[proofplan] We first record the [transformation law for the Bergman kernel](/theorems/3707) under the biholomorphism $F$, deriving it from the unitary change-of-variables map between Bergman spaces. Restricting this identity to the diagonal gives an identity for the logarithmic Bergman potentials. The Jacobian determinant term is locally the real part of a [holomorphic function](/page/Holomorphic%20Function), hence is killed by $\partial \overline{\partial}$. Applying the chain rule to the remaining term gives exactly the pullback formula for the Hermitian tensor. [/proofplan]

[step:Derive the Bergman kernel transformation law from the unitary change of variables]Let $A^2(\Omega)$ denote the [Hilbert space](/page/Hilbert%20Space) of holomorphic functions $f:\Omega \to \mathbb{C}$ such that \begin{align*} \|f\|_{A^2(\Omega)}^2 := \int_\Omega |f(z)|^2\, d\mathcal{L}^{2n}(z) < \infty. \end{align*} Define the map \begin{align*} U:A^2(\widetilde{\Omega}) &\to A^2(\Omega) \\ h &\mapsto \bigl(z \mapsto h(F(z))\,JF(z)\bigr). \end{align*} Since $F$ is biholomorphic, $JF(z)\neq 0$ for every $z\in\Omega$, and the real Jacobian determinant of $F$ is $|JF(z)|^2$. Therefore the change-of-variables formula gives \begin{align*} \|Uh\|_{A^2(\Omega)}^2 &= \int_\Omega |h(F(z))|^2 |JF(z)|^2\, d\mathcal{L}^{2n}(z) \\ &= \int_{\widetilde{\Omega}} |h(\zeta)|^2\, d\mathcal{L}^{2n}(\zeta) = \|h\|_{A^2(\widetilde{\Omega})}^2. \end{align*} Thus $U$ is an isometry. Its inverse is \begin{align*} U^{-1}:A^2(\Omega) &\to A^2(\widetilde{\Omega}) \\ f &\mapsto \left(\zeta \mapsto f(F^{-1}(\zeta))\,J(F^{-1})(\zeta)\right), \end{align*} so $U$ is unitary. Let $K_\Omega:\Omega\times\Omega\to\mathbb{C}$ and $K_{\widetilde{\Omega}}:\widetilde{\Omega}\times\widetilde{\Omega}\to\mathbb{C}$ be the Bergman kernels. For fixed $w\in\Omega$, define \begin{align*} H_w:\Omega &\to \mathbb{C} \\ z &\mapsto JF(z)\,\overline{JF(w)}\,K_{\widetilde{\Omega}}(F(z),F(w)). \end{align*} The function $H_w$ is holomorphic in $z$ and belongs to $A^2(\Omega)$ because it is the image under $U$ of the Bergman kernel function $\zeta\mapsto \overline{JF(w)}K_{\widetilde{\Omega}}(\zeta,F(w))$. For every $f\in A^2(\Omega)$, put $h:=U^{-1}f\in A^2(\widetilde{\Omega})$. Using the reproducing property on $\widetilde{\Omega}$, \begin{align*} (f,H_w)_{A^2(\Omega)} &= (U^{-1}f,U^{-1}H_w)_{A^2(\widetilde{\Omega})} \\ &= \left(h,\overline{JF(w)}K_{\widetilde{\Omega}}(\cdot,F(w))\right)_{A^2(\widetilde{\Omega})} \\ &= JF(w)\,h(F(w)). \end{align*} Since $f(w)=h(F(w))JF(w)$, this gives \begin{align*} (f,H_w)_{A^2(\Omega)}=f(w). \end{align*} By uniqueness of the reproducing kernel vector at $w$, $H_w(z)=K_\Omega(z,w)$ for all $z,w\in\Omega$. Hence \begin{align*} K_\Omega(z,w)=JF(z)\,\overline{JF(w)}\,K_{\widetilde{\Omega}}(F(z),F(w)). \end{align*}[/step]

