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. 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). The proof of Zeeman described in this note is by a substantial margin the easiest and most conceptual proof I know. 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. 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. (It took me a long time to thus actually understand the argument of Zeeman.) 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. Especially in the topological setting, this is a subtle point, requiring even in dimension 2 a kind of Schönflies theorem (and being a very difficult theorem in dimension 4). The second part is essentially equivalent to the well-definedness of the connected sum. The isomorphism type only depends on the number of handles and cross-caps attached.īoth the Zip-proof and the Zeeman proof refered to above in the comments only prove the first part and not the second part. But even if we choose such a setting, there are two statements one has to prove (at least in one approach):Įvery closed surface is isomorphic to a sphere with handles or cross-caps attached. Of course, there are the differentiable, triangulated and topological setting. The first thing to note is that there are different strenghts of the classification theorem for surfaces. This is more an extended comment than an answer to the question. The triangulability of surfaces is of course not so easy, its proof in Moise uses the Schoenflies theorem. (Uniformization has, besides the classical proof via the Dirichlet principle, by now more conceptual proofs via circle packing or via Ricci flow.)įinally, if you believe that every surface can be triangulated, then the classification of surfaces is proven by a not so difficult combinatorial inductive argument. 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). (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.) 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 $P^1C$. Thus the proof of classification boils down to show that a (closed, orientable) surface with $\chi(S)=2$ must be $S\cong S^2$. 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 S‘ with $\chi(S^\prime)=\chi(S) 2$. By induction, each component of $S^\prime$ is $S^2$ with finitely many handles added, so the same is true for $S$. The Morse function extends (constant on the disks). Consider the (possibly disconnected) surface $S^\prime$ obtained by cutting off the pair of pants and gluing in three disks. A neighborhood of a saddle point is a pair of pants, with the Morse function constant on boundary curves. For the inductive step, consider a Morse function with k saddle points. Assume by induction that a surface with k-1 saddle points is $S^2$ with finitely many handles added. If it has no saddle points, then (using the gradient flow) $S\cong S^2$. Take a Morse function on the (closed, orientable) surface S. I guess the most conceptual proof is the one using Morse theory:
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