Superintegrable models on riemannian surfaces of revolution with integrals of any integer degree (I)

Transcription

1 arxiv: v2 [math-ph] 30 Jun 207 Superintegrable models on riemannian surfaces of revolution with integrals of any integer degree I Galliano VALENT LPMP: Laboratoire de Physique Mathématique de Provence Avenue Marius Jouveau, 3090 Aix en Provence, France Abstract We present a family of superintegrable SI sytems living on a riemannian surface of revolution and which exhibits one linear integral and two integrals of any integer degree larger or equal to 2 in the momenta. When this degree is 2 one recovers a metric due to Koenigs. The local structure of these systems is under control of a linear ordinary differential equation of order n which is homogeneous for even integrals and weakly inhomogeneous for odd integrals. The form of the integrals is explicitly given in the so-called simple case see definition 2. Some globally defined examples are worked out which live either in H 2 or in R 2. MSC 200 numbers: 32C05, 8V99, 37E99, 37K25.

2 Contents Introduction 2 I Integrals of even degree in the momenta 3 2 The local structure of the first integral 4 3 The local structure of the second integral 4 Cascading 4 5 Some globally defined examples 6 II Integrals of odd degree in the momenta 23 6 The local structure of the first integral 23 7 The local structure of the second integral 27 8 Some globally defined examples 29 9 Conclusion 35 Appendices 36 A The cases of a non simple F 36 A. A multiple real zero A.2 Multiple complex conjugate zeroes B Symmetric functions of the roots 40

3 Introduction The possibility of integrable or SI dynamical systems with integrals of degree larger than 4 in the momenta is such a difficult problem that the conjecture that they could not exist was put forward, see for instance [][p. 663]: Conjecture Kozlov and Fomenko On the two dimensional sphere, there are no riemannian metrics whose geodesic flows are integrable by means of an integral of degree n > 4 and do not admit integrals of degree 4. Up to now most of the integrable systems on riemannian surfaces, be explicit or not, do exhibit integrals of degree 4 in the momenta. This apparent barrier opens the challenging question of the existence of integrable systems with integrals of strictly higher degree than 4. The first example of an integrable system, on S 2, with integrals of sixth degree, emerging from astrophysics, was found by Gaffet[6]. His results are described in a more attractive form in [] and in [2]. More recently Tsiganov gave a new example of this kind [2] on the two-sphere. In fact Kiyohara [7] was the first to show the existence of integrable systems with globally defined riemannian metrics having integrals of arbitrary integer degree which, furthermore, are Zoll metrics. The study of SI models, generalizing Kepler and Hooke problems [3], has led also to interesting examples exhibiting integrals of any integer degree. For instance, considering the following generalization of the two dimensional Kepler problem H = 2 p 2 r + p2 φ r 2 + k r + k 2 +k 3 cosλφ r 2 sin 2 λφ λ = m n the SI follows from Bertrand integral f and the complex integrals of degree m+n:. Φ m,n = 2f pr i2f/r +k m 2f pφ sinλφ+i2f cosλφ+k 3 n..2 The linear real span of these integrals is four dimensional, but of course they must be functionally related. The aim of this article is to construct SI systems defined on riemannian surfaces of revolution, starting from the framework laid down by Matveev and Shevchishin [9] for cubic integrals. Their analysis, as we will show later on, can be generalized to integrals of any degree, starting from degree 2. In this last simple case it was proved in [5] that one recovers Koenigs metrics [8] studied and generalized in [0]. Koenigs systems exhibit the following integrals H P y S S 2.3 where H, S, S 2 are of second degree in the momenta. In our generalization the integrals S and S 2 will be of any integer degree 3. Of course these four quantities, as we will show explicitly, are algebraically related. However the explicit form of these metrics is not known. 2

4 Since the analysis required for even and odd degrees integrals display some differences we have divided the article in two Parts. In Part I we consider the case of integrals S and S 2 of degree 2n 2. Their local structure is constructed and shown to be determined by a linear and homogeneous ODE of order n. Globally defined examples are given either on M = H 2 or on M = R 2. In Part II the case of integrals S and S 2 of degree 2n+ 3 is considered. Their local structure is constructed and shown to be determined by a linear and weakly inhomogeneous ODE of order n. Globally defined examples give rise to the same manifolds as in Part I. The last Section is devoted to some concluding remarks. Part I Integrals of even degree in the momenta We will consider the cases for which the observables have for degree in the momenta Q = 2n where n 2. The case n = is marginal since the corresponding SI models were discovered by Koenigs in [8] and generalized in [0]. Taking for hamiltonian we have the obvious result: H = Π 2 +axp 2 y Π = axp x.4 Proposition The system H,P y is integrable in Liouville sense. Proof: We have {H,P y } = 0 and since H and P y are generically independent the proposition follows. This dynamical system will become SI if we can construct at least one more, generically independent integral, of degree 2n in the momenta. To this aim let us first define G = A k H k P 2n k y A n 0 G = 2n n..5 Since A n cannot vanish, we will set A n =. Therefore the function G defines a string of n real constants: A 0, A,,A n. The definition of the two observables S, S 2 by S = Q +yg entail the relations S 2 = Q 2 +yq + y2 2 G Q = Q 2 = 2n,.6 {P y,s } = G {P y,s 2 } = S..7 These observables will become integrals if we impose: 3

5 Proposition 2 The observables S and S 2 are integrals iff {H,Q }+2aP y G = 0 {H,Q 2 }+2aP y Q = 0..8 Proof: The first relation is nothing but {H,S } = 0 and the second one is {H,S 2 } = 0. Let us observe that we get in fact two maximally SI systems which are For further use let us define the sets I = {H, P y, S } & I 2 = {H, P y, S 2 }..9 S p n = {p,p+,...,n} 0 p n as well as the Pochammer symbols z 0 = n z n = zz + z +n. 2 The local structure of the first integral The first step will be Proposition 3 The most general form of Q being Q = b k xπ 2k P 2n k+ y Q = 2n n 2, 2. the constraint {H,S } = 0 is equivalent to the differential system 2 : where Fa = 0 = 2 a b F b k = k + 2 a b k+ Dk af k! b n = Dn a F, n! A k a k. k {, 2,..., n } 2.2 Proof: The first relation in.8 is 2 A prime stands for a derivation with respect to the variable x. {H,Q }+2aP y G =

