Homology of $n$-sheeted covering space of Klein bottle

Question

Let $X$ be the Klein bottle, that is $X=\mathbb{R}^2/G$ with $$G=\langle a,b\mid a^{-1}b ab=1\rangle,$$

acting via $a: \mathbb{R}^2\to \mathbb{R}^2, (x,y)\mapsto (x+1,y)$, $b: \mathbb{R}^2\to \mathbb{R}^2,(x,y)\mapsto (-x,y+1)$.

Let $n$ be an odd positive integer and $Y\to X$ be an $n$-sheeted covering. What is $H_*(Y;\mathbb{Z})$?

How to solve this question?

Answer 1

Letting $\chi$ denote Euler characteristic, we have $\chi(Y) = n \cdot \chi(X)=0$. So $Y$ is a compact surface of Euler characteristic zero.

I'll add the assumption that $Y$ is connected, with a few words on the disconnected case afterwards.

By the classification of surfaces, $Y$ is either a torus or a Klein bottle, and we just have to figure out which one. I'll prove that $Y$ is a Klein bottle, whose homology one calculates to be $$H_0(Y;\mathbb{Z})=\mathbb{Z}, \,\, H_1(Y;\mathbb{Z}) = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}, \,\, H_2(Y;\mathbb{Z}) = 0 $$

The group $G$ acts on $\mathbb{R}^2$ by deck transformations with quotient $X$. Some elements of $G$ act preserving orientation of $\mathbb{R}^2$, and some act reversing orientation. Since the quotient $X$ is nonorientable, the orientation preserving elements form a normal subgroup $N < G$ of index $2$, in fact there is a surjective homomorphism $G \mapsto \mathbb{Z}/2\mathbb{Z}$ whose kernel is $N$.

By the covering space theory, there is an index $n$ subgroup $H < G$ such that the universal covering map $\mathbb{R}^2 \mapsto X$ lifts to a universal covering map $\mathbb{R}^2 \mapsto Y$ with deck transformation group $H$. The quotient $Y = \mathbb{R}^2/H$ is a torus if and only if $H$ has no orientation reversing elements, and it is a Klein bottle otherwise. So it remains to decide whether $H$ has any orientation reversing elements.

Restrict the homomorphism $G \mapsto \mathbb{Z}/2\mathbb{Z}$ to the subgroup $H$. If $H$ has no orientation reversing elements then this restriction is the trivial homomorphism and so $H < N$, implying the following equation of subgroup indices: $$n = [G:H] = [G:N] \, [N:H] = 2 \, [N:H] $$ and so $n$ is even, a contradiction.

It follows that $H$ contains orientation reversing elements, so $Y$ is a Klein bottle.

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Now a few words on the case when $Y$ is disconnected. By the same Euler characteristic argument, each component of $Y$ is a torus or Klein bottle. The degrees of the restrictions to the components partition the total degree $n$ as a sum of positive integers. Since $n$ can be partitioned in more than one way, there is not a unique solution to the problem in the disconnected case. Nonetheless one can still enumerate the possibilities, if that is desired: all of the components that cover $X$ with odd degree are Klein bottles, by the above argument; the remaining components, that cover $X$ with even degree, can be either tori or Klein bottles (each of which have a covering map over the Klein bottle of any given even degree).

Answer 2

Here is an outline on how one might approach this:

• The Klein bottle $X$ is a model for $K(\pi_1(X),1)$, whose only non-trivial homotopy group is its fundamental group.
• A covering space projection induces isomorphisms on homotopy groups of degree $n\geq 2$.
• Deduce that $Y$ is an Eilenberg-Maclane space $K(\pi_1(Y),1)$.

What do you know about $\pi_1(Y)$? How does one compute homology of Eilenberg-Maclane spaces?

How one approaches this last question depends on your background.