对极小曲面的研究一直是微分几何研究领域中的一个重要课题, 特别是关于极小曲面的几何性质以及分类问题的研究。当外围空间是空间形式的时候, 一些重要的分类结果已经先后被提出。更为一般的情况是外围空间是对称空间的时候, 对称空间中的极小曲面研究也有不错的进展, 但依旧有许多值得关注的问题。所以, 本文主要关注外围空间是四元素射影空间的情形, 这对于四元素射影空间中极小曲面的几何及分类的研究是具有重要意义的。

在近几十年中, 国内外对极小曲面的研究都取得了许多重要成果。

1982年, Bryant^[1]通过扭映射π:${\mathbb{C}}{P^3} \to {\mathbb{H}}{P^1}$证明${\mathbb{C}}{P^3}$中水平全纯曲面的投影为${\mathbb{H}}{P^1}$中的极小曲面, 并且构造出${\mathbb{C}}{P^3}$中紧致的非分歧的全纯水平曲面。1986年Aithal^[2]构造${\mathbb{H}}{P^2}$中所有的调和二维球面。Bolton等^[3]也在1988年给出${\mathbb{C}}{P^n}$中的Veronese序列即常曲率极小二维球面。1991年, Bahy-El-Dien和Wood^[4]构造${\mathbb{H}}{P^n}$中的所有的调和二维球面。2014年, He和Jiao^[5]给出${\mathbb{H}}{P^2}$中线性满、全非分歧的常曲率共形极小球面的分类。同年, He和Jiao^[6]给出${\mathbb{H}}{P^n}$中第二基本形式平行的共形极小球面的分类。

近几十年中, 关于复射影空间中极小曲面的几何研究, 许多学者已经给出很多重要的结论。而由扭映射π:${\mathbb{C}}{P^3} \to {\mathbb{H}}{P^1}$可知四元素射影空间与复射影空间的水平分量有一个自然等同, 所以我们希望通过复射影空间研究四元素射影空间。

1 预备知识

首先介绍四元数、扭映射以及相关的一些知识。

四元数${\mathbb{H}}$是以1, i, j, k为基的一个四维实向量空间, 即${\mathbb{H}} = \{ a + b{\rm{i}} + c{\rm{j}} + d{\rm{k}}|a, b, c, d \in {\mathbb{R}}\} $, 其中1, i, j, k满足:

$ \begin{array}{c}{\mathrm{i}^{2}=\mathrm{j}^{2}=\mathrm{k}^{2}=-1, \mathrm{ij}=\mathrm{k}=-\mathrm{ji}} \\ {\mathrm{jk}=\mathrm{i}=-\mathrm{kj}, \mathrm{ki}=\mathrm{j}=-\mathrm{ik}}.\end{array} $

由此可见${\mathbb{H}}$为一个不可交换环。

因而, 对于${h_1} = {a_1} + {b_1}{\rm{i}} + {c_1}{\rm{j}} + {d_1}{\rm{k}}$, ${h_2} = {a_2} + {b_2}{\rm{i}} + {c_2}{\rm{j}} + {d_2}{\rm{k}} \in {\mathbb{H}}$有

$ \begin{array}{c}{h_{1} h_{2}=\left(a_{1} a_{2}-b_{1} b_{2}-c_{1} c_{2}-d_{1} d_{2}\right)+} \\ {\quad\left(a_{1} b_{2}+a_{2} b_{1}+c_{1} d_{2}-c_{2} d_{1}\right) \mathrm{i}+} \\ {\left(a_{1} c_{2}+a_{2} c_{1}+d_{1} b_{2}-d_{2} b_{1}\right) \mathrm{j}+} \\ {\quad\left(a_{1} d_{2}+a_{2} d_{1}+b_{1} c_{2}-b_{2} c_{1}\right) \mathrm{k}}.\end{array} $

与复数${\mathbb{C}}$类似, 四元数上也有个自然的共轭作用^*:若${\rm{h = }}a + b{\rm{i}} + c{\rm{j}} + d{\rm{k}}$, 则${{\rm{h}}^*}{\rm{ = }}a - b{\rm{i}} - c{\rm{j}} - d{\rm{k}}$.

更为经常的, 我们是把${\mathbb{H}}$看成以1, j为基的复数域${\mathbb{C}}$上的二维右模, 则$\forall h \in {\mathbb{H}}$, 可写成$\mathit{\boldsymbol{h}} = \mathit{\boldsymbol{u}} + {\rm{j}}\mathit{\boldsymbol{v}}, \mathit{\boldsymbol{u}}, \mathit{\boldsymbol{v}} \in {\mathbb{C}}$。对于${\mathit{\boldsymbol{h}}_1} = {\mathit{\boldsymbol{u}}_1} + {\rm{j}}{\mathit{\boldsymbol{v}}_1}$, $\mathit{\boldsymbol{h}}{\mathit{\boldsymbol{h}}_1} = \left( {\mathit{\boldsymbol{uv}} - {{\mathit{\boldsymbol{\bar u}}}_1}{\mathit{\boldsymbol{v}}_1}} \right) + {\rm{j}}\left( {{\mathit{\boldsymbol{u}}_1}\mathit{\boldsymbol{v}} + \mathit{\boldsymbol{\bar u}}{\mathit{\boldsymbol{v}}_1}} \right)$, 且知上述两种定义等价。

一般来说, 用${{\mathbb{H}}^n}$表示$ n $个四元数组成的列向量, ${\mathbb{H}}{P^n}$表示$ {{\mathbb{H}}^{n + 1}} $中所有通过原点的直线组成的集合, 则$ {[{\mathit{\boldsymbol{h}}_1}]_{\mathbb{H}}} $和$ {[{\mathit{\boldsymbol{h}}_2}]_{\mathbb{H}}} \in {\mathbb{H}}{P^n} $, $ {[{\mathit{\boldsymbol{h}}_1}]_{\mathbb{H}}} = {[{\mathit{\boldsymbol{h}}_2}]_{\mathbb{H}}}$当且仅当存在$ \mathit{\boldsymbol{h}} \in {\mathbb{H}} $, 使得$ {\mathit{\boldsymbol{h}}_1} = {\mathit{\boldsymbol{h}}_2}\mathit{\boldsymbol{h}} $。

若令$ Sp(n) = \{ \mathit{\boldsymbol{H}} \in GL(n;{\mathbb{H}})|{\mathit{\boldsymbol{H}}^{{\rm{*T}}}} \cdot \mathit{\boldsymbol{H}} = {\mathit{\boldsymbol{I}}_n}\} $, 其中$ {\mathit{\boldsymbol{I}}_n} $表示$ n $阶单位矩阵, 可在${\mathbb{H}}{P^n}$定义一个作用使得${\mathbb{H}}{P^n}$有齐性表示

$ \mathbb{H} P^{n}=S p(n+1) /(S p(1) \times S p(n)). $

根据文献[7]可知有如下的一个交换图：

$ \begin{matrix} {} & Sp(n+1) & {} \\ {{\tau }_{2}}\swarrow & {} & \searrow {{\tau }_{1}} \\ {\mathbb{H}}{{P}^{n}} & \xleftarrow{\pi } & {\mathbb{C}}{{P}^{2n+1}} \\ \end{matrix} $

其中τ₁:$Sp(n + 1) \to {\mathbb{C}}{P^{2n + 1}}$, $\mathit{\boldsymbol{H}} = \mathit{\boldsymbol{U}} + j\mathit{\boldsymbol{V}} \mapsto \left[ {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{U}}_0}}\\ {{\mathit{\boldsymbol{V}}_0}} \end{array}} \right]$为自然投影, $\mathit{\boldsymbol{U}} = \left( {{\mathit{\boldsymbol{U}}_0}, \cdots , {\mathit{\boldsymbol{U}}_n}} \right)$, $\mathit{\boldsymbol{V}} = \left( {{\mathit{\boldsymbol{V}}_0}, \cdots } \right.{\mathit{\boldsymbol{V}}_n}) \in GL(n + 1;{\mathbb{C}})$, 其中将$ Sp(n) $中的元素$ \mathit{\boldsymbol{U}} + j\mathit{\boldsymbol{V}} $看作$ SU(n) $中的元素$\left( {\begin{array}{*{20}{c}} \mathit{\boldsymbol{U}}&{ - \mathit{\boldsymbol{\bar V}}}\\ \mathit{\boldsymbol{V}}&{\mathit{\boldsymbol{\bar U}}} \end{array}} \right)$。τ₂:$ Sp(n + ) \to {\mathbb{H}}{P^n} $, $\mathit{\boldsymbol{H}} = \left( {{\mathit{\boldsymbol{H}}_0}, \cdots , {\mathit{\boldsymbol{H}}_n}} \right) \to {\left[ {{\mathit{\boldsymbol{H}}_0}} \right]_{\mathbb{H}}}$。

定义1.1  在上述交换图中我们称π:${\mathbb{C}}{P^{2n + 1}} \to {\mathbb{H}}{P^n}$, $ \left[ {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{U}}_0}}\\ {{\mathit{\boldsymbol{V}}_0}} \end{array}} \right] \to {\left[ {{\mathit{\boldsymbol{U}}_0} + j{\mathit{\boldsymbol{V}}_0}} \right]_{\mathbb{H}}} $为扭映射.