6 Expanding the left hand side leads to n b k k +/2a b k+ + Dk a F k! + F 2 a b aπ 2k P 2n k+ y apy 2n+ + b n + Dn x F n! aπ 2n P y = 0 which proves the Proposition. To reduce this system to a tractable form we will use now, instead of the coordinate x, the coordinate a. This is legitimate since our considerations are purely local. The hamiltonian becomes H = Π 2 +ap 2 y Π = ȧ x P a. 2.4 Transforming the equations in Proposition 3 gives Proposition 4 The constraint {H,S } = 0 is equivalent to the differential system 3 : a b = 2F ẋ b ḃ k = k + 2 b k+ Dk a F ẋ k! c ḃ n = Dn a F ẋ = ẋ n! k {, 2,..., n } 2.5 Let us notice that we have n+ equations for n unknown functions b k a, k Sn. The definition will be useful: Definition The linear differential operator Op n [F] is defined as Op n [F] = s=0 F n s D n s!/2 a. s 2.6 s The Leibnitz formula gives for it the following property: Op n [FG] = s=0 We are now in position to solve the system for the b k : F n s n s! Op s[g]. 2.7 Proposition 5 Linearizing ODE The local structure of Q is given by Q = b k Π 2k P 2n k+ y, A dot stands for a derivation with respect to the variable a. 5

7 with k S n b k [F] = k s= F k s k s! Da sx = Op /2 k [F]x Fk x 2.9 s k! where x is a solution of the linear and homogeneous ODE of order n 2: Op n [F]x = Proof: To determine the functions b k we need to solve the differential system 2.5: we will proceed recursively. The relation 2.5a gives b = F D ax /2 in agreement with 2.9 for k =. Let it be supposed that the relation for b k in 2.9 is true for any k < n and let us prove that it is also true for k + n. To this aim we will start from relation 2.5b written k +/2b k+ = ḃk + Dk a F k! ẋ = k s= Da k+ s F k s! In the second sum let us shift s s = s+ so that k Da k+ s F Da s k +/2b k+ = x k+ + k s! /2 s= s s = The relations give for final result = F Dk+ a x + /2 k k [ s= + k + s /2 s /2 s D s a x /2 s + k s=0 D k+ s a F k + s! Da k s F k s! D s a x /2 s ] D k+ s a F k + s! Ds ax. Da s+ x. /2 s + k + s = k +/2 /2 k+ = k +/2/2 k /2 s /2 s /2 s b k+ = k+ s= Da k+ s F k + s! Da sx /2 s which concludes the recurrence proof of the relation 2.9 for b k. We are left with relation 2.5c which integrates up to b n = A n x x 0 and this last relation must agree with the b n obtained from 2.5b for k = n. This produces the linear ODE s= Da n s F n s! Da sx + Dn a F /2 s n! x x 0 = 0. We can set x 0 = 0 because the metric does depend solely on ẋ and this proves 2.0. Remark: Let us observe that the relation 2.9 is valid for any choice of F. Let us define: 6

8 Definition 2 We will say that F is simple if all of its zeroes are simple, with symbol F: n Fa A k a k = a a i A n R\{0}. Then we have Proposition 6 For a simple F, if one takes x = i = ǫ i a a i ǫ 2 i = a i R 2. where the are n arbitrary real parameters, the relation /2 i F Op n [ F]x = Da n n! implies that x is the general solution of the ODE 2.0. Proof: The linearity of the ODE allows to check term by term for which Hence so that using i S n x i = /2 i i = ǫ i a a i ǫ 2 i =, s N Op n [ F]x i = /2 i s N and Leibnitz formula we end up with i 2.2 D s ax i /2 s = ǫ i s s /2 i. 2.3 s=0 Da n s F n s! ǫ i s s ǫ i s s i Op n [ F]x i = /2 i D n a n! = s! Ds a i Adding all the terms gives 2.2. The proposition follows since each term in this sum does vanish. Remarks:. The previous formula is in fact valid for which does vanish only for k = n. F k S 0 n : Op k[ F]x i = /2 i 7 i D k a k!. F i, 2.4

9 2. The forms taken by xa when F is not simple are given in Appendix A. As a bonus the metrics of constant scalar curvature are excluded: Proposition 7 The metric g = ẋ2 a 2 da2 + dy2 2.5 a is never of constant curvature if xa is a solution of the ODE 2.0. An embedding in R 2, is given by g = dx 2 +dy 2 dz where X = y a Y Z = a Proof: We have seen that the scalar curvature is Y +Z = y2 +2 a ẋ 2 a da R = 2aẍ+ẋ ẋ so if we take R to be a constant, defining u = we have to solve ẋ2 a u u = 2R = u = Ka 2R. According to the value of the integration constant K, and omitting an additive constant, we have K = 0 = x = ± a 2R, K 0 = x = ± 2 K Ka 2R and both functions are never solutions of 2.0. The embedding formulas are easily checked. For the next step we need Lemma One has the following identity k l : l s=k s l s!s k! = k δ kl. 2.9 Proof: For k = l this sum is just k. For k < l defining N = l k and t = s k we have l s N l s!s k! = k+t N t!n t! = k N t = 0 N! t s=k t=0 by the binomial theorem. Let us first compute the coefficients b k : 8 t=0

10 Proposition 8 For a simple F we have Q = b k [ F]Π 2k Py 2n k+, 2.20 with k S n Proof: Relation 2.9 b k [ F] = k s= and the identity 2.3 give first b k [ F] = b k [ F] = F k s k s! The change of index s = t+ gives Using the identity b k [ F] = ǫ i ξ i Da k F. 2.2 k! D s a x /2 s x = ξ i k s= ǫ i ξ i k t=0 ǫ i t t i F k s k s! ǫ i s s i. F k t i k t! ǫ i t t i. = Dt a i t! and Leibnitz formula, we are led to 2.2. Using this form of Q it is rather difficult to obtain Q 2. To solve this problem we need to transform Q according to Q = b k [F]Π 2k P 2n k+ y = Let us determine the new coefficients b k : Proposition 9 In general we have and in the simple case 4 k S n : bk [F] = k S n : 4 The symbol σk i is defined in Appendix B. k s= bk [F]H n k ΠP 2k y n s a k s b n s+ [F] 2.23 k s bk [ F] = k 9 σ i k. 2.24

11 Proof: If, inthefirst formofq, oneuses Π 2 = H apy 2 andinterchanges thesummations order, one gets the relation Using formula 2.2 we have b n+ s [ F] = ǫ i ξ i F Da n s s Sn n s! a a. i Expanding the term inside the bracket using relation B.3 one has b n+ s [ F] = n s! s l σl i and inserting this formula in the definition, given above, of b k we get bk [ F] = k+ k s= s l= l= l +s k s! n l! n k! Reversing the first and the second summations we end up with bk [ F] = k+ k l σl i l= ak l n l! n k! n l! s l! as l, k s=l a k l s l! σi l. s k s!s l! and the identity 2.9 concludes the proof. It is interesting, in order to check the result obtained in Proposition 9, to write down the differential system for the b k. We have: Proposition 0 Defining for any choice of F the relations k S n : bk [F] = k β k [F], 2.25 β = ẋ k n : βk+ = a β k 2 β k +σ k ẋ 2.26 k = n : 0 = a β n 2 β n +σ n ẋ, ensure the conservation of S. For a simple F the formula obtained for b k [ F] in Proposition 9 is indeed a solution of this system. Proof: A routine computation leads to 0 = {H,Q }+2aP y G = 2ȧ x { n + k βk+ H n k P 2k+ y a β k k + 2 β k k A n k ẋ H n k Py 2k+ } 0

