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Given a positive real number $l$. Does there exist a closed hyperbolic surface $X$ so that injectivity radius not less than $l$?

Mikhail Katz
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Bidyut Sanki
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2 Answers2

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Let $Y$ be a compact hyperbolic surface. There are only finitely many closed geodesics in $Y$ whose lengths are less that $\ell$. Since $\pi_1(Y)$ is residually finite, there is a normal subgroup of finite index in $\pi_1(Y)$ that does not contain anything in the conjugacy classes corresponding to these geodesics. The corresponding finite covering space $X$ of $Y$ has no closed geodesics of length less than $\ell$.

Autumn Kent
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    Ditto for all compact hyperbolic manifolds. – Misha Apr 03 '13 at 13:32
  • Yes, I should've said that. Thanks, Misha. – Autumn Kent Apr 03 '13 at 13:35
  • Richard, maybe you should also mention that you use normal finite index subgroups, this will take care of the base point business. – Misha Apr 03 '13 at 13:59
  • I didn't find the word "congruence subgroup" on this page, so I thought I would comment that using congruence subgroups may be more elementary than appealing to residual finiteness of 3-manifolds. – Mikhail Katz Apr 15 '13 at 13:44
  • One (usually) proves residual finiteness of hyperbolic 3-manifolds using congruence quotients, so this is just the same thing. – HJRW Apr 15 '13 at 14:12
  • Exactly my point. Congruence subgroups provide both lower bounds for systole and proof of RF for arithmetic manifolds. I don't remember if RF is also true for nonarithmetic ones, but I assume it is harder. – Mikhail Katz Apr 15 '13 at 14:17
  • I see that the paper

    Thurston, William P. Three-dimensional manifolds, Kleinian groups and hyperbolic geometry. Bull. Amer. Math. Soc. (N.S.) 6 (1982), no. 3, 357–381

    proves RF in the Haken case.

     – Mikhail Katz Apr 15 '13 at 14:23
  • Hempel, John Residual finiteness for 3 -manifolds. Combinatorial group theory and topology (Alta, Utah, 1984), 379–396, Ann. of Math. Stud., 111, Princeton Univ. Press, Princeton, NJ, 1987 proved this again in the Haken case. I guess he wasn't convinced by Thurston. Does anybody know the history of this? – Mikhail Katz Apr 15 '13 at 14:30
  • In response to your first comment, no, it's not significantly harder. In any Jacobson ring you have enough maximal ideals to prove residual finiteness. Malcev observed this in the 40s; the upshot is that any fg linear group is residually finite. – HJRW Apr 15 '13 at 15:12
  • Re: your later comments, no, I don't think that Hempel didn't believe Thurston; rather, Thurston characteristically doesn't give all the details. Hempel's paper shows how to deduce residual finiteness from geometrization. – HJRW Apr 15 '13 at 15:17
  • Do you have a reference where your remarks on Jacobson rings and linear groups are treated in more detail? – Mikhail Katz Apr 15 '13 at 16:13
  • The proof was written up by Steve D in this MO answer: http://mathoverflow.net/questions/9628/finitely-presented-sub-groups-of-gln-c/14244#14244 . Roger Alperin wrote up an elementary proof in his paper 'An elementary account of Selberg's lemma'. – HJRW Apr 15 '13 at 19:28
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Buser in 1992 gives a lower bound on the Bers constant for surfaces of genus $g$, and it goes to infinity as $g$ goes to infinity. So this means there's compact hyperbolic surfaces whose injectivity radius is arbitrarily large, but you have to go to high genus to realize them.

P. Buser. Geometry and spectra of compact Riemann surfaces Prog. Math. Vol 106. (1992)

The genus needs to grow like the square of the injectivity radius. Probably a good way to find this surface would be to use Fenchel-Nielsen coordinates, where the pants correspond to points on a planar square lattice and you want to connect the vertices in some (likely non-planar) tri-valent graph configuration, so that it represents pants. Then you want to connect the edges in a way so that there's no short closed loops in the resulting graph. So I'm describing a surface that's less the gluing-together of pants, but more the gluing together of quadratically-many "short shorts".

Ryan Budney
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    @Ryan: Actually, don't we have $g\geq \cosh^2(l/2)$ by Gauss-Bonnet? This is a much faster growth of $g$ in terms of $l$ than you indicate. – Robert Bryant Apr 04 '13 at 20:32
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    @Robert: I should have thought of that. I'm teaching Gauss-Bonnet in my differential geometry class this week. – Ryan Budney Apr 04 '13 at 22:39