首先证明必要性("$\Rightarrow$")
由于曲线位于球面 $S^2(R)$ 上, 因此 $|r(s)|^2=R^2$. 对其求导, 得 $\langle\dot{r}(s),r(s)\rangle=0$, 即 $\dot{r}(s)\perp r(s)$. 对 $\langle\dot{r}(s),r(s)\rangle=0$ 两边再求导, 得
\[\langle\ddot{r}(s),r(s)\rangle+\langle\dot{r}(s),\dot{r}(s)\rangle=0.\]
注意到 $|\dot{r}(s)|=1$, 因此有
\[\langle\ddot{r}(s),r(s)\rangle=-1.\tag{$1$}\]
对上式进一步求导, 得
\[\langle\dddot{r}(s),r(s)\rangle+\langle\ddot{r}(s),\dot{r}(s)\rangle=0,\]
要知道 $\ddot{r}(s)\perp\dot{r}(s)$ (因为 $|\dot{r}(s)|=1$), 所以得到
\[\langle\dddot{r}(s),r(s)\rangle=0.\tag{$2$}\]
这说明 $\dddot{r}(s)$ 与 位于球面的切平面内.
另一方面, 由 $r(s)\perp\dot{r}(s)$, 可知 $r(s)$ 可由 $e_2(s)$ 和 $e_3(s)$ 线性表示. 等价的, $r(s)$ 可由 $\ddot{r}(s)$ 和 $\dot{r}(s)\times\ddot{r}(s)$ 线性表示. 因此不妨设
\[r(s)=a\ddot{r}(s)+b\dot{r}(s)\times\ddot{r}(s).\]
将此式代入 $(2)$ 式, 得
\[a\langle\ddot{r}(s),\dddot{r}(s)\rangle+b\langle\dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle=0.\]
而根据 $(1)$ 式,
\[-1=\langle r(s),\ddot{r}(s)\rangle=a\langle \ddot{r}(s),\ddot{r}(s)\rangle=a\kappa^2(s),\]
因此 $a=-\frac{1}{\kappa^2(s)}$, 从而也可解出 $b$(当 $\langle \dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle\neq 0$ 时):
\[b=\frac{1}{\kappa^2(s)}\frac{\langle \ddot{r}(s),\dddot{r}(s)\rangle}{\langle \dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle},\]
因此
\[
\begin{split}
r(s)&=-\frac{\ddot{r}(s)}{\kappa^2(s)}+\frac{1}{\kappa^2(s)}\frac{\langle \ddot{r}(s),\dddot{r}(s)\rangle}{\langle \dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle}\dot{r}(s)\times\ddot{r}(s)\\
&=-\frac{1}{|\ddot{r}(s)|}\frac{\ddot{r}(s)}{|\ddot{r}(s)|}+\frac{1}{|\ddot{r}(s)|}\frac{\langle \ddot{r}(s),\dddot{r}(s)\rangle}{\langle \dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle}\frac{\dot{r}(s)\times\ddot{r}(s)}{{|\ddot{r}(s)|}}\\
&=-\frac{1}{\kappa(s)}e_2(s)+\frac{|\ddot{r}(s)|\'_s}{\langle \dot{r}(s)\times\ddot{r}(s),\dddot{r}(s)\rangle}e_3(s)\\
&=-\frac{1}{\kappa(s)}e_2(s)+\frac{\dot{\kappa}(s)}{\tau(s)\kappa^2(s)}e_3(s),
\end{split}
\]
从而
\[R^2=|r(s)|^2=\frac{1}{\kappa^2(s)}+\frac{(\dot{\kappa}(s))^2}{\tau^2(s)\kappa^4(s)},\]
整理得
\[R^2=\frac{1}{\kappa^2(s)}\biggl[1+\frac{(\dot{\kappa}(s))^2}{\tau^2(s)\kappa^2(s)}\biggr].\]
