two curves filling a surface

Let $S$ be a closed surface of genus $g \geq 2$. There exist two simple closed curves filling $S$?

Definitions:

Two closed curves $\alpha, \beta$ fill $S$ if they have minimal intersection and $S \setminus (\alpha \cup \beta)$ is a union of topological disks.

Two closed curves $\alpha, \beta$ have minimal intersection if $card(\alpha \cap \beta) \leq card(\alpha^' \cap \beta^')$ for all $\alpha^'$ in the same homotopy class of $\alpha$ and for all $\beta^'$ in the same homotopy class of $\beta$.

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Yes, otherwise curve complex would have diameter at most 2, while in fact it has infinite diameter. Pick your favorite simple nontrivial loops $a$ and $b$ and your favorite pseudoanosOv homeomorphism $f$. Now, apply a high power of $f$ to $b$ and do not change $a$. The result is a filling pair of loops. – Misha Oct 21 '12 at 11:45
Sorry: S(α∪β) in the question is a typo for S∖(α∪β), right? – Pietro Majer Oct 22 '12 at 13:15
This question is probably more appropriate for math.stackexchange.com – Ian Agol Oct 22 '12 at 14:57

Here is a quick but nonelementary proof; however, if you are interested in geometry and topology of surfaces, or Teichmuller theory, you should learn about curve complex in any case, this is a very powerful tool for studying mapping class groups, etc.

Let $X=C(S)$ be the curve complex of the surface $S$, see e.g. Schleimer's notes here. Vertices of $X$ are isotopy classes of simple nontrivial loops on $S$; two vertices are connected by an edge iff the loops can be made disjoint. Equip $X$ with the path-metric where every edge has unit length. If $\alpha, \beta$ are vertices of $X$ so that the pair $(\alpha, \beta)$ does not fill in the surface $S$, then the distance between $\alpha, \beta$ in $X$ is at most $2$, since you can find a nontrivial loop disjoint from both $\alpha$ and $\beta$. On the other hand, $X$ is connected and has infinite diameter, see the same source as above. Thus, there are pairs of loops in $S$ which fill.

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thank you for the very useful notes – Mario Oct 22 '12 at 11:00

A first obvious remark: if there are two curves separating the surface as required, up to a generic perturbation they have an intersection of finite cardinality. So by well-ordering, there are two such curves with intersection of minimal cardinality. In other words, the requirement that $\mathrm{card}(\alpha\cup\beta)$ be be minimal can always be ensured, if there are two not necessarily minimal separating curves.

If two surfaces have this property, so does their connected sum (w.r.to the connected sum of the disks suitably arranged). Also, the property is certainly true for the sphere, for the torus, and for the real projective plane, by direct decomposition. Therefore the property it is true for any closed surface, since it is a connected sum of some copies of the preceding.

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@Pietro: This induction (on genus) argument does not work without some extra efforts: You have to specify how you perform the connected sum relative to the curves you constructed for lower genus surfaces (say, tori). If you try to do so naively, say, by removing small disks near the double intersection points, then you may loose the "filling" condition after the connected sum (just consider the case of connected sum of two tori). – Misha Oct 21 '12 at 20:08
yes, one may loose the filling condition if the connected sum is made naively; but if the disks and the small hole for the sum are "suitably arranged", I think one should obtain a decomposition into two disks for the connected sum. However, I didn't bother to go into the details. – Pietro Majer Oct 21 '12 at 21:26
Pietro: One could also (naively) try to take connected sum outside of the curves, but then you loose filling property as well (even worse, instead of 2 curves, you may end up with four). So, "suitably arranged" needs a real explanation. – Misha Oct 21 '12 at 23:16
sure, that's all clear. I think the questioner is able to fix these details. – Pietro Majer Oct 22 '12 at 6:51
Prof. Majer: suppose that the surface minus the filling curves is a connected disk (it happens for the torus, for example). I can't figure out how to do the connected sum in a suitable way. – Mario Oct 22 '12 at 10:55