$\mathbb{C}^{n}$中一个域$D$的自同构群是$D$上的双全纯映射的集合.自同构群在多复变函数论的研究中已成为一个非常有力的工具.在文[7]中, Thullen研究了所谓的Thullen域

$\mathbb{E}^{\mu}:=\{(z, \zeta)\in\mathbb{C}^{2}:|z|^{2}+|\zeta|^{2\mu}<1\}, $

其中$\mu>1$, 并得到${\rm Aut}(\mathbb{E}^{\mu})$是下列所有映射构成的集合:

$(z, \zeta)\longmapsto\left(e^{i\theta_{1}}\frac{z-a}{1-\overline{a}z}, e^{i\theta_{2}}\frac{(1-|a|^{2})^{1/2\mu}} {(1-\overline{a}z)^{1/\mu}}\zeta \right), $

其中$a\in\mathbb{C}$且$|a|<1$, $\theta_{1}, \theta_{2}\in\mathbb{R}$.同样在文[2, 3]中, Bedford和Pinchuk证明了如下定理:如果$D\subset\mathbb{C}^{2}$是一个带有实解析边界的有界(拟凸)域且它的自同构群是非紧的, 那么$D$双全纯等价于$\mathbb{E}^{\mu}$ (其中$\mu$为整数).

有界对称域代表了全体非紧型的Hermitian对称空间.每一个对称有界域, 都有一个对应的一般范数$N_{\Omega}(z, z)$ (generic norm), 满足对任意的$z\in\Omega$都有$0＜N_{\Omega}(z, z)\leq 1$并且$N_{\Omega}(z, z)=1$当且仅当$z=0$. $N_{\Omega}(z, z)$与$\Omega$的Bergman核$K_{\Omega}(z, z)$有如下关系: $N_{\Omega}(z, z)=(V(\Omega)K_{\Omega}(z, z))^{-\frac{1}{\gamma}}$, 其中$\gamma$为$\Omega$的亏格而$V(\Omega)$为$\Omega$在Lebsgue测度下的体积.对一个不可分解的有界对称域$\Omega$, 定义Hartogs型域

$\begin{equation}\label{1} \widehat{\Omega}_{m}:=\{(z, \zeta)\in\Omega\times\mathbb{C}^{m}:\|\zeta\|^{2\mu}＜N_{\Omega}(z, z)\}. \end{equation}$

(1.1)

如果$\Omega$是单位圆盘且$m=1$, 那么该Hartogs型域就变成了Thullen域.

设$G$由具有下列形式的的映射$\Phi$生成

$\begin{equation}\label{2} \Phi(z, \zeta)=(\Phi_{1}(z, \zeta), \Phi_{2}(z, \zeta)), \;\; (z, \zeta)\in\widehat{\Omega}_{m}, \end{equation}$

(1.2)

其中$\Phi_{1}(z, \zeta)=\varphi(z)\in {\rm Aut}(\Omega)$且

$\Phi_{2}(z, \zeta)=U(\zeta)\frac{N_{\Omega}(z_{0}, z_{0})^{1/2\mu}}{N_{\Omega}(z, z_{0})^{1/\mu}}, \; z_{0}=\varphi^{-1}(0), \; U\in \mathcal{U}(m), $

这里$\mathcal{U}(m)$是一个$m$阶酉矩阵.论文[10]证明了$G$为${\rm Aut}(\widehat{\Omega}_{m})$的一个子群.文[1]给出了如下结果:

定理1.1 设$\widehat{\Omega}_{m}$是一典型域$\Omega$上的Hartogs型域.假设$\widehat{\Omega}_{m}$不是单位球, 那么$G$即为${\rm Aut}(\widehat{\Omega}_{m})$.

最近, 涂振汉和王磊^[9]研究了华域之间的逆紧全纯映照的刚性问题, 并且得到了全部华域的自同构群, 作为其特例得到了全部不可分解的有界对称域$\Omega$上的Hartogs型域$\widehat{\Omega}_{m}$的自同构群.

作为不可分解的有界对称域上的Hartogs型域的一种自然推广, 我们考虑一类稍微广泛的Hartogs型域.设$\Omega: =\Omega_{1}\times\Omega_{2}$ (其中$\Omega_{1}$和$\Omega_{2}$都是不可分解的有界对称域), 定义

$\begin{equation}\label{3} \widetilde{\Omega}_{m}:=\{(z, \eta, \zeta)\in \Omega_{1}\times\Omega_{2} \times\mathbb{C}^{m}\subset\mathbb{C}^{d_{1}}\times\mathbb{C}^{d_{2}}\times\mathbb{C}^{m}:\;\; \|\zeta\|^{2\mu}＜N_{\Omega}(z, z; \eta, \eta)\}, \end{equation}$

(1.3)

其中$N_{\Omega}(z, z;\eta, \eta):=N_{\Omega_1}(z, z)N_{\Omega_2}(\eta, \eta)$.

