# possible number of sheets for a Moebius band covering

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.

• Something of a side note, but that isomorphism you write is not correct. The left hand side is isomorphic to the circle, while the right is $n$ disjoint circles. – WSL Feb 23 at 2:50
• This is closely related to what we talked about in your other question. Remember that the universal cover of $M$ is $\tilde{M}=\mathbb{R}\times [0,1]$ and the deck transformations act by translating and flipping. All the connected coverings of $M$ are quotients of $\tilde{M}$ by subgroups of $\mathbb{Z}$, so you should try to determine for which values of $n$ the quotient $\tilde{M}/n\mathbb{Z}$ is homeomorphic to $M$. – William Feb 23 at 4:23

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.

• Thanks @William for your thorough answer. I see now why you say connected instead of path-connected. It boild down to the same as local path-connectedness is inherited by the coverings – PerelMan Feb 23 at 17:45
• Yes that's true. Since we already know $M$ is locally path-connected, so will be any covering. Therefore every connected covering of $M$ is already path-connected. – William Feb 23 at 19:40

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 $$p(z, x) = (z^2, x)$$.

• Thank you @William interesting alternate argument! I only studied some elementary De Rham cohomology, so not able to understand all of it but will read about the ring cohomology. I am interested in any references. – PerelMan Feb 23 at 21:09