Integration by parts for laplacian on a kahler manifold without boundary

Question

Given a kahler manifold $(X, w)$ without boundary, we can define the laplacian $\Delta$ acting on a smooth function $f$ to be $\Delta f = g^{j \bar i} \partial_i \partial_{\bar j} f$, where $g^{j\bar i}$ is the component of the inverse of the metric $w$. Now I want to prove the following integration by parts formula(integrating with respect to the kahler volume form $w^n$):

$\int \Delta f h \omega^n = \int \Delta h f \omega^n$

Starting by writing the left hand side in local coordinates:

$\int g^{j \bar i} \partial_i \partial_{\bar j} f g = -\int \partial_i g^{i\bar j} h \partial_{\bar i} f = \int \partial_{i} \partial_{\bar j}( g^{j \bar i} h)f$

Some of the derivatives will hit on the metric producing extra terms. How do I bring it into the form on the right hand side?

Answer 1

Your definition of the Laplacian is incorrect. I suggest you try to modify it in order for the integration by parts to work, but you can look out the correct expression in coordinates on the Wikipedia page for the Laplace-Beltrame operator: https://en.m.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator

Answer 2

That formula for the Laplacian seems strange to me, too. I agree with Dario. It should be $$\Delta f=\frac{1}{\sqrt g} \partial_i (\sqrt g g^{ij}\partial_j f).$$ Also, the volume element plays a role here, but it disappeared from your computations. The correct integral to consider is $$\int_M \Delta f_1 f_2\, \omega^n, $$ which in coordinates should read $$ \int_M \frac{1}{\sqrt g} \partial_i (\sqrt g g^{ij}\partial_j f_1)f_2 \sqrt{g}dx^1\ldots dx^n.$$ You can see in this expression that $\sqrt g$ simplifies and that you can immediately integrate by parts. This proves that $$ \int_M \Delta f_1 f_2\, \omega^n=\int_M \langle \nabla f_1, \nabla f_2\rangle_g\, \omega^n, $$ and now you can redo the computation reversing the roles of $f_1$ and $f_2$ and you conclude that $$\int_M \Delta f_1 f_2\, \omega^n=\int_M f_1 \Delta f_2\, \omega^n.$$

This is how I would do this computation, but I am thinking of a Riemannian manifold, I don't know a thing about Kähler.