[Math] Cone angles for Riemannian metrics in polar coordinates

dg.differential-geometrygeometrygt.geometric-topology

This is the simplest case of a question that's been bugging me for a while: say we have a Riemannian metric in polar coordinates on a (2-d) surface:
g=dr2+f2(r, θ)dθ2, such that the θ parameter runs from 0 to 2π. Assume that f is a smooth function on (0,∞)X S1 such that f(0, θ)=0.

Define the cone angle at the pole to be
$ C=\lim_{r\rightarrow 0^{+}} \frac{L(\partial B(r))}{r} $, where B(r) is the geodesic disc of radius r centered at the origin. Then it's fairly easy to see(by switching into Cartesian coordinates) that a necessary condition for the metric to be smooth is that C=2π. If C<2π, there is a cone point at the origin. One can write out a cone metric, and show that the triangle inequality holds, so there is a singular metric, but which still induces a metric space structure.

Now, if C>2π, it seems pretty clear that we'll end up with a space which violates the triangle inequality; it will be shorter to take a broken segment through the origin than to follow the shortest geodesic(in the sense of a curve γ(t) such that Dγ'γ'=0.) One can show this directly for some simple cases, eg a flat metric with a cone angle greater than 2π.

But there must be an elementary proof of the general case! I can't seem to find one though, and I spent the afternoon playing around with the Topogonov and Rauch comparison estimates to no avail. The basic problem I'm having is that the cone angle condition is essentially a condition on metric balls, but we expect a violation of the triangle inequality, which is a condition on distances.

This is not really related to anything I'm working on, but it's driving me crazy, so I'd appreciate any insight.

Best Answer

Although this isn't quite the answer you want, the distinction between the 'shortest path' an the 'shortest geodesic' (in the "no acceleration" sense) has been observed in discrete settings.

Your limiting conditions are akin to the case of the pole being a flat, convex or "saddle" point. It's been known for a while that on convex polytopes (specifically where there are no points with C > 2 pi), the shortest path between two points can be found by identifying the relevant facets the shortest path goes through, unfolding these facets, and then drawing a straight line. This is precisely the geodesic criterion you're referring to. And it's also known in general that this might not be true for non-convex polyhedra

The paper that does is by Sharir and Schorr: "On shortest paths in polyhedral spaces, SIAM J. Comp. 15, 193-215, 1986". Although this is all in discrete land, the underlying phenomenon is likely the same.