---- Let $X$ be a [[topological space]]. Suppose $G$ is a [[topological group|topological]] [[group]] [[group action|acting]] on $X$ via $\begin{align} G &\xrightarrow{\sigma} \text{Homeo}(X) \\ g & \mapsto g \cdot \_ \end{align}.$ We call the action $\sigma$ of $G$ on $X$ **properly discontinuous** if for all $x \in X$ there exists a [[neighborhood]] $U \ni x$ such that $g \cdot U \cap U=\emptyset$ for any nontrivial $g \in G$. (This is the 'covering map version') ^ this is not always the convention, need to write a bit concerning how to disambiguate https://math.stackexchange.com/questions/1082834/properly-discontinuous-action-equivalent-definitions > [!definition] Definition. ([[properly discontinuous subgroup of the homeomorphism group]]) > Let $X$ be a [[topological space]] and $\text{homeo}(X)$ the [[homeomorphism]] of $X$. A [[subgroup]] $G \subset \text{homeo}(X)$ is called **properly discontinuous** if for all $x \in X$ there exists a [[neighborhood]] $U \ni x$ s.t. $g(U) \cap U = \emptyset$ > for any nontrivial $g \in G$. Let $p: E \to B$ be a [[covering space|covering map]] with $E$ [[connected]] and $G \subset \text{homeo}(E)$ be the set of [[homeomorphism|homeomorphisms]] $g$ s.t. $p \circ g=p$. **(a) Convince yourself that $G$ is a [[subgroup]].** Clearly $\id_{X} \in G$. And given $g_{1},g_{2} \in G$ we have $p \circ (g_{1} \circ g_{2})=(p \circ g_{1}) \circ g_{2}=p \circ g_{2}=p$. Finally, given $g \in G$ we have $p \circ g = p$, for which rearrangement yields $p=p \circ g ^{-1}$. **(b) Prove that $G$ is properly discontinuous.** I can prove this given the additional condition that $E$ is [[locally connected, locally path-connected|locally connected]]. I do not know how tricky it is to remove this assumption. If $E$ is locally connected: Let $e_{0} \in E$ and consider $b_{0}=p(e_{0})$. Since $p$ a [[covering space|covering map]], $b_{0}$ admits a [[neighborhood]] $U \ni b_{0}$ that is [[evenly covered]] by $p$: $p ^{-1}(U)=\bigsqcup_{\alpha}V_{\alpha}, \text{ where } p |_{_{V_{\alpha}}} \text{ is a homeomorphism onto }U \ (*).$ Obtain an open [[connected]] [[neighborhood]] $U_{LC}$ of $b_{0}$ satisfying $b_{0} \in U_{LC} \subset U$. Since $U$ is [[homeomorphism]] to each $V_{\alpha}$ via $p |_{V_{\alpha}}$, $U_{LC}$ lifts to an open, disjoint union $p ^{-1}(U_{LC}) = \bigsqcup_{\alpha}V_{\alpha_{LC}} \ (* *)$ where $p |_{V_{\alpha_{LC}}}$ is a [[homeomorphism]] onto $U_{LC}$ and each $V_{\alpha_{LC}}$ is [[connected]]. Obtain the $\alpha$ for which $V_{\alpha_{LC}} \ni e_{0}$. Let $g \in G$ be nontrivial. We know $p \circ g(V_{\alpha_{LC}})=p(V_{\alpha_{LC}})=U_{LC}$, and claim that this implies $g(V_{\alpha_{LC}})=V_{\alpha'_{LC}}$ for some $\alpha' \neq \alpha$. Indeed: - For *any* $S \subset E$ satisfying $p(S)=U_{LC}$, we have $p ^{-1} p(S) \subset p ^{-1}(U_{LC})=(* *)$. Thus $S \subset \bigsqcup_{\alpha}V_{\alpha_{LC}}$. $g(V_{\alpha})$ is such a set; by [[connected subspace of