I would like to know if there is a nice proof of the fact that every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without use of the Lefschetz fixed point theorem.
Every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without Lefschetz theorem.
algebraic-topologydifferential-topologygeneral-topology
Best Answer
Through any two distinct points in $\mathbb{C}P^n$ there is a unique (complex) geodesic. Therefore from any fix point free self map $f$ , we have a 1 dimensional complex subbundle of the tangent bundle by taking the subspace above a point $p$ to be the tangent space of the geodesic from $p$ to $f(p)$ at $p$.
This implies that the total Chern class of $\mathbb{C}P^n$ has a linear factor. If $n=2$ this implies that the total Chern class $1+3x+3x^2$ has two real (integer) roots. However, this is easily checked to be false since the discriminant is negative.
As a sanity check, this should be different if $n=3$. In that case the total Chern class is $1+4x +6x^2 +4x^3$ and this equals $(2 x + 1) (2 x^2 + 2 x + 1)$, as expected.
I imagine if you are better with polynomials than I, you can get this to work for any even $n$.