实例 $S^2$ 球面

球面 $S^2 \subset \mathbb{R}^3$ 定义如下

$$ \bbox[5pt]{ S^2 \triangleq \{(x,y,z)\in\mathbb{R}^3 \big| x^2+y^2+z^2=1\} } $$

切映射与切空间

取局部参数化表示 $\varphi : D^2 \to S^2_+$ 满足 $\varphi(x,y) = (x,y,\sqrt{1-x^2-y^2})$
对 $p = (x,y) \in D^2\backslash\partial{D^2}$ 有切映射 $d\varphi_p : \mathbb{R}^2 \to \mathbb{R}^3$

$$ \bbox[5pt]{ \begin{aligned} & d\varphi_p(h_1,h_2) \\ = & \lim_{t\to0}{\frac{\varphi(x+th_1,y+th_2)-\varphi(x,y)}{t}} \\ = & \lim_{t\to0}{\frac{(x+th_1,y+th_2,\sqrt{1-(x+th_1)^2-(y+th_2)^2})-(x,y,\sqrt{1-x^2-y^2})}{t}} \\ = & \left(h_1,h_2,-\frac{xh_1+yh_2}{\sqrt{1-x^2-y^2}}\right) \end{aligned} } $$

注意到

$$ \bbox[5pt]{ \begin{aligned} & d\varphi_p(h_1,h_2) \cdot \varphi(p) \\ = & \left(h_1,h_2,-\frac{xh_1+yh_2}{\sqrt{1-x^2-y^2}}\right) \cdot (x,y,\sqrt{1-x^2-y^2}) \\ = & xh_1 + yh_2 - (xh_1+yh_2) \\ = & 0 \end{aligned} } $$

故有切空间 $T_{\varphi(p)}S^2$

$$ \bbox[5pt]{ T_{\varphi(p)}S^2 = \{(h_1,h_2,h_3) \in \mathbb{R}^3 \big| (h_1,h_2,h_3)\cdot\varphi(p)=0\} } $$

此时可视作

$$ \bbox[5pt]{ d\varphi_p : T_pD^2 = \mathbb{R}^2 \to T_{\varphi(p)}S^2 \subsetneq \mathbb{R}^3 } $$

光滑映射

取光滑映射 $f : S^2 \to S^2$ 满足 $f(x,y,z) = (-y,x,z)$ 和光滑映射 $g : S^2 \to S^2$ 满足 $g(x,y,z) = (-x,-y,z)$
存在光滑映射 $F : S^2\times[0,1] \to S^2$ 满足

$$ \bbox[5pt]{ F(x,y,z,t) = \left(x\cos{\left(\frac{\pi}{2}+\frac{\pi}{2}t\right)}-y\sin{\left(\frac{\pi}{2}+\frac{\pi}{2}t\right)},x\sin{\left(\frac{\pi}{2}+\frac{\pi}{2}t\right)}+y\cos{\left(\frac{\pi}{2}+\frac{\pi}{2}t\right)},z\right) } $$

使得

$$ \bbox[5pt]{ \begin{aligned} F(x,y,z,0) & = (-y,x,z) = f(x,y,z) \\ F(x,y,z,1) & = (-x,-y,z) = g(x,y,z) \end{aligned} } $$

故有同伦 $f \sim g$

又对 $p = (x,y,z) \in S^2$ 有切映射

$$ \bbox[5pt]{ \begin{aligned} & df_p(h_1,h_2,h_3) \\ = & \lim_{t\to0}{\frac{f(x+th_1,y+th_2,z+th_3)-f(x,y,z)}{t}} \\ = & \lim_{t\to0}{\frac{(-y-th_2,x+th_1,z+th_3)-(-y,x,z)}{t}} \\ = & (-h_2,h_1,h_3) \end{aligned} } $$ $$ \bbox[5pt]{ \begin{aligned} & dg_p(h_1,h_2,h_3) \\ = & \lim_{t\to0}{\frac{g(x+th_1,y+th_2,z+th_3)-g(x,y,z)}{t}} \\ = & \lim_{t\to0}{\frac{(-x-th_1,-y-th_2,z+th_3)-(-x,-y,z)}{t}} \\ = & (-h_1,-h_2,h_3) \end{aligned} } $$