Julie Déserti

THE EMBEDDINGS OF THE HEISENBERG GROUP INTO THE CREMONA GROUP

Keywords: 2010 Mathematics Subject Classification. 14E07, 14E05 Cremona group, birational map

In this note we describe the embeddings of the Heisenberg group into the Cremona group.

INTRODUCTION

The Heisenberg group is the non-abelian nilpotent group given by

H = f, g | [f, g] = h, [f, h] = [g, h] = id .
It has two generators, f and g, and h is the generator of the center of H . The Cremona group is the group Bir(P 2 C ) of birational maps of the projective plane P 2 C into itself. Such maps can be written in the form (x : y : z) P 0 (x, y, z) : P 1 (x, y, z) : P 2 (x, y, z)

where P 0 , P 1 , P 2 ∈ C[x, y, z] are homogeneous polynomials of the same degree, and this degree is the degree of the map, if the polynomials have no common factor (of positive degree). Recall that if φ is a birational self map of the complex projective plane, then one of the following holds ([Giz80, Can01, DF01, BD15]):

⋄ the sequence (deg(φ n )) n∈N is bounded, and φ is said to be elliptic; ⋄ the sequence (deg(φ n )) n∈N grows linearly with n, and φ is said to be a Jonquières twist;

⋄ the sequence (deg(φ n )) n∈N grows quadratically with n, and φ is said to be a Halphen twist;

⋄ (deg(φ n )) n∈N grows exponentially fast with n, and φ is said to be hyperbolic.

Proposition A. Let ρ be an embedding of H into the Cremona group. Then ρ(H )

does not contain hyperbolic birational maps. More precisely ρ(f) and ρ(g) are either elliptic birational maps, or Jonquières twists.

We describe the embeddings of H into the Cremona group. In [D 06] we already looked at such embeddings but with the following assumption: the images of f and g are elliptic birational self maps. There are other embeddings of H into the Cremona group:

Theorem B. Let ρ be an embedding of H into the Cremona group. Then up to birational conjugacy

⋄ either ρ(H ) is a subgroup of PGL(3, C) and ρ(f) = (x + αy, y + β) ρ(g) = (x + γy, y + δ)
with α, β, γ, δ in C such that αδβγ = 1; ⋄ or ρ(H ) is a subgroup of the group of polynomial automorphisms of C 2 and (ρ(f), ρ(g)) is one of the following pairs

(ax + Q(y), y + c), (αx + P(y), y + γ) , ax + Q(y), by + γ(b -1) β -1 , αx + P(y), βy + γ with β ∈ C * {1}, a, α, b in C * , c, γ in C and P, Q in C[y]; ⋄ or ρ(f) is a Jonquières twist and (ρ(f), ρ(g)) is one of the following pairs x, δx ±1 y , γx, ya(x) ,
x, δx ±2 y , γx, ya(x) , (-x, δx ±1 y), (γx, yb(x)) , (λx, yc(x)), (δx, yd(x))

with δ, γ ∈ C * , λ ∈ C * {1, -1} and a, b, c, d ∈ C(x) * such that b(x) b(-x) ∈ C * , c(δx)d(x) c(x)d(λx) ∈ C * .
Remark C. Note that the two last families are not empty. For instance (-x, αx ±1 y), (βx, γx 2 y) , (λx, αx p y), (γx, βx q y) with α, β, γ ∈ C * , λ ∈ C * {1, -1}, p, q ∈ N are such pairs.

1. SOME RECALLS 1.1. About birational maps of the complex projective plane. Let φ be a birational self map of the complex projective plane. Then one of the following holds ([Giz80, Can01, DF01, BD15]):

⋄ φ is elliptic if and only if the sequence (deg(φ n )) n∈N is bounded. In this case there exist a birational map ψ : S P 2 C and an integer k ≥ 1 such that ψ -1 • φ k • ψ belongs to the connected component of the identity of the group Aut(S). Either φ is of finite order, or φ is conjugate to an automorphism of P 2 C , which restricts to one of the following automorphisms on some open subset isomorphic to C 2 :

• (x, y) → (αx, βy) where the kernel of the group morphism

Z 2 → C 2 (i, j) → α i β j
is generated by (k, 0) for some k ∈ Z;

there is a constant c(φ) such that deg(φ n ) = c(φ)λ(φ) n + O(1).
1.2. About distorted elements. If G is a group generated by a finite subset F ⊂ G the F-length |g| F of an element g of G is defined as the least non-negative integer ℓ such that g admits an expression of the form g = f 1 f 2 . . . f ℓ where each f i belongs to

F ∪ F -1 . We say that g is distorted if lim k→+∞ |g k | F k
= 0 (note that the limit lim k→+∞ |g k | F k always exists and is a real number since the sequence k → |g k | F is subadditive). This notion actually does not depend on the chosen F, but only on the pair (g, G).

