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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.

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    $\begingroup$ In order to have a covering space $Y\to Y/G$, the group $G$ must act properly discontinuously without fixed points. $\endgroup$ – paf Aug 12 '16 at 14:01
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    $\begingroup$ $\Bbb CP^n$ (for $n\geq 1$) can be given a CW-complex structure with $\Bbb CP^1$ as its 2-skeleton (see any algebraic topology book). Since the fundamental group of a CW-complex is isomorphic to the fundamental group of its 2-skeleton (see any algebraic topology book), $\Bbb CP^n$ is also simply connected. $\endgroup$ – PVAL-inactive Aug 12 '16 at 14:02
  • $\begingroup$ Yes, but is it true that $PGL_{n+1}(\Bbb Z)$ acts properly discontinuously on $\Bbb C\Bbb P^n$? $\endgroup$ – paf Aug 12 '16 at 14:06
  • $\begingroup$ @PVAL yeah $CP^n$ is simply connected but i dont understand the connection of the properly discontinuous action with the simply coneectedness of $CP^n$ $\endgroup$ – Shivani Sengupta Aug 14 '16 at 16:04
  • $\begingroup$ You should specify your definition of a properly discontinuous action since there are two competing definitions in the literature. I think, the definition you have in mind amounts to a free action of a finite group of homeomorphisms on $CP^n$. $\endgroup$ – Moishe Kohan Aug 15 '16 at 5:44
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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 $G\cong {\mathbb Z}_2$.

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  • $\begingroup$ i am very very sorry because I couldn't reply.still now I am hospitalized.here i am using the definition of covering space action of Allan Hatcher algebraic topology book. $\endgroup$ – Shivani Sengupta Aug 18 '16 at 13:29
  • $\begingroup$ I will not be able to read your answer right now .I'll read it later.Thanks $\endgroup$ – Shivani Sengupta Aug 18 '16 at 13:31
  • $\begingroup$ @prakriti oh, do not worry about this and get better soon! $\endgroup$ – Moishe Kohan Aug 19 '16 at 6:26
  • $\begingroup$ For $n=1$, it's easy to verify the involution $[z_0:z_1]\mapsto [-\overline{z}_1: \overline{z}_0]$ acts freely. Then you can just copy this two coordinates at a time for any $\mathbb{C}P^n$ with $n$ odd. For example, for $n=3$, one can use $[z_0:z_1:z_2:z_3]\mapsto [-\overline{z}_1:\overline{z}_0: -\overline{z}_3: \overline{z}_2].$ These involutions are obviously real linear and not complex linear, but it's not clear to me why they can't be topologically conjugate to a complex linear involution. $\endgroup$ – Jason DeVito Oct 9 at 12:55
  • $\begingroup$ @JasonDeVito: Good point! As for complex-linear actions: Every element of $GL(n+1,C)$ has an eigenvector in $C^{n+1}$, hence, has a fixed point in $CP^n$. $\endgroup$ – Moishe Kohan Oct 9 at 15:25

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