7.02 Path-lifting, monodromy

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

Notes

Monodromy: intuition

(0.21) Consider the 3-to-1 cover of the circle by the circle p(eiθ)=ei3θ. 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 π1(S1,1)→Perm(p−1(1))=S3 (which sends our loop to (abc)). Since (abc) is nontrivial in S3, this means that our loop must be nontrivial in π1.

(4.40) More generally, monodromy is an action of π1(X,x) on p−1(x), that is a homomorphism π1(X,x)→Perm(p−1(x)). It gives a way to detect nontrivial elements in π1(X,x) (if they act nontrivially). Moreover, there is a covering space (the universal cover) in which all elements of π1 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:Y→X is a covering map, let δ:[0,1]→X be a path with δ(0)=x and let y∈p−1(x). Then there exists a unique path γ:[0,1]→Y which lifts δ, in the sense that p∘γ=δ, and satisfies the initial condition γ(0)=y.

In our earlier example, p(eiθ)=ei3θ, the path δ is the loop in the base, the initial condition a gives a path γ which ends at b, the initial condition b gives a path γ which ends at c and the initial condition c gives a path γ which ends at a.

(9.54) Let 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 {δ−1(U) : U∈U} is an open cover of [0,1], so has a finite subcover. There is therefore a finite subdivision 0=t0≤t1≤⋅≤tn=1 such that δk:=δ|[tk,tk+1] lands in Uk for some elementary neighbourhood Uk.

(12.44) We will construct γ by induction on k.

k=0: We need to have γ(0)=y in the end. Since p is a covering map, we have a local inverse q0:U0→Y to p such that q0(x)=y and p∘q0=id|U0, so if we define γ0=q0∘δ0. Now γ0 is a lift of δ0.

(14.52) Suppose we have constructed γ0,…,γk−1 and we wish to construct γk:[tk,tk+1]→Y. In order for γ to be continuous, we need γk(tk)=γk−1(tk). There exists qk:Uk→Y such that qk(δ(tk))=γk−1(tk), so define γk=qk∘δk. This extends γ as a lift of δ continuously to the interval [tk,tk+1].

(16.35) Define γ(t)=γk(t) if t∈[tk,tk+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∘γ=p∘qk∘δk=δk on [tk,tk+1] (as p∘qk=idUk).

This gives existence of lifts. Uniqueness will be proved in the next video.

Pre-class questions

  1. Suppose that p:Y→X is a covering map. Is it true that for all x0,x1∈X, there is a bijection p−1(x0)→p−1(x1)? Give a proof or a counterexample. What if x0 and x1 are connected by a path?

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