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Consider the free action of $\mathbb{Z}_2$ on $\mathbb{C}P^n$( $n$ is odd ) by $$[z_0,\dots, z_n]\rightarrow [-\overline z_1,\overline z_0,\dots,-\overline z_{n-1},\overline z_n].$$ Consider the orbit space $\mathbb{C}P^n/\mathbb{Z}_2$. My question is it possible to define a free group action by some group $G$ on $S^{2n+1}$ such that $S^{2n+1}/G$ is homeomorphic to $\mathbb{C}P^n/\mathbb{Z}_2$? Obviously $G$ cannot be a finite group.

Thanks in advance.

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    $\begingroup$ At least this is true for $S^{4n-1}$, which carries the action by unit quarternions, in which case what you want is $\text{Pin}(2)$. $\endgroup$
    – mme
    Commented Jul 26, 2018 at 17:29
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    $\begingroup$ Oh, of course you assumed the dimension of your projective space is singly even to define the action (and indeed you said so). So it is indeed Pin(2). $\endgroup$
    – mme
    Commented Jul 26, 2018 at 17:38

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