设集合$\widetilde{G}$由下列形式的映射生成

$\begin{equation}\label{4} \Phi(z, \eta, \zeta)=(\Phi_{1}(z, \eta, \zeta), \Phi_{2}(z, \eta, \zeta), \Phi_{3}(z, \eta, \zeta)), \;\;(z, \eta, \zeta)\in\widetilde{\Omega}_{m}, \end{equation}$

(1.4)

其中$\Phi_{1}(z, \eta, \zeta)=\phi(z)\in {\rm Aut}(\Omega_{1}), \Phi_{2}(z, \eta, \zeta)=\psi(z)\in {\rm Aut}(\Omega_{2})$且

$\Phi_{3}(z, \eta, \zeta)=U(\zeta)\frac{N_{\Omega}(z_{0}, z_{0};\eta_{0}, \eta_{0})^{1/2\mu}}{N_{\Omega}(z, z_{0};\eta, \eta_{0})^{1/\mu}}, \;\;z_{0}=\phi_{1}^{-1}(0), \;\; \eta_{0}=\psi_{2}^{-1}(0), \;\; U\in \mathcal{U}(m), $

这里$\mathcal{U}(m)$是一个$m$阶酉矩阵.

应用涂振汉和王磊^[9]的推理, 本文给出了如下结果:

定理1.2 假定$\widetilde{\Omega}_{m}$不是单位球, 那么集合$\widetilde{G}$即为${\rm Aut}(\widetilde{\Omega}_{m})$.

定理1.2的证明类似于定理1.1.我们的证明主要分为两步:首先需要证明${\rm Aut}(\widetilde{\Omega}_{m})$的每个元素保持零截面$\Omega\times\{0\}$, 此处用到了$\widetilde{\Omega}_{m}$中只有边界部分$b_{0}\widetilde{\Omega}_{m}$是强拟凸的(其中$b_{0}\widetilde{\Omega}_{m}$是强拟凸的证明来源于文献[9]的推理), $\widetilde{\Omega}_{m}$是齐性域当且仅当$\widetilde{\Omega}_{m}$是单位球; 其次使用Cartan定理来得到在原点的迷向自同构必为线性形式.这就完成了定理1.2的证明.相关的研究也可见文献[4, 6, 8].

按照华罗庚[5]的定义列出四类典型域$\Omega$和相应的一般范数$N_{\Omega}(z, w)$如下:

(ⅰ)类型$I_{p, q}(1\leq p\leq q):$设$V$是由$p\times q$复矩阵构成的空间,

$\Omega=\{z\in V: \; I-z\overline{z}^{t}>0\}, \; N_{\Omega}(z, w)=\det(I-z\overline{w}^{t}), $

这里记号$z>0$表示方阵$z$是正定的, 且$I$表示单位矩阵.

(ⅱ)类型$II_{n}:$设$V$是由反对称$n\times n$复矩阵构成的空间,

$\Omega=\{z\in V:\; I+z\overline{z}>0\}, \; N_{\Omega}(z, w)^{2}=\det(I+z\overline{w}^{t}).$

(ⅲ)类型$III_{n}:$设$V$是由对称$n\times n$复矩阵构成的空间,

$\Omega=\{{z\in V:\; I-z\overline{z}>0}\}, \; N_{\Omega}(z, w)=\det(I-z\overline{w}^{t}).$

(ⅳ)类型$IV_{n}:$

$\begin{eqnarray*} &&\Omega=\{z=(z_{1}, \cdots, z_{n})\in \mathbb{C}^{n}:\; 1-2Q(z, \overline{z})+|Q(z, z)|^{2}>0, \; Q(z, z)＜1\}, \\ &&N_{\Omega}(z, w)=1-2Q(z, \overline{w})+Q(z, z)Q(\overline{w}, \overline{w}), \end{eqnarray*}$

其中$Q(z, w)=\sum\limits_{j=1}^{n}z_{j}w_{j}.$另外, 不可分解的有界对称域还包含两个例外域.在本文中, $\Omega_{i} (i=1, 2)$总表示不可分解的有界对称域中的一种且用$N_{i}(z, w)$来表示$N_{\Omega_{i}}(z, w)$.

下面的命题2.1的证明中的关键部分是自同构群下的下列一般范数的变换公式(见文[10]).设$\Omega$为一个不可分解的有界对称域, $\gamma$是$\Omega$的亏格且$J\phi(z)$是$\phi$的Jacabian行列式.对$\phi\in {\rm Aut}(\Omega)$和$z, t\in \Omega, $总成立$N(\phi(z), \phi(t))^{\gamma}=J\phi(z)N(z, t)^{\gamma}\overline{J\phi(t)}, $并且如果$z_{0}=\phi^{-1}(0), $则有$\frac{N(\phi(z), \phi(z))}{N(z, z)}=\frac{N(z_{0}, z_{0})}{|N(z, z_{0})|^{2}}.$

命题2.1  $\widetilde{G}$是${\rm Aut}(\widetilde{\Omega}_{m})$的子群.

