Real projective space has fixed-point property iff it has even dimension

Statement

Suppose $n$ is a natural number. Consider the Real projective space (?) $\mathbb {P} ^{n}(\mathbb {R} )$ (also denoted $R\mathbb {P} ^{n}$) of dimension $n$. Note that this can be interpreted as the set of nonzero vectors, up to scalar multiplication equivalence, in $\mathbb {R} ^{n+1}$, and its elements can be written in the form $[a_{1}:a_{2}:\dots :a_{n}:a_{n+1}]$ with all $a_{i}\in \mathbb {R}$ and not all of them simultaneously zero, where:

$[a_{1}:a_{2}:\dots :a_{n}:a_{n+1}]=[b_{1}:b_{2}:\dots :b_{n}:b_{n+1}]\iff a_{i}b_{j}=a_{j}b_{i}\ \forall \ i,j\in \{1,2,3,\dots ,n+1\}$

The claim is the following:

• If $n$ is odd (i.e., $n=1,3,5,\dots$), then $\mathbb {P} ^{n}(\mathbb {R} )$ does not have the Fixed-point property (?), i.e., we can find a continuous map $f:\mathbb {P} ^{n}(\mathbb {R} )\to \mathbb {P} ^{n}(\mathbb {R} )$ such that $f$ does not have any fixed point. In fact, we can choose the continuous map to be a self-homeomorphism and even to be an algebraic automorphism.
• If $n$ is even (i.e., $n=2,4,6,\dots$), then $\mathbb {P} ^{n}(\mathbb {R} )$ has the fixed-point property, i.e., for any continuous map from $\mathbb {P} ^{n}(\mathbb {R} )$ to itself, there is a fixed point.

Related facts

Similar facts that distinguish real projective spaces of even and odd dimension

• Odd-dimensional real projective spaces are orientable, even-dimensional real projective spaces are non-orientable. See homology of real projective space and cohomology of real projective space.
• Odd-dimensional real projective spaces also possess an orientation-reversing homeomorphism. The question doesn't even arise for even-dimensional real projective space.

Similar facts about complex and quaternionic projective space

• Complex projective space has fixed-point property iff it has even complex dimension
• Complex projective space has orientation-reversing self-homeomorphism iff it has odd complex dimension

Facts used

1. Rationally acyclic compact polyhedron has fixed-point property, which in turn follows from the Lefschetz fixed-point theorem.
2. Homology of real projective space, whereby it's clear that the even-dimensional real projective spaces are rationally acyclic.
3. Real projective spaces are compact polyhedra.

Proof

Even dimension

The proof in even dimensions follows directly by combining Facts (1), (2), and (3).

Odd dimension

In odd dimension, we can explicitly construct an algebraic self-map $f:\mathbb {P} ^{n}(\mathbb {R} )\to \mathbb {P} ^{n}(\mathbb {R} )$ that works:

$[a_{1}:a_{2}:a_{3}:a_{4}:\dots :a_{n}:a_{n+1}]\mapsto [-a_{2}:a_{1}:-a_{4}:a_{3}:\dots :-a_{n+1}:a_{n}]$

Note that we are using that $n$, or equivalently, that $n+1$ is even, to make sense of the above.

We verify:

What to check	Why it's true
The map is well defined	The image of $\lambda$ times a vector is $\lambda$ times the image of the vector, so the map does indeed descend to the projective space.
The map is continuous	This follows from the fact that it is given algebraically.
The map has no fixed points	The key idea is that if 0 is a sum of squares of real numbers, all summands must equal 0. [SHOW MORE] Recall the equality condition $[a_{1}:a_{2}:\dots :a_{n}:a_{n+1}]=[b_{1}:b_{2}:\dots :b_{n}:b_{n+1}]$ iff $a_{i}b_{j}=a_{j}b_{i}$ for all $i,j\in \{1,2,3,\dots ,n\}$. Apply this to $(i,j)$ being each of the pairs $(1,2)$, $(3,4)$, and so on. We get $a_{1}^{2}=-a_{2}^{2}$, $a_{3}^{2}=-a_{4}^{2}$, and so on till $a_{n}^{2}=-a_{n+1}^{2}$. Solving, we get that $a_{1}=a_{2}=\dots =a_{n+1}=0$. But this violates the condition that the vector is a nonzero vector, so there are in fact no fixed points.