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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.

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Hmm. Cone angles for what? –  Mariano Suárez-Alvarez Dec 29 '09 at 0:11
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2 Answers 2

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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.

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EDITED.

Since the queston is modifyied a bit, here is the modified answer. Let us start with an example. Conisder the flat cone $C$ with cone angle $4\pi$. It can be seen as a double cover of flat $R^2$ with ramification at point $(0,0)$. Let $x$ and $y$ be two preimages of point $(1,0)$. Let us first show that on $C$ there is no smooth geodesic that joins $x$ and $y$ and does not contain $0$. Indeed if there were such a geodesic $\gamma$, its projection to $R^2$ would be a straight segment with bough ends at point $1,0$, so such a segment would have zero length, which is an obvious contradiction.

On the contrary, we have the "broken geodesic" that joins first $x$ with $0$ and then $0$ with $y$, it has length 2. The moral of the story, is that if would be wrong to define the distance between two points as the lenght of the shortes SMOOTH geodesic. Such geodesic may not exist. We are dealing with a singular metric and so should admit that some geodesic should be "singular" too. The correct defintion of a geodesic in this case can be "shortest continuous path that joins two points". Notice that for a flat cone with angle $2\pi$ and more there is always a unique geodesic in this sense, because this is a space of non-positive curvature. Sometimes the geodesic is a "broken geodesic", sometimes just a smooth segment. And surelly this is a metric space, so triangle inequality holds.

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I'll edit the question to clarify this point, but there's a differential-geometric notion of geodesic(curves whose velocity is parallel with respect to the connection on the manifold,) which is distinct from the metric-space notion. For example, a Lorentzian metric doesn't induce a metric space structure on a manifold, but geodesics are still well-defined. –  Gordon Craig Dec 29 '09 at 1:12
    
The Lorentzian metric is not really a metric, but a pseudo-metric. Also, it is not clear that the injectivity radius at (0,*) is zero. –  Steve Huntsman Dec 29 '09 at 2:19
    
@Steve Huntsman: Thanks for the point on the injectivity radius. I think that resolves the major point that was bugging me. –  Gordon Craig Jan 4 '10 at 23:23
    
By the way for a Riemannian manifold it is standard to define the length to be the infimum over "picewise smooth" "paths" joining the two points. No geodesics mentioned, check [Petersen,p122]. Realizing the distance with a geodesic is another matter and needs completeness etc. A second thing you can do: take the cone point out and now you have a smooth manifold, use standard RG to define everything and then take the completion of this metric space to recover the singular point. PS:It seems to me that the answer of Dmitri is on the right track though the other one is chosen to be the best. :) –  kalafat Mar 20 '13 at 22:57
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