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Suppose $Y$ is a pair of pants with a hyperbolic structure and $\gamma_i; i = 1, 2, 3$ are the geodesic boundaries of length $l_i; i=1, 2, 3$ respectively. Now consider a essential simple arc $\sigma$ in $Y$ with end points on a same boundary component of $Y$ and $l$ denote the length of unique geodesic in the homotopy class of $\sigma$. Then, my question is weather the inequality, $l$ $\geq$ $\frac{1}{2}$ min{$l_i | i= 1, 2, 3$} hold or not for every pair of pants $Y$.

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up vote 4 down vote accepted

If you keep the lengths of two cuffs fixed (say equal to L) and let the third one go to infinity (say equal to R), then the length $l$ of the ortho-geodesic with endpoints on the third cuff goes to zero. There is a trigonometric formula to see this. Cut the pants into two right-angled hexagons then further into four pentagons with the ortho-geodesic. Formula 2.6.17 for right-angled pentagons in Chapter 2 of Thurston's notes gives $\sinh(R/4) \sinh (l/2) = \cosh(L/2)$. Therefore no such bound holds.

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No it does not. Suppose that the three boundary components have equal and very long length $R$. Then the pair of pants is almost isometric to a graph having two vertices $V,W$ and three edges of length $R/2$ each connecting $V$ to $W$; by "almost isometric" I mean that there is a function from the pants to the graph which distorts length of closed geodesics and of geodesics hitting the boundary at right angles by an additive amount which is bounded for $R$ near $+\infty$. Each of the three boundary curves has its length preserved in this graph. However, the arc you ask about maps to a single edge, of length $R/2$, so back in the pants this arc has length as close to $R/2$ as you like.

This can be made rigorous using formulas of hyperbolic trigonometry found, for example, in Thurston's book The geometry and topology of 3-manifolds.

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I have edited the question. sorry for inconvenience. –  Bidyut Sanki May 24 '13 at 4:51

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