Every compact surface is diffeomorphic to $S^2$, $\underbrace{T^2\#\ldots \#T^2}_n$, or $\underbrace{RP^2\#\ldots \#RP^2}_n$ for some $n\ge 1$.

Is there a conceptual proof of this classification theorem?

Here, the term "conceptual" a little bit up for interpretation...

One may take it to mean: that doesn't rely on unintuitive facts. For example, a proof that doesn't need $T^2\#RP^2\cong RP^2\#RP^2\#RP^2$ as an ingredient, but which has the property that this diffeomorphism comes out as a consequence would definitely count as conceptual.

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    $\begingroup$ The Zip proof is probably what you are looking for. $\endgroup$ – Ben McKay Feb 20 at 12:47
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    $\begingroup$ Take a look at this explanation of the Zip proof: maths.ed.ac.uk/~v1ranick/papers/francisweeks.pdf $\endgroup$ – Ben McKay Feb 20 at 12:48
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    $\begingroup$ You might find this helpful: Gallier, Jean, and Dianna Xu. A guide to the classification theorem for compact surfaces. Springer Science & Business Media, 2013. PDF download. $\endgroup$ – Joseph O'Rourke Feb 20 at 13:50
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    $\begingroup$ @Ben McKay: Lemma2 of the ZIP proof is exactly what I wanted to avoid... $\endgroup$ – André Henriques Feb 20 at 14:37
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    $\begingroup$ I'm not sure if this is André's motivation, but one reason why one might want to better understand conceptually where 3 cross caps = crosscap + handle comes from, is that this relation makes the classification of non-oriented 2d TQFTs extremely weird. In particular, this doesn't look like a "homotopy fixed point" condition, in contrast to the intuition for fully extended TFTs. $\endgroup$ – Noah Snyder Feb 20 at 19:15

I guess the most conceptual proof is the one using Morse theory:

Take a Morse function on the (closed, orientable) surface S. If it has no saddle points, then (using the gradient flow) $S\cong S^2$. Assume by induction that a surface with k-1 saddle points is $S^2$ with finitely many handles added. For the inductive step, consider a Morse function with k saddle points. A neighborhood of a saddle point is a pair of pants, with the Morse function constant on boundary curves. Consider the (possibly disconnected) surface $S^\prime$ obtained by cutting off the pair of pants and gluing in three disks. The Morse function extends (constant on the disks). By induction, each component of $S^\prime$ is $S^2$ with finitely many handles added, so the same is true for $S$.

There are other nonclassical proofs.

It is actually not hard, to prove by purely topological arguments that for a (closed, orientable) surface S with $\chi(S)<2$, there is another surface Sโ€˜ with $\chi(S^\prime)=\chi(S)+2$. Thus the proof of classification boils down to show that a (closed, orientable) surface with $\chi(S)=2$ must be $S\cong S^2$. This can be seen from the Riemann-Roch theorem, which for $\chi(S)=2$ implies existence of a meromorphic function with only one pole of order 1, thus a biholomorphic map to $P^1C$. (For this argument you need that every orientable surface is complex; this follows from the homotopy equivalence $GL^+(2,R)\sim GL(1,C)$ and the obvious vanishing of the Nijenhuis tensor.)

Yet another approach would be to use uniformization and thus reduce the classification of surfaces to classification of discrete, torsion-free subgroups of PSL(2,R). (Uniformization has, besides the classical proof via the Dirichlet principle, by now more conceptual proofs via circle packing or via Ricci flow.)

Finally, if you believe that every surface can be triangulated, then the classification of surfaces is proven by a not so difficult combinatorial inductive argument. The triangulability of surfaces is of course not so easy, its proof in Moise uses the Schoenflies theorem.

