Which groups can act properly discontinuously on $\Bbb C\Bbb P^n$? My question is: which group can act properly discontinuously on $\Bbb C\Bbb P^n$ ?
I know that if any group $G$ acts properly discontinuously on a space then the quotient map $p:Y \to Y/G$ is a covering space and as $\Bbb C\Bbb P^2$ does not cover any other space except itself so for $\Bbb C\Bbb P^2$ only trivial group will act properly discontinuously. But I dont have any idea for $\Bbb C\Bbb P^n$ for any $n$.
Thanks in advance for any help.
 A: This is an answer to your question which I interpret as asking for a classification of free actions of finite groups of homeomorphisms on $CP^n$. (Unfortunately, there are several inequivalent notions of a properly discontinuous action $G\times X\to X$ in the literature. The standard one would imply that every finite subgroup of $Homeo(CP^n)$ acts properly discontinuously and there is no way to classify such subgroups. The less standard one amounts to the requirement that the action is a faithful covering action, i.e. the quotient map $X\to X/G$ is a covering map. My answer assumes the latter definition.)  
Claim. If $G< Homeo(CP^n)$ is a finite nontrivial subgroup acting freely, then $n$ is odd and $G$ has order 2 and its generator reverses orientation on $CP^n$. 
Remark. For $n=1$, $CP^1= S^2$ admits a fixed-point free orientation-reversing involution (the antipodal map). I do not know if such involutions exist for $n=2k+1>1$, I only can tell that such involutions cannot be linearizable, i.e. they are not topologically conjugate to elements of $PGL(n+1, {\mathbb C})$. 
Proof. Let $x\in H^2(CP^n, {\mathbb Q})$ denote a/the generator of the cohomology ring of $CP^n$, $H^*(CP^n, {\mathbb Q})\cong {\mathbb Q}[x]/(x^{n+1})$. (From now on, all cohomology is with rational coefficients which are suppressed.) Consider $g\in Homeo(CP^n)$ and its action on $H^*(CP^n)$. There are two cases to consider. 
(a) $g^*(x)=x$. Then, since $x$ generates the cohomology ring, $g$ acts trivially on $H^*(CP^n)$ and, hence, its Lefschetz number equals
$$
\Lambda_g= \sum_{k=0}^n (-1)^{2k} Tr\left(g^*: H^{2k}(CP^n)\to H^{2k}(CP^n)\right)= n+1\ne 0. 
$$
Therefore, by the Lefschetz fixed point theorem, the fixed point set of $g$ is nonempty. 
(b) $g^*(x)=-x$. Then for each $k=0,...,n$, we obtain $g^*(x^k)=(-1)^k x^k$ and, hence, 
$$
\Lambda_g= \sum_{k=0}^n (-1)^k, 
$$
and the latter equals $1$ if $n$ is even and $0$ if $n$ is odd. Moreover, $g^*(x^n)=-x^n$ if $n$ is odd, i.e. $g$ is orientation reversing. 
We conclude that if either $g$ is orientation-preserving or $n$ is even,  then $g$ has a nonempty fixed point set. Now, if $n$ is odd and $G$ acts freely it follows that its orientation preserving index 2 subgroup has to be trivial, i.e. $G=Z_2$. qed
Edit. As Jason noticed, for every even $n=2k$ the projective space $CP^n$ admits a fixed-point free involution
$$
[z_0:z_1:...:z_{2k-1}:z_{2k}]\mapsto [-\bar{z}_1: \bar{z}_0:...:-\bar{z}_{2k}: \bar{z}_{2k-1}]. 
$$
Hence, the conclusion is that $CP^n$ admits a free action of a nontrivial finite group $G$ if and only if $n$ is even and  $G\cong {\mathbb Z}_2$.  