12 which gives 2.26 using the relation B.2. Let us now check that the formula k S n β k [ F] = proved in Proposition 9 does solve this system. For k = we have β = σ0 i = i For k S 2 n, using the relation B.4, we have a β k 2 β k +σ k ẋ = ǫ i 2 3/2 i σ i k = x. a i σ i k σ k = ǫ i σ 2 3/2 k i. i If k {,2,...,n } we do recover β k+, while for k = n the result vanishes. Remarks:. Using symbolic computation we checked the conservation of S = Q +yg using for Q its form 2.24 for n = 2 and n = An interesting exercise, left for the reader, is to derive the differential system 2.26 from the relation 2.23 and the differential system for the coefficients b k. As we will see now this last form of Q will allow for a simple construction of Q 2. 3 The local structure of the second integral Let us proceed with Q 2. From Proposition 2 we need to solve Let us prove {H,Q 2 }+2aP y Q = 0. Proposition For a simple F, the observable is given by k S n c k [ F] = k+ 2 Q 2 = c k [ F]H n k P 2k y 3. ξi 2 σk i + i i j= ξ j σ ij k k 2 +aσij 3.2 j where the σ ij k are defined in Appendix B.

13 Proof: An elementary computation gives {H,Q 2 }+2aP y Q = 2aΠ Since we are working locally, this is equivalent to or explicitly Expanding into D a c k = k 2 and integrating up to 5 c k = k 2 k S n D a c k = k 2 The second piece can be written 2 i>j= Using the relation B.4 gives ξ j j Da c k ẋ + b k H n k P 2k y. D a c k = b k ẋ σ i k j= ǫ i ξi 2 a a i 2 σi k + ξ 2 i i σ i k +2 i j= i j= ǫ j ξ j j 3/2. ξ j ǫ j j 3/2 σi k ξ j a a i a i a j σk i. i j σk i a a i σ j k a a j. a i a j σ i k a a i σ j k a a j = aσ i k σj k +σi k σj k and thanks to B.8 we obtain 3.2. Remark: For k = we have D a c = b ẋ = xẋ which integrates up to c = x2. It does 2 agree with formula 3.2 since we have σ ij 0 = and σij = 0. Let us add an important algebraic relation: Proposition 2 The integrals S and S 2 are algebraically related by S 2 2GS 2 = A 2 n Q kl H 2n k l Py 2k+l 3.3 where k,l= Q kl = k+l+ ǫ i ξi 2 σi k σi l The integration constants can be omitted since they would add trivially conserved terms. 2

14 Proof: We have first X S 2 2GS 2 = Q 2 2GQ 2. Expanding this expression in powers of H and of P y and upon use of the identities B.4 and B.0 leads, after some hairy computations to the given formula. Summarizing the results obtained up to now we have: Theorem The hamiltonian H = Π 2 +ap 2 y Π = ȧ x P a a > for a simple F = F and x = i = ǫ i a a i ǫ 2 i = R, 3.6 exhibits two integrals where and G = k S n : S = Q +yg A n k H n k P 2k y Q = bk = k c k = k+ 2 σ i k S 2 = Q 2 +yq + y2 2 G 3.7 bk H n k ΠP 2k y Q 2 = ξi 2 σk i + i i j= These two integrals generate two maximally SI systems: c k H n k P 2k y 3.8 ξ j σ ij k k 2 +aσij. j 3.9 I = {H, P y, S } and I 2 = {H, P y, S 2 }. 3.0 Proof: We just need to check the functional independence of the integrals. Let us define We have J = dh dp y ds J 2 = dh dp y ds 2. J = dh dp y ds = ydh dp y dq +Q H P x dp x dp y dy which cannot vanish everywhere due to the last term. Differentiating the identity 3.3 we have ds 2 2GdS 2 = 2dGS 2 +dxh,p y which implies that does not vanish everywhere. 2GJ 2 = dh dp y 2GdS 2 = 2S J 3

15 4 Cascading Let us first introduce some notations which make explicit the degree 2n of the integrals. Notice that we have to bring back the coefficient A n which was taken to be. We have first G n = A k H k P 2n k y F n = n A k a k = A n a a l 4. and similarly for the symmetric functions of the roots: l= The hamiltonian is while the integrals are σ n k H n = Π n 2 +ap 2 y σ in k σ ijn k Π n = a ẋ n P a 4.3 with S n = Q n +yg n S n 2 = Q n 2 +yq n + y2 2 Gn 4.4 Q n = We have the relations bn k H n k ΠPy 2k n lim A Fn = F n = F n = A n a a k n 0 It follows from the ODE for xa that x n = n x n = Q n 2 = c n k H n k P 2k y. 4.5 lim A Gn = Py 2 Gn. 4.6 n 0 = H n H n. 4.7 Hence to take the limit properly we have to let A n 0 and ξ n 0. Let us denote this limit by the symbol LIM. We will now prove: Proposition 3 In the limit A n 0 and ξ n 0 the integrals Q n and Q n 2 become reducible according to the relations: LIM Q n = P 2 y Q n LIM Q n 2 = P 2 y Q n Factoring out by Py 2 systems: the integrals, we have exhibited the cascading between the following SI {H n, P y, Q n } {Hn, P y, Q n } {H n, P y, Q n 2 } {H n, P y, Q n 2 }

16 Proof: From Theorem we have LIM Q n = LIM n b k H n n k Py 2k. The first term in the sum vanishes: bn = A n = LIM b n = 0. Substituting l = k in the remaining sum we get LIM Q n = P 2 y n l= LIM n b l+ H n n l Py 2l bn l+ = l+ A n σ in l. Using relation B.5 in Appendix B we have lim A nσ in l = l A n 0 which leads to l s= l a l s i A n s = l a l t i A n t = A n σ in l 4.0 t=0 LIM n b l+ = n l A n σ in l = b n l = LIM Q n = P 2 y Qn. From Theorem we have also LIM Q n 2 = LIM c n k H n n k P 2k y. The first term vanishes again: c n = A n 2 i + i j= ξ j j = LIM c n = 0. Substituting l = k in the remaining sum we get LIM Q n 2 = P 2 y n LIM c n l+ l= H n n l P 2l y with c n l l+ a = 2 ξ 2 i A n σ in l i + i j= 5 ξ j A n σ ijn l j +aa n σ ijn l 4.

