极小曲面高斯映射的值分布理论研究历史悠久, 硕果累累. 用Nevanlinna值分布理论研究极小曲面的值分布问题起源于1988年, Fujimoto在文献[4]中证明了Nirenberg猜想, 即$ {\Bbb R}^3 $中非平凡的完备极小曲面的高斯映射至多有$ 4 $个例外值, 紧接着Fujimoto将其推广到$ {\Bbb R}^n $中的极小曲面的情形. 从此许多学者开始关注紧黎曼曲面上高斯映射的值分布问题. 例如, Jin和Ru^[6]建立了从黎曼面到$ n $维射影空间代数曲线分担处于一般位置超平面的第二基本定理. 最近, Thai和Thoan^[12]用Nochka权的技巧将Jin和Ru的工作进行了如下推广.

定理1.1^[12]  设$ V $为$ {\Bbb P}^{n}({\Bbb C}) $中$ k(k<n) $维复射影子簇, $ {\Bbb P}^{n}({\Bbb C}) $中超曲面$ Q_{1}, \cdots, Q_{q} $在$ V $上处于$ N $次一般位置, $ d_i: = \deg Q_i $, $ d $为$ d_{1}, \cdots, d_{q} $的最小公倍数. 设$ S $为亏格为$ g $的紧黎曼曲面, $ E $为$ S $的任意一个有限子集. 如果$ f $为从$ S $到$ {\Bbb P}^{n}({\Bbb C}) $的代数非退化全纯曲线，并且$ f(S)\subseteq V $, 则对任意的$ \varepsilon>0 $, 有

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M\}+\frac{(N-k+1)(k+1)M(2(g-1)+|E|)}{2ud}, \end{eqnarray*} $

其中$ \nu_{Q_{j}(f)} = \nu_{f^{*}Q_{j}} $$ (1\leq j\leq q) $为$ Q(f) $的零点重数, $ H_{V}(d) $是$ V $的希尔伯特函数.

最近, Quang^[8]运用构造辅助函数的方法得到了从复平面到复射影簇的亚纯映射第二基本定理. 受此启示, 我们将证明如下结论.

定理1.2  设$ V $为$ {\Bbb P}^{n}({\Bbb C}) $中$ k $$ (k<n) $维复射影子簇, $ Q_{1}, \cdots, Q_{q} $是$ {\Bbb P}^{n}({\Bbb C}) $中关于$ V $处于$ N $次一般位置的超曲面, $ d_i = {\rm deg} Q_i $, $ d $为$ d_{1}, \cdots, d_{q} $的最小公倍数. 设$ S $为亏格为$ g $的紧黎曼曲面, $ E $为$ S $的任意一个有限子集. 如果$ f $为从$ S $到$ {\Bbb P}^{n}({\Bbb C}) $的代数非退化全纯曲线，并且$ f(S)\subseteq V $, 则对任意的$ \varepsilon>0 $和所有的满足$ u> \frac{(N-k+1)(k+1)^{2}(2k+1)d^{k}q!\deg(V)}{\varepsilon} $的正整数, 我们有

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M\}+\frac{(N-k+1)(k+1)M(2(g-1)+|E|)}{2ud} \end{eqnarray*} $

成立, 其中$ M = [\deg(V)^{k+1}e^{k}d^{k^{2}+k}(N-k+1)^{k}(2k+4)^{k}l^{k}\varepsilon^{-k}] $, $ [x] $表示实数$ x $的整数部分.

特别地, $ V = {\Bbb P}^{n}({\Bbb C}) $, 我们证明将下面涉及更小截断重数的第二基本定理.

定理1.3  设$ f $为$ S $到$ {\Bbb P}^{n}({\Bbb C}) $的代数非退化的全纯曲线, $ Q_{1}, \cdots, Q_{q} $是$ {\Bbb P}^{n}({\Bbb C}) $中处于$ N $次一般位置的超曲面, $ d_i = {\rm deg} Q_i $, $ d $为$ d_{1}, \cdots, d_{q} $的最小公倍数. 设$ S $为亏格为$ g $的紧黎曼曲面, $ E $为$ S $的任意一个有限子集. 则任意的$ \varepsilon >0 $, $ u\geq (n+1)d+p(n+1)^{3}dI(\varepsilon)^{-1} $的正整数, 我们有

$ \begin{eqnarray*} & &(q-p(n+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M_{0}-1\} +\frac{p(n+1)(M_{0}-1)(2(g-1)+|E|)}{2ud} \end{eqnarray*} $

成立, 其中$ p = N-n+1 $且$ M_{0} = 4(edp(n+1)^{2}I(\varepsilon^{-1}))^{n} $, $ [x] $记为不超过实数$ x $的最小正整数.

全曲率有限的完备极小曲面共形等价于一个挖掉有限点的紧黎曼曲面, 从而其广义的高斯映射可以延拓到紧黎曼曲面上. 作为第二基本定理的应用, Thai, Thoan和Vangty^[12]得到了如下分歧定理.

定理1.4^[12]  设$ V $为$ {\Bbb P}^{n}({\Bbb C}) $中$ k $$ (k<n) $维复射影子簇, $ {\Bbb P}^{n}({\Bbb C}) $中超曲面$ Q_{1}, \cdots, Q_{q} $是在$ V $上处于$ N $次一般位置, 且$ d = lcm(d_{1}, \cdots, d_{q}) $, 即$ d $为$ (d_{1}, \cdots, d_{q}) $的最小公倍数. 令$ x: S\rightarrow {\Bbb R}^{m} $为总曲率有限的非平坦的完备紧黎曼曲面, $ G:S\rightarrow {\Bbb P}^{m-1}({\Bbb C}) $为它上面的广义高斯映射. 假设$ G(S) $被包含在$ V $中且映射$ G $在$ V $中是线性非退化的, 也就是像集$ f(S) $是不包含在任何一个维数低于$ k $的$ {\Bbb P}^{m-1}({\Bbb C}) $复射影子空间中. 对于每一个$ j $, 假设$ G $交超曲面$ Q_{j} $的分歧重数至少是$ m_{j} $. 则

$ \sum\limits^{q}_{j = 1}\left(1-\frac{H_{V}(d)-1}{m_{j}}\right)< \frac{(2N-k+1)H_{V}(d)(H_{V}(d)-1+2d)}{2(k+1)d}. $

相应地, 我们将证明:

