I've heard this result bandied about many times, and I know that it follows from e.g. the theory of divisors, but I'd like to see some simpler, straightforward ways of proving this fact.
What is the easiest way to see that there are no nonconstant holomorphic forms on the Riemann Sphere?
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complex-analysis
1 Answers
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Suppose there were a nonzero holomorphic 1-form $\omega$ on $S^2$. Then we have two charts $U_1, U_2$, each isomorphic to $\mathbb{C}$, on which coordinates are given by $z$ and z' = 1/z. In each representation, we have $\omega = f(z) dz$ and \omega = g(z') dz'. On the overlap, we must have $f(z) dz = g(1/z) (-1/z^2) dz$ by the transition formulas. That is, $f(z) = g(1/z) (-1/z^2)$ on $\mathbb{C}-\{0\}$. If you write out the Laurent series expansion (note that both $f,g$ are entire functions!) this is absurd if $g$ is nonzero. This also works in the algebraic category (and gives an easy proof that the projective line has genus zero).
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0Thanks Akhil. I also encountered the following argument which I quite like: just notice that dz has two poles (an order 2 pole actually), and that any two meromorphic forms have the same number of poles, since their ratio is a meromorphic function. So any holomorphic form would have to have two more poles than zeros...whoops! – 2011-03-20