7.06 Group actions and covering spaces, 1

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

Previous video: 7.05 Fundamental group of the circle.
Next video: 7.07 Group actions and covering spaces, 2.
Index of all lectures.

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

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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Next video: 7.07 Group actions and covering spaces, 2.
Index of all lectures.