Regular polygonal equilibrium configurations on S^1 and stability of the associated relative equilibria
aa r X i v : . [ m a t h . D S ] O c t REGULAR POLYGONAL EQUILIBRIUM CONFIGURATIONSON S AND STABILITY OF THE ASSOCIATED RELATIVEEQUILIBRIA
Xiang Yu and Shuqiang Zhu , In memoriam of Florin Diacu School of Economic and Mathematics, Southwestern University of Finance andEconomics, Chengdu 611130, P.R. China School of Mathematical Sciences, University of Science and Technology ofChina, Hefei 230026, P.R. [email protected], [email protected]
Abstract.
For the curved n -body problem in S , we show that a regularpolygonal configuration for n masses on a geodesic is an equilibrium config-uration if and only if n is odd and the masses are equal. The equilibriumconfiguration is associated with a one-parameter family (depending on the an-gular velocity) of relative equilibria, which take place on S embedded in S .We then study the stability of the associated relative equilibria on two invariantmanifolds, T ∗ (( S ) n \ ∆) and T ∗ (( S ) n \ ∆). We show that they are Lyapunovstable on S , they are Lyapunov stable on S if the absolute value of angularvelocity is larger than a certain value, and that they are linearly unstable on S if the absolute value of angular velocity is smaller than that certain value. Key Words: curved n -body problem; equilibrium configurations; regular polyg-onal configurations; Lyapunov stability; Jacobi coordinates.1. introduction The curved n -body problem studies the motion of particles interacting under thecotangent potential in three-dimensional sphere and three-dimensional hyperbolicsphere. It is a natural extension of the Newtonian n -body problem. It roots inthe research of Bolyai and Lobachevsky. There are many researches in this areaover the past two decades on the Kepler problem, two-body problem, relativeequilibria, stability of periodic orbits, etc. For history and recent advances, onecan refer to Arnold et al. [1], Borisov et al. [2] and Diacu et al [5, 6].In classical mechanics, a particle is in mechanical equilibrium configuration if thenet force on that particle is zero. An equilibrium configuration is a configuration for which all particles are in mechanical equilibrium. Equilibrium configurationsdo not exist in the Newtonian n -body problem. However, they do exist in thecurved n -body problem in S . They are critical points of the potential. They leadto the equilibrium solutions, as well as families of relative equilibria.The purpose of this paper is to study regular polygonal equilibrium configura-tions in S and the stability of the associated relative equilibria. We show that aregular polygonal configuration for n masses on the equator (denoted by S ) of a2-dimensional great sphere (denoted by S ) is an equilibrium configuration if andonly if n is odd and the masses are equal. Each of the equilibrium configurationsleads to a one-parameter family (depending on the angular velocity) of relativeequilibria on S . Both T ∗ (( S ) n \ ∆) and T ∗ (( S ) n \ ∆) are invariant manifoldsof the Hamiltonian system. We show that the family of relative equilibria areLyapunov stable on S , they are Lyapunov stable on S if the absolute value ofangular velocity is larger than a certain value, and that they are linearly unstableon S if the absolute value of angular velocity is smaller than that certain value.In the Newtonian n -body problem, relative equilibria are related to the planarcentral configurations. The symmetry of central configurations always associateswith the symmetry of the masses. For example, Perko-Walter [13] shows thatthe regular n -gon is a central configuration if and only if all masses are equal.The regular n -gon central configurations always lead to linearly unstable relativeequilibria (cf. Moeckel [11], Roberts [14]).In the curved n -body problem in S , the stability of regular polygonal relativeequilibria was first studied by Mart´ınez-Sim´o [9]. They consider relative equilibriaof three equal masses moving on upper half of S embedded in R . The massesare moving on a circle x + y = r , r ∈ (0 ,
