PomeProposition 11.28.3.37B317ABBC08 The two metrics satisfy
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11.28.4
\begin{equation} \int _ X\bigl (|S|_{g,h}^2+I_ g(h)-\mathcal R_ g(h)\bigr )\, \mathrm dV_ g=0. \end{equation}
Proposition 11.28.337B317ABBC08
Proof. We first compute the exact term that distinguishes the two curvature traces. Define
\begin{align*} D_{i\bar j k\bar l} ={}& \partial _ i\partial _{\bar j}h_{k\bar l} +\partial _ k\partial _{\bar l}h_{i\bar j}\\ & -\partial _ i\partial _{\bar l}h_{k\bar j} -\partial _ k\partial _{\bar j}h_{i\bar l}. \end{align*}
This is a tensor: putting holomorphic differentials first gives
\[ \mathrm i\partial \bar\partial \omega _ h =\sum _{i<k,\, j<l}D_{i\bar j k\bar l} \, \mathrm dz^ i\wedge \mathrm dz^ k \wedge \mathrm d\bar z^ j\wedge \mathrm d\bar z^ l. \]
For $D_ g=\sum _{i,k}D_{i\bar i k\bar k}$ in a $g$-unitary frame, expansion of the wedge product gives
\[ D_ g\, \mathrm dV_ g =2\mathrm i\partial \bar\partial \omega _ h \wedge \frac{\omega _ g^{n-2}}{(n-2)!}. \]
Indeed, each pair $i<k$ contributes $D_{i\bar i k\bar k}\mathrm dV_ g$ to the wedge product and occurs twice in the contraction. Closedness of $\omega _ g$ and Stokes' theorem therefore give $\int _ XD_ g\, \mathrm dV_ g=0$.
The four-term antisymmetrization of the curvature formula is
\begin{align*} & R^ h_{i\bar j k\bar l}+R^ h_{k\bar l i\bar j} -R^ h_{i\bar l k\bar j}-R^ h_{k\bar j i\bar l}\\ & \hspace{3em}=h_{p\bar q}T^ p_{ik}\overline{T^ q_{jl}} -D_{i\bar j k\bar l}. \end{align*}
The second derivatives give $-D$, and the quadratic terms give the squared connection difference in the two skew indices. Contracting gives $2(Q_ g(h)-B_ g(h))=|T|_{g,h}^2-D_ g$, hence
\[ Q_ g(h)=\mathcal R_ g(h)+\tfrac 14|T|_{g,h}^2-\tfrac 14D_ g. \]
In $g$-normal holomorphic coordinates, the first derivatives of $g$ vanish. Differentiating $u=g^{k\bar l}h_{k\bar l}$ twice, the inverse-metric term is $I_ g(h)$ by the Kähler curvature symmetries of $g$, while the second derivatives of $h$ give $-Q_ g(h)+|A|_{g,h}^2$. Thus
\begin{align*} P_ g u& =|A|_{g,h}^2+I_ g(h)-Q_ g(h)\\ & =|S|_{g,h}^2+I_ g(h)-\mathcal R_ g(h)+\tfrac 14D_ g, \end{align*}
where symmetric and skew tensors are orthogonal and $|A|_{g,h}^2=|S|_{g,h}^2+|T|_{g,h}^2/4$. Integrating removes both $P_ g u$ and $D_ g$, proving (11.28.4). $\square$
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