等参超曲面是高维微分几何中非常重要的研究对象, 在三维空间称为等参曲面, 这些曲面是一些平行曲面族.最简单的情况, 就是一个点波源诱发的球面波, 波前就是同心球面族; 然后是一个线波源, 诱发的同轴柱面族; 或者是一个面波源诱导的平行平面族.在三维欧氏空间$ \mathbb{R}^3 $中, 平行曲面之间的性质研究已出现在相关文献中, 参见文献[1].在三维双曲空间$ \mathbb{H}^3 $中, 平行曲面族的相关特征可参看文献[2, 3].自然的想法是, 在三维球空间中, 平行曲面族之间又有些什么样的特征?本文将给出一个初步的探讨.

设$ x:M^2\rightarrow S^3(c) $为三维球空间中的曲面, $ e_i $是$ x $的局部单位切标架, $ \omega_i $为其对偶标架, $ e_3 $为$ x $的单位法向量, 做$ M $的平行曲面族$ \{M_t\} $为

$ \begin{equation} x_t = x\cos t-e_3\sin t, \ \ -\epsilon <t<\epsilon. \end{equation} $

(2.1)

为方便起见, 我们约束指标范围$ 1\leq i,j\leq 2 $.对于$ x $而言, 有

$ \begin{equation} \begin{cases} dx = \omega_i e_i,\\ de_i = \omega_{ij}e_j+h_{ij}\omega_j e_3-c\omega_i x, \\ de_3 = -h_{ij}\omega_j e_i, \end{cases} \end{equation} $

(2.2)

其中$ h_{ij} $为$ M $的第二基本形式的系数.对(2.1)外微分并利用(2.2)可得

$ \begin{equation} dx_t = (\omega_i \cos t+h_{ij}\omega_j \sin t)e_i, \end{equation} $

(2.3)

即$ \{e_i\} $也是$ T_{x_t}M_t $的标准基, 其对偶基为

$ \begin{equation} \omega_{i}^t = (\cos t \delta_{ij}+\sin t h_{ij})\omega_j. \end{equation} $

(2.4)

而

$ \begin{equation} e_{3}^t = x\sin t+e_3 \cos t \end{equation} $

(2.5)

为$ x_t $的单位法向量.一方面,

$ \begin{equation} de_{3}^t = ( \omega_i\sin t- h_{ij}\omega_j\cos t)e_i; \end{equation} $

(2.6)

另一方面,

$ \begin{equation} de_{3}^t = -h_{ik}^t\omega_{k}^te_i = -h_{ik}^t(\cos t \delta_{kj}+\sin t h_{kj})\omega_je_i. \end{equation} $

(2.7)

比较(2.6)和(2.7)式可得

$ \begin{equation} h_{ik}^t(\cos t \delta_{kj}+\sin t h_{kj}) = h_{ij}\cos t-\delta_{ij}\sin t. \end{equation} $

(2.8)

引理2.1 $ S^3 $中的平行曲面$ M_t $的主曲率为

$ \begin{equation} \lambda_{i}(t) = \cot(\theta_i+t), \end{equation} $

(2.9)

其中$ \theta_i $由$ \cot\theta_i = \lambda_{i} $确定.

证 由(2.8)式, 令$ h_{ij} = \lambda_i \delta_{ij} $可得

$ \begin{equation*} \lambda_{i}^t = \frac{\cos\theta_i\cos t-\sin\theta_i\sin t}{\sin\theta_i\cos t+\cos\theta_i\sin t} = \frac{\cos (\theta_i+t)}{\sin (\theta_i+t)} = \cot (\theta_i+t). \end{equation*} $

为便于主要定理的证明, 我们需要下面的引理.

引理2.2 设$ x:M^2\rightarrow S^3(c) $是三维球面上的光滑曲面, $ \lambda_1,\lambda_2 $是曲面上的两个主曲率, 则

(1) 两个主曲率$ \lambda_1,\lambda_2 $是曲面上的连续函数;

(2) 两个主曲率$ \lambda_1,\lambda_2 $是曲面上在非脐点处的光滑函数.