定义1.2  扭映射π:${\mathbb{C}}{P^{2n + 1}} \to {\mathbb{H}}{P^n}$给出$ {\mathbb{C}}{P^{2n + 1}} $的水平切空间与${\mathbb{H}}{P^n}$的切空间一个自然等同，所以定义$ {\mathbb{C}}{P^{2n + 1}} $在点$ [u] $的水平部分$ {{\rm{T}}_{[u]}} $为$ \pi ([u]) $处纤维的正交补, 其中${\mathbb{C}}{P^{2n + 1}}$上的度量为Fubini-Study度量.若Ω:$N \to {\mathbb{C}}{P^{2n + 1}} $称为水平的, Ω的切映射的像落在$ {{\rm{T}}_{[u]}} $中。

任意的$ \mathit{\boldsymbol{u}} = \left[ {\left( {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{u}}_1}}\\ {{\mathit{\boldsymbol{u}}_2}} \end{array}} \right)} \right] \in {{\mathbb{C}}^{2n + 2}} $, 由Yang^[8]的结论可知

$ \begin{array}{*{20}{c}} {{{\text{T}}_{[u]}} \leftrightarrow \left\{ {\mathit{\boldsymbol{v}} \in {\mathit{\boldsymbol{u}}^ \bot }|} \right.{\sigma _u}(\mathit{\boldsymbol{v}}) = 0,} \\ {{\sigma _u} = - \mathit{\boldsymbol{v}}_2^{\text{T}}{\text{d}}{\mathit{\boldsymbol{u}}_1} + \mathit{\boldsymbol{u}}_1^{\text{T}}{\text{d}}{\mathit{\boldsymbol{u}}_2},{\mathit{\boldsymbol{u}}_1},{\mathit{\boldsymbol{u}}_2} \in {\mathbb{C}^{n + 1}}\} .} \end{array} $

(1)

2 ${\mathbb{H}}{P^3}$中的共形极小曲面 2.1 ${\mathbb{H}}{P^3}$中的共形极小曲面

命题2.1  设$ \Omega = [\mathit{\boldsymbol{\omega }}] = \left[ {\left( {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{u}}_1}}\\ {{\mathit{\boldsymbol{u}}_2}} \end{array}} \right)} \right]:N \to {{\mathbb{C}}^{2n + 2}} $为浸入, 则Ω为水平的当且仅当$\left\{ {\begin{array}{*{20}{l}} {\langle \mathit{\boldsymbol{\omega }}, \partial \mathit{\boldsymbol{v}}\rangle = 0}\\ {\langle \mathit{\boldsymbol{\omega }}, \overline \partial \mathit{\boldsymbol{v}}\rangle = 0} \end{array}} \right.$, 其中$ [\mathit{\boldsymbol{v}}] = \left[ {\left( {\begin{array}{*{20}{c}} { - {{\mathit{\boldsymbol{\bar u}}}_2}}\\ {{{\mathit{\boldsymbol{\bar u}}}_1}} \end{array}} \right)} \right] $.

证明  由式(1)知Ω水平$ \Leftrightarrow \left\langle {\left( {\begin{array}{*{20}{l}} {{\rm{d}}{\mathit{\boldsymbol{u}}_1}}\\ {{\rm{d}}{\mathit{\boldsymbol{u}}_2}} \end{array}} \right), \left( {\begin{array}{*{20}{c}} { - {{\mathit{\boldsymbol{\bar u}}}_2}}\\ {{{\mathit{\boldsymbol{\bar u}}}_1}} \end{array}} \right)} \right\rangle = \langle {\rm{d}}\mathit{\boldsymbol{\omega }}, \mathit{\boldsymbol{v}}\rangle = 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {\langle \partial \mathit{\boldsymbol{\omega }}, \mathit{\boldsymbol{v}}\rangle = 0}\\ {\langle \bar \partial \mathit{\boldsymbol{\omega }}, \mathit{\boldsymbol{v}}\rangle = 0} \end{array}} \right.$, 又由$\langle \mathit{\boldsymbol{\omega }}, \mathit{\boldsymbol{v}}\rangle = 0$, 在等式两边分别作用$ \partial $和$ {\bar \partial } $, 即可得所证。

根据Bolton等^[3]给出的结论, 有如下定义：

定义2.1  称$ \left\{ {{\Phi _0}, \cdots , {\Phi _n}} \right\} $为${\mathbb{C}}{P^n}$中的Veronese序列, 其中$ {\Phi _s}:{S^2} \to {\mathbb{C}}{P^n}, \mathit{\boldsymbol{u}} \mapsto \left[ {\left( {{\phi _{s, 0}}(u)} \right.} \right., \cdots , {\phi _{s, n}}(u){)^{\rm{T}}}] $为曲率为$ \frac{4}{{n + 2s(n - s)}} $的共形极小二维球面, 其中$ s = 0, 1, \cdots , n, u $为$ {S^2} $上的全纯坐标。对于$ t = 0, 1, \cdots , n $,

$ {\phi _{s,t}}(u) = \frac{{s!}}{{{{(1 + u\bar u)}^s}}}\sqrt {C_n^t} {u^{t - s}}\sum\limits_k {{{( - 1)}^k}} C_t^{s - k}C_{n - t}^k{(u\bar u)^k}. $

例1  通过上述可算得${\mathbb{C}}{P^7}$的Veronese序列如下：

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_0^{(7)} = \left[ {{{\left( {1,\sqrt 7 z,\sqrt {21z} {z^2},\sqrt {35} {z^3},\sqrt {35} {z^4},\sqrt {21{z^5}} ,\sqrt 7 {z^6},{z^7}} \right)}^{\rm{T}}}} \right]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_1^{(7)} = \left[ {\left( { - 7\bar z,\sqrt 7 (1 - 6zz),\sqrt {21} \left( {2z - 5{z^2}z} \right),\sqrt {35} \left( {3{z^2} - 4{z^3}z} \right),\sqrt {35} \left( {4{z^3} - 3{z^4}\bar z} \right)} \right.} \right.,\\ \;\;\;\;\;\;\;\;\;\sqrt {21} \left( {5{z^4} - 2{z^5}z} \right),\sqrt 7 {\left( {6{z^5} - {z^6}z} \right)^{\rm{T}}}]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_2^{(7)} = \left[ {\left( {21{{\bar z}^2},\sqrt 7 \left( { - 6\bar z + 15z{{\bar z}^2}} \right),\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right),\sqrt {35} \left( {3z - 12{z^2}\bar z + 6{z^3}{{\bar z}^2}} \right),} \right.} \right.\\ \;\;\;\;\;\;\;\;\;{\left. {\sqrt {35} \left( {6{z^2} - 12{z^3}\bar z + 3{z^4}{{\bar z}^2}} \right),\sqrt {21} \left( {10{z^3} - 10{z^4}\bar z + {z^5}{{\bar z}^2}} \right),\sqrt 7 \left( {15{z^4} - 6{z^5}z} \right),21{z^5}} \right)^{\rm{T}}}]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_3^{(7)} = \left[ {\left( { - 35{{\bar z}^3},\sqrt 7 \left( {15{{\bar z}^2} - 20z{{\bar z}^3}} \right),\sqrt {21} \left( { - 5\bar z + 20{z^2}{{\bar z}^2} - 10{z^2}{{\bar z}^3}} \right),\sqrt {35} \left( {1 - 12z\bar z - 18{z^2}{{\bar z}^2} - 4{z^3}{{\bar z}^3}} \right),} \right.} \right.\\ \;\;\;\;\;\;\;\;\;\left. {{{\left. {\sqrt {35} \left( {4z - 18{z^2}\bar z + 12{z^3}{{\bar z}^2} - {z^4}{{\bar z}^3}} \right),\sqrt {21} \left( {10{z^2} - 20{z^3}\bar z + 5{z^4}{{\bar z}^2}} \right),\sqrt 7 \left( {20{z^3} - 15{z^4}\bar z} \right),35{z^4}} \right)}^{\rm{T}}}} \right]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_4^{(7)} = \left[ {\left( {35{{\bar z}^4},\sqrt 7 \left( { - 20{{\bar z}^4} + 15z{{\bar z}^4}} \right),\sqrt {21} \left( { - 10{{\bar z}^2} - 20z{{\bar z}^3} + 5{z^2}{{\bar z}^4}} \right),\sqrt {35} \left( { - 4\bar z + 18z{{\bar z}^2} - 12{z^2}{{\bar z}^3} + {z^3}{{\bar z}^4}} \right),} \right.} \right.\\ \;\;\;\;\;\;\;\;\;\sqrt {35} \left( {1 - 12z\bar z + 18{z^2}{{\bar z}^2} - 4{z^3}{{\bar z}^3}} \right),\sqrt {21} \left( {5z - 20{z^2}\bar z + 10{z^3}{{\bar z}^2}} \right),\sqrt 7 \left( {15{z^2} - 20{z^3}\bar z} \right),35{z^3}{)^{\rm{T}}}]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_5^{(7)} = \left[ {\left( { - 21{{\bar z}^5},\sqrt 7 \left( {15{{\bar z}^4} - 6z{{\bar z}^5}} \right),\sqrt {21} \left( { - 10{{\bar z}^3} + 10z{{\bar z}^4} - {z^2}{{\bar z}^5}} \right),\sqrt {35} \left( {6{{\bar z}^2} - 12z{{\bar z}^3} + 3{z^2}{{\bar z}^4}} \right)} \right.,} \right.\\ \;\;\;\;\;\;\;\;\;\sqrt {35} \left( { - 3\bar z + 12z{{\bar z}^2} - 6{z^2}{{\bar z}^3}} \right),\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right),\sqrt 7 \left( {6z - 15{z^2}z} \right),21{z^2}{)^{\rm{T}}}]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_6^{(7)} = \left[ {\left( {7{{\bar z}^6},\sqrt 7 \left( { - 6{{\bar z}^5} + z{{\bar z}^6}} \right),\sqrt {21} \left( {5{{\bar z}^4} - 2z{{\bar z}^5}} \right),\sqrt {35} \left( { - 4{{\bar z}^3} + 3z{{\bar z}^4}} \right),\sqrt {35} \left( {3{{\bar z}^2} - 4z{{\bar z}^3}} \right)} \right.,} \right.\\ \;\;\;\;\;\;\;\;\;\;\sqrt {21} \left( { - 2\bar z + 5z{{\bar z}^2}} \right),\sqrt 7 (1 - 6z\bar z),7z{)^{\rm{T}}}]\\ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_7^{(7)} = \left[ {{{\left( { - {{\bar z}^7},\sqrt 7 {{\bar z}^6}, - \sqrt {21} {{\bar z}^5},\sqrt {35} {{\bar z}^4}, - \sqrt {35} {{\bar z}^3},\sqrt {21} {{\bar z}^2}, - \sqrt 7 \bar z,1} \right)}^{\rm{T}}}} \right] \end{array} \right. $

