This is a preprint of: “Invariant Fibrations for some Birational Maps of C2”, Anna Cima, Sundus Zafar, J. Difference Equ. Appl., vol. 25(8), 2019. DOI: [10.1080/10236198.2019.1654464] Invariant Fibrations for some Birational Maps of C2∗ Anna Cima(1) and Sundus Zafar(1) (1) Departament de Matemàtiques, Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain cima@mat.uab.cat, sundus@mat.uab.cat Abstract In this article we extract and study the zero entropy subfamilies of a certain family of birational maps of the plane. We find these zero entropy mappings and give the invariant fibrations associated to them. Mathematics Subject Classification 2010: 14E05, 26C15, 34K19, 37B40, 37C15, 39A23. Keywords: Birational maps, Algebraic entropy, First Integrals, Fibrations, Blowing-up, Integrability, Periodicity. 1 Introduction A mapping f = (f1 , f2 ) : C2 −→ C2 is said to be rational if each coordinate function is rational, that is, fi is a quotient of polynomials for i = 1, 2. These maps can be naturally extended to the projective plane P C2 by considering the embedding (x1 , x2 ) ∈ C → [1 : x1 : x2 ] ∈ P C2 . The induced mapping F : P C2 −→ P C2 has three components Fi [x0 : x1 : x2 ] which are homogeneous polynomials of the same degree. If F1 , F2 , F3 have no common factors and have degree d, we say that f or F has degree d. Similarly we can define the degree of F n = F ◦ · · · ◦ F for each n ∈ N. We are interested in birational maps. It is said that a rational mapping f : C2 −→ C2 is birational if it exists an algebraic curve and another rational map g such that f ◦g = g◦f = id in C2 \ V. The study of the dynamics generated by birational mappings in the plane has been growing in recent years, see for instance [2, 3, 6, 8, 13, 15, 16, 17, 18, 19, 20, 21]. ∗ Acknowledgements. The authors are supported by Ministry of Economy and Competitiveness of the Spanish Government through grants MTM2013-40998-P. They are also supported by the grant 2014-SGR- 568 from AGAUR, Generalitat de Catalunya. http://www.gsd.uab.cat 1 It can be seen that if f (x1 , x2 ) is a birational map, then the sequence of the degrees of F n satisfies a homogeneous linear recurrence with constant coefficients (see [12] for instance). This is governed by the characteristic polynomial X (x) of a certain matrix associated to F. The other information we get from X (x) is the dynamical degree, δ(F ), which is defined as 1 δ(F ) := lim (deg(F n )) n . (1) n→∞ The logarithm of this quantity has been called the algebraic entropy of F . It is known that the algebraic entropy is an upper bound of the topological entropy, which in turn is a dynamic measure of the complexity of the mapping. For instance, periodic or integrable birational mappings have zero algebraic entropy. Birational mappings with zero algebraic entropy have been characterized, see [12] and [4]. From its results we know the existence of some fibrations associated to the mapping, which give almost a complete dynamical information of the mapping. In this paper we consider the family of fractional maps f : C2 → C2 :   β0 + β1 x + β2 y f (x, y) = α0 + α1 x + α2 y, , α1 6= 0 , β1 6= 0 , γ2 6= 0. (2) γ0 + γ2 y This family is part of a more general family studied in [10] and [11], which in turn is a generalization of the birational mappings studied by Bedford and Kim in [2]. The goal of this paper is to extract, under affine equivalence, all mappings of type (2) having zero algebraic entropy and give the corresponding invariant fibrations associated to them. The methodology involves the implementation of the blowing-up technique and the extension of the mappings at the Picard group (see Section 2). In general, given a parametric family of mappings, to decide for which values of the parameters the mappings are periodic, is not an easy problem (see [9], [5], for instance). When the mapping is a plane birational mapping it is possible to face that problem (see [2]) and it is fascinating to see how these cases arise, and not only the periodic ones, also all the zero entropy cases. The paper is organized as follows. In Section 2 we give some preliminary results and we explain how we proceed to find the invariant fibrations associated to zero entropy maps. Section 3 deals with the subfamily α2 6= 0. The main result is Theorem 11. Similarly in Section 4 we consider the subfamily α2 = 0 getting Theorem 13. 2 Preliminary results Rational mappings F : P C2 → P C2 have an indeterminacy set I(F ) of points where F is ill-defined as a continuous map. This set is given by: I(F ) = {[x0 : x1 : x2 ] ∈ P C2 : F1 [x0 : x1 : x2 ] = 0 , F2 [x0 : x1 : x2 ] = 0 , F3 [x0 : x1 : x2 ] = 0]}. 2 If F is birational then we can also consider the indeterminacy points of its inverse F −1 . On the other hand, if we consider one irreducible component V of the determinant of the Jacobian of F , it is known (see Proposition 3.3 in [13]) that F (V ) reduces to a point which belongs to I(F −1 ). The set of these curves which are sent to a single point is called the exceptional locus of F and it is denoted by E(F ). It is known that the dynamical degree depends on the orbits of the indeterminacy points of the inverse of F under the action of F, see [12, 14]. Indeed, the key point is whether the iterates of such points coincide with any of the indeterminacy points of F. When it happens, this orbit is finite. Sometimes some orbit collision appears. The expression orbit collision refers to the following: Let S ∈ E(F ) which collapses at the point A ∈ I(F −1 ) (we will write S  A to describe this behaviour). Following the orbit of A, assume that it ends at a point O ∈ I(F ). It can happen that S  A → ∗ → · · · → σ → · · · → O , σ ∈ S̄ ∈ E(F ). Then, being f birational, (see [12]) it exists Ā ∈ I(F −1 ) with S̄  Ā. When it happens it is said that the orbits of A and Ā collides. This is exactly the behaviour that we get in family (1) and what makes the family so interesting. 2.1 Blow-up’s and the Picard group Given a point p ∈ C2 , let (X, π), be the blowing-up of C2 at the point p. Then,translating p at the origin, π −1 p = π −1 (0, 0) = {((0, 0), [u : v])} := Ep ' P C1 and if q = (x, y) 6= (0, 0), then π −1 q = π −1 (x, y) = ((x, y), [x : y]) ∈ X. Given the point ((0, 0), [u : v]) ∈ Ep (resp. ((x, y), [x : y])) we are going to represent it by [u : v]Ep (resp. by (x, y) ∈ C2 or by [1 : x : y] ∈ P C2 if it is convenient). After every blow up we get a new expanded space X and the induced map F̃ : X → X. And then F̃ induces a morphism of groups, F̃ ∗ : Pic(X) → Pic(X) just by taking classes of preimages, where Pic(X) is the Picard group of X (see [1, 2]). It is proved  that after a ∗ n finite number of blowing-up’s we get a map F̃ which satisfies F̃ n = F̃ ∗ . Maps F̃ satisfying this equality are called Algebraically Stable Maps (AS for short), (see [12]). The characteristic polynomial of the matrix of F̃ ∗ is the one associated to the sequence of degrees dn := degree F n . 