separated set lies in one constituent]] we therefore have $g(V_{\alpha})=V_{\alpha'}$ for some $\alpha'$. - We can't have $\alpha=\alpha'$ because we can't have $g(V_{\alpha})=V_{\alpha}$. It follows that $g(V_{\alpha_{LC}}) \cap V_{\alpha_{LC}}$ is nontrivial if and only if $g=\id_{E}$. Thus $G$ is properly discontinuous. --- Let $E$ be a [[topological space]] and $G \subset \text{homeo}(E)$ a [[properly discontinuous subgroup of the homeomorphism group|properly discontinuous]] [[subgroup]]. Let $\sim$ be the [[equivalence relation]] on $E$ obtained by declaring $x \sim y$ if there exists $g \in G$ s.t. $y =g(x)$; note that under $\sim$ we have $[x]=\{ g(x): g \in G \}$. We note $B=E / \sim$ and $p: E \to B$ the [[topological quotient map|quotient map]]. **(a) Convince yourself that $\sim$ is an equivalence relation.** - That $x \sim x$ is witnessed, by $g=\id_{X}$, since $\id_{X}(x)=x$. - Suppose $x \sim y$; fix $g \in G$ satisfying $g(x)=y$. Now the map $g^{-1}$ witnesses that $y \sim x$. - Suppose $x \sim y$, witnessed by $g_{1}$ and $y \sim z$, witnessed by $g_{2}$, i.e., $g_{1}(x)=y$ and $g_{2}(y)=z$. Then $g_{2}\circ g_{1}(x)=z$ and so $g_{2} \circ g_{1}$ witnesses that $x \sim z$. **b. Prove that for any open set $V \subset E$, the image $U=p(V)$ is open in $B$.** By the [[quotient topology]] definition, $p(V)$ is open in $B$ iff $p ^{-1}(p(V))$ is open in $E$. We see that $\begin{align} p ^{-1} (p(V)) = &p ^{-1}\{ \{ g(v): g \in G \} : v \in V \} \\ = & \bigcup_{g \in G, v \in V}^{}[g(v)] \\ = & \bigcup_{g \in G}^{}g(V) \end{align}$ which is a open is a union of open sets in $V$ ($V$ is open and $g$ is a [[homeomorphism]], hence an [[open map]], and so $g(V)$ is open too). Thus, $p ^{-1}\big( p(V) \big)$ is open in $E$, which by definition implies $p(V)$ is open in $B$. **c. Prove that the quotient map $p: E \to B$ is a [[covering space|covering map]].** $p$ is a [[continuous]] [[surjection]] by construction. Let $b_{0} \in B$; we want to find a [[neighborhood]] $U \ni b_{0}$ that is [[evenly covered]] by $p$. We choose $U$ as follows. Fix any $e_{0} \in p ^{-1}(b_{0})$. Using that $G$ is [[properly discontinuous subgroup of the homeomorphism group|properly discontinuous]], obtain an open [[neighborhood]] $V \ni e_{0}$ such that $g(V)=V \iff g=\id_{E}$ for all $g \in G$. Then set $U:=p(V) \ni b_{0}$. Note that $U$ is open in $B$ since (b) showed that $p$ is an [[open map]]. Now, $\begin{align} p ^{-1}(U)= & p ^{-1}(\{ [v] : v \in V \}) \\ = & \bigcup_{g \in G}^{} g(V), \text{ where }g(V) \text{ open because }g \in \text{homeo}(E); \end{align}$ we claim that because $G$ is properly discontinuous, this union is disjoint. Indeed, observe that, for any $g_{1},g_{2} \in G$, $\begin{align} g_{1}(v)=g_{2}(v) \iff & v=g_{1}^{-1}g_{2}^{}(v) \\ \iff & g_{1}^{-1}g_{2} = \id_{E} \\ \iff & g_{1}=g_{2}, \end{align}$ where in the penultimate step we used that $G$ is properly discontinuous. This implies that $g_{1}(V) \cap g_{2}(V)\neq \emptyset \iff g_{1}=g_{2}$. Thus, the collection $\{ g(V): g \in G \}$ [[evenly covered|partitions]] $p ^{-1}(U)$ into slices. We claim that $p |_{g(V)}$ is a [[homeomorphism]] onto $U$ for all $g \in G$, with [[inverse map]] $\begin{align} h: U \to & g(V) \\ [v] \mapsto & p ^{-1}([v]) \cap g(V) . \end{align}$ First verify that $\im p |_{g(V)}=U$: $p |_{g(V)}(g(V))=\{ [g(v)] : v \in V \}=\{ [v]: v \in V \}=U,$ where we used the fact that $g(x) \sim x$ for all $x$ (as witnessed by $g ^{-1}$). [[continuous|Continuity]] follows immediately from the fact $p$ is a [[topological quotient map|quotient map]] and the fact that, in general, the [[restriction of continuous function is continuous]]. And $\begin{align} p |_{g(V)}\big( h ([v]) \big)= & p |_{g(V)}\big( p ^{-1} ([v] ) \cap g(V) \big) \\ = & p |_{g(V)} \big( p ^{-1}([v]) \big) \text{ ($p$ is restricted to $g(V)$ anyway}) \\ = & [v] \end{align}$ and $\begin{align} h\big( p |_{g(V)}(g(v)) \big)= & h([g(v)]) \\ = & p ^{-1}([g(v)]) \cap g(V) \\ = & (\bigsqcup_{g_{\alpha} \in G} g_{\alpha}(v) ) \cap g(V) \\ = & g(v), \end{align}$ where we again used the general fact that $g(x) \sim x$ for all $x$, and that $g$ is properly discontinuous. This completes the proof. **d.** Prove that if $E$ is [[connected]] then $G=\{ g \in \text{homeo}(E) : p \circ g = p \}$. One inclusion follows from the fact that we always have $[g(x)]=[x]$. For the other inclusion, let $h$ satisfy $p \circ h=p$. Write $h(x_{1})=x_{2}$. Then $p \circ h(x_{1})=p(x_{1})=p(x_{2})$, thus there exists $g \in G$ for which $g(x_{1})=x_{2}$. Now the uniqueness part of Munkres' theorem 79.2 implies $h=g$. --- For any pair of [[integers]] $(k,n) \in \mathbb{Z}^{2}$, let $g_{k,n}:\mathbb{R}^{2} \to \mathbb{R}^{2}$ be the function defined by $g_{k,n}(x,y):=(x+k, (-1)^{k}y+n)$. We claim that $G:=\{ g_{k,n}: (k,n) \in \mathbb{Z}^{2} \}$ is a [[properly discontinuous subgroup of the homeomorphism group|properly discontinuous]] [[subgroup]] of $\text{homeo}(\mathbb{R}^{2})$. First we show that $g_{k,n}$ is a [[homeomorphism]] for all $(k,n) \in \mathbb{Z}^{2}$, with inverse $g_{-k,-n}$. Clearly $g_{k,n}$ and $h_{k,n}$ are both [[continuous]] on $\mathbb{R}^{2}$, for both [[continuous on product space iff continuous on coordinates|coordinate functions]] are sums of [[continuous]] functions on $\mathbb{R}$. And since $\begin{align} g_{-k,(-1)^{k+1}n} \circ g_{k,n}(x,y)= & g_{-k,(-1)^{k+1}n}\big( \ x+k, (-1)^{k}y +n \ \big) \\ = & \big( \ (x+k)-k, (-1)^{-k}\big( (-1)^{k}y+n \big) - (-1)^{k+1}n \ \big) \\ = & \big( \ x , y +(-1)^{-k}n - (-1)^{k+1}n\ \big) \\ = & (x, y) \end{align}$and $\begin{align} g_{k,n} \circ g_{-k,(-1)^{k+1}n}(x,y) = & \ g_{k,n}\big(\ x-k, (-1)^{-k}y + (-1)^{k+1}n \ \big) \\ = & \ \big(\ (x-k)+k, (-1)^{k}\big((-1)^{-k}y + (-1)^{k+1}n\big) + n \ \big) \\ = & \ \big(\ x , y - n + n \ \big) \\ = & \ (x, y) \end{align}$ we conclude $g_{k,n} \in \text{homeo}(\mathbb{R}^{2})$. Next we show it is stable under composition: given $x,y \in \mathbb{R}^{2}$, and $(a,b), (k,n) \in \mathbb{Z}^{2}$, we have $\begin{align} g_{a,b} \circ g_{k,n}(x,y)= & g_{a,b}\big( x+k, (-1)^{k}y+n \big) \\ = & \big( \ (x+k) +a , (-1)^{a}\big( (-1)^{k}y+n \big) +b \ \big) \\ = & \big( \ x+(k+a), (-1)^{k+a}y + (-1)^{ a}n + b \ \big) \\ = & g_{k+a, (-1)^{a}n+b}(x,y) \end{align}$ so $g_{a,b} \circ g_{k,n} \in G$. Since $g_{k,n}^{-1}=g_{-k, (-1)^{k+1}n} \in G$ and $g=\id_{\mathbb{R}^{2}}$ when $k=n=0$, we conclude that $G$ is a [[subgroup]] of $\text{homeo}(\mathbb{R}^{2})$. Next we show $G$ is [[properly discontinuous subgroup of the homeomorphism group|properly discontinuous]]. Given $(x_{0},y_{0}) \in \mathbb{R}^{2}$, define $U:=(x_{0}-\varepsilon, x_{0}+\varepsilon) \times (y_{0}-\varepsilon, y_{0} +\varepsilon)$, where $\varepsilon$ is any real number satisfying $0<\varepsilon \leq \frac{1}{2}$ (could simply be $\frac{1}{2}$). Let $g_{k,n}(x,y)\in g_{k,n}(U)$ be arbitrary, $g \neq \id$. We want to show that $g_{k,n}(x,y) = (g_{k,n}^{(1)}(x,y), g_{k,n}^{(2)}(x,y)) \notin (x_0 - \varepsilon, x_{0}+\varepsilon) \times (y_{0}-\varepsilon, y_{0}+\varepsilon)$, i.e., that $\begin{align} d(g_{k,n}^{(1)}(x,y), x_{0})= & |g_{k,n}^{(1)}(x,y)-x_{0}|\geq\varepsilon \text{ and } \\ d(g_{k,n}^{(2)}(x,y), y_{0})= & |g_{k,n}^{(2)}(x,y)-y_{0}|\geq \varepsilon. \end{align}$ We have $\begin{align} |g_{k,n}^{(1)}(x,y) - x| \geq & 1 (\text{since }k \in \mathbb{Z}), \text{ as well as } \\ |x - x_{0}| < & \frac{1}{2}. \\ \end{align}$ It follows that $\begin{align} \varepsilon \leq \frac{1}{2} \leq & | \ |g_{k,n}^{(1)}(x,y)-x| - |x_{0}-x| \ | \\ \leq & |g_{k,n}^{(1)}(x,y)- x - (x_{0} - x)| \\ = & |g_{k,n}^{(1)}(x,y)- x_{0}| \end{align}$ and so we are done with the first coordinate. The proof for the second coordinate is the exact same as above if $k$ is even, so assume $k$ is odd, i.e., that $g_{k,n}^{(2)}(x,y)=-y+n$. We have $\begin{align} |g_{k,n}^{(1)}(x,y) + y| \geq & 1 (\text{since }n \in \mathbb{Z}), \text{ as well as } \\ |y-y_{0}| < & \frac{1}{2}. \\ \end{align}$ It follows that $\begin{align} \varepsilon \leq\frac{1}{2} \leq & | \ |g_{k,n}^{(2)}(x,y)+y| - |y_{0}-y| \ | \\ \leq & |g_{k,n}^{(2)}(x,y)+ y - (y_{0} - y)| \\ = & |g_{k,n}^{(2)}(x,y)- y_{0}| \end{align}$ as desired. Since $(x_{0},y_{0})\in \mathbb{R}^{2}$, $(x,y) \in U$ and $(k,n)\neq(0,0)$ were all arbitrary, we conclude the result. ![[CleanShot 2024-04-18 at 20.34.49.jpg]] Looks like a [[klein bottle]]. Let $G= \{ g_{i} \}_{i=1}^{n} \subset \text{homeo}(X)$ be a finite subgroup. If every $g \in G - \{ \id_{X} \}$ has no fixed point then $G$ is [[properly discontinuous subgroup of the homeomorphism group|properly discontinuous]]. Let $x_{0} \in X$. Consider its orbit $\{ g_{i}(x_{0}) \}_{i=1}^{n}$, note that $g_{i}(x_{0}) \neq x_{0}$. ---- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further ReadEN file.tags GROUP BY file.tags as Tag