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  • $\begingroup$ By purely what arguments? (also, if you could give a reference or a sketch of those arguments I would love to see them) $\endgroup$ – Denis Nardin Feb 20 at 13:20
  • $\begingroup$ Sorry, purely topological arguments. You just get Sโ€˜ from S by cutting along a homological nontrivial curve without self-intersections. The existence of a homologically nontrivial curve follows from the Euler characteristic. If that curve has self-intersections, you decompose it into closed curves without self-intersections. At least one of this curves is homologically nontrivial. Cutting along a homologically nontrivial curve yields a connected surface Sโ€˜. Thus S is the connected sum of Sโ€˜ with a torus. $\endgroup$ – ThiKu Feb 20 at 13:34
  • $\begingroup$ @ThiKu: Does ๐‘‡^2#๐‘…๐‘ƒ^2โ‰…๐‘…๐‘ƒ^2#๐‘…๐‘ƒ^2#๐‘…๐‘ƒ^2 follow from the type of arguments that you describe? $\endgroup$ – André Henriques Feb 20 at 14:40
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    $\begingroup$ Once you have a smooth structure (needed for Morse theory, hyperbolization, uniformization, etc), finding a triangulation is trivial. $\endgroup$ – Andy Putman Feb 20 at 18:10
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    $\begingroup$ I know that every smooth manifold has a triangulation, but trivial? $\endgroup$ – ThiKu Feb 20 at 18:37

This is more an extended comment than an answer to the question. The first thing to note is that there are different strenghts of the classification theorem for surfaces. Of course, there are the differentiable, triangulated and topological setting. But even if we choose such a setting, there are two statements one has to prove (at least in one approach):

Every closed surface is isomorphic to a sphere with handles or cross-caps attached.

The isomorphism type only depends on the number of handles and cross-caps attached.

Both the Zip-proof and the Zeeman proof refered to above in the comments only prove the first part and not the second part. The second part is essentially equivalent to the well-definedness of the connected sum. Especially in the topological setting, this is a subtle point, requiring even in dimension 2 a kind of Schรถnflies theorem (and being a very difficult theorem in dimension 4). Out of frustration about this state of affairs I wrote up a variant of Zeeman's proof (based on a treatment by Thomassen), but including the well-definedness of attaching handles/cross-caps. (It took me a long time to thus actually understand the argument of Zeeman.)

This is not really an answer to the original question, both because this proof is in the triangulated setting and does not avoid the isomorphism of the original question. But I really want to stress the point that well-definedness of attaching handles or connected sum is something one has to prove and not just hide by abuse of notation.

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  • $\begingroup$ With the Zip proof the problems you are worrying about do not actually arise. To begin, fix a set of standard model surfaces $M(a,b,c,d)$ with $a$ handles (tubes), $b$ cross-handles, $c$ cross-caps, and $d$ boundary circles. One just has to show that if one takes a collection of disjoint model surfaces and identifies two oriented boundary arcs, then the result is again homeomorphic to a collection of disjoint model surfaces. (continued) $\endgroup$ – Allen Hatcher Mar 2 at 14:24
  • $\begingroup$ Once one specifies the two oriented arcs to be identified, the choice of an orientation-preserving homeomorphism identifying them doesn't matter since each orientation-preserving homeomorphism of one of the arcs extends to a homeomorphism of the model surface (before identification) using a collar neighborhood of the boundary. (continued) $\endgroup$ – Allen Hatcher Mar 2 at 14:25
  • $\begingroup$ After all the pairs of boundary arcs have been identified one has a model surface $M(a,b,c,d)$ with $d=0$, so all that remains to show is that if $b$ or $c$ is nonzero then all handles and cross-handles can be replaced by cross-caps, which isn't hard for the standard models. The operation of forming a connected sum plays no role in this proof, nor does the induction step involve attaching handles, cross-handles, or cross-caps in the interior of a surface. $\endgroup$ – Allen Hatcher Mar 2 at 14:25

The proof of Zeeman described in this note is by a substantial margin the easiest and most conceptual proof I know. To simplify the exposition I restrict to orientable surfaces in the note, but it is trivial to also do the non-orientable case (and see the edit below for one description of how to arrange this to avoid using the fact that three cross caps is a handle plus a cross cap).

