This is a preprint of: “Meromorphic first integrals of analytic diffeomorphisms”, Antoni Ferragut, Armengol Gasull, Xiang Zhang, J. Math. Anal. Appl., vol. 519, 126796, 2023. DOI: [10.1016/j.jmaa.2022.126796] MEROMORPHIC FIRST INTEGRALS OF ANALYTIC DIFFEOMORPHISMS ANTONI FERRAGUT1 , ARMENGOL GASULL2 , XIANG ZHANG3 Abstract. We give an upper bound for the number of functionally independent mero- morphic first integrals that a discrete dynamical system generated by an analytic map f can have in a neighborhood of one of its fixed points. This bound is obtained in terms of the resonances among the eigenvalues of the differential of f at this point. Our approach is inspired on similar Poincaré type results for ordinary differential equations. We also apply our results to several examples, some of them motivated by the study of several difference equations. 1. Introduction and statement of the main results One of the first steps to study the dynamics of a discrete dynamical system (DDS) is to determine the number m of functionally independent first integrals that it has. It is clear that each new first integral reduces the region where any orbit can lie, so the bigger is m, the simpler will be the dynamics. For instance, if this DDS is n-dimensional, this number m is at most n, and in the case m = n the DDS is called integrable and it has extremely simple dynamics: in most cases it is globally periodic, that is, there exists p ∈ N, such that f p = Id, where f is the invertible map that generates it, see [4]. Similarly, DDS having m = n − 1 are such that all their orbits lie in one-dimensional manifolds, see some examples in [5]. The aim of this paper is to give an upper bound of the number of meromorphic first integrals that a DDS generated by an invertible analytic map can have in a neighborhood of a fixed point. We follow the approach of Poincaré for studying the same problem for continuous dynamical systems given by analytic ordinary differential equations. It is based on the study of the resonances among the eigenvalues of the differential of the vector field at one of its critical points, see for instance [9] and their references. We will use similar tools that the ones introduced in that paper. Consider analytic diffeomorphisms in (Cn , 0), a neighborhood of the origin, (1) y = f (x), x ∈ (Cn , 0), with f (0) = 0. A function R(x) = G(x)/H(x) with G and H analytic functions in (Cn , 0) is a meromorphic first integral of the diffeomorphism (1) if G(f (x))H(x) = G(x)H(f (x)), for all x ∈ (Cn , 0). Notice that the above condition implies that R(f (x)) = R(x), for all x ∈ (Cn , 0) where both functions are well defined. Specially if G and H are polynomial functions, then R(x) is a rational first integral of (1). If H is a non–zero constant, then R(x) is an analytic first integral of (1). So meromorphic first integrals include rational and analytic first integrals as particular cases. Denote by A = Df (0) the Jacobian matrix of f (x) at x = 0. Let µ = (µ1 , . . . , µn ) be the n–tuple of eigenvalues of A. Notice that since f is a diffeomorphism at 0, we have µ1 µ2 · · · µn 6= 0. 2010 Mathematics Subject Classification. Primary: 37C05. Secondary: 37C25; 37C80. Key words and phrases. Discrete dynamical system, Integrability, Meromorphic first integrals. 1 2 A. FERRAGUT, A. GASULL, X. ZHANG We say that the eigenvalues µ satisfy a resonant condition if µk = 1, for some k ∈ Zn with kkk = 6 0, where Z is the set of integers, and kkk = |k1 | + · · · + |kn |. The aim of this paper is to prove the next result and give some applications. Theorem 1. Assume that the analytic diffeomorphism (1) satisfies f (0) = 0 and let µ = (µ1 , . . . , µn ) be the eigenvalues of Df (0). Then the number of functionally independent generalized rational first integrals of the analytic diffeomorphism (1) in (Cn , 0) is at most the dimension of the Z–linear space generated from {k ∈ Zn : µk = 1}. The above theorem also extends some results proved for n = 2 in [7]. In section 2 we prove Theorem 1. Section 3 is devoted to applications of Theorem 1. 2. Proof of Theorem 1 For an analytic or a polynomial function R(x) in (Cn , 0), we denote by R0 (x) its ho- mogeneous term of the lowest degree. For a rational or a meromorphic function R(x) = G(x)/H(x) in (Cn , 0), we denote by R0 (x) the rational function G0 (x)/H 0 (x). We expand the analytic functions G(x) and H(x) as ∞ X ∞ X G0 (x) + Gi (x) and H 0 (x) + H i (x), i=1 i=1 where Gi (x)and H i (x) are homogeneous polynomials of degrees deg G0 (x)+i and deg H 0 (x)+ i, respectively. Then we have ∞ ! ∞ !