# 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 \(y\).

## 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?

# Navigation

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