3 2.2 Lists of orbits. We derive our results by using Theorem 1 below, established and proved in [1, 2]. The proof of that is based in the same tools explained in the above paragraph. In order to determine the matrix of the extended map in the Picard group, it is necessary to distinguish between different behaviors of the iterates of the map on the indeterminacy points of its inverse. The theorem is written for a general family G of quadratic maps of the form G = L ◦ J. As we will see the maps of family (18), when the triangle is non-degenerate, are linearly conjugated to such a maps. Here L is an invertible linear map and J is the involution in P C2 as follows: J[x0 : x1 : x2 ] = [x1 x2 : x0 x2 : x0 x1 ]. We find that the involution J has an indeterminacy locus I = {0 , 1 , 2 } and a set of T exceptional curves E = {Σ0 , Σ1 , Σ2 }, where Σi = {xi = 0} for i = 0, 1, 2, and i = Σj Σk with {i, j, k} = {0, 1, 2} and i 6= j 6= k , i 6= k. Let I(G−1 ) := {a0 , a1 , a2 }, the elements of this set are determined by ai := G(Σi − I(J)) = L i for i = 0, 1, 2; see [1]. To follow the orbits of the points of I(G−1 ) we need to understand the following defini- tions and construction of lists of orbits in order to apply the result of Theorem 1. We assemble the orbit of a point p ∈ P C2 under the map G as follows. For a point p ∈ E(G) ∪ I(G) we say that the orbit O(p) = {p}. Now consider that there exits a p ∈ P C2 such that its nth − iterate belongs to E(G) ∪ I(G) for some n, whereas all the other n − 1 iterates of p under G are never in E(G) ∪ I(G). This is to say that for some n the orbit of p reaches an exceptional curve of G or an indeterminacy point of G. We thus define the orbit of p as O(p) = {p, G(p), ..., Gn (p)} and we call it a singular orbit. If for some p ∈ P C2 in turns out that p and all of its iterates under G are never in E(G) ∪ I(G) for all n, we set as O(p) = {p, G(p), G2 (p)...} and O(p) is non singular orbit. We now make another characterization of these orbits. Consider that a singular orbit reaches an indeterminacy point of G, this is to say that Gn (p) ∈ I(G) but its not in E(G). We call such orbits as singular elementary orbits and we refer them as SE-orbits. To apply Theorem 1 we need to organize our SE orbits into lists in the following way. Two orbits O1 = {a1 , ..., j1 } and O2 = {a2 , ..., j2 } are in the same list if either j1 = 2 or j2 = 1, that is, if the ending index of one orbit is the same as the beginning index of the other. We say that a list of orbits  L = Oi = {ai , ..., τ (i) }, . . . , Oj = {aj , ..., τ (j) } 4 is closed if τ (j) = i. Otherwise it is an open list. For instance, L1 = {O1 = {a1 , ..., 1 }} , L2 = {O0 = {a0 , ..., 2 }, O2 = {a2 , ..., 0 }} , L3 = {O0 = {a0 , ..., 1 }, O1 = {a1 , ..., 2 }, O2 = {a2 , ..., 0 }}, are closed lists. We now define two polynomials TL and SL which we will use to state Theorem 1. Let ni denote the sum of the number of elements of an orbit Oi and let NL = nu + ... + nu+µ denote the sum of the numbers of elements of each list |L| . If L is closed then TL = xNL − 1 and if L is open then TL = xNL . Now we define SL for different lists as follows:    1 if |L| = {n1 } ,     n x +x +2 1 n2 if L is closed and |L| = {n1 , n2 } ,      n x +x +1 1 n2 if L is open and |L| = {n1 , n2 } , SL (x) = P 3     xNL −ni + xni + 3 if L is closed and |L| = {n1 , n2 , n3 } ,     i=1   P 3 P ni   xNL −ni + x +1 if L is open and |L| = {n1 , n2 , n3 } .  i=1 i6=2 Theorem 1. ([2]) If G = L ◦ J, then the dynamical degree δ(G) is the largest real zero of the polynomial Y X Y X (x) = (x − 2) TL (x) + (x − 1) SL (x) TL0 (x). L∈Lc ∪Lo L∈Lc ∪Lo L0 6=L Here L runs over all the orbit lists. This theorem enables us to calculate the characteristic polynomial associated to dn . To this end we have to perform the lists of the orbits of the points in I(F −1 ), but for this we have to do the necessary blow-up’s to get an AS mapping. In order to get AS maps we will use the following useful result showed by Fornaess and Sibony in [14] (see also Theorem 1.14) of [12]: The map F̃ is AS if and only if for every exceptional curve C and all n ≥ 0 , F̃ n (C) ∈ / I(F̃ ). (3) 2.3 Zero entropy The following result is quiet useful in our work. It is a direct consequence of Theorem 0.2 of [12]. 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 5 Theorem 2. (See [12]) Let F : P C2 → P C2 be a birational map, F̃ be its regularized map and let dn = deg(F n ). Then up to bimeromorphic conjugacy, exactly one of the following holds: • The sequence dn grows quadratically, F̃ is an automorphism and f preserves an elliptic fibration. • The sequence dn grows linearly and f preserves a rational fibration. In this case F̃ cannot be conjugated to an automorphism. • The sequence dn is bounded, F̃ is an automorphism and f preserves two generically transverse rational fibrations. • 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. We recall that f : C2 → C preserves a fibration V : C2 → C if f sends level curves of V to level curves of V. If f sends each level curve of V to itself, it is said that f is integrable and that V is a first integral of f. When the sequence dn is bounded it can happen that it is periodic or not. For mappings which are not periodic, we have the following result (Theorem A of [4]): Theorem 3. (See [4]) Let F : P C2 → P C2 be a non-periodic birational map such that the corresponding sequence of degrees is bounded. Then F is conjugate to an automorphism of P C2 , which restricts to one of the following automorphisms on some open subset isomorphic to C2 : (1) (x, y) 7→ (α x, β y), where α, β ∈ C∗ , and where the kernel of the group homomorphism Z2 → C∗ given by (i, j) 7→ α β j is generated by (k, 0) for some k ∈ Z. (2) (x, y) 7→ (α x, y + 1), where α ∈ C∗ . 2.4 Invariant fibrations From Theorem 2 we know the existence of rational invariant fibrations depending on the P (x,y) growth of dn . To find them, we consider V (x, y) = Q(x,y) for some polynomials P (x, y), Q(x, y) without common factors. If V is an invariant fibration, then f sends V = k to V = k 0 . ω1 k+ω2 In this work we consider that the relation between k, k 0 is of type ψ(k) = ω3 k+ω4 for some ω1 , ω2 , ω3 , ω4 ∈ C. In particular we will have the following cases: (a) V (f ) = V, the integrable case. 6 (b) V (f ) = ω1 V, the scaled fibration case. (c) V (f ) = ω1 V + ω2 , the scaled translated fibration case. Note that in case (a) in general the functions P and Q are invariant under f as they satisfy the equation P ·Q(f ) = Q·P (f ) (unless that the denominators of P (f ) or Q(f ) are simplified with Q or P respectively). Similarly for case (b) it follows. In case (c) only Q is invariant as it satisfies the relation Q · P (f ) = (ω1 P + ω2 Q) · Q(f ). Hence we always begin finding invariant algebraic curves. To find them, we introduce the following definition. Given a birational map and given a curve C ⊂ P C2 we define F (C) := F (C \ I(F ) to be the proper transform of C by F. When C ∩ I(F ) = ∅, we have that deg F (C) = d· deg(C) where d is the degree of F. In general, X deg F (C) = d · deg (C) − mO (C), (4) O∈I(F ) where mO (C) is the algebraic multiplicity of C at O (see (1), pg. 416, [?]). The approach is the following. Take an arbitrary curve C and impose that deg F (C) = P deg C, that is, (d − 1) deg (C) = O∈I(F ) mO (C). For instance if d = 2 and we consider C of degree 3, then a necessary condition for C to be invariant under f is that C passes through three indeterminacy points of F of multiplicity one or through one indeterminacy point with multiplicity two and another of multiplicity one or through one indeterminacy point with multiplicity three. In the first case, for instance, if O1 , O2 , O3 ∈ I(F ), then there exist T1 , T2 , T3 ∈ E(F ) such that F −1 : Ti  Oi . Also, if C = {P = 0} for some polynomial P, then F (C) ⊂ {P ◦ F −1 = 0} and we have that P ◦ F −1 = T1 · T2 · T3 · P̄ for a certain polynomial P̄ , with F (C) = {P̄ = 0}. Then imposing that P − k · P̄ = 0 we will find, if we get a solution, an invariant curve of degree three. As we will see our particular mappings, sometimes depend on a number α which is a C[α] zero of certain polynomial P . Then all the calculations have to be made in (P (α)) [x, y], which in fact make them more complicated ( (PC[α] (α)) is the quotient ring C[α] over the ideal generated by the polynomial P (α)). 3 The subfamily α2 6= 0. Taking into account that α1 , β1 and γ2 are not zero, it can be proved that when α2 6= 0, after an affine change of coordinates f (x, y) can be written as   x f (x, y) = α0 + α1 x + y, , α1 6= 0. (5) γ0 + y 7 We consider the imbedding (x, y) 7→ [1 : x : y] ∈ P C2 into projective space and consider the induced map F : P C2 → P C2 given by: F [x0 : x1 : x2 ] = [x0 (γ0 x0 + x2 ) : (α0 x0 + α1 x1 + x2 )(γ0 x0 + x2 ) : x0 x1 ]. (6) The indeterminacy locus of F is I(F ) = {O0 , O1 , O2 } , where O0 = [1 : 0 : −γ0 ] , O1 = [0 : 1 : −α1 ] , O2 = [0 : 1 : 0] , and the indeterminacy locus of F −1 is I(F −1 ) = {A0 , A1 , A2 } , where A0 = [0 : 1 : 0] , A1 = [0 : 0 : 1] , A2 = [−α1 : −α1 (α0 − γ0 ) : 1] . The set of exceptional curves is given as E(F ) = {S0 , S1 , S2 } , where S0 = {x0 = 0} , S1 = {γ0 x0 + x2 = 0} , S2 = {γ0 x0 + x2 + α1 x1 = 0} , and the set of exceptional curves of F −1 is given as E(F −1 ) = {T0 , T1 , T2 } , where T0 = {(α0 − γ0 )x0 − x1 = 0} , T1 = {x0 − α1 x2 } , T2 = {x0 = 0} . Theorem 4. Let F [x0 : x1 : x2 ] be defined by F [x0 : x1 : x2 ] = [x0 (γ0 x0 + x2 ) : (α0 x0 + α1 x1 + x2 )(γ0 x0 + x2 ) : x0 x1 ] and let F̃ be the induced map after blowing up the point A0 . Then the following hold: • If F̃ 2k (A1 ) 6= O1 for all k ∈ N and F̃ p (A2 ) = O0 for some p ∈ N then the characteristic polynomial associated with F is given by Xp = xp+1 (x2 − x − 1) + x2 , and – for p = 0, p = 1 the sequence of degrees dn is bounded, – for p = 2 the sequence of degrees dn grows linearly, – for p > 2 the sequence of degrees dn grows exponentially. • Assume that F̃ 2k (A1 ) = O1 for some k ∈ N. Let F̃1 be the induced map after we blow-up the points A0 , A1 , F̃ (A1 ), . . . , F̃ 2k (A1 ) = O1 . If F̃1p (A2 ) 6= O0 for all p ∈ N, then the characteristic polynomial associated with F is given by Xk = x2k+1 (x2 − x − 1) + 1, and the sequence of degrees grows exponentially. Furthermore δ(F ) → δ ∗ as k → ∞. 8 • If F̃ 2k (A1 ) = O1 and F̃ p (A2 ) = O0 for some p, k ∈ N then the characteristic polyno- mial associated with F is given by X(k,p) = xp+1 (x2k+3 − x2k+2 − x2k+1 + 1) + x2k+3 − x2 − x + 1, and 2 (1+k) – for p > k the sequence of degrees dn grows exponentially. – for (k, p) ∈ {(2, 3), (1, 4)} the sequence of degrees dn either, it is periodic or it grows quadratically; – for (k, p) ∈ {(k, 0), (k, 1), (k, 2), (1, 3)} the sequence of degrees dn is periodic. • Assume that F̃ 2k (A1 ) 6= O1 and F̃ p (A2 ) 6= O0 for all k, p ∈ N. Then the characteristic polynomial associated with F is given by X (x) = x2 − x − 1, and the sequence of degrees grows exponentially with δ(F ) = δ ∗ . Proof. Observe that S0  A0 = O2 . The orbit of A0 is SE. By blowing up A0 we get the exceptional fibre E0 and the new space X. The induced map F̃ : X → X sends the curve S0 → E0 → S0 . Observe that now I(F̃ ) = {O0 , O1 } and E(F̃ ) = {S1 , S2 } . We see that A1 6= O1 and the exceptional curve S1  A1 ∈ S0 . We observe that the collision of orbits discussed in preliminaries is happening here. The orbit of A1 under F̃ is as follows: S1  A1 → [γ2 : β2 ]E0 → [0 : α1 (γ0 + β2 ) : β1 ] ∈ S0 → · · · After some iterates we can write the expression of F̃ 2k (A1 ) for all k > 0 ∈ N as F̃ 2k (A1 ) = [0 : α1 (γ0 + β2 )(1 + α1 + α12 + · · · + α1k−1 ) : β1 ] ∈ S0 . Observe that for some value of k ∈ N it is possible that F̃ 2k (A1 ) = O1 . This happens when the following condition k is satisfied for some k. α12 (γ0 + β2 )(1 + α1 + α12 + · · · + α1k−1 ) + α2 β1 = 0. (7) For such k ∈ N the orbit of A1 is SE. By blowing up the points of this orbit we get the new space X1 and the induced map F̃1 . Then under the action of F̃1 we have S1 → G0 → G1 → G2 → · · · → G2k−1 → G2k → T1 . Then I(F̃1 ) = {O0 } and E(F̃1 ) = {S2 }. Now if the orbit of A1 is SE and if F̃1p (A2 ) = O0 that is the orbit of A2 is also SE for some p ∈ N then we have three SE orbits. If condition k is not satisfied then with the 9 extended map F̃ we have I(F̃ ) = {O0 , O1 }. Therefore we have two options: F̃ p (A2 ) = O0 or F̃ p (A2 ) = O1 . We claim that for all p ∈ N, F̃ p (A2 ) 6= O1 . Assume that F̃ p (A2 ) = O1 and assume that F j (A2 ) ∈ / S0 for j = 1, 2, . . . , p − 1. F̃ p (A2 ) = F p (A2 ) = O1 . Since O1 ∈ S0 and A2 ∈ / S0 if F p (A2 ) = O1 then p would be greater than zero and since S0 = T2 , it would imply that O1 = A1 or O1 = A2 , which is not the case (recall that the only points in T2 which have a preimage are A1 and A2 ). Contrarily, if it exists some l ∈ N, l < p such that F j (A2 ) ∈ / S0 for j = 1, 2, . . . , l − 1 but F l (A2 ) ∈ S0 \ {O1 } then F l (A2 ) must be equal to A1 or A2 that is, F l (A2 ) = A1 or F l (A2 ) = A2 . The second case is not possible as A2 is a fixed point. In the first case F̃ p (A2 ) = F̃ p−l (F l (A2 )) = F̃ p−l (A1 ) = O1 which implies that p = l + 2r and F̃ 2r (A1 ) = O1 . Hence the orbit of A1 must be SE and that condition k must be satisfied for k = r which is a contradiction. It implies that the only available possibility for O2 to be SE is to have that for some p, F̃ p (A2 ) = O0 . After the blow up process we get S2 → E1 → E2 → · · · → Ep → Ep+1 → T0 . The extended map F̃2 is an automorphism when we have three SE orbits. The above discussion gives us three different cases. • One SE orbit: This happens when A0 = O2 with the conditions that F̃ 2k (A1 ) 6= O1 and F̃ p (A2 ) 6= O0 for all k, p ∈ N. Therefore we have only one list Lo which is open √ 5+1 that is Lo = {O0 = {A0 = O2 }}. By using Theorem 1 we find that δ(F ) = 2 , which is given by the greatest root of the polynomial X(x) = x2 − x − 1. Therefore it has exponential growth. • Two SE orbits (a): It is the case when A0 = O2 , F̃ p (A2 ) = O0 and F̃ 2k (A1 ) 6= O1 for all k ∈ N. By organizing the orbits into lists we have one closed list Lc = {O0 = {A0 = O2 }, O2 = {A2 , F̃ (A2 ) , ..., F̃ p (A2 ) = O0 }}. By utilizing Theorem 1 we find that the characteristic polynomial associated to F is Xp = xp+1 (x2 − x − 1) + x2 . For p = 0 and p = 1 the sequence of degrees satisfies dn+3 = dn and dn+4 = dn+3 respectively which corresponds towards boundedness of dn . For p = 2 we get the polynomial X2 = x2 (x + 1)(x − 1)2 . Looking at the first degrees we get that the sequence of degrees is dn = −1 + 2 n. For p > 2, we observe that Xp (1) = 0, X 0 p (1) = 2 − p < 0 and limx→+∞ Xp (x) = +∞. Hence Xp always has a root λ > 1 and the result follows. 