7.04 Homotopy lifting, monodromy

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

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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

$y$ .

$\gamma_1\simeq\gamma_2$ 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

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?
Why does monodromy around a loop depend only on the homotopy class of the loop?

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Previous video: 7.03 Path-lifting: uniqueness.
Next video: 7.05 Fundamental group of the circle.
Index of all lectures.