I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):
Remark 2.1: The Cohn-Vossen inequality says that $\int_S K \leq 2\pi \chi(S)$, where [$K$ is the Gauss curvature and] $\chi(S)$ is the Euler characteristic of $S$. Combining this with (2.18) we see immediately that $S$ is homeomorphic to $\mathbb{R}^2$
Here, (2.18) refers to the inequality $\int_S K > 0$ derived just before.
How can I see that the surface $S$ is homeomorphic to $\mathbb{R}^2$? The immediate implication of (2.18) obviously is that $\chi(S) > 0$, so $\chi(S) \geq 1$. Apart from that, I know that $S$ is non-compact but I'm unsure about other properties$^\dagger$ of $S$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $b_i$ vanish for $i \geq 1$?
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$^\dagger$ For instance, I suspect that $S$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $N$ but since I only have a rough idea of how $S$ is constructed as a limit of surfaces $S_\sigma$ (namely by representing the minimal surfaces $S_\sigma$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $S$ that could help prove the claim.
Best Answer
A noncompact surface has no homology groups above dimension 1. So $\chi(S) = 1 - \dim H_1(S)$. If $\chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(\Sigma_{g,n}) = \Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $\Bbb R^2$, as desired.