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Linear isometric invariants of bounded domains Fusheng Denga, Jiafu Ningb, Zhiwei Wangc, Xiangyu Zhoudea a School of Mathematical Sciences, University of the Chinese Academy of Sciences, Beijing, P. R. China b School of Mathematics and Statistics, HNP-LAMA, Central South University, Changsha, Hunan, P. R. China c Laboratory of Mathematics and Complex Systems (Ministry of Education), School of Mathematical Sciences, Beijing Normal University, Beijing, P. R. China d Institute of Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing, P. R. China e Hua Loo-Keng Key Laboratory of Mathematics, Chinese Academy of Sciences, Beijing, P. R. China
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 4, DOI: https://doi.org/10.4213/im9542 
Bibliographic databases:
  § 1. IntroductionIt is known that two pseudoconvex domains (or even Stein manifolds) $D_1$ and $D_2$ are biholomorphic if and only if $\mathcal O(D_1)$ and $\mathcal O(D_2)$, the spaces of holomorphic functions, are isomorphic as $\mathbb{C}$-algebras with unit. This implies that the holomorphic structure of a pseudoconvex domain is uniquely determined by the algebraic structure of the space of holomorphic functions on the domain. In the above result, the multiplicative structure on the spaces of holomorphic functions plays an essential role. In the present work, we prove some results in a related but different direction. Let $ D$ be a domain in $\mathbb C^n$. Let $z=(z_1,\dots,z_n)$ be the natural holomorphic coordinates of $\mathbb C^n$, and let $d\lambda_n:=(i/2)^n\, dz_1\wedge d\overline{z}_1\wedge\dots\wedge dz_n\wedge d\overline{z}_n$ be the canonical volume form on $ D$. For $p>0$, we denote by $A^p(D)$ the space of all holomorphic functions $\phi$ on $D$ with finite $L^p$-norm
$$
\begin{equation*}
\|\phi\|_p:=\biggl(\int_ D |\phi|^p \, d\lambda_n\biggr)^{1/p}.
\end{equation*}
\notag
$$
It is a standard fact that, for $p\geqslant 1$, $A^p(D)$ are separable Banach spaces, and for $0<p<1$, $A^p(D)$ are complete separable metric spaces with respect to the metric The $p$-Bergman kernel is defined as follows:
$$
\begin{equation*}
B_{D,p}(z):=\sup_{\phi\in A^p(D)}\frac{|\phi(z)|^2}{\|\phi\|^2_p}.
\end{equation*}
\notag
$$
For $p=2$, $B_{D,p}$ is the ordinary Bergman kernel. By a standard argument of the Montel theorem, one can prove that $B_{D,p}$ is a continuous plurisubharmonic function on $D$. We say that $B_{D,p}$ is exhaustive if for any real number $c$ the set $\{z\in D\mid B_{D,p}(z)\leqslant c\}$ is compact. A bounded domain $ D$ is called hyperconvex if there is a plurisubharmonic function $\rho\colon D\to [-\infty, 0)$ such that, for any $c<0$, the set $\{z\in D\mid \rho(z)\leqslant c\}$ is compact. In this paper, instead of the $\mathbb{C}$-algebra $\mathcal O(D)$, we will consider the space $A^p( D)$ (for some fixed $p>0$) as linear invariants of bounded domains. The main purpose is to prove, under certain conditions, that two bounded domains $D_1$ and $D_2$ are holomorphically equivalent if there exists a linear isometry $T\colon A^p(D_1)\to A^p(D_2) $ between $A^p(D_1)$ and $A^p(D_2)$, that is, $T$ is a linear isomorphism and $\|T(\phi)\|_p=\|\phi\|_p$ for all $\phi\in A^p( D_1)$. Note that $A^p(D)$ carries a norm structure (a geometric structure) but there is no natural multiplicative structure on it. In some sense, the norm structure on $A^p(D)$ plays a similar role to the multiplicative structure on $\mathcal O(D)$ in our consideration. The following is proved in [5]. Theorem 1.1. Let $ D_1\subset\mathbb{C}^n$ and $ D_2\subset\mathbb{C}^m$ be hyperconvex bounded domains. Suppose that there exists $p>0$, $p\neq 2,4,6,\dots$, such that (1) there is a linear isometry $T\colon A^p(D_1)\to A^p(D_2)$, and (2) the $p$-Bergman kernels of $D_1$ and $D_2$ are exhaustive. Then $m=n$, and there exists a unique biholomorphic map $F\colon D_1\to D_2$ such that where $J_F$ is the holomorphic Jacobian of $F$. If $n=1$, the assumption of hyperconvexity can be dropped.For the case that $n=m=1$ and $p=1$, Theorem 1.1 was proved by Lakic [8] and Markovic in [9]. An application of spaces of pluricanonical forms with pseudonorms to birational geometry of projective algebraic manifolds was proposed by Chi and Yau in [4] and was studied in [1], [3], [13]. A relative version of Theorem 1.1 was established by Inayama [6]. The purpose of the present paper is to generalize Theorem 1.1 to more general domains. The first observation is to replace the $p$-Bergman kernel exhaustion condition in Theorem 1.1 by the new $A^p$-completeness condition. Definition 1.1. A bounded domain $D\subset\mathbb{C}^n$ is $A^p$-complete if there does not exist a domain $\widetilde D$ with $D\subsetneqq \widetilde D$ such that the restriction map $i\colon A^p(\widetilde{D})\to A^p(D)$ is a linear