证 现取曲面$ x $的局部参数化$ (U,u,v)\in M^2,\ \ x:U\rightarrow S^3,\ x = x(u,v)\in S^3. $它的诱导度量和第二基本形式局部上分别表示为

$ \begin{equation*} ds^2 = g_{11}du^2+2g_{12}dudv+g_{22}dv^2,\ II = h_{11}du^2+2h_{12}dudv+h_{22}dv^2. \end{equation*} $

由于曲面的光滑性, 因此度量张量的系数$ \{g_{11},g_{12},g_{22}\} $和第二基本形式的系数$ \{h_{11},h_{12},h_{22}\} $都是曲面$ x $上关于参数$ u,v $的光滑函数.这样矩阵

$ \begin{equation*} A = \left(\begin{array}{cccc} g_{11} & g_{12} \\ g_{21} & g_{22} \end{array}\right),\ \ B = \left(\begin{array}{cccc} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right) \end{equation*} $

是两个光滑矩阵.既然$ A $是正定矩阵, 那么$ C = BA^{-1} $是一个光滑矩阵.由于曲面上的高斯曲率$ K $和平均曲率$ H $可以表达为

$ \begin{equation*} K = |C|,\ \ H = \text{tr}(C). \end{equation*} $

这样曲面上的高斯曲率$ K $和平均曲率$ H $都是曲面上的光滑函数, 即是参数$ u,v $的光滑函数.

现设$ \lambda_1,\lambda_2 $是曲面上的两个主曲率, 则$ \lambda_1+\lambda_2 = 2H, \ \ \lambda_1\lambda_2 = K, $从而有

$ \begin{equation*} \lambda_1 = H+\sqrt{H^2-K}, \ \ \lambda_2 = H-\sqrt{H^2-K}. \end{equation*} $

注意到$ \sqrt{H^2-K} $在$ H^2-K>0 $处是光滑的, 即在非脐点处主曲率是光滑的, 从而引理得证.

定理3.1 $ M $有常主曲率的充要条件是$ \sum\limits_{i = 1}^2\lambda_{i}^k(t) $只是$ t $的函数($ k = 1,2,3 $).

证 必要性是显然的.故只需证明充分性即可, 先证明结论对$ k = 3 $时成立.设

$ \begin{equation} a(t) = \sum\limits_{i = 1}^2\cot^3(\theta_i+t) = \sum\limits_{i = 1}^2\lambda_{i}^3(t). \end{equation} $

(3.1)

对(3.1)式关于$ t $求导, 得

$ \begin{equation} \begin{cases} a^{'}(t) = -3\sum\limits_{i = 1}^2\left(\lambda_{i}^4(t)+\lambda_{i}^2(t)\right),\\ a^{''}(t) = 3\sum\limits_{i = 1}^2\left(4\lambda_{i}^5(t)+6\lambda_{i}^3(t)+2\lambda_{i}(t)\right), \\ a^{'''}(t) = -6\sum\limits_{i = 1}^2\left(10\lambda_{i}^6(t)+19\lambda_{i}^4(t)+10\lambda_{i}^2(t)+1\right). \end{cases} \end{equation} $

(3.2)

在(3.1)和(3.2)式中令$ t = 0 $, 从而得

$ \begin{align} \lambda_{1}^{3}+\lambda_{2}^{3} = c_1, \end{align} $

(3.3)

$ \begin{align} \lambda_{1}^{4}+\lambda_{2}^{4}+\lambda_{1}^{2}+\lambda_{2}^{2} = c_2, \end{align} $

(3.4)

$ \begin{align} 4(\lambda_{1}^{5}+\lambda_{2}^{5})+6(\lambda_{1}^{3}+\lambda_{2}^{3})+2(\lambda_1+\lambda_2) = c_3, \end{align} $

(3.5)

$ \begin{align} 10(\lambda_{1}^{6}+\lambda_{2}^{6})+19(\lambda_{1}^{4}+\lambda_{2}^{4})+10(\lambda_{1}^{2}+\lambda_{2}^{2})+1 = c_4. \end{align} $

(3.6)

由(3.3)和(3.5)式得

$ \begin{equation} 2(\lambda_{1}^{5}+\lambda_{2}^{5})+(\lambda_{1}+\lambda_{2}) = c_5. \end{equation} $

(3.7)

再由(3.4)和(3.6)式可得

$ \begin{equation} 10(\lambda_{1}^{6}+\lambda_{2}^{6})-9(\lambda_{1}^{2}+\lambda_{2}^{2}) = c_6. \end{equation} $

(3.8)

上面的$ c_p,\ p = 1,2,\cdots, 6 $为常数.