17 Let use relation B.8 A n σ ijn l which, combined with 4.0, leads to = A nσ in l+ A nσ jn l+ a i a j lim A nσ ijn l A n 0 = A n σ in l A n σ jn l a i a j = A n σ ijn l. Plugging this last relation as well as 4.0 into 4. we get LIM c n l+ = cn l = LIM Q n 2 = P 2 y Q n 2 which concludes the proof. Remarks:. The cascading process reduces the degree of the integrals from 2n to 2n. 2. Let us start from the iff equations.8 for integrals of degree 2n {H,Q n }+2aP y G n = 0 {H,Q n 2 }+2aP y Q n = 0. The cascading substitutions Q n Py 2 Q n and Q n 2 Py 2 Q n 2 in these relations and factoring by Py 2 lead to iff equations for integrals of degree 2n. 3. We have given a direct proof of the cascading phenomenon based on the explicit form of the b n k and c n k. This constitutes a check of the formulas obtained above for these coefficients when F is simple. 5 Some globally defined examples A look at the metric shows that ẋ 2 g = da 2 + a a dy2 = dx2 a + 2 a dy2 5. Proposition 4 The metric is riemannian iff a > 0. In general a will take values in some interval a m, a M with a m > 0. The end-points of this interval may be of two kinds:. True singularities, namely curvature singularities, which cannot be disposed of by some coordinates change and which do prevent the metric to be defined on a manifold. They can be detected from the behaviour in a neighbourhood of these points of the scalar curvature given by

18 2. Apparent singularities, as for instance g dχ 2 +χ 2 dy 2 χ 0+ y S which can be removed using cartesian coordinates Let us prove x = χ cosy y = χ siny = g dx 2 +dy 2 Proposition 5 We have the following possibilities: a If ẋa a a α the point a = a 0 is a curvature singularity if α >. b If ẋa a α then a = 0+ is a curvature singularity if α > 0. c If ẋa a α then a + is a curvature singularity if α, /2 /2,0. Proof: In the case a the computation of the curvature gives which proves the statement. In the case b we have 2R 2a α a a 2α 2R 2α+ a 2α which is not continuous for a 0+ if α > 0. In the case c the same formula holds but now a + and the curvature must remain bounded, which is excluded iff α < 0 and α /2. The case α = /2 needs a specific analysis for each metric. As a first check let us consider the simplest case n = for which the integrals are merely quadratic: we should recover one of the Koenigs metrics [5]. Proposition 6 For n =, i. e. quadratic integrals, there is a single SI metric, globally defined g. d. on H 2 : it is the Koenigs metric of type 3. Proof: Here we have Fa = a a so we need to order the discussion according to the values taken by a. If a > 0, we may take a = and ξ =. Hence we have xa = a+ a > 0, and by Proposition 3 this metric is singular for a +. Indeed the coordinate change t = a a+ = g = t 2 dt2 +dy 2 t 2 t 0, y R 7

19 shows that t i.e. a + is a curvature singularity of the metric because the conformal factor does vanish. The same argument applies if a = 0. If a < 0, up to scalings, we may take a = and ξ =. A first possible case is x = a a,+ and here too Proposition 3 shows that this metric is singular for a +. The last possible case is x = a 0, = g = da 2 a a 4a a 3 +dy2 a The coordinate change u = shows that a g = +u 2 du2 +dy 2 u 2 = +u 2 g 0 H 2,P u > 0 y R where g 0 is the Poincaré half-plane model of H 2. The resulting hamiltonian H = u2 P 2 +u 2 u +Py 2 is nothing but Koenigs metric of type 3 as it is written in Theorem 7 of [5] setting ξ = 0 which was shown to be globally defined on the manifold M = H 2. This Koenigs metric of type 3 suggests the following generalization to SI systems with integrals of any even degree larger than 4: Proposition 7 Let us consider, for n 2 with Fa = a Fa a i < 0 a i > The associated SI system produces the metric where with n Fa = a a k 0 < a < 5.2 i=2 i = 2,...n. g = +u 2 µ2 udu 2 +dy 2 u 2 u 0,+ y R, 5.3 µu = c+ i=2 +ρ i u 2 3/2 5.4 c > 0 > 0 ρ i 0,,+. This metric and the related integrals S, S 2 are globally defined on M = H 2. 8.

20 Proof: The choice made for Fa and Proposition 6 imply that we may take xa = c a It follows that, under the coordinate change i=2 ǫ i ǫ i a i 3/2 ǫi a a i. 5.5 u = a a a 0, u 0,+, = dx a = µudu where µ is given in 5.4. Under this substitution the metric takes the form given in the formula 5.3. The parameters ρ i which appear are given by Defining the coordinate change ρ i = a i = ρ i 0,,+. t = uωu Ωu = c+ i=2 +ρi u 2 : u 0,+ t 0,+, since dt du = µ > 0 it follows that the inverse function ut is increasing and C [0,+. The metric becomes g = +u 2 tω 2 t dt2 +dy 2 t 2 = +u 2 tω 2 t g 0 H 2,P and since the conformal factor +u 2 tω 2 t never vanishes we conclude that M = H 2. We have seen that the integrals are S = Q +yg S 2 = Q 2 +yq + y2 2 G G = A n H k P 2n k y. 5.6 The global structure of these integrals is easy to study because H, P y, Π, hence G, are globally defined on M. Let us consider Q = bk H n k ΠP 2k y c σ bk = k k a i I ǫ i ǫ i a i 3/2 σk i which becomes in terms of the coordinate t k Sn : bk = k +u 2 t cσk + i I a i σ i k +ρi u 2 t, showingthatallthesecoefficientsarec [0,+ sothatq hences aregloballydefined on M = H 2. 9

21 Let us consider Q 2. We have and the coefficients become k S n : with Q 2 = c k = k +u 2 t 2 + i I c k H n k P 2k y c 2 σ k +2c i I a 2 i ξi 2 +ρ i u 2 t σi k + i j I a i +ρi u 2 t τi k + a i a j ξ j +ρi u 2 t+ρ j u 2 t τij k τ ij k = σij k + u2 +u 2 σij k 2. From this formula it follows that Q 2 hence S 2 are globally defined on M = H 2. As a second example we have: Proposition 8 For n 2 the choice with 6 Fa = a a a a 2 Fa 0 < a < a < a 2 n Fa = a a i : i=3 leads to a SI system with the metric a i < a a i > a 2 i = 3,...n g = At dt2 +dy 2 t,y R globally defined on the manifold M = R 2 as well as the integrals S and S 2. Proof: Let us consider 7 with xa = ξ a a + ξ 2 a2 a i=3 ǫ i ξi ǫi a a i ξ = a a2 a ξ ξ2 = a 2 a2 +a ξ 2 ξi = a i a2 a and ǫ i = + a i < a & ǫ i = a i > a 2. 6 If n = 2 we have Fa =. 7 For n = 2 the last sum is absent. 20

22 The coordinate change a = a +a 2 a s 2 s sinθ : a a,a 2 θ 0,π/2 gives xθ = ξ a s So differentiating we get + ξ 2a 2 s 2 These relations show that D θ x > 0. Defining dt = dx a one gets i=3 ǫ i a i ǫi ρ i +s 2 a c D θ x = ξ s +ξ a 2 s 2 2 s 2 + ts = a +a 2 a s 2 i=3 a i sc ρ i +s 2 3/2 ξ s + ξ 2 s 2 i=3 ρ i = a a i a 2 a. ǫ i. ρi +s 2 It follows that θ 0,π/2 is mapped into t R and that the function t = hs is a C increasing bijection from θ 0,π/2 R, hence its inverse function s = h t is also a C increasing bijection. We obtain the metric given by 5.7: g = At dt2 +dy 2 A = a h t,y R 2 and since the conformal factor At never vanishes, we conclude that the manifold is M = R 2. As in the previous case considered above, let us consider with the coefficients Q = bk = k a ξ σ k At bk H n k ΠP 2k y + a 2ξ 2 σk 2 A2 t ǫ i σk i ǫi ρ i +A 2 t Since θ 0,π/2 this proves that Q is indeed globally defined. The check for Q 2 is similar. Let us conclude with the following negative result: Proposition 9 The choice F = a 2 + a 2 0 n for n =,2,... never leads to a globally defined SI system. 2 i=3.