isometry. It is obvious that if the $p$-Bergman kernel of $D$ is exhaustive, then $D$ is $A^p$-complete. But the converse is not true in general. In fact, $D$ is $A^p$-complete for any $p>0$ if $\mathring{\overline{D}}= D$, namely, the interior of the closure of $D$ is $D$ itself. Note that $D$ must be pseudoconvex if the $p$-Bergman kernel on $D$ is exhaustive, so the condition of $A^p$-completeness is much weaker than the $p$-Bergman exhaustion condition. It was proved in [10] that, for $0<p<2$, the $p$-Bergman kernel on any bounded pseudoconvex domain is exhaustive, so such domains are $A^p$-complete for all $p\in (0,2)$. In our discussion, the importance of the property of $A^p$-completeness is encoded in the following result. For $A^p$-complete domains $D_1$ and $D_2$, if there is a linear isometry from $A^p(D_1)$ to $A^p(D_2)$, then $D_1$ and $D_2$ are almost holomorphically equivalent in the sense that there exist hypersurfaces $A_1$ and $A_2$ in $D_1$ and $D_2$, respectively, such that $D_1\setminus A_1$ and $D_2\setminus A_2$ are holomorphically equivalent. (See Theorem 4.1 for details.) Our second observation is to replace the hyperconvexity condition in Theorem 1.1 by a weaker new condition called of not being of boundary blow down type (BBDT for short, see § 5 for definition). The main result of the present paper is the following. Theorem 1.2. Assume that $D_1$ and $D_2$ are $A^p$-complete domains which are not of boundary blow down type, and $T\colon A^p(D_1)\to A^p(D_2)$ is a linear isometry, for some non-even integer $p>0$. Then there exists a unique biholomorphic map $F\colon D_1\to D_2$ such that $J_F(z)^{2/p}$ has a single-valued branch on $D_1$, and where $\lambda$ is a constant with $|\lambda|=1$. For $p\in (0,2)$, we believe that the non-BBDT condition in the above theorem is not necessary, namely, we pose the following. Conjecture 1.1. Let $D_1$ and $D_2$ be $A^p$-complete bounded domains for some $p\in (0,2)$. If there is a linear isometry between $A^p(D_1)$ and $A^p(D_2)$, then $D_1$ and $D_2$ are holomorphically equivalent. In the last section, we will construct an example to show that this conjecture is not true for $p>2$. Here we point out that a domain $D$ is not of BBDT if $D$ can be represented as $\widetilde D\setminus K$ for some hyperconvex domain $\widetilde D$ and compact subset $K$ (possibly empty) of $\widetilde D$. Note that $D$ can not be pseudoconvex if $K\neq \varnothing$. So, one of the novelties of Theorem 1.2 is that we can handle some domains that are not pseudoconvex, which is a point that is beyond the scope in [5]. It is known that bounded pseudoconvex domains with Hölder boundary are hyperconvex [2]. Of course, such a domain $D$ must satisfy $\mathring{\overline{D}}=D$ and hence is $A^p$-complete. So we get a corollary of Theorem 1.2 as follows, which can not be deduced from Theorem 1.1 if $p>2$, since we do not know if the $p$-Bergman kernels of $D_1$ and $D_2$ are exhaustive. Corollary 1.1. Assume that $D_1$ and $D_2$ are bounded pseudoconvex domains in $\mathbb{C}^n$ with Hölder boundaries. If there exists a linear isometry between $A^p(D_1)$ and $A^p(D_2)$ for some non-even integer $p>0$, then $D_1$ and $D_2$ are holomorphically equivalent. For further study, it is possible to generalize the methods and results in the present paper to complex manifolds equipped with Hermitian holomorphic vector bundles, and to develop some relative version of them. AcknowledgementsThe authors would like to thank Dr. Zhenqian Li for helpful discussions on related topics. § 2. A measure theoretic preparationAs in [9], [5], one of the key tools for our discussions in the present paper is the following result of Rudin. Lemma 2.1 (see [11]). Let $\mu$ and $\nu$ be finite positive measures on two sets $M$ and $N$, respectively. Assume $0<p<\infty$ and $p$ is not even. Let $n$ be a positive integer. If $f_i\in L^p(M,\mu)$, $g_i\in L^p(N,\nu)$ for $1\leqslant i\leqslant n$ satisfy The following lemma is a direct corollary of Lemma 2.1. Lemma 2.2 (see [5], Lemma 2.2). Let $ D_1$ and $ D_2$ be two bounded domains in $\mathbb C^n$ and $\mathbb C^m$, respectively. Suppose that $\phi_k$, $k=0,1,2,\dots,N$, $N\in \mathbb N$, are elements of $A^p( D_1)$, and suppose that $\psi_k$, $k=0,1,2,\dots, N$, are elements of $A^p(D_2)$ such that, for every $N$-tuple of complex numbers $\alpha_k$, $k=1,\dots, N$, If neither $\phi_0$ nor $\psi_0$ is constantly zero, then for every real valued nonnegative Borel function $u\colon \mathbb C^N\to \mathbb C$, § 3. The holomorphic map associated with a linear isometryLet $D$ be a bounded domain in $\mathbb C^n$. For $p>0$, as in the introduction, we denote by $A^p( D)$ the space of all holomorphic functions $\phi$ on $D$ with finite $L^p$-norm
$$
\begin{equation*}
\|\phi\|_p:=\biggl(\int_ D |\phi|^p \, d\lambda_n \biggr)^{1/p}.
