# Show that if $p\neq 2$, then $\mathbb{Z}_p$ cannot act freely on $\mathbb{C}P^n$

If $$p\neq 2$$, then the cyclic group $$\mathbb{Z}_p$$ has no free continuous action on $$\mathbb{C}P^n$$. My question is how to prove the above fact using Leray-Serre spectral sequence associated to the Borel fibration $$\mathbb{C}P^n\hookrightarrow X_{\mathbb{Z}_p}\rightarrow B_{\mathbb{Z}_p}$$.

From the Euler Characteristic argument, $$p$$ divides $$n+1$$. Also, $$\pi_1(B_{\mathbb{Z}_p})={\mathbb{Z}_p}$$ acts trivially on $$H^*(\mathbb{C}P^n;\mathbb{Z}_p)$$ by using Lefschetz fixed point theorem. $$H^*(\mathbb{C}P^n;\mathbb{Z}_p)=\mathbb{Z}_p[b]/\langle b^{n+1}\rangle$$ and $$H^*(\mathbb{Z}_p;\mathbb{Z}_p)=\bigwedge(s)\otimes\mathbb{Z}_p[t]$$. So the only possibility is $$d_3(b)=st$$. It follows $$d_3(sb)=0$$ and $$d_3{(tb)}=st^2$$. After that, I am unable to deduce any contradiction.

Thank you so much in advance.

• Do you mean the $p$-adics or the cyclic group of order $p$? I guess the second one (although I initially spent a time reading as if it were the $p$-adics, where the question is meaningful) – YCor Nov 16 '18 at 19:45
• @YCor Here I am considering the cyclic group of order p. – Shivani Sengupta Nov 17 '18 at 4:25

Consider the cohomology with $$\mathbb{Z}$$ coefficients (and reduce the 0-th term modulo $$p$$ to get uniform description of it). Then we have a spectral sequence starting from $$\mathbb{F}_p[x,y]/x^{n+1}$$ with $$deg(x)=deg(y)=2$$ and converging to $$H^*(\mathbb{C}P^n/\mathbb{Z}_p,\mathbb{Z})$$. Since the $$E^2$$ term is concentrated in even degrees, the spectral sequence degenerates at the $$E^2$$-term, and so $$H^*(\mathbb{C}P^n/\mathbb{Z}_p)$$ has arbitrarily high non-zero cohomologies. But is is a finite dimensional manifold (being the quotient of a manifold by a free action), and this is a contradiction.
• @scarmeli Instead of taking $CP^n$ if we take $Y = CP^{n_1} \times CP^{n_2} \times ... \times CP^{n_k}$, then how to show that $Z_p$ does not act freely on $Y$ using spectral sequence where $p\neq 2$? I have edited the question. Please have a look. – Shivani Sengupta Nov 17 '18 at 6:10