23 Proof: In this case, setting a 0 =, we have xθ = µ + k cosθk /2 cosk 3/2θ+µ k cosθk /2 sink 3/2θ. with a = tanθ and θ 0, π. Let us consider the metric behavior for θ π for a fixed 2 2 value of k which can always be obtained by an appropriate choice of the coefficients µ ± k. We have the following equivalent: g c 2 k cosθ2k dθ 2 +cosθdy 2. Using the coordinate change v = v 0 sinθ 2k+/4 for an appropriate constant v 0 leads to g c 2 k dv 2 +v 2/2k+ dy 2 v 0+ k {,2,...,n} where Y is merely homothetic to y. Even restricting Y S and no matter one chooses k the exponent of v will never be equal to 2 so we have a true singularity for v

24 Part II Integrals of odd degree in the momenta We will consider the cases for which the observables have for degree in the momenta Q = 2n+ where n. The case n = was first analyzed in [9] and [3]. The hamiltonian remains unchanged H = Π 2 +ap 2 y Π = ȧ x P a, 5.8 while G = A k H k P 2n k+ y A n 0 G = 2n+ n S n 5.9 is still built up from the n constants: A 0, A,,A n. The SI stems from the two observables S = Q +yg S 2 = Q 2 +yq + y2 G Let us begin with the determination of Q. 6 The local structure of the first integral Proposition 20 Taking Q = the observable S will be an integral iff a b 0 = 2F ẋ b ḃ k = k + 2 b k Dk af k! c ḃ n = 0 b k aπ 2k+ P 2n k y, 6. ẋ k =,...,n 6.2 where Fa = A k a k. Proof: The equation {H,S } = {H,Q }+2aP y G = 0 23

25 expands into n+ ab k Π 2k P 2n k+ y k +/2a b k aπ 2k P 2n k+ y + giving the differential system 0 = 2 a b 0 Fa b k = k + 2 a b k Dk af k! b n = 0 Switching to the new variable a, instead of x, gives 6.2. We can proceed to D k a F k! k =,...,n aπ 2k P 2n k+ y = 0 Proposition 2 Linearizing ODE The differential system 6.2 has for unique solution k S 0 n \{n} : b k = k+ s= Da k s+ F k s+! D s ax /2 s b n = const = ν n R\{0} 6.3 where xa is a solution of the ODE Op n [F]xa = n+ ν n a+β n 2 β n R. 6.4 Its solution in the simple case, up to an additive constant, is given by x = ν n 2 a+ ξ i i = ǫ i a a i ǫ 2 i = R. 6.5 Proof: The recursive proof giving b k for any k S 0 n is similar to the one given for Proposition 5. Integrating equation b in 6.2 for k = n leads to which we combine with b n + Dn a F n! b n = s= x = n+/2ν n a+β n D a n s F /2 s n s! Ds ax to get 6.4. The homogeneous equation was already solved for in Proposition 6 and looking for an affine solution gives ν n 2 a+β n ν n A n in which the constant term may be deleted. Remarks: 24

26 . The transition from integrals of degree 2n to 2n+ is strikingly simple: one just adds a linear term in x! This was observed in [3] for the cubic case but was not expected to be so general. 2. In Proposition 7 we have seen that if x = ± a 2R the metric is of constant negative curvature. In order to avoid such a case we must exclude the trivial possibility that all the be vanishing. Let us conclude this section by giving a useful form of Q : Proposition 22 For a generic choice of F one can write Q = bk [F]ΠH n k P 2k y 6.6 with For a simple F we have k S 0 n : bk [F] = k s=0 k S 0 n : bk [ F] = k ν n σ k + n s a n k b n s [F]. 6.7 k s σ i k. 6.8 Proof: By the same argument used in Part I for Q one gets relation 6.7. For a simple F we have bk [ F] = where k s=0 n s n a k s b n s [ F] = a k ν n + n k n k D n s n s+ a F b n s [ F] = ν n n s! + l= k s= Da n s+ l F /2 l n s+ l! Dl ax 0 n s a k s b n s [ F] n k where x 0 is just the dependent part of x. So we have to compute two pieces: bk [ F] = ν n k s=0 n s a n k F k n s! + k s Dn s a s=0 n s a k s b n s+ even n k where b n s+ even is the same as in the proof of the Proposition 5 and therefore gives the same result as in Proposition 9: k 25 σ i k.

27 The first piece gives l=n k ν n k s=0 l=n s n s l A l a k s k s a k+l n. n k n s Reversing the summations we conclude to ν n A l a k+l n l! k s k n k! k s!l+s n! = ν na n k = k ν n σ k s=n l and use of the identity B.2 concludes the proof. As in Part I, let us check the result obtained for b k [ F] using its differential system. We have Proposition 23 Defining for any choice of F the relations k S 0 n : bk [F] = k β k [F], 6.9 β 0 = 0 0 k n : βk+ = a β k 2 β k +σ k ẋ 6.0 k = n : 0 = a β n 2 β n +σ n ẋ imply that S is an integral. For a simple F the formula obtained for b k [ F] in Proposition 22 is indeed a solution of this system. Proof: a routine computation leads to 0 = {H,Q }+2aP y G = ȧ x 2 β 0 H n+ ȧ n k 2 x β k+ H n k Py 2k+ + ȧ β k k 2a β k +2 k A n k ẋ H n k Py 2k+ x from which we deduce 6.0 using the identity B.2. Letuscheck thatforasimple F therelation6.8doesgiveasolutionofthisdifferential system. We have first β 0 = ν n which is fine. Then computing a β k 2 β k +σ k ẋ = 2 ǫ i a 3/2 i σk i σ k = 2 i ǫ i σ 3/2 k i i So for 0 k n we get β k+ while for k = n it indeed vanishes. In fact there is a simple structural relation between the x, β k [F] with k = 0,,...,n for odd degree integrals and x =, β =k [F] with k =,...,n for even degree integrals given by: 26

