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A famous theorem of Birman and Series says that if $S$ is a compact hyperbolic surface, then the set of points that lie on simple geodesics is nowhere dense and has Hausdorff dimension one; in particular, it has measure zero. This is proved in

Birman, Joan S.; Series, Caroline, Geodesics with bounded intersection number on surfaces are sparsely distributed. Topology 24 (1985), no. 2, 217–225.

Question: Is this true for compact oriented surfaces equipped with metrics of variable negative curvature?

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  • $\begingroup$ I deleted the word "closed" from the definition of the Birman-Series set, because it should not be there. See the definition of G_0 in the second paragraph of their paper "Geodesics with bounded intersection number are sparsely distributed". $\endgroup$
    – Sam Nead
    Jul 16, 2021 at 7:58

2 Answers 2

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Here is another argument which reduces the general result to the constant curvature case:

For any negatively curved metric $g$ on $S$, the set of geodesics of the universal cover $\tilde{S}$ is canonically identified with $\partial_\infty \tilde{S}^{(2)}$, the set of pairs of distinct points in the Gromov boundary of $\tilde{S}$, in which $\overline{\Lambda}$ (the subset of geodesics that are embedded in $S$) is a closed subset. Note that the Gromov boundary and the set $\Lambda$ are quasi-isometric invariants, so they don't really depend on the choice of the metric $g$. What choosing the metric $g$ does is it puts a $\mathcal C^1$ structure on $\partial_\infty \tilde{S}^{(2)}$.

Let $g_0$ be another metric of constant curvature of minus one. Then Birman and Series' theorem tells you that $\overline{\Lambda} \subset \partial_\infty \tilde{S}^{(2)}$ has Hausdorff dimension $0$ with respect to the $\mathcal C^1$ structure defined by $g_0$. Now, one can prove that the two $\mathcal C^1$ structures on $\partial_\infty \tilde{S}^{(2)}$ induced respectively by $g$ and $g_0$ are Hölder regular with respect to each other, which implies that $\overline{\Lambda}$ also has Hausdorff dimension zero for the $\mathcal C^1$ structure defined by $g$.

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  • $\begingroup$ This answer really clarifies what is going on here -- thanks! $\endgroup$ Jul 20, 2021 at 17:31
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Yes, Birman-Series is true in variable negative curvature. Here is a sketch.

  1. Suppose that $S$ is a closed, connected, oriented surface of genus $g > 1$. Fix a metric $g$ of variable negative curvature on $S$. Let $\Lambda$ be the Birman-Series set - the union of all simple geodesics.

  2. The theory of train tracks is "soft", and so goes through in the variable negative curvature case. (For example, see Bonahon's 1996 paper Shearing hyperbolic surfaces, bending pleated surfaces and Thurston’s symplectic form.) So there are a finite number of almost geodesic train tracks in $S$ that, between them, carry all simple geodesics (with very small, and very well-spaced, "errors"). In any one track $\tau$, the set of carried curves of length at most $L$ has size at most $O(L^k)$ for some $k \leq 6 g - 6$. Let $\Lambda_\tau$ be the subset of $\Lambda$ obtained by taking the union of simple geodesics carried by $\tau$. Let $\Lambda_\tau(L)$ be the union of simple closed geodesics carried by $\tau$ with length at most $L$.

  3. In variable negative curvature we still have "exponential convergence" of geodesics. So, let $b$ be a branch of $\tau$, one of the tracks given above. Let $I$ be a small arc dual to $b$. We examine the intersection of $I$ with $\Lambda_\tau$. These cluster about the intersections of $I$ with $\Lambda_\tau(L)$. We deduce that we can cover the intersection of $I$ with $\Lambda_\tau$ with polynomially many (in $L$) exponentially small (in $L$) intervals. Thus the cross-sectional Cantor set (to $\Lambda_\tau$) has Hausdorff dimension zero.


Here is a related example that confused me for a little while.

Suppose that $S$ is the surface of genus two. Consider a sequence $(S, g_t)_{t \to 0}$ of variable negative curvature metrics that Gromov-Hausdorff converge to the octagon surface $(S, g_0)$: the translation surface obtained by gluing opposite sides of a regular octagon. Let $\Lambda_t$ be the Berman-Series set for $(S, g_t)$ - the union of all simple geodesics (or equivalently, the closure of the union of all simple closed geodesics). Then the sets $\Lambda_t$ Hausdorff converge to $\Lambda_0$. However, by the above, if $t > 0$ then $\Lambda_t$ has Hausdorff dimension one while $\Lambda_0 = S$ has Hausdorff dimension two.

So we have a natural example of Hausdorff dimension jumping under taking Hausdorff limits. However, this sort of thing happens often enough that we should not be surprised. (Or at least, we should not be as surprised as I was when I realised the above was not an obstruction to the train track theory going through!)

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  • $\begingroup$ I do not see how this answers the question. Train tracks of course exist in variable curvature but they capture topology of laminations, not their geometry. $\endgroup$ Jul 16, 2021 at 11:34
  • $\begingroup$ I've added details to the argument. But, more directly towards your comment, train tracks can be chosen to represent the geometry of laminations - for example arrange matters so that all branches are very long geodesic segments, and the geodesic curvatures are very small at all switches. $\endgroup$
    – Sam Nead
    Jul 16, 2021 at 12:25

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