Some properties of the k-dimensional Lyness' map
SSome properties of the k -dimensional Lyness’ map Anna Cima (1) , Armengol Gasull (1) and V´ıctor Ma˜nosa (2) (1)
Dept. de Matem`atiques, Facultat de Ci`encies,Universitat Aut`onoma de Barcelona,08193 Bellaterra, Barcelona, [email protected], [email protected] (2)
Dept. de Matem`atica Aplicada III (MA3),Control, Dynamics and Applications Group (CoDALab)Universitat Polit`ecnica de Catalunya (UPC)Colom 1, 08222 Terrassa, [email protected]
October 25, 2018
Abstract
This paper is devoted to study some properties of the k -dimensional Lyness’ map F ( x , . . . , x k ) = ( x , . . . , x k , ( a + (cid:80) ki =2 x i ) /x ) . Our main result presentes a rationalvector field that gives a Lie symmetry for F. This vector field is used, for k ≤ , togive information about the nature of the invariant sets under F. When k is odd, we alsopresent a new (as far as we know) first integral for F ◦ F which allows to deduce in a verysimple way several properties of the dynamical system generated by F. In particular forthis case we prove that, except on a given codimension one algebraic set, none of thepositive initial conditions can be a periodic point of odd period.
PACS numbers
Keywords:
Lyness’ difference equations, Lie symmetry, first integral, integrable and non–integrable discrete dynamical system.
The second and more specially the third order Lyness’ difference equations y n +2 = a + y n +1 y n and y n +3 = a + y n +1 + y n +2 y n , with a ≥ , have been considered as emblematic examples of integrable discrete systems, see for instance[15, 20, 22, 23]. The dynamics of the above equations, or their associated maps, has been1 a r X i v : . [ m a t h . D S ] J a n he objective of recent intensive investigation. Nowadays the behaviour of the orbits is wellknown when positive initial conditions are considered, see [2, 5, 7, 10, 24]. However fewresults have been obtained for negative initial conditions, see [6, 10].On the other hand, the study of the 2 and 3-dimensional Lyness’ maps is the startingpoint for the study of other birational integrable maps, see for instance [3, 8, 10, 13, 16, 17,18, 23], and [9] for a general paper on this topic.For k ≥ , very few results, apart of the ones obtained recently by Bastien and Rogalsky[4], are known for the k -th order Lyness’ equation y n + k = a + (cid:80) k − i =1 y n + i y n . (1)The main difference between the k = 2 , k ≥ integrable in the sense that the associated maps have 1 and 2 functionally independentfirst integrals, respectively (in this paper we say that a map F is integrable if it has k − k ≥ . Indeed, consider the k -dimensional Lyness’ map associated to the difference equation (1), F ( x , . . . , x k ) = (cid:32) x , . . . , x k , a + (cid:80) ki =2 x i x (cid:33) , with a ≥ , (2)which is a diffeomorphism from Q + := { x = ( x , . . . , x k ) ∈ R k : x > , x > , . . . , x k > } into itself. It is well known that it has the following couple of functionally independent firstintegrals V ( x ) = (cid:32) a + k (cid:88) i =1 x i (cid:33) (cid:32) k (cid:89) i =1 ( x i + 1) (cid:33) / ( x · · · x k ) (3)and V ( x ) = (cid:32) a + k (cid:88) i =1 x i + x x k (cid:33) (cid:32) k − (cid:89) i =1 (1 + x i + x i +1 ) (cid:33) / ( x · · · x k ) . (4)A third functionally independent first integral for k ≥ k, the Lyness’ map has up to E (cid:0) k +12 (cid:1) functionally independent first integrals, where E ( · ) denotes the integer part function. Theconjecture seems to be true for k ≤ , see again [11].The integrable structure for k = 2 , k ≥ − E (cid:0) k +12 (cid:1) , would be, generically, the dimension of the invariant manifold given by thelevel sets of the first integrals.One geometrical object that has played a key role to understand the dynamics of a largeclass of 2 and 3-dimensional maps is the Lie symmetry of the map, [7, 8]. Recall, that avector field X is said to be a Lie symmetry of a map G if it satisfies the condition X ( G ( x )) = ( DG ( x )) X ( x ) . (5)The vector field X is related with the dynamics of G in the following sense: G maps anyorbit of the differential system determined by the vector field, to another orbit of thissystem, see [8]. In the integrable case, where the dynamics are in fact one dimensional, theexistence of a Lie symmetry fully characterizes the dynamics. In [8, Thm. 1] it is provedthat if G : U → U is a diffeomorphism having a Lie symmetry X , and such G preserves γ, asolution of the differential equation ˙ x = X ( x ) , then the dynamics of G | γ is either conjugatedto a rotation, conjugated to a translation of the line, or constant, according whether γ ishomeomorphic to S , R , or a point, respectively. Other properties of the Lie symmetries ofdiscrete systems are studied in [14].One of the main results of this paper, is the following theorem where the expression ofa Lie symmetry for the k -dimensional Lyness’ map is given. Theorem 1.
