Let $X = \mathbb{RP}^n \times \mathbb{RP}^n$.

I know the following:

  • the universal cover of $X$ is $Y = \Bbb S^n \times \Bbb S^n$
  • the fundamental group of $X$ is $G = \Bbb Z/2 \Bbb Z \times \Bbb Z/2 \Bbb Z = \{(0,0), (0,1), (1,0), (1,1)\}$
  • Covering spaces of $X$ are defined by actions of subgroups of $G$ on $Y$.

Each of the elements of $G$ generates a subgroup of order two. Clearly the covering spaces defined by the action of $\langle(0,1)\rangle$ and $\langle(1,0)\rangle$ on $Y$ are $\Bbb S^2 \times \mathbb{RP}^2$. But what about the action of $\langle(1,1)\rangle$? What covering space does this define? And finally, which of the covering spaces are equivalent? And which are homeomorphic? Thanks.


2 Answers 2


By fundamental theorem of covering spaces, there is a bijective correspondence between conjugacy classes of subgroups of $\pi_1(\Bbb RP^n \times \Bbb RP^n)$ and covers of $\Bbb RP^n \times \Bbb RP^n$.

$\pi_1(\Bbb RP^n \times \Bbb RP^n) \cong \Bbb Z/2 \times \Bbb Z/2$, subgroups of which are precisely the trivial group, the whole group, the two factors isomorphic to $\Bbb Z/2$ and an extra copy of $\Bbb Z/2$ coming from the diagonal group. None of these are conjugate to any other, since $\Bbb Z/2 \times \Bbb Z/2$ is an abelian group.

Thus, any covering space of $\Bbb RP^n \times \Bbb RP^n$ is homeomorphic to either $S^n \times S^n$, $S^n \times \Bbb RP^n$, $S^n \times S^n/(x, y) \sim (-x, -y)$ or itself, and all of these are distinct.

  • $\begingroup$ I don't understand, there are 5 subgroups but four covering spaces? $\endgroup$
    – Twink
    Jan 16, 2020 at 21:54
  • $\begingroup$ Whay do you mean withthe two factos isomorphic to $\Bbb Z/2$? Wouldn't that give the same covering space obtained with the extra copy of $\Bbb Z/2$? $\endgroup$
    – Twink
    Jan 16, 2020 at 21:56
  • 1
    $\begingroup$ @Twink: There are coverings $\mathbb{RP}^n\times S^n \to \mathbb{RP}^n\times \mathbb{RP}^n$ and $S^n\times\mathbb{RP}^n \to \mathbb{RP}^n\times\mathbb{RP}^n$ corresponding to the subgroups $\mathbb{Z}_2\times\{0\}$ and $\{0\}\times\mathbb{Z}_2$ respectively. They are different coverings, but have homeomorphic total spaces. That is, there are five different coverings, but only four distinct covering spaces. $\endgroup$ Jun 20, 2021 at 19:46
  • $\begingroup$ @MichaelAlbanese: Is that really important to distinguish $A\times B$ and $B\times A$? it look like a joke to me!! $\endgroup$
    – C.F.G
    Jun 21, 2021 at 3:20
  • 1
    $\begingroup$ @C.F.G: The two coverings are not isomorphic, so it is important to distinguish them. As I mentioned in my comment, the total spaces are homeomorphic though. $\endgroup$ Jun 21, 2021 at 3:39

The nonzero element $1$ of each Cartesian product factor $\Bbb Z / 2 \Bbb Z$ in $G$ acts by the antipodal map, $x \mapsto -x$ on the corresponding sphere. So, by definition, the element $(1, 1) \in G$ corresponds to the space quotient space $$(\Bbb S^2 \times \Bbb S^2) / \!\sim,$$ where $(x, y) \sim (-x, -y)$.

  • $\begingroup$ tahnks a lot for your comments. I got it. $\endgroup$ May 11, 2015 at 2:47
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ May 11, 2015 at 3:24

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