on $S^3$ such that $K$ is a geodesic (for some parametrisation).

Is there a Riemannian metric on $S^3$ such that all its simple closed geodesics are topologically knotted?

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  • $\begingroup$ Perhaps relevant is: arxiv.org/abs/math-ph/9906021 where Etnyre and Grist establish existence of analytic contact flows on the three-sphere whose solution curves realize every knot type. Geodesic flow on the three-sphere however is a contact flow on a two-sphere bundle OVER the three-sphere so is a different matter ... somehow feels easier to me. (? but is it?) $\endgroup$
    – Richard Montgomery
    Commented Jul 28, 2015 at 23:43
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    $\begingroup$ This feels a strange conjecture, because contrapositively you are asking whether, in some metric, every unknot is has intrinsic curvature. $\endgroup$
    – Jesse C. McKeown
    Commented Mar 24, 2016 at 18:35
  • $\begingroup$ Then I go ahead and calculate Euler-Lagrange equations for the action functional $\int \langle \nabla_X X, \nabla_X X\rangle dt $, and indeed there are Riemann curvature terms as well as $\nabla_X^3 X$ terms... $\endgroup$
    – Jesse C. McKeown
    Commented Mar 28, 2016 at 2:22

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