# Integration by parts on a Kähler manifold

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?

• 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. – Ben McKay Dec 3 '20 at 9:16
• The formula that you wrote is correct, and the RHS is real! The integrand is not always real-valued, but the integral is. – YangMills Dec 3 '20 at 13:55
• Is it obvious that the integral has to be real?@YangMills – penny Dec 3 '20 at 14:05
• 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). – 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 $$(i-1)$$-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 $$\int_X h \Delta_\omega f \omega^n = \int_X g(\bar{\partial} h, \bar{\partial} f) \omega^n.$$
• Generally people use $\langle u, v\rangle$ rather than $<u, v>$. The former is given by $\langle u, v\rangle$. – Michael Albanese Dec 3 '20 at 12:10
• I made the edit @MichaelAlbanese mentions. Note also that \quad or \qquad or one of the other spacing commands is probably preferable to repeated \ \ \ . – LSpice Dec 3 '20 at 13:58
• 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. – penny Dec 3 '20 at 14:31