How do we get from $\Delta f = \rho$ to $\partial\bar{\partial}f = \text{Const.} \rho\,dz\wedge d\bar{z}$?

Question

I asked this question about the Kähler potential on MathOverflow. Donu Arapura left a comment saying

Classically, a potential satisfies $\Delta f = \rho$. In the plane, this can be rewritten as $\partial\bar{\partial}f = \text{Const.}\rho \, dz\wedge d\bar{z}$.

When I then asked him to explain how he went from $\Delta f = \rho$ to $\partial\bar{\partial} f = \text{Const.}\rho \, dz\wedge d\bar{z}$, he stated

This is pretty standard. $\bar{\partial}f = 1/2(f_x − if_y)dz$ etc. So $\partial\bar{\partial}f$ should work out to $\Delta f$ times a nonzero constant.

Which I still do not get, so I thought to ask about this part on this forum. Can someone explain this?

Answer 1

If $\Omega \subseteq \mathbb{C}$ is open, and $f : \Omega \to \mathbb{C}$ is suitably differentiable, we have two partial derivatives, $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial\bar{z}}$ which are defined as

\begin{align*} \frac{\partial f}{\partial z} &= \frac{1}{2}\left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right)\\ \frac{\partial f}{\partial \bar{z}} &= \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right). \end{align*}

Then we have the operators $\partial$ and $\bar{\partial}$ which act on functions as follows:

\begin{align*} \partial f &= \frac{\partial f}{\partial z}\, dz\\ \bar{\partial} f &= \frac{\partial f}{\partial \bar{z}}\, d\bar{z}. \end{align*}

These operators act on one-forms as well:

\begin{align*} \partial(g\, dz + h\, d\bar{z}) &= \frac{\partial g}{\partial z}\, dz\wedge dz + \frac{\partial h}{\partial z}\, dz\wedge d\bar{z}\hspace{1.5mm} = \frac{\partial h}{\partial z}\, dz\wedge d\bar{z}\\ \bar{\partial}(g\, dz + h\, d\bar{z}) &= \frac{\partial g}{\partial \bar{z}}\, d\bar{z}\wedge dz + \frac{\partial h}{\partial \bar{z}}\, d\bar{z}\wedge d\bar{z} = -\frac{\partial g}{\partial \bar{z}}\, dz\wedge d\bar{z}. \end{align*}

Now we can compute $\partial\bar{\partial}f$:

$$\partial\bar{\partial}f = \partial\left(\frac{\partial f}{\partial \bar{z}}\, d\bar{z}\right) = \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial\bar{z}}\right)\, dz\wedge d\bar{z}.$$

Note that

\begin{align*} \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial\bar{z}}\right) &= \frac{\partial}{\partial z}\left(\frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)\right)\\ &= \frac{1}{2}\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}\right) + \frac{i}{2}\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial y}\right)\\ &= \frac{1}{2}\left[\frac{1}{2}\left(\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) - i\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\right)\right] + \frac{i}{2}\left[\frac{1}{2}\left(\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) - i\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)\right)\right]\\ &= \frac{1}{4}\frac{\partial^2 f}{\partial x^2} - \frac{i}{4}\frac{\partial^2 f}{\partial y\partial x} + \frac{i}{4}\frac{\partial^2 f}{\partial x\partial y} + \frac{1}{4}\frac{\partial^2 f}{\partial y^2}. \end{align*}

It is now important to discuss the differentiability of $f$. So far, all we needed in the above calculations was the existence of the second order partial derivatives of $f$. However, for the result you're looking for, we need a stronger hypothesis on $f$. The usual one is that $f$ is $\mathcal{C}^2$ (not only do the second order partial derivatives exist, they are continuous). Under this assumption we have

$$\frac{\partial^2 f}{\partial y\partial x} =\frac{\partial^2 f}{\partial x\partial y},$$

so

\begin{align*} \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial\bar{z}}\right) &= \frac{1}{4}\frac{\partial^2 f}{\partial x^2} - \frac{i}{4}\frac{\partial^2 f}{\partial y\partial x} + \frac{i}{4}\frac{\partial^2 f}{\partial x\partial y} + \frac{1}{4}\frac{\partial^2 f}{\partial y^2}\\ &= \frac{1}{4}\frac{\partial^2 f}{\partial x^2} - \frac{i}{4}\frac{\partial^2 f}{\partial x\partial y} + \frac{i}{4}\frac{\partial^2 f}{\partial x\partial y} + \frac{1}{4}\frac{\partial^2 f}{\partial y^2}\\ &= \frac{1}{4}\frac{\partial^2 f}{\partial x^2} + \frac{1}{4}\frac{\partial^2 f}{\partial y^2}\\ &= \frac{1}{4}\Delta f. \end{align*}

Therefore, we have the following result.

If $f$ is $\mathcal{C}^2$, $\partial\bar{\partial}f = \frac{1}{4}\Delta f\, dz\wedge d\bar{z}$.

In particular, if $f$ is $\mathcal{C}^2$, then $\Delta f = \rho$ if and only if $\partial\bar{\partial}f = \frac{1}{4}\rho\, dz\wedge d\bar{z}$.