

中国科学院大学数学科学学院, 北京 100049
2015年11月07日 收稿; 2016年01月07日 收修改稿
基金项目: 国家自然科学基金(11471308)资助
通信作者: 魏志强 ?E-mail:wei1988321@163.com
摘要: HCMU度量是紧黎曼面上带奇点的extremal度量.研究它的存在性十分重要.通过研究Chen和Wu(Pacific J Math,2009,240(2):267-288)给出的S2上HCMU度量存在的充分必要条件,证明当S2上至少有(N-1)个鞍点时,一定存在non-CSC HCMU度量,其中N是所有锥奇点的个数.
关键词: 极值度量紧黎曼面HCMU度量锥奇点
Existence of a class HCMU metric on S2
WEI Zhiqiang


School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Abstract: HCMU metric is an extremal K?hler metric with singularities on a compact Riemann surface. It is important to study the existence of HCMU metrics. Through studying the sufficient and necessary condition of Chen and Wu(Pacific J Math, 2009, 240(2):267-288) for the existence of HCMU metrics on S2, we show that there must exist a non-CSC HCMU metric on S2 which has N conical singularities and at least (N-1) saddle points.
Key words: extremal metriccompact Riemann surfaceHCMU metricconical singularity
一般意义上的extremal K?hler度量最早由Calabi[1]提出, 目的是在一个紧K?hler流形的固定K?hler类中找到“最好”的度量.具体地, 设M为一个紧K?hler流形, 在一个固定的K?hler类中, extremal K?hler度量是下述Calabi能量的临界点
$\mathscr{C}\left( g \right) = \int_M {{R^2}{\rm{d}}g, } $ |
${{R}_{, \alpha \beta }}\text{=}0, 1\le \alpha, \beta \le di{{m}_{\mathbb{C}}}M, $ | (1) |
本文将利用上述这些结果研究这样一个问题:设p1, …, pN是S2给定的N个锥点, 对应的锥角度分别为2πα1, …, 2παN, 其中αk>0(k=1, …, N)且不为1.设α1, …, αN中有I个整数, 当N-I满足什么条件时, S2上一定存在non-CSC HCMU度量.
1 预备知识1.1 HCMU度量, 锥奇点和cusp奇点设M是一个紧致无边的Riemann面, p1, …, pN是M上的N个点.设g是M\P上的光滑保角度量, 其中P={p1, …, pN}.如果g满足
${\Delta _g}K + {K^2} = C, $ | (2) |
$\frac{{\partial {K_{, zz}}}}{{\partial \;\bar z}} = 0, $ | (3) |
${K_{, zz}} = 0, $ | (4) |
$\int_{M\backslash P} {{\rm{d}}g < + \infty, \int_{M\backslash P} {{K^2}{\rm{d}}g < + \infty .} } $ | (5) |
定义1.1 ??设M是Riemann面, p∈M.又设(U, z)为p附近的复坐标系且z(p)=0,g为U\{p}上的光滑度量.如果
定义1.2 ??设M是Riemann面,p∈M.又设(U, z)为p附近的复坐标系且z(p)=0, g为U\{p}上的光滑度量.如果
当HCMU度量满足(5), 则它的奇点或者是锥奇点或者是cusp奇点[7-8].
1.2 带锥奇点的non-CSC HCMU度量的基本性质, football分解与粘合设M是一个紧致无边的Riemann面, p1, …, pN是M上的N个点, 记P:={p1, …, pN}.设g是M\P上的光滑保角度量.设(U, z)为M\P上的局部复坐标系, 则g在U上可以写成
$g = {e^{2\varphi }}{\left| {{\rm{d}}z} \right|^2}, $ |
$\nabla K = \sqrt {-1} {{\rm{e}}^{-2\varphi }}{K_{\overline z }}\frac{\partial }{{\partial z}}$ |
$V = \frac{1}{2}\left( {\sqrt {-1} {{\rm{e}}^{-2\varphi }}{K_{\overline z }}\frac{\partial }{{\partial z}}-\sqrt { - 1} {{\rm{e}}^{ - 2\varphi }}{K_z}\frac{\partial }{{\partial \;\bar z}}} \right)$ |
现在假设g为non-CSC HCMU度量, p1, …, pN为g的锥奇点, 并且g在这些锥奇点的锥角度分别为2πα1, …, 2παN, 这里α1, …, αN都为正实数并且都不为1, 因为Lin和Zhu[5]证明如果HCMU度量在某点处的锥角度为2π, 则度量在该点处是光滑的, 即该点实际上不是度量的奇点. Chen[3]通过研究V以及V的积分曲线得到下列重要结论:
命题1.1[3]??Gauss曲率可以连续地延拓到整个M上.