−1 G(x) G0 (x) X Gi (x) X H i (x) R(x) = = + 1+ H(x) H 0 (x) H 0 (x) H 0 (x) i=1 i=1 ∞ X G0 (x) Ai (x) (2) = + , H 0 (x) B i (x) i=1 where Ai (x) and B i (x) are homogeneous polynomials. Clearly, deg G0 (x) − deg H 0 (x) < deg Ai (x) − deg B i (x) for all i ≥ 1. In what follows we will say that − deg Ai (x) deg B i (x) is the degree of Ai (x)/B i (x), and 0 0 G (x)/H (x) is the lowest degree term of R(x) in the expansion (2). For simplicity we denote d(G) = deg G0 (x), d(R) = d(G) − d(H) = deg G0 (x) − deg H 0 (x), and call d(R) the lowest degree of R. The following known result, first proved by Ziglin [21] in 1983, see also [1, 9, 12, 20], will be used in the proof of Theorem 1. Lemma 2. Let G1 (x) Gm (x) R1 (x) = , . . . , Rm (x) = , H1 (x) Hm (x) be functionally independent meromorphic functions in (Cn , 0). Then there exist polynomials Pi (z1 , . . . , zi ) for i = 2, . . . , m such that R1 (x), R e2 (x) = P2 (R1 (x), R2 (x)), . . . , R em (x) = Pm (R1 (x), . . . , Rm (x)) are functionally independent meromorphic functions, and that R10 (x), e0 (x), . . . , R R em0 (x) are functionally independent rational functions. 2 Next we give some properties that meromorphic first integrals of diffeomorphism (1) must have. A rational monomial is by definition the ratio of two monomials, i.e. of the form xp /xq with p, q ∈ (N0 )n , where N0 = N ∪ {0} and N is the set of positive integers. The rational monomial xp /xq is resonant if µp−q = 1. A rational function is homogeneous if its denominator and numerator are both homogeneous polynomials. A rational homogeneous MEROMORPHIC FIRST INTEGRALS OF DIFFEOMORPHISMS 3 function is resonant if the ratio of any two elements in the set of all its monomials in both denominator and numerator is a resonant rational monomial. For the diffeomorphism (1) we assume without loss of generality that A = Df (0) is in its Jordan normal form and is a lower triangular matrix. Lemma 3. Let R(x) = G(x)/H(x) be a meromorphic first integral of the analytic diffe- omorphism (1). By changing R by R − a, for some suitable constant a ∈ C, if needed, it is not restrictive to assume that R0 (x) = G0 (x)/H 0 (x) is non–constant. Moreover R0 is a resonant rational homogeneous first integral of the linear part of f (x). For proving this last lemma we will use the following result, for a proof see any of the references [3, 13, 19, 20]. Lemma 4. Let Hnp be the linear space of complex coefficient homogeneous polynomials of degree p in n variables. For any constant c ∈ C, define a linear operator on Hnp by Lc (h)(x) = h(Ax) − c h(x), h(x) ∈ Hnp . Then the spectrum of Lc is {µk − c : k ∈ (N0 )n , |k| = k1 + . . . + kn = p}, where µ are the eigenvalues of A. Proof of Lemma 3. If R0 (x) = G0 (x)/H 0 (x) ≡ a is constant, the function R(x) − a is also a meromorphic first integral and (R(x)−a)0 is not constant. Hence, without loss of generality, we can assume that R is a meromorphic first integral and that R0 is not identically constant. Let us prove that R0 is a resonant rational homogeneous first integral. As in (2) we write R(x) as ∞ X 0 R(x) = R (x) + Ri (x), i=1 where R0 (x) is the lowest order rational homogeneous function and Ri (x) for i ∈ N are rational homogeneous functions of order larger than R0 (x). Since R(x) is a first integral of the diffeomorphism f (x) in a neighborhood of 0 ∈ Cn , we have R(f (x)) = R(x), x ∈ (Cn , 0). Equating the lowest order rational homogeneous functions we get G0 (Ax) G0 (x) (3) R0 (Ax) = R0 (x), i.e. = . H 0 (Ax) H 0 (x) This implies that R0 (x) is a rational homogeneous first integral of the linear part of the analytic diffeomorphism (1). Next we shall prove that R0 (x) is resonant. From the equality (3) we can assume without loss of generality that G0 (x) and H 0 (x) are relative prime. Now equation (3) can be written as H 0 (x)G0 (Ax) = G0 (x)H 0 (Ax). Since G0 and H 0 are relatively prime, and C[x] is a unique factorization domain (see e.g. [11, p. 2]), we get that G0 (x) divides G0 (Ax) and H 0 (x) divides H 0 (Ax). In addition, G0 (Ax) and G0 (x) have the same degree, so there exists a constant c such that G0 (Ax) − c G0 (x) ≡ 0, H 0 (Ax) − c H 0 (x) ≡ 0. We remark that if G0 (x) ≡ 1 or H 0 (x) ≡ 1, we have c = 1. Set deg G0 (x) = l, deg H 0 (x) = m and Lc the linear operator defined in Lemma 4. Recall from Lemma 4 that Lc has respectively the spectrums on Hnp Sp := {µp − c : p ∈ (Z+ )n , |p| = p}, and on Hnq Sq := {µq − c : q ∈ (Z+ )n , |q| = q}. 