1) and form an equilateral triangleviewed from the ambient space R , [4]. The angular velocity is determined by r , see Remark 12. They find that the linear stability depends on the angularvelocity. The stability of three-body relative equilibria on the equator of S wasstudied by Diacu-S´anchez Cerritos-Zhu [7]. They find that relative equilibriaof three masses (not necessarily equal) on the equator are Lyapunov stable onthe equator, Lyapunov stable on S if the absolute value of the angular velocityis larger than a certain value. On the other hand, Stoica [15] investigated thegeneral n -body problem on surface of revolution. For the equal masses case, ifthe potential is attractive, she finds that regular polygonal relative equilibria ongeodesic circles are unstable if the angular momentum is smaller than a certainvalue, and are stable otherwise within a four-dimensional invariant manifold. Shealso finds that there is typically a pitchfork bifurcation. Our work can be viewedas an extension of her stability result in the four-dimensional invariant manifoldto the full phase space for the case of gravitational n -body problem on S . The curved n -body problem in S and main results In this section, we review the curved n -body problem in S , discuss the equi-librium configurations and state the main results of this paper. Vectors are allcolumn vectors, but written as row vectors in the text.2.1. The curved n -body problem in S and equilibrium configurations. The curved n -body problem in the three-dimensional sphere studies the mo-tion of n particles interacting under the so-called cotangent potential. Thereare researches in which the problem was set up with other models of the three-dimensional sphere. Following Diacu [6], we use the unit sphere in R . That is, S = { ( x, y, z, w ) ∈ R | x + y + z + w = 1 } . The metric on S is inducedfrom the standard metric of R . The distance between two point q i and q j , d ij ,is computed by cos d ij = q i · q j , where · is the scalar product in R .The curved n -body problem in S is a Hamiltonian system in ( R ) n with holo-nomic constraints. The Hamiltonian is H = n X i =1 || p i || m i − U ( q ) , where q = ( q , ..., q n ), q i ∈ R , p i = m i ˙ q i , U ( q ) is the potential defined by U = P m i m j cot d ij , and the constraints are q i · q i = 1 , i = 1 , ..., n . The potentialimplies that the singularity set of the configuration space is ∆ = ∪ ≤ i A configuration q ∈ ( S ) n \ ∆ is called an equilibrium configurationif it is a critical point of the potential, i.e., ∇ q i U ( q ) = , i = 1 , ..., n, where ∇ U ( q ) means the gradient of U . That is, q is an equilibrium configuration if (2) ∇ q i U ( q ) = n X j =1 ,j = i m i m j [ q j − cos d ij q i ]sin d ij = , i = 1 , ..., n. Those configurations are first introduced by Diacu [6] in the name fixed-points .Then they are called special central configurations by Diacu-Stoica-Zhu [8]. Sincethe name “central configuration” does not suit them very well, namely, they donot lead to total collision motion as in the Newtonian n -body problem, we callthem equilibrium configurations . Note that the set of equilibrium configurationshave O (4) symmetry. The system (2) can be written in another equivalent form. Proposition 1. An n -body configuration q in S is an equilibrium configurationif there are n real constants λ , ..., λ n such that (3) n X j = i,j =1 m j q j sin d ij − λ i q i = 0 , i = 1 , ..., n. Proof. Suppose that system (3) hold. Multiplying the i -th one with q i , we obtain λ i = λ i q i · q i = n X j = i,j =1 m j q j · q i sin d ij = n X j = i,j =1 m j cos d ij sin d ij . Thus, system (3) are equivalent to system (2). (cid:3) The equilibrium configurations obviously lead to equilibrium solutions q ( t ) = q (0) , t ∈ R . They also lead to other simple motions. Relative equilibrium isa phase curve that is at the same time a one-parameter orbit of the action ofthe symmetry group of the system. For the curved n -body problem in S , withthe unite sphere model, the symmetry group is O (4). Each of one-parametersubgroups of O (4) is a conjugate to the following subgroup A α,β ( t ) = cos αt − sin αt αt cos αt βt − sin βt βt cos βt , α, β ∈ R , which justifies the following definition. Definition 2. For the curved n -body problem in S , a solution in the form of A α,β ( t ) q (0) is called a relative equilibrium . Proposition 2 ([8]) . For any equilibrium configuration q , there is a one-parameterfamily of relative equilibria associated to it, namely, A α,α ( t ) q for any α ∈ R . Fur-ther more, if the equilibrium configuration lies on the union of the two great circles, x + y = 1 , and z + w = 1 , then there is a two-parameter family of relativeequilibria associated to it, namely, A α,β ( t ) q for any α, β ∈ R . There are actually more relative equilibria related to one equilibrium configura-tions q . Let τ ∈ O (4). Then obviously, τ q = ( τ q , ..., τ q n ) is also an equilibriumconfiguration. Thus, A α,α ( t ) τ q is