\end{equation*}
\notag
$$
For two given bounded domains $D_1\subset\mathbb{C}^n$ and $D_2\subset\mathbb{C}^m$ and a linear isometry $T\colon A^p(D_1)\to A^p(D_2)$ for some $p>0$, we try to construct an associated biholomorphic map $F\colon D_1\to D_2$. By definition, a hyperplane in $A^p(D)$ is the kernel of a non-zero continuous linear functional on $A^p(D)$. For $z\in D$, the evaluation map is a continuous linear functional on $A^p(D)$. So, a natural way to connect points in $D$ and $A^p(D)$ is as follows: any point $z\in D$ corresponds to a hyperplane of $A^p(D)$. Since $D$ is bounded, $A^p(D)$ separates points on $D$ and hence $z$ is uniquely determined by $H_{D,z}$. Let $\mathbb P(A^p(D))$ be the set of all hyperplanes in $A^p(D)$, we then get an injective map
$$
\begin{equation*}
\sigma_D\colon D\to \mathbb P(A^p(D)),\qquad z\mapsto H_{D,z}.
\end{equation*}
\notag
$$
From the linear isometry $T\colon A^p(D_1)\to A^p(D_2)$, we get a natural bijective map
$$
\begin{equation*}
T_*\colon \mathbb P(A^p(D_1))\to \mathbb P(A^p(D_2)),\qquad H\mapsto T(H).
\end{equation*}
\notag
$$
We define subsets $D'_1\subset D_1$ and $D'_2\subset D_2$ as follows: for $z\in D_1$, $w\in D_2$, by definition, we say $z\in D'_1$ if and only if there exists $w\in D_2$ such that $T(H_{D_1,z})=H_{D_2,w}$, namely, $T$ maps the hyperplane in $A^p(D_1)$ associated with $z$ to the hyperplane in $A^p(D_2)$ associated with $w$. Similarly, we say $w\in D'_2$, if and only if there exists $z\in D_1$, such that $T^{-1}(H_{D_2,w})=H_{D_1,z}$. Now we can define a bijective map $F\colon D'_1\to D'_2$ by setting $F(z)=w$ provided $T(H_{D_1,z})=H_{D_2,w}$. In other words, $D'_1=\sigma^{-1}_{D_1}\circ T^{-1}_*\circ\sigma_{D_2}(D_2)$, $D'_2=\sigma^{-1}_{D_2}\circ T_*\circ \sigma_{D_1}(D_1)$, and $F=\sigma^{-1}_{D_2}\circ T_*\circ \sigma_{D_1}$. Other equivalent formulations of $D'_1$, $D'_2$ and $F$ are as follows. Lemma 3.1. For $z\in D_1$, $w\in D_2$, the following conditions are equivalent: (1) $z\in D'_1$, $w\in D'_2$, and $F(z)=w$, (2) $\phi_1(z)T\phi_2(w)=T\phi_1(w)\phi_2(z)$ for all $\phi_1,\phi_2\in A^p(D_1)$, (3) there exists $\lambda\in\mathbb{C}^*$ such that $T\phi(w)=\lambda\phi(z)$ for all $\phi\in A^p(D_1)$. Proof. Let $l_z\colon A^p(D_1)\to\mathbb{C}$ and $l_w\colon A^p(D_2)\to\mathbb{C}$ be the evaluation maps. Then $T_*(H_{D_1,z})=H_{D_2,w}$ if and only if $l_w(T\phi)=\lambda l_z(\phi)$ for some $\lambda\in\mathbb{C}^*$. So (1) implies (3). Other implications are obvious. Lemma is proved. The following lemma is a direct consequence of Lemma 3.1, which implies that $F$ is a closed map. Lemma 3.2. (1) Let $z_j$ be a sequence in $D'_1$ with $z_j\to z\in D_1$ as $j\to \infty$. If $w_j:=F(z_j)$ converges to some $w\in D_2$, then $z\in D'_1$, $w\in D'_2$ and $F(z)=w$. (2) For $z\in D'_1$ and $\phi\in A^p(D_1)$, $\phi(z)=0$ if and only if $T\phi(F(z))=0$. To study $D'_1$, $D'_2$ and $F$ concretely, following the idea in [9], we take an arbitrary countable dense subset of $A^p(D_1)$ and express $F$ in term of it as follows. Let $\{\phi_j\}^{+\infty}_{j=0}$ be a countable dense subset of $A^p(D_1)$ and let $\psi_j=T\phi_j$, then $\{\psi_j\}^{+\infty}_{j=0}$ is a countable dense subset of $A^p(D_2)$. We assume that $\phi_0$ is not identically $0$. Roughly speaking, we may view $[\phi_0(z):\phi_1(z):\dots]$ as the homogenous coordinate on $\mathbb P(A^p(D_1))$, and consider $(\phi_1(z)/\phi_0(z),\phi_2(z)/\phi_0(z),\dots)$ as the inhomogenous coordinate on $\mathbb P(A^p(D_1))$. For $N\in \mathbb N_+$, we define maps $I_N, J_N$ as follows:
