Necessary conditions for partial and super-integrability of Hamiltonian systems with homoge

a r X i v :n l i n /0701057v 1 [n l i n .S I ] 29 J a n 2007Necessary conditions for partial and

super-integrability of Hamiltonian systems with

homogeneous potential

Andrzej J.Maciejewski 1,Maria Przybylska 2,3and Haruo Yoshida 41Institute of Astronomy,University of Zielona Góra,Podgórna 50,PL-65–246Zielona Góra,Poland (e-mail:maciejka@astro.ia.uz.zgora.pl)2Institut Fourier,UMR 5582du CNRS,Universitéde Grenoble I,100rue des Maths,BP 74,38402Saint-Martin d’Hères Cedex,France 3Toru′n Centre for Astronomy,N.Copernicus University,Gagarina 11,PL-87–100Toru′n,Poland,(e-mail:Maria.Przybylska@astri.uni.torun.pl)4National Astronomical Observatory,2-21-1Osawa,Mitaka,181-8588Tokyo,Japan,(e-mail:h.yoshida@nao.ac.jp)Abstract.We consider a natural Hamiltonian system of n degrees of freedom with a homoge-neous potential.Such system is called partially integrable if it admits 1

1Introduction

The fundamental problem in Hamiltonian mechanics is to decide whether a given sys-tem is integrable.Integrability in this context usually means the integrability in the Liouville sense [1],but it is also important to consider the non-commutative integra-bility as it was de?ned in [14].Moreover,there exist examples of systems which are super-integrable,i.e.,systems with n degrees of freedom admitting m >n independent ?rst integrals such that n of them commute.Such super-integrable systems attract much attention,see e.g.[8,22,10,7].On the other hand,even if a considered system is not integrable,it is anyway important to know if it admits one or more ?rst integrals.These additional ?rst integrals can be used,e.g.to reduce the dimension of the system.

In this paper,we consider Hamiltonian systems with n degrees of freedom given by a natural Hamiltonian function

H=

1

d t q=p,

d

?q

,(1.2)

admit a straight line solution of the form

q(t)=?(t)d,p(t)=˙?(t)d,(1.3) where d is a non-zero vector in C n,and?(t)is a scalar function.Such solution exists iff at d gradient V′(d)is parallel to d,i.e.,V′(d)=γd for a non-zeroγ.Such d is called the

Darboux point of potential V.The length of d can be?xed arbitrarily and traditionally d is normalised in such a way that it satis?es the following non-linear equation

V′(d)=d.(1.4) R emark1.1If V′(d)=γd,then d:=αd satis?es V′( d)=γαk?2 d.Thus,for k=2we can?nd suchαthat V′( d)= d.For k=2we cannot generally assume that for a straight line solution(1.3)d satis?es(1.4).

Accepting the above convention,it is easy to see that(1.3)is a solution of(1.2) iff?(t)satis?es¨?=??k?1.For further considerations we choose a phase curveΓcorresponding to a non-zero energy level

e=1

k

?k,e=0.(1.5)

The necessary conditions of the Morales-Ramis theorem were obtained by an analysis of the variational equations along the considered phase curve.These equations are of the form

¨x=??(t)k?2V′′(d)x,(1.6)

where V′′(d)is the Hessian of V calculated at d.Let us assume that V′′(d)is diagonal-isable.Then,in an appropriate base,equations(1.6)split into a direct product of second order equations

¨y i=?λi?(t)k?2y i,1≤i≤n,(1.7) whereλ1,...,λn are eigenvalues of V′′(d).One of these eigenvalues,let us sayλn is k?1.We call this eigenvalue trivial.

In[16]J.J.Morales-Ruiz and J.P.Ramis proved the following theorem.

Theorem1.1(Morales-Ramis).If the Hamiltonian system de?ned by Hamiltonian(1.1)with a homogeneous potential of degree k∈Z?is integrable in the Liouville sense,then each pair (k,λi)belongs to an item of the following list

case kλ

2

p(p?1)

3.k 1

k

+p(p+1)k

4.3?1

6

(1+3p)2,?

1

32

(1+4p)2

?150(1+5p)2,?150(2+5p)2

5.4?1

9

(1+3p)2

6.5?9

18

(1+3p)2,?