For k ≥ , the vector field X k = (cid:80) ki = i X i ∂∂x i , is a Lie symmetry for the k -dimensional Lyness’ map (2), where X ( x ) = ( x + 1) (cid:104)(cid:81) k − i =2 (1 + x i + x i +1 ) (cid:105) ( a + (cid:80) k − i =1 x i − x x k ) (cid:81) ki =2 x i , (6) X m ( x ) = ( x m + 1) (cid:104)(cid:81) k − i =1 ,i (cid:54) = m − ,m (1 + x i + x i +1 ) (cid:105) ( a + (cid:80) ki =1 x i + x x k )( x m − − x m +1 ) (cid:81) ki =1 ,i (cid:54) = m x i , (7) for all ≤ m ≤ k − , and X k ( x ) = − ( x k + 1) (cid:104)(cid:81) k − i =1 (1 + x i + x i +1 ) (cid:105) ( a + (cid:80) ki =2 x i − x x k − ) (cid:81) k − i =1 x i . (8)Once we have the candidate X k to be a Lie symmetry of the Lyness’ map F the proof ofTheorem 1 only will consist in checking that (5) holds. We give here some hints of how wehave found the above X k . Observe that if there exists a vector field X k , satisfying equation35) for the k -dimensional Lyness’ map (2), then X ( F ) X ( F )... X k ( F ) = · · ·
00 0 1 0 · · · − a + P ki =2 x i x x · · · x X X ... X k . Hence it is necessary that X i +1 = X i ( F ) , for i = 1 , . . . , k − , (9)and the “compatibility condition” : X k ( F ) = − (cid:32) a + (cid:80) ki =2 x i x (cid:33) X + 1 x (cid:34) k (cid:88) i =2 X i (cid:35) . Thus, the construction of a Lie symmetry is straightforward once the right expression of X , as a seed of equations (9), is obtained. In [5, 7] and [8] the expressions of X and X are given. The idea of these papers is that these vector fields have to be multiples of ∇ V and ∇ V × ∇ V , respectively. These constructions are used to force F and X k to share thesame set of first integrals. When k ≥ X and X , are ( x + 1)( a + x − x ) x and ( x + 1)(1 + x + x )( a + x + x − x x ) x x , respectively. Thus it seems natural to try with X as the expression given in (6) and,indeed, it works!. From this starting point, the proof for a given small k is only a matter ofcomputations, while the proof for a general k is long and tedious, but straightforward. Itis done in Section 2. We suggest to skip this section in a first reading of the paper.Another result that helps for understanding the dynamics generated by F, when k isodd, is given in next proposition. In this result, the key point is the existence of a new (asfar as we know) first integral for F = F ◦ F for any odd k ≥ . As we will see, our proof ofthe existence of this function is inspired in the paper [22], where this first integral is givenfor k = 3 . Theorem 2.
Set k = 2 (cid:96) + 1 , x = ( x , . . . , x (cid:96) +1 ) and consider F from Q + into itself. Then(a) The function W ( x ) = (cid:81) (cid:96)j =0 ( x j +1 + 1) (cid:81) (cid:96)j =1 x j , (10) is a first integral of F . b) For any (cid:96) ≥ , the function V := W + W ( F ) , which is, V ( x ) = (cid:81) (cid:96)j =0 x j +1 ( x j +1 + 1) + ( a + (cid:80) (cid:96) +1 j =1 x j ) (cid:81) (cid:96)j =1 x j ( x j + 1) (cid:81) (cid:96) +1 i =1 x i , (11) is a first integral of F which is functionally independent with the first integrals V and V given in (3) and (4), respectively.(c) The function W · W ( F ) coincides with the first integral of F, V given in (3). In otherwords, V = W · W ( F ) . (d) The algebraic set G := { x ∈ Q + : W ( x ) − W ( F ( x )) = 0 } is invariant by F. (e) If the map F has some periodic point of odd period then it has to be contained in G . (f ) Setting G ± := { x ∈ Q + : ± ( W ( x ) − W ( F ( x ))) > } , the map F sends G + into G − and viceversa, and both sets are invariant by F . Furthermore, the dynamics of F on each of these sets are conjugated, being the map F itself the conjugation.(g) The measure m ( B ) := (cid:90) B ± x )( W ( x ) − W ( F ( x ))) d x , where Π( x ) = (cid:81) ki =1 x i , is an invariant measure for F , i.e. m ( F ( B )) = m ( B ) , where B is any measurable set in G ± . (h) The measure m ( B ) := (cid:90) B x ) d x , is an invariant measure for F , i.e. m ( F ( B )) = m ( B ) , where B is any measurableset in Q + . We remark that, when k is odd, the first integral V given above coincides with the onegiven recently in [11]. Observe also that the invariant algebraic surface G was already foundin [7], but only for k = 3 . Also, as we will see in Subsection 3.1, the function W is useful tomake an explicit simple order reduction when we study the dynamics of F for k odd.Although, by using both theorems we have not been able to present a complete studyof the higher dimensional Lyness’ map, in next results we give some information about theinvariant sets in the phase space when k = 4 , . We prove:
Proposition 3.
The vector field X given by equations (6)–(8) for k = 4 , is a Lie symmetryfor the -dimensional Lyness’ map. Moreover, X ( V i ) = 0 , for i = 1 , , and then the sets I h,k := { V = h } ∩ { V = k } ∩ Q + are invariant by F and by the flow of X . Furthermore, if we assume that both first integrals intersect transversally on C h,k , aconnected component of I h,k , then C h,k is diffeomorphic to a torus. roposition 4. The vector field X given by equations (6)–(8) for k = 5 is a Lie symmetryfor the -dimensional Lyness’ map. Moreover, X ( V i ) = 0 , for i = 1 , , , where V ( x ) = 1 x x x x x (cid:16) x x x (1 + x ) (1 + x ) (1 + x )+ x x (1 + x ) (1 + x ) ( a + x + x + x + x + x ) (cid:17) , and the sets I h,k,l := { V = h } ∩ { V = k } ∩ { V = (cid:96) } ∩ Q + , which generically have at leasttwo connected components, are invariant by F and by the flow of X . Furthermore, if we assume that the three first integrals intersect transversally on C h,k,(cid:96) ,a connected component of I h,k,(cid:96) , then C h,k,(cid:96) is diffeomorphic to a (two–dimensional) torus. It is important to notice that in the above results we do not assert that most of theconnected components C h,k and C h,k,(cid:96) are tori, since we did not succeed to prove that overthem the intersection of the energy levels at the whole sets C h,k and C h,k,(cid:96) are transversal.Hence it remains open to decide whether they are two–dimensional differentiable manifoldsor not, and in the case that they are not differentiable manifolds to decide which are theirtopology. In any case, our result reduces the problem to a computational question.Figure 1. Projections into R of the flow associated to the Lie symmetry X , and the orbitof the Lyness’ map, for k = 4 and a = 4 , both with initial condition (1,2,3,4).Our numerical simulations seem to indicate that for k = 4 , all the generic level curves I h,k , are connected. On the other hand, for k = 5 , generically the sets I h,k,l seem to haveexactly two connected components. In Figures 1 and 2 we give a projection in R of thesesurfaces. Indeed, in Figure 1 we represent, both an orbit of F and an orbit of X startingwith the same initial condition, and in Figure 2 an orbit of F for k = 5 . Notice that inboth cases the behavior of the orbits seems to indicate that two, (respectively three) is themaximum number of independent first integrals for F when k = 4 (resp. k = 5), as it issuggested in [11]. 6igure 2. Projection into R of the first 5000 iterates of the Lyness’ map for k = 5 and a = 1 , staring at (1,2,3,4,5). Odd and even iterates are in different connected components.A remarkable fact that Figure 1 shows is that, although for k = 4 the manifold I h,k isinvariant for both the map F and the flow of X , the map F seems to send an orbit of X to a different orbit of the vector field. Further numerical experiments seems to confirm thisfact. Under this situation, we can not use the techniques developed in [8]. This fact makesmore difficult the knowledge of the behavior of F restricted to each I h,k and is one of theimportant differences between the cases k = 2 , k = 4 . As we have already explained in the introduction, the existence of a vector field X k = (cid:80) ki = i X i ∂∂x i satisfying equation (5) for the k -dimensional Lyness’ map (2), is equivalent tothe set of equations X i +1 = X i ( F ) , for i = 1 , . . . , k − , (12)together with the compatibility condition: X k ( F ) = − (cid:32) a + (cid:80) ki =2 x i x (cid:33) X + 1 x (cid:34) k (cid:88) i =2 X i (cid:35) . (13)The proof will consist in checking that the choice of X k given in the statement satisfiesequations (12) and (13). The result is straightforward for k = 3 , , k ≥ . We proceed in two steps:
First step:
We will show that from the expression (6) of X as a seed of equations (12)we obtain the expressions of X m for m = 2 , . . . ,k − X k given by equations (7) and(8), respectively. 7 econd step: We will prove that the compatibility condition (13) is satisfied.
First step:
We start with some preliminary observations:
Observation 1.
Set K i := x i + 1 for i = 1 , . . . , k − . Then K i ( F ) = x i +1 + 1. Observation 2.
If we set L i := 1 + x i + x i +1 , then for all 1 ≤ i ≤ k − L i ( F ) =1 + x i +1 + x i +2 = L i +1 , and L k − ( F ) = 1 + x k + x k +1 = 1 + x k + a + (cid:80) ki =2 x i x = a + (cid:80) ki =1 x i + x x k x . For this reason(a) If M := (cid:81) k − i =1 (1 + x i + x i +1 ), then M ( F ) = ( a + (cid:80) ki =1 x i + x x k ) (cid:16)(cid:81) k − i =2 (1 + x i + x i +1 ) (cid:17) x . (b) Setting M m := (cid:81) k − i =1 ,i (cid:54) = m − ,m (1 + x i + x i +1 ) for 2 ≤ m ≤ k −
2, we obtain M m ( F ) = k − (cid:89) i =2 ,i (cid:54) = m,m +1 (1 + x i + x i +1 ) ( a + k (cid:88) i =1 x i + x x k ) /x . (c) If M k − := (cid:81) k − i =1 (1 + x i + x i +1 ), then M k − ( F ) = (cid:81) k − i =2 (1 + x i + x i +1 ). Observation 3.
Set N = a + k − (cid:80) i =1 x i − x x k , then N ( F ) = a + k (cid:88) i =2 x i − x x k +1 = a + k (cid:88) i =2 x i − x a + (cid:80) ki =2 x i x = a + (cid:80) ki =2 x i x ( x − x ) = x k +1 ( x − x ) . Observation 4.
Set R = a + k (cid:80) i =1 x i + x x k , then R ( F ) = a + k (cid:88) i =2 x i + x k +1 + x x k +1 = a + (cid:80) ki =2 x i x (1 + x + x ) = x k +1 (1 + x + x ) . Observation 5. (a) For all 2 ≤ i ≤ k − S i = x i − − x i +1 , then S i ( F ) = x i − x i +2 .(b) Set S k − = x k − − x k , then S k − ( F ) = x k − − x k +1 = x k − − a + (cid:80) ki =2 x i x = − a + (cid:80) ki =2 x i − x x k − x .
8f we now consider the seed X given by equation (6), using Observations 1, 2a, and 3we obtain that X = ( x + 1) (cid:104)(cid:81) k − i =3 (1 + x i + x i +1 ) (cid:105) ( a + (cid:80) ki =1 x i + x x k )( x − x ) (cid:81) ki =1 ,i (cid:54) =2 x i . Now applying systematically Observations 1, 2b, 4 and 5a, we obtain that for 2 ≤ m ≤ k − X m = X m − ( F ) is given by equation (7).Observe that in particular X k − = X k − ( F ) = ( x k − + 1) (cid:104)(cid:81) k − i =1 (1 + x i + x i +1 ) (cid:105) ( a + (cid:80) ki =1 x i + x x k )( x k − − x k ) (cid:81) k − i =1 x i , hence the term L k − = 1 + x k − + x k does not appear and so, in order to compute X k weneed to use Observations 1, 2c, 4 and 5b, obtaining the expression of X k given by (8). Second step (compatibility condition (13)).
