8.02 Covering transformations

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

Previous video: 8.01 Lifting criterion.
Next video: 8.03 Normal covering spaces.
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Notes

Definition, existence and uniqueness

(0.12) Given two covering spaces $p_1\colon Y_1\to X$ and $p_2\colon Y_2\to X$ of the same space $X$ , a covering transformation $F\colon(Y_1,p_1)\to (Y_2,p_2)$ is a continuous map $F\colon Y_1\to Y_2$ such that $p_1=p_2\circ F$ [NOTE: I got this formula back-to-front in the video].
A covering transformation is called a covering isomorphism if $F$ is also a homeomorphism.
A deck transformation or covering automorphism of a covering space $(Y,p)$ is a covering isomorphism $(Y,p)\to (Y,p)$ .
The deck group $Deck(Y,p)$ is the group of all deck transformations of $(Y,p)$ .

(3.32) Let

$p_1\colon Y_1\to X$ and

$p_2\colon Y_2\to X$ be covering spaces, let

$x\in X$ and let

$y_1\in p_1^{-1}(x)$ and

$y_2\in p_2^{-1}(x)$ . Then there exists a covering transformation

$F\colon (Y_1,p_1)\to (Y_2,p_2)$ such that

$F(y_1)=y_2$ if and only if

$(p_1)_*\pi_1(Y_1,y_1)\subset (p_2)_*\pi_1(Y_2,y_2)$ as subgroups of

$\pi_1(X,x)$ . Moreover, if such a covering transformation exists then it is unique (uniquely determined by the condition

$F(y_1)=y_2)$ ).

(6.38) There is a covering isomorphism

$F\colon (Y_1,p_1)\to (Y_2,p_2)$ if and only if

$(p_1)_*\pi_1(Y_1,y_1)=(p_2)_*\pi_1(Y_2,y_2)$ as subgroups of

$\pi_1(X,x)$ .

(7.40) The theorem follows immediately from the lifting criterion applied to the following situation. Take

$T=Y_1$ and

$f=p_1\colon Y_1\to X$ . A covering transformation

$F\colon Y_1\to Y_2$ is precisely a lift of

$p_1$ to a map

$Y_1\to Y_2$ . We see that the statement of the theorem is just the lifting criterion applied to this siutation.

Example

(9.44) Let

$X=S^1$ and take

$Y_1=S^1$ ,

$Y_2=S^1$ with covering maps

$p_1(z)=z^m$ and

$p_2(z)=z^n$ . When is there a covering transformation

$F\colon Y_1\to Y_2$ ? The pushforward

$(p_1)_*\pi_1(Y_1,1)$ is the subgroup

$m\mathbf{Z}\subset\mathbf{Z}$ . The pushforward

$(p_2)_*\pi_1(Y_2,1)$ is the subgroup

$n\mathbf{Z}\subset\mathbf{Z}$ . Therefore the criterion from the theorem holds if and only if

$n$ divides

$m$ .

(11.54) For example, if $m=6$ and $n=2$ then we can take $F(z)=z^3$ and we get $p_1(z)=z^6=(z^3)^2=F(p_1(z))$ so $F$ is a covering transformation. We could also take $F(z)=-z^3$ because $(-z^3)^2=z^6$ too. This exhausts the possible covering transformations $Y_1\to Y_2$ because a covering transformation is determined by its value at a single point, $F(1)$ , and we need $F(1)=\pm 1$ because $1=p_1(1)=p_2(F(1))=(F(1))^2$ .

(14.47) Note that $z\mapsto z^3$ is again a covering map. We will now see that this is a general feature.

Covering transformations are covering maps

(15.00) Assume that

$p_1\colon Y_1\to X$ and

$p_2\colon Y_2\to X$ are covering spaces and that

$Y_2$ is path-connected. A covering transformation

$F\colon Y_1\to Y_2$ is a covering map.

(16.04) Each

$x\in X$ has an elementary neighbourhood

$U_1$ for

$p_1$ and

$U_2$ for

$p_2$ , so

$U:=U_1\cap U_2$ is an elementary neighbourhood for both simultaneously.

(17.55) $F$ is surjective. To see this, pick a point $y_2\in Y_2$ . We want to find a point $z\in Y_1$ such that $F(z)=y_2$ . Pick a point $y_1\in Y_1$ and a path $\alpha$ in $Y_2$ from $F(y_1)$ to $y_2$ . This projects along $p_2$ to give a path $p_2\circ\alpha$ from $p_1(y_1)$ to $p_2(y_2)$ . Path-lifting for $p_1$ yields a path $\widetilde{p_2\circ\alpha}$ in $Y_1$ from $y_1$ to some point $z$ . Applying $F$ to this gives a path $F\circ\widetilde{p_2\circ\alpha}$ in $Y_2$ from $y_1$ to $F(z)$ .

(20.15) We have $p_2\circ(F\circ\widetilde{p_2\circ\alpha})=p_1\circ\widetilde{p_2\circ\alpha}=p_2\circ\alpha$ , so $F\circ\widetilde{p_2\circ\alpha}$ and $\alpha$ are both paths lifting $p_2\circ\alpha$ along $p_2$ satisfying the initial condition $F(\widetilde{p_2\circ\alpha}(0))=\alpha(0)=y_2$ . Therefore they agree by uniqueness of lifts, so their endpoints, $F(z)$ and $y_2$ , agree, so $F(z)=y_2$ and we see that $F$ is surjective.

(22.32) $F$ is a covering map. Fix a point $y_2\in Y_2$ . We want to find an elementary neighbourhood around $y_2$ together with a local inverse for $F$ . Let $x=p_2(y_2)$ and let $x\in U\subset X$ be an elementary neighbourhood simultaneously for $p_1$ and $p_2$ . For each $y_1\in F^{-1}(y_2)$ (which is non-empty because $F$ is surjective) we have an elementary sheet $y_1\in V\subset Y_1$ for $p_1$ living over $U$ and a local inverse $q\colon U\to V$ with $q(x)=y_1$ . Let $y_2\in W\subset Y_2$ be an elementary sheet for $p_2$ over $U$ . The map $q\circ (p_2)|_W\colon W\to V$ is a local inverse for $F$ . If we do this for all our elementary sheets, we deduce that $F$ is a covering map.

Pre-class questions

1. In the proof that covering transformations are covering maps, why is $q\circ (p_2)|_W$ a local inverse for $F$ ? 2. Because I was using the lifting criterion, I should have added an assumption about my spaces in the first theorem. What should I have said?

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Index of all lectures.