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基础拓扑学讲义 1.10 道路提升引理

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记号来自《基础拓扑学》《基础拓扑学讲义》

道路提升引理

定义

\[\begin{aligned} \text{指数映射 }\pi:~ \R &\to S^1 \\ x &\mapsto e^{2\pi ix} \end{aligned} \]

\[\begin{aligned} \R \text{上道路 }\gamma_n:~&I \to nI\\ &s\mapsto ns\\ \gamma_n(s) &= ns,0\le s\le 1 \end{aligned} \]

\[\begin{aligned} \text{整数加群到基本群 }\phi:\Z &\to \pi_1(S^1, 1)\\ n&\mapsto\langle \pi \circ \gamma_n\rangle\\ \phi(n) &=\langle\pi \circ \gamma_n\rangle, n\in \Z \end{aligned} \]

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首先是

要证明 \(\phi\) 是同构,需要

  • \(\phi\) 是同态
  • \(\phi\) 是满同态
  • \(\phi\) 是单同态

尤承业的结论是 \(\pi_1(S^1, z_0)\) 是自由循环群,也即 \(\pi_1(S^1, z_0) \cong \Z\)

同态

\[\begin{aligned} \phi(n+m) &= \langle \pi \circ \gamma_{n+m}\rangle\\ &= \langle \pi \circ (\gamma_n\gamma_m)\rangle\\ &= \langle (\pi \circ \gamma_n)(\pi \circ \gamma_m)\rangle\\ &= \langle\pi \circ \gamma_n\rangle\langle\pi \circ \gamma_m\rangle\\ &= \phi(n)\phi(m) \end{aligned} \]

只需 \(\forall \alpha \in \pi_1(S^1, 1), \exist x\in \Z, s.t. \phi(x)=\langle \alpha \rangle\)

因此需要做的就是根据 \(\alpha\) 构造出 \(\phi^{-1}(\alpha)\),事实上 \(\phi\) 的定义在上面已经给出

道路提升引理

若 \(\sigma\) 为 \(S^1\) 内以 \(1\) 为起点的一条道路,则存在 \(\R\) 内唯一道路 \(\widetilde{\sigma}\) 以 \(0\) 为起点,并满足 \(\pi \circ \widetilde{\sigma}=\sigma\)

\[\begin{aligned} \sigma: I &\to S^1, \sigma(0)=1\\ &\Updownarrow\\ \widetilde{\sigma} :I &\to \R, ~\widetilde{\sigma}(0)=0 \end{aligned} \]

    1. 记 \(U=S^1-\{1\}, V=S^1-\{-1\}\)
      则 \(\sigma^{-1}(S^1) = \sigma^{-1}(U\cup V) = \sigma^{-1}(U)\cup \sigma^{-1}(V)\) 构成 \(I\) 开覆盖
    2. 将 \(I=\sigma^{-1}(S^1)\) 分割为小段闭区间 \(\{I_a\}\) (Lebesgue引理),分别落在 \(\sigma^{-1}(U)\cup \sigma^{-1}(V)\) 当中
    3. 有: 限制在每一小段 \(I_a\) 上的覆叠映射 \(\pi|_{I_a}\) 同胚。将 \(\widetilde{\sigma}|_{I_a} = (\pi|_{I_a})^{-1}\circ (\sigma|_{I_a})\) 拼接起来得到完整的 \(\widetilde{\sigma}\)
  • 由复叠空间提升唯一性定理可知每一小段唯一性,因而 \(\widetilde{\sigma}\) 唯一

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\(\alpha\) 的提升

利用上面引理,得到 \(\alpha\) 在覆叠空间 \((\R, \pi)\) 上有且有唯一以 \(0\) 为起点的提升 \(\widetilde{\alpha}: I\to \R\),且由于 \(\alpha(1)=1\),所以 \(\widetilde{\alpha}(1)\in \Z\)

\[\alpha(1)=\pi \circ \widetilde{\alpha}(1) = 1\\ \Downarrow\\ \widetilde{\alpha}(1) \in \pi^{-1}(1)\in \Z \]

取 \(n'=\widetilde{\alpha}(1)\),则有 \(\widetilde{\alpha}: I \to \R\) 是 \(\R\) 上以 \(0\) 为起点,以 \(n'\) 为终点的道路

那么 \(\pi \circ \gamma_{n'}\in \langle \pi \circ \widetilde{\alpha} \rangle = \langle \alpha \rangle\)

\(\phi(n')=\langle \pi \circ \gamma_{n'} \rangle = \langle \alpha \rangle\),这就证明了 \(\phi\) 是满同态

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设 \(n\ne m\in \Z\),那么要证 \(\phi(n) \ne \phi(m)\),只需 \(\phi^{-1}(m)\phi(n) \ne \langle e \rangle\)

