Properly Discontinuous Action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:34:44Z http://mathoverflow.net/feeds/question/55726 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55726/properly-discontinuous-action Properly Discontinuous Action Martin David 2011-02-17T12:23:06Z 2013-01-30T21:55:53Z <p>When looking definition, and theorems related to <strong>Properly discontinuous action</strong> of a group $G$ on a topological space $X$, it is different in different books (Topology and Geometry-Bredon, Complex Functions-Jones, Three Dimensional Geometry and Topology- Thurston). Therefore, it will be clarified, if we write these definitions separately, and see which is stronger or which are equivalent? I will name them as "Type A", "Type B"..)</p> <p>Let $X$ be a topological space and $G$ be a group acting on $X$.</p> <p><strong>Definition 1:</strong> The action is of "<strong>Type A</strong>" if the map $G \times X \rightarrow X \times X$, given by $(g,x)\mapsto (x,g.x)$ is proper, i.e. inverse image of any compact set under this map is compact.</p> <p><strong>Definition 2:</strong> The action is of "<strong>Type B"</strong> if for any compact set $K\subseteq X$, $K\cap g.K=\phi$ for all but finitely many $g\in G$.</p> <p><strong>Definition 3:</strong> The action is of "<strong>Type C</strong>" if for each $x\in X$ has an open neighbourhood $U$ such that $g.U\cap U=\phi$ for all but finitely many $g\in G$. </p> <p><strong>Definition 4:</strong> The action is of "<strong>Type D</strong>" if for each $x\in X$, there exist an open neighbourhood $U$ of $x$, such that $g.U\cap U\neq \phi$ for $g\in G$ implies $g.x=x$. </p> <p><strong>Definition 5:</strong> The action is of "<strong>Type E</strong>" if each $x\in X$ has a neighbourhood $U$ such that the set {$g\in G \colon g.x\in U$} is finite. </p> <p><strong>Q.1</strong> Which type of actions imply which other type of action? </p> <p><strong>Q.2</strong> If $X$ is Hausdorff, then under which type action, the quotient $X/G$ is Hausdorff?</p> <p>(These are required, when studying action of a group on a compact Riemann surface, its quotient, whether quotient map is branched or unbranched, etc,) </p> <p>(This question may be not applicable to post for MO; but when reading a paper related to enumeration of equivalent coverings of a space, with given (finite) transformation group, I came across this notion, and when looked into details, the different definitions puzzled.) </p> http://mathoverflow.net/questions/55726/properly-discontinuous-action/56490#56490 Answer by Theo Buehler for Properly Discontinuous Action Theo Buehler 2011-02-24T06:10:07Z 2011-02-24T06:10:07Z <p>Below <em>locally compact</em> spaces are assumed to be <em>Hausdorff</em>. The following is essentially a distillate of results from Bourbaki's <em>Topologie G&eacute;n&eacute;rale</em>, Chapitres II and III.</p> <hr> <p><strong>Definition.</strong> A continuous function $f: X \to Y$ is called <em>proper</em> if $f$ maps closed sets to closed sets and $f^{-1}(K)$ is compact for all compact $K \subset Y$.</p> <p><strong>Remark.</strong> If $X$ is Hausdorff and $Y$ is locally compact then a continuous function $f: X \to Y$ is proper if and only if $f^{-1}(K)$ is compact for all compact $K \subset Y$. Moreover, $X$ must be locally compact.</p> <p>To see this, cover $Y$ with open and relatively compact sets $U_{\alpha}$. Then $f^{-1}(U_{\alpha})$ is an open covering of $X$ by relatively compact sets, hence $X$ is locally compact. If $F \subset X$ is closed then $f(F)$ is closed. Indeed, if $(y_{n}) \subset f(F)$ is a net converging to $y$, then we may assume that all $y_{n}$ are in a compact neighborhood $K$ of $y$. Pick a pre-image $x_{n}$ of each $y_{n} \in f^{-1}(K)$, which is compact by assumption. If $x_{i} \to x \in f^{-1}(K)$ is a convergent subnet of $(x_{n})$ then $(f(x_{i}))$ is a subnet of $(y_{n})$, hence $f(x) = y$ by continuity and thus $y \in f(F)$. </p> <p><strong>Remark.</strong> In the definition of properness it would suffice to require that $f$ is closed and $f^{-1}(y)$ is compact for all $y \in Y$, but the definition above is good enough for the present purposes.</p> <hr> <p><strong>Definition.