对应的Gaussian曲率分别为4/7, 4/19, 4/27, 4/31, 4/31, 4/27, 4/19, 4/7。

接下来给出两个引理：

引理2.1^[7]  对于水平极小曲面$ \Omega = [\mathit{\boldsymbol{\omega }}] $:$ N \to {{\mathbb{C}}^{2n + 2}} $，则$ \pi \circ \Omega :N \to {\mathbb{H}}{P^n} $为${\mathbb{H}}{P^n}$中的共形极小曲面, 其中

$ \pi : \mathbb{C} P^{2 n+1} \rightarrow \mathbb{H} P^{n} 为扭映射。$

引理2.2^[3]  设$ \Phi :{S^2} \to {\mathbb{C}}{P^n} $为线性满的常曲率共形极小浸入, 那么在差一个全纯等距意义下, $ \Phi $与${\mathbb{C}}{P^n}$中的Vernoese序列可以看作等同。

这样根据上述两个引理，可以得到以下命题。

命题2.2  设$ \mathit{\boldsymbol{U}} \in \mathit{\boldsymbol{U}}(2n + 2) $, 若$ U \cdot {\Phi _s} $为水平的, 其中$ \Phi :{S^2} \to {\mathbb{C}}{P^{2n + 1}} $为Veronese序列中的一个元素, 则$\pi \circ \left( {\mathit{\boldsymbol{U}} \cdot {\Phi _s}} \right):{S^2} \to {\mathbb{H}}{P^n}$为曲率是$ \frac{4}{2 n+1+2 s(2 n+1-s)} $的共形极小曲面, 其中$\pi :{\mathbb{C}}{P^{2n + 1}}\to {\mathbb{H}}{P^n}$为扭映射。

接下来利用Veronese序列介绍文献[7]给出的构造${\mathbb{H}}{P^n}$中常曲率极小二维球面的构造方法。

定理2.1  设$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = [\mathit{\boldsymbol{\omega }}] = \left[ {{{\left( {{\omega _1}, \cdots , {\omega _{2n + 2}}} \right)}^{\rm{T}}}} \right]:{S^2} \to {\mathbb{C}}{P^{2n + 1}} $为浸入, 若存在$ \mathit{\boldsymbol{U}} \in \mathit{\boldsymbol{U}}(2n + 2) $使得$ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} $水平当且仅当

$ \sum\limits_{p, q=1}^{2 n+2} A_{p q} \omega_{p} \partial \omega_{q}=0, \sum\limits_{p, q=1}^{2 n+2} A_{p q} \omega_{p} \overline{\partial} \omega_{q}=0, $

(2)

其中$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_{n + 1}}}\\ { - {\mathit{\boldsymbol{I}}_{n + 1}}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} = \left( {{A_{pq}}} \right) $为反对称矩阵。

证明  由命题2.1知, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} $水平$ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {\langle \mathit{\boldsymbol{U\omega }}, \partial \mathit{\boldsymbol{v}}\rangle = 0}\\ {\langle \mathit{\boldsymbol{U\omega }}, \bar \partial \mathit{\boldsymbol{v}}\rangle = 0} \end{array}} \right. $, 其中$ \mathit{\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}} 0&{ - {\mathit{\boldsymbol{I}}_{n + 1}}}\\ {{\mathit{\boldsymbol{I}}_{n + 1}}}&0 \end{array}} \right) \cdot (\mathit{\boldsymbol{\bar U\bar \omega }}) $.利用复内积运算将上述展开, 可得$ \left\{ {\begin{array}{*{20}{l}} {\partial {\mathit{\boldsymbol{\omega }}^{\rm{T}}}{\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_{n + 1}}}\\ { - {\mathit{\boldsymbol{I}}_{n + 1}}}&0 \end{array}} \right)\mathit{\boldsymbol{U\omega }} = 0}\\ {\bar \partial {\mathit{\boldsymbol{\omega }}^{\rm{T}}}{\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_{n + 1}}}\\ { - {\mathit{\boldsymbol{I}}_{n + 1}}}&0 \end{array}} \right)\mathit{\boldsymbol{U\omega }} = 0} \end{array}} \right., $记$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_{n + 1}}}\\ { - {\mathit{\boldsymbol{I}}_{n + 1}}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} = \left( {{A_{pq}}} \right) $, 则第一个等式展开

$ \begin{array}{*{20}{c}} {\left( {\partial {\omega _1}, \cdots ,\partial {\omega _{2n + 2}}} \right)\left( {\begin{array}{*{20}{c}} {{A_{1\;1}}}& \cdots &{{A_{1\;2n + 2}}}\\ \vdots &{}& \vdots \\ {{A_{2N + 2\;1}}}& \cdots &{{A_{2n + 2\;2n + 2}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{\omega _1}}\\ \vdots \\ {{\omega _{2n + 2}}} \end{array}} \right) = }\\ {\left( {\sum\limits_{p = 1}^{2n + 1} {{A_{p1}}} \partial {\omega _p}, \cdots ,\sum\limits_{p = 1}^{2n + 1} {{A_{p2n + 2}}} \partial {\omega _p}} \right)\left( {\begin{array}{*{20}{c}} {{\omega _1}}\\ \vdots \\ {{\omega _{2n + 2}}} \end{array}} \right) = 0.} \end{array} $

同理，将第2个等式展开, (2)式得证。

2.2 ${\mathbb{H}}{P^3}$中的共形极小二维球面

接下来利用上述定理, 以$ n=3 $为例, 构造${\mathbb{H}}{P^3}$中的共形极小二维球面例子。而根据命题2.2, 若$ {\Phi _s} = \Phi _s^{(7)} $为例1中给出的$ {\mathbb{C}}{P^7} $中的Veronese序列中一个元素, 能找到$ \mathit{\boldsymbol{U}} \in \mathit{\boldsymbol{U}}(8) $使得$ {\Phi _s} \cdot \mathit{\boldsymbol{U}} $为水平的, 则$ \pi ^\circ \left( {\mathit{\boldsymbol{U}} \cdot {\Phi _s}} \right):{S^2} \to {\mathbb{H}}{P^3}$为曲率是$ \frac{4}{{7 + 2s(7 - s)}} $的共形极小二维球面, 而使得$ {\Phi _s} \cdot \mathit{\boldsymbol{U}} $为水平的条件已由定理2.1给出。

情形1：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_0^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_0^{(7)} = \left[ {\left( {1, \sqrt 7 z, \sqrt {21} {z^2}, \sqrt {35} {z^3}, } \right.} \right.\sqrt {35} {z^4}, \sqrt {21} {z^5}, \sqrt 7 {z^6}, {z^7}{)^{\rm{T}}}] $, 则式(2)可展开如下：

$ \begin{array}{l} \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = \sqrt 7 {A_{12}} + 2\sqrt {21} {A_{13}}z + \left( {3\sqrt {35} {A_{14}} + 7\sqrt 3 {A_{23}}} \right){z^2} + \left( {4\sqrt {35} {A_{15}} + 14\sqrt 5 {A_{24}}} \right){z^3} + \left( {5\sqrt {21} {A_{16}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;21\sqrt 5 {A_{25}} + 7\sqrt {15} {A_{34}}){z^4} + \left( {6\sqrt 7 {A_{17}} + 28\sqrt 3 {A_{26}} + 14\sqrt {15} {A_{35}}} \right){z^5} + \left( {7{A_{18}} + 35{A_{27}} + 63{A_{36}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {35{A_{45}}} \right){z^6} + \left( {6\sqrt 7 {A_{28}} + 28\sqrt 3 {A_{37}} + 14\sqrt {15} {A_{46}}} \right){z^7} + \left( {5\sqrt {21} {A_{38}} + 21\sqrt 5 {A_{47}} + 7\sqrt {15} {A_{56}}} \right){z^8} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {4\sqrt {35} {A_{48}} + 14\sqrt 5 {A_{57}}} \right){z^9} + \left( {3\sqrt {35} {A_{58}} + 7\sqrt 3 {A_{67}}} \right){z^{10}} = 0, \end{array} $

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\overline \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以可算得

