14

I recently learned about the Classification Theorem for compact 2-manifolds. Is there a similar classification theorem for ALL 2-manifolds, not just the compact ones?

Moreover, is there a theorem which classifies the 2-manifolds with boundary?

Jesse Madnick
  • 32,819
  • 3
    Any open subset of R^2 is a 2-manifold, so my impression is that this classification is hopeless in general. – Qiaochu Yuan Sep 27 '10 at 18:33
  • 5
    It's pretty fussy but not quite hopeless. That's one of the most interesting things about doing mathematics -- in that you get to see the blurry line where "hard" problems live evolve and take shape, some problems become "doable" and some become "hopeless", some stay mysteriously "hard" and never budge... – Ryan Budney Sep 27 '10 at 19:00
  • For the classification of compact connected 2-manifolds with boundary, you can look at Massey's "Algebraic Topology. An introduction". Notice that "the other" Massey, "Basic course in Algebraic Topology", though merging the former with his "Singular homology theory", does no longer contain the aforementioned classification. – Agustí Roig Sep 27 '10 at 23:30

2 Answers2

19

Yes, there's a classification theorem for non-compact 2-manifolds.

This paper gives the classification for triangulable 2-manifolds:

http://www.jstor.org/stable/1993768

That an arbitrary (2nd countable, Hausdorff) topological 2-manifold admits a triangulation is fairly classical. Ahlfors book "Riemann Surfaces" has a proof. There are others available, see for example this list:

https://mathoverflow.net/questions/17578/triangulating-surfaces

If all you're interested in is compact manifolds with boundary, you get that classification immediately from the closed manifold case. Because if you have a compact manifold with boundary, its boundary is a disjoint union of circles. So cap those circles off with discs to produce a closed manifold. So compact manifolds with boundary are classified by the closed manifold you get by "capping off" and the number of boundary circles you started with.

Non-compact manifolds have a more delicate classification -- think for example about the complement of a Cantor set in a compact surface.

Community
  • 1
Ryan Budney
  • 23,509
17

Here's a summary of the situation regarding noncompact $2$-manifolds with boundary, thanks to Moishe Kohan and Jacques Darné.

Originally, I posted an answer pointing to the 2007 paper Classification of noncompact surfaces with boundary by A. O. Prishlyak and K. I. Mischenko, Methods Funct. Anal. Topology 13 (2007), no. 1, 62–66. However, Kohan pointed out that the classification had actually been completed much earlier by E. Brown and R. Messer (The classification of two-dimensional manifolds, Trans. Amer. Math. Soc. 255 (1979), 377–402). Then more recently Darné pointed out that the theorem claimed by Prishlyak and Mischenko is false, because it contradicts the one of Brown and Messer. See Darné's comment below for details.

So the upshot is that the correct classification of noncompact $2$-manifolds with boundary was completed in 1979 by Brown and Messer in the paper cited above.

Sumanta
  • 9,777
Jack Lee
  • 50,850
  • 4
    Actually, this classification of noncompact surfaces with boundary was completed in 1979: E. Brown and R. Messer, "The classification of two-dimensional manifolds", Trans. Amer. Math. Soc., vol. 255 (1979), 377–402. Just Prishlyak and Mischenko were unaware of that paper. – Moishe Kohan Nov 13 '17 at 12:04
  • 1
    @MoisheCohen: Wow. I was unaware of it too. Thanks for pointing that out. – Jack Lee Nov 13 '17 at 23:55
  • 2
    @JackLee For further reference, you should edit your answer: the theorem of Prishlyak and Mischenko is false. In fact, it contradicts the one of Brown and Messer, and examples in the latter paper show that their invariants are not strong enough. Precisely, they only take connected components of the graph of non-compact connected components, when they should consider the whole graph, with some chosen (equivalence class of) orientation. Also, their claim that connected components of the graph can only be circles is not true at all (cf. the introduction of Brown and Messer). – J. Darné Oct 26 '20 at 12:32
  • 1
    @J.Darné -- at this point, maybe it will be best if you write a separate answer and I delete mine, since my answer is now mainly disinformation. – Jack Lee Oct 26 '20 at 14:50
  • 1
    @JackLee Well, you can edit yours if you like, to turn it into information! After all, it has been useful to me: I am just beginning to read about this stuff, and I was unaware of the paper of Brown and Messer. But if you prefer, I can also write an answer. As you wish :) – J. Darné Oct 27 '20 at 11:04