Topological Transformation Groups

这篇文章的背景是我 ODE 课的一道习题:

Problem

设 $\phi_t: \RR^n \to \RR^n$ 为一连续流. 设 $p \in \RR^n$ 不是 $\phi_t$ 的不动点. 证明: 如果 $\operatorname{Orb}(p)$ 紧致, 那么这条轨道一定是周期轨道.

所谓的连续流, 就是指加法群 $\RR$ 作用在一个流形 $M$ 上. 这道题并不简单, 我首先找到的是 MSE 3433255, 之后我顺藤摸瓜去看了看 [1] . 将 $\RR$ 推广为更一般的拓扑群, 引入 topological transformation group 的概念, 这道习题便成为了 Theorem (704b110) 的特例.

$$ \DeclareMathOperator{\id}{id} \newcommand{\ol}{\overline} \DeclareMathOperator{\Hom}{Hom} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} $$

Topological Transformation Groups

Definition

一个 topological transformation group is a pair $(G, M)$, 这里 $M$ 是一个 Hausdorff 空间, $G$ 是一个拓扑群, together with a continuous map

$$ G \times M \to M, \quad (g, m) \mapsto g\cdot m $$

satisfying the following property:

$(g_1g_2)\cdot m = g_1\cdot (g_2\cdot m)$ for all $g_1, g_2 \in G, m \in M$.
$1_G \cdot m = m$ for all $m \in M$.

每一个 $g \in G$ 对应 $M$ 上的一个自同胚, 所以我们可以将 $g\cdot x$ 写成 $g(x)$, 将 $x$ 的轨道写成 $G(x)$. The transformation group $G$ is called transitive on $M$ if for every $x, y \in M$ there is at least one $g \in G$, such that $g(x) = y$. In other words, $M$ 有且只有一条轨道. We say $M$ is a homogeneous space in this case. 对 $x \in M$, 我们用 $G_x = \{g \in G: g(x) = x\}$ 表示点 $x$ 的稳定子群.

Lemma

$G_x$ is closed.

Proof. 设 $g \in G$ 使得 $g(x) =: y \ne x$. 因为 $M$ is Hausdorff, 存在 $y$ 的邻域 $Y$ not containing $x$. 因为群作用连续, 存在 $g$ 的邻域 $U$ 使得 $g'(x) \in Y$ for all $g' \in U$. 特别的 $g'(x) \ne x$, 这就说明了 $G_x$ 的补集是开集. $\square$

Theorem

Let $G$ be a topological transformation group which is transitive on $M$.

对任意两个点 $x, y \in M$, 他们对应的稳定子群 $G_x, G_y$ 互相共轭.
对任意 $x \in M$, the map defined by

$$ T: G/G_x \to M, \quad gG_x \mapsto g(x). $$

is continuous and bijective.

Proof. 设 $g \in G$ 满足 $g(x) = y$. 对任意 $h \in G_x$, 我们有 $h(x) = x$, $ghg^{-1}(y) = y$, 所以 $ghg^{-1} \in G_y$.

$T$ 显然是双射. $T$ 连续是拓扑范畴里商空间的 universal property. $\square$

注意这里我们只能保证 $T$ 是连续的双射, 不一定是同胚.

Baire Category Theorem

Recall that a compact subset of a Hausdorff space is closed. Recall that a compact Hausdorff space is regular.

Lemma

设 $X$ 是一个紧致的 Hausdorff 空间. 设 $F_n$ 是一列闭集. 如果每一个 $F_n$ 的内部都是空集, the same is true of their union.

Proof. 设 $F = \bigcup_n F_n$, 设 $O_0 \subseteq F$ is an non-empty open. 因为 $F_1$ 没有内点, $O_0$ 不可能是 $F_1$ 的子集, 所以存在 $x_1 \in O_0 \cap F_1^c$, 再利用 regularity, 存在 $x_1$ 的开邻域 $O_1$ 使得 $\ol{O_1} \subseteq O_0 \cap F_1^c$.

以此类推, 对任意正整数 $n$ 我们可以找到非空开集 $O_n \subseteq O_{n-1}$ 满足 $\ol{O_n} \cap (F_1 \cup \cdots \cup F_n) = \varnothing$. Let $O = \bigcap_n \ol{O_n}$, then $O \cap F = \varnothing$. 另一方面我们有 $O \subseteq O_0 \subseteq F$, 所以只能是 $O = \varnothing$.

