# What is the easiest way to see that there are no nonconstant holomorphic forms on the Riemann Sphere?

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.

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## 1 Answer

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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Thanks 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! – Tony Mar 20 '11 at 4:18
@Akhil: which transition formula do you use here? – noctusraid Apr 5 at 15:19