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In a question, I am asked to compute the homology of $\Sigma_g\#S_n$ (via Mayer-Vietoris), and deduce that this space is in fact $S_{n+2g}$. I did so by assuming the classification of compact surfaces, but now I am wondering whether this is all circular: how is the classification of compact (or even triangulable) surfaces proved and does it not use the result $\Sigma_g\# S_n=S_{n+2g}$ in its proof? I saw here that $\Sigma_1\#S_1=S_3$ can be showed directly. Is the general case done in a similar fashion? Or is there another argument?

Edit: $\Sigma_g$ is the connected sum of g tori, while $S_n$ is the connected sum of n $\mathbb{RP}^2$'s.

Guillaume Berlat
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  • What is the meaning of your notation $\Sigma_g$ and $S_n$? – Lee Mosher Mar 20 '24 at 19:53
  • @LeeMosher Sorry for the confusion I edited the question – Guillaume Berlat Mar 20 '24 at 19:56
  • The standard proofs of classification of surfaces do not use that general fact, only the special case of the connected sum of torus and projective plane. – Moishe Kohan Mar 20 '24 at 20:01
  • @MoisheKohan I see, using induction and the special case proves the general case (or not even needed for the classification if some other argument is used). Then I guess proving the general case using homology as in the question is silly as it must be proven by induction from the special case anyway. – Guillaume Berlat Mar 20 '24 at 20:04

1 Answers1

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When classifying any kind of mathematical structure (up to equivalence), there are two steps:

  • Make a list which includes representatives of every equivalence class (and try to winnow your list down by eliminating duplicates)
  • Use invariants to prove that no two members of your list are equivalent.

In the case of surfaces, a direct argument that $\Sigma_1 \#\, S_1 = S_3$ can be used to eliminate $\Sigma_1 \#\, S_1$ from your list. Using similar such arguments you can winnow your list down to $$\Sigma_0, S_1, \Sigma_1, S_2, \Sigma_2, S_3, \Sigma_3,... $$ And now you have to prove that any two members of this list are not homoemorphic. That's where the homology calculations come in.

Lee Mosher
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