问题及解答

记 $S^n=\{(x^1,\ldots,x^{n+1})\in\mathbb{R}^{n+1}\mid \sum_{i=1}^{n+1}(x^i)^2=1\}$, 包含映射 $i:S^n\rightarrow\mathbb{R}^{n+1}$ 是一个嵌入. 试在 $S^n$ 的局部坐标系中把 $i^*g_0$ 表达出来, 此处 $g_0=\sum_{i=1}^{n+1}dx^i\otimes dx^i$ 是 $\mathbb{R}^{n+1}$ 上的标准度量.

Hint:

比较简单的办法是通过变换直接将度量张量计算出来.

另一种比较笨的办法是通过切映射将度量张量计算出来.

令 $U=S^n-\{(0,\ldots,0,-1)\}$, $V=S^n-\{(0,\ldots,0,1)\}$.

考虑球极投影:

\[
\begin{array}{rcl}
\varphi_U:\ U&\rightarrow&\mathbb{R}^n\\
(x^1,\ldots,x^n,x^{n+1})&\mapsto&\Bigl(\frac{x^1}{1+x^{n+1}},\ldots,\frac{x^n}{1+x^{n+1}}\Bigr)=:(u^1,\ldots,u^n)
\end{array}
\]

从而经过计算得到

\[
i^*g_0=\frac{4}{\bigl(1+\sum_{i=1}^{n}(u^i)^2\bigr)^2}\sum_{i=1}^{n}du^i\otimes du^i.
\]

(这里的做法是使用切映射, 根据度量的定义去验证, 比较繁琐.)

令 $U=S^n-\{(0,\ldots,0,-1)\}$, $V=S^n-\{(0,\ldots,0,1)\}$.

考虑球极投影(以南极点投影为例):

\[
\begin{array}{rcl}
\varphi_U:\ U&\rightarrow&\mathbb{R}^n\\
(x^1,\ldots,x^n,x^{n+1})&\mapsto&\Bigl(\frac{x^1}{1+x^{n+1}},\ldots,\frac{x^n}{1+x^{n+1}}\Bigr)=:(u^1,\ldots,u^n)
\end{array}
\]

切映射 $i_{*_p}: T_p S^n\rightarrow T_{i(p)}\mathbb{R}^{n+1}$ 为
\[
\frac{\partial}{\partial u^i}\bigr|_{p}\mapsto i_{*_p}(\frac{\partial}{\partial u^i}\bigr|_{p})=\frac{\partial x^k}{\partial u^i}(i(p))\cdot\frac{\partial}{\partial x^k}\bigr|_{p}.
\]
由于
\[
\begin{aligned}
x^k=(1+x^{n+1})u^k,\quad k=1,2,\ldots,n\\
1-(x^{n+1})^2=\sum_{i=1}^{n}(x^i)^2,
\end{aligned}
\]
推出
\[
x^{n+1}=\frac{1-\sum\limits_{i=1}^{n}(u^i)^2}{1+\sum\limits_{i=1}^{n}(u^i)^2}.
\]
若记 $b=\sum\limits_{i=1}^{n}(u^i)^2$, 则 $x^{n+1}=\frac{1-b}{1+b}$. 且
\[
x^k=(1+x^{n+1})u^k=\frac{2u^k}{1+\sum\limits_{i=1}^{n}(u^i)^2}=\frac{2}{1+b}u^k,\quad k=1,2,\ldots,n.
\]
因此, 当 $1\leqslant k\leqslant n$ 时,
\[
\frac{\partial x^k}{\partial u^i}=
\begin{cases}
2u^k\cdot\dfrac{-1}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}\cdot 2u^i=\dfrac{-4u^k u^i}{(1+b)^2}, & k\neq i,\\
\dfrac{2\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)-2u^i\cdot 2u^i}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}=\dfrac{2b-4(u^i)^2}{(1+b)^2},& k=i.
\end{cases}
\]
当 $k=n+1$ 时, 偏导数为
\[
\frac{\partial x^{n+1}}{\partial u^i}=\frac{-2u^i\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)-\bigl(1-\sum\limits_{i=1}^{n}(u^i)^2\bigr)\cdot 2u^i}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}=\frac{-4u^i}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}=\frac{-4u^i}{(1+b)^2}.
\]
因此, 切向量为
\[
\begin{split}
i_{*_p}(\frac{\partial}{\partial u^i}\bigr|_{p})&=\frac{\partial x^k}{\partial u^i}(i(p))\cdot\frac{\partial}{\partial x^k}\bigr|_{p}\\
&=\sum_{k=1,k\neq i}^{n}\frac{-4u^i u^k}{(1+b)^2}\frac{\partial}{\partial x^k}\bigr|_{p}+\frac{2(1+b)-4(u^i)^2}{(1+b)^2}\frac{\partial}{\partial x^i}\bigr|_{p}+\frac{-4u^i}{(1+b)^2}\frac{\partial}{\partial x^{n+1}}\bigr|_{p}.
\end{split}
\]