4 A. FERRAGUT, A. GASULL, X. ZHANG Separate Hnp = Hn,1 p p + Hn,2 in such a way that for any p(x) ∈ Hn,1 l its monomial xp p satisfies µp − c = 0, and for any Q(x) ∈ Hn,2 its monomial xp satisfies µp − c 6= 0. Separate p p G0 (x) in two parts G0 (x) = G01 (x) + G02 (x) with G01 ∈ Hn,1 and G02 ∈ Hn,2 . Since A is in its Jordan normal form and is lower triangular, it follows that p p p p Lc Hn,1 ⊂ Hn,1 , and Lc Hn,2 ⊂ Hn,2 . Hence Lc G0 (x) ≡ 0 is equivalent to Lc G01 (x) ≡ 0 and Lc G02 (x) ≡ 0. p p Since Lc has the spectrum without zero element on Hn,2 and so it is invertible on Hn,2 , 0 0 the equation Lc G2 (x) ≡ 0 has only the trivial solution, i.e. G2 (x) ≡ 0. This proves that G0 (x) = G01 (x), i.e. each monomial, say xp , of G0 (x) satisfies µp − c = 0. Similarly we can prove that each monomial, say xq , of H 0 (x) satisfies µq − c = 0. This implies that µp−q = 1. The above proofs show that R0 (x) = G0 (x)/H 0 (x) is a resonant rational homogeneous first integral of the linear part of f (x).  Having the above lemmas we can prove Theorem 1. Proof of Theorem 1. Let G1 (x) Gm (x) R1 (x) = , . . . , Rm (x) = , H1 (x) Hm (x) be the m functionally independent meromorphic first integrals of the diffeomorphism f . Since the polynomial functions of Ri (x) for i = 1, . . . , m are also meromorphic first integrals of the diffeomorphsim f , by Lemma 2 we can assume without loss of generality that G01 (x) G0m (x) R10 (x) = , . . . , R 0 m (x) = , H10 (x) Hm0 (x) are functionally independent. Lemma 3 shows that R10 (x), . . . , Rm 0 (x) are resonant rational homogeneous first integrals of the linear part Ax of f (x). So these first integrals can be written as rational functions in the variables given by resonant rational monomials. Write A = AS + AN with AS = diag(µ1 , . . . , µn ) and AN nilpotent. Then direct calculations show that any resonant rational monomial is a first integral of AS x. For instance, let xk be a resonant rational monomial, then µk = 1. Hence, we have (AS x)k = µk xk = xk . This implies that R10 (x), . . . , Rm 0 (x) are also first integrals of AS x. We claim that m is less than or equal to the number of elements in a basis of the Z–linear space generated from {k ∈ Zn : µk = 1}. We denote by γ this last number. 0 (x) are the first integrals of A x, we only need to prove Indeed, since R10 (x), . . . , Rm S that the number of functionally independent resonant rational homogeneous first integrals of AS x is equal to γ. In fact, from Lemma 3 and its proof it follows that any resonant rational homogeneous first integral consists of resonant rational monomials. This means that the maximum number of functionally independent resonant rational homogeneous first integrals is equal to the maximum number of functionally independent resonant rational monomials. Whereas, by definition a resonant rational monomial, saying xk for k ∈ Zn , satisfies µk = 1. This proves the claim. After the claim, the proof of the theorem is completed.  3. Applications 3.1. Some simple examples. We start studying the integrability of two simple examples coming from second order difference equations. Recall that the study of the sequences generated by second order difference equations xn+2 = g(xn , xn+1 ) can be reduced to the study of the DDS generated by the planar map f (x, y) = (y, g(x, y)). Moreover, the first integrals H for the DDS are usually called invariants for the difference equation. Then H(xn , xn+1 ) = H(xn+1 , xx+2 ) for all n ∈ N. MEROMORPHIC FIRST INTEGRALS OF DIFFEOMORPHISMS 5 As a first example we study the integrability of the planar map (4) f (x, y) = (y, −bx + c/y), b, c ∈ C, b 6= 0. coming from the difference equation xn+2 = −bxn + c/xn+1 . We prove the following propo- sition. Proposition 5. If the planar map (4) has a meromorphic first integral then b is a root of the unity. In particular, if b, c ∈ R and the map has a meromorphic first integral then b ∈ {−1, 1}. Somehow the result for the real case is sharp because when b = 1 the function H(x, y) = x2 y 2 − cxy is a first integral of (4). Moreover when b = −1 and c = 0, clearly the map is linear and integrable. The first integral for the case b = 1 (in fact for its inverse) and also for many other rational maps are given in [17]. For the complex case, observe that when c = 0 and b is a root of the unity the map is linear and globally periodic, i.e. f n = Id for some n ∈ N. For instance, by using the results of [4] it is easy to construct two functionally independent rational first integrals for each one of them. Proof of Proposition √ 5. When c = 0 the map has (0, 0) as a fixed point and the eigenvalues of Df (0, 0) are ± b. Then, by Theorem√ 1, to√have