If G is any group, an element g ∈ G is distorted if it is distorted in some finitely generated subgroup of G.

The element h of

H = f, g| [f, g] = h, [f, h] = [g, h] = id
satisfies the following property:

∀ k ∈ Z h k 2 = [f k , g k ] = f k g k f -k g -k so ||h k 2 || ≤ 4k and lim k→+∞ ||h k || k 2 = 0. Hence h is distorted. An element φ ∈ Bir(P 2 C
) is said to be algebraic if it is contained in an algebraic subgroup of Bir(P 2 C ). By [BF13, §2.6] the map φ ∈ Bir(P 2 C ) is algebraic if and only if the sequence (deg(φ n )) n∈N is bounded. In other words elliptic elements and algebraic elements coincide. By [BD15, Proposition 2.3] this is also equivalent to say that φ is of finite order or conjugate to an element of Aut(P 2 C ). A straightforward computation shows that every element of Aut(P 2 C ) is distorted in Bir(P 2 C ) (see [BF19, Lemma 4.40]). As a consequence every algebraic element of Bir(P 2 C ) is distorted. The converse statement also holds:

Theorem 1.1 ([BF19, CC19]). Any distorted element of Bir(P 2 C ) is elliptic.
Corollary 1.2. Let ρ be an embedding of H into Bir(P 2 C ). Then ρ(h) is elliptic. We will use this corollary and the following description of the centralizer of hyperbolic birational maps to prove Proposition A:

Proposition 1.3 ([BC16]
). Let φ be a birational map of the complex projective plane. If φ is hyperbolic, then the infinite cyclic group generated by φ is a finite index subgroup of the centralizer of φ in Bir(P 2 C ).

Proof of the first part of Proposition A. Assume that ρ(H ) contains an hyperbolic element φ. Since ρ(h) is the generator of the center of ρ(H ), ρ(h) commutes with φ.

Proposition 1.3 implies that either ρ(h) is hyperbolic, or ρ(h) is of finite order. But ρ(h) is not hyperbolic Corollary (1.
2) and by definition ρ(h) is of infinite order. As a result ρ(H ) does not contain hyperbolic element.

1.3. About centralizers of elliptic birational maps. Let us recall the description of the centralizers of the elliptic birational self maps of infinite order of the complex projective plane obtained in [START_REF] Blanc | Degree growth of birational maps of the plane[END_REF].

Consider φ an elliptic element of Bir(P 2 C ). Assume that φ is of infinite order. As recalled in §1.1 the map φ is conjugate to an automorphism of P 2

C which restricts to one of the following automorphisms on some open subset isomorphic to C 2 :

(1) (αx, βy) where α, β belong to C * ;

(2) (αx, y + 1) where α ∈ C * .

In case (1) the centralizer of φ in Bir

(P 2 C ) is (η(x), ya(x k )) | a ∈ C(x), η ∈ PGL(2, C), η(αx) = αη(x) ;
in particular the elements of the centralizer of φ are elliptic birational maps or Jonquières twists. In case (2) the centralizer of φ in Bir

(P 2 C ) is (η(x), y + a(x)) | η ∈ PGL(2, C), η(αx) = αη(x), a ∈ C(x), a(αx) = a(x) ;
in particular the elements of the centralizer of φ are elliptic birational maps.

