Local formula of Laplacian on Kähler manifolds Let $M$ be a Kähler manifold with Kähler form $\omega=g_{j\bar{k}}\,dz^j\wedge d\bar{z}^k$ in local holomorphic coordinates. I want to show that the associated Laplacian $\Delta:=2(\bar{\partial}^*\bar{\partial}+\bar{\partial}\bar{\partial}^*)$ (one could take $d,\partial$ instead, but I think $\bar{\partial}$ is more convenient here) has the following expression acting on functions: $$\Delta f=-2g^{\bar{j}k}\frac{\partial^2f}{\partial z^j\partial\bar{z}^k}.$$ It is remarkable that this formula does not involve derivatives of the metric!
Here is my attempt:
Given functions $f,\phi$ compactly supported on a holomorphic chart, we compute the $L^2$ inner product $$\frac{1}{2}(\Delta f,\phi)=(\partial f,\partial\phi)=\int_{\mathbb{C}^n}\frac{\partial f}{\partial\bar{z}^j}\frac{\partial\bar{\phi}}{\partial z^k}g^{\bar{j}k}G,$$ where $G=\det(g_{j\bar{k}})$ (coefficient of the volume form). After integration by parts, this becomes $$-\int\frac{\partial^2f}{\partial z^j\partial\bar{z}^k}g^{\bar{j}k}\bar{\phi}G-\int\frac{\partial f}{\partial\bar{z}^j}\bar{\phi}\,\frac{\partial(g^{\bar{j}k}G)}{\partial z^k}.$$ Now I have to show that the latter term vanishes, but I can't see why. I guess I have to use the Kähler condition. However, it seems that expanding the determinant and inverse matrix leads to nowhere. How do I proceed? Or is there any other way to show this?
 A: There is another way to get the same result. I think this method is indeed useful, when dealing with some integration by parts in Kähler geometry context. First, note that for a vector field $X$
$$
\DeclareMathOperator{\dv}{div} \DeclareMathOperator{\tr}{tr} \dv X = \tr (Y \mapsto \nabla_Y X).
$$
Thus, in a local holomorphic coordinate system,
$$
\newcommand{\cframe}[1]{\frac{\partial}{\partial {#1}}} \dv X = \left(\nabla_{\cframe{z^i}} X\right) (dz^i) +  \left(\nabla_{\cframe{\bar{z}^i}} X\right) (d\bar{z}^i).
$$
Here, the summation convention is used. Given $(0,1)$-form $\alpha$, define a vector field $X$ so that
$$
X^i = g^{i \bar{j}} \alpha_{\bar{j}} \bar{f}.
$$
That means $X = \bar{f} g(\cdot, \alpha)$ as a contravariant tensor. Now compute $\left(\nabla_{\cframe{z^i}} X\right) (dz^i)$ and $\left(\nabla_{\cframe{\bar{z}^i}} X\right) (d\bar{z}^i)$, then one can show that
$$
\left(\nabla_{\cframe{z^i}} X\right) (dz^i) = \frac{\partial \bar{f}}{\partial z^i} g^{i \bar{k}} \alpha_{\bar{k}} + \bar{f} g^{i \bar{k}} \frac{\partial \alpha_{\bar{k}}}{\partial z^i} \quad \text{and} \quad \left(\nabla_{\cframe{\bar{z}^i}} X\right) (d\bar{z}^i) = 0.
$$
Thus,
$$
\dv X = \frac{\partial \bar{f}}{\partial z^i} g^{i \bar{k}} \alpha_{\bar{k}} + \bar{f} g^{i \bar{k}} \frac{\partial \alpha_{\bar{k}}}{\partial z^i}.
$$
Note that $(\dv X )dV$ is an exact form, where $dV$ is the Riemmanian volume form. Therefore, by Stokes' theorem, if $f$ is compactly supported in the coordinate neighborhood,
$$
\int_M \frac{\partial \bar{f}}{\partial z^i} g^{i \bar{k}} \alpha_{\bar{k}} dV = - \int_M \bar{f} g^{i \bar{k}} \frac{\partial \alpha_{\bar{k}}}{\partial z^i} dV.
$$
This shows that
$$
\bar{\partial}^* \alpha = - g^{i \bar{k}} \frac{\partial \alpha_{\bar{k}}}{\partial z^i}.
$$
Now for a smooth function $u$, take $\alpha = \bar{\partial} u$. Then, you can get the local formula of the Laplacian of $u$.
I learned this approach form Székelyhidi's Introduction to Extremal metrics.
A: Let $A = A_{\bar{i}} \overline{dz^{i}}$ be any $(0,1)$-form (we want to use $A = \overline\partial f$ later). Using the definition of $\overline\partial ^*$: for all test function $\varphi$, write $A^i = g^{\bar j i} A_{\bar j}$,
\begin{align*}
\int_M \varphi\overline{\overline\partial ^* A} dV &=\int_M (\overline\partial \varphi)_{\bar i} \overline{A^{i}}dV \\
&=\int_M \frac{\partial \varphi}{\partial \bar{z^i}}\overline{A^{i}} (\sqrt{-1})^n G dz^N \wedge \overline{dz^N},
\end{align*}
Where $dz^N = dz^1 \wedge \cdots \wedge dz^n$ and $G = \det (g_{i\bar j})$. Integration by part gives
\begin{align*}
\int_M \varphi\overline{\overline\partial ^* A} dV 
&= - \int_M \varphi \partial_{\bar i}(\overline{A^{i}}  G) (\sqrt{-1})^n dz^N \wedge \overline{dz^N}\\
&= -\int_M \varphi \left(\overline{\partial_i A^i + A^i G^{-1} \partial_i G} \right) (\sqrt{-1})^n G dz^N \wedge \overline{dz^N} \\
&= -\int_M \varphi \left(\overline{\partial_i A^i + A^i \partial_i(\log G)} \right)dV.
\end{align*}
Thus
\begin{equation}\tag{1}
\overline\partial ^* A= - (\partial_i A^i+ (\partial_i \log G) A^i),
\end{equation}
Now calculate:
\begin{align*}
\partial_i A^i + \partial_i \log G A^i &= \partial_i (g^{\bar j i} A_{\bar j}) + g^{ m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} A_{\bar j}\\
&= g^{\bar j i} \frac{\partial A_{\bar j}}{\partial z^i} + \left(\frac{\partial g_{\bar j i}}{\partial z^i} +  g^{m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} \right)A_{\bar j} 
\end{align*}
Now we use the Kähler condition: in particular, we have
$$ \frac{\partial g_{m\bar k}}{\partial z^i} = \frac{\partial g_{i\bar k}}{\partial z^m}, $$
thus
\begin{align*}
g^{m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} &= g^{m \bar k} \frac{\partial g_{i\bar k}}{\partial z^m} g^{\bar j i} \\
&= - g^{m \bar k} \frac{\partial g^{\bar j i}}{\partial z^m} g_{i\bar k} \\
&=-\frac{\partial g^{\bar j i}}{\partial z^i}
\end{align*}
Then we have
$$\overline\partial ^* A = - g^{\bar j i} \frac{\partial A_{\bar j}}{\partial z^i}$$
and setting $A = \overline\partial f$ gives
$$ \Delta f = 2 \overline\partial^* \overline\partial f = -2 g^{\bar j i} \frac{\partial ^2 f}{\partial z^i \partial \bar z^j}.$$