定理1.5  设$ Q_{1}, \cdots, Q_{q} $是$ {\Bbb P}^{n}({\Bbb C}) $中处于$ N $次一般位置的超曲面, 且$ d = lcm(d_{1}, \cdots, d_{q}) $, 即$ d $为$ (d_{1}, \cdots, d_{q}) $的最小公倍数. 令$ x: S\rightarrow {\Bbb R}^{m} $为总曲率有限的非平坦的完备紧黎曼曲面. 设$ G:S\rightarrow {\Bbb P}^{m-1}({\Bbb C}) $为它上面的广义高斯映射. 对于每一个$ j $, 假设$ G $是代数非退化的, $ G $交超曲面$ Q_{j} $的分歧重数至少是$ m_{j} $, 则

$ \sum\limits^{q}_{j = 1}\bigg(1-\frac{M_{0}-1}{m_{j}}\bigg) <\frac{p(n+1)(2ud+M_{0})}{2ud}. $

设$ {\Bbb P}^{n}({\Bbb C}) $的齐次坐标为$ (\omega_{0}:\cdots:\omega_{m}) $, $ Q $为$ {\Bbb P}^{n}({\Bbb C}) $中次数是$ d $的超曲面. 本文中，在不产生混淆的情况下超曲面由多项式给出, 即

$ Q(\omega) = \sum\limits_{I\in {\cal T}_{d}}a_{I}\omega^{I}, $

其中$ {\cal T}_{d} = \{(i_{0}, \cdots, i_{n})\in {\Bbb Z}^{n+1}_{\geq 0};i_{0}+\cdots+i_{n} = d\} $, $ \omega = (\omega_{0}, \cdots, \omega_{n}) $, $ \omega^{I} = \omega_{0}^{i_{0}}\cdots\omega_{n}^{i_{n}} $, $ I = (i_{0}, \cdots, i_{n})\in {\cal T}_{d} $且$ a_{I}(I\in {\cal T}_{d}) $是非零常数.

定义2.1  若函数$ Q(f) $的所有零点重数至少是e, 我们称映射$ f $交超曲面$ Q $的分歧重数至少是e. 如果$ f $与超曲面$ Q $相交为空, 我们称$ f $交超曲面$ Q $的分歧重数为$ \infty $.

定义2.2  设$ V $为$ {\Bbb P}^{n}({\Bbb C}) $中$ k $$ (k\leq n) $维复射影子簇, $ {\Bbb P}^{n}({\Bbb C}) $中$ q $个超曲面$ Q_{1}, \cdots, Q_{q} $$ (q\geq k+1) $称为在$ V $上处于$ N $次一般位置是指, 对任意的$ 1\leq i_{1}<\cdots<i_{N+1}\leq q $, 都有

$ V\cap (\bigcap\limits^{N+1}_{j = 1}Q_{i_{j}}) = \emptyset. $

特别地, 当$ N = n $时, 我们就称之为处于一般位置.

定义2.3  设$ f:{\Bbb C}^{m}\rightarrow V $是一个亚纯映射, 如果存在$ Q\in I_{d}(V)\backslash \{0\} $, 满足$ Q(f)\equiv 0 $, 我们称$ f $在域$ I_{d}(V) $上是代数退化的, 否则我们说$ f $在域$ I_{d}(V) $上非退化. 显然地, 当$ f $是代数非退化的, 则对任意的$ d\geq 1 $, $ f $退化到域$ I_{d}(V) $.

2.1  假设$ f:S\rightarrow {\Bbb P}^{n}({\Bbb C}) $代数非退化的代数曲线, 对于每一点$ P\in S $, 取$ P $点坐标邻域$ z(P) = 0 $内, $ f $在$ P $点的精简表达式为$ f(z) = (f_{0}(z), \cdots, f_{n}(z)) $, $ f_{0}, \cdots, f_{n} $是没有公共零点的全纯函数. 令$ Q:\sum\limits_{{\cal I}}a_{{\cal I}}z^{{\cal I}} = 0 $是$ {\Bbb P}^{n}({\Bbb C}) $中的超曲面, 且

$ Q(f) = \sum\limits_{{\cal I}}a_{{\cal I}}f^{{\cal I}}, $

则$ \frac{1}{\deg Q}\sum\limits_{P\in S}\nu_{Q(f)}(z) $不依赖于超曲面$ Q $的选择, 其中$ \nu_{Q(f)}(z) $是$ f $将超曲面$ Q $拉回在$ z $点的分歧重数. 我们定义$ f $的度为

$ \deg(f) = \frac{1}{\deg Q}\sum\limits_{z\in S}\nu_{Q(f)}(z). $

若$ f^{-1}(H) = \{P_{1}, \cdots, P_{r}\} $, 则

$ \deg(f) = \frac{1}{\deg Q}\sum\limits^{r}_{j = 1}\nu_{Q(f)}(P_{j}). $

在$ {\Bbb C}^{n+1} $中选择合适的局部坐标系满足: $ f(0) = (1, 0, \cdots, 0) $, $ f_{1}(0) = \cdots = f_{n}(0) = 0, $$ (f_{1}(z), \cdots, f_{n}(z)) = z^{\delta_{1}}(f^{1}_{1}(z), \cdots, f^{1}_{n}(z)) $且$ (f^{1}_{1}(0) = \cdots = f^{1}_{n}(0))\neq 0 $; $ (f^{1}_{1}(0) = \cdots = f^{1}_{n}(0)) = (1, 0, \cdots, 0) $, $ (f^{1}_{2}(z) = \cdots = f^{1}_{n}(z)) = z^{\delta_{2}-\delta_{1}}(f^{2}_{2}(z) = \cdots = f^{2}_{n}(z)) $以及$ (f^{2}_{2}(0) = \cdots = f^{2}_{n}(0))\neq 0, \ldots $, 依次变换下去, 我们得到以下形式

(2.1) $ \begin{equation} f(z) = (z^{\delta_{0}}+\cdots, z^{\delta_{1}}+\cdots, \cdots, z^{\delta_{n}}+\cdots), \end{equation} $

其中$ 0 = \delta_{0}<\delta_{1}<\cdots<\delta_{n} $. 令$ \nu_{i} = \delta_{i+1}-\delta_{i}-1 $, $ 0\leq i\leq n-1 $. 因为$ P\in S $, 我们有

(2.2) $ \begin{equation} \sum\limits^{n}_{i = 0}(n-i)\nu_{i}(P)+\frac{1}{2}n(n+1) = \delta_{0}(P)+\delta_{1}(P) +\cdots+\delta_{n}(P). \end{equation} $

令

(2.3) $ \begin{equation} \sigma_{i} = \sum\limits_{P\in S}\nu_{i}(P). \end{equation} $

由Plücker公式, 也就是推广的Riemann-Hurwitz定理, 我们得

(2.4) $ \begin{equation} \sum\limits^{n}_{i = 0}(n-i)\sigma_{i} = (n+1)\deg(f)+n(n+1)(g-1), \end{equation} $

其中$ g $表示$ S $的亏格.

