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I've had a few undergraduate students ask me for references for the classical fact (due to Rado) that closed topological surfaces can be triangulated. I know two sources for this, namely Ahlfors's book on Riemann surfaces and Moise's book "Geometric topology in dimensions 2 and 3". Both of these strike me as being a bit much for a bright undergraduate. Question : in the 30+ years since Moise's book, has anyone written a more accessible account?

edited Mar 9 2010 at 6:19

Qiaochu Yuan
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asked Mar 9 2010 at 5:36

Andy Putman
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Tony Pantev · Accepted Answer · 2010-03-09 06:32:23Z UTC

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Try this book by Jean Gallier and Diana Xu. It is aimed at undergraduates and has a nice account of Thomassen's elementary proof of the triangulation theorem in the last appendix. Or you can refer the students to Thomassen's original paper which is also quite readable.

edited Mar 9 2010 at 6:32

answered Mar 9 2010 at 6:13

Tony Pantev
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Thanks, I didn't know anyone had simplified the existence of triangulations proof that much. – Ryan Budney Mar 9 2010 at 6:23

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This is exactly what I was looking for! I had never seen that paper of Thomassen, which from the looks of it is a great little gem to share with my students. The bibliography also mentions a very short paper of Doyle and Moran that triangulates surfaces. I had figured that classical material like this had easy proofs by now... – Andy Putman Mar 9 2010 at 6:28

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Thomassen's paper is very nice, except that I had a lot of trouble with Theorem 3.1, the Jordan-Schoenflies theorem. Can someone give a crisp explanation of his argument here? – Greg Kuperberg Mar 9 2010 at 7:04

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Thomassen's paper can be downloaded from Andrew Ranicki's website at maths.ed.ac.uk/~aar/jordan/index.htm . I think there's a fixable error in his main proof. He assumes that if a point is accessible from a region, then after one puts in a polygonal path based at the point then it is still accessible from the new regions. This is false: consider the "cuspidal cubic" curve. But one can retrieve the situation by redfining accessibility by insisting that a positive angle's worth of segments starting at the given point and going into the given region. – Robin Chapman Mar 10 2010 at 8:03

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Rado's original triangulation paper is now also available from the website, along with sundry other items. – Andrew Ranicki Nov 24 2011 at 21:33

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Marvin Jay Greenberg · Answer 2 · 2013-03-01 00:48:52Z UTC

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The notes (in French) for Gramain's 1969-70 course, available at

http://www.math.u-psud.fr/~biblio/numerisation/docs/G_GRAMAIN-55/pdf/G_GRAMAIN-55.pdf

seem to include a proof using Morse theory, based on a quick glance I made.

answered Mar 1 at 0:48

Marvin Jay Greenberg
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Rafe Mazzeo · Answer 3 · 2010-03-09 06:24:23Z UTC

There is a forthcoming undergraduate differential geometry book by Christian Baer (at the University of Potsdam, Germany) which is very nice; it gives a careful account of this theorem and a very accessible proof. If the book is not publically available, it should be soon.

Ryan Budney · Answer 4 · 2010-03-09 05:51:32Z UTC

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If you're okay going the extra step and assuming a smooth structure, the standard argument of Whitehead goes like this: take a smooth embedding of your manifold (of any dimension) into euclidean space. Triangulate Euclidean space, perturb the embedding to make it transverse to the skeleta of the triangulation. Refine the triangulation (barycentric subdivision) to the point where the embedding "looks linear" in each top-dimensional simplex. The triangulations of the simplices pulls-back to a polyhedral decomposition of the manifold, which you can subdivide to be a triangulation.

If you insist on going the extra step to topological manifolds you could smooth the topological structure. I believe much of that argument appears in Thurston's 3-dimensional geometry and topology book but I don't have it at home at the moment, and I don't remember.

answered Mar 9 2010 at 5:51

Ryan Budney
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The main reason they want to see a proof for a topological surface is that it is the first step in classifying surfaces. The way I like to arrange that proof immediately thickens the triangulation up to a handle decomposition; if I assumed that the manifold was smooth, then I could dispense with the triangulation and apply Morse theory. Do you know a proof that topological surfaces can be smoothed that doesn't pass through a triangulation? – Andy Putman Mar 9 2010 at 5:57

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And thanks for pointing me towards Thurston's book! There are an amazing number of things in there, though I am loath to recommend it to undergraduates given Thurston's cavalier attitude toward rigor... – Andy Putman Mar 9 2010 at 5:58

I'll check tomorrow or at latest Wednesday. I thought he was doing something more along the lines of Kirby-Siebenmann, smoothing the transition maps from the atlas. But it's been a long time and I haven't read that part of the book with any focus. – Ryan Budney Mar 9 2010 at 6:14

Oh, sorry. Thurston only talks about the PL <-> Smooth relations. – Ryan Budney Mar 10 2010 at 23:44

Yeah, that's what I thought I remembered, but there's so much stuff in Thurston's book that that I thought I might have missed it. – Andy Putman Mar 16 2010 at 3:49

Triangulating surfaces

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