2k p • Two SE orbits (b): When we have A0 = O2 , F˜1 (A1 ) = O1 and F˜1 (A2 ) 6= O0 for all p ∈ N then there is one open and one closed list and Xk = x2k+1 (x2 − x − 1) + 1. 10 We observe that for all the values of k ∈ N , k ≥ 1 the polynomial Xk has always a root λ > 1. Therefore f has exponential growth. • Three SE orbits: In this case we have A0 = O2 , F̃ 2k (A1 ) = O1 , F̃ p (A2 ) = O0 , for a certain p, k ∈ N. We have two closed lists as follows: Lc = {O0 = {A0 = O2 }, O2 = {A2 , F̃ (A2 ) , ..., F̃ p (A2 ) = O0 }}, Lc = {O1 = {A1 , F̃ (A1 )E0 , ..., F̃ 2k (A1 )S0 = O1 }}. From Theorem 1 we can write X(k,p) = xp+1 (x2k+3 − x2k+2 − x2k+1 + 1) + x2k+3 − x2 − x + 1. The map F̃2 is an automorphism for all the values (k, p). According to Diller and Favre in [12] the degree growth of iterates of an automorphism could be bounded, quadratic or exponential but it cannot be linear as in such a case the map is never an automorphism. For this we observe the behavior of X(k,p) around x = 1. We consider it’s Taylor expansion near x = 1 : X(k,p) (x) = 2(2 − kp + 2k)(x − 1)2 + O(|x − 1|3 ). 2(1+k) Thus X(k,p) vanishes at x = 1 and has a maximum on it if p > k . Since 2(1+k) limx→+∞X(k,p) (x) = +∞, always exists a root greater than one. If p ≤ k , k≥1 then the pairs (k, p) are in the set: A(k,p) = {((k ≥ 1), 0), ((k ≥ 1), 1), ((k ≥ 1), 2), (1, 3), (2, 3), (1, 4)}. For (k, p) = (k, 0), X(k,0) (x) = (x2k+2 − 1) (x − 1) (x + 1), and hence the sequence of degrees is dn = c0 + c1 n + c2 (−1)n + c3 (−1)n n + c4 λn1 + c5 λn2 + ... + c2k+3 λn2k , where ci are constants and λ’s are the roots of polynomial x2k+2 = 1 different from ±1. Since F̃2 is an automorphism for all (k, p), using [12] we have c1 = 0 = c3 . This implies that d2k+2+n = dn , i. e., the sequence of degrees is periodic with period 2k +2. The argument for the proof of other values of (k, p) ∈ A(k,p) follows accordingly. From the above theorem we see that zero entropy cases only appear when F̃ p (A2 ) = O0 , F̃ 2k (A1 ) 6= O1 for p ∈ {0, 1, 2} and ∀k ∈ N and when F̃ p (A2 ) = O0 , F̃ 2k (A1 ) = O1 for (k, p) ∈ {(k, 0), (k, 1), (k, 2), (1, 3), (2, 3), (1, 4)}. We are going to study the dynamics of each case separately. Recall that condition k is given by α12 γ0 (1 + α1 + α12 + · · · + α1k−1 ) + 1 = 0. (8) 11 The following proposition considers the case when p = 0. From the above theorem we know that if condition k is not satisfied the sequence dn is bounded and when it is satisfied, dn is a periodic sequence of period 2k + 2. In any case we have to find two generically trans- verse fibrations. In the second case we present two first integrals functionally independent. We also prove that when dn is periodic, the mapping f (x, y) is itself periodic. Proposition 5. Assume that A2 = O0 . Then f (x, y) can be written as ! 1 x f (x, y) = + α1 x + y, 1 , α1 6= 0 (9) α1 α1 + y and the following hold: • If α1 6= 1 then f (x, y) preserves the two generically transverse fibrations √ √ √ √ α1 − α1 (α1 + α1 )x + α1 (1 + 2 α1 )y + α12 (1 + α1 )y 2 V1 (x, y) = 1 + α1 y √ √ √ −1 + α1 (1 − α1 )x + ( α1 − 2α1 )y + α1 ( α1 − α1 )y 2 V2 (x, y) = 1 + α1 y √ √ with V1 (f (x, y)) = − α1 V1 (x, y) and V2 (f (x, y)) = α1 V2 (x, y). If α1k+1 = 1 then f is a (2k + 2)−periodic map. In this case W1 (x, y) and W2 (x, y) are two independent first integrals, where Wi (x, y) := (Vi (x, y))2k+2 .   x • If α1 = 1 then f (x, y) = 1 + x + y, 1+y and it preserves the two generically trans- verse fibrations 1 − 2 x + 3 y + 2 y2 V1 (x, y) = 1+y 1 + 2 x + 3 y + 2 y2 V2 (x, y) = 2 (1 + y) with V1 (f (x, y)) = −V1 (x, y) and V2 (f (x, y)) = V2 (x, y) + 1. Furthermore f (x, y) is integrable being W (x, y) = V12 (x, y) a first integral. 1 Proof. Condition A2 = O0 gives α0 = γ0 = α1 . From Theorem 4 we know that f has two invariant fibrations. To find them follow the procedure explained in subsection 2.4. We consider an arbitrary cubic projective curve: C[x0 : x1 : x2 ] = r0 x0 3 + r1 x0 2 x1 + r2 x0 2 x2 + r3 x0 x1 2 + r4 x0 x2 2 +r5 x0 x1 x2 + r6 x1 3 + r7 x1 2 x2 + r8 x1 x2 2 + r9 x2 3 and we force that C is zero over the indeterminacy points of F, that is, C(O0 ) = C(O1 ) = C(O2 ) = 0. Then C(F −1 [x0 : x1 : x2 ]) = T0 · T1 · T2 · C̄[x0 : x1 : x2 ], 12 where {T0 , T1 , T2 } = E(F −1 ) and C̄[x0 : x1 : x2 ] is as follows: (r2 α12 − 2r4 α1 + 3r9 )x30 + (r4 α12 − 3r9 α1 )x20 x1 + (r2 α13 + r1 α12 − 2r4 α12 − r5 α1 + 3r9 α1 + r8 )x20 x2 + α12 r9 x0 x21 + (r5 α1 ]2 − 2r8 α1 )x0 x1 x2 + (r1 α13 − r5 α12 + r8 α1 )x0 x22 + (−r9 α13 + r8 α12 )x21 x2 + (r9 α13 + r3 α12 − r3 α12 )x1 x22 . The curve C̄ is a degree three algebraic curve. We now impose that C[x0 : x1 : x2 ] = k C̄[x0 : x1 : x2 ], then after some calculations we found (in affine coordinates) √ √ √ √ Q1 := α1 − α1 (α1 + α1 )x + α1 (1 + 2 α1 )y + α12 (1 + α1 )y 2 , √ √ √ Q2 := −1 + α1 (1 − α1 )x + ( α1 − 2α1 )y + α1 ( α1 − α1 )y 2 , L := 1 + α1 y. The curves Q1 and Q2 are invariant algebraic curves while L is an exceptional curve. Taking √ V1 = Q1 /L and V2 = Q2 /L, simple computations prove that V1 (f (x, y)) = − α1 V1 (x, y), √ V2 (f (x, y)) = α1 V2 (x, y) and that V1 (x, y), V2 (x, y) are generically transverse. Now considering the mapping ϕ(x, y) := (V1 (x, y), V2 (x, y)), we see that it is a birational √ √ mapping and it has the property that (ϕ−1 ◦ f ◦ ϕ)(x, y) = (− α1 x, α1 y). From this we deduce that if α1k+1 = 1 , α1 6= ±1 then f (x, y) is a (2k + 2)−periodic map. For α1 = −1, f is a 4-periodic map.Furthermore since Wi (f (x, y)) = Wi (x, y) for i = 1, 2 we get that W1 (x, y), W2 (x, y) are first integrals. When α1 = 1 we see that V1 or V2 is a constant function and that it is the unique √ value of the parameters which has this behaviour. If we take 1 = 1 we get the invariant 1−2 x+3 y+2 y 2 fibration V1 (x, y) = 1+y with V1 (f (x, y)) = −V1 (x, y). To find V2 we consider a k0 +k1 x+k2 y+k3 y 2 rational function of type V (x, y) = 1+y where ki ∈ C for i ∈ {0, 1, 2, 3} and imposing V (f (x, y)) = V (x, y) + 1 after some calculations we find V2 (x, y). Also in this case f (x, y) is birationally conjugated to (−x, y + 1), see Theorem 3 again. To deal with the case p = 1, that is F (A2 ) = O0 , we notice that this condition is equivalent to α12 (α0 − γ0 ) + α0 α1 − 1 = 0 , α1 γ0 (γ0 − 1) + α0 α1 − γ0 = 0 , γ0 α1 − 1 6= 0. It is easy to see that it is true if and only if 1 1 + α1 + α12 γ0 = , α0 = , α1 ∈ / {0, −1}. 