命题1.2[3]?? V的奇点个数有限, 并且V的奇点集一定是K的鞍点集和K的局部极值点集的并集, 这里K的鞍点是指存在2条以上V的积分曲线在该点相交.
命题1.3[3]??K的鞍点必为g的锥奇点, 并且K的鞍点锥角度为2πα形式, 这里α为整数, 进一步, V在K的鞍点处的旋转指标为α-1.
命题1.4[3]?? g的锥奇点除了K的鞍点外, 都是K的局部极值点. K的光滑临界点都是局部极值点, 即K的局部极值点分成2类:一类是g除鞍点外的锥奇点, 另一类是K的全部光滑临界点. V在K的任何一个局部极值点处的旋转指标都是1.
综合命题1.2,1.3和1.4对V应用Poincaré-Hopf指标定理有
定理1.1[3] ??设
${\chi ^{\left( M \right)}} = \sum\limits_{i = 1}^I {\left( {1-{\alpha _i}} \right) + \left( {N-I} \right) + s, } $ | (6) |
由定理1.1, 如果K没有鞍点, 即I=0, 则χ(M)=N+S>0, 因此, M一定是为球面, 于是2=χ(M)=N+S, 即K只有2个局部极值点分别在最大、最小值点处.这种HCMU度量实际上是旋转对称的, 称为football.如图 1所示.
Fig. 1
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图 1 Football Fig. 1 Football |
其中, p和q分别表示K的最大、最小值点, 且度量在这2点处的锥角度分别为2πα和2πβ, α>β>0:数字1表示纬线, 即V的积分曲线.具体地,
$g = {\rm{d}}{u^2} + {f^2}\left( u \right){\rm{d}}{\theta ^2}\;\;\left( {0 \le u \le l, 0 \le \theta \le 2{\rm{\pi }}} \right)$ |
Fig. 2
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图 2 2个football粘合 Fig. 2 Gluing of two footballs |
分别从K的最小值点沿经线剪开相同的距离, 然后分别将经线P和P′, Q和Q′粘在一起, 于是得到一个新的non-CSC HCMU度量, 这个HCMU度量有锥角度
在文献[4]中, Chen等继续了Chen在文献[3]中的研究并得到下面一些结果.
命题1.5[4]??带锥奇点的non-CSC HCMU度量, 任何K的局部极值点都是K的最大值点或者最小值点.如果记K1和K2分别为K的最大值和最小值, 则K1>0, K1>K2>-(K1+K2).
定理1.2[4]??带锥奇点的non-CSC HCMU度量都可以沿着有限多条从K的最大值点到K的最小值点的测地线(这些测地线与V垂直)将底流形剖分成有限多片, 而每一片都与某个football等距.
定理1.3[4]?? 2个football
$\frac{{{\alpha _1}}}{{{\beta _1}}} = \frac{{{\alpha _2}}}{{{\beta _2}}}, \frac{{{\alpha _1}}}{{{\alpha _2}}} = \frac{{{A_1}}}{{{A_2}}}.$ |
1.3 球面S2上non-CSC HCMU度量的存在与具体构造Lin和Zhu[5]研究
定理1.4[5] ??设
$\alpha = {\alpha _1} + \sum\limits_{i = 2}^N {\left( {{\alpha _i}-1} \right)} .$ |
$\sum\limits_{l = 1}^k {\frac{{-2a}}{{z-{\gamma _l}}}} + \sum\limits_{l' = k = 1}^{\alpha + 1} {\frac{2}{{z-{\gamma _{l'}}}}} = \frac{{B\prod\nolimits_{i = 2}^n {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\prod \begin{array}{l}\alpha + 1\\l = 1\end{array} \left( {z - {\gamma _l}} \right)}}, $ |
$\begin{array}{l}\frac{1}{{{{\left( {K + \frac{{\left( {a-2} \right)\beta }}{{2a-1}}} \right)}^a}}}\left( {\beta-K} \right){\left( {K + \frac{{\beta \left( {a + 1} \right)}}{{2a - 1}}} \right)^{a - 1}} = \\A\prod\limits_{l = 1}^k {{{\left| {z - {\gamma _l}} \right|}^{ - 2a}}\prod\limits_{l' = k + 1}^{\alpha + 1} {{{\left| {z - {\gamma _{l'}}} \right|}^2}, } } \end{array}$ |
$\varphi = \frac{1}{2}\ln \left[{\left( {\frac{1}{{{{\left| {F\left( z \right)} \right|}^2}}}} \right)\left( {-\frac{1}{3}{K^3} + cK + c'} \right)} \right], $ |
