# 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.

• In order to have a covering space $Y\to Y/G$, the group $G$ must act properly discontinuously without fixed points. – paf Aug 12 '16 at 14:01
• $\Bbb CP^n$ (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. – PVAL-inactive Aug 12 '16 at 14:02
• Yes, but is it true that $PGL_{n+1}(\Bbb Z)$ acts properly discontinuously on $\Bbb C\Bbb P^n$? – paf Aug 12 '16 at 14:06
• @PVAL yeah $CP^n$ is simply connected but i dont understand the connection of the properly discontinuous action with the simply coneectedness of $CP^n$ – Shivani Sengupta Aug 14 '16 at 16:04
• 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 $CP^n$. – Moishe Kohan Aug 15 '16 at 5:44

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 $$G\cong {\mathbb Z}_2$$.

• 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. – Shivani Sengupta Aug 18 '16 at 13:29
• I will not be able to read your answer right now .I'll read it later.Thanks – Shivani Sengupta Aug 18 '16 at 13:31
• @prakriti oh, do not worry about this and get better soon! – Moishe Kohan Aug 19 '16 at 6:26
• 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. – Jason DeVito Oct 9 at 12:55
• @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 $CP^n$. – Moishe Kohan Oct 9 at 15:25