2.2  设$ X\subset {\Bbb P}^{n}({\Bbb C}) $是维数为$ k $的射影簇且度为$ \Delta $. 若$ X $上的多项式

$ F_{X}({\bf u}_{0}, \cdots, {\bf u}_{k}) = F_{X}(u_{00}, \cdots, u_{0n}; \cdots;u_{k0}, \cdots, u_{kn}), $

$ {\bf u}_{i} = (u_{i0}, \cdots, u_{in}) $, $ i = 0, \cdots, k $, 满足以下条件: $ F_{X} $是$ {\Bbb C}[u_{00}, \cdots, u_{kn}] $中的不可约的齐次多项式, 每个分量$ {\bf u}_{i} $, $ i = 0, \cdots, k $的次数为$ \Delta $, 并且$ F_{X}({\bf u}_{0}, \cdots, {\bf u}_{k}) = 0 $当且仅当$ X\cap H_{{\bf u}_{0}}\cap H_{{\bf u}_{k}}\neq\emptyset, $这里$ H_{{\bf u}_{i}} $, $ i = 0, \cdots, k $是由$ {\bf u}_{i0}x_{0}+\cdots+{\bf u}_{in}x_{n} = 0 $所定义的超平面. 我们称$ F_X $为$ X $上的Chow形式.

设$ F_{X} $为$ X $上的Chow形式, 令$ {\bf c} = (c_{0}, \cdots, c_{n}) $为一实数数组. 我们作如下分解

$ \begin{eqnarray*} F_{X}(t^{c_{0}}u_{00}, \cdots, t^{c_{n}}u_{0n};\cdots; t^{c_{0}}u_{k0}, \cdots, t^{c_{n}}u_{kn}) & = &t^{e_{0}}G_{0}({\bf u}_{0}, \cdots, {\bf u}_{k})+\cdots+ t^{e_{r}}G_{r}({\bf u}_{0}, \cdots, {\bf u}_{k}), \end{eqnarray*} $

其中$ G_{0}, \cdots, G_{r}\in {\Bbb C}[u_{00}, \cdots, u_{0n};\cdots;u_{k0}, \cdots, u_{kn}] $, $ e_{0}>e_{1}>\cdots>e_{r} $, 则$ X $上的Chow形式关于$ c $的权定义为

$ e_{X}({\bf c}): = e_{0}. $

对$ \{0, \cdots, n\} $的任意子集$ J = \{j_{0}, \cdots, j_{k}\} $, $ j_{0}<j_{1}<\cdots<j_{k} $, 定义

$ [J] = [J]({\bf u}_{0}, \cdots, {\bf u}_{n}): = \det(u_{ij_{t}}), i, t = 0, \cdots, k, $

其中$ {\bf u}_{i} = (u_{i0}, \cdots, u_{in}) $是$ n+1 $个变量的块. 令$ \beta: = \left( \begin{array}{c} n+1 \\ k+1 \end{array} \right) $, $ J_{1}, \cdots, J_{\beta} $为$ \{0, \cdots, n\} $的子集, 且每个$ J_i $中元的个数是$ k+1 $. 则$ X $的Chow形式$ F_{X} $在$ [J_{1}], \cdots, [J_{\beta}] $中可以写成度为$ \Delta $的齐次多项式. 对$ J_{1}, \cdots, J_{\beta} $中的任意$ J $, $ {\bf c} = (c_{0}, \cdots, c_{n})\in {\Bbb R}^{n+1} $,

$ \begin{eqnarray*} [J](t^{c_{0}}u_{00}, \cdots, t^{c_{n}}u_{0n};\cdots; t^{c_{0}}u_{k0}, \cdots, t^{c_{n}}u_{kn}) & = &t^{\sum\limits_{j\in J}c_{j}}[J](u_{00}, \cdots, u_{0n};\cdots; u_{k0}, \cdots, u_{kn}). \end{eqnarray*} $

设$ {\bf a} = (a_{0}, \cdots, a_{n})\in {\Bbb Z}^{n+1} $, 记$ {\bf x}^{{\bf a}}: = x^{a_{0}}_{0}\cdots x^{a_{n}}_{n} $. 令$ I = I_{X} $是$ {\Bbb C}[x_{0}, \cdots, x_{n}] $中由$ X $生成的理想, $ {\Bbb C}[x_{0}, \cdots, x_{n}]_{m} $表示$ {\Bbb C}[x_{0}, \cdots, x_{n}] $中次数为$ m $(包括0次)的齐次多项式组成的向量空间, $ I_{m}: = {\bf C}[x_{0}, \cdots, x_{n}]_{m}\cap {I} $, $ m = 1, 2, \cdots $. 我们定义$ X $中的希尔伯特函数$ H_{X}(m) $为

$ H_{X}(m): = \dim \left(\frac{{\Bbb C}[x_{0}, \cdots, x_{n}]_{m}}{I_{m}}\right). $

由希尔伯特多项式性质得

$ H_{X}(m) = \Delta\cdot\frac{m^{n}}{n!}+O(m^{n-1}). $

对$ {\bf c} = (c_{0}, \cdots, c_{n})\in {\Bbb R}^{n+1} $, 我们定义$ X $的$ m $次希尔伯特函数权重$ S_{X}(m, {\bf c}) $如下

$ S_{X}(m, {\bf c}): = \max\left(\sum\limits^{H_{X}(m)}_{i = 1}{\bf a}_{i}\cdot{\bf c}\right), $

其中最大值是在所有满足$ {\bf x}^{{\bf a}_{1}}, \cdots, {\bf x}^{{\bf a}_{H_{X}(m)}} $是$ \frac{{\Bbb C}[x_{0}, \cdots, x_{n}]_{m}}{I_{m}} $的基底的集合里取.

依据Mumford的工作, 有

$ S_{X}(m, {\bf c}) = e_{X}({\bf c}).\frac{m^{n+1}}{(n+1)!}+O(m^{n}), $

即

$ \lim\limits_{m\rightarrow \infty}\frac{1}{mH_{X}(m)}\cdot S_{X}(m, {\bf c}) = \frac{1}{(n+1)\Delta}\cdot e_{X}({\bf c}). $

我们称$ \frac{1}{mH_{X}(m)}\cdot S_{X}(m, {\bf c}) $为$ m $次希尔伯特函数标准权, $ \frac{1}{(n+1)\Delta}\cdot e_{X}({\bf c}) $为$ X $中关于$ {\bf c} $的Chow权.

