I would like to ask a very basic/naive question. Given a Riemannian or pseudo-Riemannian manidold equipped with the Levi-Civita connection, is it known that all solutions of the geodesics equation are embedded curves? Is it possible to get immersed geodesic curves as well (are there immersed, but not embedded, solutions of the geodesics equation)?
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4$\begingroup$ An irregular donut in $\mathbb{R}^3$ has self-intersecting geodesics which are only immersed. $\endgroup$– user44143Commented Dec 5, 2022 at 15:19
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3$\begingroup$ All geodesics on the round sphere are immersed curves $\endgroup$– NealCommented Dec 5, 2022 at 16:03
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2 Answers
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Take a long strip of paper, and draw a long straight line down the middle of it. Now without folding the paper, glue two far apart pieces together in such a way that the two glued parts of the line cross each other. This is an example of a flat 2-manifold with a self-intersecting geodesic.
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Embedded geodesics are usually very rare. On a flat torus there are only countably many embedded geodesics.
For a generic metric on a 2-dimensional sphere there are only three such geodesics. Since $S^2$ is compact, for a geodesic to be embedded is the same as to be simple (closed without self-intersections). It is known that a triaxial ellipsoid admits exactly three simple geodesics (it follows from Jacobi's integrability of the geodesic flow on an ellipsoid). On the other hand, the number of simple geodesics is an upper-semicontinuous function of metric (this, hopefully, follows from the continuous dependence on parameters for solutions of ODE), so most of the metrics admit not more than three simple geodesics.
answered Dec 5, 2022 at 17:54
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1$\begingroup$ 'For a generic metric on $S^2$ there are only three such geodesics.' Is this accurate? Certainly there's at least three, but only three is a much stronger claim. (You can find arbitrarily $C^0$-small perturbations of any metric with many geodesics.) Also, is the number of simple closed geodesics really upper semicontinuous? What about something like a rotationally symmetric surface obtained by rotating $f =1+x^2(1+\epsilon \operatorname{sin}(1/\epsilon x))$ and letting $\epsilon \to 0$? $\endgroup$– Leo MoosCommented Jan 11, 2023 at 8:54
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$\begingroup$ @LeoMoos thank you, you are right. the statement that an $C^0$ open dense set of metrics can't be as i claimed. nevertheless i wonder if one still can make sense of the quustion "what is average/generic number of simple geodesics" [refining topology, extending the notion of genericity etc.] $\endgroup$ Commented Jan 12, 2023 at 4:10
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$\begingroup$ @LeoMoos also a compact counterexample to the upper-semicontinuity can be like this: take an ellipsoid with exactly three simple geodesics and modify it in a small region outside of the union of simple geodesics such that there is one more simple geodesic and shrink this modified part. $\endgroup$ Commented Jan 12, 2023 at 6:10
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$\begingroup$ OK, thanks for the clarification! It seems like an interesting question to ask how many simple, closed geodesics there are say $C^1$ or $C^2$ generically. (This means that you can't glue in small 'mushroom-like' caps with short geodesic loops anymore.) The rotationally symmetric example you can also make closed by capping it on both ends. I included this because it can be made to converge smoothly, unlike examples with small caps. (In higher dimensions, manifolds generically contain infinitely many embedded minimal surfaces; the contrast would be interesting.) $\endgroup$– Leo MoosCommented Jan 12, 2023 at 9:13
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$\begingroup$ @LeoMoos >>the rotationally symmetric example you can also make closed by capping it on both ends.<< but all the meridians of a surface of revolution are geodesics, so there are infinitely many closed simple geodesics at each step. $\endgroup$ Commented Jan 12, 2023 at 9:24