$ \left\{ \begin{array}{l} {A_{12}} = {A_{13}} = {A_{68}} = {A_{78}} = 0,\\ 3\sqrt {35} {A_{14}} + 7\sqrt 3 {A_{23}} = 0,\\ 4\sqrt {35} {A_{15}} + 14\sqrt 5 {A_{24}} = 0,\\ 5\sqrt {21} {A_{16}} + 21\sqrt 5 {A_{25}} + 7\sqrt {15} {A_{34}} = 0,\\ 6\sqrt 7 {A_{17}} + 28\sqrt 3 {A_{26}} + 14\sqrt {15} {A_{35}} = 0,\\ {\rm{7}}{A_{18}} + 35{A_{27}} + 63{A_{36}} + 35{A_{45}} = 0,\\ 6\sqrt 7 {A_{28}} + 28\sqrt 3 {A_{37}} + 14\sqrt {15} {A_{46}} = 0,\\ 5\sqrt {21} {A_{38}} + 21\sqrt 5 {A_{47}} + 7\sqrt {15} {A_{56}} = 0,\\ 4\sqrt {35} {A_{48}} + 14\sqrt 5 {A_{57}} = 0,\\ 3\sqrt {35} {A_{58}} + 7\sqrt 3 {A_{67}} = 0. \end{array} \right. $

所以, 由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_0^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $,

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&{{A_{14}}}\\ 0&0&{ - \frac{{\sqrt {105} }}{7}{A_{14}}}&{ - \frac{{2\sqrt 7 }}{7}{A_{15}}}\\ 0&{\frac{{\sqrt {105} }}{7}{A_{14}}}&0&{ - \frac{{\sqrt {35} }}{7}{A_{16}} - \frac{{\sqrt 3 }}{9}{A_{25}}}\\ { - {A_{14}}}&{\frac{{2\sqrt 7 }}{7}{A_{15}}}&{\frac{{\sqrt {35} }}{7}{A_{16}} + \frac{{\sqrt 3 }}{9}{A_{25}}}&0\\ { - {A_{15}}}&{ - {A_{25}}}&{\frac{{\sqrt {105} }}{{35}}{A_{17}} + \frac{{2\sqrt 5 }}{5}{A_{26}}}&{\frac{1}{5}{A_{18}} + {A_{27}} + \frac{9}{5}{A_{36}}}\\ { - {A_{16}}}&{ - {A_{26}}}&{ - {A_{36}}}&{\frac{{\sqrt {105} }}{{34}}{A_{28}} + \frac{{2\sqrt 5 }}{5}{A_{37}}}\\ { - {A_{17}}}&{ - {A_{27}}}&{ - {A_{37}}}&{ - {A_{47}}}\\ { - {A_{18}}}&{ - {A_{28}}}&{ - {A_{38}}}&{ - {A_{48}}} \end{array}} \right) $

$ {\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} {{A_{15}}}&{{A_{16}}}&{{A_{17}}}&{{A_{18}}}\\ {{A_{25}}}&{{A_{26}}}&{{A_{27}}}&{{A_{28}}}\\ { - \frac{{\sqrt {105} }}{{35}}{A_{17}} - \frac{{2\sqrt 5 }}{5}{A_{26}}}&{{A_{36}}}&{{A_{37}}}&{{A_{38}}}\\ { - \frac{1}{5}{A_{18}} - {A_{27}} - \frac{9}{5}{A_{36}}}&{ - \frac{{\sqrt {105} }}{{35}}{A_{28}} - \frac{{2\sqrt 5 }}{5}{A_{37}}}&{{A_{47}}}&{{A_{48}}}\\ 0&{ - \frac{{\sqrt {35} }}{7}{A_{38}} - \sqrt 3 {A_{47}}}&{ - \frac{{2\sqrt 7 }}{7}{A_{48}}}&{{A_{58}}}\\ {\frac{{\sqrt {35} }}{7}{A_{38}} + \sqrt 3 {A_{47}}}&0&{ - \frac{{\sqrt {105} }}{7}{A_{58}}}&0\\ {\frac{{2\sqrt 7 }}{7}{A_{48}}}&{\frac{{\sqrt {105} }}{7}{A_{58}}}&0&0\\ { - {A_{58}}}&0&0&0 \end{array}} \right) $

情形2：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_1^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_1^{(7)} = [( - 7\bar z, $, $\sqrt 7 (1 - 6z\bar z)$$\sqrt {21} \left( {2z - 5{z^2}\bar z} \right), $$, \sqrt {35} \left( {3{z^2} - 4{z^3}\bar z} \right), $$\sqrt {35} \left( {4{z^3} - 3{z^4}\bar z} \right), $$\sqrt {21} \left( {5{z^4} - 2{z^5}\bar z} \right), $$\sqrt 7 \left( {6{z^5} - {z^6}\bar z} \right), $$7{z^6}{)^{\rm{T}}}]$则式(2)可展开如下：

$ \begin{array}{l} \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 14\sqrt 3 {A_{23}} + 70\sqrt {21} {A_{13}}z{{\bar z}^2} + 42\sqrt 5 {A_{24}}z + 42\sqrt 7 {A_{12}}{{\bar z}^2} - \left( {42\sqrt {35} {A_{14}} + 70\sqrt 3 {A_{23}}} \right)z\bar z + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {84\sqrt 5 {A_{25}} + 42\sqrt {15} {A_{34}}} \right){z^2} - 14\sqrt {21} {A_{13}}\bar z + \left( {140\sqrt 3 {A_{26}} + 112\sqrt {15} {A_{35}}} \right){z^3} + \left( {210{A_{27}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;630{A_{36}} + 420{A_{45}}){z^4} - \left( {84\sqrt {35} {A_{15}} + 210\sqrt 5 {A_{24}}} \right){z^2}\bar z - \left( {140\sqrt {21} {A_{16}} + 420\sqrt 5 {A_{25}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;112\sqrt {15} {A_{34}}){z^3}\bar z + \left( {84\sqrt {35} {A_{14}} + 210\sqrt 3 {A_{23}}} \right){z^2}{{\bar z}^2} + \left( {84\sqrt {35} {A_{15}} + 336\sqrt 5 {A_{24}}} \right){z^3}{{\bar z}^2} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {210\sqrt 7 {A_{17}} + 700\sqrt 3 {A_{26}} + 266\sqrt {15} {A_{35}}} \right){z^4}\bar z + \left( {42\sqrt 7 {A_{28}} + 336\sqrt 3 {A_{37}} + 210\sqrt {15} {A_{46}}} \right){z^5} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {70\sqrt {21} {A_{16}} + 378\sqrt 5 {A_{25}} + 140\sqrt {15} {A_{34}}} \right){z^4}{{\bar z}^2} - \left( {294{A_{18}} + 1050{A_{27}} + 1386{A_{36}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;630{A_{45}}){z^5}z + \left( {70\sqrt {21} {A_{38}} + 378\sqrt 5 {A_{47}} + 140\sqrt {15} {A_{56}}} \right){z^6} + \left( {42\sqrt 7 {A_{17}} + 336\sqrt 3 {A_{26}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;210\sqrt {15} {A_{35}}){z^5}{{\bar z}^2} - \left( {140\sqrt {21} {A_{38}} + 420\sqrt 5 {A_{47}} + 112\sqrt {15} {A_{56}}} \right){z^7}\bar z + \left( {84\sqrt {35} {A_{48}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;336\sqrt 5 {A_{57}}){z^7} + \left( {210{A_{27}} + 630{A_{36}} + 420{A_{45}}} \right){z^6}{{\bar z}^2} - \left( {210\sqrt 7 {A_{28}} + 700\sqrt 3 {A_{37}} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;266\sqrt {15} {A_{46}}){z^6}\bar z + \left( {84\sqrt {35} {A_{58}} + 210\sqrt 3 {A_{67}}} \right){z^8} + \left( {140\sqrt 3 {A_{37}} + 112\sqrt {15} {A_{46}}} \right){z^7}{{\bar z}^2} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {84\sqrt {35} {A_{48}} + 210\sqrt 5 {A_{57}}} \right){z^8}\bar z + 70\sqrt {21} {A_{68}}{z^9} + \left( {84\sqrt 5 {A_{47}} + 42\sqrt {15} {A_{56}}} \right){z^8}{{\bar z}^2} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {42\sqrt {35} {A_{58}} + 70\sqrt 3 {A_{67}}} \right){z^9}\bar z + 42\sqrt 7 {A_{78}}{z^{10}} + 42\sqrt 5 {A_{57}}{z^9}{{\bar z}^2} - 14\sqrt {21} {A_{68}}{z^{10}}\bar z + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;14\sqrt 3 {A_{67}}{z^{10}}{z^{ - 2}} = 0, \end{array} $

$ \begin{array}{l} \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\overline \partial {\omega _q} = - 7\sqrt 7 {A_{21}} - 14\sqrt {21} {A_{31}}z - \left( {49\sqrt 3 {A_{32}} + 21\sqrt {35} {A_{41}}} \right){z^2} - \left( {98\sqrt 5 {A_{42}} + 28\sqrt {35} {A_{51}}} \right){z^3} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {49\sqrt {15} {A_{43}} + 147\sqrt 5 {A_{52}} + 35\sqrt {21} {A_{61}}} \right){z^4} - \left( {98\sqrt {15} {A_{53}} + 196\sqrt 3 {A_{62}} + 42\sqrt 7 {A_{71}}} \right){z^5} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {245{A_{54}} + 441{A_{63}} + 245{A_{72}} + 49{A_{81}}} \right){z^6} - \left( {98\sqrt {15} {A_{64}} + 196\sqrt 3 {A_{73}} + 42\sqrt 7 {A_{82}}} \right){z^7} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {49\sqrt {15} {A_{65}} + 147\sqrt 5 {A_{74}} + 35\sqrt {21} {A_{83}}} \right){z^8} - \left( {98\sqrt 5 {A_{75}} + 28\sqrt {35} {A_{84}}} \right){z^9} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {49\sqrt 3 {A_{76}} + 21\sqrt {35} {A_{85}}} \right){z^{10}} - 14\sqrt {21} {A_{86}}{z^{11}} - 7\sqrt 7 {A_{87}}{z^{12}} = 0. \end{array} $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以可算得