引理2.1^[3]   设$ X\subset {\Bbb P}^{n}({\Bbb C}) $是维数为$ k $且次数为$ \Delta $的代数簇. 令$ m>\Delta $是一个整数, $ {\bf c} = (c_{0}, \cdots, c_{n})\in {\Bbb R}^{n+1}_{\geq 0} $. 则

$ \frac{1}{mH_{X}(m)}S_{X}(m, {\bf c})\geq \frac{1}{(k+1)\Delta}e_{X}({\bf c})-\frac{(2k+1)\Delta}{m} \cdot\left(\max\limits_{i = 0, \cdots, n}c_{i}\right). $

引理2.2^[11]  设$ Y $是$ {\Bbb P}^{q-1}({\Bbb C}) $中维数是$ n $且次数是$ \Delta $的代数子簇. 设$ {\bf c} = (c_{1}, \cdots, c_{q}) $为实数数组, $ \{i_{0}, \cdots, i_{n}\} $是$ \{1, \cdots, q\} $中的子集且满足

$ Y\cap \{y_{i_{0}} = \cdots = y_{i_{n}} = 0\} = \emptyset, $

则

$ e_{Y}({\bf c})\geq (c_{i_{0}}+\cdots+c_{i_{n}})\Delta. $

引理2.3^[12]  设$ V $为$ {\Bbb P}^{n}({\Bbb C}) $中维数是$ k $的光滑射影子簇, 令$ Q_{1}, \cdots, Q_{N+1} $为$ {\Bbb P}^{n}({\Bbb C}) $中有相同次数$ d\geq 1 $的超曲面, 且满足

$ \left(\bigcap\limits^{N+1}_{i = 1}Q_{i}\right)\cap V = \emptyset, $

则存在$ k+1 $个超曲面

$ P_{1} = Q_{1}, \ P_{t} = \sum\limits^{N-k+t}_{j = 2}c_{tj}Q_{j}, c_{tj}\in {\Bbb C}, t = 2, \cdots, k+1, $

使得$ \left(\bigcap\limits^{k+1}_{t = 1}P_{t}\right)\cap V = \emptyset $成立.

定理1.2的证明  首先我们证明该定理对于所有超曲面$ Q_{i}(1\leq i\leq q) $有相同次数$ d $的情况. 我们假设$ q>(N-k+1)(k+1) $, 记$ {\cal I} $为$ \{1, \cdots, q\} $所有排列组合情况的集合. $ n_{0} $是集合$ {\cal I} $的个数, 显然$ n_{0} = q! $. 我们记集合$ {\cal I} = \{I_{1}, \cdots, I_{n_{0}}\} $, 其中$ I_{i} = (I_{i}(1), \cdots, I_{i}(q))\in {\Bbb N}^{q} $, $ I_{1}< I_{2}<\cdots<I_{n_{0}} $是按照字典顺序排序. 对于每一个$ I_{i}\in{\cal I} $, 我们记$ P_{i, 1}, \cdots, P_{i, k+1} $由引理2.3得到的超曲面. 构造映射$ \Phi: V\rightarrow {\Bbb P}^{l-1}({\Bbb C})(l = n_{0}(k+1)) $,

$ \Phi(x) = (P_{1, 1}(x):\cdots:P_{1, k+1}(x):\cdots :P_{n_{0}, 1}(x):\cdots:P_{n_{0}, k+1}(x)). $

置$ Y = \Phi(V) $. 因为$ V\cap \bigcap\limits^{k+1}_{j = 1}P_{1, j} = \emptyset $, 所以$ \Phi $是一个$ V $中的有限态射, $ Y $是$ {\Bbb P}^{l-1}({\Bbb C}) $中次数为$ \dim Y = k $的复射影子簇且$ \Delta: = \deg Y\leq d^{k}\cdot \deg V $. 对于每一个

$ {\bf a} = (a_{1, 1}, \cdots, a_{1, k+1}, a_{2, 1}, \cdots, a_{2, k+1}, \cdots, a_{n_{0}, 1}, \cdots, a_{n_{0}, k+1}), $

$ {\bf y} = (y_{1, 1}, \cdots, y_{1, k+1}, y_{2, 1}, \cdots, y_{2, k+1}, \cdots, y_{n_{0}, 1}, \cdots, y_{n_{0}, k+1}), $

记$ {\bf y}^{{\bf a}} = y^{a_{1, 1}}_{1, 1}\cdots y^{a_{1, k+1}}_{1, k+1}\cdots y^{a_{n_{0}, 1}}_{n_{0}, 1}\cdots y^{a_{n_{0}, k+1}}_{n_{0}, k+1} $. 令

$ n_{u}: = H_{Y}(u)-1, l_{u}: = \left(\begin{array}{c}l+u-1 \\ u\end{array}\right), \ Y_{u} = \frac{{\Bbb C}[y_{1}, \cdots, y_{l}]_{u}}{(I_{Y})_{u}}, $

则$ Y_{u} $是有限维的向量空间, 设$ \{\upsilon_{0}, \cdots, \upsilon_{n_{u}}\} $是它的一组基. 考虑全纯映射

$ \widetilde{F} = (\upsilon_{0}(\Phi\circ\widetilde{f}), \cdots, \upsilon_{n_{u}}(\Phi\circ\widetilde{f})):S\rightarrow {\Bbb C}^{n_{u}+1}, \ F = \rho(\widetilde{F}):S\rightarrow {\Bbb P}^{n_{u}}({\Bbb C}). $

因为$ f $是代数非退化的, 因此$ F $是线性非退化的.

