7.02 Path-lifting, monodromy


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


Monodromy: intuition

(0.21) Consider the 3-to-1 cover of the circle by the circle \(p(e^{i\theta})=e^{i3\theta}\). The preimage of each point consists of three points, and monodromy means looking at what happens to those three points as we move around the circle. For example, suppose that \(p^{-1}(1)=\{a,b,c\}\); as we go anticlockwise around the loop in the base, we see that \(a\) moves around a path and ends up at \(b\), \(b\) moves around a path to \(c\) and \(c\) moves around a path and ends up at \(a\), so we get a cyclic permutation \((a,b,c)\). Iterating the loop in the base means iterating the permutation, so we get a homomorphism \(\pi_1(S^1,1)\to Perm(p^{-1}(\{a,b,c\}))=S_3\) (which sends our loop to \((abc)\)). Since \((abc)\) is nontrivial in \(S_3\), this means that our loop must be nontrivial in \(\pi_1\).

(4.40) More generally, monodromy is an action of \(\pi_1(X,x)\) on \(p^{-1}(x)\), that is a homomorphism \(\pi_1(X,x)\to Perm(p^{-1}(x))\). It gives a way to detect nontrivial elements in \(\pi_1(X,x)\) (if they act nontrivially). Moreover, there is a covering space (the universal cover) in which all elements of \(\pi_1\) act nontrivially.

(6.20) We need to explain what it means for \(a\) to go around a path that ends up at \(b\) (this will be justified by the path-lifting lemma). We also need to explain why the monodromy around a loop only depends on the homotopy class of that loop (which will be justified by the homotopy-lifting lemma).


(6.45) Suppose that \(p\colon Y\to X\) is a covering map, let \(\delta\colon[0,1]\to X\) be a path with \(\delta(0)=x\) and let \(y\in p^{-1}(x)\). Then there exists a unique path \(\gamma\colon[0,1]\to Y\) which lifts \(\delta\), in the sense that \(p\circ\gamma=\delta\), and satisfies the initial condition \(\gamma(0)=y\).

In our earlier example, \(p(e^{i\theta})=e^{i3\theta}\), the path \(\delta\) is the loop in the base, the initial condition \(a\) gives a path \(\gamma\) which ends at \(b\), the initial condition \(b\) gives a path \(\gamma\) which ends at \(c\) and the initial condition \(c\) gives a path \(\gamma\) which ends at \(a\).

(9.54) Let \(\mathcal{U}\) be a collection of elementary neighbourhoods for \(p\) covering the whole of \(X\) (recall that elementary neighbourhoods are the neighbourhoods on which local inverses are defined). Then \(\{\delta^{-1}(U)\ :\ U\in\mathcal{U}\}\) is an open cover of \([0,1]\), so has a finite subcover. There is therefore a finite subdivision \(0=t_0\leq t_1\leq \cdot\leq t_n=1\) such that \(\delta_k:=\delta|_{[t_k,t_{k+1}]}\) lands in \(U_k\) for some elementary neighbourhood \(U_k\).

(12.44) We will construct \(\gamma\) by induction on \(k\).

\(k=0\): We need to have \(\gamma(0)=y\) in the end. Since \(p\) is a covering map, we have a local inverse \(q_0\colon U_0\to Y\) to \(p\) such that \(q_0(x)=y\) and \(p\circ q_0=id|_{U_0}\), so if we define \(\gamma_0=q_0\circ\delta_0\). Now \(\gamma_0\) is a lift of \(\delta_0\).

(14.52) Suppose we have constructed \(\gamma_0,\ldots,\gamma_{k-1}\) and we wish to construct \(\gamma_k\colon[t_k,t_{k+1}]\to Y\). In order for \(\gamma\) to be continuous, we need \(\gamma_{k}(t_k)=\gamma_{k-1}(t_k)\). There exists \(q_k\colon U_k\to Y\) such that \(q_k(\delta(t_k))=\gamma_{k-1}(t_k)\), so define \(\gamma_k=q_k\circ\delta_k\). This extends \(\gamma\) as a lift of \(\delta\) continuously to the interval \([t_k,t_{k+1}]\).

(16.35) Define \(\gamma(t)=\gamma_k(t)\) if \(t\in[t_k,t_{k+1}]\). The result is continuous because a piecewise-defined function which is continuous on pieces and agrees on overlaps is continuous. It is a lift because \(p\circ\gamma=p\circ q_k\circ\delta_k=\delta_k\) on \([t_k,t_{k+1}]\) (as \(p\circ q_k=id_{U_k}\)).

This gives existence of lifts. Uniqueness will be proved in the next video.

Pre-class questions

  1. Suppose that \(p\colon Y\to X\) is a covering map. Is it true that for all \(x_0,x_1\in X\), there is a bijection \(p^{-1}(x_0)\to p^{-1}(x_1)\)? Give a proof or a counterexample. What if \(x_0\) and \(x_1\) are connected by a path?


CC-BY-SA, Jonny Evans 2017