a meromorphic first integral there must exist (m, n) ∈ Z \ (0, 0) such that ( b) (− b)m = 1 and, as a consequence b must be a 2 n root of the unity. From now on we can assume that c 6= 0. Then, the fixed points of f are (z, z) where (b + 1)z 2 = c. If b = −1 we are done because b2 = 1. Otherwise, the map has two (real or complex) fixed points. Anyhow, for c(b + 1) 6= 0,     0 1 0 1 Df (z, z) = = . −b −c/z 2 −b −(b + 1) Hence, its eigenvalues satisfy p(µ) = µ2 + (b + 1)µ + b = 0. Since b 6= 0 we can write them as µ and b/µ. By Theorem 1, if f is meromorphically integrable, there exists (m, n) ∈ Z2 \(0, 0) such that µn (b/µ)m = 1. If m = n, then bm = 1 and we are done, again. Hence, from now on, m − n 6= 0. Moreover, µ = bm/(m−n) . Thus 2m m p(µ) = µ2 + (b + 1)µ + b = b m−n + (b + 1)b m−n + b = 0. If we introduce a such that b = am−n , the above equality writes as   a2m + am−n + 1 am + am−n = am−n am+n + am + an + 1 = am−n (am + 1)(an + 1) = 0. Hence a must be a root of the unity, and as a consequence b = am−n also must, as we wanted to prove.  As a second example, next lemma studies conditions for a planar map to have two functionally independent meromorphic first integrals. Lemma 6. Let f be a real analytic planar map with a fixed point x ∈ R2 such that the characteristic polynomial of Df (x) is p(µ) = µ2 + bµ + c ∈ R[µ]. Assume that f has two functionally independent meromorphic first integrals. Then, the eigenvalues of Df (x) are roots of the unity. In particular, if b2 − 4c < 0 then c = 1 and otherwise (b, c) ∈ {(±2, 1), (0, −1)}. Proof. Let u, v ∈ C be the two eigenvalues of Df (x). By Theorem 1, there exist (n, m) and (n0 , m0 ) in Z2 such that 0 0 un v m = 1, un v m = 1, and mn0 − m0 n 6= 0. 0 0 0 0 Hence unm −n m = 1 and v mn −m n = 1. Therefore, u and v are roots of the unity. When u and v are complex we are done, because v = ū and c = uū = |u|2 = 1. When u, v ∈ R then u and v are either 1 or −1 and the lemma follows.  6 A. FERRAGUT, A. GASULL, X. ZHANG We apply the above lemma to study the map (5) f (x, y) = (y, xp y q ), p, q ∈ Z, p 6= 0, that describes the difference equation xn+2 = xpn xqn+1 . It has the fixed point (1, 1) for all values of p and q, and provides a good test for our result because it can also be studied by another approach. It can be linearized on {(x, y) ∈ R2 : x > 0, y > 0}, because with the new variables (u, v) = (ln x, ln y) the DDS generated by f is conjugated to the DDS generated by g(u, v) = (v, pu + qv). As we have already commented, we will center our attention in finding the values of p and q such that the DDS generated by f can have two functionally independent first integrals. Notice that f (1, 1) = (1, 1) and     0 1 0 1 Df (1, 1) = = . pxp−1 y q qxp y q−1 (x,y)=(1,1) p q Hence the eigenvalues µ of Df (1, 1) satisfy µ2 − qµ − p = 0. We can use Lemma 6: since (p, q) ∈ Z2 , the only cases for which the map (5) can have two functionally independent first integrals are (p, q) ∈ {(−1, −2), (−1, 2), (1, 0)} when q 2 + 4p ≥ 0, and p = −1 when q 2 + 4p < 0. Moreover, in this last case all roots of P (µ) = µ2 − qµ + 1 must be roots of the unity. This only happens when q ∈ {−1, 0, 1}, because these values are the only ones for which P is a quadratic cyclotomic polynomial. Recall that the p-th cyclotomic polynomial Φp is the monic polynomial with integer coefficients, and irreducible in Q(x), such that its roots are the primitive p-roots of the unity, that is e2nπi/p , for p and n relatively prime. Recall also that only the 3rd , 4th and 6th roots of the unity have quadratic cyclotomic polynomials. Hence, from our tools, we have seen that there are six maps of the form (5) that are candidates to have two functionally independent first integrals:  1   y2   f1 (x, y) = y, 2 , f2 (x, y) = y, , f3 (x, y) = y, x , xy x  1   1  y f4 (x, y) = y, , f5 (x, y) = y, , f6 (x, y) = y, . xy x x From these candidates it is not difficult to see, by using the tools of [4], that fj (x, y) j = 3, 4, 5, 6 have the desired property. For instance, two functionally independent first integrals for f6 are 1 1 x y 1 x2 y x y2 H1 (x, y) = x + + y + + + , H2 (x, y) = xy + + + 2+ 2+ . x y y x xy y x y x In fact, these four maps are globally periodic, with periods 2, 3, 4 and 6, respectively. 