9

10

(2+5p)2

7.?325

6

(1+3p)2,

25

32

(1+4p)2 25

50

(1+5p)2,

25

50

(2+5p)2

8.?49

9

(1+3p)2

9.?549

18

(1+3p)2,

49

10

(2+5p)2

Theorem1.3.If a Hamiltonian system de?ned by Hamiltonian(1.1)with a homogeneous po-tential of degree k∈Z?admits n+l,1

?if|k|≤2,then at least l pairs(k,λi),whereλi is a non-trivial eigenvalue,belong to the following list

case kλ

(1.9)

where r∈Q?;

?if|k|>2,then at least l pairs(k,λi)whereλi is a non-trivial eigenvalue,belong to items 3–9of table(1.8).

R emark1.2Assume that d satis?es V′(d)=γd withγ=1then,working with straight line solution(1.3),we arrive to the same results.However,to apply the above three theorems,we have to make the following modi?cation.If?λ1,...,?λn are eigenvalues of V′′(d),then we put λi=?λi/γ,for i=1,...n,see[13]for details.This remark is important only for k=2,as for k=2we can always assume that for a straight line solution(1.3)Darboux point d satis?es V′(d)=d.

The rest of this paper,except for the last section,is devoted to present proofs of the above theorems.To this end,in the next section we recall basic facts from the Ziglin the-ory[27,28]and its differential Galois extension developed by A.Baider,R.C.Churchill, J.J.Morales,J.-P.Ramis,D.L.Rod,C.Simóand M.F.Singer,see[4,15,19,17]and refer-ences therein,which is called the Morales-Ramis theory.Section3contains a derivation of variational equations and their reduction to an algebraic form.It appears that these equations are a direct sum of a certain type of hypergeometric equations.In section4we give a detailed analysis of the differential Galois group of this type of hypergeometric equation.Obstructions for partial and super-integrability follow from the fact that the differential Galois group of variational equations must have an appropriate number of invariants.This problem,reformulated into the language of Lie algebra of the differen-tial Galois group,is analysed in Section5.Short proofs of Theorem1.2and1.3are given in Section6.In the last section we present an application of our theorems.We analyse three and four body problems on a line and a radial potential.To make the paper self-contained,we collect in Appendix several known facts concerning the differential Galois group of a general second order equation with rational coef?cients and the Riemann P equation.

2Basic facts from the general theory

In this section we recall several basic facts from the Ziglin and Morales-Ramis theory in the setting needed in this paper.For detailed expositions,see e.g.[2,3,17,18,15,4].

Thus let us consider a complex holomorphic system of differential equations

d

where U is an open and connected subset of C n.The Ziglin and Morales-Ramis theory are based on the linearization of the original system around a particular non-equilibrium solution.Hence,let?(t)be a non-equilibrium solution of(2.1).Usually it is not a single-valued function of the complex time t.Thus,we associate with?a Riemann surfaceΓwith t as a local coordinate.The variational equations alongΓhave the form

˙ξ=A(t)ξ,A(t)=?v

g(ξ)=g(˙ξ)=g(A(t)ξ)=A(t)g(ξ),

d t

as g does not change elements of K.Thus,ifΞ∈M(n,L)is a fundamental matrix of(2.2),i.e.,its columns are linearly independent solutions of(2.2),then g(Ξ)=ΞM g,

where M g∈GL(n,C).In other words,we can look at the differential Galois group as a matrix group.

It is known that the monodromy group is contained in the differential Galois group. Moreover,in a case when equations(2.1)are Hamiltonian,then both these groups are subgroups of Sp(n,C).

Now we explain why the monodromy and differential Galois groups of variational equations are important in a study of integrability.At?rst,we introduce a few de?ni-tions.Let us consider a holomorphic function F de?ned in a certain connected neigh-bourhood of solution?(t).In this neighbourhood we have the expansion

F(?(t)+ξ)=F m(ξ)+O( ξ m+1),F m=0.(2.3) Then the leading term f of F is the lowest order term of the above expansion i.e.,f(ξ):= F m(ξ).Note that f(ξ)is a homogeneous polynomial of variablesξ=(ξ1,...,ξn)of degree m;its coef?cients are polynomials in?(t).If F is a meromorphic function,then it can be written as F=P/Q for certain holomorphic functions P and Q.Then the leading term f of F is de?ned as f=p/q,where p and q are leading terms of P and Q, respectively.In this case f(ξ)is a homogeneous rational function ofξ.

