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.

  • 1
    $\begingroup$ 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) $\endgroup$ – YCor Nov 16 '18 at 19:45
  • 1
    $\begingroup$ @YCor Here I am considering the cyclic group of order p. $\endgroup$ – 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.

  • $\begingroup$ @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. $\endgroup$ – Shivani Sengupta Nov 17 '18 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.