$\begin{array}{*{20}{c}}{\beta = \sqrt c \left( {2a - 1} \right)/\sqrt {{a^2} - a + 1} ,}\\{\;c' = \frac{{\left( {a + 1} \right)\left( {a - 2} \right){\beta ^3}}}{{3{{\left( {2a - 1} \right)}^2}}},}\\{F\left( z \right) = {{\left( {z - {z_2}} \right)}^{ - \left( {{\alpha _2} - 1} \right)}} \cdots {{\left( {z - {z_N}} \right)}^{ - \left( {{\alpha _N} - 1} \right)}}f\left( z \right),}\\{\;\;f\left( z \right) = {B^{ - 1}}\frac{{ - 3a\left( {a - 1} \right){\beta ^2}}}{{{{\left( {2a - 1} \right)}^2}}}\prod\limits_{l = 1}^{\alpha + 1} {\left( {z - {\gamma _l}} \right).} }\end{array}$ |
定理1.5[6] ??设
· S2存在一个HCMU度量使得p1, p2, …, pN是锥奇点, 锥角度分别为2πα1, …, 2παN且
· p1, p2, …, pj是K的所有鞍点.
等价于
·
·存在j+1, …, N的置换δ, 整数l满足0≤l≤N-j, 整数t满足0≤t≤S和
$\begin{array}{c}{\alpha _{\delta \left( 1 \right)}} + {\alpha _{\delta \left( 2 \right)}} + \cdots + {\alpha _{\delta \left( l \right)}} + t > {\alpha _{\delta \left( {l + 1} \right)}} + \\\; \cdots + {\alpha _{\delta \left( {N-j} \right)}} + S-t, \end{array}$ |
$\begin{array}{*{20}{c}}{\sum\limits_{k = 1}^l {\frac{{{\alpha _{\delta \left( k \right)}}}}{{z - {z_{\delta \left( k \right)}}}} + \sum\limits_{k' = l + 1}^{N - j} {\frac{{\left( {a - 1} \right){\alpha _{\delta \left( {k'} \right)}}}}{{z - {z_{\delta \left( {k'} \right)}}}} + } } }\\{\;\;\sum\limits_{m = 1}^t {\frac{1}{{z - {\beta _m}}}} + \sum\limits_{m' = t + 1}^s {\frac{{a - 1}}{{z - {\beta _{m'}}}}} }\\{ = \frac{{B\prod\nolimits_{i'}^j {{{\left( {z - {z_{i'}}} \right)}^{{\alpha _{i'}} - 1}}} }}{{\prod\nolimits_{k = j + 1}^N {\left( {z - {z_k}} \right)\prod\nolimits_{m = 1}^s {\left( {z - {\beta _m}} \right)} } }},}\end{array}$ |
$\begin{array}{c}{\alpha _{\max }} = {\alpha _{\delta \left( 1 \right)}} + {\alpha _{\delta \left( 2 \right)}} + \cdots + {\alpha _{\delta \left( l \right)}} + t, \\{\alpha _{\min }} = {\alpha _{\delta \left( {l + 1} \right)}} + \cdots + {\alpha _{\delta \left( {N-j} \right)}} + S-t.\end{array}$ |
定理2.1 ??设
证明 ??因为α1, …, αI为整数, 其中I=N-1.由障碍定理知若满足条件的non-CSC HCMU度量存在, 则一定存在鞍点.现在我们证明S2上一定存在以p1, …, pI为鞍点的non-CSC HCMU度量.由障碍定理, 有
$2 = \sum\limits_{i = 1}^I {\left( {1-{\alpha _i}} \right) + \left( {N-I} \right) + s, } $ |
$S = \sum\limits_{i = 1}^I {\left( {{\alpha _i}-1} \right) + 1 \ge 3.} $ |
引理2.1 ??若存在ω1, …, ωS-1, B使得方程
$\begin{array}{*{20}{c}}{\frac{{{\alpha _N}}}{{z - {z_N}}} + \frac{\lambda }{{z - \beta }} + \sum\limits_{m = 1}^{S - 1} {\frac{1}{{z - {\omega _m}}}} }\\{ = \frac{{B\prod\nolimits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\left( {z - {z_N}} \right)\left( {z - \beta } \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {\omega _m}} \right)} }},}\end{array}$ | (7) |
1)若对任意m,ωm≠zi,i=2, …, I, 则ωm≠zN, m=1, …, S-1, 且ω1, …, ωS-1互不相等.