证 从(1.4) 式中容易证明$\widetilde{G}$的每个元素是$\widetilde{\Omega}_{m}$的自同构.为了证明$\widetilde{G}$是一个群, 将证$\widetilde{\Phi}\circ\Phi\in\widetilde{G}$, 这里$\Phi$是由(1.4) 式定义且$\widetilde{\Phi}$具有下列形式

$\begin{eqnarray*} &&\widetilde{\Phi}(z, \eta, \zeta)=(\widetilde{\phi}(z), \widetilde{\psi}(\eta), \widetilde{U}(\zeta)\frac{N(\widetilde{z}_{0}, \widetilde{z}_{0};\widetilde{\eta}_{0}, \widetilde{\eta}_{0})^{1/2\mu}}{N(z, \widetilde{z}_{0};\eta, \widetilde{\eta}_{0})^{1/\mu}}), \\ &&\widetilde{\phi}\in {\rm Aut}(\Omega_{1}), \widetilde{\psi}\in {\rm Aut}(\Omega_{2}), \; \widetilde{z}_{0}=\widetilde{\phi}^{-1}(0), \; \widetilde{\eta}_{0}= \widetilde{\psi}^{-1}(0), \; \widetilde{U}\in \mathcal{U}(m), \end{eqnarray*}$

继而断言存在$z_{1}:=(\widetilde{\phi}\circ\phi)^{-1}(0), \; \eta_{1}:=(\widetilde{\psi}\circ\psi)^{-1}(0)$且$U_{1}\in \mathcal{U}(m)$满足

$\begin{equation}\label{11} \widetilde{\Phi}\circ\Phi(z, \eta, \zeta)=(\widetilde{\phi}\circ\phi(z), \widetilde{\psi}\circ\psi(\eta), U_{1} (\zeta)\frac{N(z_{1}, z_{1};\eta_{1}, \eta_{1})^{1/2\mu}}{N(z, z_{1};\eta, \eta_{1})^{1/\mu}}). \end{equation}$

(2.1)

注意到

$\widetilde{\Phi}(\Phi(z, \eta, \zeta))=(\widetilde{\phi}(\phi(z)), \widetilde{\psi}(\psi(\eta)), \widetilde{U}(U(\zeta)) \frac{N(z_{0}, z_{0};\eta_{0}, \eta_{0})^{1/2\mu}}{N(z, z_{0};\eta, \eta_{0})^{1/\mu}}\frac{N(\widetilde{z}_{0}, \widetilde{z}_{0};\widetilde{\eta}_{0}, \widetilde{\eta}_{0})^{1/2\mu}}{N(\phi(z), \widetilde{z_{0}};\psi(\eta), \widetilde{\eta_{0}})^{1/\mu}}), $

由(2.1) 式有

$\begin{eqnarray*} &&N(\phi(z_{1}), \phi(z_{1});\psi(\eta_{1}), \psi(\eta_{1}))=N(z_{1}, z_{1};\eta_{1}, \eta_{1})|J\phi(z_{1})J\psi(\eta_{1})|^{2/\gamma}, \\ &&N(\phi(z), \phi(z_{1});\psi(\eta), \psi(\eta_{1}))=N(z, z_{1};\eta, \eta_{1})(J\phi(z) \overline{J\phi(z_{1})})^{1/\gamma}(J\psi(\eta)\overline{J\psi(\eta_{1})})^{1/\gamma}, \end{eqnarray*}$

并且由于$N(z, 0;\eta, 0)=N(0, w;0, v)=1$对任意的$z, w, \eta, v\in\widetilde{\Omega}_{m}$成立, 有

$\begin{eqnarray*} N(z_{0}, z_{0};\eta_{0}, \eta_{0})&=&N(\phi(z_{0}), \phi(z_{0});\psi(\eta_{0}), \psi(\eta_{0}))|J\phi(z_{0})J\psi(\eta_{0})|^{-2/\gamma}\\ &=&|J\phi(z_{0})J\psi(\eta_{0})|^{-2/\gamma}, \\ N(z, z_{0};\eta, \eta_{0})&=&N(\phi(z), \phi(z_{0});\psi(\eta), \psi(\eta_{0}))(J\phi(z)\overline{J\phi(z_{0})})^{-1/\gamma} (J\psi(\eta)\overline{J\psi(\eta_{0})})^{-1/\gamma}\\ &=&(J\phi(z)\overline{J\phi(z_{0})})^{-1/\gamma}(J\psi(\eta)\overline{J\psi(\eta_{0})})^{-1/\gamma}. \end{eqnarray*}$

由于$\phi(z_{1})=\widetilde{z_{0}}, \psi(\eta_{1})=\widetilde{\eta_{0}}$, 综合上述等式, 得到

