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Suppose f : P 2 P 2 is rational such that f f = I 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 1 with a rational function g : C C with the same property. Does it always rise to an f on 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.

Answer & Explanation



Beginner2022-06-26Added 17 answers

Here is an involution
( x , y , z ) ( x 2 y z , y 2 x z , z 2 x y )
It is the inversion wr to the conic x y + x z + x z = 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 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 P 2 of degree > 2. Also, I can't find rational involutions of P n of degree > n.
Tristian Velazquez

Tristian Velazquez

Beginner2022-06-27Added 7 answers

As Mohan mentioned in the comments, there exist non-linear functions whose power is I d
For example [ x : y : z ] [ y z : x z : x y ] has order two, and [ x : y : z ] [ x y : y z : z x ] has order six.

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