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There is a high-powered proof of the fact that orientable noncompact surfaces have free fundamental group here that invokes the ability to put a complex structure on any such surface. But why should your favorite noncompact orientable surface actually have a complex structure? I seem to remember a theorem that noncompact orientable surfaces can be embedded in the plane, which would certainly do the job, but I can't find a reference.

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    Every oriented connected noncompact surface admits an immersion (not an embedding) into the plane. This is a special case of the Hirsh - Smale theory but can be also easily proven directly. – Moishe Kohan Aug 24 '14 at 16:30

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Given a Riemannian metric, giving a compatible almost complex structure is equivalent to giving a reduction of the holonomy group from $O(n)$ to $U(n)$. An orientable surface must have holonomy group in $SO(2)$, which is just $U(1)$, so the parallel transport automatically preserves a compatible almost complex structure chosen at some point. And in two dimensions, the Nijenhuis tensor automatically vanishes so an almost complex structure is complex. None of this called on compactness.

The countable connect sum of tori is an example of a noncompact orientable surface that doesn't embed in the plane, unless I'm being silly.

Kevin Carlson
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    Good to know. Could you also recommend an intro to complex structure? – Troy Woo Aug 24 '14 at 07:46
  • I have a rather overwrought proof that the punctured torus doesn't embed into the plane, but I couldn't manage to prove yours doesn't. –  Aug 24 '14 at 08:52
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    The punctured torus doesn't embed because that would extend to an embedding of the torus, since every circle in the plane bounds a disk. Similarly an embedding of the surface of countable genus would restrict to an embedding of the punctured torus. @Troy Woo Huybrechts has the best book on complex manifolds. – Kevin Carlson Aug 24 '14 at 16:19
  • Of course you mean $O(2n)$ here. –  Dec 22 '14 at 17:14
  • I think you should edit $O(n)$ to be $O(2n)$. –  Mar 31 '15 at 05:52
  • Note that $O(n)$ and $O(2n)$ are two different groups. –  Dec 16 '15 at 14:12
  • @MikeMiller I don't want to rush into any rash modifications... – Kevin Carlson Dec 16 '15 at 17:20
  • You're right, my mistake. What I said only holds for $n>0$. –  Dec 16 '15 at 17:20
  • Your answer is incorrect for $n \neq 0$. –  Apr 09 '16 at 01:10
  • @Mike Miller let us continue this conversation in chat. – Kevin Carlson Apr 09 '16 at 18:54
  • $O(2n)$ has a split surjection onto $\Bbb Z/2$, while $O(n)$ doesn't, at least for $n$ odd and positive (see previous comments). So they are provably not the same. You might edit your answer appropriately. –  Jun 06 '16 at 01:22