$\begin{eqnarray*} &&\frac{N(z_{0}, z_{0};\eta_{0}, \eta_{0})}{N(z, z_{0};\eta, \eta_{0})^{2}}\frac{N(\widetilde{z}_{0}, \widetilde{z}_{0}; \widetilde{\eta}_{0}, \widetilde{\eta}_{0})}{N(\phi(z), \phi(\widetilde{z}_{0});\psi(\eta), \psi(\widetilde{\eta}_{0}))^{2}}\\ &=&\frac{N(z_{0}, z_{0};\eta_{0}, \eta_{0})}{N(z, z_{0};\eta, \eta_{0})^{2}}\frac{N(\phi(z_{1}), \phi(z_{1});\psi(\eta_{1}), \psi(\eta_{1}))}{N(\phi(z), \phi(z_{1}); \psi(\eta), \psi(\eta_{1}))^{2}}\\ &=&\frac{\overline{J\phi(z_{0})J\psi(\eta_{0})}^{2/\gamma}}{|J\phi(z_{0})J\psi(\eta_{0})|^{2/\gamma}} \frac{|J\phi(z_{1})J\psi(\eta_{1})|^{2/\gamma}}{\overline{J\phi(z_{1})J\psi(\eta_{1})}^{2/\gamma}} \frac{N(z_{1}, z_{1};\eta_{1}, \eta_{1})}{N(z, z_{1};\eta, \eta_{1})^{2}}, \end{eqnarray*}$

取$U_{1}\in\mathcal{U}(m)$使得

$U_{1}=(\frac{\overline{J\phi(z_{0})J\psi(\eta_{0})}^{2/\gamma}}{|J\phi(z_{0})J\psi(\eta_{0})|^{2/\gamma}} \frac{|J\phi(z_{1})J\psi(\eta_{1})|^{2/\gamma}}{\overline{J\phi(z_{1})J\psi(\eta_{1})}^{2/\gamma}})^{1/2\mu}\widetilde{U}\circ U.$

即为等式(2.1).命题2.1得证.

现在要研究$\widetilde{\Omega}_{m}$的边界$b\widetilde{\Omega}_{m}$的强拟凸性, 其边界$b\widetilde{\Omega}_{m}$可以进行如下分解:

$b\widetilde{\Omega}_{m}=b_{0}\widetilde{\Omega}_{m}\cup b(\Omega_1\times \Omega_2\times \{0\}), $

其中$b(\Omega_1\times \Omega_2\times \{0\})=（b\Omega_{1}\times\Omega_{2}\times\{0\})\cup(\Omega_{1} \times b\Omega_{2}\times\{0\})\cup(b\Omega_{1}\times b\Omega_{2}\times\{0\}), $而$b_{0}\widetilde{\Omega}_{m}:=\{(z, \eta, \zeta)\in\Omega_{1}\times\Omega_{2}\times\mathbb{C}^{m} \subset\mathbb{C}^{d_{1}}\times\mathbb{C}^{d_{2}}\times\mathbb{C}^{m}:\rho:=\|\zeta\|^{2\mu}-N(z, z;\eta, \eta)=0\}$是$b\widetilde{\Omega}_{m}$的光滑部分.注意到对任意$(z, \eta, \zeta)\in b_{0}\widetilde{\Omega}_{m}$有$(z, \eta)\in \Omega_1\times \Omega_2$, 从而$N(z, z;\eta, \eta)>0$, 故$\zeta\neq 0$.

下列的命题2.2是显然的.

命题2.2 假设$D_{i}\subset\mathbb{C}^{n_{i}}$是有界域, 其Bergman核函数记为$K_{i}(i=1, 2)$, 则$D:=D_{1}\times D_{2}$的Bergman核函数$K$满足

$K((\xi_{1}, \xi_{2}), (z_{1}, z_{2}))=K_{1}(\xi_{1}, z_{1})K_{2}(\xi_{2}, z_{2}), (\xi_{1}, \xi_{2}), (z_{1}, z_{2})\in D_{1}\times D_{2}.$

现在给出如下的关键引理, 它是论文涂振汉、王磊文[9]中的一个引理的稍微推广.

命题2.3 设$\Omega_{i}\subset\mathbb{C}^{d_{i}} (i=1.2)$ (在Harish-Chandra嵌入下)是亏格为$\gamma^{i}$的不可分解的有界对称域, 则有如下结论成立:

1. $\widetilde{\Omega}_{m}$在$b_{0}\widetilde{\Omega}_{m}$的每一个点都是强拟凸的;

2. $\widetilde{\Omega}_{m}$在$b(\Omega\times\{0\})$的任意一点都不是强拟凸的, 这里$b(\Omega\times\{0\})=(b\Omega_{1}\times\Omega_{2}\times\{0\}) \cup(\Omega_{1}\times b\Omega_{2}\times\{0\})\cup(b\Omega_{1}\times b\Omega_{2}\times\{0\})$.

