1
$\begingroup$

Let $X=\mathbb{T} \times \mathbb{RP}^2$. I want to find all covering scpaces (up to isomorphism) of $X$.

First, suppose we know all covering spaces of toro $\mathbb{T}$ and projective plane $\mathbb{RP}^2$. We know that if $(\tilde A,p)$ is a covering space of $\mathbb{T}$ and $(\tilde B,q)$ is a covering space of $\mathbb{RP}^2$, then $(\tilde A \times \tilde B, p \times q)$ is a covering space of $X$. By this information we can find a universal covering space of $X$ (for example $\mathbb{R}^2 \times \mathbb{S}^2$) and so we can conclude that all covering spaces of $X$ are a quotient between the universal covering space and subgroups of $\pi_1(X)=\mathbb{Z}^2 \times (\mathbb{Z}/2\mathbb{Z})$ which is abelian. Using this corrispondence we observe again that all covering sapces in the form $(\tilde A \times \tilde B, p \times q)$ are associated with subgroups $A \times B < \mathbb{Z}^2 \times (\mathbb{Z}/2\mathbb{Z})$, but this is not enough to complete the list because there exist subgroups of $\mathbb{Z}^2 \times (\mathbb{Z}/2\mathbb{Z})$ which are not product of a subgroup of $\mathbb{Z}^2$ with once of $\mathbb{Z}/2\mathbb{Z}$. How can we actually find out the covering spaces of $X$ linked to these particular subgroups?

(Stupid question: Is true that all covering spaces of $\mathbb{T}$ are also of $X$? I think so, but it looks quite weird...)

$\endgroup$
  • 1
    $\begingroup$ Topologically, I think you have found all the covering spaces (I don't think the other subgroups give you different spaces up to homeomorphism, only up to covering isomorphism). For your last question, that can't be true, as the coverings you care about are 4-manifolds. $\endgroup$ – Steve D Jul 9 '18 at 12:26
  • $\begingroup$ Is there a way to show that these are the only covering spaces up to homeomorphism? (Basically can we show that any subgroup of the fundamental group is conjugate to the subgroups we get from taking direct products of subgroups of the factors $\mathbb{Z}^2$ and $\mathbb{Z}_2$?) $\endgroup$ – Osama Ghani Jul 9 '18 at 14:43
  • $\begingroup$ @OsamaGhani: They are not conjugate (indeed, the fundamental group is abelian, conjugacy means nothing). This is why they are different covering spaces. As topological spaces, I don't know the cleanest way to see they are the same. My thinking here is that the quotients we are taking even with a "diagonal" group action still gives us circles and $S^2$ quotients. $\endgroup$ – Steve D Jul 9 '18 at 15:27
  • $\begingroup$ My bad, I confused the classification of conjugacy classes with homeomorphism classes of covering spaces instead of (unbased) covering space isomorphism. I asked my professor and I didn’t read it completely but the gist was there are diffeomorphisms of $T^2$ and $\mathbb{R}P^2$ that convert any subgroup to the sort of “standard” subgroups for which we know what the classification is. $\endgroup$ – Osama Ghani Jul 9 '18 at 18:50
  • $\begingroup$ But now that I look, OP asked for all covering spaces up to covering space isomorphism so the right thing to look at IS conjugacy classes of subgroups, but since the group is abelian, this means we just look at subgroups up to...equality? I think I’ve confused myself. How would one show what the “non-obvious” subgroups yield have as corresponding covers? $\endgroup$ – Osama Ghani Jul 9 '18 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.