中国科学院大学数学科学学院, 北京 100049
2017年03月13日 收稿; 2017年04月17日 收修改稿
基金项目: 国家自然科学基金(11331002,11471308)资助
通信作者: 陈晓东, E-mail:chenxiaodong12@mails.ucas.ac.cn
摘要: 通过扭映射π:
关键词: 四元数射影空间共形极小曲面扭映射
Construction of conformal minimal surfaces in
ZHANG Wenjuan, CHEN Xiaodong
School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Abstract: In this paper, we construct ten families of conformal minimal surfaces immersed into
Key words: quaternionic projective spaceconformal minimal surfacetwistor map
极小曲面的研究一直是微分几何子流形研究领域中的一个重点, 通常对于极小曲面的研究, 会选取具有特殊性质的外围空间, 如空间形式、齐性空间等。而四元数射影空间是对称空间, 因此四元数射影空间中极小曲面的研究具有重要意义。但是因为四元数射影空间本身的结构比较复杂, 所以四元数射影空间中极小曲面的研究也颇为困难。
这些年来, 在国内外****的研究下, 四元数射影空间中极小曲面的研究也取得了一些重要的研究成果。
由文献[1], 我们知道四元数射影空间
由于扭映射给出
1 预备知识用符号
$\begin{array}{*{20}{c}}{{{\rm{i}}^2} = {{\rm{j}}^2} = {{\rm{k}}^2} = - 1,{\rm{ij}} = {\rm{k}} = - {\rm{ji,}}}\\{{\rm{jk}} = {\rm{i}} = - {\rm{kj}},{\rm{ki}} = {\rm{j}} = - {\rm{ik}}{\rm{.}}}\end{array}$ |
$\mathit{\boldsymbol{zw}} = \left( {{\mathit{\boldsymbol{z}}_1}{\mathit{\boldsymbol{w}}_1} - {{\mathit{\boldsymbol{\bar z}}}_2}{\mathit{\boldsymbol{w}}_2}} \right) + {\rm{j}}\left( {{\mathit{\boldsymbol{z}}_2}{\mathit{\boldsymbol{w}}_1} + {{\mathit{\boldsymbol{\bar z}}}_1}{\mathit{\boldsymbol{w}}_2}} \right),$ |
在四元数
${\left( {{\mathit{\boldsymbol{z}}_1} + {\rm{j}}{\mathit{\boldsymbol{z}}_2}} \right)^ * } = {{\mathit{\boldsymbol{\bar z}}}_1} - {\rm{j}}{\mathit{\boldsymbol{z}}_2}.$ |
${\mathit{\boldsymbol{z}}_1} + {\rm{j}}{\mathit{\boldsymbol{z}}_2} \mapsto \left( {\begin{array}{*{20}{c}}{{\mathit{\boldsymbol{z}}_1}}\\{{\mathit{\boldsymbol{z}}_2}}\end{array}} \right).$ |
辛群Sp(n)={A∈GL(n;
$\mathit{\boldsymbol{A}} \cdot {\left[ \mathit{\boldsymbol{v}} \right]_\mathbb{H}} = {\left[ {\mathit{\boldsymbol{Av}}} \right]_\mathbb{H}},$ |
$\mathbb{H}{P^n} = Sp\left( {n + 1} \right)/Sp\left( 1 \right) \times Sp\left( n \right).$ |
如果A=X+jY∈Sp(n+1), 记A=(A0, …, An), X=(X0, …, Xn), Y=(Y0, …, Yn), 那么π1(A)=
定义1.1??由于扭映射是一个纤维化, 因此
根据文献[1], 对于[z]∈