One can prove that if F is a meromorphic(holomorphic)?rst integral of equa-tion(2.1),then its leading term f is a rational(polynomial)?rst integral of variational equations(2.2).If system(2.1)has m≥2functionally independent meromorphic?rst integrals F1,...,F m,then their leading terms can be functionally dependent.However, by the Ziglin Lemma[27,2,4],we can?nd m polynomials G1,...,G m∈C[z1,...,z m] such that leading terms of G i(F1,...,F m),for1≤i≤m are functionally independent.

Additionally,if G?GL(n,C)is the differential Galois group of(2.2),and f is its rational?rst integral,then f(g(ξ))=f(ξ)for every g∈G,see[2,15].This means that f is a rational invariant of group G.Thus we have a correspondence between the?rst integrals of the system(2.1)and invariants of G.

Lemma2.1.If equation(2.1)has k functionally independent?rst integrals which are meromor-phic in a connected neighbourhood a non-equilibrium solution?(t),then the differential Galois group G of the variational equations along?(t),as well as their monodromy group,have k functionally independent rational invariants.

As mentioned above,a differential Galois group is a linear algebraic group,thus,in particular,it is a Lie group,and one can consider its a Lie algebra.This Lie algebra re?ects only the properties of the identity component of the group.It is easy to show that if a Lie group has an invariant,then also its Lie algebra has an integral.Let us explain what the last expression means.Let g?GL(n,C)denote the Lie algebra of G. Then an element Y∈g can be considered as a linear vector?eld:x→Y(x):=Y x,for x∈C n.We say that f∈C(x)is an integral of g,iff Y(f)(x)=d f(x)·Y(x)=0,for all Y∈g.

Proposition2.1.If f1,...,f k∈C(x)are algebraically independent invariants of an algebraic group G?GL(n,C),then they are algebraically independent?rst integrals of the Lie algebra g of G.

The above facts are the starting points for applications of differential Galois methods to a study of integrability.

If the considered system is Hamiltonian,then we have additional constrains.First of all,the differential Galois group of variational equations is a subgroup of the symplectic group.Secondly,commutation of?rst integrals imposed by the Liouville integrability implies commutation of variational?rst integrals.The following lemma plays the crucial role and this is why it was called The Key Lemma see Lemma III.3.7on page72in[2].

Lemma2.2.Assume that Lie algebra g?sp(2k,C)admits k functionally independent and commuting?rst integrals.Then g is Abelian.

Hence,if g in the above lemma is the Lie algebra of a Lie group G,then the identity component G?of G is Abelian.

Using all these facts Morales and Ramis proved the following theorem[15,17]. Theorem2.1(Morales-Ramis).Assume that a Hamiltonian system is meromorphically inte-grable in the Liouville sense in a neighbourhood of a phase curveΓ,and that variational equations alongΓare Fuchsian.Then the identity component of the differential Galois group of the varia-tional equations is Abelian.

Generally,it is dif?cult to determine the differential Galois group of a given system of variational equations when its dimension is greater than two.This is a reason why, instead of the variational equations,it is convenient to work with their reduced form called the normal variational equations.For a general de?nition of this notion see e.g.

[15].Here we de?ne the normal variational equations for a case when the considered Hamiltonian system is de?ned on C2n with z=(q1,p1,...,q n,p n)as canonical coordi-nates.Let us assume that the system admits a two dimensional symplectic invariant plane

Π:= z∈C2n|q i=p i=0for i=1,...,n?1 .(2.4) Thus,if H is the Hamiltonian of the system,then

?H

(0,...0,q n,p n)=0for i=1,...,n?1.(2.5)

?p i

Now,for a particular solution?(t)=(0,...0,q n(t),p n(t)),the matrix of the variational equations has a block diagonal form

A(t)= N(t)0

B(t)T(t) ,(2.6) where N(t),B(t)and T(t)are2(n?1)×2(n?1),2×2(n?1)and2×2matrices, respectively.Hence,the variational equations are a product of two systems

d

η=B(t)ξ+T(t)η,η∈C2.(2.7)

d t

The?rst of them is called the normal variational equations.

