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I often see $\partial \bar{\partial}$ arising in complex geometry, what is it (explain in relatively simpler language)?

And sometimes $\partial \bar{\partial}T$ is non-zero (for example, the "current" in a complex dynamical system). I thought that $\partial \bar{\partial}$ is by definition always zero?

Yan King Yin
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    If $f = z \overline z$ then $\partial \overline \partial (z \overline z) = 1 $ so for general $C^{\infty}$ function there is no reason that this operator is zero. –  Dec 10 '15 at 08:30
  • I thought it is analogous to the $dd = 0$ in (de Rham) homology? – Yan King Yin Dec 10 '15 at 11:32

3 Answers3

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$\partial \bar\partial$ is essentially the Laplace operator, but written more conventiently with complex derivatives. To keep it simple, let's assume that we are in the setting of a domain in the complex plane. Then

$$ \partial f = \frac{\partial f}{\partial z}\,dz $$ and $$ \bar\partial f = \frac{\partial f}{\partial \bar z}\,d\bar z $$ Hence, $$ \partial\bar\partial f = \partial\left( \frac{\partial f}{\partial \bar z} \, d\bar z \right) = \frac{\partial^2 f}{\partial z\partial \bar z} \,dz \wedge d\bar z $$ (since the $d\bar z \wedge d\bar z$-term vanishes). Furthermore, since $$ \frac{\partial}{\partial z} = \frac12 \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) \quad\text{and}\quad \frac{\partial}{\partial \bar z} = \frac12 \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) $$ it follows that $$ \frac{\partial^2}{\partial z \partial \bar z} = \frac14 \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right). $$

In higher dimension, and acting on currents, the computations are similar but a little more cumbersome.

mrf
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For simplicity assume the dimension is 1. $\partial \overline \partial$ is simply the operator $\frac{1}{4}\Delta$, where $\Delta$ is the usual Laplace operator $\Delta = \frac{\partial^2}{\partial^2x} + \frac{\partial^2}{\partial^2y}$. It's zero where restricted to holomorphic or anti-holomorphic function but for general $C^{\infty}$ function there is no reason that $\partial \overline \partial f = 0$.

In fact there is a beautiful theorem (the zero mean theorem, see the book of Rick Miranda on Riemann surface, page 318) : A $C^{\infty}$ $2$ form $\eta$ on an algebraic curve $X$ can be written as $\partial \overline \partial f$ for some smooth function $f$ if and only if $$ \int \int_X \eta =0 $$

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When $f$ is a smooth function, then $$\partial f = \sum \frac{\partial f}{\partial z^i} dz^i, \ \ \overline\partial f = \sum \frac{\partial f}{\partial \bar z^i}d\bar z^i. $$

In general if $\omega $ is a $(p, q)$ form, locally written as $$\omega = \sum_{|I|=p,|J| = q} f_{IJ} \ dz^I\wedge d\bar z^J,$$ then $$\partial \omega = \sum_{|I|=p,|J| = q} \partial f_{IJ} \wedge dz^I\wedge d\bar z^J$$ and similar for $\bar \partial \omega$. Note that $d = \partial +\bar\partial$ (direct checking), so $$0 = d^2 = (\partial + \bar\partial)(\partial +\bar \partial) = \partial ^2 + (\partial \bar\partial + \bar\partial \partial) + \bar\partial^2,$$

this implies that $$\partial^2 = \partial \bar\partial + \bar\partial \partial = \bar\partial^2 = 0.$$ It is not true that $\partial \bar\partial =0$ though, as shown in the other answers.

This is the definition of $\partial, \bar\partial$ when acting on forms. But if you want the definition on current, it's just the dual: Given a current $T$, then it is just a linear functional on the space of $k$-forms. Define

$$\partial T(\omega) := -T(\partial \omega).$$

You might find more information in the book "Principal of Algebraic Geometry" by Griffiths and Harris, Chapter 3.