$\begin{array}{*{20}{c}}{{\mathscr{H}_{\left[ z \right]}} \leftrightarrow \left\{ {\mathit{\boldsymbol{w}} \in {\mathit{\boldsymbol{z}}^ \bot }\left| {{\sigma _z}\left( \mathit{\boldsymbol{w}} \right) = 0} \right.,{\sigma _z} = } \right.}\\{\left. { - \mathit{\boldsymbol{z}}_2^{\rm{T}}{\rm{d}}{\mathit{\boldsymbol{z}}_1} + \mathit{\boldsymbol{z}}_1^{\rm{T}}{\rm{d}}{\mathit{\boldsymbol{z}}_2}} \right\},}\end{array}$ | (1) |
2
2.1
用Ω表示Sp(n+1)上的Maurer-Cartan形式, 即Ω是Sp(n+1)上的一个取值在
由
$\mathscr{g} = {\mathscr{g}_0} \oplus \mathscr{m},$ |
因此,
$\sum\limits_{a = 1}^n {\mathit{\Omega }_0^{\alpha * } \cdot \mathit{\Omega }_0^\alpha } $ |
${\text{d}}s_{\mathbb{H}{P^n}}^2 = \sum\limits_{a = 1}^n {\omega _0^{\alpha * } \cdot \omega _0^\alpha } ,$ | (2) |
在预备知识中, 我们知道Sp(n+1)
类似于
${\text{d}}s_{\mathbb{C}{P^{2n + 1}}}^2 = \sum\limits_{A = 1}^{2n + 1} {\bar {\tilde \omega} _0^A \cdot \tilde \omega _0^A} .$ | (3) |
$\mathit{\boldsymbol{ \boldsymbol{\varOmega} }} = {\mathit{\boldsymbol{g}}^{ - 1}}{\rm{d}}\mathit{\boldsymbol{g}},$ | (4) |
$\mathit{\boldsymbol{ \boldsymbol{\tilde \varOmega} }}= {\left( {\begin{array}{*{20}{c}}\mathit{\boldsymbol{X}}&{ - \mathit{\boldsymbol{\bar Y}}}\\\mathit{\boldsymbol{Y}}&{ \mathit{\boldsymbol{\bar X}}}\end{array}} \right)^{ - 1}}{\rm{d}}\left( {\begin{array}{*{20}{c}}\mathit{\boldsymbol{X}}&{ - \mathit{\boldsymbol{\bar Y}}}\\\mathit{\boldsymbol{Y}}&{ \mathit{\boldsymbol{\bar X}}}\end{array}} \right).$ | (5) |
则式(4)和式(5)可以写成(为书写方便, 本文仅给出n=3的情形, 一般的情况只是矩阵的维数不同, 并没有本质区别):
$\mathit{\boldsymbol{ \boldsymbol{\varOmega} }} = \left( {\begin{array}{*{20}{c}}{\mathit{\Omega }_0^0}&{\mathit{\Omega }_1^0}&{\mathit{\Omega }_2^0}&{\mathit{\Omega }_3^0}\\{\mathit{\Omega }_0^1}&{\mathit{\Omega }_1^1}&{\mathit{\Omega }_2^1}&{\mathit{\Omega }_3^1}\\{\mathit{\Omega }_0^2}&{\mathit{\Omega }_1^2}&{\mathit{\Omega }_2^2}&{\mathit{\Omega }_3^2}\\{\mathit{\Omega }_0^3}&{\mathit{\Omega }_1^3}&{\mathit{\Omega }_2^3}&{\mathit{\Omega }_3^3}\end{array}} \right),$ | (6) |
$\mathit{\boldsymbol{ \boldsymbol{\tilde \varOmega} }} = \left( {\begin{array}{*{20}{c}}{\mathit{\Gamma }_0^0}&{\mathit{\Gamma }_1^0}&{\mathit{\Gamma }_2^0}&{\mathit{\Gamma }_3^0}&{ - \mathit{\Sigma }_0^0}&{ - \mathit{\Sigma }_1^0}&{ - \mathit{\Sigma }_2^0}&{ - \mathit{\Sigma }_3^0}\\{\mathit{\Gamma }_0^1}&{\mathit{\Gamma }_1^1}&{\mathit{\Gamma }_2^1}&{\mathit{\Gamma }_3^1}&{ - \mathit{\Sigma }_0^1}&{ - \mathit{\Sigma }_1^1}&{ - \mathit{\Sigma }_2^1}&{ - \mathit{\Sigma }_3^1}\\{\mathit{\Gamma }_0^2}&{\mathit{\Gamma }_1^2}&{\mathit{\Gamma }_2^2}&{\mathit{\Gamma }_3^2}&{ - \mathit{\Sigma }_0^2}&{ - \mathit{\Sigma }_1^2}&{ - \mathit{\Sigma }_2^2}&{ - \mathit{\Sigma }_3^2}\\{\mathit{\Gamma }_0^3}&{\mathit{\Gamma }_1^3}&{\mathit{\Gamma }_2^3}&{\mathit{\Gamma }_3^3}&{ - \mathit{\Sigma }_0^3}&{ - \mathit{\Sigma }_1^3}&{ - \mathit{\Sigma }_2^3}&{ - \mathit{\Sigma }_3^3}\\{\mathit{\Sigma }_0^0}&{\mathit{\Sigma }_1^0}&{\mathit{\Sigma }_2^0}&{\mathit{\Sigma }_3^0}&{\mathit{\Gamma }_0^0}&{\mathit{\Gamma }_1^0}&{\mathit{\Gamma }_2^0}&{\mathit{\Gamma }_3^0}\\{\mathit{\Sigma }_0^1}&{\mathit{\Sigma }_1^1}&{\mathit{\Sigma }_2^1}&{\mathit{\Sigma }_3^1}&{\mathit{\Gamma }_0^1}&{\mathit{\Gamma }_1^1}&{\mathit{\Gamma }_2^1}&{\mathit{\Gamma }_3^1}\\{\mathit{\Sigma }_0^2}&{\mathit{\Sigma }_1^2}&{\mathit{\Sigma }_2^2}&{\mathit{\Sigma }_3^2}&{\mathit{\Gamma }_0^2}&{\mathit{\Gamma }_1^2}&{\mathit{\Gamma }_2^2}&{\mathit{\Gamma }_3^2}\\{\mathit{\Sigma }_0^3}&{\mathit{\Sigma }_1^3}&{\mathit{\Sigma }_2^3}&{\mathit{\Sigma }_3^3}&{\mathit{\Gamma }_0^3}&{\mathit{\Gamma }_1^3}&{\mathit{\Gamma }_2^3}&{\mathit{\Gamma }_3^3}\end{array}} \right),$ | (7) |
定理2.1??设Φ:M→
证明??取e:U?M→Sp(n+1)是沿Φ的活动标架, 则π1°e=Φ, 其中π1:Sp(n+1)→
${\pi _2} \circ e = \pi \circ {\pi _1} \circ e = \pi \circ \mathit{\Phi },$ |
由式(2), 式(3), 式(6)和式(7), 有
$\begin{array}{*{20}{c}} {{{\left( {\pi \circ \mathit{\Phi }} \right)}^ * }{\text{d}}s_{\mathbb{H}{P^n}}^2 = \sum\limits_{a = 1}^n {\bar \gamma _0^a \cdot \gamma _0^a} + \sum\limits_{a = 1}^n {\bar \sigma _0^a \cdot \sigma _0^a} ,} \\ {{\mathit{\Phi }^ * }{\text{d}}s_{\mathbb{C}{P^{2n + 1}}}^2 = \sum\limits_{a = 1}^n {\bar \gamma _0^a \cdot \gamma _0^a} + \sum\limits_{a = 0}^n {\bar \sigma _0^a \cdot \sigma _0^a} } \\ { = {{\left( {\pi \circ \mathit{\Phi }} \right)}^ * }{\text{d}}s_{\mathbb{H}{P^n}}^2 + \bar \sigma _0^0 \cdot \sigma _0^0,} \end{array}$ |
根据题设Φ水平, 则Φ*ds
所以, Φ水平当且仅当e*Σ00=0.
定理2.1是从度量的角度来刻画曲面水平的条件的, 它与预备知识中的式(1)是从两个不同角度说明曲面水平, 它们是等价的.
2.2
设
由预备知识中的式(1), 我们知道Φ在M中任意一点水平当且仅当Φ*σ=0, 其中σ=-z2Tdz1+z1Tdz2.
