Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Maybe this question be very simple, but I don't know why it is hard for me. Thanks for any guide and help.

We say a surface $S$ (2-dimensional metric(compact) Riemannian surface) is good (denote by $GS$), if every $2n$, $n\geq1$, points on surface can be separate by some geodesic to two distinct subsets $V_1$ and $V_2$, where $|V_1|=|V_2|=n$. Also, if a surface $S$ is not good, we say it is bad an denote it by $BS$.

For example, it is not difficult to show that plane is a $GS$. Also, a sphere is $GS$.

1) Do we have some $BS$ examples(class of examples)?

2) Can we characterize the $GS$ and $BS$ surfaces?

I can't find any $BS$ examples and also I can't prove that they are $GS$.

For example, is Klein Bottle $GS$ or $BS$?

Is there any related works and questions about this post?

share|improve this question
    
By a surface, do you mean a two-dimensional Riemannian manifold? Or a 2-dimensional topological space (in which case, you are asking your question for ALL riemannian metrics)? –  Igor Rivin Jan 9 '12 at 12:14
    
I am not a professional in geometry. But I am thinking about two-dimensional metric(compact) Riemannian manifold.I can understand the Klein Bottle and the geodesics on it. Also, I believe that the Klein Bottle is $GS$. But until now, I don't know what kind of surfaces the Klein Bottle is? –  Shahrooz Jan 9 '12 at 13:16
7  
A flat torus can't be cut in two by a closed geodesic, so this gives an easy family of bad surfaces. Small perturbations of the flat metric should behave similarly. I think there are lots of higher-genus surfaces with the same inseparability property. Flat Klein bottles are good (you can parametrize the separating geodesics relatively easily), but I don't know about general metrics. –  S. Carnahan Jan 9 '12 at 14:27
2  
@S. Carnahan: Every closed oriented higher genus surface with a metric of negative curvature has a separating geodesic simple geodesic, and an infinite number of such. One way to attempt to do this is to take a hyperbolic surface $S$ (say of genus 3, for simplicity) and a very large equidistributed point set on it. A simple separating geodesic cuts $S$ into two pieces of unequal area, so it is plausible that for a sufficiently dense set there is no simple closed geodesic with half the points on each side. Of course, since the curves can be arbitrarily complicated, this is not clear. –  Igor Rivin Jan 9 '12 at 21:48
add comment

1 Answer

up vote 11 down vote accepted

Assuming that "geodesic" in this question means "simple closed geodesic", then every complete hyperbolic surface $S$ of finite area is "bad": You cannot even separate an arbitrary pair of points. The reason is that the union of simple closed geodesics on $S$ is nowhere dense (even more, its closure has Hausdorff dimension 1) by the result of Birman and Series, "Geodesics with bounded intersection numbers on surfaces are sparsely distributed", Topology 24 (1985). The paper is available at: http://www.math.columbia.edu/~jb/bdd-int.no-sparce.pdf

In view of this theorem, there exists an open disk $D\subset S$ which is disjoint from all simple closed geodesics in $S$. Now, take two points from this disk. I did not check it, but it is quite likely that Birman-Series result also holds in the case of negatively pinched variable curvature.

Hyperbolic surfaces are probably still "bad" if you allow non-simple closed geodesics, but pairs of points no longer suffice; one could try to use Hausdorff dimension arguments in the products of hyperbolic surface by itself to get a contradiction.

share|improve this answer
    
Dear Misha, thank you very much for your answer. Would you please give me a short reference related to your answer? –  Shahrooz Apr 4 '12 at 18:58
    
@Shahrooz: I updated my answer to include the link to Birman-Series paper. –  Misha Apr 5 '12 at 16:43
    
@ Dear Misha: thank you for the paper. –  Shahrooz Apr 8 '12 at 9:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.