It can be shown that if the Hamiltonian system possesses a?rst integral F,then the normal variational equations also have a?rst integral which is an invariant of their differential Galois group G N?Sp(2(n?1),C).Moreover,if F1and F2are commuting ?rst integrals functionally independent together with H,then we can assume that the corresponding?rst integrals f1and f2of the normal variational equations are indepen-dent and commuting,see[4,15].These facts imply that the statement of Theorem2.1 remains valid if in its formulation the normal variational equations are used instead of the variational equations.

3Necessary conditions for Liouville integrability

Let G(k,λ)denote the differential Galois group of equation

¨y=?λ?(t)k?2y.(3.1)

It is a subgroup of SL (2,C )?Sp (2,C ).

It is clear that the differential Galois group G ,of equations (1.7)is a

direct

product

G

=

G

(

k ,λ1)×···×G (k ,λn )?Sp (2n ,C ).(3.2)

Hence,G ?is Abelian if and only if groups G (k ,λi )?are Abelian,for i =1,...,n .It follows that we know for which values of k and λthe identity component G (k ,λ)?of the differential Galois group G (k ,λ)of equation (3.1)is Abelian.

To solve this problem we introduce a new independent variable in equation (3.1),as it was proposed in [26],namely,assuming that k =0and e =0we put

t →z :=1

k ?3k ?2

2k y =0,(3.4)

where prime denotes the differentiation with respect to z .It is the Gauss hypergeometric equation

z (1?z )y ′′+[c ?(a +b +1)z ]y ′?aby =0,(3.5)with parameters

a +

b =k ?2

2k

,c =1?1k ,σ=c ?a ?b =1

2k

k ?3k ?2

2k y i =0,1≤i ≤n ,(3.8)

whose differential Galois group G

is a direct product G

= G (k ,λ1)×···× G (k ,λn ).A necessary condition for the integrability is now following:all groups G

(k ,λi )?have to be Abelian,and thus solvable.Exactly this reasoning was used in the proof of Theo-rem 1.1given in [15,16].4Group G (k ,λ)?

From the previous section it follows that it is important to know precisely the iden-tity component of the differential Galois group of hypergeometric equation (3.5)with parameters a ,b and c given by (3.6).This is the aim of this section.

As we have already mentioned the differential Galois group of(3.5)is not a sub-group of SL(2,C).It causes some technical problems.To avoid them,we transform equation(3.5)to the normal form putting

w=y exp p d z,p:=c?(a+b+1)z

w.(4.2)

4z2(z?1)2

For this equation exponents at0,1and at the in?nity are

12(1+ρ) , 12(1+σ) , ?12(1+τ) ,(4.3)

respectively.Its monodromy and differential Galois groups are now subgroups of SL(2,C).It is important to remark here that the identity components of the differ-ential Galois groups of(3.5)and(4.2)are the same.Notice also that the differences of exponents at singular points were unchanged.

Assuming thatρ,σandτare de?ned by(3.7),we denote by G(k,λ)the differential Galois group of equation(4.2).In what follows we describe properties of G(k,λ)?,but, as we explained,groups G(k,λ)?, G(k,λ)?and G(k,λ)?are isomorphic,so,as a result, we obtain a characterisation of G(k,λ)?.

At?rst,we recall the following fact which explains the origin of table(1.8)given in Theorem1.1.

Proposition4.1.Group G(k,λ)?is solvable if and only if(k,λ)belongs to an item in table(1.8). Proof.Equation(4.2)is the Riemann P equation.The Kimura theorem A.1gives neces-sary and suf?cient conditions for the solvability of the identity component of its differ-ential Galois group.Table(1.8)is just a speci?cation of these conditions forρ,σandτgiven by(3.7).

A necessary condition for the integrability is that G(k,λ)?is Abelian.As not all solv-able groups are Abelian,one can think that conditions of Theorem1.1can be sharpened. We show that it is not like that,i.e.,we prove that if G(k,λ)?is solvable,then it is Abelian. Suppose that G(k,λ)?is solvable but not Abelian.Then,as it is explained in Appendix, there is only one possibility:G(k,λ)=G(k,λ)?=T,where T is the triangular subgroup of SL(2,C).So,such case can appear only if the considered equation is reducible.Let us recall,see Appendix,that equation(4.2)is reducible iff it has a solution w=exp[ ω] whereω∈C(z).

