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Let $S$ be a hyperbolic surface, which is not the punctured torus or $4$-holed sphere. I am interested in finding a ``geometrically optimal'' pants decomposition on $S$.

Here is a candidate definition. Given a pants decomposition $P$, order the curves of $P$ from longest to shortest (in the hyperbolic metric). Then, pants decompositions $P$ and $P'$ can be compared by comparing the lengths of their curves lexicographically. That is: if the longest curve $c_1$ of $P$ is shorter than the longest curve $c'_1$ of $P'$, then $P$ is better. Or, if $\ell(c_1) = \ell(c'_1)$ and $\ell(c_2) < \ell(c'_2)$, where $c_2, c'_2$ are the second-longest curves of $P$ and $P'$, then $P$ is better. And so on, lexicographically.

With this definition, the induced ordering on pants decompositions becomes a well-ordering. More precisely: given a fixed pants decomposition $P$, there are finitely many curves shorter than the longest curve of $P$, hence finitely many better pants decompositions. In particular, there exists an ``optimal'' decomposition, whose longest curve is no longer than the Bers constant.

It is clear that optimal decompositions are not necessarily unique (otherwise, the optimal decomposition would never change as we move in Teichmuller space). But if $P$ and $P'$ are both optimal on a given surface, what can be said about how far apart they are? For example: are their shortest curves necessarily disjoint?

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most likely not going to be disjoint shortest curves. Check out Schmutz-Schaller's papers: – Ian Agol Oct 1 '11 at 0:12
Thanks! The discussion of systoles seems relevant. – Dave Futer Oct 1 '11 at 13:52
up vote 4 down vote accepted

A pair of optimal pants decompositions $A, B$ need not have disjoint shortest curves. Here is an example.

Let $S$ be the genus two hyperbolic surface built from four equilateral right-angled hexagons by "doubling". That is, let $H$ be such a hexagon. Let $a = 2 \cosh^{-1}(\sqrt{3/2}) = 1.31695\ldots$ denote the side-length of $H$. (See Beardon.) Let $\alpha$ be the union of three non-adjacent sides of $H$ and let $\beta$ be the other three non-adjacent sides. Doubling $H$ across $\alpha$ gives a pants $P$. So $\alpha$ gives the seams of $P$ and $\beta$ doubles to give $\partial P$, each component having length $2a$. Now double across $\beta$ to get $S$.

Let $B = \partial P$ be the double of $\beta$ in $S$. So $B$ is a pants decomposition of $S$, each curve having length $2a$. Likewise $A$, the double of $\alpha$, is a pants decomposition.

Claim: All curves of $A$ and $B$ are systoles of $S$.

Proof: Any geodesic in a genus two surface is preserved by the hyperelliptic and so double covers an arc or a loop in the orbifold $O = S^2(2,2,2,2,2,2)$. In our situation $O$ is the double of $H$ across $\partial H$. The shortest arc is an edge of $H$. The shortest loop divides $O$ in half and has length $4\cosh^{-1}(\sqrt{2}) = 3.525494\ldots$. QED

Thus both $A$ and $B$ are optimal, yet every curve of $A$ crosses every curve of $B$.

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Very elegant - thank you! – Dave Futer Oct 10 '11 at 14:29
You are welcome. – Sam Nead Oct 10 '11 at 18:18

My paper arXiv:cs/0604034 provides an approximately-optimal pants decomposition for the multiply-punctured hyperbolic plane, with a somewhat different optimality condition in which we add the curve lengths rather than compare them lexicographically. I don't know about extensions to other hyperbolic surfaces, though.

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Thanks! I'll take a look. By the way, it's quite cool to get an answer to a math question in the form of a paper under arXiv:cs/ ! – Dave Futer Sep 30 '11 at 22:06

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