$ \left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{A_{23}} = {A_{13}} = {A_{24}} = {A_{12}} = {A_{78}} = 0}\\ {{A_{86}} = {A_{48}} = {A_{58}} = {A_{75}} = {A_{76}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {2\sqrt 5 {A_{25}} + \sqrt {15} {A_{34}} = 0}\\ {5\sqrt {21} {A_{16}} + 15\sqrt 5 {A_{25}} + 4\sqrt {15} {A_{34}} = 0}\\ {5\sqrt {21} {A_{16}} + 27\sqrt 5 {A_{25}} + 10\sqrt {15} {A_{34}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {5\sqrt 3 {A_{26}} + 4\sqrt {15} {A_{35}} = 0}\\ {15\sqrt 7 {A_{17}} + 50\sqrt 3 {A_{26}} + 19\sqrt {15} {A_{35}} = 0}\\ {\sqrt 7 {A_{17}} + 8\sqrt 3 {A_{26}} + 15\sqrt {15} {A_{35}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {{A_{27}} + 3{A_{36}} + 2{A_{45}} = 0}\\ {294{A_{18}} + 1050{A_{27}} + 1386{A_{36}} + 630{A_{45}} = 0}\\ {{A_{18}} + 5{A_{27}} + 9{A_{36}} + 5{A_{45}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {5\sqrt {21} {A_{38}} + 21\sqrt 5 {A_{47}} + 7\sqrt {15} {A_{56}} = 0}\\ {5\sqrt {21} {A_{38}} + 21\sqrt 5 {A_{47}} + 10\sqrt {15} {A_{56}} = 0}\\ {5\sqrt {21} {A_{38}} + 15\sqrt 5 {A_{47}} + 4\sqrt {15} {A_{56}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {3\sqrt 7 {A_{28}} + 14\sqrt 3 {A_{37}} + 7\sqrt {15} {A_{46}} = 0}\\ {\sqrt 7 {A_{28}} + 8\sqrt 3 {A_{37}} + 5\sqrt {15} {A_{46}} = 0}\\ {105\sqrt 7 {A_{28}} + 350\sqrt 3 {A_{37}} + 133\sqrt {15} {A_{46}} = 0} \end{array} \end{array} \right. $

所以, 由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_1^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $,

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&{\frac{{2\sqrt {35} }}{7}{A_{16}}}\\ 0&0&{ - \frac{{2\sqrt {35} }}{7}{A_{16}}}&0\\ 0&{\frac{{\sqrt {105} }}{7}{A_{16}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{17}}}&{ - {A_{18}} - 2{A_{27}}}\\ { - {A_{16}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{\frac{5}{3}{A_{27}} + \frac{2}{3}{A_{18}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{28}}}\\ { - {A_{17}}}&{ - {A_{27}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{28}}}&{\frac{{\sqrt {105} }}{7}{A_{38}}}\\ { - {A_{18}}}&{ - {A_{28}}}&{ - {A_{38}}}&0 \end{array}} \right), $

$ {\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} 0&{{A_{16}}}&{{A_{17}}}&{{A_{18}}}\\ { - \frac{{\sqrt {105} }}{7}{A_{16}}}&{ - \frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{{A_{27}}}&{{A_{28}}}\\ {\frac{{\sqrt {105} }}{{21}}{A_{17}}}&{ - \frac{5}{3}{A_{27}} - \frac{2}{3}{A_{18}}}&{ - \frac{{4\sqrt {21} }}{{21}}{A_{28}}}&{{A_{37}}}\\ {{A_{18}} + 2{A_{27}}}&{\frac{{\sqrt {105} }}{{21}}{A_{28}}}&{ - \frac{{\sqrt {105} }}{7}{A_{38}}}&0\\ 0&{\frac{{2\sqrt {35} }}{7}{A_{38}}}&0&0\\ { - \frac{{2\sqrt {35} }}{7}{A_{38}}}&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right). $

情形3：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_2^{(7)} $。

由例1知, $\mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_2^{(7)} = \left[ {\left( {21{{\bar z}^2}, \sqrt 7 \left( { - 6\bar z + 15{{\bar z}^2}} \right)} \right.} \right.$, $\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right), \;\;\, \sqrt {35} \left( {3z - 12{z^2}\bar z + } \right.$$ 6{z^3}{\bar z^2}), \sqrt {35} \left( {6{z^2} - 12{z^3}\bar z + 3{z^4}{{\bar z}^2}} \right), \sqrt {21} \left( {10{z^3} - } \right. $$ 10{z^4}\bar z + {z^5}{\bar z^2}), \sqrt 7 \left( {15{z^4} - 6{z^5}\bar z} \right), 21{z^5}{)^{\rm{T}}}] $, 则由式(2)得

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 0,\sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\bar \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以可算得

$ \left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{A_{12}} = {A_{13}} = {A_{14}} = {A_{15}} = {A_{16}} = {A_{17}} = 0}\\ {{A_{23}} = {A_{24}} = {A_{25}} = {A_{26}} = {A_{28}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {{A_{34}} = {A_{35}} = {A_{37}} = {A_{38}} = 0}\\ {{A_{46}} = {A_{47}} = {A_{48}} = {A_{56}} = {A_{57}} = {A_{58}} = 0}\\ {{A_{67}} = {A_{68}} = {A_{78}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {7{A_{18}} + 19{A_{27}} + 27{A_{36}} + 15{A_{45}} = 0}\\ {{A_{27}} + 2{A_{36}} + {A_{45}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {7{A_{18}} + 25{A_{27}} + 33{A_{36}} + 15{A_{45}} = 0}\\ {{A_{36}} + {A_{45}} = 0} \end{array} \end{array} \right. $

所以, 由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_2^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中

$ \mathit{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&{{A_{18}}}\\ 0&0&0&0&0&0&{ - {A_{18}}}&0\\ 0&0&0&0&0&{{A_{18}}}&0&0\\ 0&0&0&0&{ - {A_{18}}}&0&0&0\\ 0&0&0&{{A_{18}}}&0&0&0&0\\ 0&0&{ - {A_{18}}}&0&0&0&0&0\\ 0&{{A_{18}}}&0&0&0&0&0&0\\ { - {A_{18}}}&0&0&0&0&0&0&0 \end{array}} \right). $

情形4：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_3^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_3^{(7)} = \left[ {\left( { - 35{{\bar z}^3}, \sqrt 7 \left( {15{z^2} - 20z{{\bar z}^3}} \right)} \right.} \right. $, $ \sqrt {21} \left( { - 5\bar z + 20z{{\bar z}^2} - 10{z^2}{{\bar z}^3}} \right), \sqrt {35} (1 - 12z\bar z - $$ 18{z^2}{\bar z^2} - 4{z^3}{\bar z^3}), \sqrt {35} \left( {4z - 18{z^2}\bar z + 12{z^3}{{\bar z}^2} - {z^4}{{\bar z}^3}} \right) $, $ \sqrt {21} \left( {10{z^2} - 20{z^3}\bar z + 5{z^4}{{\bar z}^2}} \right), \sqrt 7 \left( {20{z^3} - 15{z^4}\bar z} \right) $, $ 35{z^4}{)^{\rm{T}}}] $, 则由式(2)得

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 0,\sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\bar \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以可算得

$ \left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{A_{12}} = {A_{13}} = {A_{15}} = {A_{16}} = {A_{17}} = {A_{18}} = 0}\\ {{A_{24}} = {A_{25}} = {A_{26}} = {A_{27}} = {A_{28}} = 0}\\ {{A_{34}} = {A_{35}} = {A_{36}} = {A_{37}} = {A_{38}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {{A_{45}} = {A_{46}} = {A_{47}} = {A_{48}} = {A_{56}} = {A_{57}} = 0}\\ {{A_{68}} = {A_{78}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {42\sqrt {35} {A_{14}} + 140\sqrt 3 {A_{23}} = 0}\\ {126\sqrt {35} {A_{14}} + 210\sqrt 3 {A_{23}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {42\sqrt {35} {A_{58}} + 140\sqrt 3 {A_{67}} = 0}\\ {126\sqrt {35} {A_{58}} + 210\sqrt 3 {A_{67}} = 0} \end{array} \end{array} \right. $

所以, 由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_3^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}}{\rm{ = 0}} $.