Lemma

Let $X$ be a compact space. If $\{F_\alpha: \alpha \in I\}$ is a collection of closed subsets with finite intersection property, then all these subsets have a common point¹.

Proof. Assume $\bigcap_\alpha F_\alpha = \varnothing$. Define $U_\alpha = F_\alpha^c$, then $\bigcup_\alpha U_\alpha = X$. Since $X$ is compact, we have finite collection $\{\alpha_1, \cdots, \alpha_n\} \subseteq I$ such that $X = U_{\alpha_1} \cup \cdots \cup U_{\alpha_n}$. Taking complements we get $F_{\alpha_1} \cap \cdots \cap F_{\alpha_n} = \varnothing$. $\blacksquare$

我们定义的 $\ol{O_n}$ 就满足 finite intersection property, 从而 $O = \bigcap_n \ol{O_n} \ne \varnothing$, 这与我们前面说明的 $O = \varnothing$ 矛盾. $\square$

A set is dense if its closure is the entire space. A set is nowhere dense if its closure has empty interior. Recall that the interior operator is dual to the closure operator

$$ S^\circ = X - \ol{S^c} $$

It follows that $U$ is dense $\iff U^c$ has empty interior and that $U$ is dense open $\iff U^c$ is nowhere dense closed.

Definition

A topological space $X$ is called a Baire space if it satisfies any of the following equivalent conditions:

Every countable intersection of dense open sets is dense.
Every countable union of nowhere dense closed sets has empty interior.

The theorem below gives sufficient conditions for a topological space to be a Baire space.

Theorem (Baire Category Theorem)

Every complete pseudometric space is a Baire space.
Every locally compact Hausdorff space is a Baire space.

Proof. (1) 的证明见 Wiki 条目, 下面我们证明 (2). 设 $F_n$ 是一列 nowhere dense 的闭集, 设 $F = \bigcup_n F_n$. 假如存在非空开集 $O \subseteq F$, choose non-empty open $K \subseteq O$ with compact closure, restrict everything on $\ol{K}$ and use Lemma (4f5a1bf). $\square$

A Survey on Locally Compact Group

A topological space $X$ is totally disconnected if the connected components in $X$ are the one-point sets. In other words, 除了单点集以外 $X$ 没有其他的连通子集. 一个基本的性质是: 对于任意不同的两点 $x, y$, 因为 $\{x, y\}$ is disconnected, 所以存在开集 $U$ 包含 $x$ 但不包含 $y$. 也就是说, totally disconnected space is necessarily a $T_1$ space.

设 $G$ 是一个拓扑群. Recall that every open subgroup of $G$ is also closed (hence $G/H$ is discrete). Recall that $G$ is Hausdorff if and only if $G$ is $T_1$ if and only if $\{e\}$ is closed, so a totally disconnected topological group is necessarily Hausdorff.

Lemma

Let $G$ be a topological group. Let $G_0$ be the connected-component of the identity element $e \in G$. Let $H$ be an open subgroup of $G$. Then

$G_0 \subseteq H$. 也就是说任意一个开子群一定包含单位元所在的连通分支.
$G_0$ is a closed normal subgroup of $G$.
The quotient group $G/G_0$ is totally disconnected.

Proof. [1, Page 39] (1) 因为 $H$ is clopen, 所以 $\varnothing \ne H \cap G_0$ is clopen in $G_0$. 因为 $G_0$ is connected, 所以只能是 $H \cap G_0 = G_0$.

(2) If $M$ is a connected subset of $G$, then $M^{-1}, gM, Mg$ is connected for all $g \in G$. Since $e \in G_0 \cap G_0^{-1}$ (两个连通分支有交集), 所以 $G_0 = G_0^{-1}$, 也就是说 $G_0$ 关于取逆运算封闭. 对任意 $g, h \in G_0$, 注意到 $G_0$ 是 $h$ 的连通分支, 从而 $gG_0$ 是 $gh$ 的连通分支, 而 $g \in gG_0 \cap G_0$, we therefore conclude that $gh \in gG_0 = G_0$. 这就说明了 $G_0$ is a group. 因为连通分支一定是闭集, 所以 $G_0$ is closed. Since $h^{-1}G_0h$ is connected and contains $e$ for all $h \in G$, we see that $G_0$ is normal.