1 + α1 α1 (1 + α1 )2 We note that for these maps condition (8) reads as 1 + α1 + α12 + · · · + α1k+1 = 0, which implies that α1k+2 = 1. Proposition 6. Assume that F (A2 ) = O0 . then f (x, y) can be written as: ! α1 2 + α1 + 1 x f (x, y) = + α1 x + y, 1 , α1 ∈ / {0, −1} (10) α1 (1 + α1 )2 1+α1 + y 13 and • If α1 6= 1 and α1 2 + α1 + 1 6= 0, then the map f (x, y) preserves the two generically transverse fibrations B0 + B1 x + B2 y + B3 y 2 V1 (x, y) = C0 + C1 x + C2 y + C3 x y + C4 y 2 D0 + D1 x + D2 y + D3 y 2 V2 (x, y) = C0 + C1 x + C2 y + C3 x y + C4 y 2 where B0 = α1 2 + α1 + 1 C0 = α1 2 + α1 + 1 √ √ B1 = − (1 + α1 )2 ( α1 − α1 − 1) α1 C1 = α1 2 (1 + α1 )2  3  B2 = − (1 + α1 ) α1 2 − 2 α1 2 − 2 α1 − 1 C2 = 2 α1 3 + 3 α1 2 + 2 α1 + 1 √ B3 = −α1 (1 + α1 )2 ( α1 − α1 − 1) C3 = α1 (α1 − 1) (1 + α1 )3 C4 = α1 2 (1 + α1 )2 and D0 = α1 2 + α1 + 1, √ √ D1 = − (1 + α1 )2 ( α1 + α1 + 1) α1 ,   , D2 = (1 + α1 ) α1 3/2 + 2 α1 2 + 2 α1 + 1 , √ D3 = α1 (1 + α1 )2 ( α1 + α1 + 1) . with V1 (f (x, y)) = √1 V1 (x, y) and V2 (f (x, y)) = − √1α1 V2 (x, y). α1 If 1 + α1 + α12 + · · · + α1k+1 = 0 then f (x, y) is a 2 (k + 2)−periodic map. In this case Wi (x, y) for i ∈ {1, 2} are two independent first integrals, where Wi (x, y) := Vi (x, y) · Vi (f (x, y)) · Vi (f 2 (x, y)) · · · Vi (f 2k+3 (x, y)). • If α1 = 1 then f (x, y) preserves the two generically tranverse fibrations 16 xy + 4 x − 6 y − 3 V1 (x, y) = 4 y2 + 4 x + 8 y + 3 12 y 2 − 12 x + 12 y + 3 V2 (x, y) = 4 y2 + 4 x + 8 y + 3 with V1 (f (x, y)) = V1 (x, y)+1 and V2 (f (x, y)) = −V2 (x, y). Hence f (x, y) is integrable being W (x, y) = V22 (x, y) a first integral. • If α12 +α1 +1 = 0 then f (x, y) is a 6-periodic mapping. It preserves the two generically tranverse fibrations 2 α1 + 2 − x + (2 α1 − 1) y − (1 + α1 ) y 2 V1 (x, y) = (α1 + 1) x + y + (α1 − 1) xy + (α1 + 1) y 2 14 α1 x − α1 y + y 2 V2 (x, y) = (α1 + 1) x + y + (α1 − 1) xy + (α1 + 1) y 2 with V1 (f (x, y)) = −α1 V1 (x, y) and V2 (f (x, y)) = α1 V2 (x, y). Furthermore W1 (x, y) := V16 (x, y) and W2 (x, y) := V26 (x, y) are two independent first integrals. Proof. From Theorem 4 we know that when 1 + α1 + α12 + · · · + α1k+1 = 0, dn is a periodic sequence while when 1 + α1 + α12 + · · · + α1k+1 6= 0, dn is bounded. In any case we have to find two generically transverse foliations. We first search for invariant curves C(x, y) = C0 + C1 x + C2 y + C3 x y + C4 y 2 . Then P (x,y) we consider a rational function V (x, y) = C(x,y) , where P (x, y) is a second degree polyno- mial. The imposition of condition V (f (x, y)) = k · V (x, y) gives two invariant fibrations V1 (x, y), V2 (x, y) for k ∈ { √1α1 , − √1α1 }. Also we see that V1 , V2 are generically transverse provided that α1 6= ±1 , α12 + α1 + 1 6= 0. Let ϕ(x, y) be defined as ϕ(x,  y) = (V1 (x, y),  V2 (x, y)) . Then ϕ(x, y) is a birational map −1 1 1 and ϕ ◦ f ◦ ϕ gives the map √α1 x, − √α1 y . Hence if condition (8) is accomplished, i. e., if 1 + α1 + α12 + · · · + α1k+1 = 0, then f (x, y) is a (2k + 4)−periodic map. √ Now assume that α1 = 1. Substituting this value with 1 = 1 in the maps V1 , V2 in the above paragraph we find that the first fibration is a constant function while the second one 12 y 2 −12 x+12 y+3 is V2 (x, y) = 4 y 2 +4 x+8 y+3 , hence it satisfies V2 (f (x, y)) = −V2 (x, y). To find the other fibration V (x, y) we consider a rational map with the same denominator of V2 (x, y) and a degree two polynomial in the numerator and imposing V (f (x, y)) = V (x, y) + 1 we find the announced V1 (x, y). The fibrations when α12 + α1 + 1 = 0 are encountered in a similar way. Proposition 7. Assume that F (F (A2 )) = O0 . Then f (x, y) can be written as  ! ω3 − ω2 + 1 2 ω ω2 − ω + 1 x f (x, y) = + ω x + y, , ω (ω + 1) (ω 2 − ω + 1) 6= 0, (ω + 1) (ω 2 − ω + 1)2 ω − 1 + (ω 3 − ω 2 + ω) y and it preserves the fibration B0 + B1 x + B2 y + B3 y 2 V (x, y) = (ω + (ω 3 + 1) y) (ω − 1 + (ω 5 − ω 4 + ω 3 + ω 2 − ω + 1) x + (ω 3 − ω 2 + ω) y) where B0 = (ω 3 − ω 2 + 1)(ω − 1), 2 B1 = −ω 2 (ω + 1) ω 2 − ω + 1 ,   B2 = ω ω 2 − ω + 1 2 ω 3 − ω 2 − ω + 1 , 2 B3 = ω 3 (ω + 1) ω 2 − ω + 1 , 15 with V (f (x, y)) = − ω1 V (x, y). If ω 4k+6 6= 1 for all k ∈ N this fibration is unique. If ω m = (−1)m for some m ∈ N, then f (x, y) is integrable being W (x, y) = V (x, y)m a first integral. P2k+2 When i=0 (−1)i ω i = 0 for a certain k ∈ N then f (x, y) is a (4k + 6)−periodic map. Proof. Now we assume that F 2 (A2 ) = O0 . It is easy to see that it is equivalent to F̃ 2 (A2 ) = O0 . For the simplification of calculations we consider α1 = ω 2 . It implies that the coefficients have to satisfy: E1 := w6 γ02 − (α0 w6 + (α0 + 1) w4 + (α0 − 2) w2 ) γ0 + w4 α0 + α0 − 1 = 0, E2 := w4 γ03 − (w6 + w4 + w2 ) γ02 + (α0 w6 + (2α0 + 1) w4 − w2 ) γ0 − w4 α0 − w2 α0 + 1 = 0 γ0 w2 − 1 6= 0 , w2 γ02 − (w2 + 1) γ0 + w2 α0 6= 0. Taking into account some resultants of E1 and E2 we find that the condition F (F (A2 )) = O0 gives the maps which appears in (a). When ω 2 − ω + 1 = 0, that is, when α12 + α1 + 1 = 0 we get the mappings (b). We note that for the parametric family (a) condition (8) is 1 − ω + ω 2 − ω 3 + ω 4 + · · · − ω 2k+1 + ω 2k+2 = 0, which implies that ω is a (4 k + 6)−root of unity, while for the two mappings (b), condition k never is satisfied. Consider f (x, y) that satisfies (a). By looking for invariant curves we find that V (x, y) can be written as shown in statement of (a). A calculation shows that V (f (x, y)) = − ω1 V (x, y). From this equality, we see that if ω m = (−1)m then W (x, y) := V (x, y)m is a first integral of f (x, y). P2k+2 If i=0 (−1)i ω i = 0 for a certain k ∈ N then we know that the sequence of degrees is periodic of period 4k + 6. We are going to prove that, the map itself is periodic of period 4k + 6. Since d4k+6 = d0 = 1, the mapping F 4k+6 is linear, that is: F 4k+6 [x0 : x1 : x2 ] = [r0 x0 + r1 x1 + r2 x2 : p0 x0 + p1 x1 + p2 x2 : q0 x0 + q1 x1 + q2 x2 ], for some constants ri , pi , qi ∈ R. As S0 is invariant under the action F 2 , it is invariant under the action of F 4k+6 as well. This implies that we can write f 4k+6 (x, y) = (p0 + p1 x + p2 y, q0 + q1 x + q2 y), (11) for some p0 , p1 , p2 , q0 , q1 , q2 ∈ N. 16 We find that the following two are the fixed points of f and the third one is fixed by f 2 .   