Corollary 1.2 and the previous description imply:

Lemma 1.4. Let ρ be an embedding of H into Bir(P 2 C ). Then ρ(h) is elliptic and up to birational conjugacy ⋄ either ρ(h) = (αx, βy), where the kernel of the group morphism

Z 2 → C 2 (i, j) → α i β j
is generated by (k, 0) for some k ∈ Z and both ρ(f), ρ(g) belong to

(η(x), ya(x k )) | a ∈ C(x), η ∈ PGL(2, C), η(αx) = αη(x) ;
⋄ or ρ(h) = (αx, y + 1), and both ρ(f), ρ(g) belong to

(η(x), y + a(x)) | η ∈ PGL(2, C), η(αx) = αη(x), a ∈ C(x), a(αx) = a(x) .
In particular ρ(f) and ρ(g) are elliptic birational maps or Jonquières twists.

It ends the proof of Proposition A.

PROOF OF THEOREM B

2.1. Assume that all the generators of ρ(H ) are elliptic. The group Aut(C 2 ) of polynomial automorphisms of C 2 is a subgroup of Bir(P 2 C ). It is generated by the group

A = (a 0 x + a 1 y + a 2 , b 0 x + b 1 y + b 2 ) | a i , b i ∈ C, a 0 b 1 -a 1 b 0 = 0 and E = (αx + P(y), βy + γ) | α, β ∈ C * , γ ∈ C, P ∈ C[y] .
Let us recall the following result obtained when we study the embeddings of SL(n, Z) into the Cremona group: Proof. The first assertion follows from Lemma 2.1. Let us focus on the second one.

If ρ(h) belongs to E and ρ(h 2 ) = (x + P(y), y), then ρ(h) = (εx + Q(y), η(y)) with ε 2 = 1, Q ∈ C[y] and η(y) ∈y + γ, y . But ρ(f) and ρ(g) belong to E so [ρ(f), ρ(g)] = ρ(h) implies that ε = 1 and η(y) = y, i.e. ρ(h) = (x + Q(y), y). Set ρ(f) = (ax + R(y), by + c), ρ(g) = (αx + P(y), βy + γ).

The second component of ρ(f)ρ(g) has to be equal to the second component of

ρ(h)ρ(g)ρ(f), that is βby + bγ + c = βby + βc + γ;
in other words either β = b = 1, or c = γ(b-1) β-1 . 2.2. Assume that ρ(f) is a Jonquières twist with trivial action on the basis of the fibration. Since ρ(h) is elliptic, then up to birational conjugacy either ρ(h) = (αx, βy), or ρ(h) = (αx, y + 1) (see §1.1). But ρ(f) belongs to the centralizer of ρ(h) and is a Jonquières twist; therefore according to §1.3 one has: ρ(h) = (αx, βy), ρ(f) can be written as (x, ya(x)) and ρ(g) as (µ(x), yb(x)) with µ ∈ PGL(2, C) and a, b ∈

C(x) * . Let us remark that if µ = id, then [ρ(f), ρ(g)] = ρ(h) implies α = β = 1 so µ = id.
The relation [ρ(f), ρ(g)] = ρ(h) implies that α = 1, and a(µ(x)) = βa(x). Let us first look at polynomials P such that P(µ(x)) = βP(x):

Claim 2.3. If P is a non-zero polynomial such that P(µ(x)) = λ 2 P(x), λ 2 = 1, then one of the following holds:

⋄ P(x) = δ γ λ 2 -1 + x , µ(x) = γ + λ 2 x with a, δ ∈ C; ⋄ P(x) = δ γ λ+1 -x 2 , µ(x) = γ -λx with γ ∈ C, and δ ∈ C * ; ⋄ P(x) = δ γ λ-1 + x 2 , µ(x) = γ + λx with γ ∈ C, and δ ∈ C * .
Proof. Let us consider the set Z P = z | P(z) = 0 of roots of P. It is a finite set invariant by µ. As a result µ n |Z P = id for some integer n.

If #Z P ≥ 3, then µ n |Z P = id implies µ n = id. Recall that ρ(f) = (x, ya(x)), ρ(g) = (µ(x), yb(x)), ρ(h) = (αx, βy) so ρ(f) n = (x, yA(x)), ρ(g) n = (µ n (x), yB(x)) = (x, yB(x)), ρ(h) n 2 = (α n 2 x, β n 2 y). Then [ρ(f) n , ρ(g) n ] = ρ(h) n 2 implies α n 2 = β n 2 = 1, that is ρ(h) is of finite order: contradiction.
Hence #Z P ≤ 2 so deg P ≤ 2. A straightforward computation implies the statement.