$g_0=\sum_{i=1}^{n+1}dx^i\otimes dx^i$ 是 $\mathbb{R}^{n+1}$ 上的标准度量. 则当 $i\neq j$ 时,
\[
\begin{split}
&i^{*}g_0(\frac{\partial}{\partial u^i}\bigr|_{p},\frac{\partial}{\partial u^j}\bigr|_{p})\\
=&g_0(i_{*}\frac{\partial}{\partial u^i}\bigr|_{p},i_{*}\frac{\partial}{\partial u^j}\bigr|_{p})\\
=&g_0\biggl(\sum_{k=1,k\neq i}^{n}\frac{-4u^i u^k}{(1+b)^2}\frac{\partial}{\partial x^k}\bigr|_{p}+\frac{2(1+b)-4(u^i)^2}{(1+b)^2}\frac{\partial}{\partial x^i}\bigr|_{p}+\frac{-4u^i}{(1+b)^2}\frac{\partial}{\partial x^{n+1}}\bigr|_{p},\\
\quad &\sum_{k=1,k\neq j}^{n}\frac{-4u^j u^k}{(1+b)^2}\frac{\partial}{\partial x^k}\bigr|_{p}+\frac{2(1+b)-4(u^j)^2}{(1+b)^2}\frac{\partial}{\partial x^j}\bigr|_{p}+\frac{-4u^j}{(1+b)^2}\frac{\partial}{\partial x^{n+1}}\bigr|_{p}\biggr)\\
=&\frac{16(u^1)^2u^i u^j}{(1+b)^4}+\frac{16(u^2)^2u^i u^j}{(1+b)^4}+\cdots+\frac{[2(1+b)-4(u^i)^2](-4u^i u^j)}{(1+b)^4}+\cdots\\
&+\frac{(-4u^i u^j)[2(1+b)-4(u^j)^2]}{(1+b)^4}+\cdots+\frac{16(u^n)^2u^i u^j}{(1+b)^4}+\frac{16u^i u^j}{(1+b)^4}\\
=&\frac{16(u^i)^2}{(1+b)^4}\bigl[(u^1)^2+(u^2)^2+\cdots+(u^i)^2+\cdots+(u^j)^2+\cdots+(u^n)^2\bigr]\\
 &+\frac{-8(1+b)u^i u^j}{(1+b)^4}+\frac{-8(1+b)u^i u^j}{(1+b)^4}+\frac{16u^i u^j}{(1+b)^4}\\
=&\frac{16u^i u^j}{(1+b)^4}\bigl[(u^1)^2+(u^2)^2+\cdots+(u^n)^2-(1+b)+1\bigr]\\
=&0.
\end{split}
\]
当 $i=j$ 时,
\[
\begin{split}
&i^{*}g_0(\frac{\partial}{\partial u^i}\bigr|_{p},\frac{\partial}{\partial u^j}\bigr|_{p})\\
=&\frac{16(u^i)^2(u^1)^2}{(1+b)^4}+\frac{16(u^i)^2(u^2)^2}{(1+b)^4}+\cdots+\frac{[2(1+b)-4(u^i)^2]^2}{(1+b)^4}+\cdots\\
&+\cdots+\frac{16(u^i)^2(u^n)^2}{(1+b)^4}+\frac{16(u^i)^2}{(1+b)^4}\\
=&\frac{16(u^i)^2}{(1+b)^4}\bigl[(u^1)^2+(u^2)^2+\cdots+(u^{i-1})^2+(u^{i+1})^2+\cdots+(u^n)^2+1\bigr]\\
 &+\frac{1}{(1+b)^4}\bigl[4(1+b)^2-16(1+b)(u^i)^2+16(u^i)^4\bigr]\\
=&\frac{16(u^i)^2}{(1+b)^4}\bigl[(u^1)^2+(u^2)^2+\cdots+(u^n)^2+1-(1+b)\bigr]+\frac{4(1+b)^2}{(1+b)^4}\\
=&\frac{4}{(1+b)^2}.
\end{split}
\]
因此
\[
i_{*}g_0=\frac{4}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}\sum_{i=1}^{n}du^i\otimes du^i.
\]