充分性 ($\Leftarrow$)
我们假设三维空间中的曲线 $r(s)$ 满足
\[R^2=\frac{1}{\kappa^2(s)}\biggl[1+\frac{(\dot{\kappa}(s))^2}{\tau^2(s)\kappa^2(s)}\biggr].\tag{$3$}\]
而从上面看到, 这来源于对
\[r(s)=-\frac{1}{\kappa(s)}e_2(s)+\frac{\dot{\kappa}(s)}{\tau(s)\kappa^2(s)}e_3(s)\]
模长的计算. 现在的问题是我们仅仅已知 ($3$) 式的成立, 要去证明 $|r(s)|=R$. 因此不妨考虑下面的曲线
\[\gamma(s):=r(s)+\frac{1}{\kappa(s)}e_2(s)-\frac{\dot{\kappa}(s)}{\tau(s)\kappa^2(s)}e_3(s).\]
直觉告诉我们要证明 $\gamma(s)\equiv\text{const.}$ (即希望证明 $r(s)$ 是上面必要性中的球面曲线的平移), 为此需要对上式求导, 希望处处为零. 实践表明, 使用下面的记号会大大简化计算, 记
\[\mathcal{R}(s):=\frac{1}{\kappa(s)},\quad\mathcal{T}(s):=\frac{1}{\tau(s)},\]
于是
\[
\begin{split}
\gamma(s)&=r(s)+\frac{1}{\kappa(s)}e_2(s)-\frac{\dot{\kappa}(s)}{\tau(s)\kappa^2(s)}e_3(s)\\
&=r(s)+\mathcal{R}(s)e_2(s)+\mathcal{T}(s)\mathcal{R}\'(s)e_3(s),
\end{split}
\]
对上式两边求导, 得
\[
\begin{split}
\dot{\gamma}(s)&=e_1(s)+\mathcal{R}\'(s)e_2(s)+\mathcal{R}(s)\dot{e}_2(s)+\bigl[\mathcal{T}(s)\mathcal{R}\'(s)\bigr]\'_s e_3(s)+\mathcal{T}(s)\mathcal{R}\'(s)\dot{e}_3(s)\\
&=e_1(s)+\mathcal{R}\'(s)e_2(s)+\mathcal{R}(s)\bigl(-\kappa(s)e_1(s)+\tau(s)e_3(s)\bigr)+\bigl[\mathcal{T}(s)\mathcal{R}\'(s)\bigr]\'_s e_3(s)\\
&\quad +\mathcal{T}(s)\mathcal{R}\'(s)\bigl(-\tau(s)e_2(s)\bigr)\\
&=\Bigl(\mathcal{R}(s)\tau(s)+\bigl[\mathcal{T}(s)\mathcal{R}\'(s)\bigr]\'_s\Bigr)e_3(s)
\end{split}
\]
另一方面, 根据刚才的记号, 条件
\[R^2=\frac{1}{\kappa^2(s)}\biggl[1+\frac{(\dot{\kappa}(s))^2}{\tau^2(s)\kappa^2(s)}\biggr]\]
变为
\[
\mathcal{R}^2(s)+\bigl(\mathcal{T}(s)\mathcal{R}\'(s)\bigr)^2=R^2.
\]
对此式两边求导, 得
\[2\mathcal{R}(s)\mathcal{R}\'(s)+2\mathcal{T}(s)\mathcal{R}\'(s)\bigl(\mathcal{T}(s)\mathcal{R}\'(s)\bigr)\'_s=0,\]
整理得
\[\mathcal{T}(s)\mathcal{R}\'(s)\Bigl[\mathcal{R}(s)\tau(s)+\bigl(\mathcal{T}(s)\mathcal{R}\'(s)\bigr)\'_s\Bigr]=0.\]
代入刚才计算的 $\dot{\gamma}(s)$, 得 $\dot{\gamma}(s)\equiv 0$. 因此“曲线” $\gamma(s)$ 实际上是一个不动的点, 记为 $P_0$. 于是
\[|r(s)-P_0|=|-\frac{1}{\kappa(s)}e_2(s)+\frac{\dot{\kappa}(s)}{\tau(s)\kappa^2(s)}e_3(s)|=R,\]
这说明曲线 $r(s)$ 位于以 $P_0$ 为中心, 半径为 $R$ 的球面上.
References:
Manfredo P. do Carmo, Differential Geometry of Curves and Surfaces. (中译本, 曲线与曲面的微分几何, 田畴、忻元龙、姜国英、彭家贵、潘养廉 译.) P.20, 习题1.5, 13题.