TITLE: Cremona group of $\mathbb{P}^n$
QUESTION [0 upvotes]: I know that the complex conjugation $\tau: \mathbb{P}^n \mapsto \mathbb{P}^n$ that sends any point $x$ to the point with complex conjugate coordinates $\tau(x)$ is a homeomorphism. In order to show that it isn't an automorphism of the algebraic variety of $\mathbb{P}^n$, I have to show that it doesn't belong to the Cremona group. But I can't describe the Cremona group of $\mathbb{P}^n$.
Can you help me?

REPLY [1 votes]: Suppose that $\tau$ is equal to a birational transformation (or even a rational map) on a non-empty open subset $U\subset \mathbb{P}^n$ and find a contradiction.
For each $[z_0:\dots:z_n]\in U$ you would have
$$\tau([z_0:\dots:z_n])=[f_0(z_0,\dots,z_n):\dots:f_n(z_0,\dots,z_n)]$$
so you get $f_i(z_0,\dots,z_n) \overline{z_j}-f_j(z_0,\dots,z_n) \overline{z_i}_{|_U}=0$. It remains to see that this is only possible when both $f_i$ and $f_j$ are zero. For this, you can for example restrict to real points and then to points where one coordinate is not real.
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