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I want to find all the connected covering spaces of $\mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$.

My attempt:

I know that $\pi_1(\mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}) \cong \pi_1(\mathbb{R}\mathbb{P}^{15}) * \pi_1(\mathbb{R}\mathbb{P}^{14}) \cong \mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}$. I also know that connected covering spaces of $B$ are in one-to-one correspondence with subgroups of $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z} = \langle a \rangle * \langle b \rangle$.

The full group $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}$ corresponds to the trivial cover $p:\mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14} \rightarrow \mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$.

The trivial subgroup $\{ e \}$ corresponds to the universal cover $p:S^{15} \vee S^{14} \rightarrow \mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$.

The subgroup generated by $a$, $\langle a \rangle \cong \mathbb{Z}_2 * \{ e \}$, corresponds to the cover $p:\mathbb{R}\mathbb{P}^{15}\vee S^{14}\rightarrow \mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$.

Similarly, the subgroup generated by $b$, $\langle b \rangle \cong \{ e \} * \mathbb{Z}_2$, corresponds to the cover $p:S^{15}\vee \mathbb{R}\mathbb{P}^{14} \rightarrow \mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$.

I think there are two subgroup of $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}$ that is isomorphic to $\mathbb{Z}$, namely the subgroup generated by $(ab)$ and $(ba)$. However, I am not sure what these two covering spaces would look like.

Am I missing anything else? Thank you in advance.

mathlover314
  • 619
  • You can take a look at https://math.stackexchange.com/questions/2760711/find-the-universal-covering-of-mathbbrp2-vee-mathbbrp2. The double cover of $\mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$ is not $S^{15} \vee \mathbb{R}\mathbb{P}^{14}$ but $S^{15} \vee \mathbb{R}\mathbb{P}^{14}\vee \mathbb{R}\mathbb{P}^{14}$ (yes, you need two copies of $\mathbb{R}\mathbb{P}^{14}$) . And the fundamental group is coined infinite dihedral group https://handwiki.org/wiki/Infinite_dihedral_group . – Peter Wu Jul 04 '23 at 08:47
  • This question might also be helpful. – Kevin.S Jul 05 '23 at 12:32
  • There is no covering map $S^{15} \vee S^{14} \to \mathbb{R}\mathbb{P}^{15} \vee \mathbb{R}\mathbb{P}^{14}$: only the wedge point can be mapped to the wedge point by a local homeomorphism. Since $\pi_1$ is infinite, the universal cover has to be infinite. – ronno Aug 22 '23 at 09:40
  • Also note that in any group $G$, if $x$ generates a copy of $\mathbb{Z}$ then so does $a^n$ for each $n > 1$. – ronno Aug 22 '23 at 09:42

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