The antipodal map $\mathbb{C}P^1\to\mathbb{C}P^1$ in coordinates

Question

I know that on $S^2$, we have the antipodal map $a:S^2\to S^2, (x,y,z)\mapsto(-x,-y,-z)$. But I know that we can identify $S^2$ with $\mathbb{CP}^1$, using the diffeomorphism $$ \Psi:\mathbb{CP}^1\to S^2, \Psi([w_0:w_1])=\frac{1}{|w_0|^2+|w_1|^2}\left(2\text{Re}(w_1\overline{w_0}), 2\text{Im}(w_1\overline{w_0}),|w_0|^2-|w_1|^2\right) $$ But now, I want to see how the antipodal map looks like in coordinates on $\mathbb{CP}^1$. I know it should be something like $[z_0:z_1]\mapsto[\overline{z_1}:\overline{z_0}]$, but I have been trying to compute it all day now and did not really progress. Could anyone help please?

Answer 1

$\mathbb{CP}^1$ is defined as the quotient space $$(\mathbb C^2 \setminus \{0\})/\sim$$ where $(w_0,w_1) \sim (w'_0,w'_1)$ if there exists $\lambda \in \mathbb C$ such that $\lambda (w_0,w_1) = (\lambda w_0,\lambda w_1) = (w'_0,w'_1)$. We have $$\Psi([w_0 :w_1]) = \frac{1}{|w_0|^2+|w_1|^2}\left(2w_1\overline{w_0},|w_0|^2-|w_1|^2\right) \in \mathbb C \times \mathbb R = \mathbb R^3 .$$ You want to see what $\Psi^{-1}(-\Psi([w_0:w_1]))$ looks like. Define $$A : \mathbb C^2 \setminus \{0\} \to \mathbb C^2 \setminus \{0\}, A(w_0,w_1) = (-\overline{w_1},\overline{w_0}) .$$ Clearly $A$ is a diffeomorphism. We have $$A(\lambda w_0,\lambda w_1) = (-\overline\lambda \overline{w_1},\overline\lambda \overline{w_0}) = \overline\lambda (-\overline{w_1},\overline{w_0}) = \overline\lambda A(w_0,w_1)$$ which means that $(w_0,w_1) \sim (w'_0,w'_1)$ implies $A(w_0,w_1) \sim A(w'_0,w'_1)$. Therefore $A$ induces a diffeomorphism $\bar A : \mathbb{CP}^1 \to \mathbb{CP}^1$ given by $\bar A ([w_0:w_1]) = [-\overline{w_1}: \overline{w_0}]$. We get $$\Psi(\bar A ([w_0:w_1])) = \Psi([-\overline{w_1}: \overline{w_0}]) = \frac{1}{|-\overline{w_1}|^2+|\overline{w_0}|^2}\left(2\overline{w_0}(-w_1),|-\overline{w_1}|^2-|\overline{w_0}|^2\right) \\ = \frac{1}{|w_0|^2+|w_1|^2}\left(-2w_1\overline{w_0},|w_1|^2-|w_0|^2\right) = - \Psi([w_0:w_1]) .$$

This shows that $\Psi^{-1}(-\Psi([w_0:w_1])) = [-\overline{w_1}: \overline{w_0}]$.