Some identity in Kähler geometry

Question

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!

Answer 1

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 $x$.

Answer 2

I could produce another proof using the permutations. I think the calculation is essentially the same with Arctic Char's answer, but this would be more instructive when someone is needed to compute a similar expression.

First, for $j=1, \dots, m$ with $m\leq n$, define $2$-forms $\eta^j$ as follows: $$ \DeclareMathOperator{\sgn}{sgn} \newcommand{\fpartial}[2]{\frac{\partial #1}{\partial #2}} \eta^j = \sum_{k,l =1}^nc^{j}_{kl} \alpha^k \wedge \beta^l, $$ where $\alpha, \beta$ are $1$-forms. Also, given subsets $K, L$ of $\lbrace 1, \dots, n \rbrace$ with $\vert K\vert = \vert L \vert = m$, write $K = \lbrace k_1 < \dots < k_m \rbrace$ and $L = \lbrace l_1 < \dots < l_m \rbrace$. Then, \begin{align} \eta^1 \wedge \dots \wedge \eta^m &= \sum_{\vert K\vert = \vert L \vert = m} \sum_{\sigma, \tau \in S_m} c^1_{k_{\sigma(1)\tau(1)}} \dots c^m_{k_{\sigma(m)\tau(m)}} \alpha^{k_{\sigma(1)}} \wedge \beta^{l_{\tau(1)}} \wedge \dots \wedge \alpha^{k_{\sigma(m)}} \wedge \beta^{l_{\tau(m)}} \\ &= \sum_{\vert K\vert = \vert L \vert = m} \sum_{\sigma, \tau \in S_m} (\sgn \sigma) (\sgn \tau) c^1_{k_{\sigma(1)\tau(1)}} \dots c^m_{k_{\sigma(m)\tau(m)}} \alpha^{k_1} \wedge \beta^{l_1} \wedge \dots \wedge \alpha^{k_m} \wedge \beta^{l_m}.\end{align} In particular, if $m=n$, then $$ \eta^1 \wedge \dots \wedge \eta^n = \sum_{\sigma, \tau \in S_n} (\sgn \sigma) (\sgn \tau)c^1_{k_{\sigma(1)\tau(1)}} \dots c^m_{k_{\sigma(n)\tau(n)}}\alpha^1 \wedge \beta^1 \wedge \dots \wedge \alpha^n \wedge \beta^n. $$ Apply this to the Kähler form $\omega_g = \sqrt{-1} g_{i \bar{j}} dz^i \wedge d \bar{z}^j$. Then, we obtain $$ \omega_g^n = (\sqrt{-1})^n n! \det(g_{i \bar{j}}) dz^1 \wedge d \bar{z}^1 \wedge \dots \wedge dz^n \wedge d \bar{z}^n. $$ Now write $$ \phi = \phi_{\bar{j}} d \bar{z}^j, $$ then $$ \partial \phi = \fpartial{\phi_{\bar{j}}}{z^i} dz^i \wedge d \bar{z}^j \quad \text{and} \quad \bar{\partial} \bar{\phi} = - \fpartial{\bar{\phi}_{\bar{i}}}{\bar{z}^j} dz^i \wedge d \bar{z}^j $$ By taking $\eta^1 = \partial \phi$, $\eta^2 = \bar{\partial} \bar{\phi}$ and $\eta^3 = \dots = \eta^n =\omega_g$, we have $$ (\sqrt{-1}^2 \partial \phi \wedge \bar{\partial} \bar{\phi} \wedge \omega_g^{n-2} = (\sqrt{-1})^n S dz^1 \wedge d \bar{z}^1 \wedge \dots \wedge dz^n \wedge d \bar{z}^n, $$ where $$ S = \sum_{\sigma, \tau} \fpartial{\phi_{\overline{\tau(1)}}}{z^{\sigma(1)}} \fpartial{\overline{\phi_{\overline{\sigma(2)}}}}{\bar{z}^{\tau(2)}} g_{\sigma(3) \overline{\tau(3)}} \dots g_{\sigma(n) \overline{\tau(n)}}. $$ Now let $z$ be a holomorphic normal coordinate system around $p$. From now on, all the calculation will be done at $p$. First, $$ (\sqrt{-1}^n) dz^1 \wedge d \bar{z}^1 \wedge \dots \wedge dz^n \wedge d \bar{z}^n = \frac{\omega_g^n}{n!}. $$ In addition to this, observe that in the sum $S$, the nonvanishing terms occur only when $\sigma(3) = \tau(3), \dots $ and $\sigma(n) = \tau(n)$. Given $\sigma$, let $\tau_{\sigma}$ be defined by $\tau(1) = \sigma(2)$, $\tau(2) = \sigma(1), \sigma(3) = \tau(3), \dots, \sigma(n) = \tau(n)$. Then, $\sgn \tau_{\sigma} = - \sgn \sigma$. Thus, we have \begin{align} S &= \sum_{\sigma \in S_n } \fpartial{\phi_{\overline{\tau(1)}}}{z^{\sigma(1)}} \fpartial{\overline{\phi_{\overline{\sigma(2)}}}}{\bar{z}^{\tau(2)}} - \sum_{\sigma \in S_n} (\sgn \sigma) (\sgn \tau_\sigma) \fpartial{\phi_{\overline{\sigma(2)}}}{z^{\sigma(1)}} \fpartial{ \overline{\phi_{\overline{\sigma(2)}}}}{\bar{z}^{\sigma(1)}} \\ &= (n-2)!\sum_{i \ne j}\left[ \fpartial{\phi_{\bar{i}}}{z^i} \fpartial{\overline{\phi_{\bar{j}}}}{\bar{z}^j} - \fpartial{\phi_{\bar{j}}}{z^{i}} \fpartial{\overline{\phi_{\bar{j}}}}{\bar{z}^i}\right] \\ &= (n-2)! \sum_{i,j } \fpartial{\phi_{\bar{i}}}{z^i} \fpartial{\overline{\phi_{\bar{j}}}}{\bar{z}^j} - (n-2)! \sum_i \left\vert\fpartial{\phi_{\bar{i}}}{z^i} \right\vert^2 - (n-2)! \sum_{i \ne j} \left\vert\fpartial{\phi_{\bar{j}}}{z^i} \right\vert^2 \\ &= (n-2)! \left\vert\ \sum_i \fpartial{\phi_{\bar{i}}}{z^i}\right\vert^2 - (n-2)! \sum_{i,j} \left\vert\fpartial{\phi_{\bar{j}}}{z^i} \right\vert^2. \end{align} Observe that $$ \bar{\partial}^* \phi = - \sum_{i} \fpartial{\phi_{\bar{i}}}{z_i} \quad \text{and} \quad \left\vert\partial \phi\right\vert^2 = \sum_{i, j} \left\vert\fpartial{\phi_{\bar{j}}}{z_i}\right\vert^2. $$ Therefore, \begin{align} \left(\sqrt{-1}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi} \wedge \omega_g^{n-2} &=(n-2)! \left(\sqrt{-1} \right)^n \left( \left\vert\bar{\partial}^* \phi\right\vert^2 - \left\vert\partial \phi\right\vert^2\right) dz_1 \wedge d \bar{z}_1 \wedge \dots \wedge dz_{n} \wedge d \bar{z}_{n} \\ &= \frac{1}{n(n-1)} \left( \vert\bar{\partial}^* \phi\vert^2 - \vert\partial \phi\vert^2\right) \omega_g^n. \end{align} The reason for the absence of the factor $2^2$ is due to the normalization of the Kähler form.