Why does every noncompact orientable surface have a complex structure?

Question

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.

Answer 1

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.