This is a preprint of: “Zero entropy for some birational maps of C 2 ”, Anna Cima, Sundus Zafar, J. Math. Anal. Appl., vol. 474, 765–781, 2019. Zero entropy for some birational maps of C2∗ Anna Cima(1) and Sundus Zafar(1) (1) Dept. de Matemàtiques, Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain cima@mat.uab.cat, sundus@mat.uab.cat Abstract This work deals with a special case of family of birational maps f : C2 → C2 dynamically classified in [9]. In this work we study the zero entropy sub families of f . The sequence of degrees dn associated to the iterates of f is found to grow periodically, linearly, quadratically or exponentially. Explicit invariant fibrations for zero entropy families and all the integrable and periodic mappings inside the family f are given. Mathematics Subject Classification 2010: 14E05, 26C15, 34K19, 37B40, 37C15, 39A23, 39A45. Keywords: Birational maps, Algebraic entropy, First Integrals, Fibrations, Blowing-up, Integrability, Periodicity, Chaos. 1 Introduction Consider the family of fractional maps f : C2 → C2 of the form:   β0 + β1 x + β2 y f (x, y) = α0 + α1 x + α2 y, , (γ1 , γ2 ) 6= (0, 0) (1) γ0 + γ1 x + γ2 y where the parameters αi , βi , γi , i ∈ {0, 1, 2} are complex numbers. In this work the family of mappings f (x, y) in (1) is required to be birational in general. The values of parameters αi , βi , γi , i ∈ {0, 1, 2} for which f (x, y) is a birational mapping is discussed in Lemma 1 in this article. The study of the dynamics generated by birational mappings in the plane and their classification is a well discussed topic in recent years as can ∗ Acknowledgements. The first author is supported by Ministry of Economy, Industry and Competi- tiveness of the Spanish Government through grants MINECO/FEDER MTM2016-77278 and also supported by the grant 2014-SGR-568 from AGAUR, Generalitat de Catalunya. The GSD-UAB Group is supported by the Government of Catalonia through the SGR program. It is also supported by MCYT through the grant MTM2008 − 03437. http://www.gsd.uab.cat 1 be found in [1, 3, 4, 12, 13, 15, 14, 18, 19, 20, 21, 22, 23, 26]. The family of mappings f (x, y) in (1) is dynamically classified in [9]. In this work we are going to study the mapping f in case where it shows a kind of degenerate behavior for general values of parameters. For a birational map f (x, y) the sequence of degrees dn of the iterates of f satisfies a homogeneous linear recurrence, see [16]. This is governed by the characteristic polynomial X (x) of a certain matrix associated to F, where F : P C2 → P C2 is the extension of f : C2 → C2 in the projective plane P C2 . This further provides information regarding the quantity called as the dynamical degree of F and defined as 1 δ(F ) := lim (deg(F n )) n , (2) n→∞ where F n represents the iterates of F. The logarithm of δ(F ) is the algebraic entropy of F , see [3, 4, 5, 15, 16]. Considering the embedding (x1 , x2 ) ∈ C2 7→ [1 : x1 : x2 ] ∈ P C2 into projective space, the induced map F : P C2 → P C2 has three components Fi [x0 : x1 : x2 ] , i = 1, 2, 3 which are homogeneous polynomials as F [x0 : x1 : x2 ] = [F1 [x0 : x1 : x2 ] : F2 [x0 : x1 : x2 ] : F3 [x0 : x1 : x2 ]], where F1 [x0 : x1 : x2 ] = x0 (γ0 x0 + γ1 x1 + γ2 x2 ), F2 [x0 : x1 : x2 ] = (α0 x0 + α1 x1 + α2 x2 )(γ0 x0 + γ1 x1 + γ2 x2 ), (3) F3 [x0 : x1 : x2 ] = x0 (β0 x0 + β1 x1 + β2 x2 ). The map F has degree two as the components of F do not have common factors for general values of the parameters. Similarly the degree of each iterate of F can be found in general by iterating F and removing the common homogenous components as F n = F ◦···◦F for each n ∈ N. Birational mappings F : P C2 → P C2 have an indeterminacy set I(F ) of points where F is ill-defined as a continuous map. Hence they also have a set of curves which are sent to a single point called the exceptional locus of F denoted as E(F ). Generically the mappings of the form (1) have three indeterminacy points. The exceptional locus is formed by three straight lines, each two of them intersecting on a single indeterminate point of F . We call them as non degenerate mappings. However in some cases exceptional locus is formed by only two straight lines. In this case these mappings are identified as degenerate mappings. Lemma 1 in preliminary results section discusses the conditions for birationality and degeneracy of family f in (1). This study includes all the subfamilies of f (x, y) where it shows a degenerate behavior. The cases where exceptional locus is formed by three straight lines are discussed and studied in the papers [8] and [9]. Such cases are recognized as non degenerate mappings. 