</strong> Let $G$ be a topological group acting continuously on a topological space $X$. The action is called <em>proper</em> if the map $\rho: G \times X \to X \times X$ given by $(g,x) \mapsto (x,gx)$ is proper.</p> <p><strong>Proposition.</strong> If $G$ acts properly on $X$ then $X/G$ is Hausdorff. In particular, each orbit $Gx$ is closed. The stabilizer $G_{x}$ of each point is compact and the map $G/G_{x} \to Gx$ is a homeomorphism. Moreover, if $G$ is Hausdorff then so is $X$.</p> <p><em>Proof.</em> Indeed, the orbit equivalence relation is the image of $\rho$, hence it is closed. Since the projection $X \to X/G$ is open, this implies that $X/G$ is Hausdorff. Since the pre-image of the point $[x]$ in $X/G$ is its orbit $Gx$, we have that orbits are closed. The stabilizer $G_{x}$ of a point $x$ is the projection of $\rho^{-1}(x,x)$ to $G$, hence it is compact. The map $G/G_{x} \to Gx$ is proper and $1$-to-$1$, hence a homeomorphism. Finally, if $G$ is Hausdorff, then $\{e\} \times X \subset G \times X$ is closed and therefore the diagonal $\Delta_{X} = \rho(\{e\} \times X)$ of $X \times X$ is closed, hence $X$ is Hausdorff.</p> <p><strong>Exercise.</strong> Let $G$ be a Hausdorff topological group acting properly on a locally compact space $X$. Then $G$ and $X/G$ are both locally compact. If $X$ is compact Hausdorff then so are $G$ and $X/G$.</p> <hr> <p>Replace <em>finite</em> by <em>compact</em> in Type A and Type B. Then we have the following implications for a continuous action:</p> <p>Proper $\Longrightarrow$ Type A, the converse holds if both $G$ and $X$ are locally compact.</p> <p>Type A $\Longrightarrow$ Type B.</p> <p>Let $K \subset X$ be compact. Then $K \times K \subset X \times X$ is compact. Thus, if the action is of type A, then $\rho^{-1}(K \times K) = \{(g,x) \in G \times X\,:\,(x,gx) \in K \times K\} \subset G \times X$ is compact. The projection of this set to $G$ is compact and consists precisely of the $g \in G$ for which $K \cap gK \neq \emptyset$.</p> <p>Type B $\Longrightarrow$ Type A if $X$ is Hausdorff.</p> <p>We have to show that $\rho^{-1}(L)$ is compact for every compact $L \subset X \times X$. Let $K$ be the union of the two projections of $L$. Then $(g,x) \in \rho^{-1}(K \times K)$ is equivalent to $x \in K \cap gK$. Since $\rho^{-1}(K \times K)$ is compact and $\rho^{-1}(L)$ is a closed subset of $\rho^{-1}(K \times K)$, we have that $\rho^{-1}(L)$ is compact.</p> <p><strong>Corollary.</strong> If $G$ and $X$ are locally compact, properness, Type A and Type B are all equivalent.</p> <hr> <p>Let me now show that in the locally compact setting properness is equivalent to a refinement of Type C:</p> <p><strong>Proposition.</strong> Let $G$ and $X$ be locally compact and assume that $G$ acts continuously on $X$. The following are equivalent:</p> <ol> <li>The action is proper.</li> <li>For all $x,y \in X$ there are open neighborhoods $U_{x}, U_{y} \subset X$ of $x$ and $y$ such that $C = \{g \in G\,:\,gU_x \cap U_{y} \neq \emptyset \}$ is relatively compact.</li> </ol> <p><em>Proof.</em> $1.$ implies $2.$ Let $K_{x}$ and $K_{y}$ be compact neighborhoods of $x$ and $y$. Then the set $\rho^{-1}(K_{x} \times K_{y})$ is compact and its projection to $G$ contains $C$ and is compact. Now let $U_{x}$ and $U_{y}$ be the interiors of $K_{x}$ and $K_{y}$.</p> <p>$2.$ implies $1$. Let $K \subset X \times X$ be compact. We want to show that $\rho^{-1}(K)$ is compact as well. Let $(g_{n},x_{n})$ be a universal net in $\rho^{-1}(K)$. Then $(x_{n},g_{n}x_{n})$ is a universal net in $K$ and hence converges to some $(x,y) \in K$. Let $U_{x}, U_{y}$ and $C$ be as in $2.$. Then $(x_{n},g_{n}x_{n}) \in U_{x} \times U_{y}$ eventually and thus also $(g_{n}) \subset C$ eventually. Since $(g_{n})$ is universal and $C$ is relatively compact, $(g_{n})$ converges to some $g \in G$. Hence $(g_{n},x_{n})$ converges to $(g,x) \in \rho^{-1}(K)$.</p> <p><strong>Example.</strong> To see that Type C is weaker than properness, consider $A = \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; 2^{-1} \end{pmatrix}$ and the action of $\mathbb{Z}$ on $\mathbb{R}^{2} \smallsetminus \{0\}$ given by $n \cdot x = A^{n} x$. For instance for $x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $y = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and all neighborhoods $U_{x} \ni x$ and $U_{y} \ni y$ the set $\{n \in \mathbb{Z}\,:\, U_{x} \cap n \cdot U_{y} \neq \emptyset \}$ is infinite. Thus this action isn't proper. On the other hand, it is easy to see that it is of Type C.