Does every continuous map $f : \mathbb{R} \mathbb{P}^n \to \mathbb{R} \mathbb{P}^n$ have a fixed point for $n = 2$? $n = 3$? and $n = 4$?

Question

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Does every continuous map $f : \mathbb{R} \mathbb{P}^n \to \mathbb{R} \mathbb{P}^n$ have a fixed point for $n = 2$? $n = 3$? and $n = 4$?

Answer 1

For $ \mathbb{RP}^3 $ the answer is no. Indeed this space is homeomorphic to a topological group (namely $SO(3)$) and the claim is false for groups (because multiplication by a group element is continuous and has no fixed points).

For $ \mathbb{RP}^2 $ and $ \mathbb{RP}^4 $ the answer is yes, using the Lefschetz fixed point theorem (see Theorem 2C.3, p. 179, of Hatcher, for instance). The point is that all the homology is torsion except in degree 0, where the induced map is the identity, so the Lefschetz number cannot vanish.

Answer 2

(Just for fun: a solution without Lefschetz theorem)

Any map $f:\mathbb RP^n\to\mathbb RP^n$ lifts to a map of universal covers, $\tilde f:S^n\to S^n$. But $\chi(S^n)$ is non-zero for even $n$, so $\tilde f$ has either a fixed point or point mapping to its antipode (otherwise $\tilde f(x)-x$ gives a non-vanishing vector field). So $f$ has a fixed point.

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And for $n=2k+1$ there is even a free $S^1$-action: $S^1\subset\mathbb C^\times$ acts on $S^{2k+1}\subset\mathbb C^{k+1}$ — so $S^1\cong S^1/\{\pm1\}$ acts on $S^{2k+1}/\{\pm1\}=\mathbb RP^{2k+1}$.