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Let $ M $ be a Riemannian manifold with metric $ ds^2 $. Suppose that every two points in $ (M,ds^2) $ can be joined by a minimizing geodesic. Now let $ \mu ds^2 $ be a metric on $ M $ conformal equivalent to $ ds^2 $. Is it true that every two points in $ (M,\mu ds^2 ) $ can be joined by a minimizing geodesic?

  • Why isn't this obvious as the conformal equivalence gives a distance-of-path preserving bijection between the paths connecting the points in either metric? – Alex Youcis Feb 11 '13 at 08:14
  • If the manifolds are complete, then every pair of points can be joined by a minimizing geodesic, regardless of any type of conformal relationship. – treble Feb 11 '13 at 08:25

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Consider $\mathbb R^2\setminus\{0\}$ with the metrics $\frac{dx^2+dy^2}{x^2 + y^2}$ and $dx^2 + dy^2$, respectively. These should be good candidates for a counterexample. Maybe one has to use a different scaling function to guarantee that the origin and the point at infinity are "infinitely far away" from any point in the manifold, but other than that, it should work.

Sam
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  • Your first metric is defined on $ R^2- 0 $. So i'm not sure that this kind of conformal metric could work for a counterexample. –  Feb 17 '13 at 10:37
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    @user55449: I meant it to be defined on $M = \mathbb R\setminus {0}$ and to use the first metric, to make this manifold complete (and therefore every pair of points can be joined by a minimizing geodesic). The second metric is not complete however and the points $(-1,0), , (1,0)$ do not have a distance-minimizing geodesic between them. Hence a counterexample. But I haven't checked in detail. I'd be enormously surprised if it (or a version of this idea) didn't work, though. – Sam Feb 17 '13 at 10:59
  • Yes!!! Now i've understood your idea. In $ R^2-0 $ let $ |dz|^2 $ be the standard euclidean metric. For a theorem of Nomizu Ozeki (see the article 'The existence of complete Riemannian metric', Nomizu-Ozeki) there exists $ \mu $ positive smooth function such that $ ds^2=\mu |dz|^2 $ is complete. –  Feb 17 '13 at 15:01
  • Now it is natural to ask: if we added the hypothesis of simply connectdeness could the statement be true? Or at least we could ask if the statement could be true for $ M $ equal to the unit disk. –  Feb 17 '13 at 15:03
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    @user55449: I don't think the answer is really dependent on the topology: Consider the open unit disk. You can identify it with the sphere minus a point. On the sphere minus a point (say minus the north pole), you can do a similar thing as on $\mathbb R^2 \setminus {0}$. I.e. find a metric such that geodesics between two antipodal points on the equator will want to go through the north pole to be minimizing. Now pull everything back to the unit disk. Then geodesics between two points which correspond to points on the equator will want to go straight into the boundary of the unit disc. – Sam Feb 17 '13 at 16:11
  • I've made a brief summary of your argument.Let $ \pi : D_1 \rightarrow S^2-N $ be the homeomorphism that identifies the boundary of the unit disk with the north pole. Let $ds^2$ be the standard metric of the sphere and $ \mu ds^2 $ a conformal complete metric (by Nomizu-Ozeki). If we pull back these two metrics on the unit disk $D_1$ we obtain two metric conformally equivalent, a complete metric and a metric that does not admit minimizing geodesic for every pair of points that are antipoldal with respect to the center of the disk. Is it right? –  Feb 17 '13 at 18:50
  • I've been thinking that the pullback of $ ds^2 $ on the unit disk could be not conformally equivalent to the standard euclidean metric of the disk. So perhaps, if it is true, our original statement could be true at least for conformal deformations of the standard euclidean metric. I'm thinking about it. –  Feb 17 '13 at 19:10
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    @user55449: Yeah, but all you need to do, is "make it really cheap for geodesics to run near the boundary of the disk, and even cheaper to be even closer". Then you take two points near the boundary which are antipodal and show that the geodesics really want to touch the boundary at some point. Maybe $g = (1-x^2)ds^2$ would do the job, already. – Sam Feb 17 '13 at 23:11