28 Proposition 24 The relations x a = 2 νa+x= a β 0 [F] = ν k S n : β k [F] = νσ k +β = k [F] 6. ensure that the equations 6.0 for the β k [F] imply the equations 2.26 for the β= k [F]. Proof: The first relation in 6. follows from Propositions 6 and 2. For k = 0 we have For k n we have β [F] = β = [F] = ν 2 ν +σ 0 2 +ẋ= = ẋ =. β k+ [F] = a β k [F] 2 β [F]+σ k ẋ = a β k = [F] 2 νσ k +β =k [F] +σ k ν 2 +ẋ= which is indeed equal to β k+ = [F]. The argument for k = n is similar. Now that Q is fixed up let us construct Q 2. 7 The local structure of the second integral As shown in Proposition, the structure of Q 2 follows from: {H,Q 2 }+2aP y Q = 0 Q 2 = c k [F]H n k P 2k+ y 7. and it is given by: Proposition 25 The observable S 2 is an integral iff Q 2 is determined from the differential system k S 0 n : D a c k = b k ẋ. 7.2 For a simple F these coefficients are given by { k Sn 0 : c k[ F] = k+ νn 2 2 aσ k +2ν n + ξi 2 σk i + i σ i k +aσi k + i j= ξ j σ ij k k 2 } +aσij. j 7.3 Proof: An elementary computation gives {H,Q 2 }+2aP y Q = 2aΠ Da c k ẋ + b k H n k P 2k+ y 27

29 which implies 7.2. In the computation of D a c k there appears the constant term while the terms linear in the give and an integration yields k+ ν n k+ ν n 2 k+ν2 n 2 σ k ǫ i + 3/2 i σk i 2 i σ k +a a i σk i = k+ ν n σk i +aσk i after use of B.4. The remaining terms are quadratic in the momenta and are easily seen to be the same as in Proposition. Let us add an important algebraic relation: Proposition 26 The integrals S and S 2 are algebraically related by S 2 2GS 2 = k,l= Q kl H 2n k l P 2k+l+ y where Q kl was already defined in νn 2 k,l=0 k+l σ k σ l H 2n+ k l Py 2k+l 7.4 Proof: Denoting the quantities defined in the Part by a sharp subscript, we have which implies G = P y G Q = A+P y Q Q 2 = B +P y Q 2 X S 2 2GS 2 = Q 2 2GQ 2 = A 2 +2AP y Q 2BP y G +P 2 y X. The first three terms, after several algebraic simplifications give the required formula while the last term is obvious. Summarizing the results obtained up to now we have Theorem 2 For a simple F, the hamiltonian H = Π 2 +ap 2 y Π = ȧ x P a a > where x = ν n 2 a+ i = ǫ i a a i ǫ 2 i =,

30 with all the R and A n R\{0}, exhibits two integrals where with and G = S = Q +yg A n k H n k P 2k+ y Q = k S 0 n : c k[ F] = k+ 2 S 2 = Q 2 +yq + y2 2 G 7.7 bk H n k ΠP 2k y Q 2 = k S 0 n : bk [ F] = k ν n σ k + { ν 2 n aσ k +2ν n + ξ 2 i σ i k i + σk i σ i k +aσi k + i j= These two integrals generate two possible maximally SI systems: c k H n k P 2k+ y 7.8 ξ j σ ij k k 2 } +aσij. j I = {H, P y, S } and I 2 = {H, P y, S 2 }. 7. Proof: The functional independence proof is the same as for Theorem. 8 Some globally defined examples Let us begin with the case n = cubic integrals for which the global structure was first analyzed in [3]. To compare most conveniently with our results let us first transform our metric g = ẋ2 a 2 da2 + dy2 a > 0 8. a under the coordinate change u = a. We get So, for Fa = a a we may take xa = ν 2 a+ ǫc ǫa a g = µ2 du 2 +dy 2 u 2 µ = 2ẋa = u = µ = ν c ǫu 2 a 3/2 ǫ = ± ν and since ν cannot vanish we can set ν = showing that our local form of the metric is in perfect agreement with the local form given in [3]. It may be noticed that the single 29

31 difference in the function µ with respect to the Koenigs case quadratic integrals is just this constant ν. However, as opposed to the Koenigs metrics, this form of µ allows for a much larger number of globally defined cases. Let us prove: Proposition 27 The SI systems having the metric g = µ2 udu 2 +dy 2 u 2 µu = c [ǫu 2 a ] 3/2 8.4 are globally defined on M = H 2 in the following cases: I ++ : µu = u 2 a 3/2 a, I + : µu = + u 2 a 3/2 a,0 I + : µu = a u 2 3/2 a 0, I : c µu = u 2 a 3/2 a 0,+ 8.5 Proof: The scalar curvature being R g = 2 µ+uµ µ 3 any singularity of it implies that the metric cannot be defined on any manifold. In the case I ++, for a 0, µ vanishes for u 0 = a + > a leading to a curvature singularity. For a < 0 we have u > 0 and for a the curvature is again singular for u 0 = a +. It remains to consider a <. Defining dt = µdu = t = uωu Ωu = a u 2 + a. Since µ = dt du never vanishes the inverse function ut is C [0,+. The metric is now G = Ω 2 ut dt2 +dy 2 t 2 = Ω 2 utg 0 H 2,P and since Ω[0,+ = [ / a 3/2,, the conformal factor never vanishes showing that M = H 2. In the case I +, for a > 0 we have u > a so that defining dt = µdu = t = uωu Ωu = a u2 a 30

32 but this time Ωu vanishes for u 0 = a +/a 2 > a. Since Ωut appears as a conformal factor in the metric there can be no manifold. For a = 0 we have u 0,+. In this last case, defining v = /3u 3 the metric is G du2 u 8 + dy2 u 2 = dv2 +3v 2/3 dy 2 showing that u 0+ precludes any manifold. For a < 0, hence u > 0, the change of coordinate implies for the metric dt = µdu = t = uωu Ωu = + a u 2 + a G = Ω 2 ut dt2 +dy 2 t 2 = Ω 2 utg 0 H 2,P, where ut is C [0,+ and Ω[0,+ =,+/ a 3/2 ] hence the manifold is again H 2. For ǫ = we must have a > 0 and u 0, a. In the case I +, for a, there is a curvature singularity for u 0 = a while for 0 < a < the function µu never vanishes. Let us define dt = µdu = t = uωu Ωu = a a u 2. The function Ωu is strictly decreasing with Ω[0, a =, /a 3/2 ] and the metric becomes G = Ω 2 ut dt2 +dy 2 = Ω 2 utg 0 H 2,P. Since Ωut never vanishes we get M = H 2. In the last case I, we can define t 2 dt = µdu = t = uωu Ωu = + a a u 2 where Ωu is strictly increasing with Ω[0, a = [ + /a 3/2, + and the metric becomes G = Ω 2 ut dt2 +dy 2 = Ω 2 utg 0 H 2,P. t 2 Since Ωut never vanishes we get M = H 2. Forthe proof that the integrals are also globally defined on H 2 the arguments presented in [3] do apply and need not be repeated. Remark: Our analysis does correct the Propositions 6 and 7 of the reference [3] but is in agreement with its Proposition 8. Let us generalize the previous SI systems: 3

