I know the group of automorphisms of $\mathbb{P}^n$ is equal to the degree $1$ birational transformations of $\mathbb{P}^n$ and every birational transformation of $\mathbb{P}^n$ can be written as $[f_1:…:f_{n+1}]$, where the $f_i$ are homogeneous polynomials.
Now I am considering this birational map $\phi:\mathbb{P}^2\to \mathbb{P}^2: [x:y:z]\mapsto [z:x-y:x-y]$, it's defined everywhere except $[1:1:0]$ as this formula. But this is indeed an automorphism of $\mathbb{P}^2$. I wonder how to get a formula at $[1:1:0]$ so that $\phi$ determines the automorphism?
Best Answer
The map is not an automorphism of $\mathbb P^2$, because its image has dimension $1$: it is supported in the diagonal copy of $\mathbb P^1$ given by $[s\colon t\colon t]$ as $s,t$ range over the ground field.