# is every path-connected covering of the Moebius strip a Galois cover?

Let $$p : E → M$$ be a covering of $$M$$ the Moebius Sttrip such that $$E$$ is path connected.

Is this a Galois covering?

My intuition is there must be some non locally path connected coverings that are not Galois. I know little about coverings of Moebius band. I know that its universal covering is $$\Bbb R\times [0,1]$$.

• Since $M$ deformation retracts onto $S^1$ and a space's theory of covering spaces is determined by its fundamental group, I think you can equivalently ask the same for $S^1$ to simplify things. – William Feb 22 at 18:23
• Locally, a covering will look like the Moebius strip, hence it can't be nonlocally path connected. Now by William's comment, the fundamental group is abelian, so path connected coverings are all Galois – Max Feb 22 at 18:53

Hint: The inclusion $$S^1 \to M$$ as the zero section induces a bijection

$$Cov(M)/\cong \to Cov(S^1)/\cong$$

where $$Cov(X)/\cong$$ is the set of isomorphism classes of covering spaces. Moreover a covering space of $$M$$ is connected (respectively regular) iff its restriction to $$S^1$$ is connected (resp. regular).

Solution to problem over $$S^1$$:

The connected covering spaces of $$S^1$$ are isomorphic to quotients of $$\mathbb{R}$$ by subgroups of $$\mathbb{Z}$$. The $$n$$-fold cover $$\mathbb{R}/n\mathbb{Z} \to S^1$$ has automorphism group $$\mathbb{Z}/n\mathbb{Z}$$ which acts transitively in each fibre.

I think you can run a similar argument on the universal cover $$\mathbb{R}\times [0,1]$$ of $$M$$. The covering is the quotient map of the relation $$(x, t) \sim (x + n, f^{(n)}(t))\text{ for }n\in\mathbb{Z}$$ where $$f$$ is the "flip" homeomorphism of the interval. The deck transformations are again $$\mathbb{Z}$$, which acts by $$n\cdot(x, t) = (x + n, f^{(n)}(t))$$.

Update: as per Max's comment to the original question, this also follows from the general theory: if $$X$$'s fundamental group is abelian then all its connected covers are regular.

The idea is that all the connected covers are quotients of the universal cover $$p\colon\tilde{X}\to X$$ by subgroups of $$G=\pi_1(X)$$. If $$H$$ is a subgroup of $$G$$ then the fibres of $$\tilde{X}/H \to X$$ will have an action of $$G$$ that looks like its action on the quotient set $$G/H$$; if moreover $$H$$ is normal then this action descends to the group $$G/H$$ and the fibres are "$$G/H$$-torsors" (i.e. have a free, transitive action of $$G/H$$) and in fact $$G/H$$ is isomorphic to the group of deck transformations, hence the covering is regular. Therefore if all subgroups are normal then all connected covers will be regular.

• Thank you @William. I am trying to understand why connected covers are quotients of the universal cover by subgroups of $G=π_1(X)$. Is it because of the bijection between coverings isomorphism classes and $π_1(X)$ subgroups conjugacy classes? the theory behind is not clear in my head. – PerelMan Feb 23 at 0:12
• Yes it is because of this bijection. Given a subgroup $H$ we get a connected covering $\tilde{X}/H$, and if we had chosen a conjugate subgroup $H'$ the resulting covering spaces would be isomorphic. Conversely, given a connected covering space $p\colon P\to X$ it is isomorphic to the quotient $\tilde{X}/p_*(\pi_1 P)$. For a reference on the relation between normality of $H$ and regularity of $\tilde{X}/H$, check out Proposition 1.39 in the Fundamental Group chapter of Hatcher (page 71). – William Feb 23 at 2:40
• Actually both of the sections "Classification of Covering Spaces" and "Deck Transformations and Group Actions" in Hatcher (starting page 63) are very relevant to this discussion. – William Feb 23 at 3:17