I am trying to make sense of integration by parts on a Kähler manifold $X$ equipped with a Kähler metric $\omega$. Given two smooth real functions $f$ and $h$ on $X$, I want to write down the integration by parts formula for the following:
$$\int\_{X} h \Delta\_{\omega} f \omega^n.$$
In local coordinates $\Delta\_{\omega} f = \sum\_{i, j} g^{i \bar\jmath} \partial\_{\bar\jmath} \partial\_{i}f$. My guess is that
$$\int\_{X} h \Delta\_{\omega} f \omega^n = -\int\_{X} g^{i \bar\jmath} \partial\_{\bar\jmath} h \partial\_{i} f \omega^n.$$
However, this is not quite right since the LHS is real while the right hand side is not necessarily so. What is the correct formula? Is there a general strategy for thinking of such things in the complex case?



Integration by parts on a Kähler manifold
Assume $(X, d = \partial + \bar{\partial})$ to be a compact Kähler manifold. The Kähler metric $g$ induces a metric on all differential forms, which we will also call $g$. It follows that $\omega^n$ defines a Hilbert space of $i$-forms on $X$ by
$$
\langle u, v\rangle = \int\_X g(u,v) \omega^n.
$$
For functions $u, v$, this is
$$
\langle u, v\rangle = \int\_X u \bar{v} \omega^n.
$$


The adjoints $d^\*$ and $\bar{\partial}^\*$ are the operators such that for all $(i-1)$-form $u$ and all $i$-form $v$,
$$
\langle du,v\rangle = \langle u,d^\*v\rangle,\quad\langle\bar{\partial}u,v\rangle = \langle u, \bar{\partial}^\* v\rangle.
$$
The expression $\bar{\partial}^\*$ equals $-\* \bar{\partial} \*$, where $\*$ is the Hodge-$\*$ and it involves the Kähler metric $g$.


The Laplacian is then
$$
\Delta\_d = d^\* d + d d^\* = 2 (\bar{\partial} \bar{\partial}^\* +\bar{\partial}^\* \bar{\partial}) = 2 \Delta\_{\bar{\partial}}.
$$
Since $h,f$ are functions,
we have
$$
\langle h, \Delta\_{\bar{\partial}} f\rangle = \langle h, (\bar{\partial} \bar{\partial}^\* +\bar{\partial}^\* \bar{\partial}) f\rangle = \langle\bar{\partial}^\* h, \bar{\partial}^\* f\rangle + \langle\bar{\partial} h, \bar{\partial} f\rangle = \langle\bar{\partial} h, \bar{\partial} f\rangle.
$$
In terms of your notation, this means
$$
\int\_X h \Delta\_\omega f \omega^n = \int\_X g(\bar{\partial} h, \bar{\partial} f) \omega^n.
$$


