Timeline for Two curves filling a surface
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12 events
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| Oct 24, 2012 at 10:07 | comment | added | Pietro Majer | I'd say: 1) an open subset of S bounded by a closed simple curve is always an Euclidean disk provided its diameter is smaller than some $\epsilon$ (e.g the Lebesgue number of a covering of S by Euclidean disks). 2) Two non-disjoint closed simple curves α and β produce connected components of S∖(α∪β) whose boundaries are closed simple curves (e.g. they can't produce an annulus). 3) There are two non-disjoint closed simple curves α and β such that all the components of S∖(α∪β) have diameter smaller than $\epsilon$, therefore they are disks. Then minimize card(α∩β) as said. | |
| Oct 24, 2012 at 9:25 | comment | added | Pietro Majer | I have to say that I suggested this hint of connected sums because it may possibly suggest some construction or computation to you (e.g., an upper bound on the cardinality of the intersection in terms of the genus of S). I hope it may be useful. That said, it seems to me that just the existence of the two filling curves follows from quick and elementary arguments. | |
| Oct 24, 2012 at 9:06 | comment | added | Pietro Majer | and also, third, it is not guaranteed that there is a disk whose boundary is made by an arc of α and one of β in S (and in S'). To create such a disk, one can perturb α and β near a crossing point, creating new small disks; in particular, with the two arcs not belonging to the closure of the same disk, as per your remark one. In any case, the aim of the construction was not a minimal decomposition for the connected sum (see my initial "obvious remark", so that the problem two is also not difficult). | |
| Oct 24, 2012 at 8:46 | comment | added | Mario | I think there are two problems: - the arcs of \alpha and \beta that you used can be on the boundary of the same disk (not the disk you removed, I'm saying in the remaining part of the surface) -if a pair of curves decompose the surface in a union of disks you cannot say that the corresponding pair realizing minimal intersection do the same | |
| Oct 23, 2012 at 5:44 | comment | added | Pietro Majer | Precisely, the two filling curves on the connected sum are the connected sum of α and α′, resp. β and β′. Is this description clear? | |
| Oct 22, 2012 at 12:39 | comment | added | Pietro Majer | me neither, but the idea was to do the decomposition of the connected sum by two curves without bothering to the minimality of intersection (for then there is also a pair of curves with minimal intersection). Say, for the connected sum we use as hole on $S$ a disk that is a component of $S\setminus(\alpha\cup\beta)$, and whose boundary is made by one arc of $\alpha$ and one of $\beta$; analogously on $S'$. Gluing the boundaries of these disks so as to identify the corresponding arcs of $\alpha'$ and $\beta'$ on $S'$we get 2 curves on the connected sum whose complement is a union of disks. | |
| Oct 22, 2012 at 10:55 | comment | added | Mario | 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. | |
| Oct 22, 2012 at 6:51 | comment | added | Pietro Majer | sure, that's all clear. I think the questioner is able to fix these details. | |
| Oct 21, 2012 at 23:16 | comment | added | Misha | 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. | |
| Oct 21, 2012 at 21:26 | comment | added | Pietro Majer | 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. | |
| Oct 21, 2012 at 20:08 | comment | added | Misha | @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). | |
| Oct 21, 2012 at 17:29 | history | answered | Pietro Majer | CC BY-SA 3.0 |