情形5：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_4^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_4^{(7)} = \left[ {\left( {35{{\bar z}^4}, \sqrt 7 \left( { - 20{{\bar z}^3} + 15{z^{ - 4}}} \right)} \right.} \right. $, $ \sqrt {21} \left( {10{{\bar z}^2} - 20z{{\bar z}^3} + 5{z^2}{{\bar z}^4}} \right), \sqrt {35} ( - 4\bar z + $$ 18z{\bar z^2} - 12{z^2}{\bar z^3} + {z^3}{\bar z^4}), \sqrt {35} \left( {1 - 12z\bar z + 18{z^2}{{\bar z}^2} - } \right. $$ 4{z^3}{\bar z^3}), \sqrt {21} \left( {5z - 20{z^2}\bar z + 10{z^3}{{\bar z}^2}} \right), \sqrt 7 \left( {15{z^2} - } \right. $$ 20{z^3}\bar z), 35{z^3}{)^{\rm{T}}}] $, 则由式(2)得

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 0,\sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\bar \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以可算得

$ \left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{A_{12}} = {A_{13}} = {A_{15}} = {A_{16}} = {A_{17}} = {A_{18}} = 0}\\ {{A_{24}} = {A_{25}} = {A_{26}} = {A_{27}} = {A_{28}} = 0}\\ {{A_{34}} = {A_{35}} = {A_{36}} = {A_{37}} = {A_{38}} = 0}\\ {{A_{45}} = {A_{46}} = {A_{47}} = {A_{48}} = {A_{56}} = {A_{57}} = 0}\\ {{A_{68}} = {A_{78}} = 0} \end{array}\\ \begin{array}{*{20}{l}} {42\sqrt {35} {A_{14}} + 140\sqrt 3 {A_{23}} = 0}\\ {126\sqrt {35} {A_{14}} + 210\sqrt 3 {A_{23}} = 0}\\ {42\sqrt {35} {A_{58}} + 140\sqrt 3 {A_{67}} = 0}\\ {126\sqrt {35} {A_{58}} + 210\sqrt 3 {A_{67}} = 0} \end{array} \end{array} \right. $

所以, 由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_4^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}}{\rm{ = 0}} $。

情形6：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_5^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_5^{(7)} = \left[ {\left( { - 21{{\bar z}^5}, \sqrt 7 \left( {15{{\bar z}^4} - 6z{{\bar z}^5}} \right)} \right.} \right. $, $ \sqrt {21} \left( { - 10{{\bar z}^3} + 10z{{\bar z}^4} - {z^2}{{\bar z}^5}} \right), \sqrt {35} \left( {6{z^2} - 12z{{\bar z}^3} + } \right. $$ 3{z^2}{\bar z^4}), \sqrt {35} \left( { - 3\bar z + 12z{{\bar z}^2} - 6{z^2}{{\bar z}^3}} \right), \sqrt {21} (1 - $ $ 10z\bar z + 10{z^2}{\bar z^2}), \sqrt 7 \left( {6z - 15{z^2}\bar z} \right), 21{z^2}{)^{\rm{T}}}] $, 则(2)式得

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 0,\sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\bar \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_5^{(7)} $水平当且仅当$ \mathit{\boldsymbol{A}} $满足$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中

$ \mathit{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&{{A_{18}}}\\ 0&0&0&0&0&0&{ - {A_{18}}}&0\\ 0&0&0&0&0&{{A_{18}}}&0&0\\ 0&0&0&0&{ - {A_{18}}}&0&0&0\\ 0&0&0&{{A_{18}}}&0&0&0&0\\ 0&0&{ - {A_{18}}}&0&0&0&0&0\\ 0&{{A_{18}}}&0&0&0&0&0&0\\ { - {A_{18}}}&0&0&0&0&0&0&0 \end{array}} \right). $

情形7：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_6^{(7)} $。

由例1知, $ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_6^{(7)} = \left[ {\left( {7{z^6}, \sqrt 7 \left( { - 6{{\bar z}^5} + z{{\bar z}^6}} \right)} \right.} \right. $, $ \sqrt {21} \left( {5{{\bar z}^4} - 2z{{\bar z}^5}} \right), \;\;\, \sqrt {35} \left( { - 4{{\bar z}^3} + 3z{{\bar z}^4}} \right) $, $ \sqrt {35} \left( {3{z^2} - 4z{{\bar z}^3}} \right), \sqrt {21} \left( { - 2z + 5z{{\bar z}^2}} \right), \sqrt 7 (1 - $$ 6z\bar z), 7z{)^{\rm{T}}}] $, 则由式(2)得

$ \sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\partial {\omega _q} = 0,\sum\limits_{p,q = 1}^8 {{A_{pq}}} {\omega _p}\bar \partial {\omega _q} = 0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_6^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足:$\mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $,

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&{\frac{{2\sqrt {35} }}{7}{A_{16}}}\\ 0&0&{ - \frac{{2\sqrt {35} }}{7}{A_{16}}}&0\\ 0&{\frac{{\sqrt {105} }}{7}{A_{16}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{17}}}&{ - {A_{18}} - 2{A_{27}}}\\ { - {A_{16}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{\frac{5}{3}{A_{27}} + \frac{2}{3}{A_{18}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{28}}}\\ { - {A_{17}}}&{ - {A_{27}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{28}}}&{\frac{{\sqrt {105} }}{7}{A_{38}}}\\ { - {A_{18}}}&{ - {A_{28}}}&{ - {A_{38}}}&0 \end{array}} \right), $

$ {\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} {{A_{15}}}&{{A_{16}}}&{{A_{17}}}&{{A_{18}}}\\ {{A_{25}}}&{{A_{26}}}&{{A_{27}}}&{{A_{28}}}\\ { - \frac{{\sqrt {105} }}{{35}}{A_{17}} - \frac{{2\sqrt 5 }}{5}{A_{26}}}&{{A_{36}}}&{{A_{37}}}&{{A_{38}}}\\ { - \frac{1}{5}{A_{18}} - {A_{27}} - \frac{9}{5}{A_{36}}}&{ - \frac{{\sqrt {105} }}{{35}}{A_{28}} - \frac{{2\sqrt 5 }}{5}{A_{37}}}&{{A_{47}}}&{{A_{48}}}\\ 0&{ - \frac{{\sqrt {35} }}{7}{A_{38}} - \sqrt 3 {A_{47}}}&{ - \frac{{2\sqrt 7 }}{7}{A_{48}}}&{{A_{58}}}\\ {\frac{{\sqrt {35} }}{7}{A_{38}} + \sqrt 3 {A_{47}}}&0&{ - \frac{{\sqrt {105} }}{7}{A_{58}}}&0\\ {\frac{{2\sqrt 7 }}{7}{A_{48}}}&{\frac{{\sqrt {105} }}{7}{A_{58}}}&0&0\\ { - {A_{58}}}&0&0&0 \end{array}} \right). $

情形8：$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_7^{(7)} $。

由例1知, $ \Phi _7^{(7)} = \left[ {\left( { - {{\bar z}^7}, \sqrt 7 {{\bar z}^6}, - \sqrt {21} {{\bar z}^5}} \right.} \right. $, $ \sqrt {35} {\bar z^4}, - \sqrt {35} {\bar z^3}, \sqrt {21} {\bar z^2}, - \sqrt 7 \bar z, 1{)^{\rm{T}}}] $, 则由式(2)得

$ \sum\limits_{p,q=1}^{8}{{{A}_{pq}}}{{\omega }_{p}}\partial {{\omega }_{q}}=0,\sum\limits_{p,q=1}^{8}{{{A}_{pq}}}{{\omega }_{p}}\overline{\partial }{{\omega }_{q}}=0. $

由于$ \mathit{\boldsymbol{A}} $为一个反对称矩阵, 所以由定理2.1, 所以由定理2.1, $ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_7^{(7)} $水平当且仅当$ \mathit{\boldsymbol{U}} $满足:$ \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{U}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}} 0&{{\mathit{\boldsymbol{I}}_4}}\\ { - {\mathit{\boldsymbol{I}}_4}}&0 \end{array}} \right)\mathit{\boldsymbol{U}} $, 其中$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $,

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&{{A_{14}}}\\ 0&0&{ - \frac{{\sqrt {105} }}{7}{A_{14}}}&{ - \frac{{2\sqrt 7 }}{7}{A_{15}}}\\ 0&{\frac{{\sqrt {105} }}{4}{A_{14}}}&0&{ - \frac{{\sqrt {35} }}{7}{A_{16}} - \frac{{\sqrt 3 }}{9}{A_{25}}}\\ { - {A_{14}}}&{\frac{{2\sqrt 7 }}{7}{A_{15}}}&{\frac{{\sqrt {35} }}{7}{A_{16}} + \frac{{\sqrt 3 }}{9}{A_{25}}}&0\\ { - {A_{15}}}&{ - {A_{25}}}&{\frac{{\sqrt {105} }}{{35}}{A_{17}} + \frac{{2\sqrt 5 }}{5}{A_{26}}}&{\frac{1}{5}{A_{18}} + {A_{27}} + \frac{9}{5}{A_{36}}}\\ { - {A_{16}}}&{ - {A_{26}}}&{ - {A_{36}}}&{\frac{{\sqrt {105} }}{{35}}{A_{28}} + \frac{{2\sqrt 5 }}{5}{A_{37}}}\\ { - {A_{17}}}&{ - {A_{27}}}&{ - {A_{37}}}&{ - {A_{47}}}\\ { - {A_{18}}}&{ - {A_{28}}}&{ - {A_{38}}}&{ - {A_{48}}} \end{array}} \right). $

$ {\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} {{A_{15}}}&{{A_{16}}}&{{A_{17}}}&{{A_{18}}}\\ {{A_{25}}}&{{A_{26}}}&{{A_{27}}}&{{A_{28}}}\\ { - \frac{{\sqrt {105} }}{{35}}{A_{17}} - \frac{{2\sqrt 5 }}{5}{A_{26}}}&{{A_{36}}}&{{A_{37}}}&{{A_{38}}}\\ { - \frac{1}{5}{A_{18}} - {A_{27}} - \frac{9}{5}{A_{36}}}&{ - \frac{{\sqrt {105} }}{{35}}{A_{28}} - \frac{{2\sqrt 5 }}{5}{A_{37}}}&{{A_{47}}}&{{A_{48}}}\\ 0&{ - \frac{{\sqrt {35} }}{7}{A_{38}} - \sqrt 3 {A_{47}}}&{ - \frac{{2\sqrt 7 }}{7}{A_{48}}}&{{A_{58}}}\\ {\frac{{\sqrt {35} }}{7}{A_{38}} + \sqrt 3 {A_{47}}}&0&{ - \frac{{\sqrt {105} }}{7}{A_{58}}}&0\\ {\frac{{2\sqrt 7 }}{7}{A_{48}}}&{\frac{{\sqrt {105} }}{7}{A_{58}}}&0&0\\ { - {A_{58}}}&0&0&0 \end{array}} \right). $