3.2. Planar rational maps with rational coefficients. We consider now rational maps   P1 (x, y) P2 (x, y) (6) f (x, y) = (R1 (x, y), R2 (x, y)) = , , Q1 (x, y) Q2 (x, y) with coefficients in Q, that is, with Pi , Qi ∈ Q[x, y] for i = 1, 2. We introduce some notation. As usual, when two integer numbers p and n are coprime we will write (p, n) = 1. Moreover we will denote the degree of the p-th cyclotomic polynomial Φp by φ(p). In fact, φ(p) is the number of integers between 1 and p that are relatively prime to p. Given two polynomials P, Q ∈ C[x], as usual, we denote by Resx (P (x), Q(x)) ∈ C the resultant of P and Q with respect to x. Recall that P and Q share some complex root if and only if Resx (P (x), Q(x)) = 0, see [16]. We will also need the following result, which is essentially contained in ([2, 18]). Proposition 7. For each p ∈ N there is a polynomial Mm ∈ Q[x] of minimal degree m = φ(p)/2 such that Mm (cos(2nπ/p)) = 0 for all (p, n) = 1. Moreover,  (7) Resx Φp (x), x2 − 2xv + 1 = Mm 2 (v). MEROMORPHIC FIRST INTEGRALS OF DIFFEOMORPHISMS 7 In particular, m = 1 if and only if p ∈ {1, 2, 3, 4, 6}; m = 2 if and only if p ∈ {5, 8, 10, 12}; m = 3 if and only if p ∈ {7, 9, 14, 18}; and m = (p − 1)/2 when p > 2 is prime. We only make some comments about (7). Notice that if x = cos(2nπ/p) + i sin(2nπ/p) then Φp (x) = 0. Moreover, if we define v = cos(2nπ/p) it holds that v = (x + x̄)/2 = (x + 1/x)/2, or equivalently (x2 −2xv+1)/x = 0. Hence the polynomial Resx Φp (x), x2 −2xv+1 has v as one of its roots. Moreover, if we start with x̄ instead of x we obtain again the same value v as a root of this last polynomial, making that all its roots are double. Notice that, as a consequence of the above result, the only rational values of cos(2π/p) are (8) cos(2π) = 1, cos(2π/2) = −1, cos(2π/3) = −1/2, cos(2π/4) = 0, cos(2π/6) = 1/2, √ √ and the only values where cos(2π/p) is in Q[ q] for q ∈ Q, but q 6∈ Q are √ √ √ (9) cos(2π/5) = ( 5 − 1)/4, cos(2π/10) = ( 5 + 1)/4, cos(2π/12) = 3/2. For instance, the minimal polynomial for p = 5 is M2 (x) = 4x2 + 2x − 1; the cases where Mm is cubic are, 8x3 + 4x2 − 4x − 1 when p = 7; 8x3 − 6x + 1 when p = 9; 8x3 − 4x2 − 4x + 1 when p = 14; and 8x3 − 6x − 1 when p = 18. Next result provides some computable conditions to know whether a planar rational map with rational coefficients has a meromorphic first integral. In particular, notice that for many 1-parametric families of maps it allows to prove that this first integral can only exist for finitely many values of the parameter, see for instance Lemma 11. Theorem 8. Consider the rational map (6) with rational coefficients. Assume that f has a x, yb) ∈ R2 and that Df (b real fixed point (b x, yb) has complex conjugated eigenvalues µ, µ̄ with modulus different from 1. If the map (6) has a meromorphic first integral, then µ/µ̄ is a  there exists a computable polynomial Uk ∈ Q[x], of degree k ∈ N, such root of the unity and that Uk Re(µ/µ̄) = 0. Moreover, some of the values  (10) Resx Uk (x), Vp (x) , with p such that deg(Vp (x)) ≤ k, must vanish, where Vp is the minimal polynomial of Re(µ/µ̄) = cos(2nπ/p), with (p, n) = 1, see Proposition 7. Proof. We introduce the polynomials S1 , S2 , T1 , T2 , D1 , D2 associated to f, S1 (x, y) = P1 (x, y) − xQ1 (x, y), S2 (x, y) = P2 (x, y) − yQ2 (x, y), T1 (x, y) ∂R1 (x, y) ∂R2 (x, y) T (x, y) = = + , T2 (x, y) ∂x ∂y D1 (x, y) ∂R1 (x, y) ∂R2 (x, y) ∂R1 (x, y) ∂R2 (x, y) D(x, y) = = − . D2 (x, y) ∂x ∂x ∂y ∂x Notice that if (b x, yb) is a fixed point of f then it is a solution of the system {S1 (x, y) = 0, S2 (x, y) = 0}, and moreover Q1 (b x, yb) 6= 0. Observe also that the eigenvalues x, yb)Q2 (b µ = µ(b x, yb) of Df (b x, yb) satisfy (11) P (µ) = µ2 − T (b x, yb) = µ2 − T µ + D = 0, x, yb)µ + D(b where, when there is no confusion, we omit the dependence of T, D and µ on the fixed point. With this notation, the condition of having complex eigenvalues µ and µ̄ with modulus different from one reads simply as T 2 − 4D < 0 and D 6= 1, because they satisfy (11), µµ̄ = D, and hence |µ|2 = D2 . By Theorem 1, if f has a meromorphic first integral, then there exists (0, 0) 6= (p, q) ∈ Z2 , such that µp µ̄q = 1. Taking norms, this means that |µ|p+q = 1. Since by hypothesis |µ| = 6 1, we must have q = −p. Hence, |µ/µ̄| = 1 and µ/µ̄ is a p-root of the unity. Therefore Re(µ/µ̄) = cos(2nπ/p) for some n ∈ {0, 1, . . . , p − 1} with n and p coprime. 8 A. FERRAGUT, A. GASULL, X. ZHANG Let us write r r T 4D − T 2 T 4D − T 2 µ= +i = α + iβ, µ̄ = − i = α − iβ. 2 4 2 4 Then µ α + iβ α2 − β 2 2αβ = = 2 2 +i 2 µ̄ α − iβ α +β α + β2 and, as a consequence,   α2 − β 2 T2 Re = − 1 = cos (2nπ/p) . α2 + β 2 2D If we name v = cos (2nπ/p) , then the following system of three equations is satisfied: ( S1 (x, y) = 0, S2 (x, y) = 0, (12) W (x, y, v) = T12 (x, y)D2 (x, y) − 2(1 + v)T22 (x, y)D1 (x, y) = 0. As usual, taking successive resultants, as follows   T1 (y) = Resx S1 (x, y), S2 (x, y) , T2 (y, v) = Resx S1 (x, y), W (x, y, v) ,  Uk (v) = Resy T1 (y), T2 (y, v) , where k is de degree of Uk , we know that Uk (v) = 0 and Uk ∈ Q[x], as we wanted to prove, see again [16]. Hence, by Proposition 7, v has to be a root of some Vp , with deg(Vp (x)) ≤ k. By the properties of the resultant the last statement of the theorem follows.  