由引理2.2知, $ M $上的主曲率$ \lambda_i $是局部坐标$ u,v $的函数, 因此在(3.3)和(3.4)式中分别对$ u,v $求偏导数可得

$ \begin{equation} \begin{cases} \lambda_{1}^{2}\frac{\partial\lambda_1}{\partial u}+\lambda_{2}^{2}\frac{\partial\lambda_2}{\partial u} = 0,\\ \lambda_{1}^{2}\frac{\partial\lambda_1}{\partial v}+\lambda_{2}^{2}\frac{\partial\lambda_2}{\partial v} = 0, \\ \lambda_{1}(2\lambda_{1}^2+1)\frac{\partial\lambda_1}{\partial u}+ \lambda_{2}(2\lambda_{2}^2+1)\frac{\partial\lambda_2}{\partial u} = 0,\\ \lambda_{1}(2\lambda_{1}^2+1)\frac{\partial\lambda_1}{\partial v}+ \lambda_{2}(2\lambda_{2}^2+1)\frac{\partial\lambda_2}{\partial v} = 0. \end{cases} \end{equation} $

(3.9)

(3.9)式是以$ \frac{\partial\lambda_1}{\partial u},\frac{\partial\lambda_2}{\partial u},\frac{\partial\lambda_1}{\partial v},\frac{\partial\lambda_2}{\partial v} $为变元的齐次方程组, 它的系数行列式为

$ \begin{equation} \left|\begin{array}{cccc} \lambda_{1}^{2} & \lambda_{2}^{2}& 0& 0 \\ 0& 0& \lambda_{1}^{2} & \lambda_{2}^{2} \\ \lambda_{1}(2\lambda_{1}^2+1) & \lambda_{2}(2\lambda_{2}^2+1)& 0& 0 \\ 0& 0& \lambda_{1}(2\lambda_{1}^2+1)& \lambda_{2}(2\lambda_{2}^2+1) \end{array}\right| = -( \lambda_{1} \lambda_{2}( \lambda_{1} -\lambda_{2})(1-2 \lambda_{1} \lambda_{2}))^2. \end{equation} $

(3.10)

当

$ \begin{equation} \lambda_{1} \lambda_{2}( \lambda_{1} -\lambda_{2})(1-2 \lambda_{1} \lambda_{2})\neq 0 \end{equation} $

(3.11)

时, $ \lambda_i $为常数.

同理, 对(3.3)和(3.7)式做类似于(3.9)和(3.10)式的讨论可知, 当

$ \begin{equation} (\lambda_{1}^2-\lambda_{2}^2)(10\lambda_{1}^2\lambda_{2}^2-1)\neq 0 \end{equation} $

(3.12)

时, $ \lambda_i $为常数.

综合(3.11)和(3.12)式可得, 当$ \lambda_{1}^2-\lambda_{2}^2\neq 0 $时, $ \lambda_i $为常数.若$ \lambda_{1} = \lambda_{2} $, 显然$ \lambda_i $为常数.因此只需证明当$ \lambda_{1}+\lambda_{2} = 0 $ ($ \lambda_i\neq 0 $)时, $ \lambda_i $为常数.事实上, 将$ \lambda_{2} = -\lambda_{1} $代入(3.4)和(3.8)式得

$ \begin{equation} \begin{cases} \lambda_{1}^{4}+\lambda_{1}^{2} = \frac{1}{2}c_2,\\ 10\lambda_{1}^{6}-9\lambda_{1}^{2} = \frac{1}{2}c_2. \end{cases} \end{equation} $

(3.13)

微分(3.13)式可得方程组

$ \begin{equation} \begin{cases} (2\lambda_{1}^2+1)d\lambda_1 = 0,\\ (10\lambda_{1}^4-3)d\lambda_1 = 0. \end{cases} \end{equation} $

(3.14)

由(3.14)式可知$ 2\lambda_{1}^2+1 = 0 $与$ 10\lambda_{1}^4-3 = 0 $无公共解, 则必然有$ d\lambda_1 = 0 $, 即$ \lambda_1 $为常数, 由(3.3)式即得$ \lambda_2 $为常数. $ k = 3 $的情形证毕.

