Singularities of bihamiltonian systems and the multidimensional rigid body, страница 9

But we shall emphasise that Arnoldequations are not just Euler equations generalised to the multidimensionalcase, but the equations which describe a real mechanical object: a multidimensional rigid body. And this object (but not just the equations!) is whatwe want to study.First we shall discuss how an n-dimensional body may rotate.

At eachmoment of time Rn is decomposed into the sum of m pairwise orthogonaltwo-dimensional planes Π1 , . . . , Πm and an n − 2m-dimensional space Π0 ,orthogonal to all these planes:Rn =mM!Πi⊕ Π0 .i=1There is an independent rotation in each of the planes Π1 , . . . , Πm , while Π0is fixed. This is just a reformulation of the theorem about canonical form of askew-symmetric operator. Note that Π0 may be zero in the even-dimensionalcase, which means that there are no fixed axes.A rotation is stationary if all planes Π0 , .

. . , Πm don’t change with time(this condition automatically implies that the velocities of rotations are alsoconstant). It is known that a rotation of a generic multidimensional rigidbody is stationary if and only if the corresponding planes are spanned bymain axes of inertia (provided that the angular velocities of rotations indifferent planes are pairwise distinct1 , see [14, 4]). We wonder which of theserotations are stable and which are not.In four dimensions the problem was studied from different points of viewby Oshemkov (see [32]), Feher and Marshall (see [14]), Petre Birtea, IoanCaşu, Tudor S.

Ratiu, and Murat Turhan (see [4]), Petre Birtea and Ioan1Some authors omit this condition, however it is crucial, see section 6.3 of this thesis.66Caşu (see [3]). The answer is known for a dense subset of relative equilibria.The five-dimensional case was studied by Caşu in [9]. The set of equilibriawhich were studied in this case is not dense.General even-dimensional rigid body was discussed in Spiegler’s PhD thesis [37]. A sufficient condition for an equilibrium to be stable is found.A.Spiegler approached the problem using the so-called Arnold’s energyCasimir method (see [1, 34]). The main idea of this method is the following:Suppose we have a system which is hamiltonian on some Poisson manifold.Let x be an equilibrium of the system. Then x is a critical point for theHamiltonian restricted to the symplectic leaf passing through x.

If it turnsout that this point is non-degenerate maximum or minimum, then x is astable equilibrium, provided that the corresponding symplectic leaf is regular.The Energy-Casimir method is a very powerful tool for studying equilibriaof general Hamiltonian systems.

But if our system possesses some additionalsymmetries (for example, it is completely integrable), there are more simpleand efficient ways to study stability. One may see, comparing the result ofA.Spiegler with our results, that there is an open subset in the set of stableequilibria for which the hamiltonian is not positive definite.

Therefore, forthis equilibria the energy-Casimir method doesn’t work.As it was shown by Manakov in his paper [27], our system is completelyintegrable2 . Therefore we may apply methods developed in the theory of2More precisely, Manakov has shown that the system admits an L − A pair with aspectral parameter. This allowed him to write down integrals and to show that the systemis integrable in θ-functions of Riemannian surfaces. However, Manakov didn’t prove thathis integrals are enough for Liouville integrability. Complete integrability in the Liouvillesense was proved in [30]. See also earlier paper [29], where Mischenko found the quadraticintegrals of the problem. These integrals are enough for complete integrability in thefour-dimensional case.67integrable hamiltonian systems.

For example, if we know that an equilibriumpoint is non-degenerate and know its type, then the stability problem is easilysolved. The four-dimensional case from the integrable point of view wasstudied in [32]. The integrable approach is very effective, but still involvesheavy calculations if the dimension is larger then four. However, instead ofexplicit calculations with the integrals, we can use the fact that the systempossesses a bihamiltonian structure.The bihamiltonian structure for the multidimensional rigid body equations which we are going to use was discovered by A.Bolsinov in [5] (see also[6, 31]). This structure is defined on the dual space of the Lie algebra ofskew-symmetric matrices3 .The bihamiltonian approach for studying the singularities of multidimensional rigid body was applied in [8]. In this paper Bolsinov and Oshemkovobtain a sufficient condition for non-degeneracy of zero-rank singularities.Developing their ideas and applying the constructions of the first part ofthe present thesis, we manage to solve the stability problem for almost allrelative equilibria.