1 ω f ix1 = (ω2 −ω+1) (ω+1) (ω 3 +1) , − ω 3 +1 ,   ω 3 −ω 2 +1 ω 3 −ω 2 +1 f ix2 = ω (ω2 −ω+1) (ω2 −1) (ω4 −ω3 +ω−1) , − ω (ω4 −ω3 +ω−1) ,   1 ω f ix3 = ω6 +2 ω3 +1 , − ω3 +1 . Now these points must also be fixed by f 4k+6 . Then by finding the images of f ix1 , f ix2 and f ix3 under the action of f 4k+6 using (11) such that f 4k+6 (f ix1 ) = f ix1 , f 4k+6 (f ix2 ) = f ix2 , f 4k+6 (f ix3 ) = f ix3 . Also as the sequence of degrees is periodic of period 4 k + 6 this implies that (F̃1∗ )4k+6 fixes the elements in the basis of Picard group. This implies that (F̃1∗ )4k+6 also fixes E1 that is the blown up fibre at A2 . Then F 4k+6 fixes the base point A2 in P C2 . By utilizing this information and then solving this system of four equations for the values of p0 , p1 , p2 , q0 , q1 , q2 we find that (p0 , p1 , p2 , q0 , q1 , q2 ) = (0, 1, 0, 0, 0, 1) which shows that f 4k+6 (x, y) = (x, y). Next case of zero entropy is when p = 3 and k = 1. Condition k implies α12 γ0 + 1 = −1 0, i. e., γ0 = α21 . It is easy to see that the condition F̃ 3 (A2 ) = O0 is equivalent to F 3 (A2 ) = O0 . Some computations show that it is true if and only if α16 + α13 + 1 = 0 and α0 = −2 α15 + α13 − α12 − α1 . Proposition 8. Assume that F 3 (A2 ) = O0 and that condition k is satisfied for k = 1. Then f (x, y) can be written as   x f (x, y) = −2 α15 + α13 − α12 − α1 + α1 x + y, , α16 + α13 + 1 = 0, (α1 + α14 ) + y and it is a 18−periodic map. It preserves the two following generically transverse foliations H1 (x,y) H2 (x,y) V1 (x, y) = C(x,y)2 and V2 (x, y) = C(x,y)2 where  C(x, y) = −α1 4 − α1 3 + α1 2 − 2 − α1 4 x + α1 5 − 2 α1 4 − α1 3 + 2 α1 2 − α1 − 1 y − y 2 and H1 (x, y) = A0 + A1 x + A2 y + A3 x2 + A4 xy + A5 y 2 + A6 x2 y + A7 xy 2 + A8 y 3 + A9 x3 y + A10 x2 y 2 + A11 xy 3 + A12 y 4 + 12x3 y 2 + A13 x2 y 3 + A14 xy 4 , H2 (x, y) = B0 + B1 x + B2 y + B3 x2 + B4 xy + B5 y 2 + B6 x2 y + B7 xy 2 + B8 y 3 + 3α14 x3 y + B9 x2 y 2 + B10 xy 3 + B11 y 4 + 3x2 y 3 + B12 xy 4 , 17 with A0 = 31 α1 5 + 23 α1 4 − 23 α1 3 + 35 α1 + 12 A8 = −4 α1 5 − 16 α1 4 − 4 α1 3 + 10 α1 2 − 6 α1 − 12 5 4 3 2 A1 = −18 α1 + 4 α1 + 20 α1 − 14 α1 − 14 α1 + 10 A9 = 2 α1 4 + 2 α1 3 + 4 α1 + 4 5 4 3 2 A2 = 32 α1 + 48 α1 − 14 α1 − 20 α1 + 44 α1 + 36 A10 = 12 α1 5 − 12 α1 4 − 12 α1 3 + 18 α1 2 + 6 α1 − 12 A3 = −3 α1 4 − 2 α1 3 + 2 α1 2 − 2 A11 = 16 α1 5 + 8 α1 4 − 12 α1 3 + 2 α1 2 + 16 α1 + 6 5 4 3 2 A4 = −60 α1 − 4 α1 + 58 α1 − 34 α1 − 48 α1 + 20 A12 = 16 α1 5 + 8 α1 4 − 12 α1 3 + 2 α1 2 + 16 α1 + 6 5 4 3 2 A5 = −6 α1 + 2 α1 + 8 α1 − 3 α1 − 2 α1 + 5 A13 = −16 α1 5 + 2 α1 3 − 14 α1 2 + 4 A6 = 8 α1 5 − 18 α1 4 − 16 α1 3 + 16 α1 2 − 20 A14 = −4 α1 5 − 4 α1 4 − 2 α1 2 − 2 α1 , 5 4 3 2 A7 = −24 α1 − 12 α1 + 12 α1 − 8 α1 − 20 α1 and B0 = −38 α1 5 − 20 α1 4 + 31 α1 3 − 7 α1 2 − 40 α1 − 7 B7 = 20 α1 5 − 30 α1 3 + 16 α1 2 + 24 α1 − 12 4 3 2 B1 = −11 α1 − 4 α1 + 9 α1 − 4 α1 − 11 B8 = −3 α1 5 + 7 α1 4 + 3 α1 3 − 9 α1 2 + 2 α1 + 9 5 4 3 2 B2 = −62 α1 − 51 α1 + 44 α1 + 5 α1 − 72 α1 − 29 B9 = −3 α1 5 + 6 α1 4 + 6 α1 3 − 6 α1 2 + 3 α1 + 3 B3 = 3 α1 5 − 2 α1 3 + 3 α1 2 + 3 α1 − 1 B10 = 3 α1 5 − 9 α1 4 + 6 α1 2 − 6 α1 − 3 5 4 3 2 B4 = 31 α1 − 15 α1 − 33 α1 + 32 α1 + 21 α1 − 24 B11 = 3 α1 5 + α1 4 − 3 α1 3 + 2 α1 5 4 3 2 B5 = −25 α1 − 20 α1 + 19 α1 + α1 − 28 α1 − 10 B12 = −3 α1 5 − 3 α1 2 . B6 = 6 α1 5 + 3 α1 4 − 3 α1 3 + 3 α1 2 + 6 α1 + 6 They satisfy V1 (f (x, y)) = α13 V1 (x, y) and V2 (f (x, y)) = α12 V2 (x, y). Hence, W1 (x, y) = V1 (x, y)6 and W2 = V2 (x, y)9 are two generically transverse first integrals of f (x, y). Proof. To find the foliations we began looking for degree 3 invariant curves. We only found C̄[x0 : x1 : x2 ] = x0 C h [x0 : x1 : x2 ] where C h [x0 : x1 : x2 ] is the homogeneous polynomial of degree two with C h [1 : x1 : x2 ] = C(x1 , x2 ). Then we were looking for degree six invariant curves, with the condition that they passes trough the three indeterminacy points O1 , O2 and O3 with multiplicity two.Consequently, its image has also degree six. Forcing that this image coincides with the curve itself we found some of them. For instance, the two numerators of V1 (x, y) and V2 (x, y). A computation gives that V1 (f (x, y)) = α13 V1 (x, y), V2 (f (x, y)) = α12 V2 (x, y) and that they are generically transverse. Clearly W1 (x, y) and W2 (x, y) are first integrals of f (x, y) because α118 = 1. From Theorem 4 we know that the sequence of degrees is periodic of periodic 18. To prove that the map is periodic we apply the result of [7], which says that if a map has two independent first integrals, then it is a periodic map. Proposition 9. Assume that F 3 (A2 ) = O0 and that condition k is satisfied for k = 2. Then either: (a) There exists α1 with α14 + α13 + α12 + α1 + 1 = 0 such that f (x, y) is of the form   3 2 x f (x, y) = −(α1 + 2 α1 + α1 + 2) + α1 x + y, . (12) −(1 + α12 + α13 ) + y 18 L(x,y)·P (x,y)·Q(x,y) That map f (x, y) preserves the elliptic fibration V (x, y) = R(x,y)2 where  L(x, y) = −α13 − 2α12 − 2α1 − 2 + (α12 + α1 )x + y  P (x, y) = yx + (−α12 − 1)x + α12 y + α13 + α1    Q(x, y) = α1 3 y 2 + −α1 3 − α1 2 − α1 − 1 xy + −α1 3 + α1 2 y + α1   R(x, y) = y 2 − 3 α1 3 + 3 α1 2 + 2 α1 + 2 y − xα1 2 + α1 3 − α1 2 + 1 with V (f (x, y)) = α12 V (x, y) and this fibration is unique. Furthermore f is integrable being W (x, y) = V (x, y)5 a first integral of f. (b) The map f (x, y) is: ! 1 x f (x, y) = + x + y, 1 . (13) 4 −2 + y That map f (x, y) preserves the elliptic fibration V (x, y) = 256x3 y 2 + 384x2 y 3 + 128xy 4 + 128x3 y + 192x2 y 2 + 32xy 3 − 16y 4 − 16x2 − 8 xy + 8y 2 − 8x − 1 (−4y 2 + 4x + 1)2 with V (f (x, y)) = V (x, y) and this fibration is unique. Hence f is integrable. −1 Proof. When k = 2 condition k says α12 γ0 (1 + α1 ) + 1 = 0, i. e., γ0 = α21 (1+α1 ) . Also here F̃ 3 (A2 ) = O0 is equivalent to F 3 (A2 ) = O0 . Some tedious computations show that it is true if and only if either, 1 + α1 + α12 + α13 + α14 = 0 with α0 = − (α13 + 2 α12 + α1 + 2) or α1 = 1 with α0 = 41 . For the mappings (a) we find the invariant conic:  y2 − 3 α1 3 + 3 α1 2 + 2 α1 + 2 y −α1 2 x+α1 3 −α1 2 +1 and a degree five invariant curve, the one given by L(x, y) · P (x, y) · Q(x, y) = 0. Taking the quotient of them, some calculations prove that in fact V (f (x, y)) = α12 V (x, y). To prove the uniqueness of the invariant fibration we have to see that dn is not a periodic sequence. Assume that it is, i. e., assume that dn is 30−periodic. Then F 30 has degree one: F 30 [x0 : x1 : x2 ] = [r0 x0 + r1 x1 + r2 x2 : p0 x0 + p1 x1 + p2 x2 : q0 x0 + q1 x1 + q2 x2 ]. As before, since S0 is invariant under F 2 , we can write f 30 as follows: f 30 (x, y) = (p0 + p1 x + p2 y, q0 + q1 x + q2 y) .  Now, using that the conic y 2 − 3 α1 3 + 3 α1 2 + 2 α1 + 2 y − xα1 2 + α1 3 − α1 2 + 1 = 0 must be invariant under f 30 and that the point (−α13 − α12 , 1) (which is a fixed point for f ) must also be fixed for f 30 , after some calculations we get that either, f 30 is the identity or f 30 ◦ f 30 is the identity. In any case, it would imply that f is a periodic mapping. But we claim that the mapping f itself is not periodic. If it were the case, then f k (x, y) = (x, y) for some k multiple of 30. We observe that f sends: {L(x, y) = 0} −→ {P (x, y) = 0} −→ {Q(x, y) = 0} −→ {L(x, y) = 0}. 19 In particular f 3 sends {Q(x, y) = 0} to {Q(x, y) = 0}. We see that the curve {Q(x, y) = 0} α (α2 +1+α y) can be parameterized by y, because {Q(x, y) = 0} if and only if x = ϕ(y) := 1 α12 +1−y 1 . 