A simple computations shows that X k ( F ) = A x (cid:16)(cid:81) ki =1 x i (cid:17) , where A = − (cid:32) a + k (cid:88) i =1 x i (cid:33) (cid:34) k − (cid:89) i =2 L i (cid:35) (cid:32) a + k (cid:88) i =2 x i + x (cid:32) a + k (cid:88) i =3 x i − x x k (cid:33)(cid:33) . Another computation gives that − (cid:32) a + (cid:80) ki =2 x i x (cid:33) X + 1 x (cid:34) k (cid:88) i =2 X i (cid:35) = B x (cid:16)(cid:81) ki =1 x i (cid:17) , where B = − ( x + 1) (cid:16) a + (cid:80) ki =2 x i (cid:17) (cid:104)(cid:81) k − i =2 L i (cid:105) (cid:16) a + (cid:80) k − i =1 x i − x x k (cid:17) + (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17) C − x k ( x k + 1) (cid:16) a + (cid:80) ki =2 x i − x x k − (cid:17) (cid:104)(cid:81) k − i =1 L i (cid:105) , and C = k − (cid:88) m =2 x m ( x m + 1) S m M m . Recall that L i = 1 + x i + x i +1 for all i = 1 , . . . , k − S m = x m − − x m +1 and M m = (cid:81) k − i =1 ,i (cid:54) = m − ,m (1 + x i + x i +1 ) for 2 ≤ m ≤ k −
1. Therefore we want to prove that A = B .9 tep 2a. First we show that B contains L and L as a factors. To see this, it sufficesto check that L and L are factors of C . Observe that if L = 0, then we can write L = x − x , and x = − (1 + x ) hence x − x = x + x + 1. Therefore C | { L =0 } = (cid:80) m =2 x m ( x m + 1) S m M m = (cid:80) m =2 x m ( x m + 1)( x m − − x m +1 ) (cid:16)(cid:81) k − i =1 ,i (cid:54) = m − ,m (1 + x i + x i +1 ) (cid:17) = x (1 + x )( x − x )( x − x ) (cid:16)(cid:81) k − i =4 (1 + x i + x i +1 ) (cid:17) + x (1 + x )( x − x )(1 + x + x ) (cid:16)(cid:81) k − i =4 (1 + x i + x i +1 ) (cid:17) = 0 . So it is a factor of C .If L = 0, then L = x − x , and x − x = − (1 + x + x ). Hence C | { L =0 } = (cid:80) m =3 x m ( x m + 1) S m M m == (cid:80) m =3 x m ( x m + 1)( x m − − x m +1 ) (cid:16)(cid:81) k − i =1 ,i (cid:54) = m − ,m (1 + x i + x i +1 ) (cid:17) = x (1 + x )( x − x )(1 + x + x ) (cid:16)(cid:81) k − i =4 (1 + x i + x i +1 ) (cid:17) − x (1 + x )(1 + x + x )(1 + x + x )(1 + x + x ) ×× (cid:16)(cid:81) k − i =5 (1 + x i + x i +1 ) (cid:17) = 0 . Hence L is a factor of C .The above results imply that C = L L Q k ( x , x , x , . . . , x k ) , (observe that x does not appear in the expression of Q k ). Hence L and L are factors inthe expression of B , and then B = L L (cid:104) − ( x + 1) (cid:16) a + (cid:80) ki =2 x i (cid:17) (cid:104)(cid:81) k − i =4 L i (cid:105) (cid:16) a + (cid:80) k − i =1 x i − x x k (cid:17) + (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17) Q k − x k ( x k + 1) (cid:16) a + (cid:80) ki =2 x i − x x k − (cid:17) L (cid:104)(cid:81) k − i =4 L i (cid:105)(cid:105) . Furthermore Q k = − x (1 + x ) (cid:16)(cid:81) k − i =4 L i (cid:17) + ( x − x ) L (cid:16)(cid:81) k − i =4 L i (cid:17) + x (1 + x ) L (cid:16)(cid:81) k − i =5 L i (cid:17) ++ (cid:80) k − m =5 x m ( x m + 1)( x m − − x m +1 ) L (cid:16)(cid:81) k − i =4 ,j (cid:54) = m − ,m L j (cid:17) . Step 2b.
Now we state the following claim, that will we proved at the end of the proof.
Claim:
For k ≥ , Q k ( x , x , x , . . . , x k ) = (cid:16)(cid:81) k − i =4 L i (cid:17) [ x x L k − − x k − x k L ].By using the claim, B = (cid:104)(cid:81) k − i =2 L i (cid:105) D , where D = − ( x + 1) (cid:16) a + (cid:80) ki =2 x i (cid:17) (cid:16) a + (cid:80) k − i =1 x i − x x k (cid:17) L k − + (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17) [ x x L k − − x k − x k L ] − x k ( x k + 1) (cid:16) a + (cid:80) ki =2 x i − x x k − (cid:17) L . tep 2c. Observe that − x k ( x k + 1) (cid:16) a + (cid:80) ki =2 x i − x x k − (cid:17) L − x k − x k L (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17) = − L x k (cid:104) ( x k + 1) (cid:16) a + (cid:80) ki =2 x i − x x k − (cid:17) − x k − (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17)(cid:105) = − L x k (cid:104) x k − (cid:16) a + (cid:80) ki =1 x i + x x k − x − x x k (cid:17) + (1 + x k ) (cid:16) a + (cid:80) ki =2 x i (cid:17)(cid:105) = − L x k (cid:104) x k − (cid:16) a + (cid:80) ki =2 x i (cid:17) + (1 + x k ) (cid:16) a + (cid:80) ki =2 x i (cid:17)(cid:105) = − L x k L k − (cid:16) a + (cid:80) ki =2 x i (cid:17) . Thus D also contains the factor L k − , and therefore B = (cid:104)(cid:81) k − i =2 L i (cid:105) E , where E = (cid:104) − ( x + 1) (cid:16) a + (cid:80) ki =2 x i (cid:17) (cid:16) a + (cid:80) k − i =1 x i − x x k (cid:17) + (cid:16) a + (cid:80) ki =1 x i + x x k (cid:17) x x − x k L (cid:16) a + (cid:80) ki =2 x i (cid:17)(cid:105) . Step 2d.
