# 7.02 Path-lifting, monodromy

## Video

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

## Notes

### 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).

### Path-lifting

(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?