Proposition4.2.Equation(4.2)is reducible if and only ifλ=p+kp(p?1)/2for some p∈Z.

Proof.To proof this lemma it is enough to check directly one of equivalent conditions given in Lemma A.3.

If equation(4.2)is reducible,then respective exponents at singular points0,1and in?nity are following

12k,12k , 14 , ?2+k(l+2)4k ,(4.4)

where l is an odd integer.

Now,we can show that if equation(4.2)is reducible,then the identity component of its differential Galois group G(k,λ)?is a proper subgroup of the triangular group T,and thus it is Abelian.

Lemma4.1.Assume that equation(4.2)is reducible.Then its differential Galois group G(k,λ) is a proper subgroup of the triangular group.

Proof.The difference of exponents for singular point z=1is1/2.Thus,from Lemma A.4, it follows that if equation(4.2)is reducible,then it possesses a solution of the form:

w=z r(1?z)s h(z),

where h(z)is a polynomial,and r is an exponent at z=0,and s in an exponent at z=1. As r and s are rational,there exists j∈N such that w j∈C(z).Now,by Lemma A.2, G(k,λ)is either a proper subgroup of the diagonal group,or a proper subgroup of the triangular group.

For our further analysis it is important to know the dimension of G(k,λ)?in a case when G(k,λ)is reducible.By the Lemma A.2,either G(k,λ)is a?nite cyclic group,and then G(k,λ)?={E},or G(k,λ)is a proper subgroup of the triangular group,and then

G(k,λ)?=T1:= 1c01 |c∈C .(4.5) Proposition4.3.Assume that G(k,λ)is diagonal.Then k∈{±1,±2}.

Proof.If G(k,λ)is diagonal,then the monodromy group of equation(4.2)is diagonal. This last group is generated by two elements M0,M1∈SL(2,C)which can be assumed diagonal.Then from Lemma A.4it follows that at least one of matrices M0,M1or M0M1is±E.The eigenvalues of M0are exp[2πi r1,2],where r1,2are exponents at z=0 for equation(4.2),i.e.

r1,2=1

k .

Hence,if M0=E,then k=±1,and it is impossible that M0=?E.Similar arguments show that M1=±E,and if M0M1=±E,then k=±2.

If a local solution near a singular points contains a logarithm,then the monodromy group contains the following element

M= 12πi01 ,

and,moreover it can be shown that such element belongs to the identity component of the differential Galois group of considered equation.Hence we can use this fact for checking whether G(k,λ)?=T1.

Proposition4.4.Assume that k=1and that equation(4.2)is reducible.Then singular point z=0is logarithmic except for the caseλ=0.

Proof.We apply Lemma A.6from Appendix,and we use notation introduced just before it.For k=1exponents at z=0areρ1=1andρ2=0,so m:=ρ1?ρ2=1,and m ={1}.Thus,by Lemma A.6,singularity z=0is logarithmic if an only if for

arbitrary exponents s and t at z=1and z=∞,respectively,we have1+s+t=1. Using(4.4)we obtain

1±p

2

+

p

2

(p?1)+r+s∈ m .

Using(4.4)we obtain the following condition:none of the three numbers

1

2(p+1),

1

1.k=?2;in this case G(k,λ)?={E}if and only ifλ=1?r2for some r∈Q\Z,and G(k,λ)?=D otherwise,or

2.k=2;in this case G(k,λ)?={E}if and only ifλ=r2for some r∈Q\Z,and G(k,λ)?=D otherwise,or

3.|k|>2and

λ=1

k

+p(p+1)k ,p∈Z,

and in this case G(k,λ)is?nite,so G(k,λ)?={E}.

Proof.We apply Lemma A.7from Appendix.A necessary condition for G(k,λ)to be a subgroup of DP group is following:at least two differences of exponents are half integers.As the difference of exponents at z=1isσ=1/2,we have two possibilities: either k=±2and thenρ=±1/2,or

λ=1

k

+p(p+1)k ,(4.7)

for some p∈Z.Moreover,if G(k,λ)is a subgroup of DP group,then it is a?nite group if and only if,at two singular points,the differences of exponents are half integers and exponents at the remaining point are rational,otherwise G(k,λ)=DP.Hence,under the assumption of our lemma,for|k|>2,group G(k,λ)is a subgroup of DP group iff λis given by(4.7).But,for these values ofλall exponents are rational,and this implies that the group is?nite.This proves case3.