情形1: Φ(M)∩U0≠?.
在坐标卡U0中, 令Φ=[(1, z11, z12, z13, z20, z21, z22, z23)T],
$\begin{array}{*{20}{c}}{{\mathit{\Phi }^ * }\sigma = - \left( {z_2^0,z_2^1,z_2^2,z_2^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}1\\{z_1^1}\\{z_1^2}\\{z_1^3}\end{array}} \right) + \left( {1,z_1^1,z_1^2,z_1^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}{z_2^0}\\{z_2^1}\\{z_2^2}\\{z_2^3}\end{array}} \right) = }\\{{\rm{d}}z_2^0 + z_1^1{\rm{d}}z_2^1 + z_1^2{\rm{d}}z_2^2 + z_1^3{\rm{d}}z_2^3 - z_2^1{\rm{d}}z_1^1 - z_2^2{\rm{d}}z_1^2 - }\\{z_2^3{\rm{d}}z_1^3 = {\rm{d}}\left( {z_2^0 + z_1^1z_2^1 + z_1^2z_2^2 + z_1^3z_2^3} \right) - 2z_2^1{\rm{d}}z_1^1 - }\\{2z_2^2{\rm{d}}z_1^2 - 2z_2^3{\rm{d}}z_1^3.}\end{array}$ |
令z11=a1, z12=a2, z13=a3, 因为d(z20+a1z21+a2z22+a3z23)=0, 所以z20+a1z21+a2z22+a3z23=a4(a4为常数)。假设z21=g1, z22=g2, z23=g3, 则g1, g2, g2为M上的亚纯函数(特别地, 可以为常数)因为Φ全纯。因此, 此时有
$\begin{array}{*{20}{c}}{\mathit{\Phi } = \left[ {\left( {1,{a_1},{a_2},{a_3},{a_4} - {a_1}{g_1} - {a_2}{g_2} - } \right.} \right.}\\{\left. {{{\left. {{a_3}{g_3},{g_1},{g_2},{g_3}} \right)}^{\rm{T}}}} \right].}\end{array}$ |
不失一般性, 假设z13不为常数, 否则通过一个置换矩阵作用Φ可转换为此情形, 因此在相差一个置换矩阵作用的意义下, 可以假设z13不为常数。
设z20+z11z21+z12z22+z13z32=g0, z11=g1, z12=g2, z13=g3, z21=g4, z22=g5, 则gm为M上的亚纯函数, m=0, …, 5.由于d(z20+z11z21+z12z22+z13z23)=2z21dz11+2z22dz12+2z23dz13, 因此z23=
$\begin{array}{*{20}{c}}{\mathit{\Phi } = \left[ {\left( {1,{g_1},{g_2},{g_3},{g_0} - {g_1}{g_4} - {g_2}{g_5} - } \right.} \right.}\\{{g_3}\left( {\frac{1}{2}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_3}}} - {g_4}\frac{{{\rm{d}}{g_1}}}{{{\rm{d}}{g_3}}} - {g_5}\frac{{{\rm{d}}{g_2}}}{{{\rm{d}}{g_3}}}} \right),}\\{\left. {{{\left. {{g_4},{g_5},\frac{1}{2}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_3}}} - {g_4}\frac{{{\rm{d}}{g_1}}}{{{\rm{d}}{g_3}}} - {g_5}\frac{{{\rm{d}}{g_2}}}{{{\rm{d}}{g_3}}}} \right)}^{\rm{T}}}} \right],}\end{array}$ |
情形2: Φ(M)∩U0=?, Φ(M)∩U1≠?