$$
\begin{equation*}
\begin{alignedat}{2} I_N &\colon D_1\to\mathbb{C}^N, &\qquad z &\mapsto \biggl(\frac{\phi_1(z)}{\phi_0(z)},\dots, \frac{\phi_N(z)}{\phi_0(z)}\biggr), \\ J_N &\colon D_2\to\mathbb{C}^N, &\qquad w &\mapsto \biggl(\frac{\psi_1(w)}{\psi_0(w)},\dots, \frac{\psi_N(w)}{\psi_0(w)}\biggr), \end{alignedat}
\end{equation*}
\notag
$$
which are measurable maps on $D_1$ and $D_2$, respectively. We can also define $I_\infty$ and $J_\infty$ by
$$
\begin{equation*}
\begin{alignedat}{2} I_\infty &\colon D_1\to\mathbb{C}^\infty, &\qquad z &\mapsto \biggl(\frac{\phi_1(z)}{\phi_0(z)}, \frac{\phi_2(z)}{\phi_0(z)}, \dots\biggr), \\ J_\infty &\colon D_2\to\mathbb{C}^\infty, &\qquad w &\mapsto \biggl(\frac{\psi_1(w)}{\psi_0(w)}, \frac{\psi_2(w)}{\psi_0(w)}, \dots\biggr). \end{alignedat}
\end{equation*}
\notag
$$
Indeed, $I_N$, $I_\infty$ are defined on $D_1\setminus \phi^{-1}_0(0)$, and $J_N$, $J_\infty$ are defined on $D_2\setminus \psi^{-1}_0(0)$. Since $A^p(D_i)$ seperates points in $D_i$, $i=1,2$, the maps $I_\infty$ and $J_\infty$ are injective. By Lemma 3.1, for $z\in D_1\setminus \phi^{-1}_0(0)$ and $w\in D_2\setminus \psi^{-1}_0(0)$, we have $z\in D'_1, w\in D'_2$ and $F(z)=w$ if and only if $I_\infty(z)=J_\infty(w)$. For $z\in D'_1\setminus \phi^{-1}_0(0)$, we have It is clear that
$$
\begin{equation*}
\begin{aligned} \, D'_1\setminus\phi^{-1}_0(0) &=\bigcap_N I^{-1}_N J_N(D_2\setminus\psi_0^{-1}(0)) =I_\infty^{-1}J_\infty( D_2\setminus\psi_0^{-1}(0)), \\ D'_2\setminus\psi^{-1}_0(0) &=\bigcap_N J_N^{-1}I_N(D_1\setminus\phi_0^{-1}(0)) =J_\infty^{-1}I_\infty(D_1\setminus\phi_0^{-1}(0)). \end{aligned}
\end{equation*}
\notag
$$
We now assume that $p>0$ is a real number and $p$ is not an even integer. The following results are proved in § 2.2 of [5]. § 4. $A^p$-completeness of bounded domainsWe first introduce a new notion, namely, $A^p$-completeness of bounded domains. Definition 4.1. Let $D\subset\mathbb{C}^n$ be a bounded domain and $p>0$. We say that $D$ is $A^p$-complete if there does not exist a domain $\widetilde D$ with $D\subsetneqq \widetilde D$ such that the restriction map $i\colon A^p(\widetilde{D})\to A^p(D)$ is an isometry. The following two basic facts related to $A^p$-completeness are obvious: (1) $D$ should be $A^p$-complete for any $p>0$ if $\mathring{\overline{D}}=D$, namely, the interior of the closure of $D$ is $D$ itself; (2) if the $p$-Bergman kernel $B_{D,p}$ of $D$ is exhaustive, then $D$ is $A^p$-complete. According to [10], for $p<2$, the $p$-Bergman kernel on any bounded pseudoconvex domain is exhaustive. Hence all bounded pseudoconvex domains in $\mathbb{C}^n$ are $A^p$-complete for any $0<p<2$. Assume that $D_1$ and $D_2$ are bounded domains in $\mathbb{C}^n$, and $T\colon A^p(D_1)\to A^p(D_2)$ is a linear isometry. We now take a dense countable subset $\{\phi_j\}_{j\geqslant 0}$ of $A^p(D_1)$ such that $\psi_0=T(\phi_0)=1$, $\psi_1=T(\phi_1)=w_1$, $\dots$, $\psi_n=T(\phi_n)=w_n$, and $(w_1,\dots, w_n)$ are the natural linear coordinates on $D_2$, and in the rest of the paper, we always assume $\psi_j=T(\phi_j)$ for all $j\geqslant0$. We denote the zero set $\phi_0^{-1}(0)$ of $\phi_0=T^{-1}(1)\in A^p(D_1)$ by $A_1$. Now it is clear that $F=I_n=(\phi_1(z)/\phi_0(z),\dots, \phi_n(z)/\phi_0(z))$ on $D'_1\setminus A_1$. Proposition 4.1. Let $D_1$ and $D_2$ be bounded domains in $\mathbb{C}^n$, and $T\colon A^p(D_1)\to A^p(D_2)$ be a linear isometry. If $D_2$ is $A^p$-complete, then $D'_1=D_1\setminus A_1$. Proof. We divide the proof into three steps.
