Fixed point property of Cayley plane

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.

• By "fixed point property", do you mean that any self-map contains a fixed point? I believe that since it's a coset space (quotient, even?) of a Lie group, then the answer is: no, it does admit maps without fixed points. Jan 19 '12 at 12:38
• @Aaron: All odd dimensional spheres are quotients of Lie groups $S^{2n+1} = SU(n+1)/SU(n)$, but admit maps without fixed points (Hopf actions by circles). Jan 19 '12 at 16:36
• Haha. I think there might be a confusion in my wording "no, it does admit maps...". Still, clearly your answer shows that my reasoning is faulty. I'm not sure why I was thinking that the action of the original Lie group has to be free. Jan 20 '12 at 10:25
• And clearly it's not a quotient, or it'd be a Lie group itself! I should stay away from M.SE when I'm sleep-deprived... Jan 20 '12 at 10:26
• @Aaron: When you say "quotient" vs "coset space", what distinction are you drawing? Jan 20 '12 at 13:02

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 $0$.

• Many thanks Jason. The proof is very clearly explained. Jan 20 '12 at 5:14
• I should add that this argument works equally well for $\mathbb{C}P^{2n}$ and $\mathbb{H}P^{2n}$. Jan 20 '12 at 15:28