A question on differential topology Let $\mathbb{C}P(1)$ denote the complex projective line.  I am attempting to show that there does not exist a nonzero holomorphic differential $1$-form on $\mathbb{C}P(1)$.  My intuition is as follows:  We can view $\mathbb{C}P(1)$ as the Riemann sphere, which is homeomorphic to the usual $2$-sphere $S^2$.  If it were true that every differential $1$-form on $S^2$ gives rise to a tangent vector field on $S^2$, then we would be done, for it is known that no such tangent vector field exists on $S^2$, and thus the existence of such a $1$-form would imply a contradiction.  My question, then is threefold:  firstly, does every differential $1$-form on $S^2$ give rise to a tangent vector field on $S^2$? Secondly, if so, how can I prove it?  Thirdly, if not, where ought I to go from here?  Any hints would be appreciated.
 A: Let $\omega(z)=f(z)dz$, with $f(z)$ holomorphic, be a holomorphic differential form on $\mathbb C$ .
 In the  neighbourhood $\mathbb P^1(\mathbb C)\setminus \{0\} $ of $\infty$ we choose the holomorphic cooordinate $w$ defined by  $w=\frac 1z$ (and $w(\infty)=0$) .
Our holomorphic form then satisfies $\omega (w)= - \frac {1}{w^2}f(\frac 1w) dw$  for $w\neq 0$ (recall that $dz=- \frac {1}{w^2} dw$).
But then the only possibility for $\omega$ to extend holomorphically to $w=0$  (i.e. to $\infty \in \mathbb P^1(\mathbb C)$) is to suppose that $f(z)=0$ .
So  $\omega=0$ is indeed the only holomorphic differential form on $\mathbb P^1(\mathbb C)$.
A: If $\omega$ is a non-zero differential form, holomorphic on the Riemann sphere, then by fixing a point $z_0$ we get that the path integral
$$
F(z)=\int_{z_0\to z}\omega
$$
gives a holomorphic function on the Riemann sphere. It being compact the image of $F$ is bounded. Hence by Liouville $F$ is constant. Hence $\omega=0$.
A: A way to see this is by using some results from complex geometry, which may be of use to you if you are familiar with them (otherwise ignore this):
We know that $\mathbb{C}P(1)$ is compact and of Kahler type (with the usual Fubini-Study metric), and hence the Hodge Decomposition theorem yields $$H^1(\mathbb{C}P(1),\mathbb{C})=H_{\overline{\partial}}^{1,0}(\mathbb{C}P(1))\oplus H_{\overline{\partial}}^{0,1}(\mathbb{C}P(1)),$$ where the left-hand side is De-Rham cohomology with complex coefficients, and the summands of the right-hand side are the Dolbeaut cohomology groups. But we also know that $H^1(\mathbb{C}P(1),\mathbb{C})=0,$ since $$H^1(\mathbb{C}P(1),\mathbb{C})=H^1(\mathbb{C}P(1),\mathbb{R})\otimes \mathbb{C},$$ where the right-hand side cohomology is De-Rham with real coefficients, and this group is zero (or you can also argue similarly with $\mathbb{R}$ replaced with $\mathbb{Z}$, and you get singular cohomology instead, also zero). Hence, by the above, you get $$H_{\overline{\partial}}^{1,0}(\mathbb{C}P(1))=0$$ But Dolbeaut's theorem says that you have an isomorphism $$H_{\overline{\partial}}^{1,0}(\mathbb{C}P(1))\simeq H^0(\mathbb{C}P(1),\Omega^1),$$ where $\Omega^1$ is the sheaf of holomorphic 1-forms on $\mathbb{C}P^1$, and hence you conclude that the right-hand side group is zero, which is exactly the global holomorphic 1-forms on this space. 