Step 1. Prove that $I_n(D_1\setminus A_1)\subset D_2$. By Lemma 3.1, we have
$$
\begin{equation*}
\psi(F(z))=\frac{\phi(z)}{\phi_0(z)}, \qquad z\in D'_1\setminus A_1,
\end{equation*}
\notag
$$
for all $\phi\in A^p(D_1)$ and $\psi= T\phi\in A^p(D_2)$. Since $D_1'$ is dense in $D_1$ by Lemma 3.3 and $I_n=F$ on $D'_1\setminus A_1$, by Lemma 3.6, we get
$$
\begin{equation}
|J_{I_n}(z)|^{2/p}=|\phi_0(z)|,\ z\in D_1\setminus A_1.
\end{equation}
\tag{1}
$$
In particular, $J_{I_n}(z)\neq 0$ for $z\in D_1\setminus A_1$.
We now argue by contradiction. Assume that there exists $z_0\in D_1\setminus A_1$ such that $w_0:=I_n(z_0)\notin D_2$. We have seen that $J_{I_n}(z_0)\neq 0$. So there exists a neighbourhood $U$ of $z_0$ in $D_1\setminus A_1$ and a neighbourhood $V$ of $w_0$ in $\mathbb{C}^n$ such that $I_n(U)=V$ and $I_n|_U\colon U\to V$ is a biholomorphic map. Let $\widetilde D=D_2\cup V$. Note that $I_n$ is smooth, $D'_1\setminus A_1$ has full measure in $D_1$ (by Lemma 3.3), and $I_n(D'_1\setminus A_1)\subset D_2$, it follows that $D_2\varsubsetneqq \widetilde D\subset \overline{D_2}$ and $\widetilde D\setminus D_2$ has null measure, where $\overline{D_2}$ is the closure of $D_2$ in $\mathbb{C}^n$. Let $\psi\in A^p(D_2)$ and $\phi\in A^p(D_1)$ with $\psi=T\phi$. We now continue $\psi$ holomorphically to a holomorphic function $\widetilde \psi$ on $\widetilde D$ as follows. Let $\psi'\in\mathcal O(V)$ be given as Then $\psi=\psi'$ on the dense subset $F(D'_1\setminus A_1)\cap D_2\cap V$ of $D_2\cap V$. Hence $\psi=\psi'$ on $D_2\cap V$. Thus we can glue $\psi$ and $\psi'$ together to get a holomorphic function $\widetilde\psi$ on $\widetilde D$, which is an extension of $\psi$ by definition. Since $\widetilde D\setminus D_2$ has null measure. Hence
$$
\begin{equation*}
\int_{\widetilde D}|\widetilde\psi|^p\,d\lambda_n=\int_{D_2}|\psi|^p\,d\lambda_n.
\end{equation*}
\notag
$$
Note that since $\psi\in A^p(D_2)$ is arbitrary, $D_2$ cannot be $A^p$-complete, a contradiction.
Step 2. Prove that $D_1\setminus A_1\subset D_1'$. Let $z\in D_1\setminus A_1$. Since $D'_1$ is dense in $D_1$, there exists a sequence $\{z_j\}_{j\geqslant 1}$ in $D'_1\setminus A_1$ that converges to $z$. We have $F(z_j)=I_n(z_j)$ for all $j$. By the conclusion in Step 1 and the continuity of $I_n$, we get $F(z_j)\to I_n(z)\in D_2$ as $j\to\infty$. It follows from Lemma 3.2 that $z\in D_1'$. Step 3. Prove that $D'_1=D_1\setminus A_1$. From Step 2, and Riemann’s removable singularity theorem, $F$ extends to a holomorphic map, denoted by $F_1$, from $D_1$ to $\mathbb{C}^n$, with $F(D_1)\subset\overline{D_2}$. By (1) and continuity, we have on $D_1$. It follows that $J_{F_1}(z)=0$ on $A_1$. Hence $A_1\cap D'_1=\varnothing$, and therefore, $D'_1=D_1\setminus A_1$, proving the proposition.As in Proposition 4.1, we assume that $D_1$ and $D_2$ are bounded domains in $\mathbb{C}^n$, and $T\colon A^p(D_1)\to A^p(D_2)$ is a linear isometry. We assume in addition that both $D_1$ and $D_2$ are $A^p$-complete. Let $A_1=(T^{-1}(1))^{-1}(0)=\phi^{-1}_0(0)$ and let $A_2=(T(1))^{-1}(0)$. Exchanging the roles of $D_1$ and $D_2$, we can define a holomorphic map $F_2\colon D_2\to\mathbb{C}^n$ with $F_2(D_2)\subset \overline{D_1}$. For the same reasons we have on $D_2$, and $D'_2=D_2\setminus A_2$.By Lemma 3.2, we get
$$
\begin{equation*}
F_1(A_1)\subset \partial D_2,\qquad F_2(A_2)\subset\partial D_1.
