minwaardekn

2022-06-25

Suppose $f:{\mathbb{P}}^{2}\to {\mathbb{P}}^{2}$ is rational such that $f\circ f=\mathbb{I}\mathbb{d}$. Then is it true that $f$ must be linear?
It feels true due to the degree which increases, but some things might cancel out.
Suppose we have a smooth curve $C$ of genus $g\ge 1$ with a rational function $g:C\to C$ with the same property. Does it always rise to an $f$ on ${\mathbb{P}}^{2}$ whose restriction is $g$? Does it imply that $g$ has to be linear too? This on the other side seems wrong to me.

jmibanezla

Here is an involution
$\left(x,y,z\right)↦\left({x}^{2}-yz,{y}^{2}-xz,{z}^{2}-xy\right)$
It is the inversion wr to the conic $xy+xz+xz=0$ with inversion center (1,1,1)
Projective rational involutions appear in other instances, think of the map $A↦{A}^{-1}$ for matrices, which in projective coordinates can be given as $A↦\mathrm{adj}A$, from a matrix to its adjugate. This map, restricted to certain subalgebras of matrices, again gives an involution.
Still, I cannot find any rational involutions of ${\mathbb{P}}^{2}$ of degree $>2$. Also, I can't find rational involutions of ${\mathbb{P}}^{n}$ of degree $>n$.

Tristian Velazquez

As Mohan mentioned in the comments, there exist non-linear functions whose power is $\mathbb{I}\mathbb{d}$
For example $\left[x:y:z\right]↦\left[yz:xz:xy\right]$ has order two, and $\left[x:y:z\right]↦\left[xy:yz:zx\right]$ has order six.

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