2 The first goal of this study is look for sequence of degrees dn . This is done by performing a series of blow-up’s in order to find the characteristic polynomial which determines the behaviour of dn . The second goal is to identify for which values of the parameters these mappings have zero algebraic entropy and extract dynamical consequences. For this we use the results of Diller and Favre, see [16], which characterize the growth rate of dn with the existence of invariant fibrations. We find all the prescribed invariant fibrations in each one of this cases. We emphasize which elements of the family are integrable mappings. We also distinguish all the periodic mappings giving a pair of first integrals generically transverse. The article is organized as follows: Section two is devoted to give some preliminary results on birational maps and family f , in order to describe the blow-up process and the Picard group. In Section three we study the subfamily α1 γ2 − α2 γ1 = 0, while in Section four we study the subfamily α1 β2 − α2 β1 = 0. The results that we get are the following. We have named Theorem the results on the dynamical degree and the growth of dn and Proposition the results on the zero entropy and existence of invariant fibrations. In this way in Section 3 we give Theorem 4 with Proposition 5 concerning the family α1 γ2 − α2 γ1 = 0. Section 4 which deals with mappings satisfying α1 β2 − α2 β1 = 0, is splitted in three subsections. We present Theorem 6 with Proposition 7 when γ1 γ2 6= 0, Theorem 8 with Proposition 9 (α2 6= 0) and Proposition 11 (α2 = 0) when γ1 = 0 and Theorem 12 with Proposition 13 when γ2 = 0. 2 Preliminary results Consider the mapping F [x0 : x1 : x2 ] : P C2 → P C2 in (3), then the exceptional locus of F [x0 : x1 : x2 ] is given as E(F ) = {S0 , S1 , S2 }, where S0 = {x0 = 0}, S1 = {γ0 x0 + γ1 x1 + γ2 x2 = 0}, S2 = {(α1 (βγ)02 − α2 (βγ)01 ) x0 + α1 (βγ)12 x1 + α2 (βγ)12 x2 = 0}. We have used the notation: (δ)ij = δi j − δj i . The exceptional locus of F −1 [x0 : x1 : x2 ] is E(F −1 ) = {T0 , T1 , T2 }, where T0 = {(γ0 (αβ)12 − γ1 (αβ)02 + γ2 (αβ)01 ) x0 − (βγ)12 x1 = 0} , T1 = {(αβ)12 x0 − (αγ)12 x2 = 0}, T2 = {x0 = 0}. The birational map F [x0 : x1 : x2 ] has an indeterminacy set I(F ) of points where F is ill-defined as a continuous map. This set is given by: {[x0 : x1 : x2 ] ∈ P C2 : F1 [x0 : x1 : x2 ] = 0, F2 [x0 : x1 : x2 ] = 0, F3 [x0 : x1 : x2 ] = 0]}, 3 which gives: I(F ) = {O1 , O2 , O3 }, where O0 = [(βγ)12 : (βγ)20 : (βγ)01 ], O1 = [0 : α2 : −α1 ], O2 = [0 : γ2 : −γ1 ], and (βγ)ij := βi γj − γj βi for i, j = 0, 1, 2. By calling g(x, y) the inverse of f (x, y) given in (1) and considering G[x0 : x1 : x2 ] its extension on P C2 , also a indeterminancy set I(G) exists i.e. I(G) = {A1 , A2 , A3 }, where A0 = [0 : 1 : 0], A1 = [0 : 0 : 1], A2 = [(βγ)12 (αγ)12 , (α0 (βγ)12 − α1 (βγ)02 + α2 (βγ)01 ) (αγ)12 : (αβ)12 (βγ)12 ]. We are interested in the birational mappings (1) when the corresponding F only has two distinct exceptional curves. Next lemma informs about the set of parameters which are available in this study. Recall that a birational map is a map f : C2 → C2 with rational components such that there exists an algebraic curve V and another rational map g such that f ◦ g = g ◦ f = id in C2 \ V. Lemma 1. Consider the mappings   β0 + β1 x1 + β2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0). γ0 + γ1 x1 + γ2 x2 Then: (a) The mapping f is birational if and only if the vectors (β0 , β1 , β2 ), (γ0 , γ1 , γ2 ) are linearly independent and ((αβ)12 , (αγ)12 ) 6= (0, 0), ((αγ)12 , (βγ)12 ) 6= (0, 0), and either ((αβ)12 , (βγ)12 ) 6= (0, 0) or (β1 , β2 ) = (0, 0). (b) The mapping f is degenerate if and only if (βγ)12 = 0 or (αγ)12 = 0. Proof. The conditions in (a) are necessary for f to be invertible as if the vectors (β0 , β1 , β2 ), (γ0 , γ1 , γ2 ) are linearly dependent then the second component of f is a constant, also if ((αβ)12 , (αγ)12 ) = (0, 0) or ((αγ)12 , (βγ)12 ) = (0, 0) then f only depends on α1 x1 + α2 x2 or on γ1 x1 + γ2 x2 . If ((αβ)12 , (βγ)12 ) = (0, 0) and (β1 , β2 ) 6= (0, 0) then f only depends on β1 x1 + β2 x2 . Now assume that conditions (a) are satisfied. Then the inverse of f which formally is   −1 −(αβ)02 + β2 x + (αγ)02 y − γ2 xy (αβ)01 − β1 x + (αγ)10 y + γ1 xy f (x, y) = , , (αβ)12 − (αγ)12 y (αβ)12 − (αγ)12 y 4 is well defined. Furthermore the numerators of the determinants of the Jacobian of f and f −1 are α1 (βγ)02 − α2 (βγ)01 + α1 (βγ)12 x + α2 (βγ)12 y (4) and α0 (βγ)12 − α1 (βγ)02 + α2 (βγ)01 − (βγ)12 y, (5) respectively. It is easily seen that conditions (a) imply that both (4) and (5) are not identically zero. Hence, f ◦ f −1 = f −1 ◦ f = id in C2 \ V, where V is the algebraic curve determined by the common zeros of (4) and (5). To see (b) we know that since Si maps to Ai , this implies that the points A0 , A1 , A2 are not all distinct. Since A0 6= A1 we have two possibilities: A0 = A2 or A1 = A2 . Condition A0 = A2 writes as (βγ)12 (αγ)12 = 0 and (αβ)12 (βγ)12 = 0. From (a), the vector ((αβ)12 , (αγ)12 ) 6= (0, 0). Hence (βγ)12 must be zero. In a similar way it is seen that