a relative equilibrium for any α ∈ R . Thus,there is a 7-parameter family of relative equilibria related to q . Suppose that q is an equilibrium configuration on the great circle x + y = 1. Then, A α,β ( t ) q isa relative equilibrium for any α, β ∈ R , which is equivalent to A α, ( t ) q for α ∈ R .Also, A α,α ( t ) τ ( q ) is a relative equilibrium for any α ∈ R and any τ ∈ O (4). Main results. Denote by S and S the specified great circle { ( x, y, z, w ) ∈ R | x + y = 1 , z = w = 0 } and the specified great two-sphere { ( x, y, z, w ) ∈ R | x + y + z = 1 , w = 0 } respectively. Then S is the equator of S embeddedin R = { ( x, y, z ) } . In this paper, we focus on the regular polygonal equilibriumconfigurations on a great circle and the associated relative equilibria. By symme-try, we assume that the equilibrium configurations are on S . The questions weare going to discuss are(1) Let ¯ q be a regular polygonal configuration (viewed from R ) on S . To forman equilibrium configuration, is it necessary that the masses are equal?(2) The associated relative equilibria q ( t ) = A α, ( t )¯ q take place on S , theequator of S . By the equations of motion (1), both T ∗ (( S ) n \ ∆) and T ∗ (( S ) n \ ∆) are invariant manifolds of the Hamiltonian system. Considerthe relative equilibria on T ∗ (( S ) n \ ∆), are they stable? Consider therelative equilibria on T ∗ (( S ) n \ ∆), are they stable?We only study the case that the number of vertices of the regular polygon isodd. If the number is even, then there are pair of particles opposite to eachother, i.e., q i = − q j . Then the configuration belongs to ∆. We use the spher-ical coordinate system ( ϕ, θ ) for S , ϕ ∈ [0 , π ) , θ ∈ [0 , π ). Recall that theCartesian coordinates and the spherical coordinates are related by ( x, y, z ) =(sin θ cos ϕ, sin θ sin ϕ, cos θ ). Then, ( S ) n is parametrized by ( ϕ , ..., ϕ n , θ , ..., θ n ),and S is parametrized by ( ϕ, π ). Proposition 3. An n -body ( n is not necessarily odd) configuration on S is anequilibrium configuration if (4) 0 = n X j = i,j =1 m j sin( ϕ j − ϕ i )sin d ij , i = 1 , ..., n. Proof. We identify S as the unit circle of the complex plane, i.e., ( x, y ) = e √− ϕ .Then, a configuration on S is given by q = ( e √− ϕ , ..., e √− ϕ n ), 0 ≤ ϕ <ϕ < · · · < ϕ n ≤ π, ϕ + 2 π = ϕ n . The system (3) reads P nj = i,j =1 m j e √− ϕj sin d ij = λ i e √− ϕ i , i = 1 , ..., n. They lead to P nj = i,j =1 m j e √− ϕj − ϕi ) sin d ij = λ i ∈ R , i = 1 , ..., n .That is, the imaginary part of the left hand side of each equation is zero. Thiscompletes the proof. (cid:3) Now let n be an odd number greater than 1. Consider the regular n -gon con-figuration on S given by ¯ q = ( πn , ..., nπn , π , ..., π ). Theorem 4. Let n be an odd number greater than , the regular n -gon configu-ration on S is an equilibrium configuration if and only if m = m = ... = m n Proposition 5. Let n = 2 p + 1 , p ≥ . Let ¯ q be the regular n -gon equilibriumconfiguration on S with masses m = ... = m n = 1 . In the coordinate system ( ϕ , ..., ϕ n , θ , ..., θ n ) , the Hessian matrix of U at ¯ q is block diagonal, i.e., D U (¯ q ) = diag { ∂ U∂ϕ i ∂ϕ j (¯ q ) , ∂ U∂θ i ∂θ j (¯ q ) } . The eigenvalues of the first block h ∂ U∂ϕ i ∂ϕ j i consist of one zero, and n − negativenumbers. The eigenvalues of the second block h ∂ U∂θ i ∂θ j i consist of two zeros, and n − positive numbers. The maximal eigenvalue of the second block is P pj =1 1 − cos j πn sin j πn . The stability of relative equilibria is often defined by the stability of the cor-responding equilibria of the flow on the reduced phase space. The usual practiceis to compute the eigenvalues of the unreduced system and then to skip the non-relevant eigenvalues at the end, [9, 11, 14]. However, we would like to do thecomputation in the reduced system. Denote by ~ϕ = ( ϕ , ..., ϕ n ) , ~p ϕ = ( p ϕ , ..., p ϕ n ) ,~θ = ( θ , ..., θ n ) , ~p θ = ( p θ , ..., p θ n ) . We first do the reduction for the n -body problem on S . In this case, thepotential U depends on ~ϕ only and the Hamiltonian system can be written as H ( ~ϕ, ~p ϕ ) = n X i =1 m i p ϕ i − U ( ~ϕ ) , T ∗ ( S ) n , ω = d ( n X i =1 ϕ i dp ϕ i ) . Obviously, the group SO (2) has a Hamiltonian action on the phase space by( ~ϕ, ~p ϕ ) ( ~ϕ + ~s, ~p ϕ ), where ~ϕ + ~s = ( ϕ + s, ..., ϕ n + s ). The correspondingfirst integral is J ( ~ϕ, ~p ϕ ) = P ni =1 p ϕ i . The action of SO (2) and the integral isanalogous to the action of R on R and the corresponding integral, so we use theJacobi coordinates [10] of the Newtonian n -body problem to do the reduction.Let µ k = P ki =1 m i , and M k = m k µ k − µ k , ( M k = m k + µ k − ). Denote by ~u = ( u , u ..., u n , g n ) , ~v = ( v , v , ..., v n , G n ) . Consider the canonical transformation from ( ~ϕ, ~p ϕ ) to ( ~u, ~v ) given by the gener-ating function F ( ~v, ~ϕ ) = A ~ϕ · ~v = A T ~v · ~ϕ, where A = − . . . − m µ − m µ . . . − m µ − m µ − m µ . . . m µ n m µ n m µ n m µ n . . . m n µ n . The explicit transformation is(5) ~u = ∂F ∂~v = A ~ϕ, ~p ϕ = ∂F ∂ ~ϕ = A T ~v. It is well-known that the Jacobi coordinates system has the following properties(cf. [10]):(6) ( ~p ϕ ) T M ~p ϕ = ~v T A M A T ~v = ~v T ˜ M~v, µ n g n = n X i =1 m i ϕ i , G n = n X i =1 p ϕ i , where M = diag { m , ..., m n } , ˜ M = diag { M , ..., M n , µ n } . Note that the potential U does not depend on g n . Suppose ~ϕ corresponds to ( u , ..., u n , g n ). Then usingthe transform (6), we see that ~ϕ + ~s corresponds to ( u , ..., u n , g n + s ). So ∂U∂g n = lim s → U ( u , ..., u n , g n + s ) − U ( u , ..., u n , g n ) s = lim s → U ( ~ϕ + ~s ) − U ( ~ϕ ) s = 0 . Hence, with the Jacobi coordinates, the Hamiltonian function can be written as H ( ~u, ~v ) = n X i =2 v i M i + G n µ n − U ( u , ..., u n ) . Consider the reduced space J − ( c ) /SO (2). Obviously, ( u , ..., u n , v , ..., v n ) canserve as a canonical coordinates system of the symplectic sub-manifold. Thereduced Hamiltonian system is H = n X i =2 v i M i − U ( u , ..., u n ) , J − ( c ) /SO (2) , ω = d ( n X i =2 u i dv i ) , where we have neglected the constant term. Consider a relative equilibrium on S in the ( ~ϕ, ~p ϕ ) coordinates system, ϕ ( t ) = ϕ + αt, ..., ϕ n ( t ) = ϕ n + αt, p ϕ ( t ) = m α, ..., p ϕ n ( t ) = m n α. It is easy to check that the motion corresponds to theequilibrium ϕ − ϕ , ..., ϕ n − P n − i =1 m i ϕ i µ n − , , ..., , of the reduced Hamiltonian system. For the relative equilibrium A α, ( t )¯ q we areinterested, the momentum is J = nα , and the corresponding equilibrium is X α = ( 2 n π, ..., k + 1 n π, ..., π, , ..., , α ∈ R . Theorem 6. Let n = 2 p + 1 , p ≥ . Let ¯ q be the regular n -gon equilibriumconfiguration on S . Then the equilibrium X a of the reduced Hamiltonian system ( H , J − ( nα ) /SO (2) , ω ) is Lyapunov stable for all α ∈ R . We now do the reduction for the n -body problem on S . The complete reductionfor the two-dimensional case is not easy, [2]. For our purpose, we will do a partialreduction. In this case, the Hamiltonian system is H ( ~ϕ, ~θ, ~p ϕ , ~p θ ) = n X i =1 ( p ϕ i m i sin θ i + p θ i m i ) − U ( ~ϕ, ~θ ) , T ∗ ( S ) n , ω = d ( n X i =1 ϕ i dp ϕ i + θ i dp θ i ) . Again, the group SO (2) acts on the phase space by ( ~ϕ, ~θ, ~p ϕ , ~p θ ) ( ~ϕ + ~s, ~θ, ~p ϕ , ~p θ ) , and the corresponding first integral is J ( ~ϕ, ~θ, ~p ϕ , ~p θ ) = P ni =1 p ϕ i . We again con-sider the canonical transformation given by (5), i.e., ~u = A ~ϕ, ~p ϕ = A T ~v. Therelations µ n g n = P ni =1 m i ϕ i , G n = P ni =1 p ϕ i still hold [10]. However, the kineticenergy is more complicated than the previous case. Let S = diag { θ , ..., θ n } . Then, the first part of the kinetic energy is ~p ϕT SM S ~p ϕ = ~v T A SM S A T ~v . De-note by P the matrix A SM S A T . Then, P is not diagonal, and by direct compu-tation we find that the elements are(7) P kk = µ k [ m sin θ + ... + m k sin θ k ] + m k +1 sin θ k +1 , k = n,P nn = µ n [ m sin θ + ... + m n sin θ n ] ,P kl = µ k µ l [ m sin θ + ... + m k sin θ k ] − µ l sin θ k +1 , k < l < n,P kn = − µ k µ n [ m sin θ + ... + m k sin θ k ] + µ n sin θ k +1 , k < n. As in the previous case, the potential U does not depend on g n . The reduced space J − ( c ) /SO (2) can be parametrized by ( u , ..., u n , ~θ, v , ..., v n , ~p θ ). The reducedHamiltonian system is H = 12 ~v T P ~v + n X i =1 p θ i m i − U ( u , ..., u n , ~θ ) , J − ( c ) /SO (2) , ω = d ( n X i =2 u i dv i + θ i dp θ i ) . The reduced system might not be useful for a general problem, but it works inour problem.For the relative equilibrium A α, ( t )¯ q we are studying, the momentum is J = nα = G n , and the corresponding equilibrium is Y α = ( 2 n π, n π, ..., nn π, π , ..., π , , ..., , , ..., , α ∈ R . Theorem 7. Let n = 2 p + 1 , p ≥ . Let ¯ q be the regular n -gon equilibriumconfiguration on S with masses m = ... = m n = 1 . Then the equilibrium Y a of the reduced Hamiltonian system ( H , J − ( nα ) /SO (2) , ω ) is linearly unstable if α < P pj =1 1 − cos j πn sin j πn , and is Lyapunov stable if α > P pj =1 1 − cos j πn sin j πn . Remark 8. Though the equilibria X α ( Y α ) are stable in the reduced system, thecorresponding relative equilibrium is obviously not stable in the unreduced system, since any relative equilibrium can be perturbed in such a way that the configura-tion only rotates more quickly. This is typical for relative equilibria stable in thereduced system, see Patrick [12].3. proof of the main results To prove Theorem 4 and Proposition 5, we need the following property of circulant matrices . An n × n matrix C = ( c kj ) is called circulant if c kj = c k − ,j − , where c ,j and c k, are identified with c n,j and c k,n , respectively. For any n × n circulant matrix C = ( c ij ), the eigenvectors and the corresponding eigenvalues are(8) v k = (1 , ρ k − , ρ k − , . . . , ρ n − k − ) , λ k = n X j =1 c j ρ j − k − , k = 1 , ..., n, where ρ k is the k -th root of unity ρ k = e √− kπn . We also need the identity(9) m X j =1 sin 2 ja = sin[ a ( m + 1)] sin am sin a . Let us introduce some notations . Assume that n = 2 p + 1 , p ≥ 1. Denoteby φ the angle πn , and by v k the k -th eigenvector of circulant matrix definedabove for k = 1 , ..., n . Recall that the regular n -gon configuration is given by¯ q = ( πn , ..., nπn , π , ..., π ) and that ρ k = cos kφ + √− kφ .3.1. Proof of Theorem 4. Proof. If the masses are equal, system (3) is obviously satisfied. Then, the con-figuration ¯ q is an equilibrium configuration. Now we show that it also necessary.For the regular n -gon, d ij = min {| ϕ i − ϕ j | , π − | ϕ i − ϕ j |} , so system (4) is P j = k m j sin( j − k ) φ | sin( j − k ) φ | = 0 , k = 1 , ..., n . It can be written as B m = , where m = ( m , ..., m n ) and B has elements b kj = ( sin( j − k ) φ | sin( j − k ) φ | for j = k, j = k. Note that B is circulant and skew-symmetric, so its eigenvalues can be com-puted by formula (8) and are purely imaginary. Denote the eigenvalues of B by √− ..., √− n . Then, Γ k = √− P nj =1 b j sin( k − j − φ , andΓ k = p X j =1 sin jφ | sin jφ | sin j ( k − φ + p X j = p +1 sin jφ | sin jφ | sin j ( k − φ = 2 p X j =1 sin j ( k − φ sin jφ , k = 1 , ..., n. We first show that Γ k = 0 for k = 2 , ..., n . Note that ρ k − is the complexconjugation of ρ n − ( k − = ρ n +2 − k − and that B is real. We obtain − Γ n +2 − k = Γ k . Thus, it is enough to show that none of the the following numbers is zero,Γ , Γ , Γ , ..., Γ p . We claim that the sequence { Γ , Γ , Γ , ..., Γ p } is concave. By elementarytrigonometric identities and the formula (9), we obtainΓ k +2 − Γ k = 4 p X j =1 cos kjφ sin jφ , Γ k +4 − k +2 + Γ k = − p X j =1 sin( k + 1) jφ = − ( k +1)( p +1)2 φ sin ( k +1) p φ sin k +12 φ = 4 cos ( k +1)(2 p +1)2 φ − cos ( k +1)2 φ sin k +12 φ = 4 ( − k − − cos ( k +1)2 φ sin k +12 φ . If k is even and k ∈ [2 , p − k +12 φ ∈ (0 , π ) and Γ k +4 − k +2 + Γ k < { Γ , Γ , Γ , ..., Γ p } is concave.We check that the two ends of the sequence are positive. The first one Γ = P pj =1 1sin jφ is obviously positive since 0 < jφ < π for 1 ≤ j ≤ p . The second oneΓ p = − p X j =1 sin 2 jφ sin jφ = − p X j =1 cot jφ = − p X j =1 [cot jφ + cot( p + 1 − j ) φ ] . Note that ( p + 1 − j ) φ = π − ( jφ − φ ), we obtain(10) Γ p = − p X j =1 cot jφ = − p X j =1 [cot jφ − cot( j − 12 ) φ ] > . Since that the sequence Γ , Γ , Γ , ..., Γ p is concave, and that the two ends arepositive, we conclude that none of the numbers Γ , Γ , Γ , ..., Γ p is zero. Thus,Γ k = 0 for k = 2 , ..., n .Now let us return to show that it is necessary that m = m (1 , ..., 1) to have B m = . Note that the n eigenvectors of B , v , ..., v n form a basis of C n . Thereare n complex constants δ , ..., δ n such that m = P ni =1 δ k v k . So, we have = B m = B n X i =1 δ k v k = n X i =1 √− δ k Γ k v k ⇒ Γ k δ k = 0 , k = 1 , ..., n. Since Γ k = 0 for k = 2 , ..., n , we obtain δ k = 0 for k = 2 , ..., n , i.e., m = δ v .That is, m = m = ... = m n , a remark that completes the proof. (cid:3) Remark 9. Consider a configuration of odd bodies on S that is close to a regularpolygon. We can find masses to form an equilibrium configuration by solvinga linear system ˜ B m = 0 equivalent to system (4). Since the matrix ˜ B is anti-symmetric and is close the matrix B in the above proof, we conclude that ˜ B m = 0has a one-dimensional solution space and that the masses can be all positive.Hence, there is an ( n − S ) n /SO (2). This is different from co-circular central configurations of theNewtonian n -body problem. For example, it is proved by Cors-Roberts that thefour-body co-circular central configurations form