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Feb 10, 2019 at 12:23 vote accept Alex Gavrilov
Jan 28, 2019 at 14:46 answer added Dmitri Panov timeline score: 9
Jan 23, 2019 at 12:39 comment added Ben McKay Any small bump will get hit infinitely often, and each time it will turn the geodesic a little (from the direction it would have had without the bump), so I would guess that small bumps will cause the typical geodesic to have infinitely many self intersections.
Jan 23, 2019 at 11:42 comment added quarague @MattF. That is an interesting example. I was more picturing a donut with a few small bumps where the kind of intersection you described does not occur. Intuitevely you need fairly high curvature locally around a bump to get a geodesic with a self intersection through winding around the bump. I wonder whether one can make this into a rigirous argument with a curvature bound. Looks like the problem is more complicated than I originally thought.
Jan 23, 2019 at 10:56 comment added user44143 @quarague, what if you stick a toothpick in a donut, and smooth over the result with icing? The surface is still a torus, but geodesics which wind up the toothpick one way may well intersect themselves on the way down. Since almost any geodesic will hit the toothpick multiple times, that seems to be a metric where almost every geodesic will self-intersect.
Jan 23, 2019 at 9:49 comment added quarague If you look at the flat torus $\mathbb{R}^2 / \mathbb{Z}^2$ than any line with an irrational slope is a non closed geodesic without intersections. My suspicion would be that the situation is the same for any other metric on the torus. That is starting at a point, there is an infinite density zero set of directions that leads to closed geodesics and for all other directions the geodesic is non-closed (fairly sure) and has no self intersections (somewhat sure).
Jan 23, 2019 at 8:43 history edited user44143 CC BY-SA 4.0
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Jan 23, 2019 at 5:36 history asked Alex Gavrilov CC BY-SA 4.0