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Consider the tiling of the Poincare disk $\mathbb{D}$ by identified octagons (i.e., representing a torus with genus 2). Suppose inside each octagon is a subset A such that the octagon minus A is a set of measure zero. In other words, the elements of A are scattered throughout the octagon almost everywhere. Suppose a straight ray were drawn from the origin to the edge of $\mathbb{D}$, i.e. going to infinity.

Keeping in mind that the tiled octagons are identified, so the elements of A remain in the same locations for each octagon, is it possible to have such a ray that is not periodic on the octagons AND never intersects A?

If not, can any geodesic be drawn on these tiled octagons (again, starting from the origin and ending at the edge) in order to not be periodic and avoid intersecting A?

Thank you.

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3  
Have you thought about letting $A$ be the complement of the geodesic itself, for any geodesic at all? –  Tracy Hall Aug 5 '10 at 4:04
    
That's a good point. I meant to have the geodesic not be periodic over the octagons. I will edit my question to address that. –  user8166 Aug 5 '10 at 4:27
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@gomaff. How about letting A be the complement of the geodesic together with all its images by the tiling group ? –  user6129 Aug 5 '10 at 8:40

1 Answer 1

It is already answered in comments, and I think quid edited the question for somebody write this well-known answer.

In fact, the octagon plays no role here (nor the origin -- all points are congruent on the Poincare's disc). The question is if there is a nonperiodic geodesic ray on a riemann surface of genus $2$ because its complement, of course, has a full measure.

One can easily construct such a geodesic, and, moreover, converging to a closed one. For this, take a geodesic $\Gamma$ on $\mathbb D$ that represents a closed geodesic on the riemann surface $\Sigma$ and draw another geodesic $G\subset{\mathbb D}$ tangent to $\Gamma$, i.e., $\Gamma$ and $G$ have a common point on the absolute. It is immediate that the distance between $\Gamma$ and $G$ is $0$ and that, starting from some moment, the ray of $G$ in the "direction of the commom point" on the absolute do not intersect $\Gamma$. So, on $\Sigma$, this ray does not intersect the closed geodesic, getting arbitrarily close to it.

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In fact any leaf of one of Thurston's geodesic laminations on $\Sigma$ will lift to a geodesic on $\mathbb D$. There are uncountably many such leafs. –  Lee Mosher Mar 10 at 0:26
    
@LeeMosher Sure. I tried a most elementary answer. –  Sasha Anan'in Mar 10 at 0:30

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