33 Proposition 28 The SI systems I +±, corresponding to Fa = a a, equipped with n cubic integrals, do generalize to Fa = a a i with integrals of degree 2n+ with n 2 which remains globally defined on M = H 2 under the following restrictions I ++ : < a i < a < & > 0 & I + : < a i < a < 0 & > 0. Proof: The first SI system, above I ++, is generated by a + <, 3/2 a i 3/2 i=2 8.6 xa = a 2 a a + i=2 a ai = µu = u 2 a 3/2 u 2 a i 3/2. i=2 The metric has the form 8.2. Noting that µ is strictly increasing from µ0 = a > 0 µ+ =. 3/2 a i 3/2 i=2 Let us define dt = µudu = t = uωu, Ωu = a u 2 a i=2 a i u 2 a i. Since D u t > 0 the inverse function ut is C [0,+ and Ωu is strictly increasing with Ω[0, + = [µ0, and therefore never vanishes. The metric becomes showing that M = H 2. Here we have with g = Ω 2 ut dt2 +dy 2 Q = t 2 = Ω 2 utg 0 H 2,P bk H n k ΠP 2k y bk = k 2 σ σk k + u2 t a i=2 σk i u2 t a i which are indeed C [0,+ hence Q is globally defined on M. The check for Q 2 is similar. The second SI system, above I +, is generated by xa = a 2 a a i=2 a ai = µu = + 32 u 2 a + 3/2 i=2 u 2 a i 3/2.

34 The function µ is decreasing from µ0 = + a + µ+ = 3/2 a i 3/2 i=2 hence it never vanishes. So we can define t = uωu Ωu = + a + u 2 a i=2 a i u 2 a i. ItfollowsthatΩudecreases fromω0 = µ0toω+ = andnever vanishes, showing by the same argument displayed above that M = H 2. The checks that Q and Q 2 are globally defined are again elementary. Let us add: Proposition 29 The SI systems I ±, corresponding to Fa = a a, equipped with n cubic integrals, do generalize to Fa = a a i with integrals of degree 2n+ with n 2 which remains globally defined on M = H 2 under the following restrictions I + : 0 < a < +, a i > & > 0 & I : 0 < a < +, a i > & > 0. Proof: The first system, above I +, is generated by xa = a 2 a a ξ i ai a i=2 The function µ is decreasing from So we can define µ0 = a 3/2 t = uωu i=2 a 3/2 i = µu = a + >, 3/2 a i 3/2 i=2 8.7 a u 2 3/2 a i u 2 3/2. i=2 < 0 µ a =. Ωu = a u2 a i=2 a i u2 a i. It follows that Ωu decreases from Ω0 = µ0 < 0 to Ω a = and never vanishes, showing by the same argument given above that M = H 2. The second system, above I +, is generated by xa = a 2 + a a + ξ i ai a i=2 = µu = + 33 a u 2 + 3/2 a i u 2 3/2. i=2

35 The function µ is increasing from So we can define µ0 = + a 3/2 + i=2 a 3/2 i < 0 µ a = +. t = uωu Ωu = + a a u + 2 i=2 a i ai u 2. It follows that Ωu increases from Ω0 = µ0 > 0 to Ω a = + and never vanishes, showing by the same argument as above that M = H 2. The checks that the integrals are globally defined are again elementary. Let us give another example which is a close cousin of the system considered in Proposition 8: Proposition 30 For n 2 the choice with 8 Fa = a a a a 2 Fa 0 < a < a < a 2 n Fa = a a i : i=3 leads to a SI system with the metric a i < a a i > a 2 i = 3,...n g = At dt2 +dy 2 t,y R globally defined on the manifold M = R 2 as well as the integrals S and S 2. Proof: Let us consider 9 with and xa = ν 2 a ξ a a + ξ 2 a2 a i=3 ǫ i ξi ǫi a a i ξ = a 2 a a ξ ξ2 = a 2 +a a 2 ξ 2 ξi = a 2 a a i The coordinate change ǫ i = + a i < a & ǫ i = a i > a 2. a = a +a 2 a s 2 s sinθ : a a,a 2 θ 0,π/2 8 If n = 2 we have Fa =. 9 For n = 2 the last sum is absent. 34

36 gives xθ = ν 2 a 2 a s 2 ξ a s So differentiating we get + ξ 2a 2 s 2 i=3 ǫ i a i ǫi ρ i +s 2 ρ i = a a i a 2 a. a c D θ x = νa 2 a sc+ξ s +ξ a 2 s 2 2 s 2 + This relation show that D θ x > 0. Defining dt = dx a one gets ts = a +a 2 a s 2 i=3 ν ξ s + ξ 2 s 2 i=3 a i sc ρ i +s 2 3/2 ǫ i. ρi +s 2 Fromnow onthe prooffollows exactly the same steps as in the proofof Proposition 8. 9 Conclusion Needless to say a lot of work is still necessary in the study of the models constructed here. Let us just mention a few items:. Determine the integrals for the non simple cases. 2. Find more globally defined systems with H 2 or R 2 for manifolds. It is possible that a general proof can be given that in this class of models called affine case in [3] the possible manifolds are restricted to H 2 or R When the integrals are quadratic in the momenta we are back to a metric due to Koenigs. In this case it was shown in [0] that the integrals generate a quadratic algebra. Is there a generalization for integrals of higher degrees? 4. The integrable metrics shown to exist by Kiyohara in [7] are Zoll metrics, which means that all the geodesics are closed. This suggests the conjecture that the corresponding systems are in fact SI. Could it be proved? 5. An even more difficult task would be to check whether the classical integrability survives to quantization in the sense of [4]. To end up the author is somewhat disappointed because the manifolds obtained here never meet S 2 and are of no use with respect to the Conjecture quoted in the Introduction. However, it was shown in [3] that the hyperbolic case is much more interesting since oneisled either to teardroporbifoldstannery orbifolds or S 2. Unfortunately theanalysis 35

37 went through explicitly for cubic integrals but the generalization to higher degrees remains unknown, indicating a more difficult ground. However the reward could be here plenty of Zoll metrics as is already the case for cubic integrals as shown in [4] which are quite interesting geometric objects. Appendices A The cases of a non simple F When F is simple, the general solution of the ODE 2.0 was given in Proposition 6. This appendix will deal with all the remaining cases: either F has a multiple real zero or it has multiple couples of complex conjugate zeroes. A. A multiple real zero Here we have 0 Fa = a a r Fa 2 r n A. where F is simple. One has: Proposition 3 For F = a a r Fa with 2 r n and a generic F, the solution of 2.0 is given by x = r µ k + k /2 i=r+ i = ǫ i a a i. A.2 The general solution of 6.4 is simply obtained by adding to xa the linear term ν n 2 a. Proof: Due to linearity we first check the ODE for x k a = [ǫ a a ] k+/2 k S k. Since the variable is a we will denote the derivation order by a superscript. Using 2.7 and interchanging the summations order we have first Op n [a a 0 r F]xk = l=0 F n l n l! Op l[a a 0 r ]x k 0 If r = n we take F =. If r = n the sum over i disappears. 36