Let us come back to a(µ(x)) = βa(x). As β is of infinite order and a belongs to C(x) * we can rewrite this equality as follows:

P(µ(x)) Q(µ(x)) = λ 2 1 P(x) λ 2 2 Q(x) where ⋄ λ 1 , λ 2 are two elements of C {±1} such that β = λ 2 1 λ 2 2 ;
⋄ P, Q are two polynomials without common factor.

As a result up to birational conjugacy (ρ(f), ρ(g)) is one of the following pairs x, δxy , λx, yb(x)

x, δx 2 y , λx, yb(x)

x, y δx , λx, yb(x)

x, y δx 2 , λx, yb(x) with δ ∈ C * , λ ∈ C * of infinite order and b ∈ C(x).

We can thus state Proposition 2.4. Let ρ be an embedding of H into Bir(P 2 C ). If ρ(f) is a Jonquières twist with trivial action on the basis of the fibration, then up to birational conjugacy (ρ(f), ρ(g)) is one of the following pairs x, δxy , λx, yb(x)

x, δx 2 y , λx, yb(x)

x, y δx , λx, yb(x) x, y δx 2 , λx, yb(x) with δ ∈ C * , λ ∈ C * of infinite order and b ∈ C(x).

2.3.

Assume that ρ(f) is a Jonquières twist with non-trivial action on the basis of the fibration. Since ρ(h) is elliptic and of infinite order, then up to birational conjugacy either ρ(h) = (αx, βy), or ρ(h) = (αx, y + 1) (see §1.1). But ρ(f) belongs to the centralizer of ρ(h) and is a Jonquières twist; therefore according to §1.3 one has: ρ(h) = (αx, βy), ρ(f) can be written as (η(x), ya(x)) and ρ(g) as (µ(x), yb(x)) with η, µ in PGL(2, C) and a, b in C(x). Note that a direct computation implies

Up to conjugacy by an element of

ν ∈ PGL(2, C) | ν(αx) = αν(x) =    PGL(2, C) if α = 1 βx ±1 | β ∈ C * if α = -1 βx | β ∈ C * if α 2 = 1 (2.1)
so when η is an homothety we will have to distinguish the cases λ = -1 and λ = -1.

2.3.1. Assume that η(x) = x + 1. Since ρ(f) and ρ(h) commute, ρ(h) can be written as (x + γ, βy).

If γ = 0, then [ρ(f), ρ(h)] = id leads to a(x + γ) = a(x), that is a(x) = a ∈ C: con- tradiction with the fact that ρ(f) is a Jonquières twist. If γ = 0, then ρ(h) = (x, βy) and [ρ(f), ρ(g)] = ρ(h) leads to ρ(g) = (x + µ, yb(x)) and b(x)a(x+µ) = βa(x)b(x+1). Let us write a as P Q and b as R S with P, Q, R, S ∈ C[y] then b(x)a(x + µ) = βa(x)b(x + 1) can be rewritten P(x + µ)Q(x)R(x)S(x + 1) = βP(x)Q(x + µ)R(x + 1)S(x) (2.2)
Denote by p i (resp. q ℓ , resp. r j , resp. s k ) the coefficient of the highest term of P (resp.

Q, resp. R, resp. S). The coefficient of the highest term of the left-hand side of (2.2) has to be equal to the coefficient of the highest term of the right-hand side of (2.2), that is p i q ℓ r j s k = βp i q ℓ r j s k . So β = 1, i.e. ρ(h) = (x, y): contradiction.

2.3.2. Suppose that η(x) = -x, i.e. ρ(f) = (-x, ya(x)).

Remark 2.5. The map ρ(f) 2 = (x, ya(x)a(-x)) is a Jonquières twist that preserves fiberwise the rational fibration x = cst; consequently Proposition 2.4 says that ρ(f) 2 is one of the following:

x, δxy , x, δx 2 y , x, δ y x , x, δ y x 2 with δ ∈ C * . Let us try to determine ρ(f). If ρ(f) 2 = (x, δxy), then we have to consider the equation a(x)a(-x) = δx. The right-hand side of this equation is invariant by x → -x whereas the left-hand side not, so there is no solution. The same holds if ρ(f) 2 = x, δ y

x . Consequently ρ(f) 2 is one of the following:

x, δx 2 y , x, δ y x 2 with δ ∈ C * and ρ(f) is thus one of the following:

-x, ζxy , -x, ζ y x with ζ ∈ C * .
Since f and h commute, then (2.1) implies that either ρ(h) = α

x , βy , or ρ(h) = (αx, βy). Let us consider these two cases.