1 u(y) Then f 3 (ϕ(y), y) = (ϕ(h(y)), h(y)) where h(y) = v(y) with   u(y) = −5α1 3 − 3α1 2 − α1 − 6 + 29 α1 3 + 10α1 2 + 13α1 + 27 y + −54 α1 3 − 3 α1 2 − 31 α1 − 40 y 2 +    50α1 3 − 9α1 2 + 32α1 + 20 y 3 + −22α1 3 + 7α1 2 − 23 α1 − 3 y 4 + 2α1 3 − 6 α1 2 + 5α1 − 2 y 5 and    v(y) = 5α1 3 − α1 2 + 4 α1 + 2 y + −11 α1 3 + 8α1 2 − 11α1 y 2 + 9α1 3 − 12 α1 2 + 12 α1 − 9 y 3 +   8 α1 2 − 9α1 + 8 y 4 + −3 α1 3 − 6 α1 2 − α1 − 5 y 5 . If f where a periodic mapping, h also would be periodic. But h has the fixed point ȳ = 1 + α12 and the derivative of h(y) at this points gives zero. And it is a contradiction because periodic maps have the eigenvalues of modulus one at the fixed points. To prove (b) we begin by proving that the sequence of degrees grows quadratically. Then the prescribed fibration will be unique. The characteristic polynomial associated   to dn is (x + 1) x2 + x + 1 x4 + x3 + x2 + x + 1 (x − 1)4 which implies that either, dn grows quadratically or it is periodic. It only depends on the initial conditions, that is on the values of dn for n = 1, 2, . . . , 11. For that mapping we have been able to calculate these numbers: 2, 3, 5, 8, 12, 16, 22, 28, 35, 43, 52 which implies that √ !n √ !n 97 5 n2 1 1 −1 + 3I 1 −1 − 3 I dn = + − (−1)n − − , 72 12 8 9 2 9 2 that is, dn grows quadratically. To find V (x, y) we searched for invariant curves and we found one of degree two: −4y 2 + 4x + 1 and one of degree five, the numerator of V (x, y). Taking the quotient of them, we verified that it satisfies V (f (x, y)) = V (x, y). The last class with zero entropy is when p = 4 with k = 1. The condition k = 1 says −1 that γ0 = a12 . From the proof and notations of Theorem 4 we know that: S0 −→ E0 −→ S0 = T2 , S1 −→ G0 −→ G1 −→ G2 −→ T1 , S2 −→ E1 −→ E2 −→ E3 −→ E4 −→ E5 −→ T0 . Hence, if A2 ∈ S1 , i.e., α1 = −1, then it could happen that F̃ 4 (A2 ) = O0 . Following the orbit of A2 we get: F̃ (A2 ) = [1 : 0]G0 , F̃ 2 [1 : 0]G0 = [1 : −α0 ]G2 and F̃ [1 : −α0 ]G2 = [1 : 0 : 1] = O0 . Hence we see that F̃ 4 (A2 ) = O0 for all values of α0 , provided that α1 = −1 and −1 γ0 = a12 = −1. 20 Proposition 10. Assume that F̃ 4 (A2 ) = O0 and that condition k is satisfied for k = 1, where F̃ is the mapping induced by F after blowing up the point [0 : 1 : 0]. Then either: (a) The map f (x, y) can be written as   x f (x, y) = α0 − x + y, (14) y−1 and it preserves the unique elliptic fibration α0 xy − x2 y + xy 2 V (x, y) = y−1 with V (f (x, y)) = V (x, y). Hence f is integrable. (b) The map f (x, y) can be written as   x f (x, y) = x + y, (15) y−1 and it preserves the unique elliptic fibration −2 y 2 + 2 x + y + 1 V (x, y) = xy (x + y) with V (f (x, y)) = −V (x, y). Furthermore f is integrable, being W (x, y) = V (x, y)2 a first integral of f. (c) The map f (x, y) can be written as   x f (x, y) = α1 x + y, with α12 + 1 = 0 (16) y+1 and it preserves the unique elliptic fibration xy (α1 y − x) V (x, y) = − (α1 y + α1 − 2 x − y − 1) (−1 + α1 − 2 y) with V (f (x, y)) = α1 V (x, y). Furthermore f is integrable, being W (x, y) = V (x, y)4 a first integral of f. (d) The map f (x, y) can be written as   x f (x, y) = 1 − α13 + α1 x + y, 2 with α14 + 1 = 0 (17) α1 + y and it preserves the unic elliptic fibration Q1 (x, y) Q2 (x, y) Q3 (x, y) V (x, y) = − (α1 − 1 + (α1 + 1) x + (α1 2 + α1 ) y)2 (α1 3 − 1 + α1 2 (α1 2 + 1) y)2 3 2 21 where Q1 (x, y) = α12 + 2 α1 + 1 + (2 α12 + α1 + 1) y + α13 y 2 − x y,   Q2 (x, y) = 2 α1 3 + α1 2 − 1 + −2 α1 3 + α1 + 1 x + α1 2 + 2 α1 + 1 y + α1 3 x2 + α1 2 xy, ,  Q3 (x, y) = −(α1 2 + 2 α1 + 1) + 2 α1 3 + α1 2 − 1 y − α1 2 xy. with V (f (x, y)) = α12 V (x, y). Furthermore f is integrable, being W (x, y) = V (x, y)4 a first integral of f. Proof. The mapping (a) corresponds to the case α1 = −1 and γ0 = −1 when there is collisions of orbits. Looking at the expression of F 4 (A2 ) and after tedious computations we get that F 4 (A2 ) = O0 if and only if f (x, y) is one of (b), (c) or (d). To see the uniqueness of the fibrations we have to prove that dn grows quadratically. The characteristic polynomial associated to dn is (x − 1)4 (x + 1)2 (x2 + 1)(x2 + x + 1) which implies that either, dn grows quadratically or it is periodic. It only depends on the initial conditions, that is on the values of dn for n = 1, 2, . . . , 10. For each one of the mappings which appear in the statement, we have been able to calculate these numbers. In the four cases they give 2, 3, 5, 7, 11, 15, 20, 25, 32, 39, which implies that 23 3 2 3 1 1 dn = + n − (−1)n − I n − (−I)n . 16 8 16 8 8 In order to prove (a) we find the family of invariant curves λ (α0 xy − x2 y + xy 2 ) + µ (y − P 1) = 0. Then taking V = Q with and P = α0 xy − x2 y + xy 2 and Q = y − 1 we have that V (f (x, y)) = V (x, y). To prove (b) we easily see that {x = 0} −→ {y = 0} −→ {x + y = 0} −→ {x = 0} and hence x y (y + x) is an invariant cubic. Then taking V as the quotient of a conic and the invariant cubic and imposing V (f (x, y)) = k V (x, y) we found that the conic can be taken as −2 y 2 + 2 x + y + 1 and k = −1. To prove (c) we find that the straight line α1 y +α1 −2 x−y −1 = 0 is sent to the straight line −1 + α1 − 2 y = 0 and viceversa, which implies that their product is an invariant curve of degree two. Also it can be seen that {x = 0} −→ {y = 0} −→ {α1 y − x = 0} −→ {x = 0} and hence x y (α1 y − x) is an invariant cubic. Taking V as the quotient of this invariant curves we get that V (f (x, y)) = α1 V (x, y) and the result follows To see (d) we began searching invariant curves of degree three and we found (in projective coordinates)      C[x0 : x1 : x2 ] = x0 · (α3 − 1) x0 + α2 + 1 x1 + α2 + α x2 · (α3 − 1) x0 + α2 α2 + 1 x2 22 Now we take a conic that passes through two indeterminacy points of F , and we impose that its image (a conic again) also passes through two indeterminacy points of F. This gives a third conic which we impose to be equal to the first one. With this we find Q1 , Q2 , Q3 . Then Q1 · Q2 · Q3 is an invariant curve of degree six. Taking V as the quotient of Q1 · Q2 · Q3 over C[1 : x : y]2 we get that V (f (x, y)) = α12 V (x, y). We can now state the main theorem of this section: Theorem 11. Assume that   β0 + β1 x + β2 y f (x, y) = α0 + α1 x + α2 y, , α1 6= 0 , β1 6= 0 , γ2 6= 0 , α2 6= 0. (18) γ0 + γ2 y Then it has zero entropy if and only if after an affine change of coordinates it can be written as one of the mappings which appear in the statements of Propositions 5, 6, 7, 8, 9, 10. Each one of them has the invariant fibrations which are stated in the above propositions. 4 The subfamily α2 = 0.   β0 γ2 −β2 γ0 γ0 By conjugating f (x, y) via h(x, y) = β1 γ2 x − β1 γ2 , β1 y − γ2 and renaiming the pa- rameters, we can consider   x + β2 y f (x, y) = α0 + α1 x, with α1 6= 0. y We consider the induced map in the projective plane : F : P C2 → P C2 given by F [x0 : x1 : x2 ] = [x0 x2 : (α0 x0 + α1 x1 ) x2 : x0 (x1 + β2 x2 )]. (19) The indeterminacy sets of F and F −1 are I(F ) = {O0 , O1 , O2 } , where O0 = [1 : 0 : 0] , O1 = [0 : 0 : 1] , O2 = [0 : 1 : 0] , and I(F −1 ) = {A0 , A1 , A2 } , where A0 = [0 : 1 : 0] , A1 = [0 : 0 : 1] , A2 = [1 : α0 : β2 ] . Furthermore the exceptional curves of F and F −1 are the following: S0 = {x0 = 0} , S1 = {x2 = 0} , S2 = {x1 = 0} , T0 = {α0 x0 − x1 = 0} , T1 = {β2 x0 − x2 = 0} , T2 = {x0 = 0} . 