[guided]The goal is to compare the two Bergman kernels by transporting square-integrable holomorphic functions from $\widetilde{\Omega}$ to $\Omega$. Let $A^2(\Omega)$ be the Bergman space \begin{align*} A^2(\Omega)=\left\{f:\Omega\to\mathbb{C}\ \middle|\ f \text{ is holomorphic and } \int_\Omega |f(z)|^2\,d\mathcal{L}^{2n}(z)<\infty\right\}. \end{align*} For $h\in A^2(\widetilde{\Omega})$, define \begin{align*} U h:\Omega &\to \mathbb{C} \\ z &\mapsto h(F(z))JF(z). \end{align*} The factor $JF(z)$ is exactly the correction needed for Lebesgue measure under a biholomorphic change of variables. Since $F$ is biholomorphic, $JF(z)\neq 0$ for every $z\in\Omega$, and the real $2n$-dimensional Jacobian determinant of $F$ is $|JF(z)|^2$. Thus \begin{align*} \|Uh\|_{A^2(\Omega)}^2 &= \int_\Omega |h(F(z))|^2 |JF(z)|^2\, d\mathcal{L}^{2n}(z) \\ &= \int_{\widetilde{\Omega}} |h(\zeta)|^2\, d\mathcal{L}^{2n}(\zeta) = \|h\|_{A^2(\widetilde{\Omega})}^2. \end{align*} So $U$ preserves the Hilbert norm. Its inverse is obtained by applying the same construction to $F^{-1}$: \begin{align*} U^{-1}f:\widetilde{\Omega} &\to \mathbb{C} \\ \zeta &\mapsto f(F^{-1}(\zeta))J(F^{-1})(\zeta). \end{align*} Therefore $U$ is unitary. Now fix $w\in\Omega$. We want to identify the reproducing kernel vector at $w$ in $A^2(\Omega)$. Define \begin{align*} H_w:\Omega &\to \mathbb{C} \\ z &\mapsto JF(z)\,\overline{JF(w)}\,K_{\widetilde{\Omega}}(F(z),F(w)). \end{align*} This is holomorphic in $z$ because $JF$ and $z\mapsto K_{\widetilde{\Omega}}(F(z),F(w))$ are holomorphic. It is square-integrable because it is the image under $U$ of the Bergman kernel vector $\zeta\mapsto \overline{JF(w)}K_{\widetilde{\Omega}}(\zeta,F(w))$. Let $f\in A^2(\Omega)$ and put $h:=U^{-1}f$. Since $U$ is unitary, inner products are preserved: \begin{align*} (f,H_w)_{A^2(\Omega)} &= (U^{-1}f,U^{-1}H_w)_{A^2(\widetilde{\Omega})}. \end{align*} By construction, \begin{align*} U^{-1}H_w:\widetilde{\Omega} &\to \mathbb{C} \\ \zeta &\mapsto \overline{JF(w)}K_{\widetilde{\Omega}}(\zeta,F(w)). \end{align*} The reproducing property on $\widetilde{\Omega}$ gives \begin{align*} (f,H_w)_{A^2(\Omega)} &= \left(h,\overline{JF(w)}K_{\widetilde{\Omega}}(\cdot,F(w))\right)_{A^2(\widetilde{\Omega})} \\ &= JF(w)h(F(w)). \end{align*} Finally, $f=Uh$, so $f(w)=h(F(w))JF(w)$. Hence \begin{align*} (f,H_w)_{A^2(\Omega)}=f(w) \end{align*} for every $f\in A^2(\Omega)$. This means that $H_w$ has the defining reproducing property of the Bergman kernel vector at $w$. By uniqueness of the reproducing kernel vector, \begin{align*} K_\Omega(z,w)=JF(z)\,\overline{JF(w)}\,K_{\widetilde{\Omega}}(F(z),F(w)). \end{align*}[/guided]

