Is $\langle f, h \rangle_{\omega}$ real on a Kaehler manifold with Kaehler metric $\omega$?

Question

Let $X$ be a Kaehler manifold equipper with a metric $\omega$. Suppose $f, h$ are smooth real functions on $X$, then $\int_{X} h \Delta_{\omega} f \frac{\omega^{n}}{n!} = -\int_{X} g^{i \bar j} h_{i} f_{\bar j} \frac{\omega_0^n}{n!} = -\int_{X} \langle h, f \rangle_{\omega} \frac{\omega^n}{n!}$ using integraion by parts. $\Delta_{\omega} f$ is defined to be $g^{i \bar j} \partial_{i} \partial_{\bar j} f$ which should be real(unchanged under complex conjugation). This is suggesting that $\langle h, f \rangle_{\omega}$ should be real as well. But why is that the case?

I computed $\overline{\langle h, f \rangle_{\omega}} = \overline{ g^{i \bar j} h_{i} f_{\bar{j}}} = g^{\bar i j} h_{\bar i} f_{j}$ which does not seem to be equal to $\langle h, f \rangle_{\omega}$.

Answer 1

There are several issues here.

1. First, what do you mean by $\langle h,f\rangle_{\omega}$? When $f$ and $h$ are scalar functions, the pointwise inner product $\langle h,f\rangle_{\omega}$ usually just means their product $hf$ (or, if they're complex-valued, $h\overline f$). But that is not what should appear in this integration-by-parts formula. Instead, it should involve first derivatives of $h$ and $f$ (see below).
2. Next, in order for a formula like this to hold, you need to assume either that $X$ is compact (without boundary) or that $f$ and $h$ are compactly supported.
3. There are actually three different formulas like this on a Kähler manifold, corresponding to the fact that the three Laplacians $\Delta = d^*d$, $\Delta_{\partial} = \partial^* \partial$, and $\Delta_{\overline\partial} = \overline\partial{}^*\overline\partial$ are all equal up to constant multiples: $$ \Delta = 2 \Delta_{\partial} = 2\Delta_{\overline\partial} . $$ Correspondingly, there are three integration-by-parts formulas: \begin{align*} \int_X h \Delta f\, dV_\omega &= \int_X \langle dh,df\rangle_\omega dV_\omega,\tag{1}\\ \int_X h \Delta_\partial f\, dV_\omega &= \int_X \langle \partial h,\partial f\rangle_\omega dV_\omega,\tag{2}\\ \int_X h \Delta_{\overline\partial} f\, dV_\omega &= \int_X \langle {\overline\partial} h,{\overline\partial}f\rangle_\omega dV_\omega.\tag{3} \end{align*} (Here I've written $dV_\omega = \omega^n/n!$. Note also that these Laplacians differ by a sign from the one you denoted by $\Delta_\omega$, which accounts for the lack of negative signs on the right-hand sides.) When $f$ and $h$ are real, these right-hand sides can be written in holomorphic coordinates as follows: \begin{align*} \langle dh,df\rangle_\omega &= g^{i\bar j} \left(h_i f_{\bar j} + h_{\bar j} f_i\right),\\ \langle \partial h,\partial f\rangle_\omega &= g^{i\bar j} h_i f_{\bar j},\\ \langle {\overline\partial} h,{\overline\partial}f\rangle_\omega &= g^{i\bar j} h_{\bar j} f_i. \end{align*} The first of these expressions is real, but the other two need not be.
4. Importantly, the fact that the right-hand side of an integral formula like (2) or (3) is real does not imply that the integrand is real. Just to emphasize the point, consider the real-valued functions on the torus $\mathbb C/\mathbb Z^2$ given by $h(x+iy) = \sin 2\pi x$ and $f(x+iy) = \sin 2\pi y$. Then $h\Delta_{\partial} f = \tfrac12 h\Delta f$ is certainly real, but using the Euclidean metric, for which $g^{1\bar 1} = 2$, $$ g^{i\bar j} h_i f_{\bar j} = g^{1\bar 1} \partial_z h \,\partial _{\bar z} f = 2\pi^2 i \cos(2\pi x)\cos(2\pi y), $$ which is not real.