23 Theorem 12. Let f (x, y) be a map of type (18) with γ2 6= 0, α1 6= 0, β1 6= 0 and suppose that α2 = 0. If f p (α0 , β2 ) = (0, 0) for some p ∈ N then the characteristic polynomial associated with f is given by Xp = (xp+1 + 1)(x − 1)2 (x + 1), and the sequence of degrees of f is periodic with period 2p + 2. If no such p exists then the characteristic polynomial associated with f is X = (x − 1)2 (x + 1), and the sequence of degrees dn grows linearly. Proof. Observe that S0  A0 = O2 and S1  A1 = O1 . Hence we blow up the points A0 , A1 getting the exceptional fibres E0 , E1 . Let X be the new space and let F̃ : X → X be the corresponding map on X. Then the map F̃ sends the curve S0 → E0 → S0 and S1 → E1 → T1 . We observe that no new indeterminacy points are created therefore I(F̃ ) = {O0 } and E(F̃ ) = {S2 } . Assume that there exists p ∈ N such that F̃ p (A2 ) = O0 . Then we blow up A2 , F̃ (A2 ), F˜2 (A2 ), . . . , F˜p (A2 ) = O0 getting the exceptional fibres which we call E2 , E3 , . . . , Ep+2 . Set F˜1 : X1 → X1 the extended map. Performing the blow up at O0 , since T0 is sent to O0 via F −1 , we have that F̃1−1 : T0 → Ep+2 . Then S2 → E2 → E3 → · · · → Ep+1 → Ep+2 → T0 . Hence F̃1 : X1 → X1 is an AS map and also an automorphism. Taking into account that A2 = [1 : α0 : β2 ] and O0 = [1 : 0 : 0] belong to the affine plane, it is clear that condition F̃ p (A2 ) = O0 reads as f p (α0 , β2 ) = (0, 0). Now we have two closed lists as follows Lc1 = {O0 = {A0 = O2 }, O2 = {A2 , F̃ (A2 ) , . . . , F̃ p (A2 ) = O0 }}, Lc2 = {O1 = {A1 = O1 }}. Then by using Theorem 1 we find that the characteristic polynomial associated to F is X = (xp+1 + 1)(x − 1)2 (x + 1). If p is even then xp+1 +1 has the factor x+1 and X = (x−1)2 (x+1)2 (xp −xp−1 +· · ·−x+1). Hence the sequence of degrees is dn = c0 + c1 n + c2 (−1)n + c3 n (−1)n + c4 λn1 + c5 λn2 + ... + cp+3 λnp , where ci are constants and λ1 , λ2 , ..., λp are the roots of polynomial xp − xp−1 + · · · − x + 1. By looking at dn we see that f does not grow quadratically or exponentially. As our map F̃1 is an automorphism then by using the results from Diller and Favre in [12] we see that also cannot have linear growth. Therefore we must have c1 = c3 = 0. Hence the 24 sequence of degrees must be periodic. This implies that d2p+2+n = dn i.e. the sequence of degrees is periodic with period 2p + 2. If p is odd then dn is also periodic of period 2p + 2. If F̃ p (A2 ) 6= O0 for all p ∈ N, then we have two lists which are open and closed as follows: Lo = {O0 = {A0 = O2 }} , Lc = {O1 = {A1 = O1 }}. Then δ(F ) is determined by the polynomial (x − 1)2 (x + 1), and δ(f ) = 1. The sequence of 5 + 12 n − 14 (−1)n . degrees is dn = 4   Theorem 13. Let f (x, y) = α0 + α1 x, x+β y 2y with α1 6= 0 and set h(x) = α0 + α1 x. Then the following hold: 1. If f p (α0 , β2 ) 6= (0, 0) for all p ∈ N then f preserves the fibration V1 (x, y) = x with V1 (f (x, y)) = α0 + α1 V1 (x, y), and this fibration is unique. If α1n = 1 for some n > 1 , α1 6= 1, the map is integrable being W (x, y) = x · h(x) · h(h(x)) · · · hn−1 (x) a first integral of f. Also when α1 = 1 and α0 = 0, f is integrable. 2. If f p (α0 , β2 ) = (0, 0) for some p ≥ 1, then f is a (2p + 2)−periodic map. These maps have W (x, y) = x · h(x) · h(h(x)) · · · h2p+1 (x) as a first integral. 3. If (α0 , β2 ) = (0, 0), then f (x, y) = (α1 x, xy ) and it preserves the two generically trans- verse fibrations √ x √ x V1 (x, y) = α1 y + V2 (x, y) = − α1 y + , y y √ √ with V1 (f (x, y)) = α1 V1 (x, y) and V2 (f (x, y)) = − α1 V2 (x, y). When α1n = 1 for some n then f is 2n−periodic and W1 (x, y) = V1 (x, y)2n , W2 (x, y) = V2 (x, y)2n are two independent first integrals. Remark 14. We notice that when p = 0, that is, α0 = 0 = β2 , then ϕ(x, y) := = (V1 (x, y), V2 (x, y)) is a birational map. It turns out that using ϕ(x, y) as a conjuga- √ √ tion we get the map ( α1 x, − α1 y). These result on linearizations was already pointed out on the work of Blanc and Deserti, see[4]. Furthermore, the sequence of degrees is dn = 2, 1, 2, 1, 2, 1, ... a two-periodic sequence, and avoiding the case α1n = 1 for some n, the map itself in not more periodic. For p ≥ 1 the map is periodic and hence it has two independent first integrals. There is 2 a method to find them (see [7]). For instance, when α1 = −1 and α0 = −β2 (case p = 1) x+β y i.e., f (x, y) = −x − β22 , y 2 , we have that x + β2 y x (β2 − y) x + β22 H(x, y) = y + + + y x + β2 y β2 − y 25 is a first integral of f and W (x, y) , H(x, y) are generically transverse. Proof. If f p (α0 , β2 ) 6= (0, 0) for all p ∈ N then from the above theorem we know that dn grows linearly, and hence we know that f (x, y) has a unique invariant fibration. Clearly V1 (x, y) = x is an invariant fibration and when α1 = 1 and α0 = 0, V1 (x, y) is a first integral. When α1n = 1 the function h(x) is periodic of period n and hence W (x, y) is a first integral of f (x, y). Now assume that f p (α0 , β2 ) = (0, 0) for a certain p ∈ N. From Theorem (12) we know that the sequence of degrees dn is 2p + 2 periodic. We are going to see that f (x, y) itself is a periodic map. Since the map F 2p+2 is linear, we can consider that for some constants ri , pi , qi ∈ R the map F 2p+2 can be written in the following form: F 2p+2 [x0 : x1 : x2 ] = [r0 x0 + r1 x1 + r2 x2 : p0 x0 + p1 x1 + p2 x2 : q0 x0 + q1 x1 + q2 x2 ]. We know that S0 is invariant under the action F 2 therefore it is invariant under the action of F 2p+2 as well. This implies that F 2p+2 [0 : x1 : x2 ] = [0 : x1 : x2 ], which further implies that r1 x1 + r2 x2 = 0 for all complex numbers x1 , x2 . This is only possible if r1 = r2 = 0. Then we can write   p0 p1 p2 q0 q1 q2 F 2p+2 [x0 : x1 : x2 ] = x0 : x0 + x1 + x2 : x0 + x1 + x2 , r0 r0 r0 r0 r0 r0 which in the affine plane by taking x0 = 1 and rewriting the parameters, as new parameters, the function F 2p+2 can be written as following: f 2p+2 (x, y) = (p0 + p1 x + p2 y, q0 + q1 x + q2 y), (20) for any p0 , p1 , p2 , q0 , q1 , q2 ∈ R. We find that the following two points are fixed for f (x, y) :  p  α0 β2 (1 − α1 ) ± (1 − α1 )2 β22 + 4α0 (1 − α1 ) (X, ±Y ) = , , 1 − α1 2(1 − α1 ) As these points are fixed by f so they are also fixed points of f 2p+2 . Then finding their images under the action of f 2p+2 using (20) we get a system equations such that f 2p+2 (X, Y )[1] = X, f 2p+2 (X, Y )[2] = Y, f 2p+2 (X, −Y )[2] = −Y. Also as the sequence of degrees is periodic of period 2p + 2 this implies that (F̃1∗ )2p+2 fixes the elements in the basis of Picard group. This implies that (F̃1∗ )2p+2 also fixes E2 that is the blown up fiber at A2 . Then F 2p+2 fixes the base point A2 in P C2 . By utilizing this information and then solving the system of equations for the values of p0 , p1 , p2 , q0 , q1 , q2 we find that (p0 , p1 , p2 , q0 , q1 , q2 ) = (0, 1, 0, 0, 0, 1) which implies that f 2p+2 (x, y) = (x, y). 26   Finally, p = 0 that is when A2 = O0 , by iterating the function f (x, y) = α1 x, xy we find that f 2n (x, y) = (α12n x, α1n y) and f 2n+1 (x, y) = (α12n+1 x, α1n xy ). Now observe that for α1n =   1 we have f 2n (x, y) = (x, y) and f 2n+1 (x, y) = x, xy . Therefore f is 2n−periodic. Now for α1n 6= 1 through simple calculations we find that f preserves the announced fibrations V1 (x, y) and V2 (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] Bedford, E. and Kim, K. 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