Complex projective space has fixed-point property iff it has even complex dimension

Statement

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

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

The claim is the following:

• If $n$ is odd (i.e., $n=1,3,5,\dots$, so the real dimension is $2,6,10,\dots$), then $\mathbb {P} ^{n}(\mathbb {C} )$ does not have the Fixed-point property (?), i.e., we can find a continuous map $f:\mathbb {P} ^{n}(\mathbb {C} )\to \mathbb {P} ^{n}(\mathbb {C} )$ such that $f$ does not have any fixed point. In fact, we can choose the continuous map to be a self-homeomorphism, and to be algebraic over the reals though not over the complex numbers.
• If $n$ is even (i.e., $n=2,4,6,\dots$, so the real dimension is $4,8,12,\dots$), then $\mathbb {P} ^{n}(\mathbb {C} )$ has the fixed-point property, i.e., for any continuous map from $\mathbb {P} ^{n}(\mathbb {C} )$ to itself, there is a fixed point.

Related facts

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

• Complex projective space has orientation-reversing self-homeomorphism iff it has odd complex dimension: Note that all complex projective spaces are orientable; however, the ones with odd complex dimension additionally have the property that they have an orientation-reversing self-homeomorphism. This makes connected sums involving them well defined.

Similar facts about real projective space

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

Facts used

1. Cohomology of complex projective space
2. Lefschetz fixed-point theorem
3. Complex projective spaces are compact polyhedra

Proof

Proof for even dimensions

Given: Complex projective space $\mathbb {P} ^{n}(\mathbb {C} )$ with $n$ even. A continuous map $f:\mathbb {P} ^{n}(\mathbb {C} )\to \mathbb {P} ^{n}(\mathbb {C} )$.

To prove: $f$ has a fixed point.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The cohomology ring of $\mathbb {P} ^{n}(\mathbb {C} )$ is of the form $\mathbb {Z} [x]/(x^{n+1})$ where $x$ is an additive generator of $H^{2}(\mathbb {P} ^{n}(\mathbb {C} ;\mathbb {Z} )$ and $x^{j}$ is an additive generator of $H^{2j}(\mathbb {P} ^{n}(\mathbb {C} ;\mathbb {Z} )$ for $j\in \{0,1,2,\dots ,n\}$	Fact (1)	The space is a complex projective space.
2	$f$ induces maps $H^{k}(f):H^{k}(\mathbb {P} ^{n}(\mathbb {C} ))\to H^{k}(\mathbb {P} ^{n}(\mathbb {C} ))$ that are compatible with the cohomology ring structure	Cohomology ring is a contravariant functor	$f$ is a continuous map from $\mathbb {P} ^{n}(\mathbb {C} )$ to itself
3	There is an integer $m$ such that $H^{2}(f)(x)=mx$ for the generator $x$ of $H^{2}(\mathbb {P} ^{n}(\mathbb {C} ;\mathbb {Z} )$. The induced map $H^{2j}$ is then multiplication by $m^{j}$ for $j=0,1,2,\dots ,n$. All other induced maps $H^{k}$ are zero maps between zero spaces.			Steps (1), (2)
4	The Lefschetz number of $f$ is $n+1$ if $m=1$ and is $(m^{n+1}-1)/(m-1)$ otherwise.			Step (3)	[SHOW MORE] The Lefschetz number is $\sum _{j=0}^{n}(-1)^{2j}m^{j}=\sum _{j=0}^{n}m^{j}$. We now use the formula for summing up a geometric progression to get $\left\lbrace {\begin{array}{rl}n+1,&\qquad m=1\\{\frac {m^{n+1}-1}{m-1}},&\qquad m\neq 1\\\end{array}}\right.$
5	The Lefschetz number cannot be zero.		$n$ is even.	Step (4)	[SHOW MORE] If $m=1$, the Lefschetz number is nonzero. If $m\neq 1$, the Lefschetz number is zero iff $m^{n+1}=1$. However, since $n$ is even, $n+1$ is odd, so there is no solution for $m\neq 1$. (Note: If $n$ were odd, then $m=-1$ would be a solution).
6	$f$ has a fixed point.	Facts (2), (3)	$f$ is a continuous map from $\mathbb {P} ^{n}(\mathbb {C} )$ to itself	Step (5)	[SHOW MORE] By Facts (2) and (3), any continuous map without fixed points must have Lefschetz number zero. But $f$ does not have Lefschetz number zero, and it is continuous, so it must have fixed points.

Proof for odd dimensions

In odd dimension, we can explicitly construct a real algebraic (but not complex algebraic) self-map $f:\mathbb {P} ^{n}(\mathbb {C} )\to \mathbb {P} ^{n}(\mathbb {C} )$ that works:

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

Note that in order to make sense of the expression, we need $n+1$ to be even and hence $n$ to be odd.

What to check	Why it's true
The map is well defined, i.e., it descends to the projective space.	The image of $\lambda$ times a vector is ${\overline {\lambda }}$ times the image of the vector.
The map is continuous	It is given by continuous expressions coordinate-wise
The map has no fixed points	The key idea is that if $|z|^{2}=-|w|^{2}$ then $z=w=0$, where $|z|^{2}=z{\overline {z}}$ is the complex modulus-squared. [SHOW MORE] Recall that $[a_{1}:a_{2}:a_{3}:a_{4}:\dots :a_{n}:a_{n+1}]=[b_{1}:b_{2}:b_{3}:b_{4}:\dots :b_{n}:b_{n+1}]$ if $a_{j}b_{k}=a_{k}b_{j}$ for $j,k\in \{1,2,\dots ,n+1\}$. Applying this to the pair $(j,k)$ being $(1,2),(3,4),\dots (n,n+1)$ we get $|a_{1}|^{2}=-|a_{2}|^{2}$, $|a_{3}|^{2}=-|a_{4}|^{2}$, and so on till $|a_{n}|^{2}=-|a_{n+1}|^{2}$. This forces all the $a_{j}$s to be zero. However, the projective space is defined by starting with nonzero vectors.