For any surface bounded by a sphere is there always an upper bound of the set of all lengths of all shortest length geodesics between pairs of points?

Question

Take a surface S. Suppose there is a sphere with a finite radius such that every point on S falls within the sphere. Then given two points P1 and P2 such that a geodesic on S exists betwwen them, is there an upper bound on the length of the shortest geodesic between these points? Does this upper bound apply to all points on the surface?

I'm dealing with an algorithm for finding geodesics on certain surfaces and it falls apart if there are two points such that only one geodesic exists and the length is infinite. I know with toroidal shapes that one can produce an infinite geodesic thar never repeats itself. However this path is not the only way to reach certain points and there is an upper bound on length of all shortest distance geodesics on that shape. My concern is where such "infinite" geodesics are the only option.

For context I'm working with surfaces made up of triangles only. However my curiosity demands the more general not strictly polygonal case.

score 2 · Answer 1 · edited Feb 02 '20 at 18:46

2

No. For instance, $S$ could be a slight "thickening" of a curve that spirals out towards the sphere, getting denser and denser in each new spherical shell, such that the total length of the curve is infinite. There are then points on $S$ whose distance travelling within $S$ is arbitrarily large.

edited Feb 02 '20 at 18:46

user64742

2,207

answered Jul 02 '19 at 04:13

Eric Wofsey

330,363

1

Yes, you can make such a thickening polygonal if you want. – Eric Wofsey Jul 02 '19 at 04:23
Yes, if you make the thickening get thinner and thinner fast enough so that the surface area still ends up finite. – Eric Wofsey Jul 02 '19 at 05:24

For any surface bounded by a sphere is there always an upper bound of the set of all lengths of all shortest length geodesics between pairs of points?

1 Answers1