For k=±1andλgiven(4.7)equation(4.2)is reducible,but we assumed that it is not reducible,so this case is excluded.

Let k=?2.Then exponents at in?nity are rational ifτis rational,see(4.3). From(3.7)we have

τ=?

√

λ,so G(2,λ)is a?nite subgroup of DP group iffλ=r2 for a rational r but if r is an integer,then equation(4.2)is reducible,thus we have to exclude these values.

We summarise our analysis in the following corollary.

Corollary4.2.Assume that the identity component G(k,λ)?of the differential Galois group of equation(4.2)is solvable.Then G(k,λ)?is Abelian.Moreover,G(k,λ)?={E}if and only if either

1.|k|>2and(k,λ)belongs to an item3–9of table(1.8),or

2.|k|≤2and(k,λ)belongs to an item of the following table

case kλ

(4.8)

where r∈Q?.

5Certain Poisson algebra

As was mentioned for a Hamiltonian system,the differential Galois group G of vari-ational equations along a particular solution is a subgroup of the symplectic group Sp(2n,C),thus the Lie algebra g is a Lie subalgebra of sp(2n,C).The necessary condi-tions for the integrability in the Liouville sense from Theorem2.1,are expressed in terms of the identity component of G.The properties of this component are encoded in the Lie algebra g of G.To?nd the necessary conditions for partial and super-integrability,we have to characterise Lie algebras g which admit a certain number of?rst integrals.And this is the main goal of this section.Here we follow the ideas and methods introduced in[20].

An element Y of Lie algebra sp(2n,C),considered as a linear vector?eld,is a Hamil-tonian vector?eld given by a global Hamiltonian function H:C2n→C,which is a degree2homogeneous polynomial of2n variables(x,y):=(x1,...,x n,y1,...,y n), i.e.H∈C2[x,y].In this way we identify sp(2n,C)with a C-linear vector space C2[x,y]with the canonical Poisson bracket as the Lie bracket.Thus,for a Lie alge-bra g?sp(2n,C)?C2[x,y],a rational function f∈C[x,y]is a?rst integral of g,iff {H,f}=0,for all H∈g.A?eld of rational?rst integrals of g we denote by C(x,y)g.

Now,we consider the case when g is a Lie subalgebra of sp(2,C).It is easy to show that Lie algebra sp(2,C)does not admit any non-constant?rst integral.

Proposition5.1.A rational function f∈C(x,y)is a?rst integral of sp(2,C),iff f∈C. Proof.Let f∈C(x,y)be a?rst integral of sp(2,C)?C2[x,y].Thus,{f,H}=0,for each H∈C2[x,y].Let us take H=x2.Then,

{f,H}=?2x?f

Let H1and H2be Hamiltonian functions from C2[x,y]such that linear vector?elds X H

1 have matrices h1and h2,respectively.It is easy to check that

and X H

2

1

H1=xy,H2=

=0,

?x

so,f∈C(y).However,for f∈C(y),we have

{f,H1}=?y?f

,

?y j

so f does not depend on y j.Taking H=y2j we show that f does not depend on x j.

If dim C g j=2,then we proceed in a similar way using arguments from the proof of Proposition5.3.

From the above lemma we have the following consequences.

Corollary5.1.Let g be a Lie subalgebra of s n,and g i=πi(g)for i=1,...,n.If g i is not Abelian for i=1,...,n,then C(x,y)g=C.

Corollary5.2.Let g be a Lie subalgebra of s n,and g i=πi(g)for i=1,...,n.If f∈C(x,y)\C is a?rst integral of g,then there exists1≤j≤n such that g j is Abelian.

Now,we consider a case when g admits more than one independent?rst integral.

Lemma5.2.Let g be a Lie subalgebra of s n,and g i=πi(g)for i=1,...,n.If g admits two algebraically independent and commuting?rst integrals f,g∈C(x,y),then there exist 1≤i

Proof.Neither f nor g is a constant.Thus,by Corollary5.2,there exists1≤i≤n such that g i is Abelian.Without loss of generality we can assume that g1is Abelian.Suppose that g j for2≤j≤n are not Abelian.Then,from Lemma5.1,it follows that f and g do not depend on(x j,y j)for2≤j≤n,and hence f,g∈C(x1,y1).But f and g commute,

thus

0={f,g}=?f

?y1?