令Φ=[(0, 1, z12, z13, z20, z21, z22, z23)T],
$\begin{array}{l}{\mathit{\Phi }^ * }\sigma = - \left( {z_2^0,z_2^1,z_2^2,z_2^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}0\\1\\{z_1^2}\\{z_1^3}\end{array}} \right) + \left( {0,1,z_1^2,z_1^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}{z_2^0}\\{z_2^1}\\{z_2^2}\\{z_2^3}\end{array}} \right)\\ = {\rm{d}}z_2^1 + z_1^2{\rm{d}}z_2^2 + z_1^3{\rm{d}}z_2^3 - z_2^2{\rm{d}}z_1^2 - z_2^3{\rm{d}}z_1^3 = \\{\rm{d}}\left( {z_2^1 + z_1^2z_2^2 + z_1^3z_2^3} \right) - 2z_2^2{\rm{d}}z_1^2 - 2z_2^3{\rm{d}}z_1^3.\end{array}$ |
令z12=a1, z13=a2, 因为d(z21+a1z22+a2z23)=0, 所以z21+a1z22+a2z23=a3(a3为常数)。令z20=g1, z22=g2, z23=g3, 则g1, g2, g2为M上的亚纯函数(特别地, 可以为常数)。所以, 有
$\mathit{\Phi }{\rm{ = }}\left[ {\left( {0,1,{\mathit{a}_{\rm{1}}},{\mathit{a}_{\rm{2}}},{\mathit{g}_{\rm{1}}},{\mathit{a}_{\rm{3}}} - {\mathit{a}_{\rm{1}}}{\mathit{g}_2} - {\mathit{a}_{\rm{2}}}{\mathit{g}_3},{\mathit{g}_2},{\mathit{g}_3}} \right)^{\rm{T}}} \right].$ |
类似于情形1.2, 可以假设z13不为常数。
设z21+z12z22+z13z23=g0, z12=g1, z13=g2, z20=g3, z22=g4, 则gm为M上的亚纯函数, m=0, …, 4.由于d(z21+z12z22+z13z23)=2z22dz12+2z23dz13, 因此, z23=
$\begin{array}{*{20}{c}}{\mathit{\Phi } = \left[ {\left( {0,1,{g_1},{g_2},{g_3},{g_0} - {g_1}{g_4} - } \right.} \right.}\\{\left. {{{\left. {{g_2}\left( {\frac{1}{2}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_2}}} - {g_4}\frac{{{\rm{d}}{g_1}}}{{{\rm{d}}{g_2}}}} \right),{g_4},\frac{1}{2}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_2}}} - {g_4}\frac{{{\rm{d}}{g_1}}}{{{\rm{d}}{g_2}}}} \right)}^{\rm{T}}}} \right],}\end{array}$ |
情形3: Φ(M)∩(U0∪U1)=?, Φ(M)∩U2≠?
令Φ=[(0, 0, 1, z13, z20, z21, z22, z23)T],
$\begin{array}{*{20}{c}}{{\mathit{\Phi }^ * }\sigma = - \left( {z_2^0,z_2^1,z_2^2,z_2^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}0\\0\\1\\{z_1^3}\end{array}} \right) + \left( {0,0,1,z_1^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}{z_2^0}\\{z_2^1}\\{z_2^2}\\{z_2^3}\end{array}} \right)}\\{ = {\rm{d}}z_2^2 + z_1^3{\rm{d}}z_2^3 - z_2^3{\rm{d}}z_1^3}\\{ = {\rm{d}}\left( {z_2^2 + z_1^3z_2^3} \right) - 2z_2^3{\rm{d}}z_1^3.}\end{array}$ |
令z13=a1, 因为d(z22+a1z23)=0, 所以z22+a1z23=a2(a2为常数)。令z20=g1, z21=g2, z23=g3, 则g1, g2, g3为M上的亚纯函数(特别地, 可以取常数)。所以, 有
$\mathit{\Phi } = \left[ {{{\left( {0,0,1,{a_1},{g_1},{g_2},{a_2} - {a_1}{g_3},{g_3}} \right)}^{\rm{T}}}} \right].$ |
设z22+z13z23=g0, z13=g1, z20=g2, z22=g3, 则gm为M上的亚纯函数, m=0, …, 3.由于d(z22+z13z23)=2z23dz13, 因此, z23=
$\mathit{\Phi } = \left[ {{{\left( {0,0,1,{g_1},{g_2},{g_3},{g_0} - \frac{1}{2}{g_1}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_1}}},\frac{1}{2}\frac{{{\rm{d}}{g_0}}}{{{\rm{d}}{g_1}}}} \right)}^{\rm{T}}}} \right],$ |
情形4: Φ(M)∩(U0∪U1∪U2)=?, Φ(M)∩U3≠?