Remark 9. Notice that the conditions in Theorem 8 depend on the value k given in its statement and obtained solving system (12). According to the fixed point (b x, yb) taken into account, some smaller k can be considered. To do this, let us treat this system in another way. We consider first the following two polynomials with rational coefficients,   T1 (y) = Resx S1 (x, y), S2 (x, y) , T3 (x) = Resy S1 (x, y), S2 (x, y) . Let V1 and V3 be the irreducible factors in Q[x] of T1 and T3 , respectively, such that T1 (b y) = 0 and T3 (b x) = 0. Then we can follow the same procedure that for system (12), but starting with the system (13) V1 (y) = 0, V3 (x) = 0, W (x, y, v) = 0. We arrive at a new Uk0 ∈ Q[x], of degree k 0 ≤ k and such that Uk0 (v) = 0. Then the set of conditions (10) has to be satisfied only until k 0 , taking Uk0 instead of Uk . Corollary 10. Consider the rational map (6) with rational coefficients. Assume that f has x, yb) ∈ Q2 and Df (b a fixed point (b x, yb) has complex conjugated eigenvalues, µ, µ̄, solution of µ2 − T (b x, yb)µ + D(b x, yb) = 0, and with modulus different of 1. If the map (6) has a meromorphic first integral, then T 2 (b x, yb) ∈ {0, 1, 2, 3, 4} . D(b x, yb) Proof. By using Theorem 8, Remark 9 and that (b x, yb) ∈ Q2 , we know that system (13) writes as V1 (y) = y − yb, V3 (x) = x − x b, W (x, y, v) = 0. Hence, T 2 (b x, yb) v= −1∈Q 2D(b x, yb) and k = 1 in Theorem 8. Therefore, the only possible values of v = cos(2nπ/p) are the ones that are rational. From (8) we get that   T 2 (b x, yb) 1 1 − 1 ∈ −1, − , 0, , 1 2D(b x, yb) 2 2 and the corollary follows.  MEROMORPHIC FIRST INTEGRALS OF DIFFEOMORPHISMS 9 In the next lemma we apply the above corollary to a simple example. Lemma 11. The rational map   (a + (2 − a)x)y f (x, y) = xy, , 1 + xy with a ∈ Q, a > 9/8, can have meromorhic first integrals only when a ∈ {3/2, 2, 9/4, 9/2}. Proof. It has a fixed point at (1, 1); the characteristic √ polynomial of Df (1, 1) is p(µ) = µ2 − 3/2µ + a/2 and their eigenvalues (3 ± i 8a − 9)/4 are complex. Moreover, when a 6= 2 they have modulus different from 1. Notice that T 2 (b x, yb)/D(b x, yb) = 9/(2a). Hence, by Corollary 10, when a 6= 2, the conditions for existence of meromorphic first integral are 9 ∈ {0, 1, 2, 3, 4}, 2a giving the result of the statement.  Although we have not found any meromorphic first integral for the four possible cases given in the statement it is worth to mention that when a = 1 then f has the meromorphic (polynomial) first integral H(x, y) = y(1 + x), see also [14]. In fact, let us see which conditions for existence of meromorphic first integral are consequence of Theorem 1 when a < 9/8. In this case, let r ∈ R be the non-negative solution of r2 = 9 − 8a. Then the x, yb) are µ1 = (3 − r)/4 and µ2 = (3 + r)/4. Hence, the condition for the eigenvalues at (b 2 = 1, for some non-zero (n, m) ∈ Z . existence of a meromorphic first integral is that µn1 µm 2 Equivalently, the only cases that might have a meromorphic first integral are either r = 1 or  3 − r .  3 + r  log log ∈ Q. 4 4 Notice that r = 1 precisely corresponds to the integrable case given above a = 1. Next result is similar to Corollary 10. The only difference is that each one of the coor- dinates of the fixed point is a zero of a quadratic polynomial with integer coefficients. Its proof is essentially the same, but instead of using (8) we will use (9) and, consequently, that the only values of cos(2nπ/p), that are solutions of a quadratic polynomial in Q[x], irreducible in Q[x], are √ √ √ √ −1 ± 5 2 1± 5 3 ,± , ,± . 