2)若存在m0, i0, 使得ωm0=zi0, 则ω1, …, ωS-1中有且只有αi0个等于zi0.
证明 ??若B=0,则(7)右边为零, 对左边求在点zN的系数应该有αN+l0=0, 其中l0是ω1, …, ωS-1中等于zN的个数.这是不可能的.从而B≠0.
若存在m0, 使得ωm0=β, 代入(7)并通分可知左边分母的次数比右边分母的次数低, 这是不可能的.因此结论成立.
1)若ω1, …, ωS-1中存在l1>0个与zN相等, 不妨设ω1, …, ωl1=zN.代入(7)并通分可知左边分母的次数低于右边分母的次数, 因此这是不可能的.类似的可证ω1, …, ωS-1互不相等.
2)若存在m0, j0, 使得ωm0=zj0, 不妨设若存在ω1=z2, 则ω1, …, ωS-1中有且只有α2个等于z2.因为若ω1, …, ωS-1中等于z2的个数大于α2,代入(7)通分知左边分母的次数低于右边分母的次数.若ω1, …, ωS-1中等于z2的个数小于α2, 代入(7)可知左边关于z2的留数为零, 这是不可能的.因此结论成立.
引理2.2 ??对
证明 ??令
$\begin{array}{c}\frac{{{\alpha _N}}}{{z-{z_N}}} + \frac{\lambda }{{z-\beta }} + \sum\limits_{m = 1}^{S-1} {\frac{1}{{z - {\omega _m}}}} \\ = \frac{{\sum\nolimits_{k = 1}^S {{b_k}{z^{S - k}}} }}{{\left( {z - {z_N}} \right)\left( {z - \beta } \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {\omega _m}} \right)} }}, \end{array}$ | (8) |
$\begin{array}{c}{b_1} =- \left[{{\alpha _N}\beta + \lambda {z_N} + \left( {{z_N} + \beta } \right)\left( {S-1} \right)-{\sigma _1}} \right], \\{b_2} = \left[{{\alpha _N}\beta + \lambda {z_N} + \left( {{z_N} + \beta } \right)\left( {S-2} \right){\sigma _1}-2{\sigma _2}} \right], \\3 \le k \le S - 1, {b_k} = {\left( { - 1} \right)^k}\left\{ {\left[{{\alpha _N}\beta + \lambda {z_N} + } \right.} \right.\\\;\;\;\;\;\;\;\left. {\;\left( {{z_N} + \beta } \right)\left( {S-k} \right)} \right]{\sigma _{k -1}} + \\\;\;\;\;\;\left. {{z_N}\beta \left( {S -k + 1} \right){\sigma _{k -2}} - k{\sigma _k}} \right\}\\\left. {{b_s} = {{\left( { - 1} \right)}^S}{\alpha _N}\beta {\sigma _{S - 1}} + \lambda {z_N}{\sigma _{S - 1}} + {z_N}\beta {\sigma _{S - 1}}} \right].\end{array}$ |
由于zN=0, 若令σ0=1, σS=0.则
$\begin{array}{l}{b_k} = {\left( {- 1} \right)^k}\left[{\left( {{\alpha _N} + S-k} \right)\beta {\sigma _{k-1}}-} \right.\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {k{\sigma _k}} \right], k = 1, \cdots, S.\end{array}$ | (9) |
$\begin{array}{*{20}{l}}{\prod\limits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}} = } }\\{\prod\limits_{i = 2}^I {\left[ {\sum\limits_{j = 0}^{{\alpha _i} - 1} {{{\left( { - 1} \right)}^{{\alpha _i} - 1 - j}}z_j^{{\alpha _i} - 1 - j}C_{{\alpha _i} - 1}^j{z^j}} } \right]} }\\{ = \sum\limits_{l = 0}^{\alpha - {\alpha _1}} {\left[ {{{\left( { - 1} \right)}^{\alpha - {\alpha _1} - l}}\sum\limits_{l = {j_2} + \cdots + {j_I}} {\left( {z_2^{{\alpha _2} - 1 - {j_2}}C_{{\alpha _2} - 1}^{{j_2}} \cdots z_I^{{\alpha _I} - 1 - {j_I}} \cdot } \right.} } \right.} }\\{\left. {\left. {C_{{\alpha _I} - 1}^{{j_I}}} \right)} \right]{z^l} = \sum\limits_{l = 0}^{\alpha - {\alpha _1}} {{d_l}{z^l}.} }\end{array}$ |
