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I have to show there is a covering of the Klein Bottle by the Torus. I realize this has been answered here: Two-sheeted covering of the Klein bottle by the torus.

However, by the Galois Correspondence we know that covering maps of the Klein Bottle correspond bijectively with subgroups of the fundamental group of the Klein Bottle. If we let $T$ denote the Torus and $K$ the Klein Bottle, then $\pi_1(T) \cong \mathbb{Z} \times \mathbb{Z}$ and $\pi_1(K) \cong \langle a,b: abab^{-1} = 1 \rangle$. To show that there is a covering of the Torus by the Klein Bottle would it be enough to show that $\langle a,b: abab^{-1} = 1 \rangle$ has a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$? Moreover, if this is the case, is this an easier problem to handle?

William
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Mike
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    That would only follow if spaces corresponding to certain subgroups were unique up to homeomorphism, which I don't believe is true. – Charles Hudgins Jan 16 '20 at 00:10
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    Doesn't the correspondence tell us that the spaces are unique up to homeomorphism? – Mike Jan 16 '20 at 00:12
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    I'm trying to verify that right now. – Charles Hudgins Jan 16 '20 at 00:13
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    @CharlesHudgins It is true: see https://www.math.unl.edu/~mbrittenham2/classwk/872s07/lecnotes/notes2.covering.spaces.pdf end of page 6 to beginning of page 7. – Mike Jan 16 '20 at 00:18
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    It looks like I was wrong. See Theorem 1.38 in Hatcher. In fact, more is true. These spaces are not just homeomorphic, but isomorphic as covering spaces. Edit: Sorry, didn't see that you had already found a source. – Charles Hudgins Jan 16 '20 at 00:18
  • @CharlesHudgins Okay, is it easy to see that there $\mathbb{Z} \times \mathbb{Z}$ is a subgroup of $\langle a,b: abab^{-1} = 1\rangle$? I have tried a couple of things but have not found anything, although I do not know if there are any good techniques for a problem like this. – Mike Jan 16 '20 at 00:19
  • I would try to work with presentations. $\mathbb{Z} \times \mathbb{Z} \cong \langle a,b | ab = ba \rangle$. Does that help? – Charles Hudgins Jan 16 '20 at 00:23
  • @CharlesHudgins Eh, I do not know what $a,b$ should be in this presentation – Mike Jan 16 '20 at 00:24
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    One subgroup of $\langle a, b\ |\ abab^{-1} = 1\rangle$ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$ (indeed $ab^2= b a^{-1} b = b^2 a)$. – William Jan 16 '20 at 00:37
  • Also, in an answer to a related question (https://math.stackexchange.com/questions/3125520/is-there-a-non-trivial-covering-of-the-klein-bottle-by-the-klein-bottle/3125692#3125692) I look at some families of subgroups and show how some give you coverings by tori and others give you coverings by Klein bottles, you may find it useful – William Jan 16 '20 at 00:39
  • @William Nice! Can you put your comment in the form of an answer so I can accept. – Mike Jan 16 '20 at 00:44

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One subgroup of $\langle a, b\ |\ abab^{-1} = 1 \rangle $ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$. Indeed these elements commute: $ab^2 = ba^{-1}b = b^2 a$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $a^m$ and $b^{2n}$.

In an answer to a related question (Is there a non-trivial covering of the Klein bottle by the Klein bottle?) I gave a few families of subgroups and determined the total spaces of the corresponding cover space in those cases, maybe you will find it useful.

William
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  • I do have a little concern: While I see that $a,b^2$ commute, are you sure that there are no other relations between these two elements. I know that presentations work weird and sometimes there are "hidden" relations. Is there any way to guarantee that doesn't happen here? – Mike Jan 16 '20 at 00:59
  • I think we're ok in this case. For any surface group $G\cong \pi_1(\Sigma)$ for some surface $\Sigma$, every finite-index subgroup is again a surface group, because it is the fundamental group of a covering space and a covering space of a surface is a surface. In particular, a finite-index subgroup of a surface group will be 1-relator – William Jan 16 '20 at 01:20