# Length of shortest geodesic and Cheeger's isoperimetric constant for a special genus 2 surface

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.

• 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

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.
• 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
• 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
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 $B$).