$i:\ S^n\rightarrow\mathbb{E}^{n+1}$ 是球面 $S^n$ 到欧氏空间 $\mathbb{E}^{n+1}=(\mathbb{R}^{n+1},g_0)$ 的自然嵌入, 其中
\[g_0=\sum_{i=1}^{n+1}dx^i\otimes dx^i\]
是 $\mathbb{R}^{n+1}$ 上的标准度量. 试在 $S^n$ 的局部坐标系中把 $i^* g_0$ 表达出来.
1
令 $U=S^n-\{(0,\ldots,0,-1)\}$, $V=S^n-\{(0,\ldots,0,1)\}$. 考虑球极投影:
\[
\begin{array}{rcl}
\varphi_U:\ U&\rightarrow &\mathbb{E^{n}}\\
(x^1,\ldots,x^n,x^{n+1})&\mapsto &(\frac{x_1}{1+x^{n+1}},\ldots,\frac{x_n}{1+x^{n+1}})\triangleq(u^1,\ldots,u^n)
\end{array}
\]
设 $p\in U\subset S^n$, $i(p)=q\in\mathbb{E}^{n+1}$.
\[
\begin{array}{rcl}
i_{*p}:\ T_p(S^n)&\rightarrow&T_{q}\mathbb{E}^{n+1}\\
(\frac{\partial}{\partial u^i})_p &\mapsto &i_{*p}(\frac{\partial}{\partial u^i})_p=(\frac{\partial x^k}{\partial u^i})_{\varphi_U(p)}\cdot(\frac{\partial}{\partial x^k})_q
\end{array}
\]
这里使用了 Einstein 求和约定. 即
\[i_{*p}(\frac{\partial}{\partial u^i})_p=(\frac{\partial x^k}{\partial u^i})_{\varphi_U(p)}\cdot(\frac{\partial}{\partial x^k})_q.\tag{$*$}\]
由于 $x^k=(1+x^{n+1})u^k$, $k=1,2,\ldots,n$, 且 $1-(x^{n+1})^2=\sum_{i=1}^{n}(x^i)^2$, 故得
\[x^{n+1}=\frac{1-\sum_{i=1}^{n}(u^i)^2}{1+\sum_{i=1}^{n}(u^i)^2}.\]
所以
\[x^k=(1+x^{n+1})u^k=\frac{2u^k}{1+\sum_{i=1}^{n}(u^i)^2},\quad k=1,2,\ldots,n.\]
故对 $1\leqslant k\leqslant n$, 有
\[
\frac{\partial x^k}{\partial u^i}=
\begin{cases}
2u^k\cdot\frac{-1}{(1+\sum_{i=1}^{n}(u^i)^2)^2}\cdot 2u^i,& k\neq i,\\
\frac{2(1+\sum_{i=1}^{n}(u^i)^2)-2u^i\cdot 2u^i}{(1+\sum_{i=1}^{n}(u^i)^2)^2},& k=i.
\end{cases}
\]
而
\[
\frac{\partial x^{n+1}}{\partial u^i}=\frac{-2u^i(1+\sum_{i=1}^{n}(u^i)^2)-(1-\sum_{i=1}^{n}(u^i)^2)2u^i}{(1+\sum_{i=1}^{n}(u^i)^2)^2}=\frac{-4u^i}{(1+\sum_{i=1}^{n}(u^i)^2)^2}.
\]
为方便书写, 令 $b=\sum_{i=1}^{n}(u^i)^2$. 将上面的计算代入 ($*$) 式, 得
\[
\begin{split}
i_{*p}(\frac{\partial}{\partial u^i})_p&=(\frac{\partial x^k}{\partial u^i})_{\varphi_U(p)}\cdot(\frac{\partial}{\partial x^k})_q\\
&=\frac{-4u^i u^1}{(1+b)^2}\cdot\frac{\partial}{\partial x^1}+\frac{-4u^i u^2}{(1+b)^2}\cdot\frac{\partial}{\partial x^2}+\cdots+\frac{-4u^i u^{i-1}}{(1+b)^2}\cdot\frac{\partial}{\partial x^{i-1}}+\frac{2(1+b)-4(u^i)^2}{(1+b)^2}\cdot\frac{\partial}{\partial x^i}\\
&\quad +\frac{-4u^i u^{i+1}}{(1+b)^2}\cdot\frac{\partial}{\partial x^{i+1}}+\cdots+\frac{-4u^i u^n}{(1+b)^2}\cdot\frac{\partial}{\partial x^n}+\frac{-4u^i}{(1+b)^2}\cdot\frac{\partial}{\partial x^{n+1}}\\
&=\frac{1}{(1+b)^2}\biggl[-4u^i\sum_{k=1,k\neq i}^{n}u^k\frac{\partial}{\partial x^k}+\bigl[2(1+b)-4(u^i)^2\bigr]\frac{\partial}{\partial x^i}+(-4u^i)\frac{\partial}{\partial x^{n+1}}\biggr].
