I would like to find a result telling me that two simple closed curves $\alpha$ and $\beta$ (on a non-orientable surface $S$) are in minimal position if and only if there is not a disk in $S$ whose boundary consists of an arc of $\alpha$ and an arc of $\beta$.

[By minimal position I mean that the number of $i(\alpha,\beta) :=$ intersections of $\alpha$ and $\beta$ = min{$i(\alpha',\beta')$|$\alpha'$ and $\beta'$ are isotopic to $\alpha$ and $\beta$ resp.}

I have searched previous literature on non-orientable surfaces and found Lemma 2.5 in: https://www.math.stonybrook.edu/~mlyubich/Archive/Topology/Epstein.pdf

The Lemma states that for two closed curves $\alpha$ and $\beta$ that intersect on an orientable or non-orientable surface, there always exists an ambient isotopy that separates them. This is well known for orientable surfaces, and to obtain the result for non-orientable you can just lift to the orientable double cover and obtain your ambient isotopy there.

Is there a way of extracting the result about minimal position from this lemma? Or is ambient isotopy going to be too weak?

  • 2
    $\begingroup$ I know two proofs of this for orientable surfaces (one using hyperbolic geometry that is explained in Farb-Margalit's primer on mapping class groups, the other using basic differential topology and explained in Buser's Geometry and Spectra of Compact Riemann Surfaces). Both proofs also work for non-orientable surfaces without changes. $\endgroup$ – Andy Putman Feb 13 '17 at 4:36
  • $\begingroup$ I think there is also a fairly elementary proof in "Automorphisms of Surfaces After Nielsen and Thurston" by Casson and Bleiler. The arguments should give you everything you need. Maybe nothing that isn't covered by Andy's references. $\endgroup$ – Richard Weidmann Feb 13 '17 at 18:46

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