# 7.06 Group actions and covering spaces, 1

## Video

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

## Notes

### 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.

### Examples

(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.

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CC-BY-SA, Jonny Evans 2017