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The classification theorem for surfaces is proved using triangulation of surfaces. Another way to do this is using Morse theory.

Can the classification of surfaces be done using only point set topology without using triangulation or Morse function?

Priyanka
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  • I'm pretty sure Armstrong does it without triangulation or Morse. – Gregory Grant Apr 02 '20 at 06:47
  • Can you please provide some reference @GregoryGrant ? – Priyanka Apr 02 '20 at 08:06
  • https://www.amazon.com/Basic-Topology-Undergraduate-Texts-Mathematics/dp/0387908390 – Gregory Grant Apr 02 '20 at 08:10
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    The proof given in the book uses triangulation. – Priyanka Apr 02 '20 at 08:43
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    Do you consider a classification of surfaces to be giving a list of normal forms to surfaces? I am pretty sure every proof passes through triangulation/smoothness since you are basically showing that every surface is nice ("obviously" has a triangulation/smooth structure). –  Apr 02 '20 at 12:26
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    My guess would be that no it's not possible through purely point-set topological methods. One major hurdle in using only point-set topology is that it can be very difficult to show two spaces are not homeomorphic. Even if you can argue without using triangulations or Morse theory that every surface can be reduced to some normal form you still need to be able to show that these normal forms are different, and in doing this there is usually an algebraic input such as Euler characteristic or fundamental/homology group (which I would consider to be Algebraic Topology rather than Point-Set). – William Apr 02 '20 at 14:13
  • @GregoryGrant: You are wrong: Armstrong gives a sketch of a proof of Rado's theorem that all surfaces can be triangulated but then says that much more work has to be done to give a rigorous proof. He then works with triangulated surfaces. There are few textbooks where it is proven that all surfaces admit triangulations, e.g. "Riemann surfaces" by Ahlfors and Sario. – Moishe Kohan Apr 03 '20 at 04:00

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It is hard to prove negative as absence of evidence is not evidence of absence. All complete proofs of classification of compact topological surfaces that I know follow the following lines:

Step 1. Proving that every topological surface admits either a triangulation or a smooth structure. This is quite hard and does resemble point-set topology.

Step 2. Proving that every triangulated surface is PL homeomorphic to one of the "standard models" (say, in the case of closed connected surfaces, a connected sum of tori and projective planes). For this, there are several strategies. Or working with smooth surfaces and use, say, Morse theory, as you mentioned. Here, hardly anybody bothers proving that every topological surface admits a smooth structure.

Step 3. Proving that Euler characteristic is a topological invariant, thereby concluding the proof of the classification theorem. (This step might precede 1 and requires some, at least rudimentary, algebraic topology.)

I am unaware of any textbook treatment of Step 1, going from topological to smooth surfaces. The textbook proofs I know proceed by constructing a triangulation (Rado's Theorem) or simply work with smooth surfaces to begin with.

One reference for the existence of a triangulation is given here, Thomassen's paper. See more references in this Mathoverflow discussion.

One place which has both the proof of Rado's theorem and classification of compact surfaces is Moise's book "Geometric topology in dimensions 2 and 3".

Thus, assuming that you want to see the complete proof of classification of topological surfaces, take a look at Moise's book. Or start with Thomassen's paper and then read any account of classification of triangulated compact surfaces. Or, follow the book (aimed at undergraduate students)

Jean Gallier & Dianna Xu, A Guide to the Classification Theorem for Compact Surfaces, Springer-Verlag, 2013.

which also proves both Rado's theorem and classification of surfaces.

PS. What's lost in the combinatorial proofs of classification of surfaces is the following:

Every topological (or triangulated) surface admits a unique smooth structure up to diffeomorphism. Moreover, every homeomorphism between smooth surfaces is isotopic to a diffeomorphism.

This uniqueness does not automatically follow from the combinatorial classification of surfaces even in the compact case. The reference I know is

J. Munkres, Obstructions to the smoothing of piecewise-differentiable homeomorphisms. Ann. of Math. (2) 72 (1960), 521–554.

Moishe Kohan
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  • Isn't Ricci Flow another method? – C.F.G Apr 04 '20 at 06:29
  • @C.F.G: No, it is not; not even close. RF starts with the assumption that you have a smooth surface. Then it yields a metric of constant curvature. Does not give you a connected sum decomposition (in dimension 2). – Moishe Kohan Apr 04 '20 at 13:56
  • It seems like it could be possible to go from constant curvature metric to connect sum decomposition, although I haven't thought about it much(maybe inductively cut along shortest simple closed curves). But I basically agree starting from the assumption that you have a smooth surface already defeats the point of this question as that is basically the hard part(then including non point set topology arguments to finish). –  Apr 04 '20 at 17:10
  • @PaulPlummer: It is unclear to me how one would give a proof assuming, say, a metric of constant negative curvature. It is easy in this case to find a simple closed geodesic $c$ (since $\pi_1(M)\ne 1$ there is a shortest closed geodesic which then has to be simple), then cut $M$ along this geodesic. But finding the second simple closed geodesic on $M-c$ is challenging. One can find a shortest arc on the surface with boundary and argue that the process has to terminate using Gauss-Bonnet. But then one gets a collection of polygons and faces a combinatorial problem. – Moishe Kohan Apr 05 '20 at 00:43