⋄ Assume first that ρ(h) = α

x , βy . Note that α x , βy does not commute neither tox, ζxy , nor to -x, ζ y

x : contradiction with [ρ(f), ρ(h)] = id. ⋄ Suppose now that ρ(h) = (αx, βy).

• If α 2 = 1, then [ρ(g), ρ(h)] = id and (2.1) imply that ρ(g) = (γx, yb(x)).

Then [ρ(f), ρ(g)] = ρ(h) leads to α = 1: contradiction with α 2 = 1.

• If α = -1, that is ρ(h) = (-x, βy), then according to [ρ(g), ρ(h)] = id and (2.1) we get that either ρ(g) = (γx, yb(x)), or ρ(g) = γ

x , yb(x) . In both cases the relation [ρ(f), ρ(g)] = ρ(h) leads to a contradiction.

• If α = 1, i.e. ρ(h) = (x, βy), then [ρ(f), ρ(g)] = ρ(h) implies that either ρ(g) = (γx, yb(x)) or ρ(g) = γ
x , yb(x) . First let us assume that ρ(g) = (γx, yb(x)). If ρ(f) = (-x, ζxy), then

[ρ(f), ρ(g)] = ρ(h) leads to γb(x) = βb(-x), that is b(x) b(-x) belongs to C * . If ρ(f) = -x, ζ y x , then [ρ(f), ρ(g)] = ρ(h) implies b(x) = βγb(-x), that is b(x) b(-x) belongs to C * . Finally suppose that ρ(g) = γ x , yb(x) . If ρ(f) = (-x, ζxy), then [ρ(f), ρ(g)] = ρ(h) leads to γb(x) = βx 2 b(-x). Write b as P Q with P, Q in C[x]; then γb(x) = βx 2 b(-x) is equivalent to γP(x)Q(-x) = βx 2 P(-x)Q(x)
and the degree of the left-hand side is deg P + deg Q whereas the degree of the right-hand side is deg

P + deg Q + 2: contradiction. If ρ(f) = -x, ζ y
x , then a straightforward computation implies similarly a contradiction.

Proposition 2.6. Let ρ be an embedding of H into Bir(P 2 C ). If ρ(f) is a Jonquières twist with a order 2 action on the basis of the fibration, then up to birational conjugacy (ρ(f), ρ(g)) is one of the following pairs

(-x, αxy), (βx, ya(x)) , -x, α y x , (βx, ya(x)) with α, β ∈ C * and a ∈ C(x) * such that a(x) a(-x) ∈ C * . 2.3.3. Assume that η(x) = λx, λ 2 = 1. Recall that ρ(f) = (λx, ya(x)), ρ(g) = (µ(x), yb(x)), ρ(h) = (υ(x), βy) with λ in C * {1, -1}, β in C * , µ, υ in PGL(2, C), and a, b in C(x) * .
First note that since ρ(f) and ρ(h) commute, υ(λx) = λυ(x). According to (2.1), the homography υ is an homothety (recall that λ 2 = 1): υ

(x) = γx with γ ∈ C * . The relations [ρ(f), ρ(g)] = ρ(h), [ρ(f), ρ(h)] = [ρ(g), ρ(h)] = id imply the follo- wing ones a(x) = a(γx) (2.3) b(x) = b(γx) (2.4) µ(γx) = γµ(x) (2.5) λµ(x) = γµ(λx) (2.6) b(x)a(µ(x)) = βa(x)b(λx) (2.7)
We will distinguish the cases γ = 1, γ = -1, γ 2 = 1.

⋄ Assume that γ 2 = 1. Then (2.1) and (2.5) lead to µ(x) = µx with µ ∈ C * . Equation (2.6) can be rewritten λµx = γµλx, that is γ = 1: contradiction with the assumption γ 2 = 1.

