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](http://arxiv.org/abs/0911.5149).


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?



Optimal pants decompositions of a hyperbolic surface
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 $B$. 


