习题 An Introduction to Manifold - Loring W. Tu, Problem 8.7

给定两个光滑流形 $M,N$
给定 $\pi_1 : \mathtip{M}{视作集合} \times \mathtip{N}{视作集合} \to \mathtip{M}{视作集合}$ , $\pi_2 : \mathtip{M}{视作集合} \times \mathtip{N}{视作集合} \to \mathtip{N}{视作集合}$ 为两个投影映射
证明 $\forall (p,q) \in \mathtip{M}{视作集合} \times \mathtip{N}{视作集合}$ ，映射

$$ \bbox[5pt]{ \left(\underset{\ast,(p,q)}{\pi_1},\underset{\ast,(p,q)}{\pi_2}\right) : \mathtip{T_{(p,q)}(M \times N)}{M \times N 在 (p,q) 处的全体切向量} \to \mathtip{T_pM}{M 在 p 处的全体切向量} \times \mathtip{T_qN}{N 在 q 处的全体切向量} } $$

是线性同构

Proof.
记 $M$ 为 $m$ 维光滑流形， $N$ 为 $n$ 维光滑流形
给定 $p \in U \subset \mathtip{M}{视作集合}$ , $q \in V \subset \mathtip{N}{视作集合}$ ，记光滑同胚映射

$$ \bbox[5pt]{ \begin{aligned} \varphi & : U \to \mathbb{R}^m , \varphi(p) = (x^1(p),\cdots,x^m(p)) \\ \psi & : V \to \mathbb{R}^n , \psi(q) = (y^1(p),\cdots,y^n(p)) \end{aligned} } $$

记 $T_pM$ 的一组基为

$$ \bbox[5pt]{ \left.\frac{\partial}{\partial{x^1}}\right|_p,\cdots,\left.\frac{\partial}{\partial{x^m}}\right|_p } $$

记 $T_qN$ 的一组基为

$$ \bbox[5pt]{ \left.\frac{\partial}{\partial{y^1}}\right|_q,\cdots,\left.\frac{\partial}{\partial{y^n}}\right|_q } $$

记光滑同胚映射

$$ \bbox[5pt]{ \begin{aligned} & \varphi\times\psi : U \times V \to \mathbb{R}^{m+n} \\ & \varphi\times\psi(p,q) = (u^1(p,q),\cdots,u^m(p,q),v^1(p,q),\cdots,v^n(p,q)) \end{aligned} } $$

其中 $u^k(p,q) = x^k \circ \pi_1(p,q)$ , $v^k(p,q) = y^k \circ \pi_2(p,q)$
记 $T_{(p,q)}(M \times N)$ 的一组基为

$$ \bbox[5pt]{ \left.\frac{\partial}{\partial{u^1}}\right|_{(p,q)},\cdots,\left.\frac{\partial}{\partial{u^m}}\right|_{(p,q)},\left.\frac{\partial}{\partial{v^1}}\right|_{(p,q)},\cdots,\left.\frac{\partial}{\partial{v^n}}\right|_{(p,q)} } $$

记 $a^k,b^k,c^k,d^k \in \mathbb{R}$
记 $X_{(p,q)} \in \mathtip{T_{(p,q)}(M \times N)}{M \times N 在 (p,q) 处的全体切向量}$ 为

$$ \bbox[5pt]{ X_{(p,q)} = \sum_{k=1}^{m}{a^k\left.\frac{\partial}{\partial{u^k}}\right|_{(p,q)}} + \sum_{k=1}^{n}{b^k\left.\frac{\partial}{\partial{v^k}}\right|_{(p,q)}} } $$

记 $X_p \in \mathtip{T_pM}{M 在 p 处的全体切向量}$ 为

$$ \bbox[5pt]{ X_p = \sum_{k=1}^{m}{a^k\left.\frac{\partial}{\partial{x^k}}\right|_p} } $$

记 $X_q \in \mathtip{T_qN}{N 在 q 处的全体切向量}$ 为

$$ \bbox[5pt]{ X_q = \sum_{k=1}^{n}{b^k\left.\frac{\partial}{\partial{y^k}}\right|_q} } $$

记

$$ \bbox[5pt]{ \begin{aligned} \underset{\ast,(p,q)}{\pi_1}(X_{(p,q)}) & = \sum_{k=1}^{m}{c^k\left.\frac{\partial}{\partial{x^k}}\right|_p} \\ \underset{\ast,(p,q)}{\pi_2}(X_{(p,q)}) & = \sum_{k=1}^{n}{d^k\left.\frac{\partial}{\partial{y^k}}\right|_q} \end{aligned} } $$

则 $\forall k$ 成立

$$ \bbox[5pt]{ \begin{aligned} c^k = & \underset{\ast,(p,q)}{\pi_1}(X_{(p,q)})x^k \\ \mathtip{\color{skyblue}{=}}{定义: \underset{\ast,(p,q)}{\pi_1}} & X_{(p,q)}(x^k \circ \pi_1(p,q)) \\ = & X_{(p,q)}(u^k(p,q)) \\ = & a^k \\\\ d^k = & \underset{\ast,(p,q)}{\pi_2}(X_{(p,q)})y^k \\ \mathtip{\color{skyblue}{=}}{定义: \underset{\ast,(p,q)}{\pi_2}} & X_{(p,q)}(y^k \circ \pi_2(p,q)) \\ = & X_{(p,q)}(v^k(p,q)) \\ = & b^k \end{aligned} } $$