证 本证明来源于文献[9]的推理.设${h_{i}(z)}_{i=1}^{\infty}, {g_{j}(\eta)}_{i=1}^{\infty}$分别是Hilbert空间$A^{2}(\Omega_{1}), $ $A^{2}(\Omega_{2})$的一组标准正交基, 其中$A^{2}(\Omega_{1}), A^{2}(\Omega_{2})$分别由所有平方可积的全纯函数构成.由命题2.2, 得到

$K(z, \overline{z};\eta, \overline{\eta})=K_1(z, \overline{z})K_2(\eta, \overline{\eta})=\sum h_{i}(z)\overline{h_{i}(z)} \sum g_{j}(\eta)\overline{g_{j}(\eta)}$

在$\Omega$的任一紧致子集上一致收敛.设$\rho(z, \eta, \zeta):=\|\zeta\|^{2\mu}-\delta K_1(z, \overline{z})^{-\lambda_{1}}K_2(\eta, \overline{\eta})^{-\lambda_{2}}, $其中$\delta:=(V(\Omega_1))^{-\frac{1}{\gamma_1}} (V(\Omega_2))^{-\frac{1}{\gamma_2}} $且$\lambda_{i}:=-\frac{1}{\gamma_{i}}, (i=1, 2)$均为正数.从而$\rho$是$b_{0}\widetilde{\Omega}_{m}$上的一个实值定义函数.

固定$(z_{0}, \eta_{0}, \zeta_{0})\in b_{0}\widetilde{\Omega}_{m}$, 并且令$T=(u, v, w)\in T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})\subset\mathbb{C}^{d_{1}} \times\mathbb{C}^{d_{2}}\times\mathbb{C}^{m}.$为了简便, 用$(z, \eta, \zeta)$代替$(z_{0}, \eta_{0}, \zeta_{0})$.由定义可得

$\begin{equation}\label{22} \begin{split} &\|\zeta\|^{2\mu}-\delta K_{1}(z, \overline{z})^{-\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}}=0, \\ &\mu\|\zeta\|^{2(\mu-1)}(\overline{\zeta}\cdot w)+\delta\lambda_{1}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}}K_{1} (z, \overline{z})^{-(\lambda_{1}+1)}\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)\\ +&\delta\lambda_{2}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+1)}K_{1}(z, \overline{z})^{-\lambda_{1}}\sum\overline{g_{j} (\eta)}(g_{j}(\eta)^{\prime}\cdot v)=0, \end{split} \end{equation}$

(2.2)

其中$h_{i}(z)^{\prime}\cdot u=\sum\limits_{i=1}^{\infty}\frac{\partial h_{i}}{\partial z_{k}}(z)u_{k}, \;\; g_{j}(\eta)^{\prime} \cdot v=\sum\limits_{l=1}^{\infty}\frac{\partial g_{j}}{\partial \eta_{l}}(\eta)v_{l}.$

最后通过(2.2) 式, $\rho$在$(z, \eta, \zeta)$的Levi形式计算如下:

$\begin{align*} L_{\rho}(T, T):=&\sum\limits_{i, j=1}^{d_{1}+d_{2}+m}\frac{\partial^{2}\rho}{\partial T{i}\partial \overline{T{j}}}\\ =&\mu\|\zeta\|^{2(\mu-1)}\|w\|^{2}+\mu(\mu-1)\|\zeta\|^{2(\mu-2)}|\overline{\zeta}\cdot w|^{2}\\ &+\delta\lambda_{1}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}}\big[-(\lambda_{1}+1)K_{1}(z, \overline{z})^{-(\lambda_{1}+2)} |\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &+K_{1}(z, \overline{z})^{-(\lambda_{1}+1)}\sum|(h_{i}(z)^{\prime}\cdot u)|^{2}\big]\\ &+\delta\lambda_{2}K_{1}(z, \overline{z})^{-\lambda_{1}}\big[-(\lambda_{2}+1)K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+2)} |\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\\ &+K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+1)}\sum|(g_{l}(\eta)^{\prime}\cdot v)|^{2}\big]\\ &-\delta\lambda_{1}\lambda_{2}K_{1}(z, \overline{z})^{-(\lambda_{1}+1)}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+1)} 2{\rm Re}\big[(\sum h_{i}(z)\overline{(h_{i}(z)^{\prime}\cdot u)})\\ &(\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot v))\big]\\ =&\mu\|\zeta\|^{2(\mu-2)}(\|\zeta\|^{2}\|w\|^{2}-|\overline{\zeta}\cdot w|^{2})+\mu^{2}\|\zeta\|^{2(\mu-2)}|\overline{\zeta}\cdot w|^{2}\\ &-\delta\lambda_{1}^{2}K_{1}(z, \overline{z})^{-(\lambda_{1}+2)}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}}| \sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &-\delta\lambda_{2}^{2}K_{1}(z, \overline{z})^{-\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+2)}| \sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\\ &-\delta\lambda_{1}\lambda_{2}K_{1}(z, \overline{z})^{-(\lambda_{1}+1)}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+1)} 2{\rm Re}\big[(\sum h_{i}(z)\overline{(h_{i}(z)^{\prime}\cdot u)})\\ &(\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot v))\big]\\ &+\delta\lambda_{1}K_{1}(z, \overline{z})^{-(\lambda_{1}+2)}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}}\big[K_{1}(z, \overline{z}) \sum|(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &-|\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\big]\\ &+\delta\lambda_{2}K_{1}(z, \overline{z})^{-\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+2)} \big[K_{2}(\eta, \overline{\eta})\sum|(g_{l}(\eta)^{\prime}\cdot v)|^{2}\\ &-|\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\big]\\ =&\mu\|\zeta\|^{2(\mu-2)}(\|\zeta\|^{2}\|w\|^{2}\\ &-|\overline{\zeta}\cdot w|^{2})+\Big[\mu^{2}\|\zeta\|^{4(\mu-1)}|\overline{\zeta}\cdot w|^{2}\\ &-\delta^{2}\lambda_{1}^{2}K_{1}(z, \overline{z})^{-(2\lambda_{1}+2)}K_{2}(\eta, \overline{\eta})^{-2\lambda_{2}} |\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &-\delta^{2}\lambda_{2}^{2}K_{1}(z, \overline{z})^{-2\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-(2\lambda_{2}+2)}| \sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\\ &-\delta^{2}\lambda_{1}\lambda_{2}K_{1}(z, \overline{z})^{-(2\lambda_{1}+1)}K_{2}(\eta, \overline{\eta})^{-(2\lambda_{2}+1)} 2{\rm Re}\big[(\sum h_{i}(z)\overline{(h_{i}(z)^{\prime}\cdot u)})\\ &(\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot v))\big]\Big]\Big/\|\zeta\|^{2\mu}\\ &+\delta\lambda_{1}K_{1}(z, \overline{z})^{-(\lambda_{1}+2)}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}} \big[K_{1}(z, \overline{z})\sum|(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &-|\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\big]\\ &+\delta\lambda_{2}K_{1}(z, \overline{z})^{-\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+2)} \big[K_{2}(\eta, \overline{\eta})\sum|(g_{l}(\eta)^{\prime}\cdot v)|^{2}\\ &-|\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\big]\\ =&\mu\|\zeta\|^{2(\mu-2)}(\|\zeta\|^{2}\|w\|^{2}\\ &-|\overline{\zeta}\cdot w|^{2})\\ &+\delta\lambda_{1}K_{1}(z, \overline{z})^{-(\lambda_{1}+2)}K_{2}(\eta, \overline{\eta})^{-\lambda_{2}} \big[K_{1}(z, \overline{z})\sum|(h_{i}(z)^{\prime}\cdot u)|^{2}\\ &-|\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}\big]\\ &+\delta\lambda_{2}K_{1}(z, \overline{z})^{-\lambda_{1}}K_{2}(\eta, \overline{\eta})^{-(\lambda_{2}+2)} \big[K_{2}(\eta, \overline{\eta})\sum|(g_{l}(\eta)^{\prime}\cdot v)|^{2}\\ &-|\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}\big]\\ \geq& 0. \end{align*}$

由Cauchy-Schwartz不等式, 对所有的$T=(u, v, w)\in T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})$等式成立当且仅当

$\begin{equation}\label{33} \begin{split} &\|\zeta\|^{2}\|w\|^{2}-|\overline{\zeta}\cdot w|^{2}=0, \\ &K_{1}(z, \overline{z})\sum|(h_{i}(z)^{\prime}\cdot u)|^{2}-|\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}=0, \\ &K_{2}(\eta, \overline{\eta})\sum|(g_{l}(\eta)^{\prime}\cdot v)|^{2}-|\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot u)|^{2}=0. \end{split} \end{equation}$

(2.3)

现在将证明$\rho$在$(z_{0}, \eta_{0}, \zeta_{0})$的Levi形式$L_{\rho}(T, T)$对所有$T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0} \widetilde{\Omega}_{m})$都是正定的.

情形1 假设$u\neq 0$.因为$K_{1}(z, \overline{z})=\sum h_{i}(z)\overline{h_{i}(z)}$与Hilbert空间$A^{2}(\Omega_1)$中标准正交基${h_{i}(z)}_{i=1}^{\infty}$的选取无关且$\Omega_1$有界, 从而可以挑选$h_{1}(z)$是一个非零常数并且选取$h_{2}(z)$满足$h_{2}(z)^{\prime}\cdot u\neq 0$.这就说明了$(h_{1}(z), h_{2}(z))$和$(h^{\prime}(z)\cdot u, h^{\prime}(z)\cdot u)$是线性无关的, 从而

$K_{1}(z, \overline{z})\sum|(h_{i}(z)^{\prime}\cdot u)|^{2}-|\sum\overline{h_{i}(z)}(h_{i}(z)^{\prime}\cdot u)|^{2}>0.$

这与(2.3) 式矛盾.因此$L_{\rho}(T, T)>0$对所有满足$u\neq 0$的$T=(u, v, w)\in T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})$成立.