?f

?x1

=

?(f,g)

Proof.Let us assume that h is nilpotent,i.e.,h is generated by H=y2.Then we have

0={H,f}=?2y1

?f

Q2

(Q{H,P}?P{H,Q}),

so Q{H,P}=P{H,Q}.As P and Q are relatively prime this implies that

{H,P}=γP and{H,Q}=γQ,(5.9) for a certainγ∈R[x1,y1].Comparing the degrees of both sides in the above equalities, we deduce thatγ∈C.Ifγ=0,then P and Q are polynomial?rst integrals of H,so in this case we have that f∈R(h)=C(h,x2,...,x n,y2,...,y n).

We show that caseγ=0is impossible.Let us assume thatγ=0.It is easy to see that if P∈R[x1,y1]satis?es equation

{H,P}=γP,(5.10) then its every homogeneous component also satis?es this equation.Thus let us assume that P is homogeneous of degree s.If we write

P=

s

∑

i=0

P i x i1y s?i

1

,P i∈R for i=1,...,s,(5.11)

then,equation(5.10)leads to the following equality

s

∑

i=0

P i(s?2i?γ)x i1y s?i1=0.(5.12)

This implies that if coef?cient P i=0,thenγ=s?2i and P=P i x i1y s?i1.Thus,every ho-mogeneous solution of(5.10)is a monomial of the form P i x i1y i+γ,whereγis a non-zero integer and i is a non-negative integer such that i+γ≥0.Thus a non-homogeneous solution of(5.10)is a?nite sum

P=∑

i+γ>0

p i x i1y i+γ.

But Q satis?es the same equation(5.10),so we have also

Q=∑

j+γ>0

q j x j1y j+γ.

Ifγ>0,then P and Q are not relatively prime because they have a common factor yγ1. On the other hand,ifγ<0,then they are not relatively prime either because they have a common factor x1.We have a contradiction and this?nishes the proof.

Lemma5.4.Let g be a Lie subalgebra of s n,g i=πi(g)for i=1,...,n.Assume that f∈C(x,y)\C is a?rst integral of g.If dim C g i=1for i=1,...,n,then there exist h i∈C[x i,y i]\C for i=1,...,n such that f∈C(h1,...,h n).

Proof.It is enough to apply n-times Proposition5.5taking g i as h for i=1,...,n. Lemma5.5.Let g be a Lie subalgebra of s n,g i=πi(g)for i=1,...,n.Assume that f1,...f n+1∈C(x,y)are algebraically independent?rst integrals of g and,moreover,f1,...f n commute.Then there exists1≤i≤n,such that dim C g i=0.

Proof.As g admits n commuting and independent?rst integrals,by Corollary5.3,g i is Abelian,and thus dim C g i≤1for i=1,...,n.

We prove the statement of the lemma by contradiction.Thus let us assume that dim C g i=1for i=1,...,n.Then,by Lemma5.4,f i∈C(h1,...,h n)for i=1,...,n+1. By assumption f1,...,f n+1are algebraically independent.But in C(h1,...,h n)any set of s>n elements is algebraically dependent.A contradiction?nishes the proof.

The above lemma can be generalised in the following way.

Lemma5.6.Let g be a Lie subalgebra of s n,g i=πi(g)for i=1,...,n.Assume that f1,...f n+s∈C(x,y),1≤s

This lemma can be easily proved by induction.We leave a proof to the reader.

6Proofs of Theorem1.2and1.3

Having the results collected in the two previous sections proofs of theorems1.2and1.3 are very simple.

Proof of Theorem1.2.The differential Galois group of variational equations(1.7)has the form of product(3.2),hence its Lie algebra g is a Lie subalgebra of s n.If the consid-ered system admits l functionally independent and commuting?rst integrals,then by Proposition2.1,g has l algebraically independent and commuting?rst integrals.By Lemma5.3,there exist1≤i1<···

tion(3.1)and(3.4)are the same,we have that G(k,λi

s )?are Abelian for s=1,...,l.The

statement of the theorem follows directly from Proposition4.1.