A1 = A2 if and only if (αγ)12 = 0. It is easy to see that F maps each Si to Ai and that the inverse of F maps Ti to Oi for i ∈ {0, 1, 2}. To specify this behaviour we write F : Si  Ai (also F −1 : Ti  Oi ). It is known that the dynamical degree depends on the orbits of A0 , A1 , A2 under the action of F (see Proposition of section 2). Indeed, the key point is whether the iterates of A0 , A1 , A2 coincide with any of the indeterminacy points of F. When we find such orbit of iterates of F that ends at some indeterminacy point of F we perform a series of blow up in order to remove the indeterminacy of F in the new extended space. For X = {((x, y), [u : v]) ∈ C2 × P C1 : xv = yu} and p ∈ C2 we let (X, π) to be the blowing-up of C2 at the point p. By translating p at the origin, π −1 p = π −1 (0, 0) = {((0, 0), [u : v])} := Ep ' P C1 and π −1 q = π −1 (x, y) = ((x, y), [x : y]) ∈ X for q = (x, y) 6= (0, 0). Every blow up gives a new expanded space X and a new induced map F̃ : X → X is defined on it. Indeterminacy sets and exceptional locus can also be defined by considering meromorphic functions on complex manifolds X we get after a series of blow ups. Consider the Picard group of X denoted by Pic(X), where X is the complex manifold. For a generic line L ∈ P C2 the Pic(P C2 ) is generated by the class of L. If the base points of the blow-ups are {p1 , p2 , . . . , pk } ⊂ P C2 and Ei := π −1 {pi } then it is known that Pic(X) is generated by {L̂, E1 , E2 , . . . , Ek }, see [3, 4]. The curve L̂ is the strict transform of L ∈ C2 is the adherence of π −1 (C \ {p}), in the Zariski topology. Furthermore π : X −→ P C2 induces a morphism of groups π ∗ : Pic(P C2 ) −→ Pic(X), with the property that for any complex curve C ⊂ P C2 , X π ∗ (C) = Ĉ + mi Ei , (6) 5 where mi is the algebraic multiplicity of C at pi . For F ∈ P C2 , F̃ is denoted as natural extension of F on X and it induces a morphism of groups, F̃ ∗ : Pic(X) → Pic(X) by Pk considering the classes of preimages such that F̃ ∗ (L̂) = d L̂ + i=1 ci Ei , ci ∈ Z, where d is the degree of F. By iterating F, we get the corresponding formula by changing F by F n and d by dn . To know the behavior of the sequence of degrees dn we deal with maps F̃ such that (F̃ n )∗ = (F̃ ∗ )n . (7) Maps F̃ satisfying condition (7) are called Algebraically Stable maps (AS for short), (see [16]). In order to get AS maps we will use the following useful result showed by Fornaess and Sibony in [17] (see also Theorem 1.14) of [16]: The map F̃ is AS if and only if for every exceptional curve C and all n ≥ 0 , F̃ n (C) ∈ / I(F̃ ). (8) It is known (see Theorem 0.1 of [16]) that one can always arrange for a birational map to be AS performing a finite number of blowing-up’s. If it is the case and we call P X (x) = xk + k−1 i ∗ i=0 ci x the characteristic polynomial of A := (F̃ ), then since X (A) = 0 and di is the (1, 1) term of Ai we get that dk = −(c0 + c1 d1 + c2 d2 + · · · + ck−1 dk−1 ), i.e., the sequence dn satisfies a homogeneous linear recurrence with constant coefficients. The dynamical degree is then the largest real root of X (x). The following is a direct consequence of Theorem 0.2 of [16]. It is quiet useful in our work. Given a birational map F of P C2 , let F̃ be its regularized map so that the induced map F̃ ∗ : Pic(X) → Pic(X) satisfies (F̃ n )∗ = (F̃ ∗ )n . Then Theorem 2. Let F : P C2 → P C2 be a birational map and let dn = deg(F n ). Then up to bimeromorphic conjugacy, exactly one of the following holds: • The sequence dn grows quadratically and F̃ is an automorphism preserving an elliptic fibration. • The sequence dn grows linearly and F̃ preserves a rational fibration. In this case F̃ cannot be conjugated to an automorphism. • The sequence dn is bounded and F̃ preserves a two generically transverse rational fibrations and F̃ is an automorphism. • The sequence dn grows exponentially. In the first three cases δ(F ) = 1 while in the last one δ(F ) > 1. Furthermore in the first and second, the invariant fibrations are unique. Since we only deal with degenerate maps, we have to consider two subfamiles: (βγ)12 = 0 or (αγ)12 = 0. We begin with the simplest case (αγ)12 = 0. 