a two-dimensional manifold [3].3.2. Proof of Proposition 5. Proof. We first show that the Hessian matrix of U is block diagonal at theequilibrium configuration ¯ q , i.e., ∂ U∂θ i ∂ϕ j | ¯ q = 0 for all pairs of ( i, j ). Denote by q + k i the coordinate ( ϕ , ..., ϕ n , θ , ..., θ i + k, ..., θ n ), by q + h j the coordinate( ϕ , ..., ϕ j + h, ..., ϕ n , θ , ..., θ n ). Then the mutual distances between the particlesare the same for the two configurations, q + k i and q − k i if q is a configurationon S . That is, U ( q + k i ) = U ( q − k i ). Hence, we obtain ∂ U∂θ i ∂ϕ j | ¯ q = lim k → k ( ∂U∂ϕ j | ¯ q + k i − ∂U∂ϕ j | ¯ q − k i )= lim ( h,k ) → (0 , hk ( U (¯ q + k i + h j ) − U (¯ q + k i ) − U (¯ q − k i + h j ) + U (¯ q − k i ))= 0 . We compute the elements of the two blocks of the Hessian matrix of U . Recallthat the masses are m = ... = m n = 1, U = P m i m j cot d ij and that cos d ij =cos θ i cos θ j + sin θ i sin θ j cos( ϕ i − ϕ j ). By direct computation, we obtain [7],(11) ∂U∂ϕ i = P j = i − sin θ i sin θ j sin( ϕ i − ϕ j )sin d ij , ∂U∂θ i = P j = i − sin θ i cos θ j +cos θ i sin θ j cos( ϕ i − ϕ j )sin d ij , ∂ U∂ϕ i ∂ϕ j = − d ij sin θ i sin θ j sin ( ϕ i − ϕ j )+sin d ij sin θ i sin θ j cos( ϕ i − ϕ j )sin d ij , ∂ U∂ϕ i = P j = i d ij sin θ i sin θ j sin ( ϕ i − ϕ j ) − sin d ij sin θ i sin θ j cos( ϕ i − ϕ j )sin d ij , ∂ U∂θ i ∂θ j = d ij ( − cos θ i sin θ j +sin θ i cos θ j cos( ϕ i − ϕ j ))( − sin θ i cos θ j +cos θ i sin θ j cos( ϕ i − ϕ j ))sin d ij + sin d ij (sin θ i sin θ j +cos θ i cos θ j cos( ϕ i − ϕ j ))sin d ij , ∂ U∂θ i = P j = i d ij ( − sin θ i cos θ j +cos θ i sin θ j cos( ϕ i − ϕ j )) − sin d ij cos d ij sin d ij . Part 1. The eigenvalues of h ∂ U∂ϕ i ∂ϕ j i at ¯ q . Note that θ i = π , i = 1 , ..., n and d ij = min {| ϕ i − ϕ j | , π − | ϕ i − ϕ j |} . By equation (11), the block has elements(12) ∂ U∂ϕ k ∂ϕ j = ( − j − k ) φ | sin( j − k ) φ | for j = k, P i = k i − k ) φ | sin( i − k ) φ | for j = k, Note that the block is circulant and symmetric. So its eigenvalues, denoted byΦ , ..., Φ n , can be computed by formula (8) and are real. Then,Φ k = p X j =1 jφ | sin jφ | − p X j =1 jφ | sin jφ | cos( k − jφ = 4 p X j =1 cos jφ sin jφ (1 − cos( k − jφ ) , k = 1 , ..., n. The first eigenvalue Φ is 0, which reflects the SO (2) symmetry of the equilibriumconfiguration on S . Note that ρ k − is the complex conjugation of ρ n − ( k − = ρ n +2 − k − and that h ∂ U∂ϕ i ∂ϕ j i is a real matrix. We have Φ n +2 − k = Φ k . Thus, it isenough to study just the following eigenvaluesΦ , Φ , Φ , ..., Φ p . Firstly, we claim that the sequence { Φ − Φ , Φ − Φ , ..., Φ p +2 − Φ p } isconcave. By formula (9) and other elementary trigonometric identities, we haveΦ k +2 − Φ k = 8 p X j =1 sin kjφ cos jφ sin jφ , Φ k +4 − k +2 + Φ k = 16 p X j =1 cos( k + 1) jφ cos jφ sin jφ , Φ k +6 − k +4 + 3Φ k +2 − Φ k = − p X j =1 sin( k + 2) jφ cos jφ = − p X j =1 (sin( k + 3) jφ + sin( k + 1) jφ )= − sin ( k +3)( p +1)2 φ sin ( k +3) p φ sin k +32 φ + sin ( k +1)( p +1)2 φ sin ( k +1) p φ sin k +12 φ ! = 8 cos ( k +3) n φ − cos k +32 φ sin k +32 φ + cos ( k +1) n φ − cos k +12 φ sin k +12 φ ! = 8 " ( − k +1 − cos k +32 φ sin ( k +3) πn + ( − k +1 − cos k +12 φ sin ( k +1) πn Thus, Φ k +6 − k +4 + 3Φ k +2 − Φ k < k is even and k ∈ [2 , p − { Φ − Φ , Φ − Φ , ..., Φ p +2 − Φ p } is concave.Secondly, note that the two ends of the sequence are positive. Note that Φ p +2 − Φ p = Φ − Φ = − Φ . Since Φ = P pj =1 cos jφ sin jφ (1 − cos 2 jφ ) = 2 P pj =1 cot jφ, whichis negative according to (10), we see that the second end Φ p +2 − Φ p is positive.The first end Φ − Φ can be written as p X j =1 cos jφ sin jφ (cos jφ − cos 3 jφ ) = p X j =1 cos jφ sin jφ (4 cos jφ − jφ ) = 4 p X j =1 cos jφ sin jφ . So the first end Φ − Φ is also positive.Hence, the sequence { Φ − Φ , Φ − Φ , ..., Φ p +2 − Φ p } is positive, whichimplies Φ < Φ < ... < Φ p < Φ p +2 = Φ = 0 . That is, the eigenvalues of theblock h ∂ U∂ϕ i ∂ϕ j i consist of one zero, and n − h ∂ U∂θ i ∂θ j i at ¯ q . Note that θ i = π , i = 1 , ..., n and d ij = min {| ϕ i − ϕ j | , π − | ϕ i − ϕ j |} . By equation (11), the block has elements ∂ U∂θ k ∂θ j = ( | sin( j − k ) φ | for j = k, − P i = k cos( i − k ) φ | sin( i − k ) φ | for j = k, Note that the block is circulant and symmetric. So its eigenvalues, denoted byΘ , ..., Θ n , can be computed by formula (8) and are real. Then,Θ k = p X j =1 − cos jφ | sin jφ | + p X j =1 cos( k − jφ | sin jφ | = 2 p X j =1 cos j ( k − φ − cos jφ sin jφ , k = 1 , ..., n. Obviously, Θ = Θ p +1 = 0, which reflects the symmetry, andΘ = 2 p X j =1 − cos jφ sin jφ > . Note that ρ k − is the complex conjugation of ρ n − ( k − = ρ n +2 − k − and that h ∂ U∂θ i ∂θ j i is a real matrix. We have Θ n +2 − k = Θ k . Thus, it is enough to study just thefollowing eigenvalues Θ , Θ , Θ , ..., Θ p . Firstly, we claim that the sequence { Θ − Θ , Θ − Θ , ..., Θ p +2 − Θ p } isconcave. By formula (9) and other elementary trigonometric identities, we haveΘ k +2 − Θ k = − p X j =1 sin kjφ sin jφ , Θ k +4 − k +2 + Θ k = − p X j =1 cos( k + 1) jφ sin jφ , Θ k +6 − k +4 + 3Θ k +2 − Θ k = 16 p X j =1 sin( k + 2) jφ = 16 p X j =1 sin ( k +2)( p +1)2 φ sin ( k +2) p φ sin k +22 φ = − ( k +2) n φ − cos k +22 φ sin k +22 φ = − − k − cos k +22 φ sin ( k +2) πn . Thus, Θ k +6 − k +4 + 3Θ k +2 − Θ k < k is even and k ∈ [2 , p − { Θ − Θ , Θ − Θ , ..., Θ p +2 − Θ p } is concave.Secondly, note that the two ends of the sequence are positive.Θ p +2 − Θ p = Θ − Θ = 2 p X j =1 − cos 2 jφ sin jφ > , Θ − Θ = 2 p X j =1 cos 3 jφ − cos jφ sin jφ = − p X j =1 cos jφ − cos jφ sin jφ = − p X j =1 cos jφ sin jφ , which is positive according to (10).Hence, the sequence { Θ − Θ , Θ − Θ , ..., Θ p +2 − Θ p } is positive, whichimplies 0 = Θ < Θ < ... < Θ p < Θ p +2 = Θ . That is, the eigenvalues of theblock h ∂ U∂θ i ∂θ j i consist of two zeros, and n − = 2 P pj =1 1 − cos j πn sin j πn . (cid:3) Stability on S . proof of Theorem 6. The equilibrium X α is a local minimum of the kinetic energy P ni =2 v i M i . In the coordinates ( u , ..., u n , g n ), the Hessian matrix of U at ¯ q is blockdiagonal since ∂U∂g n = 0, i.e, D U = diag { h ∂ U∂u i ∂u j i , } . The signature of the Hes-sian matrix is ( n , n + , n − ) = (1 , , n − 1) by Proposition 5. So, the first block h ∂ U∂u i ∂u j i has n − H = P ni =2 v i M i − U ( u , ..., u n ) is positive definite at the equilibrium X α , which implies that X α isLyapunov stable in the Hamiltonian system ( H , J − ( nα ) /SO (2) , ω ). (cid:3) Stability on S . Recall that the reduced Hamiltonian is H ( u , ..., u n , ~θ, v , ..., v n , ~p θ ) = F ( ~θ, v , ..., v n ) + n X i =1 p θ i m i − U ( u , ..., u n , ~θ ) , where we denote by F ( ~θ, v , ..., v n ) the function ~v T P ~v . Explicitly, it is F ( ~θ, v , ..., v n ) = X ≤ i ≤ j ≤ n − v i +1 v j +1 P ij + n − X i =1 v i +1 nαP in + n α P nn , since G n = J ( Y α ) = nα , and the equilibrium is Y α = ( 2 n π, ..., k + 1 n π, ..., π, π , ..., π , , ..., , , ..., , α ∈ R . Linearizing the flow at Y α leads to L = ∂ H ∂v i ∂u j ∂ H ∂v i ∂θ j ∂ H ∂v i ∂v j ∂ H ∂v i ∂p θj ∂ H ∂p θi ∂u j ∂ H ∂p θi ∂θ j ∂ H ∂p θi ∂v j ∂ H ∂p θi ∂p θj − ∂ H ∂u i ∂u j − ∂ H ∂u i ∂θ j − ∂ H ∂u i ∂v j − ∂ H ∂u i ∂p θj − ∂ H ∂θ i ∂u j − ∂ H ∂θ i ∂θ j − ∂ H ∂θ i ∂v j − ∂ H ∂θ i ∂p θj = Ø ∂ F∂v i ∂θ j ∂ F∂v i ∂v j ØØ Ø Ø M ∂ U∂u i ∂u j ∂ U∂u i ∂θ j Ø Ø ∂ U∂θ i ∂u j ∂ U∂θ i ∂θ j − ∂ F∂θ i ∂θ j − ∂ F∂θ i ∂v j Ø , where Ø is the zero block. Note that h ∂ U∂u i ∂θ j i | Y α = Ø. Recall that m = ... = m n = 1, θ = ... = θ n = π , v = ... = v n = 0 at Y α , and that the matrix P depends on { sin θ , ..., sin θ n } only (see equations (7)). At Y α , we get ∂ F∂v i ∂θ j =cos θ j ( ∗ ) = 0, so h ∂ F∂v i ∂θ j i = Ø; we get ∂ F∂v i +1 ∂v j +1 = P ij , i, j < n , and it is easy tocheck that P ij = 0 if i = j and P ii = M i +1 , so h ∂ F∂v i ∂v j i = diag { M , M , ..., M n } ; we get h ∂ F∂θ i ∂θ j i = n α h ∂ P nn ∂θ i ∂θ j i = α I n , since P nn = P θi n . Thus, L | Y α = Ø Ø ˜ M ØØ Ø Ø I n∂ U∂u i ∂u j Ø Ø ØØ ∂ U∂θ i ∂θ j − α I n Ø Ø , where ˜ M = diag { M , M , ..., M n } . Proposition 10. For a block matrix in the form of Ø Ø D ØØ Ø Ø EK Ø Ø ØØ Q Ø Ø , supposethat D and E are invertible, D, K (resp. Q, E )are of the same size. If u isan eigenvector of KD (resp. QE ) with eigenvalue λ = 0 , then there is a two-dimensional invariant subspace on which the matrix is similar to (cid:20) √ λ −√ λ (cid:21) .If u is an eigenvector of KD (resp. QE ) with eigenvalue , then there is a two-dimensional invariant subspace on which the matrix is similar to (cid:20) (cid:21) . Actually, if KD u = λ u and λ = 0, then the basis of the two-dimensionalinvariant subspace is { ( D u √ λ , , u , ) , ( − D u √ λ , , u , ) } . If KD u = , then the basisof the two-dimensional invariant subspace is { ( D u , , , ) , ( , , u , ) } . proof of Theorem 7. By the proof of Theorem 6, the matrix h ∂ U∂u i ∂u j i has n − M is positive definite and diagonal, so (cid:20) ∂ U∂u i ∂u j (cid:21) ˜ M = ( ˜ M ) − ˜ M (cid:20) ∂ U∂u i ∂u j (cid:21) ( ˜ M ) T ˜ M . The above equality and Proposition 