(3) 设 $T: G \to G_0$ 是典范映射, 设 $M$ 是 $G/G_0$ 的一个连通子集. It suffices to show that $T^{-1}(M)$ is connected. Suppose that $$ T^{-1}(M) = A \cup B $$ where $A$ and $B$ are relatively open in $T^{-1}(M)$ and $A \cap B = \varnothing$. Then $M = T(T^{-1}(M)) = T(A) \cup T(B)$. Let $U$ be open such that $A = U \cap T^{-1}(M)$, we claim that $T(A) = T(U) \cap M$. 注意 $\text{LHS} \subseteq \text{RHS}$ 是显然的, 下面我们证明另一个包含方向, 设 $y \in \text{RHS}$. 因为 $y \in T(U)$, 所以存在 $x \in U$ 使得 $y = T(x)$. 因为 $y \in M$, 所以 $x \in T^{-1}(M)$. Thus $x \in A$ and $y \in \text{LHS}$.

因为 $T$ 是开映射, we see that $T(A)$ is relatively open in $M$, the same is true of $T(B)$. Since $M$ is connected, it follows that $T(A) \cap T(B) \ne \varnothing$. 设陪集 $xG_0 \in T(A) \cap T(B)$, 注意到我们有分解式 $$ xG_0 = (xG_0 \cap A) \cup (xG_0 \cap B), $$ 这与 $xG_0$ 是连通集矛盾. $\square$

设 $U$ 是 $e \in G$ 的一个开邻域, 我们总是能 shrink $U$ 使得 $U$ is symmetric (a subset $S \subseteq G$ is called symmetric if $S = S^{-1}$): 取 $U’ = U \cap U^{-1}$, then $U’$ is a symmetric neighborhood of $e$. 我们还有另外一种取法是 $U’ = U \cup U^{-1}$, 只不过这种情况就没有 shrink 了, 但好处是 if $U$ is compact then $U’$ is also compact.

Theorem (van Dantzig)

If $G$ is a topological group which is locally compact and totally disconnected, then every neighborhood of the identity element $e$ contains a compact open subgroup.

Proof. To prove this theorem, we first need a lemma from point set topology, which shows that totally disconnected spaces contain enough clopen sets to separate points:

Lemma

Let $X$ be a totally disconnected compact Hausdorff space. Let $x, y$ be distinct points in $X$. Then there exists a clopen set that contains $x$ but not $y$.

Proof. [2, Lemma 5] Let $K$ be the intersection of all the clopen sets that contain $x$. Clearly $K$ is closed and contains $x$. Our objective is to show that $K$ consists solely of $\{x\}$. As $X$ is totally disconnected, it will suffice to show that $K$ is connected.

Suppose this is not the case, then we can split $K = K_1 \cup K_2$ where $K_1, K_2$ are disjoint non-empty closed sets. As $K$ itself is closed, we can assume $K_1, K_2$ are closed in $X$. We can wlog assume $x \in K_1$, as all compact Hausdorff spaces are normal, we can thus enclose $K_1, K_2$ in disjoint open subsets $U_1, U_2$ of $X$. The choice of $U$ is such that $$ K_2 \subseteq U_2 \subseteq \ol{U_2} \subseteq K_1^c, $$ so the topological boundary $\partial U_2$ is closed (hence compact) and lies outside of $K_1 \cup K_2 = K$. By definition of $K$, we see that for every $y \in \partial U_2$, we can find a clopen neighborhood of $x$ that avoids $y$; by compactness of $\partial U_2$ and the fact that finite intersections of clopen sets are clopen, we can thus find a clopen neighborhood $L$ of $x$ that is disjoint from $\partial U_2$. The choice of $L$ is such that $V := L - U = L - \ol{U}$, so $V$ is a clopen neighborhood of $x$ that is disjoint from $K_2$, contradicting the definition of $K$. $\blacksquare$

We first claim that there exists a compact clopen neighborhood $K$ of $e$.

Lemma

If $G$ is locally compact and totally disconnected, then there exists a compact clopen neighborhood $K$ of the identity element $e$.

