Borel subgroups of the plane Cremona group

Proof of Proposition 1.4.

Set U k := D 2 ⁢ n - 2 ⁢ k ⁢ ( ℬ n ) for 0 ≤ k ≤ n and V k := D ⁢ ( U k ) for 1 ≤ k ≤ n .

We could easily check by induction that U k is contained in the group of elements f = ( f 1 , … , f n ) in ℬ n satisfying f i = x i for i > k and that V k is contained in the group of elements f = ( f 1 , … , f n ) in ℬ n satisfying f i = x i for i > k and f k = x k + b k with b k ∈ ℂ ⁢ ( x k + 1 , … , x n ) . This implies U 0 = { id } . Hence the derived length of ℬ n is at most 2 ⁢ n .

Conversely, we will now prove that this derived length is at least 2 ⁢ n , i.e. that V 1 ≠ { id } . For this, we introduce the following notation. For each k ∈ { 1 , … , n } , each nonzero element a ∈ ℂ ⁢ ( x k + 1 , … , x n ) , and each element b ∈ ℂ ⁢ ( x k + 1 , … , x n ) , we define the dilatation d ⁢ ( k , a ) ∈ Bir ⁢ ( ℙ n ) and the elementary transformation e ⁢ ( k , b ) ∈ Bir ⁢ ( ℙ n ) by

and

where g k = a ⁢ x k , h k = x k + b and g i = h i = x i for i ≠ k .

If i ∈ { 1 , … , n - 1 } and if G is a subgroup of Bir ⁢ ( ℙ n ) , properties ( D i ) and ( E i ) for G are defined in the following way:

1. There exists a ∈ ℂ ⁢ ( x i + 1 ) ∖ ℂ such that d ⁢ ( i , a ) ∈ G .

2. For all b ∈ ℂ ⁢ ( x i + 1 ) we have e ⁢ ( i , b ) ∈ G .

For i = n , properties ( D n ) and ( E n ) are defined in the following slightly different way:

1. There exists a ∈ ℂ ∖ { 0 , 1 } such that d ⁢ ( n , a ) ∈ G .

2. For all b ∈ ℂ we have e ⁢ ( n , b ) ∈ G .

We will prove by decreasing induction on k that for each k ∈ { 1 , … , n } the two following assertions hold:

1. The group U k satisfies ( E i ) and ( D i ) for i ∈ { 1 , … , k } .

2. The group V k satisfies ( E i ) for i ∈ { 1 , … , k } and ( D i ) for i ∈ { 1 , … , k - 1 } .

We will use two obvious identities. The first one is

where we have either i ∈ { 1 , … , n - 1 } , a ∈ ℂ ⁢ ( x i + 1 ) ∖ { 0 } , b ∈ ℂ ⁢ ( x i + 1 ) , or i = n , a ∈ ℂ * , b ∈ ℂ .

The second one is

where i ∈ { 1 , … , n - 1 } , a ∈ ℂ ⁢ ( x i + 1 ) ∖ { 0 } , c ∈ ℂ .

The identity (A.1) implies that if a subgroup G of Bir ⁢ ( ℙ n ) satisfies ( D i ) and ( E i ) , then its derived subgroup D 1 ⁢ ( G ) also satisfies ( E i ) .

Before making use of (A.2), let us check that if a ∈ ℂ ⁢ ( x ) is nonconstant and c ∈ ℂ is nonzero, then a ⁢ ( x ) / a ⁢ ( x - c ) is nonconstant. Otherwise, we would have a ⁢ ( x ) = λ ⁢ a ⁢ ( x - c ) for some λ ∈ ℂ * . This implies that the union U ⁢ ( a ) of the zeros and poles of a is invariant by translation by c. Since U ⁢ ( a ) is finite, it must be empty, proving that a is constant. A contradiction.

It follows from (A.2) and the last observation that if a subgroup G of Bir ⁢ ( ℙ n ) satisfies ( D i ) and ( E i + 1 ) , then D 1 ⁢ ( G ) also satisfies ( D i ) .

We are now ready for the induction. We begin by noting that the hypothesis ( a n ) is obviously satisfied. It is then enough to observe that we have ( a k ) ⇒ ( b k ) for each k ∈ { 1 , … , n } and ( b k ) ⇒ ( a k - 1 ) for each k ∈ { 2 , … , n } . ∎

Example B.1.

The subgroups 𝕋 y ⁢ ( y - 1 ) and 𝕋 y of PGL 2 ⁢ ( ℂ ⁢ ( y ) ) are conjugate in Jonq since the curves x 2 = y ⁢ ( y - 1 ) and x 2 = y both have genus 0. However, since the odd supports S odd ⁡ ( y ⁢ ( y - 1 ) ) and S odd ⁡ ( y ) are respectively equal to { 0 , 1 } and { 0 , ∞ } , Proposition 10.11 asserts that 𝕋 y ⁢ ( y - 1 ) is a Borel subgroup of PGL 2 ⁢ ( ℂ ⁢ ( y ) ) ⋊ Aff 1 but that 𝕋 y is not. In fact, the unique Borel subgroup of PGL 2 ⁢ ( ℂ ⁢ ( y ) ) ⋊ Aff 1 containing 𝕋 y is 𝕋 y ⋊ 𝕋 1 , 2 ; see Theorem 12.1. This also shows that even if 𝕋 y ⁢ ( y - 1 ) is a Borel subgroup of PGL 2 ⁢ ( ℂ ⁢ ( y ) ) ⋊ Aff 1 , it is no longer a Borel subgroup of Jonq .