6 3 Subfamily (αγ)12 = 0. Lemma 3. Consider birational mappings   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the condition (αγ)12 = α1 γ2 − α2 γ1 = 0. Then either (i) The four numbers α1 , α2 , γ1 , γ2 are distinct for zero. (ii) α1 = 0, γ1 = 0 and α2 6= 0 6= γ2 . (iii) α2 = 0, γ2 = 0 and α1 6= 0 6= γ1 . Proof. From Lemma 1 we know that (α1 , α2 ) 6= (0, 0). Then if α1 (resp. γ1 ) is zero then α2 (resp. γ2 ) is not and from α1 γ2 − α2 γ1 = 0 we get that γ1 (resp. α1 ) must be zero. Theorem 4. Consider birational mappings   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the condition (αγ)12 = α1 γ2 − α2 γ1 = 0. Then the following hold: (i) If α1 6= 0, α2 6= 0, γ1 6= 0 and γ2 6= 0, then δ(F ) = 2. (ii) If α1 = γ1 = 0, then δ(F ) = δ ∗ and dn+2 = dn+1 + dn . (iii) If α2 = γ2 = 0, then δ(F ) = 1 and dn = 1 + n. Proof. From the hypothesis we have that: E(F ) = {S0 , S1 } , I(F ) = {O0 , O1 }, E(F −1 ) = {T0 , T1 } and I(F −1 ) = {A0 , A1 } with S0 = {x0 = 0} , S1 = {γ0 x0 + γ1 x1 + γ2 x2 = 0}, O0 = [(βγ)12 : (βγ)20 : (βγ)01 )] , O1 = [0 : α2 : −α1 ], T0 = {(β2 (αγ)01 − β1 (αγ)12 )x0 − (βγ)12 x1 = 0} , T1 = {x0 = 0}, A0 = [0 : 1 : 0] , A1 = [0 : 0 : 1]. When α1 , α2 , γ1 and γ2 are non zero we observe that A0 6= O0 and A0 6= O1 . Hence, since F (A0 ) = [0 : α1 γ1 : 0] = A0 and F (A1 ) = [0 : α2 γ2 : 0] = A0 we get that F is AS. It implies that dn = 2n and consequently δ(F ) = 2. To prove (ii) we observe that α1 = γ1 = 0 not only implies that (α2 , γ2 ) 6= (0, 0) but also that β1 6= 0 (if not f would only depend on y and it would not be birational). Now 7 A0 = O1 ∈ I(F ) and we have to blow-up this point. Let E0 be the principal divisor at this point and consider a point [u : v]E0 ∈ E0 . In order to extend F on E0 we see [u : v]E0 as limt→0 [tu : 1 : tv] and we are going to evaluate F [tu : 1 : tv] : F [tu : 1 : tv] = [u(γ0 u + γ2 v)t : (α0 u + α2 v)(γ0 u + γ2 v)t : β1 u + (β0 u + β2 v)ut]. Taking the limit when t tends to zero we have that when u 6= 0 , F̃ [u : v]E0 = [0 : 0 : 1] while [0 : 1]E0 becomes an indeterminacy point for F̃ . To know the action of F̃ on S0 we see the point [0 : x1 : x2 ] as limt→0 [t : x1 : x2 ]. Then for t → 0 (and x2 6= 0) lim F [t : x1 : x2 ] = lim[γ2 x2 t : α2 γ2 x22 : (β1 x1 + β2 x2 )t] = [γ2 x2 : β1 x1 + β2 x2 ]E0 . t→0 t→0 The above considerations imply that I(F̃ ) = {O0 , [0 : 1]E0 } , E(F̃ ) = {Ŝ1 , E0 } with Sˆ1  A1 and E0  A1 . To follow the orbit of A1 under F̃ we observe that A1 = [0 : 0 : 1] ∈ S0 and hence F̃ [0 : 0 : 1] = [γ2 : β2 ]E0 6= [0 : 1]E0 and which is sent to A1 again giving a two-periodic orbit. It implies that F̃ : X −→ X is AS. The Picard group of X is P ic(X) =< L̂, E0 > where L is a generic line of P C2 . Let F̃ ∗ denote the corresponding map on P ic(X), which acts just taking preimages. Hence F̃ ∗ (E0 ) = Sˆ0 . In order to write Sˆ0 as a linear combination of L̂, E0 we are going to use (6). We have that π ∗ (S0 ) = Sˆ0 + E0 = L̂ which implies that F̃ ∗ (E0 ) = L̂ − E0 . Also π ∗ (F −1 (L)) = Fˆ−1 (L) + E0 = 2L̂ which implies ! 2 1 that F̃ ∗ (L̂) = 2L̂ − E0 . Hence the matrix of F̃ ∗ on P ic(X) =< L̂, E0 > is with −1 −1 characteristic polynomial z 2 − z − 1. Hence δ(F ) = δ ∗ . To prove (iii) we observe, as before, that α2 = γ2 = 0 not only implies that (α1 , γ1 ) 6= (0, 0) but also that β2 6= 0 (if not f would only depend on x and it would not be birational). Now A1 = O1 = [0 : 0 : 1] ∈ I(F ) and we have to blow-up this point. To know the action of F̃ on S1 we see the point [γ1 x0 : −γ0 x0 : γ1 x2 ] as limt→0 [γ1 x0 : t − γ0 x0 : γ1 x2 ]. Similar computations as before give us that each point in S1 \ {O0 , [0 : 0 : 1]} is sent to the point [γ1 : (αγ)01 ]E1 . That is Sˆ1 is still exceptional for F̃ . Now consider a point [u : v]E1 ∈ E1 . It is seen as limt→0 [tu : tv : 1]. Then for t → 0, lim F [tu : tv : 1] = lim[tu(γ0 u + γ1 v) : t(γ0 u + γ1 v)(α0 u + α1 v) : β2 u]. t→0 t→0 If γ0 u + γ1 v 6= 0 and u 6= 0 then F̃ [u : v]E1 = [u : α0 u + α1 v]E1 . If γ0 u + γ1 v = 0, in the above computation with [u : v]E1 = [γ1 : −γ0 ]E1 we deal with the point [γ1 t : −γ0 t : 1] ∈ S1 and we have to apply F̃ giving F̃ [γ1 t : −γ0 t : 1] = [γ1 : (αγ)01 ]E1 . We observe that limu→γ1 ,v→−γ0 F̃ [u : v]E1 = [γ1 : (αγ)01 ]E1 , that is F̃ is well defined. But if u = 0 , F [0 : t : 1] = [0 : 1 : 0]. It implies that [0 : 1]E1 ∈ I(F̃ ). 8 We claim that after this blow-up the map F̃ is AS. It is so because S0  A0 and A0 is a fixed point of F and Sˆ1  [γ1 : (αγ)01 ]E1 and the iterates of this point never coincide with [0 : 1]E1 . The Picard group of X is now P ic(X) =< L̂, E1 > where L is a generic line of P C2 , F̃ ∗ (E1 ) = Sˆ1 + E1 , and similar computations as the ones of (ii) give the matrix ! 2 1 . −1 0 The characteristic polynomial is given by (z − 1)2 . Hence δ(F ) = 1. Furthermore, since d1 = 2 and d2 = 3 we get that dn = 1 + n. Proposition 5. Assume that   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the condition (αγ)12 = α1 γ2 − α2 γ1 = 0, has zero entropy. Then after an affine change of coordinates it can be written as   β0 + y f (x, y) = α0 + α1 x, , α1 6= 0. x This map preserves the fibration V (x, y) = x and this fibration is unique. If m(x) := α0 + α1 x is periodic of period p, that is if α1p = 1 for some p > 1 , α1 6= 1, then W (x, y) = x · m(x) · m(m(x)) · · · mp−1 (x) is a first integral of f (x, y). Also when α1 = 1 and α0 = 0, f is integrable. Proof. From Theorem 4 we know that the only zero entropy maps in the family are the ones with α2 = γ2 = 0 and we also know that in this case β2 , α1 and γ1 are  different from β2 γ0 1 β1 zero. Hence we can conjugate f (x, y) with h(x, y) = γ1 x − γ1 , β2 y + γ1 . Renaming the parameters we see that the conjugate map is of the form   β0 + y f (x, y) = α0 + α1 x, , α1 6= 0. x Clearly this map preserves the fibration V (x, y) = x and this fibration is unique from Theorem 2. If α1p = 1 for some p > 1 , α1 6= 1, then W (f (x, y)) = W (x, y) and the result follows. When α1 = 1 then we see that f (x, y) is integrable if and only if β0 = 0. Now we are going to consider the second subfamily. 