\end{split}
\]
若 $i\neq j$,
\[
\begin{split}
i^* g_0(\frac{\partial}{\partial u^i},\frac{\partial}{\partial u^j})&=g_0(i_*\frac{\partial}{\partial u^i},i_*\frac{\partial}{\partial u^j})\\
&=\frac{1}{(1+b)^4}g_0\biggl(-4u^i\sum_{k=1,k\neq i}^{n}u^k\frac{\partial}{\partial x^k}+\bigl[2(1+b)-4(u^i)^2\bigr]\frac{\partial}{\partial x^i}+(-4u^i)\frac{\partial}{\partial x^{n+1}},\\
&\quad -4u^j\sum_{k=1,k\neq j}^{n}u^k\frac{\partial}{\partial x^k}+\bigl[2(1+b)-4(u^j)^2\bigr]\frac{\partial}{\partial x^j}+(-4u^j)\frac{\partial}{\partial x^{n+1}}\biggr)\\
&=\frac{1}{(1+b)^4}\biggl[16(u^1)^2u^i u^j+16(u^2)^2u^i u^j+\cdots+\bigl[2(1+b)-4(u^i)^2\bigr](-4u^i u^j)\\
&\quad +\cdots+\bigl[2(1+b)-4(u^j)^2\bigr](-4u^i u^j)+\cdots+16(u^n)^2u^i u^j+16u^i u^j\biggr]\\
&=\frac{16u^i u^j}{(1+b)^4}\sum_{k=1}^{n}(u^k)^2+\frac{-8(1+b)u^i u^j}{(1+b)^4}+\frac{-8(1+b)u^i u^j}{(1+b)^4}+\frac{16u^i u^j}{(1+b)^4}\\
&=\frac{16u^i u^j}{(1+b)^4}\bigl[(u^1)^2+\cdots+(u^n)^2-(1+b)+1\bigr]\\
&=0.
\end{split}
\]
当 $i=j$ 时,
\[
\begin{split}
i^* g_0(\frac{\partial}{\partial u^i},\frac{\partial}{\partial u^i})&=g_0(i_*\frac{\partial}{\partial u^i},i_*\frac{\partial}{\partial u^i})\\
&=\frac{1}{(1+b)^4}\biggl[16(u^i)^2(u^1)^2+16(u^i)^2(u^2)^2+\cdots+\bigl[2(1+b)-4(u^i)^2\bigr]^2\\
&\quad +\cdots+16(u^i)^2(u^n)^2+16(u^i)^2\biggl]\\
&=\frac{16(u^i)^2}{(1+b)^4}\bigl[\sum_{k=1,k\neq i}^{n}(u^k)^2+1\bigr]+\frac{1}{(1+b)^4}\bigl[4(1+b)^2-16(1+b)(u^i)^2+16(u^i)^4\bigr]\\
&=\frac{16(u^i)^2}{(1+b)^4}\bigl[\sum_{k=1}^{n}(u^k)^2+1-(1+b)\bigr]+\frac{4(1+b)^2}{(1+b)^4}\\
&=\frac{4}{(1+b)^2}.
\end{split}
\]
因此
\[
\begin{split}
i^* g_0&=\frac{4}{(1+b)^2}\sum_{i=1}^{n}du^i\otimes du^i\\
&=\frac{4}{\Bigl[1+\sum_{i=1}^{n}(u^i)^2\Bigr]^2}\sum_{i=1}^{n}du^i\otimes du^i.
\end{split}
\]