# 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.