Proof. 设 $U$ 是 $e$ 的一个开邻域使得 $\ol{U}$ 紧致. 对任意 $y \in \partial U$, by Lemma (0c4f879) we can find a clopen neighborhood of $e$ that avoids $y$; by compactness of $\partial U$, we may thus find a clopen neighborhood $V$ of $e$ that avoids $\partial U$. Let $K := V \cap \ol{U} = V \cap U$, then $K$ is a compact clopen neighborhood of $e$. $\blacksquare$

For each $x \in K$, by continutity we can find an open $U_x \ni e$ such that $x U_x \subseteq K$ and an open $V_x \ni e$ such that $V_xV_x \subseteq U_x$. We can also wlog assume $V_x$ is symmetric. Since $K$ is compact, it suffices a finite number of elements of $K$, say $x_1, \cdots, x_n$, to have $$ K \subseteq x_1V_{x_1} \cup \cdots \cup x_nV_{x_n}. $$ Set $V = V_{x_1} \cap \cdots \cap V_{x_n}$, and we abbreviate $V_{x_i}$ to $V_i$, then we have $$ KV \subseteq \left(\bigcup_{i=1}^n x_iV_i\right)V \subseteq \bigcup_{i=1}^n x_iV_iV_i \subseteq \bigcup_{i=1}^n x_iU_{x_i} \subseteq K. $$ The inclusion $V = eV \subseteq KV \subseteq K$ implies that $V^n \subseteq K$ for all $n \in \NN$. 设 $H$ 是由 $V$ 生成的子群, 因为 $V$ is symmetric, 所以 $H$ is of the form $$ V \subseteq H = \bigcup_{n = 1}^\infty V^n \subseteq K. $$ Now $H$ is a subgroup of $G$ containing an open set $V$, we claim that such subgroup is necessarily open.

Lemma

Let $G$ be a topological group and $H$ be a subgroup of $G$. If $H$ contains an open set $U$, then $H$ is open.

Proof. 取 $g \in U \subseteq H$, then $V := g^{-1}U \subseteq H$ is an open neighborhood of $e$. 对任意 $h \in H$, 我们有 $hV \subseteq hH = H$, this is an open neighborhood of $h$. $\blacksquare$

An open subgroup $H$ is necessarily closed (in $K$), so $H$ is compact. 现在我们证明了: 对任意 $e$ 的紧致邻域 $K$, 总能找到一个 compact open subgroup $H \subseteq K$. 因为 $G$ is Hausdorff, 这样的 $K$ 构成了 $e$ 的邻域基 c.f. Proposition (9c11705). $\square$

Lemma

Let $G$ be a locally compact topological group and $H$ be a subgroup.

If $H$ is closed, then $H$ is locally compact.
Coset-space $G/H$ is locally compact.

Proof. (1) 因为紧集的闭子集紧致. (2) 设 $T: G \to G/H$ 是典范映射. 对任意 $y \in G/H$, 取 $x \in G$ 使得 $T(x) = y$, 取 $x$ 的紧致邻域 $K$. 因为 $T$ 是连续的开映射, 所以 $T(K)$ 就是 $y$ 的紧致邻域. $\square$

Lemma

Let $G$ be a locally compact group. There exists an open subgroup $G’$ of $G$ such that $G’/G_0$ is compact. For every choice of such a $G’$, there exists a compact neighborhood $W$ of $e$ such that $G’ = W\cdot G_0$ and that $G’$ is generated by $W$.

Proof. [1, Page 54] 设 $T: G \to G/G_0$ 是典范映射. By Lemma (133b5c0) and Lemma (d87f10e) we see that $G/G_0$ is locally compact and totally disconnected. By Theorem (1cad229) we can choose a compact open subgroup $D$ of $G/G_0$. Now $T^{-1}(D)$ is the required $G'$.

设 $\varphi: G' \to G'/G_0$ 是典范映射. 任取 $e$ 的一个紧致邻域 $W' \subseteq G'$, then $\varphi(W')$ is a neighborhood of $\varphi(e)$. 因为 $G'/G_0$ is compact, 所以存在一个有限集 $F \subseteq G', e \in F$ 使得 $$ G'/G_0 \subseteq \varphi\left(\bigcup_{g \in F} W'g\right). $$ Finally, put $W = \bigcup_{g \in F} W'g$, it's easy to see that $W \subseteq G'$ and $W$ is compact. Since $\varphi(W) = G'/G_0$, so $G' = W \cdot G_0$.

