Complex projective space has orientation-reversing self-homeomorphism iff it has odd complex dimension

Statement

Consider a complex projective space $\mathbb{P}^n(\mathbb{C})$, also denoted $\mathbb{C}\mathbb{P}^n$, for $n$ a natural number. This has complex dimension $n$ (as a complex manifold) and real dimension $2n$ (as a real manifold). Moreover, for all values of $n$, it is a compact connected orientable manifold.

The claim is that:

• For $n$ odd, i.e., $n = 1,3,5,\dots$ (so real dimensions $2,6,10,\dots$), $\mathbb{C}\mathbb{P}^n$ possesses an orientation-reversing self-homeomorphism. Viewed on homology, this is a homeomorphism that sends the fundamental class to its negative.
• For $n$ even, i.e., $n = 2,4,6,\dots$ (so real dimensions $4,8,12,\dots$), $\mathbb{C}\mathbb{P}^n$ does not possess an orientation-reversing self-homeomorphism. Viewed on homology, there is no homeomorphism that sends the fundamental class to its inegative.

Facts used

1. Cohomology of complex projective space

Proof

The proof relies on the cohomology ring structure of complex projective space. The key idea is that if we take a generator for $H^2(\mathbb{P}^n(\mathbb{C});\mathbb{Z})$, the $n^{th}$ power of this under the cup product gives a fundamental class.

Any self-homeomorphism of $\mathbb{P}^n(\mathbb{C})$ must send a generator of $H^2(\mathbb{P}^n(\mathbb{C});\mathbb{Z})$ either to itself or to its negative, and there are self-homeomorphisms that do both these. When $n$ is even, both these types of homeomorphisms fix the fundamental class because the $n^{th}$ power of $-1$ is $1$. When $n$ is odd, the homeomorphism that acts as negation on $H^2(\mathbb{P}^n(\mathbb{C});\mathbb{Z})$ also sends the fundamental class to its negative.