4 2 4 2 We skip the details of the proof. Corollary 12. Consider the rational map (6) with √ rational coefficients. Assume that f has x, yb) with both coordinates in Q[ s] \ Q, s ∈ Q, and Df (b a fixed point (b x, yb) has complex conjugated eigenvalues µ, µ̄ solution of µ2 − T (b x, yb)µ + D(b x, yb) = 0, and with modulus different from 1. If the map (6) has a meromorphic first integral, then ( √ √ ) T 2 (b x, yb) 3± 5 √ 5± 5 √ ∈ , 2 ± 2, ,2 ± 3 . D(b x, yb) 2 2 Similar results to Corollaries 10 and 12 could be stated assuming that the rational map f given in (6) has a fixed point (bx, yb) with both coordinates being a zero of a polynomial of higher degree and the same conditions for the eigenvalues of Df (b x, yb) hold. As an example we apply our techniques to prove the non-meromorphic integrability of a concrete rational map. Consider   2 x2 + xy + 1 (14) f (x, y) = x + y − xy, 2 , x − 3y + 1 x, yb) = (s, s), where s ≈ 4.836 is the real root of P (x) = x3 −5x2 +x−1. and its fixed point (b 10 A. FERRAGUT, A. GASULL, X. ZHANG It is easy to see that we are under the hypotheses of Theorem 8, because T 2 (s, s) − 4D(s, s) < 0 and D(s, s) 6= 1. For the sake of shortness, we omit the explicit expressions of T (x, y) and D(x, y). Moreover, in the notation of this theorem, S1 (x, y) = y(y − x), S2 (x, y) = −x2 y + x2 + xy + 3y 2 − y + 1 and W (x, y, v) is a polynomial of degree 10 that we do not explicit either. Then,  T1 (y) = Resx S1 (x, y), S2 (x, y) = −y 2 P (y),  T3 (x) = Resy S1 (x, y), S2 (x, y) = −(x2 + 1)P (x). Therefore, in the notation of Remark 9, V1 (y) = P (y) and V3 (x) = P (x). Finally,  Resx Resy (W (x, y, v), V1 (y)), V3 (x) = U3 (v)Z6 (v), where U3 , Z6 ∈ Q(x), U3 (x) = 5833x3 +16607x2 +15650x+4874 and U3 (Re(µ/µ̄)) = 0, with µ and µ̄ the eigenvalues of Df (s, s). Since U3 has degree 3, to prove that system (14) has no meromorphic first integral it suffices to prove that all the resultants Resx (U3 (x), Q(x)), Q ∈ Q, do not vanish, where Q = {x, x − 1, x + 1, 2x + 1, 2x − 1, 2x2 − 1, 4x2 − 3, 4x2 + 2x − 1, 4x2 − 2x − 1, 8x3 + 4x2 − 4x − 1, 8x3 − 6x + 1, 8x3 − 4x2 − 4x + 1, 8x3 − 6x − 1}. The above set of polynomials corresponds to the only irreducible ones of degree at most 3 that have a root cos(2nπ/p) for some n, p ∈ Z, see the comments after Proposition 7. These thirteen resultants are all different from zero, and the result follows. 3.3. Higher dimensional examples. As in the two dimensional case, the study of the sequences generated by n-th order difference equations xk+n = g(xk , xk+1 , . . . , xk+n−2 , xk+n−1 ) can be reduced to the study of the DDS generated by the map f (x1 , x2 , . . . , xn ) = (x2 , x3 , . . . , xn−1 , g(x1 , x2 , . . . , xn )), from Cn into itself. Some of the maps that we will consider here have this shape. We start studying the case of general analytic maps having the maximum possible number of functionally independent meromorphic first integrals: n. Proposition 13. Let f be an analytic map from Cn into itself, with isolated fixed points, and with n functionally independent meromorphic first integrals. If q is a fixed point of f, then all the eigenvalues µ, of Df (q) are p-roots of the unity. Moreover, there is a constructive procedure, described in the proof, to find a polynomial Pk ∈ C[x] of degree k such that, for all µ, Pk (µ) = 0. Additionally, if f is a rational map and all the numerators and denominators of its components are polynomials in Q[x1 , . . . , xn ], then Pk ∈ Q[x] and the eigenvalues of Df (q) are p-roots of the unity with p ≤ M , where M is the maximum m ∈ N such that the degree of the cyclotomic polynomial Φm is k. Furthermore, Resx (Φj (x), Pk (x)) = 0 for some cyclotomic polynomial Φj with deg(Φj ) ≤ k. Proof. By Theorem 1, the dimension of the Z–linear space generated from {k ∈ Zn : µk = 1} is n. Let ki , i = 1, 2, . . . , n be a basis of this space. Then, det(K) 6= 0, where K is the n×n | det(K)| matrix K = (ki,j ) and, moreover it can be seen that µ` = 1, for all ` = 1, 2, . . . , n, proving that p = | det(K)| ∈ N. Set x = (x1 , x2 , . . . , xn ). Then, the eigenvalues of Df at a fixed point are the soluti- ons x, µ of the system of n + 1 equations, f (x) = x, P (x, µ) = 0, where P (x, µ) is the characteristic polynomial of Df (x) at an arbitrary point x, not necessarily fixed. Instead of considering them, we take their numerators as a polynomial system of n + 1 equations and n + 1 unknowns, (15) g(x) = Num(f (x) − x) = 0, R(x, µ) = Num(P (x, µ)) = 0, MEROMORPHIC FIRST INTEGRALS OF DIFFEOMORPHISMS 11 where Num denotes the numerator of a quotient of polynomials. Doing successive resultants following [16], as in the proof of Theorem 8, and because f has no continuum of fixed points, we arrive at a non-zero polynomial Pk ∈ C[x], such that Pk (µ) = 0 for all eigenvalues µ of Df (x) at any x, fixed point of f, as we wanted to prove. Moreover, if f is rational, with numerator and denominator in Q[x], then Pk ∈ Q[x]. Finally, it is known that given a primitive p-th root of the unity, µ, the minimal degree of a polynomial S ∈ Q[x] such that S(µ) = 0 is φ(p) = deg(Φp (x)). Hence, p ≤ M, as we wanted to prove. Moreover Pk must share some root with one of the polynomials Φj with φ(j) ≤ k, and as a consequence, for this value of j, Resx (Φj (x), Pk (x)) = 0.  