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How many simple closed curves can be put on a orientable genus $g$ surface $\Sigma_g$ such that the following are true:

  1. The curves are pairwise non-homotopic
  2. The curves are pairwise set-theoretically disjoint.

I am trying to give an upper bound on this number.

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    $\begingroup$ You should google "pants decomposition". Here is one hit. brownsharpie.courtneygibbons.org/?p=144 $\endgroup$
    – Sam Nead
    Commented Apr 7, 2012 at 9:52
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    $\begingroup$ Provided $g \geq 2$ such a maximal collection is called a pant decomposition of the surface. That there's $3g-3$ curves in such a decomposition is a very old result, observed probably by many people but certainly well-known by Poincare's time. Such curve collections are used to put "coordinates" on the collection of all curves on the surface, usually called "Dehn-Thurston coordinates". $\endgroup$
    – Ryan Budney
    Commented Apr 7, 2012 at 10:06

2 Answers 2

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This is Lemma 3.2 of this paper by Juvan, Malnic and Mohar.

Theorem. Let $F$ be a family of non-null homotopic closed curves on a surface $\Sigma$ with $b \geq 0$ boundary components which are pairwise non-homotopic and pairwise disjoint. Then

$|F| \leq \max (1, 3(g_{\Sigma}-1)+2b)$,

where $g_{\Sigma}$ is the genus of $\Sigma$ (orientable or non-orientable).

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I am tacitly assuming that your curves should also be non-contractible, otherwise you have to add 1 to the answer. Let's suppose for simplicity that the surface is closed (but the argument adapts in a straightforward way to the non-closed situation) and has negative Euler characteristic.

Clearly, the best solution is to have your curves decompose the surface into pairs-of-pants (discs are forbidden; annuli are forbidden; and any other surface can be further decomposed).

Each pair-of-pants has Euler characteristic -1, and contributes 3/2 boundary curves (dividing by 2 to account for the fact that each boundary curve appears twice). The answer is then -3/2 the Euler characteristic, as described in the first answer above.

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    $\begingroup$ Except of course if both ends of the annulus are the same, in which case you are on the torus, and the correct answer is $1$. $\endgroup$
    – Will Sawin
    Commented Apr 6, 2012 at 18:37

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