Is not difficult to check that E is a quadratic polynomial in x . Now we see that E vanishes either when x = x , := − a − (cid:80) ki =1 ,i (cid:54) =3 x i (that is when a + (cid:80) ki =1 x i = 0), andwhen x = x , := − (cid:32) a + k − (cid:88) i =4 x i + x (1 − x x k )1 + x (cid:33) . In the first case we have E | { a + P ki =1 x i =0 } = x ( x + 1)( − x k − x x k ) + x x x k + L x x k = x x k [ − ( x + 1)( x + 1) + x x + 1 + x + x ]= x x k [ − − x − x − x x + x x + 1 + x + x ] = 0 . In the second case, since a + (cid:80) k − i =3 x i = x ( x x k − / ( x + 1), we have E | { x = x , } = − (cid:18) x + x ( x x k − x + 1 (cid:19) ( x + 1) (cid:18) x ( x x k − x + 1 + x + x − x k − x x k (cid:19) + x x (cid:18) x + x + x ( x x k − x + 1 + x x k (cid:19) − L x k (cid:18) x + x ( x x k − x + 1 (cid:19) =: 1 x + 1 F , where (after some computations) F = − x x ( x k + 1) L ( x − x k ) + x x L ( x k + 1) − L x x x k ( x k + 1)= x x L ( x k + 1)[ − ( x − x k ) + x − x k ] = 0 . Therefore E | { x = x , } = 0.In summary, as a quadratic polynomial in x , E factorizes as E = − ( x + 1) (cid:16) x + a + (cid:80) ki =1 ,i (cid:54) =3 x i (cid:17) (cid:16) x + a + (cid:80) k − i =4 x i + ( x (1 − x x k )) / ( x + 1) (cid:17) = − (cid:16) a + (cid:80) ki =1 x i (cid:17) (cid:16)(cid:16) x + a + (cid:80) k − i =4 x i (cid:17) ( x + 1) + x (1 − x x k ) (cid:17) = − (cid:16) a + (cid:80) ki =1 x i (cid:17) (cid:16) ( x + 1) (cid:16) a + (cid:80) k − i =3 x i (cid:17) + x − x x x k (cid:17) = − (cid:16) a + (cid:80) ki =1 x i (cid:17) (cid:16) a + (cid:80) k − i =2 x i + x (cid:16) a + (cid:80) k − i =3 x i − x x k (cid:17)(cid:17) . B = − (cid:104)(cid:81) k − i =2 L i (cid:105) (cid:16) a + (cid:80) ki =1 x i (cid:17) (cid:16) a + (cid:80) k − i =2 x i + x (cid:16) a + (cid:80) k − i =3 x i − x x k (cid:17)(cid:17) = A , as we wanted to show.To end the proof it only remains to prove the claim. We proceed by induction. That itis true when k = 6 is straightforward. Assume now that the claim is true for Q k , then Q k +1 = − x (1 + x ) (cid:16)(cid:81) ki =4 L i (cid:17) + ( x − x ) L (cid:16)(cid:81) ki =4 L i (cid:17) + x (1 + x ) L (cid:16)(cid:81) ki =5 L i (cid:17) ++ (cid:80) k − m =5 x m (1 + x m )( x m − − x m +1 ) L (cid:16)(cid:81) ki =4 ,i (cid:54) = m − ,m L i (cid:17) = L k Q k + L (cid:16)(cid:81) k − i =4 L i (cid:17) ( x k (1 + x k )( x k − − x k +1 )) . By using the hypothesis of induction, we obtain that Q k +1 = L k (cid:16)(cid:81) k − i =4 L i (cid:17) [ x x L k − − x k − x k L ] + L (cid:16)(cid:81) k − i =4 L i (cid:17) ( x k (1 + x k )( x k − − x k +1 )) = (cid:16)(cid:81) k − i =4 L i (cid:17) ( x x L k L k − + L [ x k (1 + x k )( x k − − x k +1 ) − x k − x k L k ]) . Thus Q k +1 = L k (cid:16)(cid:81) k − i =4 L i (cid:17) [ x x L k − − x k − x k L ] + L (cid:16)(cid:81) k − i =4 L i (cid:17) ( x k (1 + x k )( x k − − x k +1 ))= (cid:16)(cid:81) k − i =4 L i (cid:17) ( x x L k L k − + L [ x k (1 + x k )( x k − − x k +1 ) − x k − x k L k ]) . An easy computation shows that x k (1 + x k )( x k − − x k +1 ) − x k − x k L k = − L k − x k x k +1 ,and the result follows. Therefore, Theorem 1 is proved. Before proving Theorem 2 and Propositions 3 and 4 we need some preliminary results.Recall that a map H which is a first integral for G p := G ◦ p ) · · · ◦ G is also called sometimes a p − first integral , or simply, for short, a p − integral , of G, see [23].The following lemma, which is very easy to prove, gives light on one utility of p − firstintegrals, specially if they are not symmetric functions of their arguments. Lemma 5.
Let H be a p -integral of a map G. Then then for any symmetric function of p variables S , the function V S := S ( H, H ◦ G, H ◦ G , . . . , H ◦ G p − ) is a first integral of G . Lemma 6.
