7.04 Homotopy lifting, monodromy


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



(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?


CC-BY-SA, Jonny Evans 2017