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There are at least three distinct simple closed quasigeodesics on convex polyhedra [Mat. Sb. (N.S.), 1949, 25(67) :2, 275–306 Quasi-geodesic lines on a convex surface Pogorelov].

Is the same true for straightest geodesics?

Related Questions: the-famous-lusternik-schnirelmann-theorem-of-the-three-closed-geodesics , midpoint-geodesic-polygon-birkhoff-curve-shortening

Straightest Geodesics:straightest_preprint.pdf

--Thanks, Bryan

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up vote 1 down vote accepted

I assume by straightest geodesic you mean: has the same surface angle to either side at every point. Let the polyhedron $P$ be a doubly covered obtuse triangle $\triangle abc$. The path $abc$ is a straightest geodesic. It seems the other quasigeodesics guaranteed by Pogorelov's theorem must pass through vertices, such as that following the altitude through the obtuse vertex $c$. But this will not be straightest. The path $aba$ is a quasigeodesic, with less than $\pi$ to each side of every point, but it is not straightest.
So it seems that this "polyhedron" has just one simple (noncrossing), straightest geodesic.

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Thanks! I checked and yep there aren't any others. The followup question is: Is there always at least one? – bjwbell Jun 23 '13 at 22:43
After a little thought. I'm pretty sure the answer is no. Choose a set of vertices with angle deficits such that no two subsets have the same total angle deficit. The proof should follow. – bjwbell Jun 24 '13 at 0:04
I agree, Bryan. In fact, a generic (random) polyhedron would have the property you mention. Nice question; too bad the answer is negative! – Joseph O'Rourke Jun 24 '13 at 0:21

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