$\begin{array}{l}{d_l} = {\left( {-1} \right)^{\alpha-{\alpha _1}-l}}\\\sum\limits_{l = {j_2} + \cdots + {j_I}} {\left( {z_2^{{\alpha _2} - 1 - {j_2}}C_{{\alpha _2} - 1}^{{j_2}} \cdots z_I^{{\alpha _I} - 1 - {j_I}}C_{{\alpha _I} - 1}^{{j_I}}} \right), } \\\alpha = {\alpha _1} + \sum\limits_{i = 2}^I {\left( {{\alpha _i} - 1} \right) = S.} \end{array}$ |
$B\sum\limits_{l = 0}^{\alpha-{\alpha _1}} {{d_l}{z^l} = \sum\limits_{k = 1}^S {{b_k}{z^{S-k}}.} } $ |
$B\sum\limits_{l = 0}^{S-{\alpha _1}} {{d_l}{z^l} = \sum\limits_{k = 1}^{{\alpha _1}-1} {{b_k}{z^{S-k}} + \sum\limits_{k = {\alpha _1}}^S {{b_k}{z^{S - k}}.} } } $ |
${b_k} = \left\{ \begin{array}{l}0, k = 1, \cdots, {\alpha _1}-1.\\B\;{d_{S-k}}, k = {\alpha _1}, \cdots, S.\end{array} \right.$ | (10) |
$k{\sigma _k} = \left\{ \begin{array}{l}\left( {{\alpha _N} + S-k} \right)\beta {\sigma _{k-1}}, k = 1, \cdots, {\alpha _1}-1.\\{\left( { - 1} \right)^{k + 1}}B{d_{S - k}} + \left( {{\alpha _N} + S - k} \right)\beta {\sigma _{k - 1}}, \\k = {\alpha _1}, \cdots, S.\end{array} \right.$ |
$\begin{array}{l}{\sigma _k} = \\\left\{ \begin{array}{l}C_{{\alpha _N} + S - 1}^k{\beta ^k},k = 1, \cdots ,{\alpha _1} - 1.\\C_{{\alpha _N} + S - 1}^k{\beta ^k} - B{\left( { - 1} \right)^k}\frac{{{d_{S - k}}}}{k}\\- B\sum\limits_{l = {\alpha _1}}^{k - 1} {{{\left( { - 1} \right)}^l}{d_{S - l}}{\beta ^{k - l}}} \\\frac{{\left( {{\alpha _N} + S - k} \right)\left( {{\alpha _N} + S - k + 1} \right) \cdots \left( {{\alpha _N} + s - l - 1} \right)}}{{l\left( {l + 1} \right) \cdots k}},\\k = {\alpha _1}, \cdots ,S.\end{array} \right.\end{array}$ | (11) |
$\begin{array}{l}{\left( {-1} \right)^s}\frac{{{d_0}}}{S} + \sum\limits_{l = {\alpha _1}}^{S-1} {{{\left( {-1} \right)}^l}{d_{S - l}}{\beta ^{S - l}}} \\\frac{{\left( {{\alpha _N}} \right)\left( {{\alpha _N} + 1} \right) \cdots \left( {{\alpha _N} + S - l - 1} \right)}}{{l\left( {l + 1} \right) \cdots S}} \ne 0.\end{array}$ | (12) |
引理2.3 ??对
证明 ??首先证明对不同的β1, β2使得对应的B, ω1, …, ωS-1和
$\begin{array}{l}\frac{{{\alpha _N}}}{{z-{z_N}}} + \frac{\lambda }{{z-{\beta _1}}} + \sum\limits_{m = 1}^{S-1} {\frac{1}{{z - {\omega _m}}}} \\ = \frac{{B\prod\nolimits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\left( {z - {z_N}} \right)\left( {z - {\beta _1}} \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {\omega _m}} \right)} }}, \\\frac{{{\alpha _N}}}{{z - {z_N}}} + \frac{\lambda }{{z - {\beta _2}}} + \sum\limits_{m = 1}^{S - 1} {\frac{1}{{z - {{\widetilde \omega }_m}}}} \\ = \frac{{\widetilde B\prod\nolimits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\left( {z - {z_N}} \right)\left( {z - {\beta _2}} \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {{\widetilde \omega }_m}} \right)} }}, \end{array}$ |
