So from the $\overline{\partial}$-Poincare lemma, there is a short exact sequence of sheaves on $X = \mathbb{P}^1$


$$0 \to \Omega \to A^{1,0} \to Z^{1,1} \to 0$$


where $\Omega$ is the sheaf of holomophic 1-forms, $A^{1,0}$ is the sheaf of (1,0)-forms, $Z^{1,1}$ is sheaf of closed (1,1)-forms and the surjection is apply $\overline{\partial}$. The higher cohomology of $A^{1,0}$ vanishes which gives an isomorphism: 


$$H^1(X,\Omega) \cong \frac{\Gamma(Z^{1,1})}{\overline{\partial}\Gamma(A^{1,0})} =: H\_{\overline{\partial}}^{1,1}(X)$$


Both groups are isomorphic to $\mathbb{C}$. The first group can be described explicitly via Cech cohomology and the standard cover of $X$; in suitable coordinates a generator is $\dfrac{dz}{z}$. 


What is the image of $\dfrac{dz}{z}$ in $H\_{\overline{\partial}}^{1,1}(X)$? More specifically, can you write down a global closed (1,1) form that represents the class of $\dfrac{dz}{z}$?


I know that being able to avoid descriptions such as these is one of the great virtues of Cech Cohomology, but I guess some stubborn part of me would like to see the Dolbeault description. 



Dolbeault Cohomology of $\mathbb{P}^1$
I wrote a [blog post](http://sbseminar.wordpress.com/2010/01/12/residues-and-integrals/) about almost exactly this question. I'll give a summary here:


Since $H^{1,1}(X)$ is one dimensional, I could answer your question by giving anythng with the correct integral. However, I'll try to give you the kind of cocycle which actually comes out of the proof of Dolbeaut-Cech equality.


Your cocycle isn't $dz/z$ but, rather, $dz/z$ with a specific choice of open cover of $X$. Lets say your choice is $U\_1 \cup U\_2$, where $U\_1 = \{ z : z \neq \infty \}$ and $U\_2 = \{ z : z \neq 0 \}$. Refine your cover to $V\_1 \cup V\_2$, where $V\_1 = \{ z : |z| > r \}$ and $V\_2 = \{ z : |z| < r^{-1} \}$ for some $r < 1$. 


Let $\theta\_1$ and $\theta\_2$ be $1$ forms on $V\_1$ and $V\_2$ such that $\theta\_1|\_{V\_1} - \theta\_2|\_{V\_2} = dz/z$. Then $\overline{\partial} \theta\_1$ and $\overline{\partial} \theta\_2$ have equal restrictions to $V\_1 \cap V\_2$. The $(1,1)$-form you are looking for is their common value, which I'll call $\omega$.




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Let's first do a fake solution. **A real solution would look like a $C^{\infty}$ smearing out of this one.** 


We'll work in the degenerate case $r=1$, so we are only gluing along a circle, not an annulus. We'll take $\theta\_1 = (1/2) \ \overline{z}\ dz$ and $\theta\_2 = -(1/2) \ dz / (\overline{z} z^2)$. Notice that both $\theta\_1$ and $\theta\_2$ restrict to $dz/z$ on the unit circle, but are constructed to extend smoothly to the appropriate discs. 


So $\overline{\partial} \theta\_1 = (1/2) d \overline{z} d z$ and $\overline{\partial} \theta\_2 = (1/2) dz d \overline{z} / (\overline{z}^2 z^2)$. Our $\omega$ is formed by gluing these two differential forms together.




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A **genuine smooth solution** would be like this, but would interpolate smoothly between these two. If you push forward in a brute force manner, you'll get something with [bump functions](http://en.wikipedia.org/wiki/Bump_function) in it. 


If you are more clever, you may discover the solution
$$\theta\_1 = \frac{dz}{z} \left( 1- \frac{1}{1+z \overline{z}} \right)$$
and
$$\theta\_2 = - \frac{dz}{z} \left( \frac{1}{1+z \overline{z}} \right).$$


You should check that $\theta\_1|\_{V\_1} - \theta\_2|\_{V\_2} = dz/z$ and that $\theta\_i$ is smooth and well-defined on $U\_i$. 


Then 
$$\overline{\partial} \theta\_1 = \overline{\partial} \theta\_2 = \frac{dz \ d\overline{z}} {(1+z \overline{z})^2}.$$


This is, as Scott guessed, the **[Fubini-Study](http://en.wikipedia.org/wiki/Fubini-Study_metric)** form.