9 4 Subfamily (βγ)12 = 0. We are going to consider three different cases, depending on γ1 γ2 6= 0 , γ1 = 0 and γ2 = 0. When (βγ)12 = β1 γ2 − β2 γ1 = 0, we have that E(F ) = {S0 , S1 } , I(F ) = {O0 , O1 }, E(F −1 ) = {T0 , T1 } and I(F −1 ) = {A0 , A1 } with S0 = {x0 = 0} , S1 = {γ0 x0 + γ1 x1 + γ2 x2 = 0} O0 = [0 : γ2 : −γ1 ] , O1 = [0 : α2 : −α1 ] T0 = {x0 = 0} , T1 = {(αβ)12 x0 − (αγ)12 x2 = 0} A0 = [0 : 1 : 0] , A1 = [0 : 0 : 1]. 4.1 (βγ)12 = 0 with γ1 γ2 6= 0 Theorem 6. Consider birational mappings   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the conditions (βγ)12 = 0 and γ1 γ2 6= 0. Then either (i) α1 6= 0 6= α2 and δ(F ) = 2 with dn = 2n for all n ∈ N. (ii) α1 = 0 and the dynamical degree is δ(F ) = δ ∗ with dn+2 = dn+1 + dn for all n ∈ N. (iii) α2 = 0 and the dynamical degree is δ(F ) = 1 with dn = 1 + n for all n ∈ N. Proof. To prove (i) we observe that S0  A0 and S1  A1 with F (A0 ) = [0 : α1 γ1 : 0] = A0 ∈ / I(F ) and F (A1 ) = [0 : α2 γ2 : 0] = A0 ∈ / I(F ). Thus using (8) we see that F is AS, which implies that dn = 2n and consequently δ(F ) = 2. Now consider that α1 = 0. It implies that α2 6= 0. In this case S0  A0 = O1 ∈ I(F ). Hence we blow-up A0 to obtain E0 . Similar computations as before says that F̃ sends Sˆ0 → E0 → Tˆ1 and no new indeterminacy points are created. Now we have to follow the orbit of A1 under the action of F̃ . As A1 ∈ S0 we find that F̃ (A1 ) = [γ2 : β2 ]E0 and F̃ [γ2 : β2 ]E0 = [γ1 γ2 : α0 γ2 + α2 β2 : β1 γ2 ] ∈ T1 . Observe that I(F̃ ) = {O0 } and O0 ∈ S0 = T0 . We know that the only points on T0 which have preimages are A0 and A1 which implies that if the iterates of A1 reaches O0 for some iterate of F then O0 should be equal to either A0 or A1 . But the conditions on the parameters implies that A0 6= O0 6= A1 . This implies that O0 has no preimages hence the iterates of A1 cannot reach O0 . Hence we see that F̃ is AS. 10 In this case F̃ ∗ (L̂) = 2L̂ − E0 and F̃ ∗ (E0 ) = L̂ − E0 . Hence the characteristic polynomial √ 1+ 5 of the corresponding matrix is z 2 −z−1. It implies that the dynamical degree is δ(F ) = 2 and dn+2 = dn+1 + dn for all n ∈ N. Finally to see (iii), since α2 = 0 we get that α1 6= 0. Now we observe that S0 collapses to A0 = [0 : 1 : 0] ∈ S0 and that F [0 : 1 : 0] = [0 : α1 γ1 : 0] = [0 : 1 : 0]. Hence A0 is a fixed point. The other exceptional curve S1  A1 = O1 = [0 : 0 : α1 ] = [0 : 0 : 1] ∈ I(F ). Hence we have to blow-up A1 obtaining E1 . Similar computations as before says that F̃ sends Sˆ1 → E1 → Tˆ1 and no new indeterminacy points are created. After this blow-up the mapping F̃ is AS. And we can see that F̃ ∗ (L̂) = 2L̂ − E0 and F̃ ∗ (E1 ) = L̂. Hence the matrix of Fe∗ is: ! 2 1 . (9) −1 0 The characteristic polynomial is (z − 1)2 , and hence the dynamical degree is 1. Since d1 = 2, d2 = 3 we get that the sequence of degrees is dn = 1 + n for all n ∈ N. Concerning the zero entropy we see that the only case is the third one, when α2 = 0. The result (and the proof) we get is very similar to the one stated in Proposition 5. Proposition 7. Let   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the conditions (βγ)12 = 0 and γ1 γ2 6= 0 and assume that f (x, y) has zero entropy. Then after an affine change of coordinates it can be written as   β0 f (x, y) = α0 + α1 x, , α1 6= 0. x+y This map preserves the fibration V (x, y) = x and this fibration is unique. If m(x) := α0 + α1 x is periodic of period p, that is if α1p = 1 for some p > 1 , α1 6= 1, then W (x, y) = x · m(x) · m(m(x)) · · · mp−1 (x) is a first integral of f (x, y). Also when α1 = 1 and α0 = 0, f is integrable. 4.2 (βγ)12 = 0 with γ1 = 0. Next Theorem gives the behaviour of dn in this family. As we can see below after affine change of coordinates these mappings are simple, and the sequence of degrees can be deduced by elementary methods. We have adopted this point of view in the proof of item (ii). But 11 in the first part we have preferred the blow-up approach. In fact some multiple blow-up’s are implemented and it is amazing to see how the method detects the different behaviours of dn . Theorem 8. Consider birational mappings   β 0 + β 1 x1 + β 2 x2 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , γ2 ) 6= (0, 0) γ0 + γ1 x1 + γ2 x2 with the conditions (βγ)12 = 0 and γ1 = 0. (i) Assume that α2 6= 0. Then, after an affine change of coordinates f (x, y) can be written as   β0 f (x, y) = α0 + α1 x + y, , α1 