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This question is motivated by Fixed point property of Cayley plane and idle curiosity.

In the link, it is shown that every continuous map from the Cayley plane to itself has a fixed point and the argument (with easy modification) applies equally well to $\mathbb{R}P^{2n}$, $\mathbb{C}P^{2n}$ and $\mathbb{H}P^{2n}$.

The leaves open the case of $\mathbb{R}P^{2n+1}$, $\mathbb{C}P^{2n+1}$, and $\mathbb{H}P^{2n+1}$.

In the case $\mathbb{R}P^{2n+1}$, this space covers many the lens spaces, so in particular (via the Deck group), admits many fixed point free self maps (diffeomorphisms, in fact).

In the case $\mathbb{C}P^{2n+1}$, the Lefschetz theorem shows that any continuous map without fixed points must act as $-1$ in $H^2$, and indeed, there is a such a map $(\mathbb{C}P^{2n+1}$ double covers a nonorientable manifold). So again, there is a diffeomorphism with no fixed points.

This leaves open the case of $\mathbb{H}P^{2n+1}$. The same argument as in the previous case shows that $f$ must act as $-1$ in $H^4$. This certainly can be achieved when $n=0$. In this case, $\mathbb{H}P^1 = S^4$ and the antipodal map is a diffeomorphism which acts without fixed points.

However, for $n > 0$, there is no such diffeomorphism. This comes from the fact that the first Pontrjagin class is a diffeomorphism invariant and is a nonzero multiple of the generator of $H^4(\mathbb{H}P^{2n+1})$. This implies that any diffeomorphism must act as $+1$ on $H^4$. I think Novikov's result on the rational invariance of Pontrjagin classes under homeomorphisms gives us the same result for homeomorphisms.

Thus, if there is a map $f:\mathbb{H}P^{2n+1}\rightarrow \mathbb{H}P^{2n+1}$ without fixed points, it cannot be a homeomorphism. Hence my question:

Is there a continuous function $f:\mathbb{H}P^{2n+1}\rightarrow \mathbb{H}P^{2n+1}$ without fixed points when $n > 0$? Is there an example where $f$ is a homotopy equivalence?

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up vote 6 down vote accepted

Any map $\mathbb{HP}^n \rightarrow \mathbb{HP}^n$ must have a fixed point for $n >1$. This is discussed on page 493 of Hatcher. It follows from some calculations and use of the Lefschetz fixed point theorem.

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Nice - thanks for the reference! It also motivates me to learn Steenrod powers finally. – Jason DeVito Feb 5 '12 at 0:51
The Steenrod operations are tremendously useful, and there's a lot to learn. It's nice what you can prove just knowing the axiomatic properties that define them. – Matthew Pancia Feb 5 '12 at 2:57
I've studied Steenrod squares a bit - even they are enough to prove a ton of stuff, but I've never really looked at Steenrod powers for general $p$ at all. – Jason DeVito Feb 5 '12 at 22:23

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