即

$$ \bbox[5pt]{ \begin{aligned} \underset{\ast,(p,q)}{\pi_1}(X_{(p,q)}) & = \sum_{k=1}^{m}{a^k\left.\frac{\partial}{\partial{x^k}}\right|_p} = X_p \\ \underset{\ast,(p,q)}{\pi_2}(X_{(p,q)}) & = \sum_{k=1}^{n}{b^k\left.\frac{\partial}{\partial{y^k}}\right|_q} = X_q \end{aligned} } $$

那么 $\forall a,b \in \mathbb{R}$

$$ \bbox[5pt]{ \begin{aligned} & \left(\underset{\ast,(p,q)}{\pi_1} , \underset{\ast,(p,q)}{\pi_2}\right)(aX_{(p,q)} + bY_{(p,q)}) \\ = & \left(\underset{\ast,(p,q)}{\pi_1}(aX_{(p,q)} + bY_{(p,q)}) , \underset{\ast,(p,q)}{\pi_2}(aX_{(p,q)} + bY_{(p,q)})\right) \\ = & \left(a\underset{\ast,(p,q)}{\pi_1}(X_{(p,q)}) + b\underset{\ast,(p,q)}{\pi_1}(Y_{(p,q)}) , a\underset{\ast,(p,q)}{\pi_2}(X_{(p,q)}) + b\underset{\ast,(p,q)}{\pi_2}(Y_{(p,q)})\right) \\ = & (aX_p + bY_p , aX_q + bY_q) \\ = & a(X_p , X_q) + b(Y_p , Y_q) \\ = & a\left(\underset{\ast,(p,q)}{\pi_1}(X_{(p,q)}) , \underset{\ast,(p,q)}{\pi_2}(X_{(p,q)})\right) + b\left(\underset{\ast,(p,q)}{\pi_1}(Y_{(p,q)}) , \underset{\ast,(p,q)}{\pi_2}(Y_{(p,q)})\right) \\ = & a\left(\underset{\ast,(p,q)}{\pi_1} , \underset{\ast,(p,q)}{\pi_2}\right)(X_{(p,q)}) + b\left(\underset{\ast,(p,q)}{\pi_1} , \underset{\ast,(p,q)}{\pi_2}\right)(Y_{(p,q)}) \end{aligned} } $$

又

$$ \bbox[5pt]{ \begin{aligned} & \left(\underset{\ast,(p,q)}{\pi_1} , \underset{\ast,(p,q)}{\pi_2}\right)\left(\left.\frac{\partial}{\partial{u^k}}\right|_{(p,q)}\right) \\ = & \left(\underset{\ast,(p,q)}{\pi_1}\left(\left.\frac{\partial}{\partial{u^k}}\right|_{(p,q)}\right) , \underset{\ast,(p,q)}{\pi_2}\left(\left.\frac{\partial}{\partial{u^k}}\right|_{(p,q)}\right)\right) \\ = & \left(\left.\frac{\partial}{\partial{x^k}}\right|_{(p,q)} , 0\right) \\\\ & \left(\underset{\ast,(p,q)}{\pi_1} , \underset{\ast,(p,q)}{\pi_2}\right)\left(\left.\frac{\partial}{\partial{v^k}}\right|_{(p,q)}\right) \\ = & \left(\underset{\ast,(p,q)}{\pi_1}\left(\left.\frac{\partial}{\partial{v^k}}\right|_{(p,q)}\right) , \underset{\ast,(p,q)}{\pi_2}\left(\left.\frac{\partial}{\partial{v^k}}\right|_{(p,q)}\right)\right) \\ = & \left(0 , \left.\frac{\partial}{\partial{y^k}}\right|_{(p,q)}\right) \end{aligned} } $$

而

$$ \bbox[5pt]{ \left(\left.\frac{\partial}{\partial{x^1}}\right|_{(p,q)} , 0\right) , \cdots , \left(\left.\frac{\partial}{\partial{x^m}}\right|_{(p,q)} , 0\right) , \left(0 , \left.\frac{\partial}{\partial{y^1}}\right|_{(p,q)}\right) , \cdots , \left(0 , \left.\frac{\partial}{\partial{y^n}}\right|_{(p,q)}\right) } $$

为 $T_pM \times T_qN$ 的一组基
故映射

$$ \bbox[5pt]{ \left(\underset{\ast,(p,q)}{\pi_1},\underset{\ast,(p,q)}{\pi_2}\right) : \mathtip{T_{(p,q)}(M \times N)}{M \times N 在 (p,q) 处的全体切向量} \to \mathtip{T_pM}{M 在 p 处的全体切向量} \times \mathtip{T_qN}{N 在 q 处的全体切向量} } $$

为线性同态且将 $T_{(p,q)}{(M \times N)}$ 的一组基映作 $T_pM \times T_qN$ 的一组基
故其为线性同构