6= 0 6= β0 (10) γ0 + y and the following hold: β0 (a) If the onedimensional mapping h(y) := γ0 +y is not a periodic map, then the sequence of degrees is dn = 1 + n. (b) If h(y) is a k- periodic map and 1 + α1k + α12k + · · · + α1nk 6= 0 for all n ∈ N, then dn = 1 + n for all n ≤ k − 1 and dn = k for all n ≥ k. (c) If h(y) is a k-periodic map and 1 + α1k + α12k + · · · + α1nk = 0 for some n ∈ N, then dn is a (n + 1)k−periodic sequence. (ii) Assume that α2 = 0. Then, after an affine change of coordinates f (x, y) can be written as   β0 f (x, y) = α0 + α1 x, , α1 6= 0 6= β0 (11) γ0 + y and the following hold: β0 (a) If the onedimensional mapping h(y) := γ0 +y is not a periodic map, then dn = 2 for all n ∈ N. (b) If h(y) is a k- periodic map then dn is a k−periodic sequence. Proof. We notice that since γ1 = 0 , γ2 6= 0 we can conjugate f (x, y) with   α2 1 β2 ψ(x, y) = x, y + γ2 γ2 γ2 and renaiming the coefficients if necessary, we get the desired map (10). Now we have that S0 = {x0 = 0} , S1 = {γ0 x0 + x2 = 0} , A0 = [0 : 1 : 0] , A1 = [0 : 0 : 1], 12 and T0 = {x0 = 0} , T1 = {x2 = 0} , O0 = [0 : 1 : 0] , O1 = [0 : 1 : −α1 ]. Since A0 = O0 we have to blow-up this point getting E0 . Then F̃ [u : v]E0 = [γ0 u + v : β0 u]E0 , [u : v]E0 6= [1 : −γ0 ]E0 and Ŝ0  [1 : 0]E0 . The point [1 : −γ0 ]E0 is now an indeterminacy point of F̃ . Hence, if F̃ p [1 : 0]E0 6= [1 : −γ0 ]E0 for all p ∈ N, since Ŝ1  A1 ∈ S0 , F̃ (A1 ) = [1 : 0]E0 and we get that F̃ is AS. It can be seen that the matrix of F̃ ∗ : P ic(X) → P ic(X) =< L̂, E0 > is ! 2 1 . (12) −1 0 The characteristic polynomial is (z − 1)2 , and hence the dynamical degree is 1. Since d1 = 2, d2 = 3 we get that the sequence of degrees is dn = 1 + n for all n ∈ N. Now assume that there exists some p ∈ N such that F̃ p [1 : 0]E0 = [1 : −γ0 ]E0 . We claim that in this case F̃ : E0 → E0 is a (p + 2)−periodic map. To prove the claim we distinguish between the case γ0 = 0 (which gives a 2-periodic map and corresponds to p = 0) and the case γ0 6= 0. We have that p [1 : 0]E0 −→F̃ [1 : −γ0 ]E0 −→F̃ [0 : 1]E0 −→F̃ [1 : 0]E0 . Hence F̃ p+2 which in fact is a Moebius map, fixes at least three different points. It clearly implies that F̃ p+2 is the identity map. Since the restriction of F̃ at E0 is exactly the map β0 h(y) = γ0 +y extended to the projective line we can assert that F̃ p [1 : 0]E0 = [1 : −γ0 ]E0 if ad only of h(y) is a (p + 2)−periodic map. Hence (a) is proved. Following the process, if F̃ p [1 : 0]E0 = [1 : −γ0 ]E0 , we have to blow-up all the points F̃ j [1 : 0]E0 for j = 0, 1, . . . , p. We call E0j the corresponding principal divisors, getting: E00 −→ E01 −→ E02 −→ · · · −→ E0p . (13) Calling again F̃ the map at this new variety we are going to see which is the image of S0 and which is the image of E0p . A point of coordinate k in E00 is seen as limt→0 [t : 1 : kt2 ]. Then, for any point in S0 different from the indeterminacy points and for t ∼ 0, we have that:   2 t β0 2 F (t, x1 , x2 ) ∼ [x2 t : (α1 x1 + α2 x2 )x2 : β0 t ] = :1: t . α1 x1 + α2 x2 x2 (α1 x1 + α2 x2 ) 13 h i t β0 (α1 x1 +α2 x2 ) 2 Naming T := α1 x1 +α2 x2 this point looks like T : 1 : x2 T , that is β0 (α1 x1 + α2 x2 ) F̃ [0 : x1 : x2 ] = ∈ E00 . x2 Now consider a point of coordinate k in E0p . This point is seen as limt→0 [t : 1 : −γ0 t + kt2 ]. Then for t ∼ 0, F [t : 1 : −γ0 t + kt2 ] ∼ [kt : α1 k : β0 ] →t→0 [0 : α1 k : β0 ] ∈ S0 . Hence (13) can be completed and we get the cycle: Ŝ0 −→ E00 −→ E01 −→ E02 −→ · · · −→ E0p −→ Ŝ0 . Now since S1  A1 ∈ S0 and F̃ p+2 sends Ŝ0 to itself, it could happen that for some n ∈ N , F̃ n(p+2) (A1 ) = O0 , which still is an indeterminacy point of F̃ . If it is not the case, these F̃ is AS. Let us to compute the matrix of F̃ ∗ . The Picard group of X is P ic(X) =< L̂, E00 , E01 , . . . , E0p , E0 > . To write Ŝ0 and Ŝ1 as a linear combination of basis elements, we are going to use the identity (6). For instance π ∗ (F −1 (L)) = F −1ˆ(L)+ Pp −1 (L) at generic j=1 mj E0j where the multiplicities mj are the order of vanishing of F points of E0j . If δ0 x0 + δ1 x1 + δ2 x2 = 0 is the equation of a generic straight line L, then a calculation gives δ0 F [t : 1 : wt + kt2 ][1] + δ1 F [t : 1 : wt + kt2 ][2] + δ2 F [t : 1 : wt + kt2 ][3] = P δ1 α1 (γ0 + w)t + o(t2 ) which let us to write π ∗ (F −1 (L)) = F −1ˆ(L) + p−1 j=1 E0j + 2E0p . Now ∗ −1 ∗ Pp−1 from π (F (L)) = 2L̂ we get that F̃ (L̂) = 2L̂ − j=1 E0j − 2E0p . Proceeding in this way we find that the matrix of F̃ ∗ is:   2 1 0 0 ... 0 0    −1 −1 1 0 . . . 0 0     0 0 0 1 ... 0 0     .. .. .. .. . . .. ..   . . .   . . . . .    −1 −1 0 0 . . . 1 0       −2 −1 0 0 . . . 0 0  −1 −1 0 0 . . . 