# 7.07 Group actions and covering spaces, 2

## Video

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

## Notes

### Reading off the fundamental group

(0.00) Suppose that $$X$$ is a path-connected space and that $$x$$ is a basepoint for which $$\pi_1(X,x)=\{1\}$$. Suppose that $$G$$ acts properly discontinuously on $$X$$. Then $$\pi_1(X/G,[x])=G$$.

(1.15) Let $$F\colon G\to\pi_1(X/G,[x])$$ be the map defined as follows. For each element $$g\in G$$ we get a point $$\rho(g)(x)\in X$$. Pick a path $$\gamma$$ from $$x$$ to $$\rho(g)(x)$$; the projection $$q\circ\gamma$$ of this path to the quotient $$X/G$$ is a loop in $$X/G$$ based at $$[x]$$ (because $$q(\rho(g)(x))=q(x)=[x]$$). We define $$F(g)=[q\circ\gamma]$$.

(3.33) Is this well-defined? Yes because $$X$$ is simply-connected (pre-class question: prove this).

(4.04) Is $$F$$ surjective? Given a loop $$\delta$$ in $$X/G$$ based at $$[x]$$, path-lifting gives us a lift $$\tilde{\delta}$$ in $$X$$ starting at $$x$$. The endpoint $$\tilde{\delta}(1)$$ is in $$q^{-1}([x]$$, so there exists $$g\in G$$ such that $$\tilde{\delta}(1)=\rho(g)(x)$$. This implies that $$F(g)=[\delta]$$.

(5.51) Is $$F$$ a homomorphism? Given $$g,h\in G$$, pick paths $$\gamma_g$$ (from $$x$$ to $$\rho(g)(x)$$ in $$X$$) and $$\gamma_h$$ (from $$x$$ to $$\rho(h)(x)$$ in $$X$$). The path $$\rho(g)\gamma_h$$ connects $$\rho(g)(x)$$ to $$\rho(g)\rho(h)(x)=\rho(gh)(x)$$. The concatenation $$(\rho(g)\gamma_h)\cdot\gamma_g$$ makes sense and provides a path from $$x$$ to $$\rho(gh)(x)$$. The projection $$q((\rho(g)\gamma_h)\cdot\gamma_g)$$ is then a loop in $$X/G$$ whose lift connects $$x$$ with $$\rho(gh)(x)$$, so $F(gh)=[q((\rho(g)\gamma_h)\cdot\gamma_g)].$ We have $q((\rho(g)\gamma_h)\cdot\gamma_g)=q(\rho(g)\gamma_h)\cdot q(\gamma_g)=q(\gamma_h)\cdot q(\gamma_g)=F(h)\cdot F(g),$ so $$F(gh)=F(h)F(g)$$.

(9.44) Therefore $$F$$ is an antihomomorphism. This is because I defined the concatenation of $$a$$ followed by $$b$$ to be $$b\cdot a$$. This had the advantage of making monodromy into a homomorphism. To make the current $$F$$ into a homomorphism, we should use right actions of the group, $$x\mapsto xg$$, satisfying $$(xg)h=x(gh)$$. Note that any homomorphism $$f\colon G\to H$$ can be turned into an antihomomorphism $$\bar{f}\colon G\to H$$ by $$\bar{f}(g)=f(g^{-1})$$, so homomorphisms and antihomomorphisms are equally useful.

(11.14) Is $$F$$ injective? I claim that $$\ker(F)$$ is trivial. If $$g\in\ker(F)$$ then the loop $$q(\gamma_g)$$ is nullhomotopic (where $$\gamma_g$$ is a path from $$x$$ to $$\rho(g)(x)$$ in $$X$$). By homotopy lifting, this implies that $$\gamma_g$$ is homotopic to the constant path rel endpoints, which implies that $$\rho(g)(x)=x$$. But the action of the group is properly discontinuous, so $$\rho(g)x=x$$ implies $$g=1$$. Therefore the kernel of $$F$$ is trivial.

### Examples

(14.00) From this theorem, we can read off the following fundamental groups:

• $$\pi_1(S^1)=\mathbf{Z}$$ because $$S^1=\mathbf{R}/\mathbf{Z}$$,
• $$\pi_1(T^n)=\mathbf{Z}^n$$ because $$T^n=\mathbf{R}^n/\mathbf{Z}^n$$,
• $$\pi_1(\mathbf{RP}^n)=\mathbf{Z}/2$$ because $$\mathbf{RP}^n=S^n/(\mathbf{Z}/2)$$.

(14.43) Consider the transformations $$g,h\colon\mathbf{R}^2\to\mathbf{R}^2$$ defined by

\begin{align*} g(x,y)&=(x+1,1-y)\\ h(x,y)&=(x,y+1). \end{align*}

These generate a subgroup $$G$$ of isometries of $$\mathbf{R}^2$$ and the quotient $$\mathbf{R}^2/G$$ is the Klein bottle $$K$$. I claim that the action is properly discontinuous and that the group $$G$$ has the presentation $$\langle g,h\ |\ hg=gh^{-1}\rangle$$, which implies that the fundamental group of the Klein bottle is $$\pi_1(K)=G$$.

(17.46) To see this, we check

\begin{align*} gh^{-1}(x,y)&=g(x,y-1)=(x+1,2-y)\\ hg(x,y)&=h(x+1,1-y)=(x+1,2-y). \end{align*}

This is the only relation we need in the group because it can be used to put all of the factors of $$h$$ to the right, so all elements of $$G$$ can be written as $$g^mh^n$$. This means that any other relation in the group would need to have the form $$g^mh^n=1$$, but we can check that $$g^mh^n=1$$ implies $$m=n=0$$. In fact, we can check that $g^mh^n(x,y)=g^m(x,y+n)=\begin{cases}(x+m,1-y-n)\mbox{ if }m\mbox{ odd}\$$x+m,y+n)\mbox{ if }m\mbox{ even.}\end{cases}$ This equals the identity if and only if \(m=n=0$$ and, in fact, $d\left((x,y),g^mh^n(x,y)\right)\geq 1\mbox{ if }(m,n)\neq (0,0),$ which implies that the $$G$$-action is properly discontinuous.

## Pre-class questions

1. Why is the map $$F$$ in the proof of the theorem well-defined?
2. In the computation of the fundamental group of the Klein bottle, I claimed that $d\left((x,y),g^mh^n(x,y)\right)\geq 1\mbox{ if }(m,n)\neq (0,0).$ Can you prove this?