Fixed point property of Cayley plane

Question

I want to know whether the Cayley plane has fixed point property or not. I think it is so but I am not able to prove this. It certainly does not admit maps of period two without fixed points. By fixed point property, I mean that any continuous self map has a fixed point.

Answer 1

Every continuous map from the $X =$ Cayley Plane to itself has a fixed point.

To see this, let $f:X\rightarrow X$ be any continuous map. This induces a linear map $f_*:H_*(X,\mathbb{Q})\rightarrow H_*(X,\mathbb{Q})$. Recall that the Lefschetz fixed point theorem says if this trace is nonzero, then $f$ has a fixed point.

Hence, we need to compute the trace of $f$ and show it's nonzero.

Looking at cohomology, we have $H^*(X,\mathbb{Q}) = \mathbb{Q}[x]/x^3$ where $x$ is a generator of $H^8(X,\mathbb{Q})$.

In degree $0$, we see that $f^*$ acts as multiplication by $1$ (as it always does). In $H^8$, let us assume $f^*(x) = kx$ for some integer $k$. Since $f^*$ is a ring homomorphism, we must have $f^*(x^2) = f^*(x)^2 = k^2 x^2$.

Using naturality of Poincare duality, it follows that the trace of $f_*$ is $1+k+k^2 = 1+k(k+1)$. Since either $k$ or $k+1$ is even, the trace is odd, so not $0$.