7.05 Fundamental group of the circle


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


Fundamental group of the circle

(0.00) In this section, we will see a proof that \(\pi_1(S^1,1)=\mathbf{Z}\).

Given an integer \(n\), define the loop \(\delta_n(t):=e^{2\pi int}\).

Define the homomorphism \(F\colon\mathbf{Z}\to\pi_1(S^1,1)\) by \(F(n)=[\delta_n]\). We need to check:

  • this is a homomorphism,
  • \(F\) is injective,
  • \(F\) is surjective.

\(F\) is a homomorphism

(1.36) \(F(0)=[\delta_0]\) and \(\delta_0(t)=e^0=1\) is the constant loop, so \(F\) takes the identity to the identity.

(2.23) To see that \(F(m+n)=F(m)\cdot F(n)\), we need to check that \(\delta_n\cdot\delta_m\simeq\delta_{m+n}\) [NOTE: I got my concatenation the wrong way around in the video]. Let \(p\colon\mathbf{R}\to S^1\) be the covering map \(p(x)=e^{ix}\) and let \(\tilde{\delta}_n\) be the unique lift of \(\delta_n\) with the initial condition \(\tilde{\delta}_n(0)=0\), in other words \(\delta_n(t)=e^{i\tilde{\delta}_n(t)}\).

(4.03) Let \(\tilde{\delta}_m\) be the unique lift of \(\delta_m\) to the same covering space with \(\tilde{\delta}_m(0)=\tilde{\delta}_n(1)\). The concatenation \(\tilde{\delta}_n\cdot\tilde{\delta}_m\) makes sense and is a lift of \(\delta_n\cdot\delta_m\).

(5.23) Finally, let \(\tilde{\delta}_{m+n}\) be the unique lift of \(\delta_{m+n}\) to the same covering space with \(\tilde{\delta}_{m+n}(0)=0\).

(6.28) We have

\begin{align*} \tilde{\delta}_{m+n}(t)&=2\pi(m+n)t)\\ \tilde{\delta}_m(t)&=2\pi mt\\ \tilde{\delta}_n(t)&=2\pi(nt+m) \end{align*}

so \(\tilde{\delta}_n\cdot\tilde{\delta}_m\) and \(\tilde{\delta}_{m+n}\) are paths which start at \(0\) and end at \(2\pi(m+n)\).

(9.16) Since \(\mathbf{R}\) is simply-connected, any two paths with the same endpoints are homotopic rel endpoints, so \(\tilde{\delta}_m\cdot\tilde{\delta}_n\simeq\tilde{\delta}_{m+n}\) via a homotopy \(H\). The homotopy \(p\circ H\) is then a homotopy \(\delta_n\cdot\delta_m\simeq\delta_{m+n}\).

\(F\) is injective

(10.30) As \(F\) is a homomorphism, it suffices to show that \(n\in\ker(F)\) implies \(n=0\). We therefore want to show that if \(n\neq 0\) then \(\delta_n\) is not homotopic to the constant loop.

(11.18) For this, we will show that, for a suitable covering space, the monodromy \(\sigma_{\delta_n}\) is nontrivial. Let \(p\colon\mathbf{R}\to S^1\) be the covering space \(p(x)=e^{ix}\). For each \(2\pi k\in 2\pi \mathbf{Z}=p^{-1}(1)\), the path \(\tilde{\delta}_n(t)=2\pi(nt+k)\) is a lift of the loop \(\delta_n\) with initial condition \(2\pi k\), so \[\sigma_{\delta_n}(2\pi k)=\tilde{\delta}_n(1)=2\pi(n+k).\] We see that the monodromy \(\sigma_{\delta_n}\colon 2\pi\mathbf{Z}\to 2\pi\mathbf{Z}\) is \(2\pi k\mapsto 2\pi(k+n)\), which is nontrivial if \(n\neq 0\).

\(F\) is surjective

(15.07) Given a loop \(\gamma\colon[0,1]\to S^1\) based at \(1\in S^1\), we want to find an \(n\) such that \(\delta_n\simeq\gamma\). Take the same covering space \(p\colon\mathbf{R}\to S^1\) and lift \(\gamma\) to get a path \(\tilde{\gamma}\) in \(\mathbf{R}\) with \(\tilde{\gamma}(0)=0\) (so \(\gamma(t)=\exp(i\tilde{\gamma}(t))\)). If \(n=\tilde{\gamma}(1)/2\pi\) then \(n\in\mathbf{Z}\). Let \(\tilde{\delta}_n\) be the unique lift of \(\delta_n\) with \(\tilde{\delta}_n(0)=0\). Then \(\tilde{\delta}_n\) and \(\tilde{\gamma}\) are paths in \(\mathbf{R}\) connecting \(0\) to \(2\pi n\). Because \(\mathbf{R}\) is simply-connected, these paths are homotopic via a homotopy \(H\), which implies that \(\delta_n\simeq\gamma\) via the homotopy \(p\circ H\).

Pre-class questions

  1. What was the key property of the covering space \(p\colon\mathbf{R}\to S^1\) which made this proof work?


CC-BY-SA, Jonny Evans 2017