Acting on CP(2) by conjugation

Solution 1:

(The original version of this answer had the wrong map for complex conjugation on $\mathbb{C}\cup \{\infty\}$. Thanks to Grigory M for pointing it out. This new answer is substantially different in terms of its conclusion, but the work is mostly the same.)

The usual homeomorphism $\mathbb{C}P^1\times \mathbb{C}P^1/\sim \rightarrow \mathbb{C}P^2$ is given by $$([a_0:a_1],[b_0:b_1])\rightarrow [a_0b_0: a_0b_1 + a_1b_0: a_1b_1].$$

The identifcation of $S^2$ with $\mathbb{C}P^1$ identifies $[a_0:a_1]$ with $\frac{a_0}{a_1}\in \mathbb{C}\cup\{\infty\}\cong S^2$.

Under this identification, the antipodal map $x\mapsto -x$ for $S^2$ corresponds to the map of $\mathbb{C}\cup\{\infty\}$ given by $z\mapsto -1/\overline{z}$, which, in turn, corresponds to the map on $\mathbb{C}P^1$: $[a_0:a_1]\mapsto [\overline{a}_1:-\overline{a}_0]$.

So, the corresponding map on $\mathbb{C}P^2$ maps $[a_0b_0: a_0b_1 + a_1b_0: a_1b_1]$ to $[\overline{a}_1\overline{b}_1: -\overline{a}_1\overline{b}_0 - \overline{a}_0\overline{b}_1: \overline{a}_1 \overline{b}_1]$. In other words, it corresponds to swapping the first and last coordinates of $\mathbb{C}P^2$, negating the middle coordinate, and taking the complex conjuate of everything.

In other words, if $A = \begin{bmatrix}0 & 0 & 1\\ 0 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}$, then Massey's map is given by multiplication by $A$ followed by complex conjugation. Note that $A\in U(3)$, which is path connected, so $A$ is isotopic to the identity through diffeomorphisms. That is, $A$ lies in the identity component of $\operatorname{Diff}(\mathbb{C}P^2)$.

Thus, Massey's map is isotopic to complex conjugation.