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In the plane, a Kakeya set is a set of points that contains unit line segments in every direction. Let's define a Kakeya set on an open surface (with a complete metric) to be a set that contains unit length geodesics in every direction. Here, by two geodesics having the same direction, we imply that they stay at finite distance from each other.

Clearly, there are Kakeya sets on any surface: If $\gamma$ is a geodesic so that $d(\gamma(t),x) \rightarrow \infty$, then by taking a sequence of geodesics connecting $x$ to $\gamma(t)$ and deriving a convergent subsequence, we can obtain a geodesic through $x$ in the direction of $\gamma(t)$.

Question: On the plane, there are Kakeya sets of measure zero. Can one construct Kakeya sets of measure zero on any surface as well?

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  • $\begingroup$ I don't see why not. One could stereographically project the standard fractal example from the plane to obtain such things in $S^2$ for example. $\endgroup$
    – Glen Wheeler
    Jun 8, 2012 at 19:46
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    $\begingroup$ Could you clarify what you mean by "every direction"? Are two directed geodesic segments declared to have the same direction if their extensions to geodesic rays have finite hausdorff distance? $\endgroup$
    – Autumn Kent
    Jun 8, 2012 at 20:31
  • $\begingroup$ Richard, I think my following definition is equivalent to yours: two geodesic segments have the same direction if for extensions $\gamma(t)$ and $\eta(t)$ of them, we have $\forall t: d(\gamma(t),\eta(t))\leq C$ for a constant $C$ independent of $t$. $\endgroup$
    – Hej
    Jun 8, 2012 at 22:20
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    $\begingroup$ Glen, note that for $S^2$ or any closed surface, the question is not interesting since every two geodesics have the same direction. In the noncompact case, it is not clear to me how one can project the standard example; for example wouldn't the nontrivial topology complicate things? $\endgroup$
    – Hej
    Jun 8, 2012 at 22:23
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    $\begingroup$ As Misha says, your question has sense for open complete surfaces with negative curvature. If in addition the curvature is negatively pinched, then existence of Kakeya set of arbitrary small area follows from the fact that there are triangles with arbitrary small sum of angles. Simply take three unit sectors at the vertices of this such triangle. $\endgroup$
    – Anton Petrunin
    Jun 9, 2012 at 8:50

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