0 1 It is not hard to see that the characteristic polynomial of such a matrix is (−z)p (z − 1)2 . Hence the sequence of degrees satisfies dn+p+2 = 2dn+p+1 − dn+p and its behaviour depends on the initial conditions, i.e., on the first terms d1 , d2 , . . . , dp+2 . So, if h(y) is k−periodic, then k = p + 2 and f k (x, y)[2] = y. It implies that dk = dk−1 . Since the first degrees are 2, 3, 4, . . . , k, k, from dn+k = 2dn+k−1 − dn+k−2 we get that dn = k for all n ≥ k. It remains to prove that the condition F̃ n(p+2) (A1 ) = O0 is equivalent to 1 + α1k + α12k + · · · + α1nk = 0 14 for k = p + 2. To this end, taking into account the terms of maximum degree of f k (x, y), (see (15) below) we get that: F̃ k [0 : x1 : x2 ] = [0 : α1k−1 (α1 x1 + x2 ) : x2 ] and hence (n−1)k F̃ nk [0 : 0 : 1] = [0 : α1k−1 (1 + α1k + α12k + · · · + α1 : 1]. Therefore F̃ nk [0 : 0 : 1] = O1 = [0 : 1 : −α1 ] if and only if 1 + α1k + α12k + · · · + α1nk = 0. (14) Statement (b) is now proved. To see (c) we just compute f (n+1)k . In this case since α1 6= 1 we can consider (doing a translation if necessary) that α0 = 0. Now the expression of f k is   f k (x, y) = α1k x + α1k−1 y + α1k−2 h(y) + α1k−3 h2 (y) + · · · + α1 hk−2 (y) + hk−1 (y), y (15) Hence:   (n+1)k f (n+1)k (x, y) = α1 x + (1 + α1k + α12k + · · · + α1nk )(α1k−1 y + α1k−2 h(y) + hk−1 (y)), y . (n+1)k Then since condition (14) implies that α1 = 1 we have that when condition (14) is satisfied, f is a (n + 1)k−periodic map, and hence also the sequence of degrees is (n + 1)k−periodic. We are going to prove (ii). First of all, γ1 = 0 implies γ2 6= 0, and from γ1 β2 − γ2 β1 = 0 we get β1 = 0. Doing a translation on y and renaiming the coefficients we get equation (11). Since this map is very simple we are going to prove the result on the behaviour of dn using simple arguments. We observe that the first component of f k (x, y) is ak x + bk for certains β0 ak , bk . And the second components are just the iterates of h(y) = γ0 +y , a one-dimensional Möbius map. We claim that if h(y) is not a periodic map then hk (y) is a Möbius map with non-constant denominator for all k ∈ N and also that the denominators of hi (y) and hj (y) are different for i 6= j. From the claim we can deduce that when h(y) is not a periodic map then dn = 2 for all n ∈ N. And when h(y) is a k−periodic map, then the sequence of degrees is dn = 2 for all n which is not a multiple of k and dn = 1 when n is a multiple of k. Nk To prove the claim we consider Nk and Dk with hk (z) = Dk and we see that, if we don’t perform simplifications, Nk+1 = β0 Dk and Dk+1 = γ0 Dk + Nk . Let pk , qk ∈ C such that Dk = pk +qk z. Then, Dk+2 −γ0 Dk+1 −β0 Dk = 0, which implies that qk+2 −γ0 qk+1 −β0 qk = 0. Analyzing this linear recurrence with constant coefficients and taking into account that  k this sequence is k−periodic if and only if λλ21 = 1, where λ1 , λ2 are the two different roots of λ2 − γ0 λ − β0 = 0 (see [11]), the claim follows. 15 Proposition 9. Consider the birational mappings   β0 f (x, y) = α0 + α1 x + y, , α1 6= 0 6= β0 . (16) γ0 + y Then the following hold: β0 (a) If the onedimensional mapping h(y) := γ0 +y is not a periodic map, then f (x, y) has the unique invariant fibration V1 (x, y) = y. (b) If h(y) is a k- periodic map and 1 + α1k + α12k + · · · + α1nk 6= 0 for all n ∈ N, then f (x, y) is integrable being H1 (x, y) = y + h(y) + h(h(y)) + · · · + hk−1 (y) a first integral and also has a second invariant fibration V2 (x, y) : (b1 ) If α1k 6= 1 we can assume that α0 = 0 and then V2 (x, y) = (α1k − 1)x + α1k−1 y + α1k−2 h(y) + α1k−3 h2 (y) + · · · + α1 hk−2 (y) + hk−1 (y) (17) satisfies V2 (f (x, y)) = α1 V2 (x, y). (b2 ) If α1k = 1 but α1 6= 1 we can assume that α0 = 0 and then V2 (x, y) = kx + (k − 1)α1k−1 y + (k − 2)α1k−2 h(y) + (k − 3)α1k−3 h(h(y)) + · · · + 2α12 hk−3 (y) + α1 hk−2 (y) α1k−1 y + α1k−2 h(y) + α1k−3 h(h(y)) + · · · + α1 hk−2 (y) + hk−1 (y) satisfies V2 (f (x, y)) = V2 (x, y) + 1. (b3 ) If α1 = 1 then kx + (k − 1)y + (k − 2)h(y) + (k − 3)h(h(y)) + · · · + 2hk−3 (y) + (y) V2 (x, y) = kα0 + α1k−1 y + α1k−2 h(y) + α1k−3 h(h(y)) + · · · + hk−2 (y) + hk−1 (y) satisfies V2 (f (x, y)) = V2 (x, y) + 1. (c) If h(y) is a k-periodic map and 1+α1k +α12k +· · ·+α1nk = 0 for some n ∈ N, then f (x, y) (n+1)k has a second first integral H2 (x, y) which can be given by H2 (x, y)) = V2 (x, y) being V2 (x, y) be defined by (17). The proofs are straightforward. Only say that to find the fibrations we have considered combinations of x, y, h(y), h(h(y)), . . . , hk−1 (y) or quotients of them. Remark 10. Assuming the hypothesis (b), since dn is a bounded sequence and f (x, y) is not a periodic map, from [6] we know that f (x, y) is birationally equivalent to either (x, y) → (ax, by) where a is a root of unity and b it is not or to (x, y) → (ax, y + 1). The fibrations encountered in (b) let us to construct such a conjugations. In fact, when V2 (f (x, y)) = α1 V2 (x, y) we are in the first case while when V2 (f (x, y)) = V2 (x, y) + 1 we are in the second one. 16 The invariant fibrations and first integrals corresponding to the mappings satisfying (ii) of