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Introduction to smooth manifolds by John M. Lee
定义
设有作用$\theta:G\times M\rightarrow M.$ 在$M$上定义等价关系$p\sim q$
$\Leftrightarrow$ $\,\exists\,g\in G,$ $p=g\cdot q.$
那么等价类恰为$G$在$M$中的轨道. 记$M/G$为商空间, 也称为轨道空间.
我们希望轨道空间也能是一个微分流形.
引理 1. 商映射$\pi:M\rightarrow G/M$为开映射.
称作用是逆紧作用, 若$G\times M\rightarrow M\times M,$
$(g,p)\mapsto (g\cdot p,p)$是逆紧的.
此条件比要求作用$\theta$本身是逆紧的要弱.
命题 2. 若李群连续逆紧地作用在流形上, 那么商空间是$T_2$的.
定义轨道关系$\mathcal{O}\subset M\times M,$
$$
\mathcal{O}=\{(g\cdot p,p)\in M\times M:p\in M,g\in G\}.
$$
那么$(p,q)\in \mathcal{O}$ $\Leftrightarrow$ $p,q$在同一轨道中.
由于映到局部紧$T_2$空间的连续逆紧映射总是闭映射,
$\mathcal{O}$为$M\times M$中的闭集.
容易证明这恰是商空间为$T_2$的充分条件.
然而直接判断逆紧作用并不容易. 不过我们有以下三个性质等价:
作用是逆紧的.
若$\{p_i\},$ $\{g_i\cdot p_i\}$收敛, 则$\{g_i\}$有子列收敛.
$\,\forall\,K\Subset M,$
$G_K=\{g\in G|( g\cdot K)\cap K\neq \varnothing\}$为紧集.
推论 3. 紧李群在流形上的连续作用是逆紧的.
命题 4. 设$\theta$是一个逆紧作用, 那么轨道映射$\theta^{(p)}:G\rightarrow M$是逆紧映射, 也因此轨道是闭的. 若它是单射, 那么轨道映射就是一个光滑嵌入, 轨道为逆紧嵌入子流形.
推论 5. 若李群逆紧作用在$M$上, 那么每个轨道是$M$中闭集, 每个稳定子都是紧的.
商空间
商流形
定理 6. 设$G$光滑, 自由, 逆紧地作用在$M$上, 那么轨道空间$M/G$是一个$\dim M-\dim G$维的拓扑流形, 有唯一一个光滑结构使得$\pi:M\rightarrow M/G$为光滑淹没.
覆盖映射
引理 7. 设离散李群$\Gamma$连续, 自由地作用在流形$E$上. 那么作用是逆紧的当且仅当如下条件成立: $\,\forall\,p\in E,$ $\,\exists\,U$为邻域, $\,\forall\,g\in \Gamma,$ $(g\cdot U)\cap U\neq \varnothing,$ 除非$g=e.$ 且若$p,p'$不在同一个轨道里, 那么分别存在邻域$V,V',$ $(g\cdot V)\cap V'=\varnothing,$ $\,\forall\,g\in \Gamma.$
命题 8. 设$\pi:E\rightarrow M$为覆盖映射, 那么配备离散拓扑的$\operatorname{Aut}_\pi(E)$光滑, 自由, 逆紧的作用在$E$上.
定理 9. 设$E$为连通光滑流形, $\Gamma$为离散李群, 光滑自由逆紧地作用在$E$上. 那么轨道空间$E/\Gamma$是拓扑流形, 具备唯一光滑结构使得$\pi:E\rightarrow E/\Gamma$为光滑正则覆盖.
齐性空间
若李群$G$在流形$M$上有一个光滑可迁作用, 那么称$M$为齐性空间.
定理 10. 若$H$为$G$的闭子群, 那么左陪集空间$G/H$是$\dim G-\dim H$维拓扑空间, 有唯一光滑结构使得$\pi:G\rightarrow G/H$为光滑淹没. $G$在$G/H$上的左作用为$g_1\cdot (g_2H)=(g_1g_2)H,$ 将$G/H$变为齐性空间.
定理 11. 设$G$为李群, $M$为齐性空间, $p\in M.$ 那么稳定子$G_p$为$G$的闭子群, $F:G/G_p\rightarrow M,$ $F(gG_p)=g\cdot p$给出一个等变微分同胚.
有一些典型的齐性空间:
$$
S^{n-1}\approx O(n)/O(n-1)\approx SO(n)/SO(n-1),
$$
$$
S^{2n-1}\approx U(n)/U(n-1)\approx SU(n)/SU(n-1).
$$
文章最后更新于 2022-10-18 22:50:20
Introduction to smooth manifolds by John M. Lee
definition
There is an effect $\theta:G\times M\rightarrow M.$ and an equivalence relation $p\sim q$ is defined on $M$.