Moreover, we give a non-degeneracy criteria for an equilibrium point and describe type of non-degenerate points, which makes itpossible to give a complete description of the behaviour of the system in aneighbourhood of an equilibrium.The answers we get are simple and geometric.3However, it is possible to give another bihamiltonian formulation: the bihamiltonianstructure is defined on the dual of sl(n)∗ and then our system is obtained by restrictionfrom sl(n)∗ to so(n)∗ . This structure (in different terminology) is present in the paper[30].

See also [35].686.2The equationsMotion of multidimensional rigid body is described by the Euler-Arnold equations on so(n)∗ (identified with so(n)). These equations have the formṀ = [M, Ω]M(6.1)= ΩJ + JΩ,where• M ∈ so(n)∗ is a skew-symmetric matrix, called the angular momentummatrix.• J is a symmetric matrix, called the mass tensor of the rigid body.• Ω is a skew-symmetric matrix, called the angular velocity matrix. It isuniquely defined by the relationM = ΩJ + JΩ.Remark 6.2.1. Since the map J : so(n) → so(n) given by the formulaJ (Ω) = ΩJ + JΩis invertible, our equations can be rewritten in the Ω-coordinates:Ω̇ = J −1 ([J (Ω), Ω]).However, the explicit formula for J −1 is quite complicated, therefore it isconvenient to introduce the variable M and write down the equations in theform (6.1).Note that the equations (6.1) are only equations on angular velocities of thebody.

If we want to recover the dynamics in the configuration space, weshould add Poisson equationsẊ = XΩ.69However, we will only be interested in reduced dynamics, given by equations(6.1). Note that relative equilibria of the rigid body is nothing else but theequilibrium points of the system (6.1).6.3Description of relative equilibriaTheorem 11. Consider the system of Euler-Arnold equationsṀ = [M, Ω]M= ΩJ + JΩ.Suppose that J has pairwise distinct eigenvalues. Then M is an equilibriumpoint of the system if and only if there exists an orthonormal basis such thatJ is diagonal, and Ω is block-diagonal of the following formνA00 1 1... 00 ,00 νk Ak(6.2)where Ai ∈ so ∩ SO and νi are arbitrary real numbers.Remark 6.3.1.

Relative equilibria of Euler-Arnold equations was studied bymany authors, however we could not find the final answer given by thistheorem in the literature.Proof. We have[M, Ω] = [ΩJ + JΩ, Ω] = [J, Ω2 ],therefore M is an equilibrium if and only if Ω2 commutes with J.Assume we have a basis such that J is diagonal and Ω has the form (6.2).ThenA2i = −Ai Ati = −E70and−ν12 E2Ω =000...00−νk2 E0.Therefore, [Ω2 , J] = 0, and our point is an equilibrium point.Now, let [Ω2 , J] = 0.

We shall prove that there exists a basis such that Jis diagonal and Ω has the form (6.2).First find an orthonormal basis such that J is diagonal. Ω2 is diagonal inthis basis as well, since J has pairwise distinct eigenvalues. By permutationof basis vectors bring Ω2 to the form−ν12 E 00... 0000 −νk2 E,where all νi are pairwise distinct.Note that eigenspaces of Ω2 are orthogonal with respect to Ω.

Indeed, letΩ2 x = −νi2 x, Ω2 y = −νj2 y, and νj 6= 0. ThenhΩx, yi = −11 2νi2νi22hΩx,Ωyi=hΩx,Ωyi=−hx,Ωyi=hΩx, yi.νj2νj2νj2νj2Since νi 6= νj , we have hΩx, yi = 0.Consequently, in our basis Ω has the formB00 1... 0 , 00 0 Bkwhere Bi2 = −νi2 E.We need to prove that Bi = νi Ai , where Ai ∈ so ∩ SO. If νi = 0,everything is proved. If not, setAi =1Bi .νi71On one hand, Ai ∈ so. On the other handAi Ati = −A2i = −1 2B = E,νi2 iwhich means that Ai ∈ SO, q.e.d.Remark 6.3.2. Note that so(2m) ∩ SO(2m) is the homogeneous spaceO(2m)/U(m) which is identified with the space of complex structures compatible with the standard euclidian metrics.Corollary 6.3.1. Suppose that M is a relative equilibrium, all eigenvalues ofJ are pairwise distinct.