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?

2$\begingroup$ Try to write in terms of the exterior derivative and the Hodge star, since integration by parts is Stokes' theorem applied to an exact form. $\endgroup$ – Ben McKay Dec 3 '20 at 9:16

$\begingroup$ The formula that you wrote is correct, and the RHS is real! The integrand is not always realvalued, but the integral is. $\endgroup$ – YangMills Dec 3 '20 at 13:55

$\begingroup$ Is it obvious that the integral has to be real?@YangMills $\endgroup$ – penny Dec 3 '20 at 14:05

$\begingroup$ It follows from the equation that you wrote, since the LHS is real. Proving your formula is a simple exercise using the divergence theorem (and the definition of covariant derivatives for a Kahler metric). $\endgroup$ – YangMills Dec 3 '20 at 14:36
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 $(i1)$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 $$ \int_X h \Delta_\omega f \omega^n = \int_X g(\bar{\partial} h, \bar{\partial} f) \omega^n. $$

2$\begingroup$ Generally people use $\langle u, v\rangle$ rather than $<u, v>$. The former is given by
$\langle u, v\rangle$
. $\endgroup$ – Michael Albanese Dec 3 '20 at 12:10 
1

$\begingroup$ Sorry I am a bit confused. Usually $\bar \partial^{\ast}$ would take a $(1, 1)$ form then gives back a smooth function. Here you seem to be acting it on a function directly. $\endgroup$ – penny Dec 3 '20 at 14:31

$\begingroup$ @penny, I corrected it. Thanks. I think YangMill is right. This is exactly what you had. It turns out simpler than I thought since we are only dealing with functions rather than forms. $\endgroup$ – Eugene Z. Xia Dec 3 '20 at 15:46