记 $$\begin{aligned}
P_{(a, b)}:I&\to (b-a)I\
t &\mapsto a+(b-a)t\
a, b\in \Z
\end{aligned}$$
由 \(\gamma_n\) 定义显然有 \(\gamma_n^{-1}\gamma_m = P_{(n, m)}\)

\[\begin{aligned} \phi^{-1}(m)\phi(n) &= \langle (\pi \circ \gamma_n)^{-1}\rangle \langle \pi \circ \gamma_m\rangle\\ &= \langle \pi \circ \gamma_n^{-1}\rangle \langle \pi \circ \gamma_m\rangle\\ &= \langle (\pi \circ \gamma_n^{-1}) (\pi \circ \gamma_m)\rangle\\ &= \langle \pi \circ (\gamma_n^{-1}\gamma_m)\rangle\\ &= \langle \pi \circ P_{(n, m)} \rangle\\ &= \langle \pi \circ \gamma_{n-m}\rangle \end{aligned} \]

即证 \(\pi \circ \gamma_{n-m} \not\simeq e\)

圈数(Armstrong度数)

定义 \(q(\alpha) = \widetilde{\alpha}(1)-\widetilde{\alpha}(0)\)

圈数似乎特指覆叠空间 \((\R, \pi)\) ,底空间为 \(S^1\),这样闭路圈数必为整数

引理3(尤承业p118)

设 \(a, b\) 是 \(S^1\) 上基点为 \(1\) 的闭路,使得 \(\forall t \in I, a(t) \ne -b(t)\),则 \(q(a)=q(b)\)

我确实没资格学数学。

这个引理我举例子问了两个同学,他们都略一思索就给出了和书中解答相同的思路

a, b 在圆形跑道上,从同一点同时开始同向出发,最终 a,b 同时停在同一点,但 a 跑了 2 圈,b 只跑了 1 圈,怎么证他们必有一时刻处于跑道的相反位置?
“a 最终比 b 多跑了 1 圈,所以中间必有一时刻比 b 多跑 1/2 圈”

这就是跑步的连续性?从这里可以看出来时间的连续性似乎就是 \(I\) 的连通性

取 \(a, b\) 提升 \(\widetilde{a}, \widetilde{b}\)
作差 \(f = \widetilde{a} - \widetilde{b}\),则 \(f\) 连续,且 \(f(0)=0, f(1) = \widetilde{a}(1) - \widetilde{b}(1)\)

\[\begin{aligned} f(1) &= \widetilde{a}(1) - \widetilde{b}(1)\\ &= \widetilde{a}(1) - \widetilde{b}(1) - f(0)\\ &= \widetilde{a}(1) - \widetilde{b}(1) - \widetilde{a}(0) + \widetilde{b}(0)\\ &= (\widetilde{a}(1) - \widetilde{a}(0)) - (\widetilde{b}(1) - \widetilde{b}(0))\\ &= q(a) - q(b) \end{aligned} \]

于是可以反证:

若 \(q(a) \ne q(b)\) 则有 \(f(1)\ne 0\),由 \(f\) 连续性,存在 \(f(t_0) = 1/2\)

\[\begin{aligned} \widetilde{a}(t_0) - \widetilde{b}(t_0) &= 1/2\\ &\Downarrow\\ \pi(\widetilde{a}(t_0) - \widetilde{b}(t_0)) &= \pi(1/2)\\ &\Downarrow\\ \frac{a(t_0)}{b(t_0)}&=-1 \end{aligned} \]

矛盾

引理4(尤承业p118)

设 \(a, b\) 是 \(S^1\) 上基点为 \(1\) 的闭路,则 \(q(a) = q(b)\Longleftrightarrow a\underset{\dot{}}{\simeq}b\)

  • \(\Longrightarrow\)
    取 \(a, b\) 提升 \(\widetilde{a}, \widetilde{b}\),使得 \(\widetilde{a}(0) = \widetilde{b}(0)\)
    则 \(\widetilde{a}(1) - \widetilde{b}(1) = (q(a) + \widetilde{a}(0)) - (q(b) + \widetilde{b}(0)) = 0\),因而 \(\widetilde{a} \underset{\dot{}}{\simeq}\widetilde{b}\)
    复合回去 \(\pi \circ \widetilde{a}\) 和 \(\pi \circ \widetilde{b}\) 仍同伦,同基点闭路故定端同伦

  • \(\Longleftarrow\)

    (这里书上说同伦映射 \(H:a\underset{\dot{}}{\simeq}b\) 一致连续,我没太搞懂)

综上

于是可知,\(\pi \circ \gamma_{n-m}\) 圈数为 \(n-m\ne q(e) = 0\),因而 \(\pi \circ \gamma_{n-m} \not\simeq e\)

得证单同态

下次看看同伦提升定理

标签:引理,1.10,circ,langle,widetilde,rangle,pi,拓扑学,gamma
来源: https://www.cnblogs.com/human-in-human/p/15681403.html