Set k = 2 (cid:96) + 1 , x = ( x , x , . . . , x k ) ∈ Q + and let W be the function givenin (10). Then, if we define the polynomial Z ( x ) : = ( (cid:81) ki =1 x i )[ W ( x ) − W ( F ( x ))] == (cid:81) (cid:96)j =0 x j +1 ( x j +1 + 1) − (cid:16) a + (cid:80) (cid:96) +1 i =1 x i (cid:17) (cid:81) (cid:96)j =1 x j ( x j + 1) , it holds that Z ( F ( x )) = det( DF ( x )) Z ( x ) . roof. In Theorem 2 (a) we will prove that W is a 2-integral of F. Thus, if we define˜ Z := W − W ( F ) , then ˜ Z ( F ( x )) = − ˜ Z ( x ) . (14)From the above equality it is clear that { x ∈ Q + : ˜ Z ( x ) = 0 } = { x ∈ Q + : Z ( x ) = 0 } isan invariant hypersurface by F. Notice also that if we define Π( x ) := (cid:81) kj =1 x j , it holds thatΠ( F ( x )) = a + (cid:80) ki =2 x i x Π( x ) = − det( DF ( x )) Π( x ) . (15)Since Z ( x ) = Π( x ) ˜ Z ( x ) , by using equalities (14) and (15), we obtain that Z ( F ( x )) = Π( F ( x )) ˜ Z ( F ( x )) = det( DF ( x ))Π( x ) ˜ Z ( x ) = det( DF ( x )) Z ( x ) , as we wanted to see. Proof of Theorem 2. (a) The proof of the equality W ( F ( x )) = W ( x ) is straightforward.We have obtained the expression (10) inspired in the results of [22]. In that paper it isproved that ( y n + 1)( y n +2 + 1) y n +1 = ( y n +2 + 1)( y n +4 + 1) y n +3 , where { y n } is the sequence given by the 3-rd order Lyness’ recurrence y n +3 = ( a + y n +1 + y n +2 ) /y n . Notice that this property is equivalent to say that for k = 3 , W is a 2-integralfor F. (b-c) By applying Lemma 5 with S ( u, v ) = u + v and S ( u, v ) = uv, we obtain the firstintegrals V and V , respectively. The functionally independence of V , V and V , for (cid:96) ≥ , follows from straightforward computations and it is already established in [11].(d-f) From Lemma 6 we know that Z ( F ( x )) = det( DF ( x )) Z ( x ) , where recall that Z ( x ) = ( (cid:81) ki =1 x i )[ W ( x ) − W ( F ( x ))] . Note also thatdet( DF ( x )) = ( − k a + x + · · · + x k − x k . Since when k is odd det( DF ) < Q + , equation Z ( F ) = det( DF ) Z means that G = { x ∈ Q + : W ( x ) = W ( F ( x )) } = { x ∈ Q + : Z ( x ) = 0 } is invariant by F and that F maps the region { x : Z ( x ) > } into the region { x : Z ( x ) < } and viceversa. Furthermore it implies that the dynamics of F on each of these sets areconjugated, being the map F itself the conjugation. Moreover, any periodic orbit with oddperiod must lie in G , as we wanted to see. 13g-h) By using the Change of Variables Theorem it is easy to see that if G is a dif-feomorphism of U , and on this region µ is a positive function that satisfies µ ( G ( x )) =det( DG ( x )) µ ( x ) , then m ( B ) = (cid:90) B µ ( x ) d x is an invariant measure for G. By Lemma 6 we know that Z ( F ( x )) = det( DF ( x )) Z ( x ) andby equality (15), that Π( F ( x )) = − det( DF ( x ))Π( x ) . By using these results we have that Z ( F ( x )) = det( DF ( x )) Z ( x ) and Π( F ( x )) = det( DF ( x ))Π( x ) , being both equalities inthe corresponding domains, which are invariant by F . Hence (f) and (g) follow. Recall that in the previous section we have seen that when k = 2 (cid:96) + 1, the regions { x : Z ( x ) > } and { x : Z ( x ) < } are invariant by F , and the dynamics on both regionare conjugated. This observation allows us to give a new application of the invariant W. We can reduce the study of the dynamics of F on { x : Z ( x ) (cid:54) = 0 } to the study of a new( k − n = 3, we get that W ( x, y, z ) = ( x + 1)( z + 1) /y , and hence anyadmissible level surface { x : W ( x ) = w } , w (cid:54) = 0 , can be described as y = k ( x + 1)( z + 1),where k = 1 /w . Therefore F (cid:12)(cid:12) { W = w } ( x, y, z ) = (cid:18) z, a + z + k ( x + 1)( z + 1) x , a + k + z ( k + 1) kx ( z + 1) (cid:19) , and we can reduce the study of the dynamics of F to the study of the reduced map˜ F ( x, z ) = (cid:18) z, a + k + z ( k + 1) kx ( z + 1) (cid:19) , or, equivalently, the study of the second order difference equation y n +2 = a + k + y n +1 ( k + 1) ky n ( y n +1 + 1) , (16)as in [22, Eqs. (8) and (9)]. The dynamics of this difference equation is studied in [8, Ex.3]. Equation (16) is sometimes called generalized Lyness’ recurrence , see [20].For n = 5, the integral of F is W ( x, y, z, t, s ) = ( x + 1)( z + 1)( s + 1) / ( yt ). Again anyadmissible level surface { x : W ( x ) = w } , w (cid:54) = 0 , can be described as t = k ( x + 1)( z +1)( s + 1) /y , where k = 1 /w . Therefore proceeding as before we can reduce the study of thedynamics of F to the study of the reduced map˜ F ( x, y, z, s ) = (cid:18) y, z, t, p ( z, s ; k ) x + p ( y, z, s ; a, k ) x + p ( y, z, s ; a, k ) xy (cid:19) p ( z, s ; k ) = k ( s + 1) ( z + 1), p ( y, z, s ; a, k ) = 2 k ( s + 1) ( z + 1) + y ( a + s + z ), and p ( y, z, s ; a, k ) = k ( s + 1) ( z + 1) + y ( z + s + a + y ), or equivalently to the difference equa-tion y n +4 = p ( y n +2 , y n +3 ; k ) y n + p ( y n +1 , y n +2 , y n +3 ; a, k ) y n + p ( y n +1 , y n +2 , y n +3 ; a, k ) y n y n +12 . Clearly, the described procedure can be generalized to higher dimensions.
This subsection is devoted to prove Propositions 3 and 4.Along the section we will use a straightforward consequence of the following result:
Theorem 7 (Bastien and Rogalsky, [4]) . Let ¯ x be the fixed point of F in Q + . For any h > V (¯ x ) , the level sets { V = h } ∩ Q + are homeomorphic to S k − . Corollary 8.