Proof of Theorem 1.3.The normal variational equations for the considered solution are a direct product of ?rst n ?1of equations (1.7).Hence,the Lie algebra g of their differential Galois group is a Lie subalgebra s n ?1.By assumption,Lie algebra g admits ?rst integrals f 2,...,f n +l ,such that (n ?1)of them f 2,...,f n commute.By Lemma (5.6),l among Lie algebras g i =πi (s n ?1),for i =1,...,n ?1,are zero dimensional.Without loss of the generality we can assume that dim C g i =0for i =1,...,l .Then G (k ,λi )?

?

G (k ,λi )?={E }for i =1,...,l .Assume that |k |>2.Then,by point 1.of Corollary 4.2,(k ,λi )belongs to an item 3–9in table (1.8).For |k |≤2,by point 2.of Corollary 4.2,(k ,λi )belongs to an item of table (4.8),for i =1,...,l .

7Examples

As the ?rst example,we consider the following potential

V =1

3(p 31+p 32+p 33)?p 1

+p 2(q 2?q 3)2?

p 3+p 1

(q 1?q 2)2?1(q 3

?q 1)2

?(p 1+p 2+p 3)(p 1q 1+p

2q 2+p 3q 3),F 5=(p 1+p 2+p 3) 2(p 21q 1+p 22q 2+p 23q 3)?q 1+q 2(q 2?q 3)2?

q

3+q 1

3(p 31+p 32+p 33)?p

1+p 2(q 2?q 3)2?

p 3+p 1

no further integrals and no further valules of k for which the system is integrable or super-integrable have been found.

For this system,let us see how Theorems1.2and1.3work.For this purpose we need solutions of the algebraic equation V′(d)=d.We do not know how to?nd all of them for an arbitrary k,nevertheless,it is suf?cient to know some of them.Here we use two solutions.The?rst one is

1

d1=(c,0,?c),c k?2=

.

3k?1

The eigenvalues of the Hessian matrix V′′(d1)are

(λ1,1,λ1,2,λ1,3)= 3(k?1)

,0,k?1 .

3

First,the considered system is partially integrable because of the existence of two com-muting?rst integrals,F1=H and F2.Then Theorem1.2requires that for each i=1,2 at least two pairs of(k,λi,j)belong to the list(1.8).Indeed this is the case,asλi,2=0 andλi,3=k?1for i=1,2are always in item2of the list(1.8).

We show that for values of k different from given above there is no integrable cases. Lemma7.1.Assume that k∈Z\{?2,0,1,2,4}.Then the Hamiltonian system with poten-tial(7.1)is not integrable in the Liouville sense.

Proof.We prove the statement of the lemma by a contradiction.Thus let k∈Z\ {?2,0,1,2,4}and the system is integrable in the Liouville sense.Then,from The-orem1.1or1.2,it follows that(k,λi,1)are in the list(1.8).We show that for each k∈Z\{?2,0,1,2,4}eitherλ1,1orλ2,1does not belong to the list.

Assume that k≥3is an odd integer.Thenλ1,1=3(k?1)/(1?2k?1)<0but for positive k the allowed values in the list are non-negative.

Assume k≥6is an even integer.We show thatλ2,1=(k?1)/3does not belong to an item of table(1.8).We have only two possibilities:eitherλ2,1belongs to item2or to item3.On the other hand,item2and item3with integer p gives a strictly increasing sequence of numbers

0,k?12k,k+2,3k?2,3k+k?1

2nor to item3.Indeed,item2and item3with integer p gives a strictly decreasing sequence of numbers

1,k?12k,k?1,3k+3,3k+k?1

1+(?1)k2k?1

>6k+4,

when k≤?3.Thusλ1,1does not belong to an item2and3of table(1.8).

When k∈{?5,?4,?3},we have to check also thatλ1,1does not belong to items7-9 in table(1.8).This task is reduced to checking if a certain quadratic polynomial has an integer root.Let us consider for example case k=?3.Thenλ1,1=?64/3.For k=?3 items2,3and7are allowed.Assume thatλ1,1is given by the?rst expression in item7. Then equation

25

6(1+3p)2=?

64

2√

2

√

2

y23.

Hence the phase curves of the systems lie on two dimensional cylinders.But for a maximally super-integrable system the maximal dimension of an invariant set is one.

Potential(7.1)has the following higher dimensional generalisation

V=

1