\end{equation*}
\notag
$$
We summarize the above results in the following theorem. Theorem 4.1. Assume that $D_1$ and $D_2$ are $A^p$-complete domains in $\mathbb{C}^n$, and $T\colon A^p(D_1)\to A^p(D_2)$ is a linear isometry. Then: (1) $D'_1=D_1\setminus A_1$ and $D'_2=D_2\setminus A_2$, (2) $F\colon D_1\setminus A_1\to D_2\setminus A_2$ is a biholomorphic map, (3) $|J_{F_1}(z)|^{2/p}=|\phi_0(z)|$ on $D_1$, and $|J_{F_2}(w)|^{2/p}=|T(1)(w)|$ on $D_2$, (4) $F_1(A_1)\subset \partial D_2$ and $F_2(A_2)\subset \partial D_1$. In § 5, we will prove that $A_1$ and $A_2$ are empty under certain additional conditions. § 5. Domains of boundary blow down type (BBDT)We first give an observation to the picture presented in Theorem 4.1. We preserve the notations, definitions, and assumptions as in Theorem 4.1. Gluing $D_1$ and $D_2$ by identifying $z\in D_1\setminus A_1$ and $w=F(z)\in D_2\setminus A_2$, we get a complex manifold, say $D:=D_1\sqcup_F D_2$. For the proof of this statement, we only need to check that $D$ is Hausdorff. It suffices to show that any $z_1\in A_1$ and $z_2\in A_2$ have disjoint neighbourhoods in $D$. The natural inclusions $j_1\colon D_1\to D$ and $j_2\colon D_2\to D$ are open maps. Since $F_1(z_1)\in \partial D_2$ and $F_2(z_2)\in\partial D_1$, there exist neighbourhoods $U_k$ of $z_k$ in $D_k$ for $k=1,2$, such that $F_1(U_1)\cap U_2=\varnothing$ and $F_2(U_2)\cap U_1=\varnothing$, then $j_1(U_1)\cap j_2(U_2)=\varnothing$. We can view $A_1$ and $A_2$ as hypersurfaces of $D$ which are disjoint. We have $J_{F_1}\in\mathcal{O}(D_1)$ and $J_{F_2}\in\mathcal{O}(D_2)$, and $J_{F_1}(z)=1/J_{F_{2}}(w)$ for $z\in D_1\setminus A_1$ with $w=F_1(z)$. So we can define a meromorphic function $J$ on $D$ as follows:
$$
\begin{equation*}
\begin{cases} J(j_1(z))=J_{F_1}(z), &z\in D_1, \\ J(j_2(w))= \dfrac{1}{J_{F_{2}}(w)}, &w\in D_2\setminus A_2. \end{cases}
\end{equation*}
\notag
$$
Hence $J\in\mathcal{M}(D)\cap\mathcal{O}(D\setminus A_2)$ and $J^{-1}(\infty)=A_2$, where $\mathcal M(D)$ denotes the space of meromorphic functions on $D$. We extract an abstract concept from this picture as follows. Definition 5.1. A domain $D\subset\mathbb{C}^n$ is of boundary blow down type (BBDT for short), if there exists a complex manifold $M$, a (non-empty) hypersurface $A\subset M$, $h\in\mathcal{M}(M)\cap\mathcal{O}(M\setminus A)$, and a holomorphic map $\sigma\colon M\to \mathbb{C}^n$ such that (i) $\sigma(M\setminus A)=D$ and $\sigma(A)\subset\partial D$, (ii) $\sigma|_{M\setminus A}\colon M\setminus A\to D$ is a biholomorphic map, (iii) $h^{-1}(\infty)=A$. Let us go back to our previous construction. If $A_2\neq \varnothing$, then $D_1$ is of BBDT. It is not easy to construct a BBDT-domain. We will see an example in § 6. The following notion is also tightly related to our discussion. Definition 5.2. A complex manifold $X$ is said to have punctured disc property (PDP, for short) if there is a proper holomorphic map from $\mathbb D^*$ to $X$, where $\mathbb D^*=\{z\in\mathbb{C}\mid 0<|z|\leqslant 1\}$. Recall that a domain $D\subset\mathbb{C}^n$ is called a Runge domain if $D$ is holomorphically convex and any holomorphic function on $D$ can be approximated uniformly on compact subsets of $D$ by polynomials. Lemma 5.1. A bounded domain $D$ does not have PDP if either $D$ is a Runge domain or $D$ is hyperconvex. Proof. We argue by contradiction. If $D$ has PDP, then there is a proper holomorphic map $h\colon \mathbb D^*\to D$.