</p> <p><strong>Remark.</strong> The previous example shows that properness of an action is <em>not</em> a local property.</p> <p><strong>Exercise.</strong> If the action of a locally compact group $G$ on a locally compact space $X$ is of type C and $X/G$ is Hausdorff then it is proper.</p> <hr> <p>To finish this discussion, it is evident that an action of type C is also of type E, hence type E is also weaker than properness. Finally, a trivial action is of type D, hence this property has nothing to do with properness.</p> <hr> <p>Here are some references:</p> <p>I've followed Bourbaki, Topologie G&eacute;n&eacute;rale, Ch. III, in terminology, and the proofs I've given are variants of Bourbaki's. I happen to like Koszul's <em>Lectures on groups of transformations</em>. If you're looking for a more pedestrian approach, you can find the most important facts in Lee's <em>Introduction to topological manifolds</em>.</p> http://mathoverflow.net/questions/55726/properly-discontinuous-action/69217#69217 Answer by Stefan Witzel for Properly Discontinuous Action Stefan Witzel 2011-06-30T21:15:57Z 2011-06-30T21:15:57Z <p>Just two small remarks:</p> <ul> <li>The action is properly discontinuous if it is proper and the group is is equipped with the discrete topology (compact then meaning finite, this accounts for some confusion, I guess).</li> <li>I think in Definition 4 the conclusion should be $g = 1$. Then it means that the action is properly discontinuous and free (which for example Bredon calls properly discontinuous).</li> </ul> http://mathoverflow.net/questions/55726/properly-discontinuous-action/69268#69268 Answer by David White for Properly Discontinuous Action David White 2011-07-01T14:58:13Z 2013-01-30T21:55:53Z <p>Theo Buehler did a great job relating Types A, B, C, and E. Going off of Stefan Witzel's new answer, I'd like to point out that in Munkres's <em><a href="http://www.scribd.com/doc/35324174/Munkres-Topology" rel="nofollow">Topology</a></em> in section 81 (page 505 of that link), he defines an action to be <strong>properly discontinuous</strong> if for all $x\in X$ there is a nbhd $U$ s.t. $g(U)\cap U = \emptyset$ unless $g=1$. So if we tweak Type D from the original question to conclude $g=1$ rather than just that $g$ fixes $x$, then we recover Munkres' definition and it's no longer as trivial as Theo's answer showed the old Type D definition was. I think Munkres' definition is the one which I think point-set topologists would use. It's nice because you don't need to assume a topology on $G$, but of course you could just put the discrete topology on it. Perhaps the other definitions are more popular in the literature of Riemann surfaces, and the difference may be because of standing hypotheses in that field, since they often care most about the case of Fuchsian groups. Certainly <strong>Munkres' definition implies Type E and Type C</strong></p> <p>Munkres also points out that the quotient map $\pi: X\rightarrow X/G$ is a covering map iff the action of $G$ is properly discontinuous.</p> <p>An exercise in section 81 gives: Let $X$ be locally compact Hausdorff and let $G$ act freely (i.e. fixed-point-free). Suppose that for each compact $C \subset X$ there are only finitely many $g\in G$ s.t. $C\cap g(C) \neq \emptyset$. Then the action of $G$ is properly discontinuous and $X/G$ is locally compact Hausdorff. So this <strong>tells you when Type B implies Munkres' definition</strong>.</p> <p>Now let's <strong>relate Munkres' definition to Type A</strong> and Theo's answer. Using Theo's various propositions and corollaries it's not hard to see that if $X$ is locally compact Hausdorff space and $G$ is any group (which we'll equip with the discrete topology) then Munkres' definition implies Type A. Conversely, if $X$ is locally compact then a proper action of a discrete group must be of Type B (by Theo's comment) and this implies Munkres' definition because local compactness lets us get from $g(K)\cap K = \emptyset$ to $g(U)\cap U = \emptyset$.</p>