由于$ \frac{1}{\deg Q}\sum\limits_{z\in S}\nu^{0}_{Q(f)}(z) $不依赖于$ Q $的选择. 取$ Q = \{w_{0} = 0\} $为$ {\Bbb P}^{n_{u}}({\Bbb C}) $中的超曲面, 因此

(3.1) $ \begin{equation} \deg F = \sum\limits^{r}_{j = 1} \nu_{\upsilon_{0}(\Phi\circ\widetilde{f})}(z_{j}) = ud\cdot\deg(f). \end{equation} $

情况1  当$ z\in E $, 我们令$ c_{i, j} = \nu^{0}_{P_{i, j}(f)}(z) $,

$ {\bf c} = (c_{1, 1}, \cdots, c_{1, k+1}, \cdots, c_{n_{0}, 1}, \cdots, c_{n_{0}, k+1}). $

由希尔伯特函数的定义,

$ {\bf a}_{i} = (a_{i, 1, 1}, \cdots, a_{i, 1, k+1}, \cdots, a_{i, n_{0}, 1}, \cdots, a_{i, n_{0}, k+1}), a_{i, j, s}\in \{1, \cdots, l_{u}\} $

是满足$ {\bf y}^{a_{1}}, \cdots, {\bf y}^{a_{H_{Y}(u)}} $的剩余模类$ (I_{Y})_{u} $中$ Y_{u} $的一组基, 且

$ S_{Y}(u, {\bf c}) = \sum\limits^{H_{Y}(u)}_{i = 1}{\bf a}_{i}\cdot{\bf c}. $

又因为$ {\bf y}^{{\bf a}_{i}}\in Y_{u} $, 则

$ {\bf y}^{{\bf a}_{i}} = L_{i}(\upsilon_{0}, \cdots, \upsilon_{n_{u}}), $

其中$ L_{i}(1\leq i\leq H_{Y}(u)) $是线性无关的线性形式. 对一组$ {\Bbb P}^{n_{u}}({\Bbb C}) $中的超平面的线性无关形式$ \{L_{i}(\widetilde{F})\}(1\leq i\leq H_{Y}(u)) $, 容易得到

$ \nu^{0}_{L_{i}(\widetilde{F})}(z) = \sum\limits_{1\leq s\leq k+1}\sum\limits_{1\leq j\leq n_{0}} a_{i, j, s}\nu^{0}_{P_{j, s}(\widetilde{f})}(z) = {\bf a}_{i}\cdot{\bf c}. $

因此

$ \sum\limits^{H_{Y}(u)}_{i = 1}\nu^{0}_{L_{i}(\widetilde{F})}(z) = \sum\limits^{H_{Y}(u)}_{i = 1} {\bf a}_{i}\cdot{\bf c} = S_{Y}(u, {\bf c}). $

因为$ P_{1, 1}, \cdots, P_{1, k+1} $处于一般位置, 由引理2.2我们有

$ e_{Y}({\bf c})\geq \Delta \cdot \sum\limits^{k+1}_{j = 1}c_{1, j} = \Delta \cdot \sum\limits^{k+1}_{j = 1}\nu^{0}_{P_{1, j}(f)}(z). $

结合以上不等式以及引理2.1, 得到

(3.2) $ \begin{eqnarray} \sum^{H_{Y}(u)}_{i = 1}\nu^{0}_{L_{i}(\widetilde{F})}(z) \geq\frac{uH_{Y}(u)}{k+1}\sum^{k+1}_{j = 1}\nu^{0}_{P_{1, j}(f)}(z) -(2k+1)H_{Y}(u)\Delta \max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z). \end{eqnarray} $

对于$ {\Bbb P}^{n_{u}}({\Bbb C}) $中超平面的一组线性无关的形式$ L_{i}(\widetilde{F})(z) $, 考虑局部参数化, 把$ F $写成(2.1)式的形式, 得

$ \sum\limits^{H_{Y}(u)}_{i = 1}\nu^{0}_{L_{i}(\widetilde{F})(z)}\leq \delta_{0}(z)+\delta_{1}(z)+\cdots+\delta_{H_{Y}(u)}(z). $

由(2.2)式, 得

$ \sum\limits^{H_{Y}(u)}_{i = 1}\nu^{0}_{L_{i}(\widetilde{F})(z)}\leq \sum\limits^{H_{Y}(u)-1}_{i = 0}(H_{Y}(u)-i)\nu_{i}(z)+\frac{1}{2}H_{Y}(u)(H_{Y}(u)-1). $

结合不等式(3.2), 有

(3.3) $ \begin{eqnarray} & &\sum^{H_{Y}(u)-1}_{i = 0}\sum\limits_{z\in E}(H_{Y}(u)-i)\nu_{i}(z){}\\ & \geq &\frac{u H_{Y}(u)}{k+1}\sum^{k+1}_{j = 1}\sum\limits_{z\in E}\nu^{0}_{P_{1j}(f)}(z)-(2k+1)H_{Y}(u)\Delta \sum\limits_{z\in E} \max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z){}\\ &&-\frac{1}{2}H_{Y}(u)(H_{Y}(u)-1)|E|{}\\ & \geq& \frac{u H_{Y}(u)}{(N-k+1)(k+1)}\sum^{q}_{j = 1}\sum\limits_{z\in E}\nu^{0}_{Q_{j}(f)}(z)-\frac{1}{2}H_{Y}(u)(H_{Y}(u)-1)|E|{}\\ & &-(2k+1)H_{Y}(u)\Delta \sum\limits_{z\in E} \max\limits_{ 1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z). \end{eqnarray} $

情况2  当$ z\not\in E $, 设$ c_{i, j} = \max\{0, \nu_{P_{i, j}(f)}(z)-n_{u}\} $,

$ {\bf c} = (c_{1, 1}, \cdots, c_{1, k+1}, \cdots, c_{n_{0}, 1}, \cdots, c_{n_{0}, k+1}). $

取

$ {\bf a}_{i} = (a_{i, 1, 1}, \cdots, a_{i, 1, k+1}, \cdots, a_{i, n_{0}, 1}, \cdots, a_{i, n_{0}, k+1}), a_{i, j, s}\in \{1, \cdots, l_{u}\}. $

使得$ {\bf y}^{a_{1}}, \cdots, {\bf y}^{a_{H_{Y}(u)}} $是$ Y_{u} $的一组基, 且

$ S_{Y}(u, {\bf c}) = \sum\limits^{H_{Y}(u)}_{i = 1}{\bf a}_{i}\cdot{\bf c}. $

类似于情况1, 令$ {\bf y}^{{\bf a}_{i}} = L_{i}(\upsilon_{0}, \cdots, \upsilon_{n_{u}})(1\leq i\leq q) $, $ L_{1}, \cdots, L_{H_{Y}(u)} $是变量$ y_{i, j} $$ (1\leq i\leq n_{0}, $$ 1\leq j\leq k+1) $线性无关的线性形式. 取$ \upsilon_{i}\in H_{d} $使得$ \{\upsilon_{0}, \cdots, \upsilon_{n_{u}}\} $是$ Y_{u} $在$ {\Bbb C} $上的一组基. 由$ k+1 $个超曲面$ P_{1, 1}(f), \cdots, P_{1, k+1}(f) $取法知, $ \{L_{i}(\widetilde{F})(z)\}_{i = 1}^q $线性无关. 不失一般性, 不妨设

