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A problem asks me to find all the covering spaces of a Klein bottle. This needs to calculate all the subgroups of the fundamental group of the Klein bottle. But I don't have any idea how to do it.

I googled it and an article says

The subgroups of the fundamental group of the Klein bottle are either trivial, free of rank one, free Abelian of rank two, or non-Abelian of rank two.

I don't know how to get the result and what is the concrete form of the subgroups (which is needed to calculate the covering spaces.)

Can you please help? Thank you.

RobPratt
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Roun
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    As I recall, the best way to solve this particular problem is to figure out what the covering spaces of the Klein bottle are geometrically, and then use that to deduce what the subgroups of the fundamental group are. – Charles Staats Apr 21 '11 at 16:08
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    I agree with Charles. In particular, every covering space is a quotient of the universal cover, and there aren't too many possibilities for what this can be. – Qiaochu Yuan Apr 21 '11 at 17:31
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    The Klein bottle group is a semi-direct product $\mathbb Z \rtimes \mathbb Z$ where $\mathbb Z$ acts on $\mathbb Z$ by its sole non-trivial involution. So any subgroup of the Klein bottle group is a semi-direct product $A \rtimes B$ where $A, B \subset \mathbb Z$, and there's basically just three possibilities for each of $A$ and $B$, up to isomorphism and the action of $B$ on $A$, etc.. – Ryan Budney Apr 21 '11 at 17:57
  • @Charles @Qiaochu : But the problem is to find all covering spaces. If I don't know all the subgroups, how can I know I haven't miss something? – Roun Apr 22 '11 at 00:56
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    @Ryan: Well, I'm not quite familiar with semi-direct product. But your statement is not true when it is direct product. Not all subgroups of $\mathbb{Z}\times\mathbb{Z}$ are in the form $A\times B$ where $A,B\subset\mathbb{Z}$. – Roun Apr 22 '11 at 01:06
  • @Roun: I think you mis-read my comment. In the direct product case the translation of what I said is that all subgroups of $\mathbb Z \times \mathbb Z$ are isomorphic to groups of the form $A \times B$ where $A, B \subset \mathbb Z$ are subgroups. – Ryan Budney Apr 22 '11 at 03:51
  • @Roun - I suggest you read Chapter 1.3 of Hatcher's book. In particular Proposition 1.31, Proposition 1.36 – Juan S Apr 22 '11 at 05:21
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    @Ryan: Uh..Do you mean that in the semi-direct product case, the form $A\rtimes B$ is not up to isomorphism? I think only knowing the isomorphism type is not enough to calculate the covering space. – Roun Apr 22 '11 at 10:14
  • Once you know the group type (up to isomorphism), computing the inclusion map is easy enough, so "computing" the covering map is a relatively standard step from there. – Ryan Budney Apr 22 '11 at 17:22

1 Answers1

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Write every element of $\mathbb{Z} \rtimes \mathbb{Z}$ as $x^m y^n$ for unique integers $m,n$, such that the group operation is given by $$x^{m_1} y^{n_1} \cdot x^{m_2} y^{n_2} = x^{m_1 + (-1)^{n_1} m_2} y^{n_1 + n_2}.$$ Define $X = \langle x \rangle \cong \mathbb{Z}$ and $Y = \langle y \rangle \cong \mathbb{Z}$. If $H$ is an arbitrary subgroup of $G$, it should be fairly easy to verify that we can find nonnegative $A,b$ such that $H \cap X = \langle x^A \rangle$ and $XH \cap Y = \langle y^b \rangle$. The latter set is also the set of all $y$-coordinates of $H$, namely $\{y^n \mid \exists m : x^m y^n \in H\}$. (Hint: it is the image of $H$ under the projection map $\mathbb{Z} \rtimes \mathbb{Z} \to Y$.)

With these numbers, you can prove the following things.

  1. If all of $A = b = 0$, then $H = 1$.
  2. If $b =0$ and $A > 0$ or $b > 0$ and $A = 0$, then $H = \langle x^a y^b \rangle \cong \mathbb{Z}$ for some $a \in \mathbb{Z}$.
  3. If $A,b > 0$, then $H = \langle x^A, x^a y^b \rangle$ for some $a \in \{0, \ldots, A-1\}$ and $H$ is a subgroup of index $Ab$. If $2 \mid b$, then $H \cong \mathbb{Z}^2$, with isomorphism $$x^A \mapsto (1,0), \: x^a y^b \mapsto (0,1).$$ If $2 \nmid b$, then $H \cong \mathbb{Z} \rtimes \mathbb{Z}$ with isomorphism $$x^A \mapsto x, \: x^a y^b \mapsto y.$$

The proofs of these facts are all relatively straightforward, though slighly tedious. I found it helpful to first establish an expression of the $k$-th power $(x^m y^n)^k$. If $n$ is even, then it is simply $x^{km} y^{kn}$. But if $n$ is odd, then it is $y^{kn}$ if $k$ is even and $x^m y^{kn}$ if $k$ is odd.

Neckverse Herdman
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