情形2 假设$v\neq 0$.与情形1类似, 因为$K_{2}(\eta, \overline{\eta})=\sum g_{j}(\eta)\overline{g_{j}(\eta)}$与Hilbert空间$A^{2}(\Omega_2)$中标准正交基${g_{j}(\eta)}_{j=1}^{\infty}$的选取无关且$\Omega_2$有界, 从而可以挑选$g_{1}(\eta)$是一个非零常数并且选取$g_{2}(\eta)$满足$g_{2}(\eta)^{\prime}\cdot v\neq 0$.这就说明了$(g_{1}(\eta), g_{2}(\eta))$和$(g^{\prime}(\eta)\cdot v, g^{\prime}(\eta)\cdot v)$是线性无关的, 从而

$K_{2}(\eta, \overline{\eta})\sum|(g_{j}(\eta)^{\prime}\cdot v)|^{2}-|\sum\overline{g_{j}(\eta)}(g_{j}(\eta)^{\prime}\cdot v)|^{2}>0, $

这与(2.3) 式矛盾.因此$L_{\rho}(T, T)>0$对所有满足$v\neq 0$的$T=(u, v, w)\in T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})$成立.

情形3 假设$u=0$且$v=0$, 则由$T=(u, v, w)\neq 0$得到$w\neq 0$, 从而由(2.2) 式, 有$\overline{\zeta}\cdot w=0$.于是由$w\neq 0, \zeta\neq 0$知

$\|\zeta\|^{2}\|w\|^{2}-|\overline{\zeta}\cdot w|^{2}=\|\zeta\|^{2}\|w\|^{2} >0, $

这与(2.3) 式矛盾, 故$L_{\rho}(T, T)>0$对所有满足$w\neq 0$的$T=(0, 0, w)\in T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})$成立.

这样Levi形式$L_{\rho}(T, T)$在$T_{(z_{0}, \eta_{0}, \zeta_{0})}^{1, 0}(b_{0}\widetilde{\Omega}_{m})$是正定的.这就是说$(b_{0}\widetilde{\Omega}_{m})$的每一个点都是强拟凸的.

对任意一个不可分解的有界对称域$\Omega_{i}\subset \mathbb{C}^{d_{i}}(i=1, 2)$, 有$G(\widetilde{\Omega}_{m})$在$\Omega_{1}\times\Omega_{2}\times\{0\}(\subset\widetilde{\Omega}_{m})$上是可迁的.因为$\widetilde{\Omega}_{m}$不是单位球, 由Wong-Rosay定理^[11]可知, $\widetilde{\Omega}_{m}$在$b\{\Omega_{1}\times\Omega_{2}\times\{0\}\}$的每一点都不是强拟凸的.

引理3.1 (见文[1]) $\Psi\in {\rm Aut}(\widetilde{\Omega}_{m})$把$\Omega\times\{0\}$映为自身.

证 设$\Psi\in {\rm Aut}(\widetilde{\Omega}_{m})$并在$\Omega_{1}\times\Omega_{2}\times\{0\}$中选取序列$\{(z_{j}, \eta_{j}, 0)\}$收敛到$b(\Omega_{1}\times\Omega_{2}\times\{0\})$.由(1.4) 及命题2.1可知, 可以选取$\Phi_{j}\in {\rm Aut}(\widetilde{\Omega}_{m})$满足$\Phi_{j}(0, 0, 0)=(z_{j}, \eta_{j}, 0)$.从而${\Psi(z_{j}, \eta_{j}, 0)}$可以被看为是一个自同构轨道$\{(\Psi\circ\Phi_{j})(0, 0, 0)\}$.假设${\Psi(z_{j}, \eta_{j}, 0)}$有一个子序列收敛到某个$P\in b_{0}\widetilde{\Omega}_{m}$.从命题2.3可知, $P$必须是一个强拟凸点.由Wong-Rosay定理可得$\widetilde{\Omega}_{m}$是齐性域, 从而$\widetilde{\Omega}_{m}$只能是单位球.这与$\widetilde{\Omega}_{m}$不是单位球矛盾.因此对每一个收敛到$b(\Omega_{1}\times\Omega_{2}\times\{0\})$的序列$\{(z_{j}, \eta_{j}, 0)\}$, 序列$\{\Psi(z_{j}, \eta_{j}, 0)\}$同样收敛到$b(\Omega_{1}\times\Omega_{2}\times\{0\})$的某点.

设$\Psi=(\Psi_{1}, \Psi_{2}, \Psi_{3})\in {\rm Aut}(\widetilde{\Omega}_{m})$.由$f(z, \eta)=\Psi_{3}(z, \eta, 0)$定义一个全纯映照$f:\Omega_{1}\times\Omega_{2}\longrightarrow\mathbb{C}^{m}$.则$f(z, \eta)$在$b(\Omega_{1}\times\Omega_{2})$退化, 因此也必然在$\Omega_{1}\times\Omega_{2}$退化, 从而$\Psi_{3}(z, \eta, 0)\equiv0$对所有的$z\in \Omega_{1}, \eta\in \Omega_{2}$成立.引理3.1得证.