38 and we will show that l S 0 n k S r X r l,k Op l[a a 0 r ]x k = 0. Computing the various derivatives and defining σ = s+r l one gets Xl,k r = ǫ l r r k l+/2 Γk +l r /2 r σ r! k /2 σ+l r. A.3 Γk r +/2 σ!r σ! /2 σ+l r Using the identities σ=0 a N+σ = a N a+n σ r σ = σ r! r σ! σ r shows that the sum is proportional to r σ k +l r /2 σ σ!l r +/2 σ σ=0 which is a hypergeometric function [5][p. 6] at unity r, k +l r /2 Γl r +/2Γr k + 2F ; = l r +/2 Γl+/2Γ k and does vanish for all k S r. For the proof to be complete let us check now that x i = of the ODE 2.0. We start from Op n [a a r F]xi = and use relation 2.4 which states that r σ = 0 r > σ A.4 ǫi a a i [a a r ] n k Op n k! k [ F]x i k S 0 n : Op k [ F]x i = Da k k! F i. A.5 is also a solution Inserting this second relation into the first one and using Leibnitz fromula we conclude to F which does vanish. Op n [a a r F]xi = i D n a i Remarks:. Due to the linearity of the ODE for x one can easily obtain its form in the generalized case where F = a a r a a s rs n i=r+ a a i r = r + r s. 2. Let usobserve thatthesolution obtainedremains valideven ifa becomes complex. 37

39 A.2 Multiple complex conjugate zeroes In this case we have: Proposition 32 If F has the structure Fa = a 2 +a 2 r Fa 2 2r n where F is simple, then the solution of the ODE 2.0 becomes 2 x = r µ + k P k +µ k Q k + k=2r+. A.6 Defining a = a tanθ θ π 2,+π 2 a 0 > 0 A.7 we have P k = cos θ 2 P k Q k = sin θ 2 P k ] P k = cosθ [U k /2 k cosθ U k 2 cosθ A.8 where the U n are the Tchebyshev polynomials of second kind supplemented with U = 0. The general solution of 6.4 is again obtained by adding to xa the linear term ν n 2 a. Proofs: Using the results of the previous proposition we know that if we start from Fa = a ia r a+ia r Fa 2 2r n the general solution is given by xa = r λ + k a ia k /2 + It is more convenient to write the first piece in x as λ k + a+ia k /2 r µ + k a +ia + µ k k /2 a ia k /2 i=2r+. sothatthebasisrequired isjustmadeoutoftherealandimaginarypartsofa +ia k+/2. This is most easily computed using 2 If 2r = n the second sum vanishes. a = a tanθ θ π 2, π 2 a > 0. 38

40 We have +ia/a k+/2 = cosθ k /2 e ik /2θ which gives a first expression for the real and imaginary parts: P k = cosθ k /2 cos k /2θ Q k = cosθ k /2 sin Recalling the defining relations of Tchebychev polynomials k /2θ. we have T n cosθ = cosnθ P k = cosθ k /2 cos θ 2 U n cosθ = sinn+θ sinθ [ ] T k cosθ+ cosθu k cosθ. Using the relation 3 T k cosθ cosθu k cosθ = U k 2 cosθ gives for P k the formula A.8. A similar computation gives the required formula also for Q k. Switching back to the variable a we have P k = a 2 +a 2 +a R k Q k = a a2 R k +a 2 +a with [ k Sr R k = a 2 +a 2 a a ] k/2 U k U k 2. a2 +a 2 a2 +a 2 For k = we have merely P a = a2 +a 2 +a a2 +a 2 Q a = a a2. A.9 a2 +a 2 +a 2 +a Let us give a second proof, which does not use the complexification argument given above, and which is similar to the proof given for Proposition 3. Setting a = we have to show that the function x k a = +ia k+/2 is indeed a solution of the ODE 2.0 when F = a 2 + r F. Using twice the formula 2.7 we have Op n a 2 + r F x k = s=0 F n s n s! 3 It is valid also for k = due to our convention that U = 0. s [ ia r ] s l Op s l! l +ia r x k l=0 39

41 and we will prove that l S 0 n k S r X r l,k Op l +ia r x k = 0. Computing the various derivatives one gets X r l,k l s=l r s r! k /2 s l s!r l+s! /2 s r σ=0 σ r! k /2 σ+l r. σ!r σ! /2 σ+l r This sum was already found in A.3 and shown to vanish for all k S r. B Symmetric functions of the roots Let us take a set of numbers {a,a 2,...,a n } and let us define P n a a k = k σ k a n k, B. where the σ k are the symmetric functions of the roots for the monic polynomial P. Restricting ourselves to real polynomials P this definition remains valid either if some root is multiple or if some complex conjugate couple of roots appear. It follows that n F = a a k = A k a k = k σ k = A n k k Sn. 0 B.2 In all what follows it will be supposed that P is simple, which means that all the zeroes a i are simple. So we can define P n+ = k σk i a a an k = σ i = σi n = 0. B.3 i Multiplying this last relation by a a i we get k S 0 n : σ k = σ i k +a i σ i k. B.4 An easy recurrence, using B.4 and B.2, gives the relations k S n : For i j we can define σi k = k k a i k s σ s = k s=0 s=0 a k s i A n s. B.5 n+2 P a a i a a j = k 2 σ ij k 2 an k = σ ij 2 = σij = σij n = σij n = 0. B.6 40

42 Multiplying this last relation by a a j we get k S 0 n : σi k = σij k +a j σ ij k 2 B.7 from which we deduce k S 0 n σ ij k 2 = σi k σj k a i a j. B.8 A quadratic identity follows from the relation i j : P a a i P P = P a a j a a i a a j. B.9 As a preliminary remark let us observe that any product of the form AB = k,l= A k B l H 2n k l P 2k+l y after setting s = k +l and reversing the order of the summations becomes with s=2 U s A,BH 2n s P 2s y + U s A,B = s 2 s=n+ A k B s k V s = V s A,BH 2n s P 2s y. k=s n Taking this observation into account one obtains the relations A k B s k. s k=s n σ i k σj s k = σ i k σj s k = s k=s n σ k σ ij s k 2 +σij s 2 σ k σ ij s k 2 s {2,3,...,n} s {n+,...,2n}. B.0 To conclude, the symmetric functions needed by our analysis are therefore σ 0 = σ σ 2... σ n 2 σ n σ n σ0 i = σ i σ2 i... σn 2 i σn i σ ij 0 = σ ij σ ij 2... σ ij n 2 σ i n = 0 σ ij n = 0 σ ij n = 0 B. 4

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