10 implies that there is a (2 n − L | Y α on which L | Y α is semi-simple and has only non-zeropurely imaginary eigenvalues.By Proposition 5, the matrix h ∂ U∂θ i ∂θ j i − α I n has eigenvalues: Θ − α , Θ − α , ..., Θ n − α . Recall also that Θ = 2 P pj =1 1 − cos j πn sin j πn > 0, Θ = Θ n = 0 andΘ k > , k = 1 , , n . By Proposition 10 and the fact that h ∂ U∂θ i ∂θ j i − α I n issymmetric, we obtain the Jordan normal form of L | Y α on the complementary2 n -dimensional subspace,diag np Θ , − p Θ , (cid:20) (cid:21) , p Θ , − p Θ , ..., (cid:20) (cid:21) o , if α = 0 , diag n (cid:20) (cid:21) , p Θ − α , − p Θ − α , ..., p Θ n − α , − p Θ n − α o , if α = Θ ;diag np Θ − α , − p Θ − α , ..., p Θ n − α , − p Θ n − α o , if α = Θ , . This implies that Y α is linearly unstable in the Hamiltonian system ( H , J − ( nα ) /SO (2) , ω )if α < Θ = 2 P pj =1 1 − cos jφ sin jφ . On the other hand, the form of L | Y α implies that D H | Y α = diag n − (cid:20) ∂ U∂u i ∂u j (cid:21) , α I n − (cid:20) ∂ U∂θ i ∂θ j (cid:21) , ˜ M , I n o . If α > Θ = 2 P pj =1 1 − cos jφ sin jφ , then H is positive definite at the equilibrium Y α ,which implies that Y α is Lyapunov stable in the reduced Hamiltonian system. (cid:3) Consider the relative equilibria associated with equilibrium configurations on S discussed in Remark 9, i.e., those close to the regular polygonal ones. Obviously,their stability depends on the two matrices, h ∂ U∂ϕ i ∂ϕ j i , h ∂ U∂θ i ∂θ j i . By continuity, theeigenvalues of the two blocks are close to that of the regular polygonal equilibriumconfigurations. Corollary 11. Let n = 2 p + 1 , p ≥ . Let q be an equilibrium configuration on S sufficiently close to ¯ q . Let Y α ∈ ( H , J − ( P m i α ) /SO (2) , ω ) be the equilibriumcorresponding to the relative equilibrium A α ( t ) q of the unreduced system. Then theequilibrium Y α of the reduced Hamiltonian system ( H , J − ( P m i α ) /SO (2) , ω ) islinearly unstable if α is smaller than a certain positive value, and is Lyapunovstable if α is larger then that value. In the case of three bodies on S , this stability property holds for all equilib-rium configurations, [7]. However, for n ≥ 5, we are unable to how to extendProposition 5 to all the n -body equilibrium configurations on S for now. Remark 12. Another interesting fact is that the relative equilibrium with thecritical angular velocity α = ±√ Θ is the intersection of two families of relativeequilibria for masses m = ... = m n = 1 , n = 2 p + 1. One family is those we havediscussed in this paper, namely, those on S , A α, ¯ q , α ∈ R . For each of the secondfamily, the masses are equally distributed on the circle x + y = sin θ , θ ∈ (0 , π ).Unlike the first family, the angular velocity is determined by θ . Actually, thecorresponding configuration is a critical point of U + α P ni =1 x i + y i = U + α P ni =1 sin θ i , [8]. Thus, the angular velocity is (cf. equations (11)), α ( θ ) = − ∂U∂θ / sin θ cos θ = n X j =2 d ij sin θ cos θ j − sin θ j cos θ cos( j πn − πn )sin θ cos θ = n X j =2 − cos( j πn − πn )sin d j As θ → π , the circle approaches the equator, and the angular velocity approaches ±√ Θ since d j → ( j − φ . Thus, the second family intersects the first family at A ±√ Θ , ( t )¯ q . In other words, there is a bifurcation going on. One can read Stoica[15] for more discussion on this bifurcation.4. acknowledgments The authors are deeply indebted to Juan Manuel S´anchez-Cerritos and CristinaStoica for suggesting the study of the stability problem of the regular polygo-nal configurations. Shuqiang Zhu would like to thank Florin Diacu for stimu-lating interest in mathematics, for his mentoring and constant encouragement.Xiang Yu is supported by NSFC(No.11701464) and the Fundamental ResearchFunds for the Central Universities (No.JBK1805001). Shuqiang Zhu is supportedby NSFC(No.11721101) and funds from China Scholarship Council (CSC NO.201806345013). References [1] V.I. Arnold, V.V. Kozlov, A.I. Neishtadt, Mathematical Aspects of Classical and Ce-lestial Mechanics. [Dynamical Systems. III], Translated from the Russian original byE. Khukhro, 3rd edition, Encyclopaedia of Mathematical Sciences, 3. Springer-Verlag,Berlin, 2006.[2] A.V. Borisov, I.S. Mamaev, A.A. Kilin, Two-body problem on a sphere. reduction,stochasticity, periodic orbits, Regul. Chaotic Dyn. 9 (2004), no. 3, 265-279.[3] J.M Cors, G.E Roberts, Four-body co-circular central configurations, Nonlinearity 25(2012), no. 2, 343-370.[4] F. Diacu, E. 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