§ Covering spaces

§ Covering spaces: Intuition

• Consider the map $p(z) = z^2 : \mathbb C^\times \rightarrow \mathbb C^\times$. This is a 2-to-1 map. We can try to define an inverse regardless.
• We do define a "square root" if we want. Cut out a half-line $[0, -infty)$called $B$ for branch cuts. We get two functions on $q_+, q_-: \mathbb C - B \rightarrow \mathbb C^\times$, such that $p(q_+(z)) = z$. Here, we have $q_- = - q_+$.
• The point of taking the branch cut is to preserve simply connectedness. $\mathbb C^\times$is not simply connected, while $\mathbb C/B$ is simply connected! (This seems so crucial, why has no one told me this before?!)
• Eg 2: exponential. Pick $exp: \mathbb C \rightarrow \mathbb C^\times$. This is surjective, and infinite to 1. $e^{z + 2 \pi n} = e^{iz}$.
• Again, on $\mathbb C / B$, we have $q_n \equiv \log + 2 \pi i n$, such that $exp(q_n(z)) = z$.
• A covering map is, roughly speaking, something like the above. It's a map that's n-to-1, which has n local inverse defined on simply connected subsets of the target.
• So if we have $p: Y \rightarrow X$, we have $q: U \rightarrow Y$ (for $U \subseteq X$) such that $p(q(z)) = z, \forall z \in U$.

§ Covering spaces: Definition

• A subset $U \subset X$ is a called as an elementary neighbourhood if there is a discrete set $F$ and a homeomorphism $h: p^{-1}(U) \rightarrow U \times F$such that $p|_{p^{-1}(U)}(y) = fst(h)$ or $p|_{p^{-1}(U)} = pr_1 \circ h$.
• Alternative definition A subset $U \subset X$ is called as evenly covered/elementary nbhd if $p^{-1}(U) = \sqcup \alpha V_\alpha$ where the $V_\alpha$ are disjoint and open, and $p|_{V_\alpha} : V_\alpha \rightarrow U$ is a homeomorphism for all $\alpha$.
• An elementary neighbourhood is the region where we have the local inverses (the complement of a branch cut).
• We get for each $i \in F$ , a map $q_i : U \rightarrow U \times F; q_i(x) = (x, i)$and then along $h^{-1}$ sending $h^{-1}(x, i) \in p^{-1}(U)$.
• We say $p$ is a covering map if $X$ is covered by elementary neighbourhoods.
• We say $V \subseteq Y$ is an elementary sheet if it is path connected and $p(V)$is an elementary neighbourhood.
• So, consider $p(x) = e^{ix}: \mathbb R \rightarrow S^1$. If we cut the space at $(0, 0)$, then we will have elementary neighbourhood $S^1 - \{(0, 0)\}$and elementary sheets $(2 \pi k, 2 \pi+1)$.
• The point is that the inverse projection $p^{-1}$ takes $U$ to some object of the form $U \times F$: a local product! So even though the global covering space $\mathbb R$ does not look like a product of circles, it locally does. So it's some sort of fiber bundle?

Slogan: Covering space is locally disjoint copies of the original space.

§ Path lifting and Monodromy

• Monodromy is behaviour that's induced in the covering space, on moving in a loop in a base.
• Etymology: Mono --- single, drome --- running. So running in a single loop / running around a single time.
• Holonomy is a type of monodromy that occurs due to parallel transport in a loop, to detect curvature
• Loop on the base is an element of $\pi_1(X)$.
• Pick some point $x \in X$. Consider $F \equiv \pi^{-1}(x)$ ( $F$ for fiber).
• Now move in a small loop on the base, $\gamma$. The local movement will cause movement of the elements of the fiber.
• Since $\gamma(1) = \gamma(0)$, the elements of the fiber at the end of the movement are equal to the original set $F$.
• So moving in a loop induces a permutation of the elements of the fiber $F$.
• Every element of $\pi_1(X)$ induces a permutation of elements of the fiber $F$.
• This lets us detect non-triviality of $\pi_1(X)$. The action of $\pi_1(X)$ on the fiber lets us "detect" what $\pi_1(X)$ is.
• We will define what is means to "move the fiber along the path".

