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  1. In terms of topology (homology, homotopy, etc), the most general orientable 2d surfaces are the genus $g$ Riemann surfaces.

  2. Is it true that an unorientable surfaces of genus $g$ is the most general unorientable 2d surfaces that we can encounter? (How to prove it?)

  3. How to construct such unorientable surfaces of genus $g$? I suppose $g=1$ is $RP^2$ and $g=2$ is Klein bottle. Can we have $g=0$ and all the other integer $g$?

annie marie cœur
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    What definition of genus are you using? – Michael Albanese Oct 05 '20 at 21:26
  • nontrivial cycle read from homology group $H_1(M,\mathbb{Z}/2)$ unoriented or $H_1(M,\mathbb{Z} )$ oriented? – annie marie cœur Oct 05 '20 at 21:58
  • You're off by a factor of two in the oriented case ($T^2$ has genus one, but $\operatorname{rank}H_1(T^2; \mathbb{Z}) = 2$). In the orientable case you should take the genus to be $\frac{1}{2}\operatorname{rank}H_1(M; \mathbb{Z})$. – Michael Albanese Oct 05 '20 at 22:14
  • "In the orientable case you should take the genus to be 1/2 rank1(;ℤ)." In the nonorientable case, do we take the genus to be rank1(;ℤ)? – annie marie cœur Nov 22 '20 at 23:52
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    No. In the non-orientable case, we take the genus to be $\operatorname{rank}H_1(M; \mathbb{Z}_2) = \operatorname{dim}H_1(M; \mathbb{Z}_2)$ so that the connected sum of $k$ copies of $\mathbb{RP}^2$ has genus $k$. – Michael Albanese Nov 26 '20 at 16:16
  • Thanks -- I think I got it. But can you say a few words why we use the 1/2 rank $_1(;ℤ)$ for orientable case, and why we use the rank $_1(;ℤ_2)$ for nonorientable case? – annie marie cœur Nov 26 '20 at 16:24
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    It depends on what your initial motivation for defining the genus is. A practical reason is that the basic building block for orientable surfaces is $T^2$ which has $\operatorname{rank}H_1(M; \mathbb{Z}) = 2$ and the basic building block for non-orientable surfaces is $\mathbb{RP}^2$ which has $\operatorname{rank}H_1(M; \mathbb{Z}_2) = 1$. With these conventions, the genus tells us how many copies of $T^2$ or $\mathbb{RP}^2$ are needed to construct the given surface. – Michael Albanese Nov 26 '20 at 16:28

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Classification of surfaces: every connected, closed surface (i.e. two-dimensional manifold) is diffeomorphic to either a sphere, a connected sum of $k > 0$ tori, or a connected sum of $\ell > 0$ real projective planes.

The first two possibilities corresponds to orientable manifolds, while the last possibility corresponds to non-orientable manifolds. In particular, a connected, closed, non-orientable surface is diffeomorphic to the connected sum of $\ell$ copies of $\mathbb{RP}^2$ for some $\ell > 0$.

For connected $R$-orientable surfaces $M_1, M_2$, we have $$H_1(M_1\# M_2; R) \cong H_1(M_1; R)\oplus H_1(M_2; R)$$ by Mayer-Vietoris. So if $M_1, M_2$ are connected, closed surfaces which are either both orientable or both non-orientable, then $g(M_1\# M_2) = g(M_1) + g(M_2)$ where $g$ denotes the genus function.

Note that $g(T^2) = 1$ and $g(\mathbb{RP}^2) = 1$, so a the connected sum of $k$ tori has genus $k$, and a connected sum of $\ell$ real projective planes has genus $\ell$. In particular, there are no non-orientable surfaces of genus zero (the only surface of genus zero is $S^2$, which is orientable).

Michael Albanese
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