# 7.02 Path-lifting, monodromy

## Video

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

- Previous video:
**7.01 Covering spaces**. - Next video:
**7.03 Path-lifting: uniqueness**. - Index of all lectures.

## 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

- 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?

## Navigation

- Previous video:
**7.01 Covering spaces**. - Next video:
**7.03 Path-lifting: uniqueness**. - Index of all lectures.