7.06 Group actions and covering spaces, 1

Below the video you will find accompanying notes and some pre-class questions.


Properly discontinuous group actions

(0.10) Recall that a group \(G\) acts continuously on \(X\) if for each \(g\in G\) there exists a homeomorphism \(\rho(g)\colon X\to X\) such that \(\rho(gh)=\rho(g)\circ\rho(h)\) and \(\rho(1)=id_X\). The quotient \(X/G\) is the set of equivalence classes, where \(x\sim\rho(g)x\) for all \(g\in G\), equipped with the quotient topology.

(1.36) Suppose that \(G\) acts continuously on \(X\). Suppose moreover that for all \(x\in X\) there is an open set \(U\) containing \(x\) such that \(U\cap \rho(g)(U)=\emptyset\) for all \(g\neq 1\) in \(G\). Then the quotient map \(q\colon X\to X/G\) is a covering map.

(3.10) Consider the action of \(\mathbf{Z}\) on \(\mathbf{R}\) given by \(\rho(n)(x)=x+n\). For any \(x\in\mathbf{R}\), for sufficiently small \(\epsilon\), the interval \((x-\epsilon,x+\epsilon)\) is disjoint from any of its translates under the action of \(\mathbf{Z}\). The theorem tells us that the quotient map \(q\colon\mathbf{R}\to \mathbf{R}/\mathbf{Z}\) is a covering map. This is the covering map \(x\mapsto e^{i2\pi x}\) we have been using for a while now.

(5.00) Given a point \(x\in X\) and its equivalence class \([x]\in X/G\) there exists an open neighbourhood \(U\subset X\) of \(x\) such that \(\rho(g)(U)\) is disjoint from \(U\) unless \(g=1\). Let \(V=q(U)\). In this section, we saw that the quotient map for a quotient by a group action is an open map, so \(V\) is an open set.

(6.45) The set \(q^{-1}(V)\) equals the union of all translates of \(U\) under the group action, that is \[q^{-1}(V)=\coprod_{g\in G}\rho(g)(U).\] To prove that \(q\) is a covering map, we need to show that there is a homeomorphism \(h\colon q^{-1}(V)\to V\times F\) for some discrete set \(F\) such that \(pr_1\circ h=q\) (where \(pr_1\colon V\times F\to V\) is the projection to the first factor).

(8.14) The discrete set \(F\) will be the group \(G\). Each \(\rho(g)(U)\) is homeomorphic to \(U\) (via the homeomorphism \(\rho(g)\)) and \(U\) is homeomorphic to \(V\) via the map \(q\). To see that \(q|_U\colon U\to V\) is a homeomorphism, note that it is continuous and open (so if it is bijective then it will be a homeomorphism) and that it is surjective (because \(V=q(U)\) by definition) and injective because if \(a,b\in U\) satisfy \(q(a)=q(b)\) then \(\rho(g)(a)=b\) for some \(g\in G\), so \(\rho(g)(U)\cap U\neq\emptyset\), so \(g=1\), so \(a=b\).

(11.00) Since \(q^{-1}(V)=\coprod_{g\in G}\rho(g)(U)\), we can define \(h\colon q^{-1}(V)\to V\times G\) as \[h(\rho(g)(u))=(q(u),g).\] We have now seen that this is a homeomorphism and \(q(\rho(g)(u))=q(u)=pr_1(q(u),g)\).

(13.36) The condition from the theorem is that for all \(x\in X\) there is an open set \(U\) containing \(x\) such that \(U\cap \rho(g)(U)=\emptyset\) for all \(g\neq 1\) in \(G\). An action satisfying this condition is called properly discontinuous.


(13.58) Let \(X\) be a metric space and suppose that \(G\) acts by isometries (i.e. \(d(\rho(g)(x),\rho(g)(y))=d(x,y)\) for all \(g\in G\) and \(x,y\in X\)). Suppose moreover that there exists \(c>0\) such that for all \(x\in X\) and all \(g\neq 1\) in \(G\) we have \[d(x,\rho(g)(x))\geq c.\] Then the action is properly discontinuous.

(16.12) Pick a point \(x\in X\) and take the metric ball of radius \(r\in(0,c/2)\) centred at \(x\). Then \(B_r(x)\cap\rho(g)(B_r(x))\) is empty if \(g\neq 1\). Otherwise there is some point \(y\in B_r(x)\cap\rho(g)(B_r(x))\) and so \(c/2>r>d(x,y)\) and \(2/c>r>d(\rho(g)x,y)\) so \(c>d(x,\rho(g)(x))\) by the triangle inequality, which contradicts the hypothesis.

(18.12) In our earlier example of \(\mathbf{Z}\) acting on \(\mathbf{R}\), translations are isometries for the standard metric on \(\mathbf{R}\) and the hypothesis of the theorem is satisfied by \(c=1/2\) (\(d(x,x+n)\geq 1\) for any integer \(n\neq 0\)).

More generally, \(\mathbf{Z}^n\) acts on \(\mathbf{R}^n\) via \[\rho(k_1,\ldots,k_n)(x_1,\ldots,x_n)=(x_1+k_1,\ldots,x_n+k_n)\] and \(d(\mathbf{x},\rho(\mathbf{k})(\mathbf{x}))=\sqrt{\sum k_i^2}\geq 1\). In this case, the quotient map gives a cover of the \(n\)-dimensional torus by \(\mathbf{R}^n\).

(20.23) Take \(S^n\subset\mathbf{R}^{n+1}\) to be the sphere of radius 1 and \(G=\mathbf{Z}/2\). There is a \(G\)-action on \(S^n\) where the nontrivial element acts as the antipodal map \(x\mapsto -x\). The distance \(d(x,-x)\) is always equal to 2 (using the metric that just takes distances in the ambient Euclidean space) or equal to \(\pi\) (using the metric that takes distances along paths that stay on the sphere), so with either of these metrics we could take \(c=1\). This gives a covering map \(S^n\to S^n/(\mathbf{Z}/2)\). The quotient \(S^n/(\mathbf{Z}/2)\) is called the /real projective space/ \(\mathbf{RP}^n\).

In the next video, we will see that if \(G\) acts properly discontinuously on \(X\) and \(X\) is simply-connected then the fundamental group of \(X/G\) is isomorphic to \(G\). This will allow us to say \(\pi_1(S^1)=\mathbf{Z}\), \(\pi_1(T^n)=\mathbf{Z}^n\) and \(\pi_1(\mathbf{RP}^n)=\mathbf{Z}/2\) just because these spaces are constructed as quotients by the corresponding group actions.

Pre-class questions

  1. Prove that the map \(p_n\colon S^1\to S^1\), \(p_n(e^{i\theta})=e^{in\theta}\), is a covering map by considering a suitable \(\mathbf{Z}/n\)-action on \(S^1\) and showing it is properly discontinuous.


Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.