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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 $l$?

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this should really be asked at math.stackexchange –  Ian Agol Apr 3 '13 at 18:00
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up vote 6 down vote accepted

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".

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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 4 '13 at 20:32
    
@Robert: I should have thought of that. I'm teaching Gauss-Bonnet in my differential geometry class this week. –  Ryan Budney Apr 4 '13 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 $\ell$.

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Ditto for all compact hyperbolic manifolds. –  Misha Apr 3 '13 at 13:32
    
Yes, I should've said that. Thanks, Misha. –  Richard Kent Apr 3 '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 3 '13 at 13:59
    
Thanks Misha, that is a better way to say it. Edited. –  Richard Kent Apr 3 '13 at 14:51
    
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. –  katz Apr 15 '13 at 13:44
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