If $D$ is a Runge domain, then the polynomial convex hull $\widehat{h(S^1)}$ of $h(S^1)$ lies in $D$. On the other hand, by Riemann’s removable singularity theorem and the maximum principle for holomorphic functions, we see that $h(\mathbb D^*)\subset \widehat{h(S^1)}$, which contradicts to the assumption that $h\colon \mathbb D^*\to D$ is proper. If $D$ is hyperconvex, then there is a strictly plurisubharmonic function $\rho\colon D\to [-\infty, 0)$ such that, for any $c<0$, the set $\{z\in D\mid \rho(z)\leqslant c\}$ is compact. Consider $\widetilde\rho:=\rho\circ h$, which is a subharmonic function on $\mathbb D^*$. Since $\widetilde\rho$ is bounded above, it can be extended to a subharmonic function on $\mathbb D$, which is also denoted by $\widetilde\rho$. It is clear that $\widetilde\rho$ attains its maximum $0$ at the origin, hence $\widetilde\rho$ is a constant function, which contradicts the assumption that $\rho$ is strictly plurisubharmonic. Lemma 5.1 is proved. Theorem 5.1. Let $D\subset\mathbb{C}^n\ (n>1)$ be a bounded domain that does not have PDP, then (i) $D$ is not of BBDT, (ii) $D\setminus K$ is not of BBDT for any compact set $K\subset D$ such that $D\setminus K$ is connected. Proof. The proof of (i) is trivial by definition. To prove (ii) we assume on the contrary that $\widetilde{D}:=D\setminus K$ is of boundary blow down type. By definition, there exists a complex manifold $M$, a (non-empty) hypersurface $A\subset M$, $h\in\mathcal{M}(M)\cap\mathcal{O}(M\setminus A)$, and a holomorphic map $\sigma\colon M\to \mathbb{C}^n$ such that
(i) $\sigma(M\setminus A)=\widetilde D$ and $\sigma(A)\subset\partial\widetilde D$, (ii) $\sigma|_{M\setminus A}\colon M\setminus A\to \widetilde D$ is a biholomorphic map, (iii) $h^{-1}(\infty)=A$. Note that $\partial \widetilde{D}\subset \partial D\cup K$. Since $D$ does not have PDP, we have $\sigma(A)\subset K$. Let $h'=h\circ (\sigma|_{M\setminus A})^{-1}$, then $h'\in\mathcal{O}(\widetilde{D})$. We can extend $h'$ holomorphically to $D$ by Hartog’s theorem. This contradicts the fact that $h^{-1}(\infty)=A$. Theorem 5.1 is proved. Let us go back to the setting of Theorem 4.1, by the above discussion, we see that $A_1$ and $A_2$ are both empty sets, and so we get a biholomorphic map $F\colon D_1\to D_2$, if $D_1$ and $D_2$ are not of BBDT. Combining this with Lemma 3.6, we get the following theorem. Theorem 5.2 (=Theorem 1.2). Assume that $D_1$ and $D_2$ are $A^p$-complete domains which are not of boundary blow down type, and $T\colon A^p(D_1)\to A^p(D_2)$ is a linear isometry, for some $p>0$ and $p$ is not an even integer. Then there exists a unique biholomorphic map $F\colon D_1\to D_2$ such that $J_F(z)^{2/p}$ has a single-valued branch on $D_1$, and where $\lambda$ is a constant with $|\lambda|=1$. Proof of Theorem 1.2. By the above discussion and Lemma 3.6, we get a biholomorphic map $F\colon D_1\to D_2$ such that
$$
\begin{equation*}
|T\phi(F(z))|\, |J_F(z)|^{2/p}=|\phi(z)|\quad \forall\, \phi\in A^p(D_1),\quad z\in D_1.
\end{equation*}
\notag
$$
Considering $\phi_0\in A^p(D_1)$ with $T\phi_0=1\in A^p(D_2)$, we see that Let $B$ be a ball in $D_1$. Then there exists a single-valued branch of $J_F^{2/p}$ on $B$ which is equal to $\lambda\phi_0$ for some $\lambda\in\mathbb{C}$ with $|\lambda|=1$. Note that $D_1$ is connected and $\phi_0$ is globally defined on $D_1$, $\lambda\phi_0$ must be a branch of $J_F^{2/p}$ on $D_1$. The last statement is a direct consequence of Lemma 3.1(2),
$$
\begin{equation*}
\phi_0(z)T\phi(F(z))=T\phi_0(F(z))\phi(z)\quad \forall \phi\in A^p(D_1),\quad z\in D_1.
\end{equation*}
\notag
$$
As $\phi_0(z)=J_F^{2/p}(z)/\lambda$ and $T\phi_0=1$, we have
$$
\begin{equation*}
T\phi(F(z))J_F(z)^{2/p}=\lambda\phi(z)\quad \forall\, \phi\in A^p(D_1),\quad z\in D_1.
\end{equation*}
\notag
$$
This proves Theorem 1.2. § 6. A counter-example to Conjecture 1.1 if $p>2$The aim of this section is to give an example to show that Conjecture 1.1 could be false if we drop the assumption that $p\in(0,2)$. The construction is as follows. Let $k\geqslant 1$ be an integer, and consider the map
$$
\begin{equation*}
f_k\colon \mathbb{C}^2\to\mathbb{C}^2,\quad (z_1,z_2)\mapsto (z_1,z^k_1z_2).
\end{equation*}
\notag
$$
Then $f_k$ gives a biholomorphic map from $\mathbb{C}^*\times \mathbb{C}$ onto $\mathbb{C}^*\times \mathbb{C}$, with inverse given by Let $\mathbb{B}$ be the unit ball in $\mathbb{C}^2$, and $\mathbb{B}'=\mathbb{B}\setminus\{z_1=0\}$. Set
$$
\begin{equation*}
D_1=\mathbb{B}\times f_k(\Delta^*\times \Delta),\qquad D_2=f_k(\mathbb{B}')\times\Delta^2,
\end{equation*}
\notag
$$
where $\Delta$ is the unit disc in $\mathbb{C}$. Note that
$$
\begin{equation*}
f_k(\Delta^*\times \Delta)=\{(w_1,w_2)\colon |w_2|<|w_1|^k<1\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
f_k(\mathbb{B}')=\bigl\{(w_1,w_2)\colon |w_2|<|w_1|^k\sqrt{1-|w_1|^2}\bigr\}.