§ Path lifting lemma

Theorem :Suppose $p: y \rightarrow X$ is a covering map. Let $\delta: [0, 1] \rightarrow X$ be a path such that $\delta(0) = x$, and let $y \in p^{-1}(x)$ [ $y$ is in the fiber of $x$]. Then there is a unique path $\gamma: [0,1] \rightarrow Y$ which "lifts" $\delta$. That is, $\delta(p(y)) = \gamma(y)$, such that $\gamma(0) = Y$.

Slogan: Paths can be lifted. Given how to begin the lift, can be extended all the way.

• Let $N$ be a collection of elementary neighbourhoods of $X$.
• $\{ \delta^{-1}(U) : U \in N \}$ is an open cover (in the compactness sense) of $[0, 1]$.
• By compactness, find a finite subcover. Divide interval into subintervals $0 = t_0 < t_1 < \dots t_n = 1$such that $\delta|k = \delta|_{[t_k, t_{k+1}]}$ lands in $U_k$, an elementary neighbourhood.
• Build $\gamma$ by induction on $k$.
• We know that $\gamma(0)$ should be $y$.
• Since we have an elementary neighbourhood, it means that there are a elementary sheets living over $U_0$, indexed by some discrete set $F$. $y$ lives in one of thse sheets. We have local inverses $q_m$. One of them lands on the sheet of $y$, call it $q$. So we get a map $q: U_0 \rightarrow Y$ such that $q(x) = y$.
• Define $\gamma(0) \equiv q(\delta(0)) = q(x) = y$.
• Extend $\gamma$ upto $t_1$.
• Continue all the way upto $t_k$.
• To get $\gamma$ from $(t_k, t_{k+1}$, there exists a $q_k: U_k \rightarrow Y$such that $q_k(\delta(t_k)) = \gamma(t_k)$. Define $\gamma(t_k \leq t \leq t_{k_1}) \equiv q_k(\delta(t_k))$.
• This is continuous because $\delta$ continuous by definition, $q_k$ continuous by neighbourhood, $\gamma$ is pieced together such that endpoints fit, and is thus continuous.
• Can check this is a lift! We get $p \circ \gamma = p \circ q_k \circ \delta_k$. Since $q_k$ is a local inverse of $p$, we get $p \circ \gamma = \delta_k$in the region.

§ 7.03: Path lifting: uniqueness

If we have a space $X$ and a covering space $Y$, for a path $gamma$ that starts at $x$, we can find a path $\gamma'$ which starts at $y \in p^{-1}(x)$ and projects down to $\gamma$: $\gamma(t) = p(\gamma'(t))$. We want to show that this path lift is unique

§ Lemma

Let $p: Y \rightarrow X$ be a covering space. Let $T$ be a connected space Let $F: T \rightarrow X$ be a continuous map (for us, $T \simeq [0, 1]$). Let $F_1, F_2: T \rightarrow Y$ be lifts of $F$ ( $p \circ F_1 = F$, $p \circ F_2 = F$). We will show that $F_1 = F_2$ iff the lifts are equal for some $t \in$T.

Slogan: Lifts of paths are unique: if they agree at one point, they agree at all points!