$\Leftrightarrow$ $\,\exists\,g\in G,$ $p=g\cdot q.$
Then the equivalence class is exactly the orbit of $G$ in $M$. Let $M/G$ be the quotient space, also called the orbit space.
We hope that orbit space can also be a differential manifold.
Lemma 1. The quotient map $\pi:M\rightarrow G/M$ is an open map.
The function is backtightening, if$G\times M\rightarrow M\times M,$
$(g,p)\mapsto (g\cdot p,p)$ is inversely compact.
This condition is weaker than requiring that the action $\theta$ itself be inversely compact.
Proposition 2. If the Lie group acts continuously and inversely on the manifold, then the quotient space is $T_2$.
Define orbit relationship $\mathcal{O}\subset M\times M,$
$$
\mathcal{O}=\{(g\cdot p,p)\in M\times M:p\in M,g\in G\}.
$$
Then $(p,q)\in \mathcal{O}$ $\Leftrightarrow$ $p,q$ are in the same orbit.
Since a continuous proper map onto a locally compact $T_2$ space is always a closed map,
$\mathcal{O}$ is the closed set in $M\times M$.
It is easy to prove that this is the sufficient condition for the quotient space to be $T_2$.
However, it is not easy to directly judge the inverse compaction. However, we have the following three equivalent properties:
The effect is inversely tight.
If $\{p_i\},$ $\{g_i\cdot p_i\}$ converges, then $\{g_i\}$ has a sub-column that converges.
$\,\forall\,K\Subset M,$
$G_K=\{g\in G|( g\cdot K)\cap K\neq \varnothing\}$ is a compact set.
Corollary 3. The continuous action of a compact Lie group on a manifold is inversely compact.
Proposition 4. Assume $\theta$ is a proper action, then the orbit map $\theta^{(p)}:G\rightarrow M$ is a proper map, and therefore the orbit is closed. If it is injective, then the orbit map is a smooth embedding, and the orbit is a proper embedding submanifold.
Corollary 5. If the Lie group acts inversely compactly on $M$, then every orbit is a closed set in $M$, and every stabilizer is compact.
business space
quotient manifold
Theorem 6. Assume $G$ is smooth, free, and acts inversely compactly on $M$, then the orbit space $M/G$ is a $\dim M-\dim G$-dimensional topological manifold, and there is only one smooth structure that makes $\pi:M\rightarrow M/G$ smoothly submerged.
overlay mapping
Lemma 7. Assume that the discrete Lie group $\Gamma$ is continuous and acts freely on the manifold $E$. Then the effect is inversely compact if and only if the following conditions are true: $\,\forall\,p\in E,$ $\,\exists\,U$ are neighbors, $\,\forall\,g\in \Gamma,$ $(g\cdot U)\cap U\neq \varnothing,$ unless $g=e.$ and if $p,p'$ are not in the same orbit, then there are neighbors $V,V',$ $(g\cdot V)\cap V'=\varnothing,$ $\,\forall\,g\in \Gamma.$ respectively.
Proposition 8. Assume $\pi:E\rightarrow M$ is a covering map, then $\operatorname{Aut}_\pi(E)$ equipped with discrete topology is smooth, free, and inversely compact, acting on $E$.
Theorem 9. Assume $E$ is a connected smooth manifold, $\Gamma$ is a discrete Lie group, and smooth free inverse compaction acts on $E$. Then the orbit space $E/\Gamma$ is a topological manifold with a unique smooth structure such that $\pi:E\rightarrow E/\Gamma$ is a smooth regular cover.
homogeneous space
If the Lie group $G$ has a smooth transitive effect on the manifold $M$, then $M$ is called a homogeneous space.
Theorem 10. If $H$ is a closed subgroup of $G$, then the left coset space $G/H$ is a $\dim G-\dim H$-dimensional topological space, and there is a unique smooth structure that makes $\pi:G\rightarrow G/H$ a smooth submergence. The left action of $G$ on $G/H$ is $g_1\cdot (g_2H)=(g_1g_2)H,$, turning $G/H$ into a homogeneous space.
Theorem 11. Assume $G$ is a Lie group, $M$ is a homogeneous space, $p\in M.$ then the stabilizer $G_p$ is the closed subgroup of $G$, $F:G/G_p\rightarrow M,$ $F(gG_p)=g\cdot p$ gives an equivariant diffeomorphism.
There are some typical homogeneous spaces:
$$
S^{n-1}\approx O(n)/O(n-1)\approx SO(n)/SO(n-1),
$$
$$
S^{2n-1}\approx U(n)/U(n-1)\approx SU(n)/SU(n-1).
$$
The article was last updated on 2022-10-18 22:50:20