[step:Take logarithms on the diagonal and isolate the pluriharmonic term]Set $w=z$ in the kernel transformation law. Since both Bergman metrics are assumed defined, $K_\Omega(z,z)>0$ and $K_{\widetilde{\Omega}}(F(z),F(z))>0$ on the domains under consideration. Thus \begin{align*} K_\Omega(z,z)=|JF(z)|^2K_{\widetilde{\Omega}}(F(z),F(z)). \end{align*} Taking logarithms gives \begin{align*} \log K_\Omega(z,z) = \log K_{\widetilde{\Omega}}(F(z),F(z)) + \log |JF(z)|^2. \end{align*} Because $JF:\Omega\to\mathbb{C}$ is holomorphic and non-vanishing, for every $z_0\in\Omega$ there is an open neighbourhood $V\subset\Omega$ of $z_0$ and a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} L:V&\to\mathbb{C} \end{align*} such that $e^{L(z)}=JF(z)$ for $z\in V$. On $V$, \begin{align*} \log |JF(z)|^2=L(z)+\overline{L(z)}. \end{align*} Therefore, for all $1\le i,j\le n$, \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}}\log |JF(z)|^2=0. \end{align*}[/step]

[guided]Restrict the kernel transformation law to the diagonal by putting $w=z$. This gives \begin{align*} K_\Omega(z,z)=JF(z)\overline{JF(z)}K_{\widetilde{\Omega}}(F(z),F(z)) =|JF(z)|^2K_{\widetilde{\Omega}}(F(z),F(z)). \end{align*} The assumption that the Bergman metrics are defined includes the positivity needed to take logarithms of the diagonal kernel values. Thus \begin{align*} \log K_\Omega(z,z) = \log K_{\widetilde{\Omega}}(F(z),F(z)) + \log |JF(z)|^2. \end{align*} The important point is that the second term does not contribute to the Bergman metric. Fix $z_0\in\Omega$. Since $JF$ is holomorphic and never zero, there is an open neighbourhood $V\subset\Omega$ of $z_0$ and a holomorphic logarithm \begin{align*} L:V&\to\mathbb{C} \end{align*} such that $e^{L(z)}=JF(z)$ for every $z\in V$. Hence \begin{align*} \log |JF(z)|^2=L(z)+\overline{L(z)}. \end{align*} Now $L$ is holomorphic, so $\partial_{\overline{z_j}}L=0$, and $\overline{L}$ is anti-holomorphic, so $\partial_{z_i}\overline{L}=0$. Therefore \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}}\log |JF(z)|^2 = \frac{\partial^2}{\partial z_i\partial\overline{z_j}}\bigl(L(z)+\overline{L(z)}\bigr) =0. \end{align*} Thus the Jacobian determinant changes the Bergman potential only by a pluriharmonic term, and $\partial\overline{\partial}$ removes it.[/guided]

[step:Apply the complex chain rule to identify the pullback tensor]For $1\le a,b\le n$, define the coefficient functions of the Bergman metric on $\widetilde{\Omega}$ by \begin{align*} \widetilde{g}_{a\overline{b}}:\widetilde{\Omega}&\to\mathbb{C} \\ \zeta&\mapsto \frac{\partial^2}{\partial \zeta_a\partial\overline{\zeta_b}} \log K_{\widetilde{\Omega}}(\zeta,\zeta). \end{align*} Write $F=(F_1,\dots,F_n)$. Since $F$ is holomorphic, $\partial_{\overline{z_j}}F_a=0$ and $\partial_{z_i}\overline{F_b}=0$. Applying the complex chain rule to \begin{align*} z \mapsto \log K_{\widetilde{\Omega}}(F(z),F(z)) \end{align*} gives \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}} \log K_{\widetilde{\Omega}}(F(z),F(z)) = \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} Combining this with the logarithmic identity and the vanishing of the mixed derivatives of $\log |JF|^2$, we obtain \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}} \log K_\Omega(z,z) = \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} The right-hand side is precisely the coefficient of $dz_i\otimes d\overline{z_j}$ in $F^*g_{\widetilde{\Omega}}$. Hence $g_\Omega=F^*g_{\widetilde{\Omega}}$.[/step]