• We just need to show that if $F_1$ and $F_2$ agree somewhere in $Y$, they agree everywhere. It is clear that if they agree everywhere, they must agree somewhere.
• To show this, pick the set $S$ where $F_1, F_2$ agree in $Y$: $S \equiv \{ t \in T : F_1(t) = F_2(t) \}$.
• We will show that $S$ is open and closed. Since $T$ is connected, $S$ must be either the full space or the empty set. Since $S$ is assumed to be non-empty, $S = T$ and the two functions agree everywhere.
• (Intuition: if both $S$ and $S^c$ are open, then we can build a function that colors $T = S \cup S^c$in two colors continuously; ie, we can partition it continuously; ie the spaces must be disconnected. Since $T$ is connected, we cannot allow that to happen, hence $S = \emptyset$ or $S = T$.)
• Let $t \ in T$. Let $U$ be an evenly covered neighbourhood/elementary neighbourhood of $F(t)$ downstairs (in $X$). Then we have $p^{-1}(U) = \sqcup_\alpha V_\alpha$ such that $p|_V{\alpha}: V_\alpha \rightarrow U$is a local homeomorphism.
• Since $F_1, F_2$ are continuous, we will have opens $V_1, V_2$ in $V_\alpha$, which contain $F_1(t), F_2(t)$ upstairs (mirrroring $U$ containing $F(t)$ downstairs).
• The pre-images of $V_1$, $V_2$ along $F_1, F_2$ give us open sets $t \in T_1, T_2 \subseteq T$.
• Define $T* = T_1 \cap T_2$. If $F_1(t) \neq F_2(t)$, then $V_1 \neq V_2$and thus $F_1 \neq F_2$ on all of $T*$. So, $S^c = T*$ is open.
• If $F_1(t) = F_2(t)$, then $V_1 = V_2$ and thus $F_1 = F_2$ on $T*$(since $p \circ F_1 = F = p \circ F_2$, and $p$is injective within $U$, ie within $V_1, V_2$). So $S$ is open.
• Hence we are done, as $S$ is non-empty and clopen and is thus equal to $T$. Thus, the two functions agree on all of $T$.

§ Homotopy lifting, Monodromy

• Given a loop $\gamma$ in $X$ based at $x$ , the monodromy around $\gamma$ is a permutation $\sigma_\gamma : p^{-1}(x) \rightarrow p^{-1}(x)$, where $\sigma_{\gamma}(y) \equiv \gamma^y(1)$where $\gamma^y$ is the unique lift of $\gamma$ staring at $y$. We have that $\sigma_{\gamma} \in Perm(p^{-1}(x))$.
• Claim: if $\gamma_1 \simeq \gamma_2$ then $\sigma_{\gamma_1} = \sigma_{\gamma_2}$.
• We need a tool: homotopy lifting lemma.

Slogan: permutation of monodromy depends only on homotopy type

§ Homotopy lifting lemma/property of covering spaces

Suppose $p: Y \rightarrow X$ is a covering map and $\gamma_s$ is a homotopy of paths rel. endpoints ( $\gamma_s(0)$ and $\gamma_s(1)$ are independent of $s$ / endpoints are fixed throughout the homotopy). Then there exists for each lift $\gamma'_0 : [0, 1] \rightarrow Y$ of $\gamma_0:[0,1] \rightarrow X$ (ie, $p \circ \gamma'_0 = gamma_0$), a completion of the lifted homotopy $\gamma'_s: [0, 1] \rightarrow Y$ (ie, $p \circ gamma'_s = gamma_s$). Moreover, this lifted homotopy is rel endpoints: ie, the endpoints of $gamma'$ are independent of $s$.

Slogan: homotopy lifted at 0 can be lifted for all time

• Let $H: [0, 1] \times [0, 1] \rightarrow X$ be the homotopy in $X$ such that $H(s, t) = \gamma_s(t)$. Subdivide the square into rectangles $R_{ij}$ such that $H(R_{ij})$ is contained in $U_{ij}$ for some elementary neighbourhood $U_{ij}$. We build $H': [0, 1] \times [0, 1] \rightarrow Y$ by building local inverses $q_{ij} : U_{ij} \rightarrow Y$ such that $p \circ q_{ij} = R_{ij}$. We then set $H'|_{R_{ij}} = q_{ij} \circ H$.

• Reference video
• Notes for uniqueness of path lifting