If we have a non-constant holomorphic map $ f: \mathbb C ^ p \to X $, where $ X $ is a complex manifold. Let $ \omega $ be a metric on $X$, so $ \omega $ is a positive definite $ (1,1) $-form. Is $ f ^ * (\omega ^ p) $ a measure over $ \mathbb C ^ p $?
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A construction impossible for $p > 1$ giving only zero for $\dim_\mathbb{C} X < p$ [see comments], for whichever reasonable meaning of “$\omega^p$” and pullback.
Denote by $z\in\mathbb{C}$ and $w\in\mathbb{C}^{p-1}$ coordinates on $\mathbb{C}^p$.
Let $f := z$ with $X = \mathbb{C}$ (a standard projection).
Consider $(p\!-\!1)$-dimensional (over $\mathbb{C}$) affine transformations $A$ acting on $\mathbb{C}^p$ as
$$(z,w)\quad\mapsto\quad (z,Aw).$$
Does the action preserve $f$? Obviously. May the resulting measure be changed? Impossibly. Which measure can be invariant under all affine transformations of $w$? Only the zero measure.
$\omega$ may be $i(dz\wedge d\bar z)$ or whatever you like – the argument stands for any non-zero form. Ī̲ am puzzled how can this be a research-level math question.
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$\begingroup$ The construction is not impossible; in fact it always well defined and always produces a measure, perhaps the zero measure. Your example produces the zero measure, since $\omega^2=0$ on $\mathbb{C}$. But on the other hand, other examples produce nonzero measures: if $f$ is any holomorphic map of $\mathbb{C}^p$, somewhere of full rank, and $X=\mathbb{C}^p$ and we take $\omega$ any Kaehler form, we get a measure $f^*\omega^p$ which is positive on any open set. $\endgroup$ Commented Nov 22, 2020 at 16:24
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$\begingroup$ But I agree that the question is too elementary. $\endgroup$ Commented Nov 22, 2020 at 16:24
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$\begingroup$ Oops, Ī̲ missed that $X$ is not (necessarily) one-dimensional. $\endgroup$ Commented Nov 22, 2020 at 16:27
$\mathbb C$rather than $\ C$$\ C$(a $C$ with a forced space beforehand) as you originally had.) $\endgroup$