令Φ=[(0, 0, 0, 1, z20, z21, z22, z23)T],
$\begin{array}{*{20}{c}}{{\mathit{\Phi }^ * }\sigma = - \left( {z_2^0,z_2^1,z_2^2,z_2^3} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right) + \left( {0,0,0,1} \right){\rm{d}}\left( {\begin{array}{*{20}{c}}{z_2^0}\\{z_2^1}\\{z_2^2}\\{z_2^3}\end{array}} \right)}\\{ = {\rm{d}}z_2^3.}\end{array}$ |
$\mathit{\Phi } = \left[ {{{\left( {0,0,0,1,{g_1},{g_2},{g_3},{a_1}} \right)}^{\rm{T}}}} \right].$ |
令Φ=[(0, 0, 0, 0, 1, z21, z22, z23)T], 此时Φ*σ=0恒成立。
令z21=g1, z22=g2, z23=g3, 因此
$\mathit{\Phi } = \left[ {{{\left( {0,0,0,0,1,{g_1},{g_2},{g_3}} \right)}^{\rm{T}}}} \right].$ |
令Φ=[(0, 0, 0, 0, 0, 1, z22, z23)T], 则Φ*σ=0恒成立。设z22=g1, z23=g2, 因此
$\mathit{\Phi } = \left[ {{{\left( {0,0,0,0,0,1,{g_1},{g_2}} \right)}^{\rm{T}}}} \right].$ |
令Φ=[(0, 0, 0, 0, 0, 0, 1, z23)T], 则Φ*σ=0恒成立。设z23=g1, 因此
$\mathit{\Phi } = \left[ {{{\left( {0,0,0,0,0,0,1,{g_1}} \right)}^{\rm{T}}}} \right].$ |
此时Φ=[(0, 0, 0, 0, 0, 0, 0, 1)T]退化, 这与Φ是浸入矛盾。
综合上述8个情形, 可以得到
定理2.2??假设Φ:M→
1) Φ=[(1, a1, a2, a3, a4-a1g1-a2g2-a3g3, g1, g2, g3)T];
2) Φ=[(1, g1, g2, g3, g0-g1g4-g2g5-g3(
3) Φ=[(0, 1, a1, a2, g1, a3-a1g2-a2g3, g2, g3)T];
4) Φ=[(0, 1, g1, g2, g3, g0-g1g4-g2(
5) Φ=[(0, 0, 1, a1, g1, g2, a2-a1g3, g3)T];
6) Φ=[(0, 0, 1, g1, g2, g3, g0
7) Φ=[(0, 0, 0, 1, g1, g2, g3, a1)T];
8) Φ=[(0, 0, 0, 0, 1, g1, g2, g3)T];
9) Φ=[(0, 0, 0, 0, 0, 1, g1, g2)T];
10)Φ=[(0, 0, 0, 0, 0, 0, 1, g1)T].
其中gm为M上的亚纯函数, an为常数, 其中m=0, …, 5, n=1, …, 4.
通过简单的直接计算可以验证定理2.2中所给出的10族共形极小曲面满足定理2.1中的条件。所以, 对于
定理2.2给出
2.3
引理2.1??设M,Y,N为黎曼流形,π:Y→N为黎曼淹没,f:M→Y是关于映射π的一个水平浸入,如果f调和,则π°f:M→N也调和。
由于对于共形映射来说, 映射调和等价于映射极小[5], 而
命题2.1??设Φ:M→
由命题2.1可知, 定理2.2中给出的
例1??设gm为黎曼面M上的亚纯函数, an为常数, m=0, …, 5, n=1, …, 4, 那么有
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
在上述例子中, 可以选取不同形式的亚纯函数及常数, 因此上面的10个例子非常丰富。而
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