[guided]We now compute the coefficients of the pullback metric. For $1\le a,b\le n$, define \begin{align*} \widetilde{g}_{a\overline{b}}:\widetilde{\Omega}&\to\mathbb{C} \\ \zeta&\mapsto \frac{\partial^2}{\partial \zeta_a\partial\overline{\zeta_b}} \log K_{\widetilde{\Omega}}(\zeta,\zeta). \end{align*} These are exactly the coordinate coefficients of the Bergman metric on $\widetilde{\Omega}$. Write \begin{align*} F=(F_1,\dots,F_n):\Omega&\to\widetilde{\Omega}. \end{align*} Because $F$ is holomorphic, its components satisfy \begin{align*} \frac{\partial F_a}{\partial \overline{z_j}}=0, \qquad \frac{\partial \overline{F_b}}{\partial z_i}=0. \end{align*} Apply the complex chain rule to the function \begin{align*} \Phi:\Omega&\to\mathbb{R} \\ z&\mapsto \log K_{\widetilde{\Omega}}(F(z),F(z)). \end{align*} First differentiating in the $\overline{z_j}$ direction gives \begin{align*} \frac{\partial \Phi}{\partial\overline{z_j}}(z) = \sum_{b=1}^n \frac{\partial}{\partial\overline{\zeta_b}} \log K_{\widetilde{\Omega}}(\zeta,\zeta)\bigg|_{\zeta=F(z)} \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} Differentiating this identity in the $z_i$ direction, the holomorphicity relation \begin{align*} \frac{\partial}{\partial z_i}\overline{\frac{\partial F_b}{\partial z_j}(z)}=0 \end{align*} removes the derivative of the conjugate Jacobian factor. Therefore \begin{align*} \frac{\partial^2 \Phi}{\partial z_i\partial\overline{z_j}}(z) = \sum_{a,b=1}^n \frac{\partial^2}{\partial \zeta_a\partial\overline{\zeta_b}} \log K_{\widetilde{\Omega}}(\zeta,\zeta)\bigg|_{\zeta=F(z)} \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} Using the definition of $\widetilde{g}_{a\overline{b}}$, this becomes \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}} \log K_{\widetilde{\Omega}}(F(z),F(z)) = \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} From the logarithmic diagonal identity, \begin{align*} \log K_\Omega(z,z) = \log K_{\widetilde{\Omega}}(F(z),F(z)) + \log |JF(z)|^2, \end{align*} and from the previous step, \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}}\log |JF(z)|^2=0. \end{align*} Hence \begin{align*} \frac{\partial^2}{\partial z_i\partial\overline{z_j}} \log K_\Omega(z,z) = \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} But this is exactly the coordinate formula for the pullback of a Hermitian metric: \begin{align*} (F^*g_{\widetilde{\Omega}})_{i\overline{j}}(z) = \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) \frac{\partial F_a}{\partial z_i}(z) \overline{\frac{\partial F_b}{\partial z_j}(z)}. \end{align*} Therefore every coefficient of $g_\Omega$ equals the corresponding coefficient of $F^*g_{\widetilde{\Omega}}$.[/guided]

[step:Conclude the metric identity on tangent vectors] Let $z\in\Omega$ and let $\xi,\eta\in T_z\Omega\cong\mathbb{C}^n$. The coefficient identity proved above gives \begin{align*} g_{\Omega,z}(\xi,\eta) &= \sum_{i,j=1}^n \frac{\partial^2}{\partial z_i\partial\overline{z_j}} \log K_\Omega(z,z)\,\xi_i\overline{\eta_j} \\ &= \sum_{a,b=1}^n \widetilde{g}_{a\overline{b}}(F(z)) (dF_z(\xi))_a \overline{(dF_z(\eta))_b} \\ &= g_{\widetilde{\Omega},F(z)}(dF_z(\xi),dF_z(\eta)). \end{align*} Thus $F^*g_{\widetilde{\Omega}}=g_\Omega$, which is the claimed biholomorphic invariance of the Bergman metric. [/step]