Let K (cid:54) = ∅ be by the intersection of some level sets of different first integralsof F, including V among them. Then K ∩ Q + is a compact set, invariant by F. For the sake of completeness and to compare with the cases k = 4 ,
5, we start by recallingsome results for the case k = 3 , most of them already proved in [7]. -dimensional map For k = 3 , F ( x, y, z ) = (cid:18) y, z, a + y + zx (cid:19) , the Lie symmetry given in Theorem 1, is X := (cid:104) x ( x + 1)(1 + y + z )( a + x + y − yz ) ∂∂x + y ( y + 1)( x − z )( a + x + y + z + xz ) ∂∂y + z ( z + 1)(1 + x + y )( a + y + z − xy ) ∂∂z (cid:3) / ( xyz )and since X ( V i ) = 0, for i = 1 , , the functions V and V given in (3) and (4) are firstintegrals for X and F. Also, by Theorem 2, W ( x, y, z ) := ( x + 1) ( z + 1) /y is a 2-integralof F ; G = { x ∈ Q + : Z ( x ) = 0 } is invariant by F, where Z ( x ) = x ( x + 1) z ( z + 1) − ( a + x + y + z ) y ( y + 1);and F maps G + := { x ∈ Q + : Z ( x ) > } into G − := { x ∈ Q + : Z ( x ) < } , and viceversa.Let ¯ x be the fixed point in Q + , of F . Set h > V (¯ x ), k > V (¯ x ). In [7] it is proved that { V = h } ∩ Q + and { V = k } ∩ Q + are diffeomorphic to two dimensional spheres, and their15ransversal intersections are formed by exactly two disjoints curves, both diffeomorphic tocircles. Their non transversal intersections correspond to:(a) The 2–periodic points of F (which are equilibrium points of X ) given by the curve L := { ( x, ( x + a ) / ( x − , x ) | x > } . (b) The levels placed at G . Those placed at G \ { ¯ x } are formed by exactly one curve,diffeomorphic to a circle.Finally, since X is also a Lie symmetry of F , as a consequence of [7, Thm. 18] or [8,Thm. 1], we know that the map F restricted to each of the sets { V = h }∩{ V = k } , whichis not simply a fixed point of F , is conjugated to a rotation. Further discussion about thepossible rotation numbers can be found in [7]. -dimensional map. Proof of Proposition 3 . Equations (6)–(8) give the following Lie symmetry for the4-dimensional Lyness’ map: X = (cid:2) x ( x + 1) (1 + y + z ) (1 + z + t ) ( a + x + y + z − yt ) ∂∂x + y ( y + 1) (1 + z + t ) ( a + x + y + z + t + xt ) ( x − z ) ∂∂y + z ( z + 1) (1 + x + y ) ( a + x + y + z + t + xt ) ( y − t ) ∂∂z − t ( t + 1) (1 + x + y ) (1 + y + z ) ( a + y + z + t − xz ) ∂∂t (cid:3) / ( xyzt ) . A straightforward computation shows that the Lie symmetry satisfies X ( V i ) = 0, for i = 1 ,
2. Hence, the orbits of both F and X generically lie in a two dimensional surface ofthe form I h,k := { V = h } ∩ { V = k } ∩ Q + . Let C h,k be a connected component of I h,k . From Corollary 8 we know that C h,k iscompact. If { V = h } and { V = k } intersect transversally on C h,k , then for all pointsRank (cid:32) ( V ) x ( V ) y ( V ) z ( V ) t ( V ) x ( V ) y ( V ) z ( V ) t (cid:33) = 2 . This fact implies that the dual 2–form associated to the 2–field ∇ V ∧ ∇ V = [( V ) x ( V ) y − ( V ) y ( V ) x ] ∂∂x ∧ ∂∂y + [( V ) x ( V ) z − ( V ) z ( V ) x ] ∂∂x ∧ ∂∂z +[( V ) x ( V ) t − ( V ) t ( V ) x ] ∂∂x ∧ ∂∂t + [( V ) y ( V ) z − ( V ) z ( V ) y ] ∂∂y ∧ ∂∂z +[( V ) y ( V ) t − ( V ) t ( V ) y ] ∂∂y ∧ ∂∂t + [( V ) z ( V ) t − ( V ) t ( V ) z ] ∂∂z ∧ ∂∂t , given by ω = [( V ) z ( V ) t − ( V ) t ( V ) z ] dxdy + [( V ) z ( V ) x − ( V ) x ( V ) z ] dxdz +[( V ) z ( V ) y − ( V ) y ( V ) z ] dxdt + [( V ) t ( V ) x − ( V ) x ( V ) t ] dydz +[( V ) t ( V ) y − ( V ) y ( V ) t ] dydt + [( V ) x ( V ) y − ( V ) y ( V ) x ] dzdt
16s nonzero at every point of C h,k , and therefore it is orientable, see [1, Sec. 2.5] or [21, Sec.X, § X h,k denote the restriction of the vector field X to the invariant surface C h,k . Somecomputations show that the unique equilibrium point of X in Q + is the fixed point of F. Hence X h,k has no equilibrium points, and therefore the Poincar´e–Hopf formula gives:0 = i ( X h,k ) = χ ( C h,k ) = 2 − g where i ( X h,k ) denotes the sum of the indices of the equilibrium points of X h,k in C h,k and χ ( C h,k ) and g are the Euler characteristic and the genus of the surface C h,k , respectively.Hence the genus of C h,k is one. An orientable, compact, connected surface of genus one isa torus, as we wanted to prove. -dimensional map Proof of Proposition 4.
