7.07 Group actions and covering spaces, 2


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


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.


(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?


CC-BY-SA, Jonny Evans 2017