定理2.2  设$ \Phi :{S^2} \to {\mathbb{C}}{P^7} $为$ {\mathbb{C}}{P^7} $中Veronese序列中的一个元素, 若存在$ \mathit{\boldsymbol{U}} \in \mathit{\boldsymbol{U}}(8) $使得$ \mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} $为水平当且仅当$ \mathit{\boldsymbol{U}} $满足

$ {\mathit{\boldsymbol{A}}}={{{\mathit{\boldsymbol{U}}}}^{\text{T}}}\left( \begin{matrix} 0 & {{{\mathit{\boldsymbol{I}}}}_{4}} \\ -{{{\mathit{\boldsymbol{I}}}}_{4}} & 0 \\ \end{matrix} \right){\mathit{\boldsymbol{U}}}, $

1) 当$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_0^{(7)} $或$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_7^{(7)} $, 此时$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $表达式如下：

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&{\frac{{2\sqrt {35} }}{7}{A_{16}}}\\ 0&0&{ - \frac{{2\sqrt {35} }}{7}{A_{16}}}&0\\ 0&{\frac{{\sqrt {105} }}{7}{A_{16}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{17}}}&{ - {A_{18}} - 2{A_{27}}}\\ { - {A_{16}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{\frac{5}{3}{A_{27}} + \frac{2}{3}{A_{18}}}&{ - \frac{{\sqrt {105} }}{{21}}{A_{28}}}\\ { - {A_{17}}}&{ - {A_{27}}}&{\frac{{4\sqrt {21} }}{{21}}{A_{28}}}&{\frac{{\sqrt {105} }}{7}{A_{38}}}\\ { - {A_{18}}}&{ - {A_{28}}}&{ - {A_{38}}}&0 \end{array}} \right), $

$ {\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} 0&{{A_{16}}}&{{A_{17}}}&{{A_{18}}}\\ { - \frac{{\sqrt {105} }}{7}{A_{16}}}&{ - \frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{{A_{27}}}&{{A_{28}}}\\ {\frac{{\sqrt {105} }}{{21}}{A_{17}}}&{ - \frac{5}{3}{A_{27}} - \frac{2}{3}{A_{18}}}&{ - \frac{{4\sqrt {21} }}{{21}}{A_{28}}}&{{A_{38}}}\\ {{A_{18}} + 2{A_{27}}}&{\frac{{\sqrt {105} }}{{21}}{A_{28}}}&{ - \frac{{\sqrt {105} }}{7}{A_{38}}}&0\\ 0&{\frac{{2\sqrt {35} }}{7}{A_{38}}}&0&0\\ { - \frac{{2\sqrt {35} }}{7}{A_{38}}}&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right). $

2) 当$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_1^{(7)}$或$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_6^{(7)}$, 此时$ \mathit{\boldsymbol{A}} = \left( {{\mathit{\boldsymbol{A}}_1}{\mathit{\boldsymbol{A}}_2}} \right) $表达式如下：

$ {{\mathit{\boldsymbol{A}}}_{1}}=\left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{2\sqrt{35}}{7}{{A}_{16}} \\ 0 & 0 & -\frac{2\sqrt{35}}{7}{{A}_{16}} & 0 \\ 0 & \frac{\sqrt{105}}{7}{{A}_{16}} & -\frac{\sqrt{105}}{21}{{A}_{17}} & -{{A}_{18}}-2{{A}_{27}} \\ -{{A}_{16}} & \frac{4\sqrt{21}}{21}{{A}_{17}} & \frac{5}{3}{{A}_{27}}+\frac{2}{3}{{A}_{18}} & -\frac{\sqrt{105}}{21}{{A}_{28}} \\ -{{A}_{17}} & -{{A}_{27}} & \frac{4\sqrt{21}}{21}{{A}_{28}} & \frac{\sqrt{105}}{7}{{A}_{38}} \\ -{{A}_{18}} & -{{A}_{28}} & -{{A}_{38}} & 0 \\ \end{matrix} \right),{{\mathit{\boldsymbol{A}}}_{2}}=\left( \begin{matrix} 0 & {{A}_{16}} & {{A}_{17}} & {{A}_{18}} \\ -\frac{\sqrt{105}}{7}{{A}_{16}} & -\frac{4\sqrt{21}}{21}{{A}_{17}} & {{A}_{27}} & {{A}_{28}} \\ \frac{\sqrt{105}}{21}{{A}_{17}} & -\frac{5}{3}{{A}_{27}}-\frac{2}{3}{{A}_{18}} & -\frac{4\sqrt{21}}{21}{{A}_{28}} & {{A}_{38}} \\ {{A}_{18}}+{{A}_{27}} & \frac{\sqrt{105}}{21}{{A}_{28}} & -\frac{\sqrt{105}}{7}{{A}_{38}} & 0 \\ 0 & \frac{2\sqrt{35}}{7}{{A}_{38}} & 0 & 0 \\ -\frac{2\sqrt{35}}{7}{{A}_{38}} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{matrix} \right). $

3) 当$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_2^{(7)}$或$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_5^{(7)}$, 此时$ \mathit{\boldsymbol{A}} $表达式如下

$ \mathit{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&{{A_{18}}}\\ 0&0&0&0&0&0&{ - {A_{18}}}&0\\ 0&0&0&0&0&{{A_{18}}}&0&0\\ 0&0&0&0&{ - {A_{18}}}&0&0&0\\ 0&0&0&{{A_{18}}}&0&0&0&0\\ 0&0&{ - {A_{18}}}&0&0&0&0&0\\ 0&{{A_{18}}}&0&0&0&0&0&0\\ { - {A_{18}}}&0&0&0&0&0&0&0 \end{array}} \right). $

4) 当$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_3^{(7)}$或$ \mathit{\boldsymbol{ \boldsymbol{\varPhi} }} = \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}_4^{(7)}$, 此时$ \mathit{\boldsymbol{A}} $表达式为

$ \mathit{\boldsymbol{A}} = 0. $

接下来, 将通过给出上述的解来构造${\mathbb{H}}{P^n}$中极小二维球面。

式(1)对应的解

设$ \mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{U}}_1}}&0\\ 0&{{\mathit{\boldsymbol{I}}_4}} \end{array}} \right) $, 那么等式(1)等价于$ \left( {\begin{array}{*{20}{c}} 0&{\mathit{\boldsymbol{U}}_1^{\rm{T}}}\\ { - {\mathit{\boldsymbol{U}}_1}}&0 \end{array}} \right) = \mathit{\boldsymbol{A}} $, 令$ A_{14}=A_{15}=A_{16}=A_{25}=A_{38}=A_{47}=A_{48}=A_{58}=0 $, 则$ {\mathit{\boldsymbol{U}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&{ - \frac{{\sqrt {105} }}{{35}}{A_{17}} - \frac{{2\sqrt 5 }}{5}{A_{26}}}&{ - \frac{1}{5}{A_{18}} - {A_{27}} - \frac{9}{5}{A_{36}}}\\ 0&{{A_{26}}}&{{A_{36}}}&{ - \frac{{\sqrt {105} }}{{35}}{A_{28}} - \frac{{2\sqrt 5 }}{5}{A_{37}}}\\ {{A_{17}}}&{{A_{27}}}&{{A_{37}}}&0\\ {{A_{18}}}&{{A_{28}}}&0&0 \end{array}} \right) $.

取$ A_{26}=A_{37}=A_{17}=A_{28}=0, A_{18}=A_{27}=A_{36}=1 $, 所以

$ \mathit{\boldsymbol{U = }}\left( {\begin{array}{*{20}{c}} 0&0&0&{ - 1}&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&{ - 1}&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1 \end{array}} \right) \in \mathit{\boldsymbol{U}}\left( 8 \right) $

为式(1)对应的特解。

此时可算得

$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_0^{\left( 7 \right)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {z^3}} \\ {\sqrt {21} {z^2}} \\ { - \sqrt 7 z} \\ 1 \end{array}} \right) + {\text{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} {z^4}} \\ {\sqrt {21} {z^5}} \\ {\sqrt 7 {z^6}} \\ {{z^7}} \end{array}} \right)} \right]_\mathbb{H}},\pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_7^{\left( 7 \right)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {{\bar z}^4}} \\ { - \sqrt {21} {{\bar z}^5}} \\ { - \sqrt 7 {{\bar z}^6}} \\ { - {{\bar z}^7}} \end{array}} \right) + {\text{j}}\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {{\bar z}^3}} \\ {\sqrt {21} {{\bar z}^2}} \\ { - \sqrt 7 \bar z} \\ 1 \end{array}} \right)} \right]_\mathbb{H}}. $