$ \nu^{0}_{P_{1, 1}(f)}(z)\leq\cdots\leq \nu^{0}_{P_{1, k+1}(f)}(z), $

则

(3.4) $ \begin{eqnarray} \sum^{H_{Y}(u)}_{i = 1}\nu^{0}_{L_{i}(\tilde{F})}(z)-\sum^{H_{Y}(u)}_{i = 1} \min\{\nu_{L_{i}(\tilde{F})}(z), n_{u}\} & = &\sum^{H_{Y}(u)}_{i = 1}\max\{\nu_{L_{i}(\tilde{F})}(z)-n_{u}, 0\}{}\\ & \leq&\sum^{H_{Y}(u)-1}_{i = 0}(\delta_{i}(z)-i). \end{eqnarray} $

因为

$ \max\{\nu_{L_{i}(\tilde{F})}(z)-n_{u}, 0\}\geq \sum\limits^{H_{Y}(u)}_{i = 1}a_{i, j, s}c_{j, s} = {\bf a}_{i}\cdot{\bf c}, $

结合(3.4)式, 我们得

$ \sum\limits^{H_{Y}(u)-1}_{i = 0}(\delta_{i}(z)-i)\geq \sum\limits^{H_{Y}(u)}_{i = 1}\max\{\nu_{L_{i}(\tilde{F})}(z)-n_{u}, 0\} \geq \sum\limits^{H_{Y}(u)}_{i = 1}{\bf a}_{i}\cdot{\bf c}. $

由(2.2)式, 得

(3.5) $ \begin{eqnarray} \sum^{H_{Y}(u)-1}_{i = 0}(H_{Y}(u)-i)\nu_{i}(z) = \sum^{H_{Y}(u)-1}_{i = 0}(\delta_{i}(z)-i) \geq \sum^{H_{Y}(u)}_{i = 1}{\bf a}_{i}\cdot{{\bf c}} = S_{Y}(u, {\bf c}). \end{eqnarray} $

由引理2.1, 知

(3.6) $ \begin{eqnarray} S_{Y}(u, {\bf c})\geq \frac{uH_{Y}(u)}{(k+1)\Delta}e_{Y}({\bf c})-(2k+1)\Delta H_{Y}(u) \max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}{\bf c}_{ij}. \end{eqnarray} $

而$ P_{1, 1}, \cdots, P_{1, k+1} $处于一般位置, 根据引理2.2, 可得

(3.7) $ \begin{eqnarray} e_{Y}({\bf c)}\geq\Delta\sum\limits_{1\leq j\leq k+1}{\bf c}_{1, j} = \Delta\sum\limits_{1\leq j\leq k+1}\max\{\nu_{P_{1, j}(f)}(z)-n_{u}, 0\}. \end{eqnarray} $

另外，对所有的$ 2\leq j\leq k+1 $, 有

$ \nu^{0}_{P_{1, j}(f)}(z)\geq \nu^{0}_{Q_{N-k+j}(f)}(z). $

因此

$ \begin{eqnarray*} & &(N-k+1)\sum^{k+1}_{j = 1}\max\{0, \nu_{P_{1, j}(f)}(z)-n_{u}\}\\ &\geq& (N-k+1)\max\{0, \nu^{0}_{Q_{1}(f)}(z)-n_{u}\}+ \sum^{k+1}_{j = 2}\max\{0, \nu^{0}_{Q_{N-k+1}(f)}(z)-n_{u}\}\\ & \geq&\sum^{N+1}_{i = 1}\max\{0, \nu^{0}_{Q_{i}(f)}(z)-n_{u}\} = \sum^{q}_{i = 1}\max\{0, \nu^{0}_{Q_{i}(f)}(z)-n_{u}\}. \end{eqnarray*} $

结合不等式(3.5), (3.6)和(3.7), 可得

$ \begin{eqnarray*} & &\sum^{H_{Y}(u)-1}_{i = 0}\sum\limits_{z\not\in E}(H_{Y}(u)-i)\nu_{i}(z)\\ &\geq&\frac{uH_{Y}(u)}{k+1}\sum\limits_{z\not\in E}\sum^{k+1}_{j = 1}\max \{\nu^{0}_{P_{1, j}(f)}(z)-n_{u}, 0\} -(2k+1)\Delta H_{Y}(u)\sum\limits_{z\not\in E}\max\nu^{0}_{P_{i, j}(f)}(z)\\ & \geq&\frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\max \{\nu^{0}_{Q_{j}(f)}(z)-n_{u}, 0\} -(2k+1)\Delta H_{Y}(u)\sum\limits_{z\not\in E}\max\nu^{0}_{P_{i, j}(f)}(z)\\ & \geq& \frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\nu^{0}_{Q_{j}(f)}(z) -\frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), n_{u}\}\\ &&-(2k+1)\Delta H_{Y}(u)\sum\limits_{z\not\in E}\max\nu^{0}_{P_{i, j}(f)}(z). \end{eqnarray*} $

由不等式(3.3), 得

$ \begin{eqnarray*} & &\sum^{H_{Y}(u)-1}_{i = 0}\sum\limits_{z\in S}(H_{Y}(u)-i)\nu_{i}(z)\\ & \geq&\frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\in S}\sum^{q}_{j = 1}\nu^{0}_{Q_{j}(f)}(z) -\frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), n_{u}\}\\ & &-\frac{1}{2}H_{Y}(u)(H_{Y}(u)-1)|E|-(2k+1)\Delta H_{Y}(u)\sum\limits_{z\in S} \max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z). \end{eqnarray*} $

另一方面，不等式(2.3)与(2.4)可知

$ \begin{eqnarray*} & &udH_{Y}(u) \deg(f)+H_{Y}(u)(H_{Y}(u)-1)(g-1)\\ & \geq&\frac{uH_{Y}(u)}{(N-k+1)(k+1)}\sum\limits_{z\in S}\sum^{q}_{j = 1}\nu^{0}_{Q_{j}(f)}(z) -\frac{uH_{Y}(u)}{(N-k+1)(k+1)} \sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), n_{u}\}\\ & &-\frac{1}{2}H_{Y}(u)(H_{Y}(u)-1)|E|-(2k+1)\Delta H_{Y}(u)\sum\limits_{z\in S} \max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z). \end{eqnarray*} $

