Can $S^1$ act freely on $\mathbb{C}P^2$?

Question

Prove that $S^1$ cannot act freely on the complex projective space $\mathbb{C}P^2$.

I just learned some basic concept about homology group and covering space. But I have no clue how to prove this proposition. If the group action is also propely discontinuous, then we may relate the problem to the deck transformation about $\mathbb{C}P^2$, which I have no idea how to compute it.

I have known that $\mathbb{C}P^n$ can be regarded as $$ S^{2n+1}\backslash\langle v\sim\lambda v\rangle, |\lambda|=1 $$ or $$ D^{2n}\backslash\langle v\sim\lambda v\rangle, v\in\partial D^{2n}, |\lambda|=1. $$

Appreciate any help!

Answer 1

Following @Eric Wofsey's suggestion, we can prove that any homeomorphism of $\mathbb{C}P^2$ has a fixed point. This implies that no non-trivial group can act freely on $\mathbb{C}P^2$.

Suppose $f : \mathbb{C}P^2 \to \mathbb{C}P^2$ is a homoemorphism. Then on each $H_{2i}(\mathbb{C}P^2; \mathbb{Z})$, the induced map $f_*$ is multiplication by $\pm 1$. Therefore the Lefschetz number of $f$ is a sum of $3$ integers which are each $\pm 1$, in particular odd, in particular non-zero. So by the Lefschetz fixed point theorem, $f$ has a fixed point.