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Let us take two copies of $ Y $-pieces [ or pair of pants ] with each boundary geodesic of length $ l $, and glue them together without any twisting to obtain a genus 2 closed orientable hyperbolic surface $M$. I want to find out the

a) length of the shortest simple closed geodesic and b) Cheeger's isoperimetric constant for this surface. FYI : Cheeger's constant for a closed surface $M $ is defined as

infimum of $ \frac{l(A)}{minimum ( \area of B and \area of B')} $ where B and B' are the two components of $ M \backslash A $, where infimum is taken over all 1 dimensional geodesic submanifolds A, i.e. union of simple closed geodesics A separating $M$ and boundary of A is a part of both boundary of B and that of B'. Any help ?

I believe length of the shortest geodesic should be minimum over some multiples of l, coming from measuring the length of geodesics in $M$ which "we first intuitively see" while drawing a diagram of a genus 2 surface, i.e.the ones surrounding just one "hole", and the ones surrounding "both holes" in $M$, the ones surrounding the "handles".I also beleive the geodesics in the isotopy classe of curves "winding around" a collar of $M$ should have more length.But I want to make it rigorous.

The reason I asked this question is if we can show that Cheeger's constant is say $ \geq \frac{l}{100} $, then it gives us a concrete way to construct a cosed hyperbolic surface with arbitrarily large eigenvalue, by Cheeger's inequality.

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The 3 geodesics "surrounding the handles" have length l by construction, the ones "running back and forth from one handle'surounding geodesic to the other is definitely $\geq \frac{l}{100} $ if we use the basic hyperbolic triangle identities and the same lower bound hold for the geodesics "joining the 1st "handle-surrounding" geodesic to the 3rd "handle surrounding" geodesic, then also we get a lower bound as well. –  Analysis Now Nov 5 '10 at 4:53
    
As $l$ grows, the collars shrink, and there are geodesics crossing the collars that are short. –  Sam Nead Nov 5 '10 at 9:51
    
Yes, but for this particular metric ( like any other hyperbolic metric ) on genus 2 surface, there would exist a shortest closed geodesic, I was interested in finding its length explicily or a lower bound. –  Analysis Now Nov 5 '10 at 14:50
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2 Answers

Any closed hyperbolic surface of genus $g$ admits some pants decomposition where each curve has length at most $B(g)$, the Bers constant. Thus, if I understand your question correctly, there is no way to achieve what you want in fixed genus.

Google "Bers constant" for many papers on this topic. The standard reference is Buser's book "Geometry and spectra of compact Riemann surfaces".

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Thanks, yes I read the book partly, I just remembered that for genus 2, $B_g$ is at most $21(g-1)$ . So, I cannot have a linear lower bound for the systoles for the family of the surfaces indexed by l, because then it will blow up. But honestly, I cannot see this very short geodesic in my special genus 2 surface ? –  Analysis Now Nov 5 '10 at 14:54
    
Since you glued without twisting, the very short geodesic is exactly perpendicular to the boundary curve of the pants. –  Sam Nead Nov 5 '10 at 15:06
    
Sam, if you have meant the ones perpendicular to the boundary geodesics , say of length 2y of the pants, then I guess we can explicitly find out their lengths. Considering the pants to be a union of isometric right-angled hexagons of 3 sides length l2 and other 3 sides, and decomposing the hexagons into two right-angled pentagons by dropping a perpendicular, we get, from pentagon identities: $cosh(0.25l)=sinh(0.5l)sinh(y) $.Which gives us : $sinh(y)=0.5(\frac{1}{sinh(0.25l)}, $which is small if l is big, but which is big if l is small. –  Analysis Now Nov 5 '10 at 15:59
    
by which I mean y or equivalently 2y, length of perpendiculsr geodesics. –  Analysis Now Nov 5 '10 at 16:00
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There is a universal constant $C$ such that for any hyperbolic surface $\Sigma$, $\lambda_1(\Sigma)\leq C$. This follows from Margulis' lemma and eigenvalue estimates based on the minimax principle. If $\epsilon$ is Margulis' constant for hyperbolic surfaces, then there is a disk $B$ of radius $\epsilon/2$ embedded in $\Sigma$. So one may estimate $\lambda_1(\Sigma)\leq \lambda_1^D(B)$ (the Dirichlet eigenvalue for the disk $B$).

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Thank you very much . –  Analysis Now Nov 23 '10 at 15:32
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