式(2)对应的解

设$ \mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{U}}_1}}&0\\ 0&{{\mathit{\boldsymbol{I}}_4}} \end{array}} \right) $, 等式(2)等价于$ \left( {\begin{array}{*{20}{c}} 0&{\mathit{\boldsymbol{U}}_1^{\rm{T}}}\\ { - {\mathit{\boldsymbol{U}}_1}}&0 \end{array}} \right) = \mathit{\boldsymbol{A}} $, 令$ A_{16}=A_{38}=0 $, 则$ {\mathit{\boldsymbol{U}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&{\frac{{\sqrt {105} }}{{21}}{A_{17}}}&{{A_{18}} + 2{A_{27}}}\\ 0&{ - \frac{{4\sqrt {21} }}{{21}}{A_{17}}}&{ - \frac{5}{3}{A_{27}} - \frac{2}{3}{A_{18}}}&{\frac{{\sqrt {105} }}{{21}}{A_{28}}}\\ {{A_{17}}}&{{A_{27}}}&{ - \frac{{4\sqrt {21} }}{{21}}{A_{28}}}&0\\ {{A_{18}}}&{{A_{28}}}&0&0 \end{array}} \right) $.取$ A_{17}=A_{28}=0, A_{18}=1, A_{27}=-1 $, 所以

$ \mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{c}} 0&0&0&{ - 1}&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&{ - 1}&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1 \end{array}} \right) \in \mathit{\boldsymbol{U}}\left( 8 \right) $

为式(2)对应的特解。

此时可算得

$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_1^{\left( 7 \right)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( {3{z^2} - 4{z^3}\bar z} \right)} \\ {\sqrt {21} \left( {2z - 5{z^2}\bar z} \right)} \\ { - \sqrt 7 \left( {1 - 6z\bar z} \right)} \\ { - 7\bar z} \end{array}} \right) + {\text{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {4{z^3} - 3{z^4}\bar z} \right)} \\ {\sqrt {21} \left( {5{z^4} - 2{z^5}\bar z} \right)} \\ {\sqrt 7 \left( {6{z^5} - {z^6}\bar z} \right)} \\ {7{z^6}} \end{array}} \right)} \right]_\mathbb{H}}, $

$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_6^{\left( 7 \right)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( { - 4{{\bar z}^3} + 3z{{\bar z}^4}} \right)} \\ {\sqrt {21} \left( {5{{\bar z}^4} - 2z{{\bar z}^5}} \right)} \\ { - \sqrt 7 \left( { - 6{{\bar z}^5} + z{{\bar z}^6}} \right)} \\ {7{{\bar z}^6}} \end{array}} \right) + {\text{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {3{{\bar z}^2} - 4z{{\bar z}^3}} \right)} \\ {\sqrt {21} \left( { - 2\bar z + 5z{{\bar z}^2}} \right)} \\ { - \sqrt 7 \left( {1 - 6z\bar z} \right)} \\ {7z} \end{array}} \right)} \right]_\mathbb{H}}. $

式(3)对应的解

设$ \mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{U}}_1}}&0\\ 0&{{\mathit{\boldsymbol{I}}_4}} \end{array}} \right) $, 等式(2)等价于$ \left( {\begin{array}{*{20}{c}} 0&{\mathit{\boldsymbol{U}}_1^{\rm{T}}}\\ { - {\mathit{\boldsymbol{U}}_1}}&0 \end{array}} \right) = \mathit{\boldsymbol{A}} $, 令$ {\mathit{\boldsymbol{A}}_{18}} = 1 $, 则

$ \mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{c}} 0&0&0&{ - 1}&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&{ - 1}&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1 \end{array}} \right) \in \mathit{\boldsymbol{U}}\left( 8 \right) $

为式(3)对应的特解。

此时可算得

$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_2^{\left( 7 \right)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( {3z - 12{z^2}\bar z + 6{z^3}{{\bar z}^2}} \right)} \\ {\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right)} \\ { - \sqrt 7 \left( { - 6\bar z + 15z{{\bar z}^2}} \right)} \\ {21{{\bar z}^2}} \end{array}} \right) + {\text{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {6{z^2} - 12{z^3}\bar z + 3{z^4}{{\bar z}^2}} \right)} \\ {\sqrt {21} \left( {10{z^3} - 10{z^4}\bar z + {z^5}{{\bar z}^2}} \right)} \\ {\sqrt 7 \left( {15{z^4} - 6{z^5}\bar z} \right)} \\ {21{z^5}} \end{array}} \right)} \right]_\mathbb{H}}, $

$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_5^{\left( 7 \right)}} \right) = \left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( { - 6{{\bar z}^2} - 12z{{\bar z}^3} + 3{z^2}{{\bar z}^4}} \right)} \\ {\sqrt {21} \left( { - 10{{\bar z}^3} + 10z{{\bar z}^4} - {z^2}{{\bar z}^5}} \right)} \\ { - \sqrt 7 \left( {15{{\bar z}^4} - 6z{{\bar z}^5}} \right)} \\ {} \end{array}} \right)} \right] + {\text{j}}{\left[ {\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( { - 3{{\bar z}^2} + 12z{{\bar z}^2} - 6{z^2}{{\bar z}^4}} \right)} \\ {\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right)} \\ {\sqrt 7 \left( {6z - 15{z^2}\bar z} \right)} \\ {21{z^2}} \end{array}} \right)} \right]_\mathbb{H}}. $

式(4)对应的解

由于$ \mathit{\boldsymbol{A}} = 0 $非退化, 所以此方程无解。

例2  设$ z $为球面上一个全纯坐标, 则下述为6个${\mathbb{H}}{P^3}$中常曲率共形极小二维球面：

1) $ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_0^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {z^3}}\\ {\sqrt {21} {z^2}}\\ { - \sqrt 7 z}\\ 1 \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} {z^4}}\\ {\sqrt {21} {z^5}}\\ {\sqrt 7 {z^6}}\\ {{z^7}} \end{array}} \right)} \right]_{\mathbb{H}}} $和$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_7^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {{\bar z}^4}}\\ { - \sqrt {21} {{\bar z}^5}}\\ { - \sqrt 7 {{\bar z}^6}}\\ { - {{\bar z}^7}} \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} { - \sqrt {35} {{\bar z}^3}}\\ {\sqrt {21} {{\bar z}^2}}\\ { - \sqrt 7 \bar z}\\ 1 \end{array}} \right)} \right]_{\mathbb{H}}} $为曲率是4/7的共形极小二维球面。

2) $ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_1^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( {3{z^2} - 4{z^3}\bar z} \right)}\\ {\sqrt {21} \left( {2z - 5{z^2}\bar z} \right)}\\ { - \sqrt 7 (1 - 6z\bar z)}\\ { - 7\bar z} \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {4{z^3} - 3{z^4}\bar z} \right)}\\ {\sqrt {21} \left( {5{z^4} - 2{z^5}\bar z} \right)}\\ {\sqrt 7 \left( {6{z^5} - {z^6}\bar z} \right)}\\ {7{z^6}} \end{array}} \right)} \right]_{\mathbb{H}}} $和$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_5^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( { - 4{{\bar z}^3} + 3z{{\bar z}^4}} \right)}\\ {\sqrt {21} \left( {5{{\bar z}^4} - 2z{{\bar z}^5}} \right)}\\ { - \sqrt 7 \left( { - 6{{\bar z}^5} + z{{\bar z}^6}} \right)}\\ {7{{\bar z}^6}} \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {3{{\bar z}^2} - 4z{{\bar z}^3}} \right)}\\ {\sqrt {21} \left( { - 2\bar z + 5z{{\bar z}^2}} \right)}\\ {\sqrt 7 (1 - 6z\bar z)}\\ {7z} \end{array}} \right)} \right]_{\mathbb{H}}} $为曲率是4/19的共形极小二维球面。

3) $ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_2^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( {3z - 12{z^2}\bar z + 6{z^3}{{\bar z}^2}} \right)}\\ {\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right)}\\ { - \sqrt 7 \left( { - 6z + 15z{{\bar z}^2}} \right)}\\ {21{{\bar z}^2}} \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( {6{z^2} - 12{z^3}\bar z + 3{z^4}{{\bar z}^2}} \right)}\\ {\sqrt {21} \left( {10{z^3} - 10{z^4}\bar z + {z^5}{{\bar z}^2}} \right)}\\ {\sqrt 7 \left( {15{z^4} - 6{z^5}\bar z} \right)}\\ {21{z^5}} \end{array}} \right)} \right]_{\mathbb{H}}} $和$ \pi \circ \left( {\mathit{\boldsymbol{U}} \cdot \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_5^{(7)}} \right) = {\left[ {\left( {\begin{array}{*{20}{c}} { - \sqrt {35} \left( {6{{\bar z}^2} - 12z{{\bar z}^3} + 3{z^2}{{\bar z}^4}} \right)}\\ {\sqrt {21} \left( { - 10{{\bar z}^3} + 10z{{\bar z}^4} - {z^2}{{\bar z}^5}} \right)}\\ { - \sqrt 7 \left( {15{{\bar z}^4} - 6z{{\bar z}^5}} \right)}\\ { - 21{{\bar z}^5}} \end{array}} \right) + {\rm{j}}\left( {\begin{array}{*{20}{c}} {\sqrt {35} \left( { - 3\bar z + 12z{{\bar z}^2} - 6{z^2}{{\bar z}^3}} \right)}\\ {\sqrt {21} \left( {1 - 10z\bar z + 10{z^2}{{\bar z}^2}} \right)}\\ {\sqrt 7 \left( {6z - 15{z^2}\bar z} \right)}\\ {21{z^2}} \end{array}} \right)} \right]_{\mathbb{H}}} $为曲率是4/27的共形极小二维球面。