What I particularly like about it is that it proves not only that the usual classification is the complete list of surfaces, but also at the same time that the Euler characteristic is a complete invariant of orientable surfaces. Indeed, the proof is by (descending) induction on the Euler characteristic, with the base case the 2d Poincare conjecture: all compact connected surfaces have Euler characteristic <=2, with equality iff the surface is a sphere (and you can see the sphere very clearly in the proof, which directly produces from equality a decomposition of the surface into two discs meeting along their boundary).

As evidence of how simple this proof is, on many occasions I have explained it at a chalk board (with full details) in about 10 minutes.

EDIT: I've thought a little bit more about whether or not you can avoid having to prove that the connect sum of $3$ projective planes is isomorphic to the connect sum of a torus and a projective plane. Here is one way to arrange the proof that avoids directly proving this.

As in the proof in my notes, you prove the theorem by downward induction on the Euler characteristic. More precisely, what you prove by induction is the following:

  1. Every connected surface has Euler characteristic less than or equal to $2$.

  2. If a surface is orientable, then its Euler characteristic is even and if it equals $2-2g$, then the surface is a connect sum of $g$ tori.

  3. If a surface is not orientable and if its Euler characteristic is $2-g$, then the surface is a connect sum of $g$ projective planes.

Notice that this implies as a consequence that the connect sum of $3$ projective planes is isomorphic to the connect sum of a torus and a projective plane (which will not be used directly in its proof).

Anyway, you follow the proof in my notes to deal with the base case (which combines $1$ above and the fact that the sphere is the only connected surface of Euler characteristic $2$). You then follow my notes in the inductive case to the point where you find the nonseparating simple closed curve $\gamma$. There are then several cases:

a. If your surface is orientable, then $\gamma$ is a $2$-sided curve, so you can cut and cap off to increase the Euler characteristic by $2$, and be done by induction.

b. If your surface is nonorientable and $\gamma$ is a $1$-sided curve such that cutting along $\gamma$ gives a nonorientable surface, then you can do just like in a (but cutting along $\gamma$ and capping only increases the Euler characteristic by $1$).

b. If your surface is nonorientable and $\gamma$ is a $1$-sided curve such that cutting along $\gamma$ gives an orientable surface, then you can cut cap and induct and deduce that your surface is isomorphic to $\Sigma_g \# \mathbb{P}^1$ for some $g \geq 1$. Of course, this is not what you want; however, if you draw the picture you can easily find a simple closed $1$-sided curve $\gamma'$ on $\Sigma_g \# \mathbb{P}^1$ such that cutting along $\gamma'$ gives a nonorientable surface. This is the curve you should have been using all along! Replace $\gamma$ with $\gamma'$, and go back to case b.

d. Finally, if your surface is nonorientable and $\gamma$ is a $2$-sided curve, then the complement of $\gamma$ must be nonorientable, so cutting and capping we find that the surface must be isomorphic to $\Sigma_1 \# (\#_{k} \mathbb{P}^1)$ for some $k$. Just like in case c, we can find a $\gamma'$ that is $1$-sided such that the complement is nonorientable, and then go back and use that one in case b.

