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Let $N_{k,r}$ be a non-orientable surface of genus $k$ (i.e the connected sum of $k$ projective planes) and with $p\geq 1$ punctures.

I'm looking at ideal triangulations of the surface, namely maximal collection of pairwise disjoint ideal arcs (joining punctures) on the surface.

Given a triangulation and an arc $\alpha$ in that triangulation, we can perform a flip along $\alpha$ : the arc belongs to two triangles forming a quadrilateral, the flip along $\alpha$ gives the new triangulation where the arc has been replaced by the other diagonal of the quadrilateral, and all the other arcs remains unchanged. (When the arc belongs to only one triangle, we cannot do the flip.)

A classical result in the orientable case states that given any two triangulations of an orientable surface, there is a sequence of flips that transform one into the other (Hatcher, On triangulations of surfaces). So the so-called triangulation graph, whose vertices are triangulations and edges are flips, is connected.

I am looking for the similar result for non-orientable surfaces. Does anyone have a reference ? My guess is that the result should also hold but I did not find anything on the litterature.

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Did you check whether hatcher's proof goes through? If not, where is the sticking point? –  Igor Rivin Feb 2 '12 at 19:00

3 Answers 3

up vote 10 down vote accepted

My article "Tiling the measured foliation space of a punctured surface", Trans. Math. 306 no. 1 (1988) contains a proof of this fact in the case of oriented surfaces. It is essentially the same as Hatcher's proof of contractibility, but focussing solely on the issue of connectivity, which introduces some simplifications. The proof works, with a little bit more effort, in the nonorientable case as well.

The crucial step in the proof for purposes of this discussion (carried out in the lemmas on pages 38 and 39) is to consider the situation that $\delta$ is an ideal triangulation, $h$ is an oriented ideal arc whose interior intersects $\delta$ transversely and efficiently ($h$ does not double back across the same arc in a triangle of $\delta$), $x_0$ is the first point where $h$ crosses $\delta$ transversely, and $\alpha$ is the edge of $\delta$ containing $x_0$. In this situation one wants to prove that $\alpha$ belongs to two distinct triangles of $\delta$ and so can be flipped; this is basically the inductive step for proving connectivity.

The proof of this step is to consider the possibility that $\alpha$ belongs to only a single triangle $T$ of $\delta$ --- meaning that $\alpha$ is the ideal arc obtained by identifying 2 sides of $T$ --- and to derive a contradiction. In the article this step is carried out only in the orientable category, where the result of gluing the 2 sides of $T$ must be a disc (depicted in the diagrams of that paper as a ``puncture piece'').

However, the conclusion of this argument remains true in the nonorientable category, and the proof requires just one more case to be considered, namely, when the result of gluing the 2 sides of $T$ is a Mobius band. In this situation, the efficient intersection condition would imply that $h$ is trapped in the interior of the Mobius band, winding more and more closely around its core and crossing $\alpha$ infinitely often, contradicting that the number of intersections must be finite.

With this consideration, the whole proof should go through unscathed; orientability is not otherwise used.

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Thank you very much for this detailed answer. –  Frederic Palesi Feb 27 '12 at 11:29

This is shown in MR1310882 (95m:05091) Negami, Seiya(J-YOKOED) Diagonal flips in triangulations of surfaces. (English summary) Discrete Math. 135 (1994), no. 1-3, 225–232. You can check it out. You will note that the argument uses some heavy lifting, unlike Hatcher's.

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Here is one result, by Mori and Nakamoto, not the surface in which you are interested, but perhaps it will lead elsewhere: "Diagonal flips in Hamiltonian triangulations on the projective plane" Discrete Mathematics, Volume 303, Issues 1–3, 2005, Pages 142–153:

In this paper, we shall prove that any two triangulations on the projective plane with $n$ vertices can be transformed into each other by at most $8n-26$ diagonal flips, up to isotopy.

One later (2007) paper that cites this is "Triangulating the Real Projective Plane" by Aanjaneya and Teillaud, arXiv:0709.2831v2 cs.CG.

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