下证$ k = 2 $的情形.令

$ \begin{equation} b(t) = \sum\limits_{i = 1}^{2}\lambda_{i}^2(t). \end{equation} $

(3.15)

对(3.15)式两边关于$ t $求二阶导数得

$ \begin{equation} b^{''}(t) = 6\sum\limits_{i = 1}^{2}\lambda_{i}^4(t)+8\sum\limits_{i = 1}^{2}\lambda_{i}^2(t)+4. \end{equation} $

(3.16)

在(3.15)和(3.16)式中令$ t = 0 $得

$ \begin{equation*} \lambda_{1}^2+\lambda_{2}^2 = c_7,\ \lambda_{1}^4+\lambda_{2}^4 = c_8, \end{equation*} $

这里$ c_{7},c_{8} $为常数.由$ \lambda_{1}^4+\lambda_{2}^4 = \frac{1}{2}(\lambda_{1}^2+\lambda_{2}^2)^2+\frac{1}{2}(\lambda_{1}^2-\lambda_{2}^2)^2, $可知$ \lambda_{1}^{2} $和$ \lambda_{2}^{2} $是常数, 从而$ \lambda_{1} $和$ \lambda_{2} $是常数. $ k = 1 $的情形类似于$ k = 2 $, 此处不再赘述.

定理3.2 设$ \lambda_{1}\lambda_{2}\neq 0 $, 且$ \lambda_{1}(1-\lambda_{1}^{2})(1+\lambda_{2}^{4})\neq\lambda_{2}(1-\lambda_{2}^{2})(1+\lambda_{1}^{4}) $, 则$ M $具有常主曲率的充要条件是每个$ M_{t} $有常值的高斯曲率.

证 必要性是显然的, 下证充分性即可.令$ G(t) = \lambda_{1}(t)\lambda_{2}(t), $为$ M_t $的高斯曲率, 因为$ \lambda_{1}\lambda_{2}\neq 0 $, 所以当$ |t| $充分小时, 由引理2.1可知$ G(t)\neq 0 $.作函数

$ \begin{equation} \log|G(t)| = \log|\lambda_{1}(t)\lambda_{2}(t)|, \end{equation} $

(3.17)

上式两边对$ t $求导得

$ \begin{equation} \frac{G^{'}(t)}{G(t)} = \sum\limits_{i = 1}^{2}\frac{\lambda_{i}^{'}(t)}{\lambda_{i}(t)}. \end{equation} $

(3.18)

令$ c(t) = \frac{G^{'}(t)}{G(t)} $, 则可知

$ \begin{align} c(t)& = -\sum\limits_{i = 1}^{2}\lambda_{i}(t)-\sum\limits_{i = 1}^{2}\frac{1}{\lambda_{i}(t)}, \end{align} $

(3.19)

$ \begin{align} c^{'}(t)& = \sum\limits_{i = 1}^{2}\lambda_{i}^2(t)-\sum\limits_{i = 1}^{2}\frac{1}{\lambda_{i}^2(t)}. \end{align} $

(3.20)

在(3.19)和(3.20)式中令$ t = 0 $得

$ \begin{equation} -\lambda_{1}-\lambda_{2}-\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}} = c_{9},\ \ \lambda_{1}^{2}+\lambda_{2}^{2}-\frac{1}{\lambda_{1}^{2}}-\frac{1}{\lambda_{2}^{2}} = c_{10}, \end{equation} $

(3.21)

这里$ c_9,c_{10} $是常数, 对(3.21)式做类似于(3.9)和(3.10)式的讨论可知, 在$ \lambda_{1}(1-\lambda_{1}^{2})(1+\lambda_{2}^{4})\neq\lambda_{2}(1-\lambda_{2}^{2})(1+\lambda_{1}^{4}) $的条件下, $ \lambda_{1},\lambda_{2} $是常数, 证毕.