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  • $\begingroup$ Could you say a little bit more about the non-orientable case? There's gotta be something a little bit tricky because Euler char 0 doesn't imply you can split off a torus without assuming orientability, but Euler char -1 does imply you can split off a torus despite non-orientability. So you need three different lemmas in general? Or two lemmas plus a separate proof that three cross caps = handle + cross-cap. $\endgroup$ – Noah Snyder Feb 20 at 19:07
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    $\begingroup$ You modify the proof as follows. Without assuming orientability, then there are two cases. (1) The neighborhood of the loop gamma is a cylinder (which is the case described in Andy's note) or (2) it is a Mobius strip. In case (2) you cut it out and glue in a single disk. This still gives a surface with larger Euler characteristic (increased by 1 this time), so is still covered by induction. $\endgroup$ – Chris Schommer-Pries Feb 20 at 19:47
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    $\begingroup$ Chris beat me to it! If you want to avoid proving that three cross caps is a handle and a cross cap, you have to prove that if the surface is non-orientable, then you can find a one-sided simple closed curve that also has the property that cutting along it does not give an orientable surface. This requires a small extra argument (since you can't just pick any cycle in the dual graph). Probably it is easier and more natural to just go ahead and prove that 3 cross caps = handle + cross-cap (which I don't think of as being a very hard fact). $\endgroup$ – Andy Putman Feb 20 at 21:11
  • $\begingroup$ I just edited the answer with an approach that avoids proving the three cross caps is a handle and a cross cap (basically by using the inductive hypothesis a little better). $\endgroup$ – Andy Putman Feb 20 at 22:01
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    $\begingroup$ [continued] I personally think that I should know the "best" proof of every result I routinely use, not matter how elementary. And given how many papers I have written about the mapping class group of a surface, the classification of surfaces is one of my most-used tools. $\endgroup$ – Andy Putman Feb 21 at 20:25

Using a little bit of real algebraic geometry, there is a conceptual proof at least in the critical case $\chi=-1$, i.e. the case you're talking about explicitly. Indeed, let $S$ be a compact connected smooth surface without boundary with $\chi(S)=-1$. Choose a conformal structure on $S$. Since $\chi$ is odd, $S$ is nonorientable. Let $\tilde S$ be its orientation cover. Then $\tilde S$ becomes a complex algebraic curve endowed with an antiholomorphic involution $\sigma$. Since $\chi(\tilde S)=2\chi(S)=-2$, the curve $\tilde S$ is of genus $2$. In particular, $\tilde S$ is hyperelliptic. Let $$\pi\colon \tilde S\rightarrow \mathbf P^1(\mathbf C)$$ be the hyperelliptic covering. Since the hyperelliptic covering is canonical, the image curve $\mathbf P^1(\mathbf C)$ acquires a real structure, i.e., an antiholomorphic involution $\tau$. In general, there are $2$ possibilities: either $\tau$ has fixed points, or it doesn't.

Let us prove that $\tau$ does have fixed points. If it hadn't any, the ramified double covering $\pi$ would induce a ramified double covering of the quotients $$\bar\pi\colon\tilde S/\sigma=S\rightarrow \mathbf P^1(\mathbf C)/\tau=\mathbf P^2(\mathbf R).$$ Both quotients are compact connected smooth surfaces without boundary as both $\sigma$ and $\tau$ act without fixed points. Their Euler characteristics are equal to $-1$ and $1$, respectively. It follows that $\bar\pi$ is ramfied over exactly $2\times 1-(-1)=3$ points, which is absurd. Therefore, $\tau$ does have fixed points. Hence $\tau$ is conjugate to the usual real structure, and we may assume that $\tau$ is equal to the usual real structure on $\mathbf P^1(\mathbf C)$.

It follows that the curve $\tilde S$ is isomorphic to a curve $C$ of the form $y^2=-p(x)$, for some real polynomial $p$ of degree $6$ taking only strictly positive values on $\mathbf R$ and having simple roots in $\mathbf C$, the isomorphism being an isomorphism of real algebraic curves, i.e., respecting the antiholomorphic involutions on $\tilde S$ and $C$.

Now, the fact that the set of all such polynomials $p$ is connected implies that all smooth surfaces with $\chi=-1$ are diffeomorphic to each other!

The argument works generally for any smooth surface $S$ of odd Euler characteristic, after arguing that one can always deform the conformal structure to a hyperelliptic one, which I believe can be proven without using the classification of smooth surfaces. For nonorientable surfaces of even Euler characteristic, one cannot exclude $\tau$ to act fixed point freely, and I do not see anything conceptual in this case.

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