# 7.04 Homotopy lifting, monodromy

## Video

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

## Notes

### Monodromy

(0.00) In the last couple of sections, we have seen the phenomenon of path-lifting for covering spaces $$p\colon Y\to X$$: given a path $$\gamma$$ in $$X$$ and a point $$y\in p^{-1}(\gamma(0))$$ there exists a unique lifted path $$\tilde{\gamma}$$ in $$Y$$ such that $$p\circ\tilde{\gamma}=\gamma$$ and $$\tilde{\gamma}(0)=y$$.

(1.07) If I start with a loop $$\gamma$$ in $$X$$ then there is no guarantee that the lift $$\tilde{\gamma}$$ will be a loop: it might just be a path. This leads to the idea of monodromy. For example, if $$p\colon S^1\to S^1$$ is the double cover $$p(e^{i\theta})=e^{i2\theta}$$ then the path $$e^{i\pi t}$$ lifts the loop $$e^{i2\pi t}$$.

(1.53) Given a loop $$\gamma$$ in $$X$$ based at $$x$$, define the monodromy around $$\gamma$$ $$\sigma_\gamma$$ to be the permutation $$\sigma_\gamma\colon p^{-1}(x)\to p^{-1}(x)$$ defined by $\sigma_\gamma(y)=\tilde{\gamma}(1),$ where $$\tilde{\gamma}$$ is the unique lift of $$\gamma$$ starting at $$y$$.

If $$\gamma_1\simeq\gamma_2$$ then $$\sigma_{\gamma_1}=\sigma_{\gamma_2}$$.

This will follow from the homotopy lifting lemma below.

### Homotopy lifting lemma

(4.25) Suppose that $$p\colon Y\to X$$ is a covering map and that $$\gamma_s$$ is a homotopy of paths rel endpoints in $$X$$ (i.e. $$\gamma_s(0)$$ and $$\gamma_s(1)$$ are independent of $$s$$). Then, given a lift $$\tilde{\gamma}_0$$ of $$\gamma_0$$, there exists a lifted homotopy $$\tilde{\gamma}_s$$ rel endpoints, that is a homotopy of $$\tilde{\gamma}_0$$ such that $$p\circ\tilde{\gamma}_s=\gamma_s$$ and such that $$\tilde{\gamma}_s(0)$$ and $$\tilde{\gamma}_s(1)$$ are independent of $$s$$.

(6.58) Let $$H\colon[0,1]\times[0,1]\to X$$ be the homotopy in $$X$$ (i.e. $$\gamma_s(t)=H(s,t)$$). We can subdivide the square into rectangles $$R_{ij}$$ such that $$H(R_{ij})\subset U_{ij}$$ for some elementary neighbourhood $$U_{ij}$$. Here, the index $$i$$ runs along the $$x$$-direction and the index $$j$$ runs along the $$y$$-direction.

(9.06) The strategy to construct the lifted homotopy $$\tilde{H}$$ (such that $$\tilde{H}(s,t)=\tilde{\gamma}_s(t)$$) is to pick local inverses $$q_{ij}\colon U_{ij}\to Y$$ for $$p$$ and set $$\tilde{H}|_{R_{ij}}=q_{ij}\circ H|_{R_{ij}}$$. Certainly this defines a lift of $$H$$ because $$p\circ\tilde{H}=p\circ q_{ij}\circ H=H$$, however, we need to pick $$q_{ij}$$ carefully to ensure that the map $$\tilde{H}$$ constructed in this way is continuous. We need to ensure that $$\tilde{H}|_{R_{ij}}$$ and $$\tilde{H}|_{R_{kl}}$$ agree along the overlap $$R_{ij}\cap R_{kl}$$.

(11.13) For simplicity, let's imagine that we only have a two-by-two grid of squares. We will start by constructing $$q_{1j}$$. To begin with, we know that $$\tilde{H}(0,t)=\tilde{\gamma}_0(t)$$ for a given lift $$\tilde{\gamma}_0$$. This tells us which $$q_{ij}$$ to use on the rectangles $$R_{11},R_{12}$$: namely, $$q_{11}\colon U_{11}\to Y$$ is the unique local inverse with $$q_{11}(\gamma_0(0))=\tilde{\gamma}_0(0)$$; $$q_{12}\colon U_{12}\to Y$$ is the unique local inverse with $$q_{12}(\gamma_0(t_1))=\tilde{\gamma}_0(t_1)$$ where $$t_1$$ is the value of the $$t$$-coordinate at the top of the rectangle $$R_{11}$$.

(14.15) We need to check that the results $$\tilde{H}_{11}:=q_{11}\circ H$$ and $$\tilde{H}_{12}:=q_{12}\circ H$$ agree along the overlap $$R_{11}\cap R_{12}$$ (which is an edge comprising points of the form $$(s,t_1)$$ with $$s\in[0,s_1]$$). By construction, we have $$\tilde{H}_{11}(0,t_1)=\tilde{H}_{12}(0,t_1)=\tilde{\gamma}_0(t_1)$$ and the restrictions $$\tilde{H}_{1j}(s,t_1)$$ to the edge $$R_{11}\cap R_{12}$$ define paths; these paths both lift $$H(s,t_1)$$ and have the same initial condition $$\tilde{H}_{11}(0,t_1)=\tilde{H}_{12}(0,t_1)$$. Therefore, by uniqueness of path-lifting, we have $$\tilde{H}_{11}(s,t_1)=\tilde{H}_{12}(s,t_1)$$ for all $$s\in[0,s_1]$$.

(18.02) We can do the same trick starting with $$\tilde{\gamma}_{s_1}$$ and extend our homotopy over $$R_{2j}$$ and, by induction, over the whole square.

(19.06) We need to check that $$\tilde{H}$$ is a homotopy rel endpoints, i.e. $$\tilde{H}(s,0)$$ and $$\tilde{H}(s,1)$$ should be constant. But $$\tilde{H}(s,0)$$ lifts $$H(s,0)$$, which is constant. The constant lift is also a lift, so by uniqueness of path-lifting, it must agree with $$\tilde{H}(s,0)$$, which must therefore be constant. Similarly for $$\tilde{H}(s,1)$$.

## Pre-class questions

1. In the proof of the homotopy lifting lemma, why can we subdivide the square into rectangles $$R_{ij}$$ such that each $$H(R_{ij})$$ is contained in an elementary neighbourhood?
2. Why does monodromy around a loop depend only on the homotopy class of the loop?

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