two curves filling a surface - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T06:25:34Zhttp://mathoverflow.net/feeds/question/110228http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/110228/two-curves-filling-a-surfacetwo curves filling a surfaceMario2012-10-21T10:55:05Z2012-10-22T15:55:04Z
<p>Let $S$ be a closed surface of genus $g \geq 2$. There exist two simple closed curves filling $S$?</p>
<p>Definitions:</p>
<p>Two closed curves $\alpha, \beta$ fill $S$ if they have minimal intersection and $S \setminus (\alpha \cup \beta)$ is a union of topological disks.</p>
<p>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 $\beta$. </p>
http://mathoverflow.net/questions/110228/two-curves-filling-a-surface/110251#110251Answer by Pietro Majer for two curves filling a surfacePietro Majer2012-10-21T17:29:38Z2012-10-21T17:29:38Z<p>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.</p>
<p>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.</p>
http://mathoverflow.net/questions/110228/two-curves-filling-a-surface/110294#110294Answer by Misha for two curves filling a surfaceMisha2012-10-22T03:25:34Z2012-10-22T03:25:34Z<p>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. </p>
<p>Let $X=C(S)$ be the curve complex of the surface $S$, see e.g.
<a href="http://www.math.rutgers.edu/~saulsch/Maths/notes.pdf" rel="nofollow">Schleimer's notes here</a>.
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. </p>