possible number of sheets for a Moebius band covering

Question

Let M be the Moebius band, identified by the quotient of $[0,1]\times [0,1]$ by the equivalence $(x,0) \sim (1-x,1)$.

Let $p: M\to M$ be a covering and $n$ its number of sheets.

Find the possible values of $n$.

what I did:

$M$ is path-connected so it is connected, all the fibers have the same cardinality.

$M$ is also compact so $n$ is finite.

My intuition is that any $n\in \Bbb Z$ would be valid, as the possible coverings are $\Bbb R/n\Bbb Z\cong \Bbb R/\Bbb Z\times \Bbb Z/ n\Bbb Z$.

Thank you for help and comments.

Answer 1

I think only an odd number of sheets are possible.

This is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $M\to M$ by first determining all connected coverings of $M$, and then detecting which ones have total space homeomorphic to $M$.

Classification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $X$ there is a bijection between isomorphism classes of connected covering spaces of $X$ and conjugacy classes of subgroups of $\pi_1(X)$. (See for example Theorem 1.38, page 67.)

This works by constructing a universal covering $\tilde{X}\to X$ so that the quotients $\tilde{X}/H$ represent all the connected coverings of $X$ as $H$ varies over conjugacy classes. The covering $\tilde{X}$ is characterized up to covering space isomorphism by being simply-connected.

Universal covering space of $M$: Recall $M\sim S^1$ so $\pi_1(M)\cong \mathbb{Z}$. The universal covering space of $M$ is $\mathbb{R}\times [0,1]$ and the action of $\mathbb{Z}$ is given by $n\cdot (x, t)= \big(x+n, f^{n}(t)\big)$ for $n\in\mathbb{Z}$ and where $f\colon [0,1] \to [0,1]$ is the "flip" homeomorphism given by $f(t)= 1-t$. (Visually, think about $\mathbb{R}\times[0,1]$ as an infinite strip of tape that you're applying to the Möbius strip, which is alternating "front" and "back" sides.)

Quotients of $\tilde{M}$: Every connected covering of $M$ is a quotient of the form $(\mathbb{R}\times [0,1])/n\mathbb{Z}$. For each $n$ a fundamental domain of the quotient is $[0,n]\times[0,1]$, and the quotient only depends on how we identify the subspaces $\{0\} \times [0,1]$ and $\{n\}\times [0,1]$. When $n$ is odd then $f^{n} = f$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $M$; on the other hand if $n$ is even then $f^{n}=id$ and so the quotient is actually the cylinder $(\mathbb{R}/n\mathbb{Z})\times [0,1]$.

Since these quotients make up all of the possible connected coverings of $M$, it follows that coverings of the form $M\to M$ can have any odd number of sheets.

Answer 2

Even though the question is already answered, I found an alternate argument that there are no even-sheeted covers using the first Stiefel-Whitney class of $TM$.

Recall that the tangent bundle of $M$ is non-orientable so $w_1(TM)\neq 0 \in H^1(M;\mathbb{Z}/2\mathbb{Z})$. An $n$-sheeted covering $p_n\colon M\to M$ is a local diffeomorphism and hence induces a bundle map $TM\to TM$, so by naturality of characteristic classes we have $w_1(TM) = p_n^*(w_1(TM))$. But the covering also restricts to an $n$-sheeted covering $S^1\to S^1$, which on cohomology $H^1(S^1;\mathbb{Z}) \to H^1(S^1;\mathbb{Z})$ induces multiplication by $n$. Since $S^1 \to M$ is a homotopy equivalence we also have

$$ p_n^* = n\cdot(-)\colon H^1(M;R) \to H^1(M;R) $$

for any $R$. In particular if such a cover exists when $n = 2k$ is even then $$w_1(TM) = p_n^*(w_1(TM)) = 2k \cdot w_1(TM) = 0$$ which contradicts $w_1(TM) \neq 0$.

Edit: I'm currently trying to modify this argument into a proof of the following:

Conjecture: If $M$ is a non-orientable smooth manifold then there are no even-sheeted coverings of the form $M\to M$.

Update: This general conjecture is NOT true. Consider the double-cover $$p\colon S^1\times \mathbb{RP}^2\to S^1 \times \mathbb{RP}^2$$ given by $p(z, x) = (z^2, x)$.