因此

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1))\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), n_{u}\} +\frac{1}{2ud}(N-k+1)(k+1)(H_{Y}(u)-1)\{2(g-1)+|E|\}\\ & &+\frac{1}{ud}(N-k+1)(k+1)(2k+1)\Delta\sum\limits_{z\in S}\max\limits_{1\leq i\leq n_{0}\atop 1\leq j\leq k+1}\nu^{0}_{P_{i, j}(f)}(z)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), n_{u}\} +\frac{1}{2ud}(N-k+1)(k+1)(H_{Y}(u)-1)\{2(g-1)+|E|\}\\ & &+\frac{1}{u}(N-k+1)(k+1)(2k+1)\Delta l\deg(f). \end{eqnarray*} $

对任意的$ \varepsilon>0 $, 我们选择正整数$ u $使得$ \frac{(N-k+1)(k+1)(2k+1)\Delta l}{u}< \varepsilon $, 则

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), H_{Y}(u)-1\} +\frac{(N-k+1)(k+1)(H_{Y}(u)-1)\{2(g-1)+|E|\}}{2ud}. \end{eqnarray*} $

注意到$ \deg Y = \Delta \leq d^{k} \deg(V) $,

$ \begin{eqnarray*} n_{u}& = &H_{Y}(u)-1\leq \Delta \left(\begin{array}{c} k+u \\ k \end{array} \right) \leq d^{k}\deg(V)e^{k}(1+\frac{u}{k})^{k}\\ &<& d^{k}\deg(V)e^{k}((N-k+1)(2k+4)\Delta l\varepsilon^{-1})^{k}\\ &\leq& [\deg(V)^{k+1}e^{k}d^{k^{2}+k}(N-k+1)^{k}(2k+4)^{k}l^{k}\varepsilon^{-k}] = M. \end{eqnarray*} $

从而

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1)-\varepsilon)\deg(f)\\ &\leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M\}+\frac{(N-k+1)(k+1)M\{2(g-1)+|E|\}}{2ud}. \end{eqnarray*} $

一般情况, deg$ Q_{i} = d_i $, $ i = 1, \cdots, q $, $ d: = lcm(d_{1}, \cdots, d_{q}) $. 此时超曲面$ Q^{\frac{d}{d_{i}}}_{i}(i = 1, \cdots, q) $具有相同的次数$ d $, 应用上述结论, 得

$ \begin{eqnarray*} & &(q-(N-k+1)(k+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M\}+\frac{(N-k+1)(k+1)M\{2(g-1)+|E|\}}{2ud}. \end{eqnarray*} $

定理1.2证毕.

定理1.3的证明  不妨假设所有的超曲面$ Q_{i} $$ (1\leq i\leq q) $都有相同的次数并且$ q>(N-n+1)(n+1) $. 取正整数$ u $, 记$ V_{u} $为由所有次数是$ u $的齐次多项式和零多项式生成的空间, 则$ V_{u} $是$ {\Bbb C}[x_{0}, \cdots, x_{n}] $的一个向量子空间. 设$ u $被$ d $整除, 对$ (i) = (i_{1}, \cdots, i_{n})\in {\Bbb N}^{n}_{0} $和$ \sigma(i) = \sum\limits^{n}_{s = 1}i_{s}\leq \frac{N}{d} $, 我们定义

$ W^{i_{0}}_{(i)} = \sum\limits_{(j) = (j_{1}, \cdots, j_{n})\geq (i)}P^{j_{1}}_{i_{0}, 1}\cdots P^{j_{n}}_{i_{0}, n}\cdot V_{u-d\sigma(j)}. $

显然$ W^{i_{0}}_{(0, \cdots, 0)} = V_{u} $, $ W^{i_{0}}_{(i)}\supset W^{i_{0}}_{(j)} $, $ (i)<(j) $ (按照字典顺序排序), 因此$ W^{i_{0}}_{(i)} $是$ V_{u} $的一个筛选. Corvaja和Zannier^[1]证明了如下结论:

引理3.1^[1]  令$ (i) = (i_{1}, \cdots, i_{n}) $, $ (i)^{'} = (i_{1}^{'}, \cdots, i_{n}^{'})\in {\Bbb N}^{n}_{0} $, 按照字典顺序排序$ (i^{'})<(i) $. 令

$ m^{i_{0}}_{(i)} = \dim\frac{W^{i_{0}}_{(i)}}{W^{i_{0}}_{(i)^{'}}}, $

若$ d\sigma(i)<u-nd $, 则$ m^{i_{0}}_{(i)} = d^{n} $.

设

$ V_{u} = W^{i_{0}}_{(i)_{1}}\supset W^{i_{0}}_{(i)_{2}}\supset \cdots \supset W^{i_{0}}_{(i)_{K}}, $

其中$ (i)_{s} = (i_{1s}, \cdots, i_{ns}) $, $ W^{i_{0}}_{(i)_{s+1}} $在$ W^{i_{0}}_{(i)_{s}} $后面且$ (i)_{K} = (\frac{u}{d}, 0, \cdots, 0) $. $ K $是满足$ i_{1}+\cdots+i_{n}\leq \frac{u}{d} $的$ n $元数组$ (i_{1}, \cdots, i_{n}) $, $ i_{j}\geq 0 $的个数, 所以

$ K = \left( \begin{array}{c} { } \frac{u}{d}+n \\ n \end{array} \right). $

对于所有的$ s = 1, \cdots, K-1 $, 定义$ m^{i_{0}}_{s} = \dim\frac{W^{i_{0}}_{(i)_{s}}}{W^{i_{0}}_{(i)_{s+1}}} $, $ m^{i_{0}}_{K} = 1 $.

令$ m_{u} = \dim V_{u} $, 由上述的筛选办法, 我们可以选择$ V_{u} $的一组基$ \{\psi^{i_{0}}_{1}, \cdots, \psi^{i_{0}}_{m_{u}}\} $, 使得

$ \{\psi_{m_{u}-(m^{i_{0}}_{s}+\cdots+m^{i_{0}}_{K})+1}, \cdots, \psi_{m_{u}}\} $

是$ W^{i_{0}}_{(i)_{s}} $的一组基. 对每一个$ s\in \{1, \cdots, K\} $, $ l\in \{m_{u}-(m^{i_{0}}_{s}+\cdots+m^{i_{0}}_{K})+1, \cdots, m_{u}-(m^{i_{0}}_{s+1}+\cdots+m^{i_{0}}_{K})\} $, 满足

$ \psi^{i_{0}}_{l} = P^{i_{1s}}_{i_{0}1}\cdots P^{i_{ns}}_{i_{0}n}h_{l}, $

其中$ (i_{1s}, \cdots, i_{ks}) = (i)_{s} $, $ h_{l}\in W^{i_{0}}_{u-d\sigma(i)_{s}} $.

