About integration by parts of $\partial$ operator

Question

As well known to us, operator $d$ satisfies property of integration by parts in the mathematical analysis. My question is :

(1) Does the operator $\partial$ or $\bar\partial$ satisfies property of integration by parts? (Of course, this may depends on whether does the operator $\partial$ or $\bar\partial$ satisfies the stokes formula or not? )

(2) Let $X$ be a complex manifold without boundary. Let $\varphi$ be a $(n-1,n-1)$-form, $f$ is a smooth function on $X$. Does the follow holds? $$\int_Xf\partial\bar\partial\varphi=\int_X\partial\bar\partial f\wedge\varphi$$

Answer 1

Let $X$ be a compact complex $n$-dimensional manifold, $\alpha \in \mathcal{E}^{p-1,q}(X)$ and $\beta \in \mathcal{E}^{n-p,n-q}(X)$. Then $\partial\alpha\wedge\beta \in \mathcal{E}^{n,n}(X)$, so we can integrate it. Note that $\bar{\partial}\alpha\wedge\beta \in \mathcal{E}^{n-1,n+1}(X)$ and is therefore zero, so $d\alpha\wedge\beta = \partial\alpha\wedge\beta + \bar{\partial}\alpha\wedge\beta = \partial\alpha\wedge\beta$. So by Stokes' Theorem, we have

$$\int_X\partial\alpha\wedge\beta = \int_Xd\alpha\wedge\beta = \int_X d(\alpha\wedge\beta) - (-1)^{|\alpha|} \int_X\alpha\wedge d\beta = -(-1)^{|\alpha|}\int_X\alpha\wedge d\beta$$

where $|\alpha|$ denotes its total degree, in this case, $(p - 1) + q$.

Now note that $\alpha\wedge\bar{\partial}\beta \in \mathcal{E}^{n-1, n+1}(X)$ and is therefore zero, so $\alpha\wedge d\beta = \alpha\wedge\partial\beta + \alpha\wedge\bar{\partial}\beta = \alpha\wedge\partial\beta$ and hence

$$\int_X\partial\alpha\wedge\beta = -(-1)^{|\alpha|}\int_X\alpha\wedge\partial\beta.$$

Similarly, if $\alpha \in \mathcal{E}^{p,q-1}(X)$ and $\beta \in \mathcal{E}^{n-p,n-q}(X)$, then

$$\int_X\bar{\partial}\alpha\wedge\beta = -(-1)^{|\alpha|}\int_X\alpha\wedge\bar{\partial}\beta.$$

Therefore

\begin{align*} \int_X\partial\bar{\partial}f\wedge\varphi &= -(-1)^{|\bar{\partial}f|}\int_X\bar{\partial}f\wedge\partial\varphi\\ &= \int_X\bar{\partial}f\wedge\partial\varphi\\ &= -(-1)^{|f|}\int_X f\wedge\bar{\partial}\partial\varphi\\ &= -\int_X f\wedge\bar{\partial}\partial\varphi\\ &= \int_X f\wedge\partial\bar{\partial}\varphi. \end{align*}