\end{equation*}
\notag
$$
It follows that $\mathring{\overline{D_1}}=D_1$ and $\mathring{\overline{D_2}}=D_2$. Therefore, both $D_1$ and $D_2$ are $A^p$-complete for any $p>0$. Consider the maps
$$
\begin{equation*}
F\colon D_1\to \mathbb{C}^4,\quad (z_1,z_2,z_3,z_4)\mapsto (w_1,w_2,w_3,w_4):=(z_1,z_1^kz_2,z_3,z_3^{-k}z_4),
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
G\colon D_2\to \mathbb{C}^4,\quad (w_1,w_2,w_3,w_4)\mapsto (z_1,z_2,z_3,z_4):=(w_1,w_1^{-k}w_2,w_3,w_3^kw_4).
\end{equation*}
\notag
$$
Let $D_1'=D_1\setminus\{z_1=0\}$ and $D_2'=D_2\setminus\{w_3=0\}$. Then is biholomorphic, whose inverse is $G|_{D_2'}$. We have
$$
\begin{equation*}
\begin{aligned} \, J_F^{-1}(0) &=\{J_F(z)=(z_1z_3^{-1})^k=0\}=\{z\in D_1\colon z_1=0\}, \\ J_G^{-1}(0) &=\{J_G(w)=(w^{-1}_1w_3)^k=0\}=\{w\in D_2\colon w_3=0\}, \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
F(\{z\in D_1\colon z_1=0\})\subset \partial D_2,\qquad G(\{w\in D_2\colon w_3=0\})\subset \partial D_1.
\end{equation*}
\notag
$$
It follows that $D_1$ and $D_2$ are domains of BBDT (see Definition 5.1). Let $k$ and $m$ be two positive integers, such that $p:=2k/m$ is not an even integer. We can take $J_F^{2/p}(z)=(z_1z_3^{-1})^m$ (respectively, $J_G^{2/p}(w)=(w_1^{-1}w_3)^m$) as a single-valued branch of $J_F^{2/p}$ on $D'_1$ (respectively, $J_G^{2/p}$ on $D'_2$). Then the map
$$
\begin{equation*}
T\colon A^p(D_1')\to A^p(D_2')\colon T(\phi)(w):=\phi(G(w))J_G^{2/p}(w)
\end{equation*}
\notag
$$
is a linear isometry between $A^p(D_1')$ and $A^p(D_2')$. If $p\geqslant 2$, then any $\phi \in A^p(D_j')$ can be extended to a holomorphic function, say $\widetilde{\phi}$, on $D_j$ with
$$
\begin{equation*}
\int_{D_j'}|\phi|^p\,d\lambda_4=\int_{D_j}|\widetilde\phi|^p\,d\lambda_4,\qquad j=1,2.
\end{equation*}
\notag
$$
Therefore we can identify $A^p(D_j')$ with $A^p(D_j)$ for $j=1,2$, and view $T$ as a linear isometry from $ A^p(D_1)$ to $ A^p(D_2)$. It follows that the Conjecture 1.1 is false in general for $p>2$. We now prove the claim. It suffices to show that the dimension of the automorphism groups $\operatorname{Aut}(D_1)$ and $\operatorname{Aut}(D_2)$ are not equal. By definition, we have
$$
\begin{equation*}
D_1\cong \mathbb B\times \Delta^*\times\Delta,\qquad D_2\cong\mathbb B'\times\Delta\times\Delta.
\end{equation*}
\notag
$$
By Corollary (5.4.11) in [7], we need to show
$$
\begin{equation}
\begin{aligned} \, &\dim \operatorname{Aut}(\mathbb B)+ \dim \operatorname{Aut}(\Delta^*)+ \dim \operatorname{Aut}(\Delta) \nonumber \\ &\qquad\neq \dim \operatorname{Aut}(\mathbb B') + \dim \operatorname{Aut}(\Delta)+\dim \operatorname{Aut}(\Delta). \end{aligned}
\end{equation}
\tag{2}
$$
The left-hand side of (2) is $12$ (for the reference on the automorphism group, see [12]), and the right-hand side of (2) is $ \dim \operatorname{Aut}(\mathbb B')+6$. So it suffices to show that $ \dim \operatorname{Aut}(\mathbb B')<6$. By Riemann’s removable singularity theorem, we have
$$
\begin{equation*}
\dim \operatorname{Aut}(\mathbb B')=\{f\in \operatorname{Aut}(\mathbb B)\mid f(H)=H\},
\end{equation*}
\notag
$$
where $H=\{(z,w)\in \mathbb B\mid z=0\}$. Note that a unitary map preserving $H$ must be diagonal, and a map $f\in \operatorname{Aut}(\mathbb B)$ preserving $H$ must map the origin to some point in $H$. Hence proving the claim.
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\Bibitem{DenNinWan24} |
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