⋄ Suppose that γ = 1. Then λ 2 = 1, (2.1) and (2. ρ(f) 4 can be written as (x, δxy), or (x, δx 2 y), or x, y δx , or x, y δx 2 . If ρ(f) 4 = (x, δxy), then δx = a(x)a(ix)a(-x)a(-ix); but the right-hand side of this equality is invariant by x → ix whereas the left-hand side is not. As a consequence ρ(f) 4 can not be written (x, δxy). Similarly one sees that ρ(f) 4 can not be written (x, δx 2 y), x, y δx , and x, y δx 2 . Thus λ = i. • Similarly one gets that the case λ = -i does not happen.

Proposition 2.7. Let ρ be an embedding of H into Bir(P 2 C ). If ρ(f) is a Jonquières twist with an action on the basis of the fibration that is neither trivial, nor of order 2, then up to birational conjugacy (ρ(f), ρ(g)) is one of the following pairs

  Lemma 2.1 ([D 06]). Let ρ be an embedding of H into Bir(P 2 C ). If ρ(f), ρ(g) and ρ(h) are elliptic, then up to birational conjugacy ⋄ either ρ(H ) is a subgroup of PGL(3, C), and more preciselyρ(f) = (x + αy, y + β) ρ(g) = (x + γy, y + δ) with α, β, γ, δ ∈ C such that αδβγ = 1; ⋄ or ρ(H ) is a subgroup of E and ρ(h 2 ) = (x + P(y), y) for some P ∈ C[y].This statement implies the following one: Proposition 2.2. Let ρ be an embedding from H into Bir(P 2 C ). Assume that ρ(f), ρ(g) and ρ(h) are elliptic.Then up to birational conjugacy⋄ either ρ(H ) is a subgroup of PGL(3, C), more precisely ρ(f) = (x + αy, y + β) ρ(g) = (x + γy, y + δ) with α, β, γ, δ ∈ C such that αδβγ = 1; ⋄ or ρ(H )is a subgroup of E and (ρ(f), ρ(g)) is one of the following pairs (ax + Q(y), y + c), (αx + P(y), y + γ) ax + Q(y), by + γ(b -1) β -1 , αx + P(y), βy + γ with a, α, b in C * , c, γ in C, β ∈ C * {1} and P, Q in C[y].

  C) one can assume that either η(x) = x + 1, or η(x) = λx (remark that this conjugacy doesn't preserve the first component of ρ(h)).

  6) lead to µ(x) = µx with µ ∈ C * . In other wordsρ(f) = (λx, ya(x)), ρ(g) = (µx, yb(x)) with λ in C * {1, -1}, µ in C * , a, b in C(x) such that a(µx)b(x) a(x)b(λx) belongs to C * . ⋄ Assume that γ = -1. Then (2.1) and (2.5) imply that µ(x) = µx ±1 with µ ∈ C * . If µ(x) = µx, then (2.6) can be rewritten λµx = -λµx: contradiction. If µ(x) = µ x , then (2.6) can be rewritten λµ x = -µ λx ; hence λ 2 = -1. • If λ = i, then ρ(f) = (ix, ya(x)) and ρ(f) 4 = x, ya(x)a(ix)a(-x)a(-ix) preserves fiberwise the fibration x = cst. According to Proposition 2.4

(

  (λx, ya(x)), (µx, yb(x)))with λ ∈ C * {1, -1}, µ ∈ C * , a, b ∈ C(x) * such that a(µx)b(x) a(x)b(λx) ∈ C * .

  S → B onto a curve B such that ψ -1 • φ • ψ permutes the fibers of π. If (deg(φ n )) n∈N grows linearly, then the fibration π is rational and φ is said to be a Jonquières twist. If (deg(φ n )) n∈N grows quadratically, then the fibration π is elliptic and φ is said to be a Halphen twist.⋄ φ is hyperbolic if and only if (deg(φ n )) n∈N grows exponentially fast with n:

	• (x, y) → (αx, y + 1) where α ∈ C * .
	⋄ φ is parabolic if and only if the sequence (deg(φ n )) n∈N grows linearly or
	quadratically with n. If φ is parabolic, there exist a birational map ψ : S	P 2 C
	and a fibration π :

Supported by the ANR grant Fatou ANR-17-CE40-0002-01 and the ANR grant Foliage ANR-16-CE40-0008-01. 1