How do you compute the pull-back of a complex differential (1,1)-form given its potential?

Question

Let $F: X \to Y$ be a holomorphic map. Let $\omega$ be a complex differential $(1,1)$-form on $Y$, $\omega=\partial \bar \partial f$, where $f$ is a pluri-subharmonic function. How would one explicitly compute the pull-back of this form under a holomorphic map $F: X \to Y$... i.e. what is $F^{*}\omega$?

Answer 1

Suppose the charts on $Y$ follow coordinates $w \in \mathbb{C}^n$. Then we may write $\omega$ locally in a chart as $$ \omega(w) = \frac{\partial^2 f}{\partial w \partial \bar w }(w) dw \wedge d\bar w $$ Now if we have a nice holo map $F: X \to Y$ (i.e. structure preserving), the pull back is just as you've written, $F^* \omega = \omega \circ F$. This amounts to swapping the coordinates in the charts. Suppose that the charts on $X$ follow coordinates $z \in \mathbb{C}^n$, then in the compatible charts we'll have $F(z) = \omega$ So $$F^* \omega(z) = \omega \circ F(z) =\frac{\partial^2 f}{\partial w \partial \bar w }(F(z)) d F(z) \wedge d \overline {F(z)} $$ You may rewrite the partial derivatives in terms of $z$ if you wish using chain rule.