Given a positive real number $l$. Does there exist a closed hyperbolic surface $X$ so that injectivity radius not less than $l$?
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2 Answers
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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".
answered Apr 3, 2013 at 6:42
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1$\begingroup$ @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. $\endgroup$ Apr 4, 2013 at 20:32
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1$\begingroup$ @Robert: I should have thought of that. I'm teaching Gauss-Bonnet in my differential geometry class this week. $\endgroup$ Apr 4, 2013 at 22:39
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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 $\ell$.
answered Apr 3, 2013 at 11:56
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$\begingroup$ Yes, I should've said that. Thanks, Misha. $\endgroup$ Apr 3, 2013 at 13:35
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$\begingroup$ Richard, maybe you should also mention that you use normal finite index subgroups, this will take care of the base point business. $\endgroup$– MishaApr 3, 2013 at 13:59
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$\begingroup$ 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. $\endgroup$ Apr 15, 2013 at 13:44
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$\begingroup$ One (usually) proves residual finiteness of hyperbolic 3-manifolds using congruence quotients, so this is just the same thing. $\endgroup$– HJRWApr 15, 2013 at 14:12