[proofplan] We compare the two Bergman spaces by the natural pullback map weighted by the complex Jacobian determinant. The holomorphic change-of-variables formula shows that this weighted pullback is a unitary operator from $A^2(\widetilde{\Omega})$ onto $A^2(\Omega)$. Applying this unitary map to the reproducing vector at $F(z)$ gives a reproducing vector at $z$, and uniqueness of reproducing vectors identifies it with the Bergman kernel vector on $\Omega$. Evaluating that vector identity at $w$ yields the stated transformation law. [/proofplan] [step:Define the weighted pullback between the two Bergman spaces] Let \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*} and define $A^2(\widetilde{\Omega})$ analogously. Define the weighted pullback map \begin{align*} U:A^2(\widetilde{\Omega})&\to A^2(\Omega)\\ g&\mapsto Ug, \end{align*} where \begin{align*} Ug:\Omega&\to\mathbb{C}\\ z&\mapsto g(F(z))J_F(z). \end{align*} Since $F$ is holomorphic, $g\circ F$ is holomorphic, and since $J_F:\Omega\to\mathbb{C}$ is holomorphic, $Ug$ is holomorphic on $\Omega$. [/step] [step:Use the holomorphic change of variables formula to prove that the pullback is unitary] For $g\in A^2(\widetilde{\Omega})$, the holomorphic change-of-variables formula for the biholomorphism $F$ gives \begin{align*} \int_\Omega |g(F(z))|^2 |J_F(z)|^2\,d\mathcal{L}^{2n}(z) = \int_{\widetilde{\Omega}} |g(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta). \end{align*} Thus $\|Ug\|_{A^2(\Omega)}=\|g\|_{A^2(\widetilde{\Omega})}$. Let $G:\widetilde{\Omega}\to\Omega$ denote the inverse biholomorphism $G:=F^{-1}$. Define \begin{align*} V:A^2(\Omega)&\to A^2(\widetilde{\Omega})\\ f&\mapsto Vf, \end{align*} where \begin{align*} Vf:\widetilde{\Omega}&\to\mathbb{C}\\ \zeta&\mapsto f(G(\zeta))J_G(\zeta). \end{align*} For $f\in A^2(\Omega)$, the function $f\circ G$ is holomorphic because $G$ is holomorphic, and $J_G:\widetilde{\Omega}\to\mathbb{C}$ is holomorphic. Applying the same holomorphic change-of-variables formula to the biholomorphism $G$ gives \begin{align*} \int_{\widetilde{\Omega}} |f(G(\zeta))|^2 |J_G(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) = \int_\Omega |f(\eta)|^2\,d\mathcal{L}^{2n}(\eta) <\infty. \end{align*} Thus $Vf\in A^2(\widetilde{\Omega})$, so $V$ is a well-defined map from $A^2(\Omega)$ to $A^2(\widetilde{\Omega})$. For every $z\in\Omega$, the complex chain rule applied to $G\circ F=\operatorname{id}_\Omega$ gives \begin{align*} J_G(F(z))J_F(z)=1. \end{align*} For every $\zeta\in\widetilde{\Omega}$, the same argument applied to $F\circ G=\operatorname{id}_{\widetilde{\Omega}}$ gives \begin{align*} J_F(G(\zeta))J_G(\zeta)=1. \end{align*} Therefore $VU=\operatorname{id}_{A^2(\widetilde{\Omega})}$ and $UV=\operatorname{id}_{A^2(\Omega)}$. Hence $U$ is a surjective isometry, so $U$ is unitary. Here we use the standard holomorphic change-of-variables formula, whose real Jacobian factor is $|J_F|^2$ for a biholomorphic map $F$; citing a result not yet in the wiki: Holomorphic Change-of-Variables Formula. [guided] The purpose of the Jacobian factor in the definition of $U$ is exactly to compensate for the way Lebesgue measure transforms under a biholomorphic change of variables. For $g\in A^2(\widetilde{\Omega})$, the function $Ug$ is holomorphic because it is the product of the [holomorphic function](/page/Holomorphic%20Function) $g\circ F:\Omega\to\mathbb{C}$ and the [holomorphic function](/page/Holomorphic%20Function) $J_F:\Omega\to\mathbb{C}$. We now check the $L^2$ norm. Since $F$ is a biholomorphic map, it is a real $C^1$ diffeomorphism from $\Omega\subset\mathbb{R}^{2n}$ onto $\widetilde{\Omega}\subset\mathbb{R}^{2n}$. Its real Jacobian determinant has absolute value $|J_F(z)|^2$. The holomorphic change-of-variables formula therefore gives \begin{align*} \|Ug\|_{A^2(\Omega)}^2 &= \int_\Omega |Ug(z)|^2\,d\mathcal{L}^{2n}(z)\\ &= \int_\Omega |g(F(z))|^2 |J_F(z)|^2\,d\mathcal{L}^{2n}(z)\\ &= \int_{\widetilde{\Omega}} |g(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)\\ &= \|g\|_{A^2(\widetilde{\Omega})}^2. \end{align*} This proves that $U$ preserves norms. To prove that $U$ is onto, let $G:\widetilde{\Omega}\to\Omega$ be the inverse biholomorphism $G:=F^{-1}$, and define \begin{align*} V:A^2(\Omega)&\to A^2(\widetilde{\Omega})\\ f&\mapsto Vf \end{align*} by \begin{align*} Vf:\widetilde{\Omega}&\to\mathbb{C}\\ \zeta&\mapsto f(G(\zeta))J_G(\zeta). \end{align*} We must check that this formula really defines an element of $A^2(\widetilde{\Omega})$. First, $f\circ G$ is holomorphic because $f$ and $G$ are holomorphic, and $J_G:\widetilde{\Omega}\to\mathbb{C}$ is holomorphic by the holomorphicity of the complex derivative matrix of $G$. Hence $Vf$ is holomorphic on $\widetilde{\Omega}$. Second, applying the holomorphic change-of-variables formula to the biholomorphism $G$ gives \begin{align*} \int_{\widetilde{\Omega}} |Vf(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) &= \int_{\widetilde{\Omega}} |f(G(\zeta))|^2 |J_G(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)\\ &= \int_\Omega |f(\eta)|^2\,d\mathcal{L}^{2n}(\eta)\\ &<\infty. \end{align*} Thus $Vf\in A^2(\widetilde{\Omega})$ for every $f\in A^2(\Omega)$. This well-definedness check is needed before the identities $VU=\operatorname{id}$ and $UV=\operatorname{id}$ can be interpreted as identities between Bergman spaces. The complex chain rule applied to $G\circ F=\operatorname{id}_\Omega$ gives \begin{align*} J_G(F(z))J_F(z)=1 \end{align*} for every $z\in\Omega$, while the chain rule applied to $F\circ G=\operatorname{id}_{\widetilde{\Omega}}$ gives \begin{align*} J_F(G(\zeta))J_G(\zeta)=1 \end{align*} for every $\zeta\in\widetilde{\Omega}$. Hence, for $g\in A^2(\widetilde{\Omega})$ and $\zeta\in\widetilde{\Omega}$, \begin{align*} (VUg)(\zeta) &= (Ug)(G(\zeta))J_G(\zeta)\\ &= g(F(G(\zeta)))J_F(G(\zeta))J_G(\zeta)\\ &= g(\zeta). \end{align*} Similarly, for $f\in A^2(\Omega)$ and $z\in\Omega$, \begin{align*} (UVf)(z) &= (Vf)(F(z))J_F(z)\\ &= f(G(F(z)))J_G(F(z))J_F(z)\\ &= f(z). \end{align*} Thus $V=U^{-1}$. Since $U$ is both norm-preserving and surjective, it is unitary. Here we use the standard holomorphic change-of-variables formula, whose real Jacobian factor is $|J_F|^2$ for a biholomorphic map $F$; citing a result not yet in the wiki: Holomorphic Change-of-Variables Formula. [/guided] [/step] [step:Transport the reproducing vector at $F(z)$ to a reproducing vector at $z$] For a domain $D\subset\mathbb{C}^n$ and a point $a\in D$, define the reproducing vector \begin{align*} k_a^D:D&\to\mathbb{C}\\ \xi&\mapsto K_D(\xi,a). \end{align*} Then $k_a^D\in A^2(D)$ and, by the defining reproducing property, \begin{align*} h(a)=\int_D h(\xi)\,\overline{k_a^D(\xi)}\,d\mathcal{L}^{2n}(\xi) \end{align*} for every $h\in A^2(D)$. Fix $z\in\Omega$. Define \begin{align*} q_z:\Omega&\to\mathbb{C}\\ w&\mapsto J_F(w)K_{\widetilde{\Omega}}(F(w),F(z))\overline{J_F(z)}. \end{align*} Equivalently, \begin{align*} q_z=\overline{J_F(z)}\,U k_{F(z)}^{\widetilde{\Omega}}. \end{align*} Since $U k_{F(z)}^{\widetilde{\Omega}}\in A^2(\Omega)$, we have $q_z\in A^2(\Omega)$. Let $f\in A^2(\Omega)$. Since $U$ is surjective, there exists $g\in A^2(\widetilde{\Omega})$ such that $f=Ug$. Because the $A^2$ inner products are linear in the first argument and conjugate-linear in the second argument, \begin{align*} \int_\Omega f(w)\,\overline{q_z(w)}\,d\mathcal{L}^{2n}(w) &= \left(f,q_z\right)_{A^2(\Omega)}\\ &= \left(Ug,\overline{J_F(z)}\,U k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\Omega)}\\ &= J_F(z)\left(Ug,U k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\Omega)}\\ &= J_F(z)\left(g,k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\widetilde{\Omega})}\\ &= J_F(z)g(F(z))\\ &= f(z). \end{align*} Thus $q_z$ is a reproducing vector for evaluation at $z$ in $A^2(\Omega)$. [guided] We want to identify the Bergman kernel vector at $z\in\Omega$. The unitary operator $U$ moves functions from $A^2(\widetilde{\Omega})$ to $A^2(\Omega)$, so the natural candidate is the image under $U$ of the kernel vector at the corresponding point $F(z)\in\widetilde{\Omega}$. The only point to watch is the scalar factor, because our [Hilbert space](/page/Hilbert%20Space) inner product is linear in the first argument and conjugate-linear in the second. For a domain $D\subset\mathbb{C}^n$ and a point $a\in D$, define \begin{align*} k_a^D:D&\to\mathbb{C}\\ \xi&\mapsto K_D(\xi,a). \end{align*} The reproducing property says that for every $h\in A^2(D)$, \begin{align*} h(a)=\left(h,k_a^D\right)_{A^2(D)} = \int_D h(\xi)\,\overline{k_a^D(\xi)}\,d\mathcal{L}^{2n}(\xi). \end{align*} Fix $z\in\Omega$. Define \begin{align*} q_z:\Omega&\to\mathbb{C}\\ w&\mapsto J_F(w)K_{\widetilde{\Omega}}(F(w),F(z))\overline{J_F(z)}. \end{align*} This is exactly \begin{align*} q_z=\overline{J_F(z)}\,U k_{F(z)}^{\widetilde{\Omega}}, \end{align*} because \begin{align*} (U k_{F(z)}^{\widetilde{\Omega}})(w) = k_{F(z)}^{\widetilde{\Omega}}(F(w))J_F(w) = K_{\widetilde{\Omega}}(F(w),F(z))J_F(w). \end{align*} Since $k_{F(z)}^{\widetilde{\Omega}}\in A^2(\widetilde{\Omega})$ and $U$ maps $A^2(\widetilde{\Omega})$ into $A^2(\Omega)$, the function $q_z$ belongs to $A^2(\Omega)$. Now take an arbitrary $f\in A^2(\Omega)$. Since $U$ is onto, there exists $g\in A^2(\widetilde{\Omega})$ with $f=Ug$. We compute the inner product of $f$ with $q_z$: \begin{align*} \left(f,q_z\right)_{A^2(\Omega)} &= \left(Ug,\overline{J_F(z)}\,U k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\Omega)}\\ &= J_F(z)\left(Ug,U k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\Omega)}. \end{align*} The scalar changes from $\overline{J_F(z)}$ to $J_F(z)$ because the inner product is conjugate-linear in its second argument. Since $U$ is unitary, \begin{align*} \left(Ug,U k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\Omega)} = \left(g,k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\widetilde{\Omega})}. \end{align*} Using the reproducing property in $A^2(\widetilde{\Omega})$, we get \begin{align*} \left(g,k_{F(z)}^{\widetilde{\Omega}}\right)_{A^2(\widetilde{\Omega})} = g(F(z)). \end{align*} Therefore \begin{align*} \left(f,q_z\right)_{A^2(\Omega)} = J_F(z)g(F(z)). \end{align*} But $f=Ug$, so by the definition of $U$, \begin{align*} f(z)=(Ug)(z)=g(F(z))J_F(z). \end{align*} Hence \begin{align*} \int_\Omega f(w)\,\overline{q_z(w)}\,d\mathcal{L}^{2n}(w) = \left(f,q_z\right)_{A^2(\Omega)} = f(z). \end{align*} Thus $q_z$ reproduces the value of every function $f\in A^2(\Omega)$ at $z$. [/guided] [/step] [step:Use uniqueness of reproducing vectors to identify the kernel] The Bergman kernel vector at $z$ is unique. Indeed, if $r_z,s_z\in A^2(\Omega)$ both satisfy \begin{align*} f(z)=\int_\Omega f(w)\,\overline{r_z(w)}\,d\mathcal{L}^{2n}(w) = \int_\Omega f(w)\,\overline{s_z(w)}\,d\mathcal{L}^{2n}(w) \end{align*} for every $f\in A^2(\Omega)$, then \begin{align*} \left(f,r_z-s_z\right)_{A^2(\Omega)}=0 \end{align*} for every $f\in A^2(\Omega)$. Taking $f=r_z-s_z$ gives \begin{align*} \|r_z-s_z\|_{A^2(\Omega)}^2=0, \end{align*} so $r_z=s_z$ in $A^2(\Omega)$. Since $q_z$ reproduces evaluation at $z$, uniqueness gives \begin{align*} q_z=k_z^\Omega. \end{align*} Evaluating both sides at $w\in\Omega$ yields \begin{align*} K_\Omega(w,z) = J_F(w)K_{\widetilde{\Omega}}(F(w),F(z))\overline{J_F(z)}. \end{align*} Finally, replacing $(w,z)$ by $(z,w)$ gives \begin{align*} K_\Omega(z,w) = J_F(z)K_{\widetilde{\Omega}}(F(z),F(w))\overline{J_F(w)}. \end{align*} This is the desired transformation law. [/step]