By Theorem 2 we know that W ( x, y, z, t, s ) := ( x + 1) ( z + 1) ( s + 1) yt , is a 2-integral of F. Moreover, the 3 functionally independent first integrals of F given in(3), (4) and (11), are V ( x, y, z, t, s ) = ( a + x + y + z + t + s ) ( x + 1) ( y + 1) ( z + 1) ( t + 1) ( s + 1) xyzts ,V ( x, y, z, t, s ) = ( a + x + y + z + t + s + xs ) (1 + x + y ) (1 + y + z ) (1 + z + t ) (1 + t + s ) xyzts ,V ( x, y, z, t, s ) = x ( x + 1) z ( z + 1) s ( s + 1) + ( a + x + y + z + t + s ) y ( y + 1) t ( t + 1) xyzts . Some tedious computations show that the hypersurface G is in the locus of non–trans-versality of the three level sets of the integrals V i , i = 1 , , Q + . Recall that precisely, G = { x ∈ Q + : Z ( x ) = 0 } is invariant by F, where Z ( x ) = x ( x + 1) z ( z + 1) s ( s + 1) − ( a + x + y + z + t + s ) y ( y + 1) t ( t + 1) , and that F maps G + = { x ∈ Q + : Z ( x ) > } into G − = { x ∈ Q + : Z ( x ) < } , andviceversa.Equations (6)–(8) give the following Lie symmetry for the 5-dimensional Lyness’ map:17 = (cid:2) x ( x + 1) (1 + y + z ) (1 + z + t ) (1 + t + s ) ( a + x + y + z + t − ys ) ∂∂x + y ( y + 1) (1 + t + s ) (1 + z + t ) ( a + x + y + z + t + s + xs ) ( x − z ) ∂∂y + z ( z + 1) (1 + x + y ) (1 + t + s ) ( a + x + y + z + t + s + xs ) ( y − t ) ∂∂z + t ( t + 1) (1 + x + y ) (1 + y + z ) ( a + x + y + z + t + s + xs ) ( z − s ) ∂∂t − s ( s + 1) (1 + x + y ) (1 + y + z ) (1 + z + t ) ( a + y + z + t + s − tx ) ∂∂s (cid:3) / ( xyzts ) . Again, direct computations show that X ( V i ) = 0, for i = 1 , , . Hence the orbits ofboth F , and X lie in a two dimensional surface of the form I h,k,(cid:96) := { V = h } ∩ { V = k } ∩ { V = (cid:96) } ∩ Q + . Let C h,k,(cid:96) be a connected component of I h,k,(cid:96) . From Corollary 8 we know that C h,k,(cid:96) iscompact. If { V = h } , { V = k } and { V = k } intersect transversally on C h,k,(cid:96) , then for allpoints Rank ( V ) x ( V ) y ( V ) z ( V ) t ( V ) s ( V ) x ( V ) y ( V ) z ( V ) t ( V ) s ( V ) x ( V ) y ( V ) z ( V ) t ( V ) s = 3 . Similarly than in the case k = 4 , this fact implies that the dual 2–form associated to the3–field ∇ V ∧ ∇ V ∧ ∇ V is nonzero at every point of C h,k,(cid:96) , and therefore this set is atwo-dimensional orientable manifold.It is not difficult to check that all the equilibrium points of X in Q + are the points ofthe curve L = (cid:26) x = (cid:18) x, x + ax − , x, x + ax − , x (cid:19) with x > (cid:27) , which contains a continuum of two periodic points and the fixed point of F. Moreover L belongs to the locus of non–transversality of the integrals V , and V in Q + .The above observation implies that X has no equilibrium points in any level set I h,k,(cid:96) where the three first integrals intersect transversally. Therefore the Poincar´e–Hopf formulagives 0 = i ( X h,k,(cid:96) ) = χ ( C h,k,(cid:96) ) = 2 − g. for each connected component C h,k,(cid:96) of such a level set (where X h,k,(cid:96) is the restriction of X to C h,k,(cid:96) ).Hence g = 1, which implies that C h,k,(cid:96) is a torus (since it is two-dimensional, orientable,compact, manifold of genus one), as we wanted to proof.Finally, observe that the fact that F maps G + into G − , and viceversa, implies that most I h,k,(cid:96) have at least one connected component on each set. In fact, it seems that each I h,k,(cid:96) , not included in Γ , has exactly two connected components in Q + , as it happens when k = 3 . R of the first 10 iterates of the Lyness’ map, for k = 5 and a = 4 , starting at (1 , , , , . Odd and even iterates are in different connected components.From the above proof it is clear that if some I h,k,(cid:96) cuts L then the first three integralsdo not cut transversally on it. Let us see that in general I h,k,(cid:96) ∩ L = ∅ . This will be aconsequence of the shape of the function v ( x ) := V (cid:18) x, x + ax − , x, x + ax − , x (cid:19) , x > , a ≥ . This function has a global minimum at the coordinate given by the fixed point x = 2 + √ a , and lim x → + v ( x ) = lim x → + ∞ v ( x ) = + ∞ .Thus, given h > v (2 + √ a ) there are only two solutions x ( h ) < √ a < x ( h )of the equation v ( x ) = h . Now set v i ( x ) := V i (cid:18) x, x + ax − , x, x + ax − , x (cid:19) , x > , a ≥ , for i = 2 , . Then, for any value of k and (cid:96) satisfying k (cid:54)∈ { v ( x ( h )) , v ( x ( h )) } or (cid:96) (cid:54)∈{ v ( x ( h )) , v ( x ( h )) } , we obtain that I h,k,(cid:96) , does not intersect L . Several properties for the k -dimensional Lyness’ map F have been given, like the existenceof a Lie symmetry for F and of a new and simple invariant for F . This Lie symmetrytogether with the new invariant give information for k = 4 and 5 about the dynamics andthe topology of the level surfaces where the dynamics of F is confined. Some general results19or k odd have been also presented. However, on the contrary that happens in the cases k = 2 and 3, the numerical explorations indicate that for k = 4 , with the same initial condition ,although they are placed in the same manifold, which has dimension k − E ( k +12 ). This isan obstruction to apply the theoretical tools developed in [8].Numerical simulations seem to show that for some initial conditions the projection in R of the iterates of F when k = 6 , R and R . These facts are coherent withthe conjecture of [11] about the number of independent first integrals of the Lyness’ maps,and show that for k ≥ k = 2 (cid:96), the simplest scenario that we imagine for the dynamics of the k -dimensionalLyness’ map is that most of the orbits lie on invariant manifolds which are diffeomorphic to (cid:96) -dimensional tori, S × (cid:96) ) · · · × S . On the other hand, when k = 2 (cid:96) + 1 , most of them lie ontwo diffeomorphic copies of S × (cid:96) ) · · · × S , separated by the invariant set G . Moreover theseorbits jump from one of these tori to the other one and viceversa.In any case, much more research must be done in order to have a total understandingof the dynamics and the geometrical structure of high dimensional Lyness’ maps.
Acknowledgements.
We want to thank Guy Bastien and Marc Rogalski for communi-cating their results of [4] prior to publication. The third author is grateful to ImmaculadaG´alvez for her kind help.GSD-UAB and CoDALab Groups are supported by the Government of Catalonia throughthe SGR program. They are also supported by DGICYT through grants MTM2005-06098-C02-01 (first and second authors) and DPI2005-08-668-C03-1 (third author).
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