I am learning about Kähler geometry using Tian's Canonical Metrics in Kähler Geometry. For proving the $\partial \bar{\partial}$-lemma, I needed to show the following identity, but I couldn't find a way to get this.
Let $\omega$ be the Kähler form of a Kähler manifold $(M,g)$. Let $\phi$ be a $(0,1)$-form. Then, the following hold.
$$ \left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi}\wedge \omega^{n-2} = \frac{1}{n(n-1)} \left( \vert \partial \phi\vert^2 – \vert \bar{\partial}^* \phi\vert^2\right) \omega^n.$$
I tried to show this identity by looking at the local formula, but the alternating sum is quite messy, and it does not seem to be related to the norms of $\partial \phi$ or $\bar{\partial}^* \phi$. Could you give some hint for proving the identity?
Thanks!
Best Answer
We calculate at the center $x$ of a complex coordinates with $g_{i\bar j} = \delta_{ij}$. Let $\omega_i = \frac{\sqrt{-1}}{2} dz^i \wedge d\bar z^i$. So we have $\omega = \sum_{i=1}^n \omega_i$. Write
$$\phi = \phi _{\bar j} \; d\bar z^j ,$$ then $$ \partial \phi = \partial_i \phi_{\bar j} \; dz^i \wedge d\bar z^j,\ \ \bar\partial \bar\phi = \overline{\partial_{i}\phi_{\bar j}}\; d\bar z^i \wedge dz^j,$$
which gives
$$\left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar\partial \bar \phi = \left( \frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar j} \overline{\partial_{l} \phi_{\bar k}} dz^i \wedge d\bar z^j \wedge d\bar z^l \wedge dz^k.$$
The above summation contains the following two types (and more):
$i=j$, $k=l$: $$ \left(\frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} dz^i \wedge d\bar z^i \wedge d\bar z^k \wedge dz^k = -\partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k,$$ and
$i = l$, $k=j$: $$\left(\frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar k} \overline{\partial_{i} \phi_{\bar k} }dz^i \wedge d\bar z^k \wedge d\bar z^i \wedge dz^k = |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k.$$
We care only these two types, since when $\{ i, k\} \neq \{ j, l\}$ or $i=j=k=l$ we have
$$ \left( \frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_j \overline{\partial_{l} \phi_{\bar k}} dz^i \wedge d\bar z^j \wedge d\bar z^l \wedge dz^k\wedge \omega^{n-2} = 0.$$
Hence we have \begin{align} \left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi}\wedge \omega^{n-2} = |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k \omega^{n-2}- \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k \wedge \omega^{n-2}. \end{align}
The remaining is combinatorics: since $\omega_i \wedge \omega _j = \omega _j \wedge\omega_i$, $\omega_i \wedge \omega_i = 0$,
\begin{align} \omega^{n-2} &= ( \omega_1 + \cdots + \omega_n)^{n-2} \\ &= \sum_{i_p \neq i_q} \omega_{i_1} \wedge \omega_{i_2} \wedge \cdots \wedge \omega_{i_{n-2}} \\ &= (n-2)! \sum_{i\neq k} \omega_1 \wedge \cdots \wedge \widehat{\omega_i}\wedge \cdots \wedge\widehat{\omega_k}\wedge \cdots \wedge \omega_n, \end{align}
here $\widehat{\omega_i}$ means $\omega_i$ is excluded. The last equality follows from the fact that there are $(n-2)!$ ways to form $\omega_1 \wedge \cdots \wedge \widehat{\omega_i}\wedge \cdots \wedge\widehat{\omega_k}\wedge \cdots \wedge \omega_n$.
Thus \begin{align} \left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi}\wedge \omega^{n-2} &= |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k \omega^{n-2}- \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k \wedge \omega^{n-2}. \\ &=(n-2)!\left( \sum_{i,k} |\partial_i \phi_{\bar k}|^2 - \sum_{i,k}\partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \right) \omega_1\wedge\cdots \wedge \omega^n\\ &= \frac{1}{n(n-1)} (|\partial \phi|^2 - |\bar\partial^* \phi|^2 ) \omega^n \end{align}
Since $$\omega^n = n!\; \omega_1\wedge \cdots\wedge \omega_n,$$
$$ |\partial \phi|^2 = \sum_{i,k} |\partial_i \phi_{\bar k}|^2$$ and (see here)
$$\bar\partial^* \phi = -\sum_i \partial_i \phi_{\bar i}$$ at $x$.