$\begin{array}{l}\frac{\lambda }{{z-{\beta _1}}}-\frac{\lambda }{{z-{\beta _2}}} + \sum\limits_{m = 1}^{S - 1} {\frac{1}{{z - {\omega _m}}} - \sum\limits_{m = 1}^{S - 1} {\frac{1}{{z - {{\widetilde \omega }_m}}}} } \\ = \frac{{B\prod\nolimits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\left( {z - {z_N}} \right)\left( {z - {\beta _1}} \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {\omega _m}} \right)} }} - \\\frac{{\widetilde B\prod\nolimits_{i = 2}^I {{{\left( {z - {z_i}} \right)}^{{\alpha _i} - 1}}} }}{{\left( {z - {z_N}} \right)\left( {z - {\beta _1}} \right)\prod\nolimits_{m = 1}^{S - 1} {\left( {z - {{\widetilde \omega }_m}} \right)} }}, \end{array}$ |
现证明引理2.3, 由于z2, …, zI是有限个点, 若某一个ωm0=zj0, 若这样参数β存在则只能唯一.因此使得参数β对应的ω1, …, ωS-1与z2, …, zI中某些项是相等的只能是有限多.而参数β取值范围是无限的, 从而结论成立.
设
1) αi≥2, i=1, …, I且为整数.由假设知这已经满足.
2)
3)存在
$\begin{align} & \frac{{{\alpha }_{N}}}{z-{{z}_{N}}}+\frac{\lambda }{z-\beta }+\sum\limits_{m=1}^{S-1}{\frac{1}{z-{{\omega }_{m}}}} \\ & =\frac{B\prod\nolimits_{i=2}^{I}{{{\left( z-{{z}_{i}} \right)}^{{{\alpha }_{i}}-1}}}}{\left( z-{{z}_{N}} \right)\left( z-\beta \right)\prod\nolimits_{m=1}^{S-1}{\left( z-{{\omega }_{m}} \right)}}, \\ \end{align}$ | (13) |
综合引理2.1, 2.2和2.3,定理2.1得证.
推论2.1 ??设p1, p2, …, pN是S2上任意N≥3个不同的点, α1, …, αN是N个不为1的正实数, 若α1, …, αN中有N-1整数, 不妨设α1, …, αN-1是整数.则S2\{p1, …, pN}上一定存在non-CSC HCMU度量g, 使得p1, …, pN是g的锥奇点, 锥角度分别是2πα1, …, 2παN且p1, …, pN-1是g的高斯曲率K的鞍点.
证明 ??因为S2上的任意3点都可以经过M?bius变换变成
注:由引理2.2知存在
$\begin{align} & \sum\limits_{i=2}^{k\ \ }{\frac{{{\alpha }_{i}}}{z-{{z}_{i}}}+\frac{{{\alpha }_{N}}}{z-{{z}_{N}}}+\frac{{{\lambda }'}}{z-\beta }+\sum\limits_{m=1}^{{S}'-1}{\frac{1}{z-{{{{\omega }'}}_{m}}}=}} \\ & \frac{B\prod\nolimits_{i=k+1}^{I}{{{\left( z-{{z}_{i}} \right)}^{{{\alpha }_{i}}-1}}}}{\left( z-{{z}_{N}} \right)\left( z-\beta \right)\prod\nolimits_{i=2}^{k}{\left( z-{{z}_{i}} \right)\prod\nolimits_{m=1}^{{S}'-1}{\left( z-{{{{\omega }'}}_{m}} \right)}}}, \\ \end{align}$ | (14) |
3 后续的讨论由定理2.1的证明过程我们发现只要给定的整数越多, S2上存在non-CSC HCMU度量的可能性越大.事实上当给定的鞍点与所有锥奇点的个数相差大于等于3时, 就不一定存在non-CSC HCMU度量, 文献[7]中给出不存在non-CSC HCMU度量的例子, 这个例子是锥点个数与鞍点个数的差为3, 因此我们后续将研究鞍点与锥点相差2的时候是否可以得到类似的结论.
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