This is the $\partial \bar \partial$-lemma as stated in [1]:
Proposition 6.17 Let $X$ be a Kähler manifold, and let $\omega$ be a form which is both $\partial$ and $\bar \partial$-closed. Then if $\omega$ is $d$ or $\partial$ or $\bar \partial$-exact, there exists a form $\chi$ such that $\omega = \partial \bar \partial \chi$.
Voisin only proves the case where $\omega$ is $\bar \partial$-exact, and I'm stuck with the case where $\omega$ is $d$-exact. I tried to follow the proof of Voisin, but there is a term that won't disappear. Here is the idea:
Since $\omega$ is $d$-exact, we can write $\omega = d \beta$ for a form $\beta$. By Hodge theory, there is a harmonic form $\alpha$ and a form $\gamma$ such that
$$\beta = \alpha + \Delta_d \gamma,$$
where $\Delta_d$ is the $d$-Laplacian. Since harmonic forms are $d$-closed, this implies
$$ \omega = d \Delta_d \gamma,$$
and since $X$ is Kähler we also have $\Delta_d = 2\Delta_{\bar\partial} = 2 (\bar \partial \bar \partial^* + \bar \partial^* \bar \partial)$. Hence
\begin{align*}
\omega & = 2 (\partial + \bar \partial)(\bar \partial \bar \partial^* + \bar \partial^* \bar \partial)\gamma \\
& = 2 \partial \bar \partial \bar \partial^* \gamma + 2\partial \bar \partial^* \bar \partial \gamma + 2 \bar \partial \bar \partial^* \bar \partial\gamma.
\end{align*}
Since $\partial \bar \partial = – \bar \partial \partial$, we see that everything except $2\partial \bar \partial^* \bar \partial \gamma$ is $\bar \partial$-closed, hence $2\partial \bar \partial^* \bar \partial \gamma$ is also $\bar \partial$-closed. But as $\partial \bar \partial^* = – \bar \partial^* \partial$ (which is true since $X$ is Kähler), we see that $2\partial \bar \partial^* \bar \partial \gamma$ is in the image of $\bar \partial^*$, hence it vanishes (consider the scalar product with itself, and use the adjointness of $\bar \partial$ and $\bar \partial^*$).
Thus
$$ w = 2 \partial \bar \partial \bar \partial^* \gamma + 2 \bar \partial \bar \partial^* \bar \partial \gamma.$$
In the same way as above I can deduce that the second term $2\bar \partial \bar \partial^* \bar \partial \gamma$ is $\partial$-closed, but I don't see why it should be zero.
Am I missing something? Or do I have to do something different? I could also use $\Delta_d \gamma = 2 \Delta_{\partial} \gamma$, but that won't really change much.
[1] Voisin, Claire; Hodge Theory and Complex Algebraic Geometry, I
Best Answer
Okay, so I think the case for $d$-exact forms follows from the cases for $\partial$-exact and $\bar \partial$-exact forms. So suppose those two cases are already proven (the $\partial$-exact case works the same as the $\bar \partial$-exact case).
Write $\omega = d \beta = \partial \beta + \bar \partial \beta$. We want to apply the $\partial \bar \partial$-lemma to both $\partial \beta$ and $\bar \partial \beta$. For this we have to check that $\partial \beta$ is $\bar \partial$-closed, and $\bar \partial \beta$ is $\partial$-closed. However by assumption $$ 0 = \bar \partial \omega = \bar \partial \partial \beta + \bar \partial^2 \beta = \bar \partial \partial \beta,$$ and so $\partial \beta$ is $\bar \partial$-closed. Similarly, $\partial \bar \partial \beta = 0$. So there are differential forms $\gamma_1$ and $\gamma_2$ with $$ \partial \beta = \partial \bar \partial \gamma_1 \quad \text{and} \quad \bar \partial \beta = \partial \bar \partial \gamma_2,$$ and so $$ \omega = \partial \bar \partial(\gamma_1 + \gamma_2).$$