As an example of application we prove the following lemma. Lemma 14. Consider the map   a+y+z (16) f (x, y, z) = y, z, , a ∈ Q. x If it has 3 functionally independent meromorphic first integrals, then a ∈ {−1, 1}. Proof. We will apply Proposition 13. We start constructing a reduced, but equivalent, version of system (15). Notice that the fixed points of f are (x, x, x) such that a + 2x = x2 and     0 1 0 0 1 0 Df (x, x, x) =  0 0 1  =  0 0 1 . − a+y+z x2 1 x 1 x (x,y,z)=(x,x,x) −1 x1 x1 Hence, the characteristic polynomial at any fixed point is (µ + 1)(µ2 − (1 + 1/x)µ + 1). Therefore, since µ = −1 is a root of the unity, system (15) with 4 unknowns can be simply reduced to a system of 2 equations and 2 unknowns, x and µ, and a parameter a: −x2 + 2x + a = 0, (µ2 − µ + 1)x − µ = 0. Hence P4 (µ) = Resx (−x2 +2x+a, (µ2 −µ+1)x−µ) = aµ4 −2(a−1)µ3 +3(a−1)µ2 −2(a−1)µ+a = 0. Therefore, we only have to consider the values of p for which the cyclotomic polynomial Φp has degree at most 4. They are p ∈ P = {1, 2, 3, 4, 5, 6, 8, 10, 12}. By doing all the resultants Resx (P4 , Φp (x)), with p ∈ P, we can discard all the values a ∈ Q but a = −1, 7/9, 5/4, 3, 1, that correspond to p = 1, 2, 3, 4, 8, respectively. For instance, we get that Resx (P4 , Φ3 (x)) = (4a − 5)2 and Resx (P4 , Φ8 (x)) = (a − 1)2 , but Resx (P4 , Φ10 (x)) = (a2 − a − 1)2 . Finally, the values a ∈ {7/9, 5/4, 3} are also discarded because, for them, some of the roots of P4 are not roots of the unity. For instance, P4 (µ)|a=3 = (µ2 + µ + 1)(5µ2 − 7µ + 5)/4. On the other hand, P4 (µ)|a=−1 = −(µ − 1)4 and P4 (µ)|a=1 = µ4 + 1, and all the roots of both polynomials are roots of the unity.  It is worth to comment that for a = 1 it is well known that the map (16) has effectively 3 functionally independent rational first integrals. Two of them exist for any a ∈ C. They are (x + 1)(y + 1)(z + 1)(a + x + y + z) H1 (x, y, z) = , xyz (1 + x + y)(1 + y + z)(a + x + y + z + xz) H2 (x, y, z) = , xyz see for instance [6] and its references, and a third one can be seen in [4] and it is found by using the tools introduced in that paper. It exists because this map, for a = 1, corresponds to the celebrated 3rd order Todd’s difference equation xn+3 = (1 + xn+2 + xn )/xn which is globally 8-periodic, that is xn+8 = xn for all n, whenever xk is well defined. We believe that when a = −1 the map has only the above 2 functionally independent meromorphic first integrals, but from our approach we cannot discard the existence of a third one. In 12 A. FERRAGUT, A. GASULL, X. ZHANG fact, it is known, even for a ∈ C, that the only globally periodic map corresponds to a = 1, see [8, 10]. There is also a simple case for which the non-existence of meromorphic first integral can be easily established. Corollary 15. Let f be an analytic diffeomorphism with a fixed point q ∈ Cn and assume that the eigenvalues of Df (q) are n different prime numbers. Then f has no meromorphic first integral. Proof. By Theorem 1 we have to calculate the dimension of the Z–linear space generated from {k ∈ Zn : µk = 1} where here µ = (p1 , p2 , . . . , pn ) and the pj are the n different prime numbers. Hence, from the condition pk11 pk22 · · · pknn = 1 we get k1 = k2 = · · · = kn = 0 and this dimension is 0. As a consequence, the map has no meromorphic first integral.  There are many other rational maps in dimension n > 2 that admit meromorphic (indeed rational) first integrals. As a final example we show one of them, obtained from the paper [15], dedicated to study systems of difference equations with invariants. The map   az + bt + c az + bt + c f (x, y, z, t) = z, t, , x y has the first integral (xy + ay + bx) (zt + at + bz) (ax + az + bt + by + c) H(x, y, z, t) = . xyzt By using our result we could embed the above map into a large family, with more parame- ters, and find necessary conditions among them for the cases with one or more meromorphic first integrals. 3.4. Acknowledgements. We appreciate the anonymous referee for his/her careful rea- ding and nice comments which help us to improve both mathematics and presentation. This work has received funding from the Ministerio de Ciencia e Innovación (PID2019- 104658GB-I00 grant and Severo Ochoa and Marı́a de Maeztu Program for Centers and Units of Excellence in R&D, grant CEX2020-001084-M), and the Agencia de Gestio d’Ajuts Universitaris i de Recerca (2017 SGR 1617 grant). 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