# A differential form generating the zero current is zero

This is probably a silly question but I wasn't able to find a quick answer. let $\omega$ be a smooth $(p,q)$-form on a complex manifold $X$ of dimension $d$. It generates a current of integration $T_\omega$ as follows. For any compactly supported and smooth $(d-p,n-q)$-form $\eta$ we define $$T_\omega:\eta\mapsto\int_X\omega\wedge\eta$$

Now suppose that $T_\omega$ is constantly $0$, is it true that $\omega=0$?

Many thanks

Of course, if $\omega$ was not zero, then it would have non-empty support, with non-empty interior (because it is smooth). Say on some open subset A of support of $\omega$, inside coordinate patch $(z_1,\ldots,z_d)$ we have $$\omega = f(z) dz_1\wedge\ldots \wedge dz_p \wedge d\bar{z}_1\ldots \wedge d\bar{z}_q + \ldots$$
Suppose that $f(z) > 0$ on $A$, then take
$$\eta = g(z) dz_{p+1}\wedge\ldots \wedge dz_d \wedge d\bar{z}_{q+1}\ldots \wedge d\bar{z}_d$$
with $\text{supp}(g) \subseteq A$ and $g(z) > 0$ in the interior of support. The integral is then clearly non zero.
EDIT: As Ted Shifrin points out this only works for real coefficients, in general take $g(z) = \overline{f(z)}$ as he suggests.
• Of course, in general, $f(z)$ is complex-valued, so it's probably best to choose $g(z)=\overline{f(z)}$. Mar 30, 2017 at 17:12