设 $H$ 是由 $W$ 生成的子群. 因为 $W \subseteq H$ 并且 $W$ 是一个邻域, 由 Lemma (6b10fe4) 知 $H$ is clopen and hence contains $G_0$. 由此便知 $H \supseteq W\cdot G_0 = G'$, 另一方面 $H \subseteq G'$ 是显然的. $\square$

The Main Theorem

Proposition

Let $G$ be a topological group which is locally compact Hausdorff. Let $G’$ be an open subgroup such that $G’/G_0$ is compact. If the cardinality of $G/G’$ is countable, then given any neighborhood $V$ of $e$, there is a countable set of elements $g_n \in G$ such that $G = \bigcup_n g_n V$.

Proof. [1, Page 59] By Lemma (1d7cb41) we can choose a compact neighborhood $W$ such that $G' = W\cdot G_0$. 用 $W \cup W^{-1}$ 替换我们可以不妨设 $W$ is symmetric, 于是我们有 $G' = \bigcup_n W^n$. For each $n, W^n$ is compact, 于是我们可以用有限多个 $gV$ 来覆盖 $W^n$, 从而我们可以用可数多个 $gV$ 来覆盖 $G'$. 又因为 $G/G'$ 的 cardinality 可数, we finally conclude that we can use countably many $gV$ to cover $G$. $\square$

Theorem

Let $(G, M)$ be a topological transformation group, 其中 $G, M$ 都是 locally compact Hausdorff, 并且 $G$ is transitive on $M$. 如果存在开子群 $G’$ 使得 $G/G’$ 的 cardinality 可数并且 $G’/G_0$ compact, then $G/G_x$ is homeomorphic to $M$ for all $x \in M$.

Proof. [1, Page 65] 我们要说明 Theorem (a2a9518) 中的 $T$ 是开映射. 根据商映射的 universal property, it suffices to show that the map $T_1: G \to M, g \mapsto g(x)$ is open.

Claim

对任意 $e$ 的紧致邻域 $W, x$ 都是 $W^2(x)$ 的内点.

Proof. 因为 $G$ is transitive on $M$, 所以我们有 $M = G(x) = \bigcup_n g_nW(x)$ c.f. Proposition (0085f51). 每个 $g_nW$ 都是紧致的, 所以 $g_nW(x)$ is compact and hence closed in $M$. 由 Baire Category Theorem 知至少其中一个 $g_nW(x)$ 有内点. 因为 $g$ 诱导了 $M$ 上的一个自同胚, 并且 $(gW)(x) = g(W(x))$, 所以 $g_nW(x)$ 有内点可以推出 $W(x)$ 有内点.

设 $h(x), h \in W$ 是 $W(x)$ 的一个内点, then $x$ is an inner point of $h^{-1}W(x) \subseteq W^2(x)$. $\blacksquare$

Let $V$ be an arbitrary open in $G$ and $g$ be an element in $V$. By Proposition (9c11705) we can choose a compact neighborhood $W$ of $e$ such that $gW^2 \subseteq V$. Now $x$ is an inner point of $W^2(x)$ and $V(x) \supseteq gW^2(x)$, so $V(x)$ has $g(x)$ as an inner point. $\square$

Final Words

对于连续流, $G$ 就是加法群 $\RR$, 他在 $0$ 处的连通分支就是全空间 $\RR$, 自然满足 Theorem (704b110) 所说的条件. 于是我们有同胚 $G/G_x \cong \operatorname{Orb}(p)$, 因为 $G = \RR$ 不是紧致的, 所以 $G_x$ 不可能是平凡群, 这就说明了存在非零的 $T$ 使得 $\phi_T(x) = x$, 即 $\operatorname{Orb}(p)$ 一定是周期轨道.

MSE 3433255 的回答还给出了一个只针对连续流的证明, 这个证明也使用了 Baire Category Theorem, 由此可见 Baire Category Theorem 是本质的.

维基百科对 locally compact group 的定义还要求这个群 Hausdorff, 但是 [1, Page 48] 的定义似乎没有要求 Hausdorff. 我认为 Hausdorff 这个性质还是需要的, 否则 Theorem (704b110) 的证明中, 如何保证能取出紧致邻域 $W$ 使得 $gW^2 \subseteq V$?

References

1: Deane Montgomery, Leo Zippin. Topological Transformation Groups. Courier Dover Publications, 2018.
2: Terry Tao’s Blog. https://terrytao.wordpress.com/2011/05/30/van-dantzigs-theorem/ .