Theorem 8 are very easy after the adequate affine change of coordinates. Next Proposition gives this information. Proposition 11. Consider the birational mappings   β0 f (x, y) = α0 + α1 x, , α1 6= 0 6= β0 . (18) γ0 + y These mappings preserve the two generically transverse invariant foliations V1 (x, y) = x and V2 (x, y) = y. Furthermore, β0 (a) If h(y) = γ0 +y is periodic of period k then H1 (x, y) = y + h(y) + h(h(y)) + · · · + hk−1 (y) is a first integral of f (x, y). (b) If m(x) := α0 + α1 x is periodic of period p then H2 (x, y) = x + m(x) + m(m(x)) + · · · + mp−1 (x) is a first integral of f (x, y). (c) If h(y) and m(x) are k−periodic then f (x, y) is a k−periodic mapping having two independent first integrals H1 (x, y) and H2 (x, y) with p = k. 4.3 (βγ)12 = 0 with γ2 = 0 If γ2 = 0 we know that γ1 6= 0 and from (βγ)12 = 0 we get β2 = 0. Also α2 6= 0, if not f (x, y) only depens on x. Theorem 12. Consider birational mappings   β 0 + β 1 x1 f (x1 , x2 ) = α0 + α1 x1 + α2 x2 , , (γ1 , α2 ) 6= (0, 0). (19) γ0 + γ1 x1 (a) Assume that α1 6= 0. Then the dynamical degree of F is δ(F ) = δ ∗ and dn+2 = dn + dn+1 . (b) Assume that α1 = 0. Then after an affine change of coordinates f (x, y) takes the form:   β0 f (x1 , x2 ) = x2 , (20) γ0 + x1 and the dynamical degree of F is δ(F ) = 1. Furthermore: 17 β0 (b1 ) If h(z) := γ0 +z is not a periodic map then dn = 2 for all n ∈ N. (b1 ) If h(z) is a k−periodic map then dn is a 2k−periodic sequence. Proof. To prove (a) we observe that now we have S1  A1 = O0 = [0 : 0 : 1] and F (A0 ) = [0 : α1 γ1 : 0] = A0 ∈ / I(F ). So we have to blow-up A1 = [0 : 0 : 1] getting E1 . Then F̃ sends S1 → E1  [0 : 1 : 0] = A0 . Since A1 ∈ S1 , π ∗ (S1 ) = Sˆ1 + E1 and the matrix of Fe∗ is: ! 2 1 . (21) −1 −1 Then the characteristic polynomial associated to F is z 2 − z − 1. Hence the dynamical degree is δ(F ) = δ ∗ and dn+2 = dn+1 + dn for all n ∈ N. We are going to prove (b). When α1 = 0 (19) can be transformed in (20) via the conjugation   1 α0 γ1 + α2 β1 1 β1 ψ(x, y) = x+ , y+ . γ1 γ1 γ1 α2 γ1 From (20) we have that f (f (x, y)) = (h(x), h(y)) and generally:  f 2n (x, y) = (hn (x), hn (y)) , f 2n+1 (x, y) = hn (y), hn+1 (x) . (22) From the same arguments as before if h is not periodic dn = 2 for all n ∈ N. If h is k−periodic then f 2k (x, y) = (x, y) and from (22) we get that dn = 2 for all n ∈ N such that it is not a multiple of 2k and dn = 1 for all n ∈ N such that it is a multiple of 2k. In any case case the dynamical degree of F is δ(F ) = 1. Proposition 13. Consider the family of mappings:   β0 f (x, y) = y, . γ0 + x Then (a) If γ02 + 4 β0 6= 0, let p and q be the two different roots of z 2 − γ0 z − β0 = 0, and let m such that m2 = q/p. Then f (x, y) preserves the generically transverse fibrations m2 p2 + mpx + p(m2 − m + 1)y + xy H1 (x, y) = , (x + p) (y + p) m2 p2 − mpx + p(m2 + m + 1)y + xy H2 (x, y) = (x + p) (y + p) with H1 (f (x, y)) = mH1 (x, y) , H2 (f (x, y)) = −mH2 (x, y). Furthermore f (x, y) is 2k− periodic if and only if m2k = 1 and in this case H12k (x, y) and H22k (x, y) are two independent first integrals of f (x, y). 18 (b) If γ02 + 4 β0 = 0 then it preserves the two generically transverse fibrations γ02 − 2γ0 x + 6γ0 y + 4 x y 2 γ0 (x + y + γ0 ) K1 (x, y) = , K2 (x, y) = , (2 x + γ0 ) (2 y + γ0 ) (2 x + γ0 ) (2 y + γ0 ) with K1 (f (x, y)) = −K1 (x, y) , K2 (f (x, y)) = K2 (x, y) + 1. Furthermore f (x, y) is integrable being W (x, y) = (K1 (x, y))2 a first integral. Proof. When γ02 + 4 β0 6= 0 some calculations give that in fact H1 (f (x, y)) = mH1 (x, y) and H2 (f (x, y)) = −mH2 (x, y). Furthermore H1 (x, y), H2 (x, y) are generically transverse because the determinant of the Jacobian of H1 (x, y), H2 (x, y) is 2p2 m(m2 − 1) − (p + x)2 (p + y)2 which is different from zero (if not m2 = 1 and it happens if and only if p = q, which is in contradiction with γ02 + 4 β0 6= 0). Also when γ02 +4 β0 = 0 the determinant of the Jacobian of K1 (x, y), K2 (x, y) is different from zero because it is equal to: 16c2 . (2y + c)2 (2x + c)2 Finally W (x, y) = (K1 (x, y))2 is a first integral integral of f (x, y) because W (f (x, y)) = (K1 (f (x, y)))2 = (−K1 (x, y))2 = W (x, y). Remark 14. Simple computations give that when γ02 +4 β0 6= 0, f is birationally conjugated to (mx, −my) via the conjugation ϕ(x, y) = (H1 (x, y), H2 (x, y)) and that when γ02 +4 β0 = 0, f is birationally conjugated to (−x, y + 1) via ψ(x, y) = (K1 (x, y), K2 (x, y)). Acknowledgements The first author is supported by Ministry of Economy, Industry and Competitiveness of the Spanish Government through grant MINECO/FEDER MTM2016-7 7278-P and also by the grant 2014-SGR-568 from AGAUR, Generalitat de Catalunya. References [1] Angles, JC, Maillard, JM and Viallet, C. On the complexity of some birational trans- formations, J. Phys. A: Math. Gen. 39 (2006) 36413654. 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