更简单的办法是直接算度量.

考虑球极投影
\[
\begin{array}{rcl}
\varphi:\ S^n-\{(0,\ldots,0,1)\}&\rightarrow&\mathbb{R}^n\\
(x^1,\ldots,x^n,x^{n+1})&\mapsto&\Bigl(\frac{x^1}{1-x^{n+1}},\ldots,\frac{x^n}{1-x^{n+1}}\Bigr)=:(u^1,\ldots,u^n)
\end{array}
\]
于是
\[
u_1^2+\cdots+u_n^2=\frac{\sum\limits_{i=1}^{n}(x^i)^2}{(1-x^{n+1})^2}=\frac{1+x^{n+1}}{1-x^{n+1}}.
\]
由此, 得
\[
1+u_1^2+\cdots+u_n^2=\frac{2}{1-x^{n+1}}.
\]
即
\[
1-x^{n+1}=\frac{2}{1+\sum\limits_{i=1}^{n}(u^i)^2}
\]
对 $x^i=u^i(1-x^{n+1})$ 两边作微分, 得
\[
dx^i=(1-x^{n+1})du^i-u^idx^{n+1}.
\]
代入到 $\mathbb{R}^{n+1}$ 的标准黎曼度量 $g=\sum\limits_{i=1}^{n+1}dx^{i}\otimes dx^{i}$ 中, 得
\[
\begin{split}
g&=\sum_{i=1}^{n+1}(dx^{i})^2\\
&=\sum_{i=1}^{n}\Bigl[(1-x^{n+1})du^i-u^idx^{n+1}\Bigr]^2+(dx^{n+1})^2\\
&=\frac{4}{(1+\sum\limits_{i=1}^{n}(u^i)^2)^2}\sum_{i=1}^{n}(du^i)^2-2(1-x^{n+1})dx^{n+1}\sum_{i=1}^{n}u^idu^i\\
&\quad +\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)(dx^{n+1})^2\\
&=\frac{4}{(1+\sum\limits_{i=1}^{n}(u^i)^2)^2}\sum_{i=1}^{n}(du^i)^2-(1-x^{n+1})dx^{n+1}\sum_{i=1}^{n}d(u^i)^2+(dx^{n+1})^2\cdot\frac{2}{1-x^{n+1}}
\end{split}
\]
其中
\[
\begin{split}
(1-x^{n+1})dx^{n+1}\sum_{i=1}^{n}d(u^i)^2&=(1-x^{n+1})dx^{n+1}d\Bigl(1+\sum_{i=1}^{n}(u^i)^2\Bigr)\\
&=(1-x^{n+1})dx^{n+1}d(\frac{2}{1-x^{n+1}})\\
&=(1-x^{n+1})dx^{n+1}\cdot\frac{2}{(1-x^{n+1})^2}dx^{n+1}\\
&=\frac{2}{1-x^{n+1}}(dx^{n+1})^2.
\end{split}
\]
故
\[
g=\frac{4}{\bigl(1+\sum\limits_{i=1}^{n}(u^i)^2\bigr)^2}\sum_{i=1}^{n}du^i\otimes du^i.
\]