我们固定$ V_{u} $的一组基$ \phi_{1}, \cdots, \phi_{m_{u}} $. 设$ L^{i_{0}}_{s}(\widetilde{F}) $$ (1\leq s\leq m_{u}) $是线性形式并且$ \widetilde{F} = (\phi_{1}(\widetilde{f}), \cdots, \phi_{m_{u}}(\widetilde{f})) $是全纯映射$ F $的精简形式, $ \psi^{i_{0}}_{s}(\widetilde{f}) = L^{i_{0}}_{s}(\widetilde{F}) $. 令

$ b^{i_{0}}_{j} = \sum\limits^{K}_{s = 1}m^{i_{0}}_{s}i_{js}, (1\leq j\leq k). $

与定理1.2类似, 可得

$ \begin{eqnarray*} &&(q-p(n+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), m_{u}-1\} +\frac{p(n+1)(m_{u}-1)(2(g-1)+|E|)}{2ud}. \end{eqnarray*} $

我们选择$ u = (n+1)d+p(n+1)^{3}I(\varepsilon^{-1})d $,

$ \begin{eqnarray*} m_{u}& = &\left( \begin{array}{c} u+n \\ n \end{array} \right)\leq e^{n}(1+\frac{u}{n})^{n}\\ & = &e^{n}\left(\frac{n+(n+1)d}{n}+\frac{p(n+1)^{3}I(\varepsilon^{-1})d}{n}\right)^{n}\\ & = &e^{n}(p(n+1)^{2}I(\varepsilon^{-1})d)^{n}\left(1+\frac{1}{n}+ \frac{n+(n+1)d}{np(n+1)^{2}I(\varepsilon^{-1})d}\right)^{n}\\ &\leq& (edp(n+1)^{2}I(\varepsilon^{-1}))^{n}\left(1+\frac{1}{n}+\frac{2}{n(n+1)}\right)^{n}\\ &<& 4(edp(n+1)^{2}I(\varepsilon^{-1}))^{n} = M_{0}. \end{eqnarray*} $

因此

$ \begin{eqnarray*} & &(q-p(n+1)-\varepsilon)\deg(f)\\ & \leq&\frac{1}{d}\sum\limits_{z\not\in E}\sum^{q}_{j = 1}\min\{\nu_{Q_{j}(f)}(z), M_{0}-1\} +\frac{p(n+1)(M_{0}-1)(2(g-1)+|E|)}{2ud}. \end{eqnarray*} $

定理1.3证毕.

定理1.5的证明  因为全曲率有限的完备极小曲面$ S $共形等价于紧黎曼面$ \overline{S} $挖掉有限个点$ P_{1}, \cdots, P_{r} $, 并且其广义高斯映射$ G $可全纯延拓到$ \overline{G}:S\rightarrow {\Bbb P}^{m-1}({\Bbb C}) $. 令

$ E = \{P_{1}, \cdots, P_{r}\}. $

根据Chern与Ossern^[2]证明的结论, 可以得到

$ C(S) = -2\pi \deg(\overline{G})\leq 2\pi (\chi-r) = 2\pi(2-2g-r-r), $

其中$ \chi $是$ \overline{S} $的欧拉特征数, $ g $是$ \overline{S} $的亏格. 因此

$ 2(g-1)\leq \deg(\overline{G})-2r. $

从而

(3.8) $ \begin{equation} 2(g-1)+|E|\leq \deg(\overline{G})-r<\deg(\overline{G}). \end{equation} $

由定理1.2, 得

(3.9) $ \begin{eqnarray} & &(q-p(n+1)-\varepsilon)\deg(\overline{G}){}\\ &\leq&\frac{1}{d_{j}}\sum\limits_{p\not\in E}\sum^{r_{0}}_{j = 1}\min\{\nu_{Q_{j}(\overline{G}})(z), M_{0}-1\} +\frac{1}{d_{j}}\sum\limits_{p\not\in E}\sum^{q}_{j = r_{0}+1}\min\{\nu_{Q_{j}(\overline{G}})(z), M_{0}-1\}{}\\ & &+\frac{1}{2u}p(n+1)(M_{0}-1)\frac{\deg(\overline{G})}{d}. \end{eqnarray} $

若$ P\not\in E $, 则$ Q_{r_{0}+1}, \cdots, Q_{q} $是例外超曲面, 故$ \nu^{0}_{Q_{j}(\overline{G})}(P) = 0 $, $ r_{0}+1\leq j\leq q $. 对$ P\in S $, $ 1\leq j\leq r_{0} $,

(3.10) $ \begin{equation} \min\{\nu_{Q_{j}(\overline{G})}(P), M_{0}-1\}\leq (M_{0}-1)\cdot \min \{\nu_{Q_{j}(\overline{G})}(P), 1\} \leq \frac{M_{0}-1}{m_{j}}\nu^{0}_{Q_{j}(\overline{G})}(P). \end{equation} $

由(3.1) 式, 我们有

$ \begin{eqnarray*} \sum^{r_{0}}_{j = 1}\sum\limits_{P\not\in E}\frac{M_{0}-1}{m_{j}d_{_{j}}}\nu^{0}_{Q_{j}(\overline{G})}(P) \leq \sum^{r_{0}}_{j = 1}\sum\limits_{P\in \overline{S}}\frac{M_{0}-1}{m_{j}d_{_{j}}}\nu^{0}_{Q_{j}(\overline{G})}(P). = \sum^{r_{0}}_{j = 1}\frac{(M_{0}-1)\deg(\overline{G})}{m_{j}}. \end{eqnarray*} $

结合(3.8), (3.9)与(3.10)式, 得

$ (q-p(n+1)-\varepsilon)\deg(\overline{G})\leq \sum\limits^{r_{0}}_{j = 1}\frac{(M_{0}-1)\deg(\overline{G})}{m_{j}} +\frac{1}{2u}p(n+1)(M_{0}-1) \frac{\deg(\overline{G})}{d}. $

任意的$ 1\leq k\leq m-1 $,

$ \sum\limits^{q}_{j = 1}(1-\frac{M_{0}-1}{m_{j}})\leq q-\sum\limits^{r_{0}}_{j = 1}\frac{M_{0}-1}{m_{j}} <\frac{p(n+1)[2ud+(M_{0}-1)]}{2ud}. $

定理1.5证毕.