# 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.

-
Wild guess: Fubini-Study metric. –  S. Carnahan Mar 29 '10 at 7:50

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.

-
There are some minor errors here, although they don't effect the main point. I'll edit when I get a chance later. –  David Speyer Mar 29 '10 at 13:34
OK, I think I got them all. –  David Speyer Mar 29 '10 at 15:12
Awesome thanks! –  solbap Mar 29 '10 at 15:44

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.

-
Can you explain what you are doing here? This looks like nonsense, but your end result does seem to be morally right in some sense. (I assume that $\delta$ is the $\delta$ measure at $0$.) Here are some of the things which confuse me: –  David Speyer Apr 3 '10 at 13:03
(1) your construction should remember not just the form $dz/z$, but the choice of Cech cover. If we switched the order of the open sets $\mathbb{P}^1 \setminus \{ 0 \}$ and $\mathbb{P}^1 \setminus \{ \infty \}$, we should negate the cohomology class. –  David Speyer Apr 3 '10 at 13:03
(2) I guess it really is true that $d (dz/z) = \delta_0$ in the sense that $\int_{\partial A} dz/z = \int_A \delta_0$ for any disc $A$. So this is going to be a horrible nitpick but: In what sense am I allowed to do DeRham cohomology with measures? The theorem/definition I know is that $H^2$ is (closed smooth $2$-forms)/d(smooth $1$-forms). Is there an analogous result where "smooth forms" is replaced by measures? –  David Speyer Apr 3 '10 at 13:16
(3) Why didn't you pick up a second $\delta$ function at $\infty$, where $dz/z$ also has a pole? Presumably, the answer to this is related to the answer to (1). –  David Speyer Apr 3 '10 at 13:17
Hi David, Griffiths and Harris chapter 3 has a section on how to define cohomology using distributions. I would start there. –  Clay Cordova Apr 13 '10 at 18:10

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.

-
This answer, on the other hand, is wrong. Your answer should be a nontrivial cohomology class. –  David Speyer Apr 3 '10 at 13:17