引理3.2 如果$\Psi\in {\rm Aut}(\widetilde{\Omega}_{m})$满足$\Psi(0, 0, 0)=(0, 0, 0)$, 则$\Psi$具有如下形式: $\Psi(z, \eta, \zeta)=(I_{1}(z), I_{2}(\eta), U(\zeta)), $其中$I_{i}\in {\rm Aut}(\Omega_{i})$, $I_{i}(0)=0(i=1, 2)$和$U\in \mathcal{U}_{m}$.特别$\Psi\in \widetilde{G}$.

证 因为$\widetilde{\Omega}_{m}$是有界圆形域, 由Cartan唯一性定理可知$\Psi$是线性的.由引理3.1知, $\Psi(z, \eta, 0)=(I_{1}(z), I_{2}(\eta), 0)$对$I_{i}\in {\rm Aut}(\Omega_{i})$且$I_{i}(0)=0 (i=1, 2)$成立.选取$\Psi_{0}\in\widetilde{G}$使得$\Psi_{0}(z, \eta, 0)=(I_{1}(z), I_{2}(\eta), 0)$, 则我们有$(\Psi_{0}^{-1}\circ\Psi)(z, \eta, 0)=(z, \eta, 0)$, 从而$\Psi_{0}^{-1}\circ\Psi$是线性的, 即

$(\Psi_{0}^{-1}\circ\Psi)(z, \eta, \zeta)=\left(\begin{array}{ccc}I_{01}&0&A\\0&I_{02}&B\\0&0&C\\ \end{array}\right)\left(\begin{array}{ccc}z\\\eta\\\zeta\\ \end{array}\right), $

其中$I_{0i} (i=1, 2)$单位矩阵且$A, B, C$分别为$d_{1}\times m, d_{2}\times m$阶矩阵.由于$\Psi_{0}^{-1}\circ\Psi\in {\rm Aut}(\widetilde{\Omega}_{m})$, $C$和$C^{-1}$映${\zeta\in \mathcal{U}_{m}:\|\zeta\|＜1}$为自身, 因此$C$是酉矩阵.下证$A, B=0$.由于线性映射把边界映为边界.由$\widetilde{\Omega}_{m}$的定义可得

$\begin{eqnarray*} &&\|C\zeta\|^{2\mu}=N_{1}(z+A\zeta, z+A\zeta)N_{2}(\eta+B\zeta, \eta+B\zeta), \\ &&\|\zeta\|^{2\mu}=N_{1}(z, z)N_{2}(\eta, \eta).\end{eqnarray*}$

由于$C$是酉矩阵, 有$N_{1}(z+A\zeta, z+A\zeta)N_{2}(\eta+B\zeta, \eta+B\zeta)=N_{1}(z, z)N_{2}(\eta, \eta).$令$z=0, \eta=0$, 可得$N_{1}(0, 0)N_{2}(0, 0)=1$和$N_{1}(A\zeta, A\zeta)N_{2}(B\zeta, B\zeta)=1.$考虑到$0\leq N_{1}(A\zeta, A\zeta)\leq 1, 0\leq N_{2}(B\zeta, B\zeta)\leq 1$, 得到如下结果:

$N_{1}(A\zeta, A\zeta)=1, N_{2}(B\zeta, B\zeta)=1, $

也就是说$A\zeta=0, B\zeta=0, \forall\zeta$.由此可得$A=0, B=0$且有

$\begin{eqnarray*} &&\Psi_{0}^{-1}\circ\Psi(z, \eta, \zeta)=\left(\begin{array}{ccc}I_{01}&0&0\\0&I_{02}&0\\0&0&C\\ \end{array}\right)\left(\begin{array}{ccc}z\\\eta\\\zeta\\ \end{array}\right), \\ &&\Psi(z, \eta, \zeta)=(I_{1}(z), I_{2}(\eta), C(\zeta)), \end{eqnarray*}$

这就证明了了引理3.2.

现在来证明定理1.2.设$\Psi\in {\rm Aut}(\widetilde{\Omega}_{m})$.由引理3.1, $\Psi(0, 0, 0)=(z_{0}, \eta_{0}, 0)$对某个$z_{0}\in\Omega_{1}, \eta_{0}\in\Omega_{2}$成立.选取$\Phi\in\widetilde{G}$使得$\Phi(z_{0}, \eta_{0}, 0)=(0, 0, 0)$.从而得到$(\Phi\circ\Psi)(0, 0, 0)=(0, 0, 0)$.由引理3.2得到$\Phi\circ\Psi\in \widetilde{G}$.再由命题2.1知$\widetilde{G}$是一个子群, 从而$\Psi\in \widetilde{G}$.这就证明了${\rm Aut}(\widetilde{\Omega}_{m})$由$\widetilde{G}$生成.定理1.2得证.