Dolbeault Cohomology of $\mathbb{P}^1$ 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.  
 A: I wrote a blog post 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 $\omega$.

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.

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 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 form.
A: The 2-form $\omega$ which is the derivative of $dz/z$ is
$\displaystyle\omega = \bar{\partial}\ \frac{dz}{z}=\frac{\partial}{\partial\bar{z}}\left(\frac{1}{z}\right)d\bar{z}\wedge dz = \pi\delta\ d\bar{z}\wedge dz = 2\pi i\delta\ dx\wedge dy$.
Hence $\omega$ belongs to the class of 2-forms which integrate to $2\pi i$. Another representative of this class is
$\displaystyle\frac{d\bar{z}\wedge dz}{(1+|z|^2)^2}$
corresponding to smearing out the point mass $2\pi i\delta$ evenly over the sphere.
A: The answer is zero. The image of the given 1-form belongs to the class of exact 2-forms and the simplest such 2-form is the zero 2-form.
To actually calculate the derivative of the given 1-form, first lift it to homogeneous coordinates $[z:w]$. The 1-form then reads
$$\theta=\frac{w}{z}\ d\left(\frac{z}{w}\right)=\frac{dz}{z}-\frac{dw}{w}.$$
Now choose a new affine coordinate $u=z+1=w-1$ and the 1-form in this coordinate is
$$\theta=\frac{du}{u-1}-\frac{du}{u+1},$$
which has the derivative
$$\bar{\partial}\theta=\pi\delta_1\ d\bar{u}\wedge du\ -\ \pi\delta_{-1}\ d\bar{u}\wedge du$$
and this integrates to zero like any other exact 2-form.
