Singularities of bihamiltonian systems and the multidimensional rigid body (792482), страница 11

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Note that for negative λ this function Hλ is complex. If wewant a real hamiltonian, we must take the real part of Hλ (while the complexpart is a Casimir function).6.7NotationsLet us fix the notations which will be used throughout the whole chapter.We’ll mainly discuss regular equilibria (except for the theorem 14). Therefore, we may always assume that there exists an orthonormal basis such thatJ is diagonal, while Ω and M are block-diagonal with two-by-two blocks onthe diagonal.Let us denote by λi the diagonal elements of J in this basis. Note thatthis means that λi are possibly different for different equilibria.

However,they are unique up to permutation and coincide with the eigenvalues of J.By ωi we will denote the non-zero entries of matrix Ω, i.e.0 ω1 −ω1 0...0ωmΩ=−ωm 00...0By mi = (λ2i−1 + λ2i )ωi we will denote the entries of the matrix M .n will always stand for the dimension of a body, m - for the number of nonzero ωi ’s (i.e. for the number of two-dimensional planes in the decomposition(6.3)).856.8The bad setSince we know that our system is bihamiltonian, we may apply the construction discussed in the first part of the thesis.

According to this construction,the first thing we should do is to describe the set Bad, i.e. the set of pointsin which the rank of all brackets falls. This is the set where our constructionsdo not work.Proposition 6.8.1. Let M be a regular equilibrium. Then M ∈ Bad if andonly ifdim Ker M > 2.In other words, there are more than two fixed axes in the even-dimensionalcase or more than one in the odd-dimensional case.The proof can be found in Section 6.13.The construction of Section 1.5 allows us to obtain an involutive systemof integrals in the neighbourhood of any point M ∈/ Bad.

But in our casethere are globally defined integrals (because Casimir functions of all bracketsof the pencil are globally defined). It is easy to check that the global systemof integrals locally coincides with the local one (which is not always the case,see Example 2.4.1). Therefore, it is possible to apply the theorems of thefirst part of the thesis to these global integrals.However, we still cannot say anything about the points which belong tothe Bad set (though our global integrals are defined on this set as well).866.9Complete integrabilityLiouville integrability of Euler-Arnold equations was proved in [30].

This canalso be easily done using Theorem 5. We have [[S=Sλ  ∪ Sλ  ,λ∈σ(J 2 )λ∈C\σ(J 2 )whereSλ = {x : rank Pλ (x) < rank Π}and σ(J 2 ) is the spectrum of J 2 .All sets Sλ for λ ∈/ σ(J 2 ) have codimension three, because correspondingbrackets are Lie-Poisson brackets of semisimple algebras. Consequently, theirunion has measure zero. Therefore, to prove that S has measure zero, itsuffices to check that algebras which we get for λ ∈ σ(J 2 ) have the sameindex as semisimple algebras in the pencil.

But it is easy to see that thesealgebras (after complexification) are isomorphic to e(n − 1), where n is thedimension of our body. But index of e(n − 1) equals index of so(n), q.e.d.We may also apply another argument, which is in some sense better,because it allows us to prove complete integrability on a given symplecticleaf (while the previous argument only proves it on almost all leafs):If a symplectic leaf contains a non-degenerate singular point, then there iscomplete integrability on this leaf. This is due to the fact that there alwaysexists a regular point in the neighbourhood of a non-degenerate singularpoint.

But our system is analytic, therefore regular points are everywheredense on the symplectic leaf.This may sound strange, because non-degeneracy is usually defined forsystems which are already known to be completely integrable. But actuallythe definition of non-degeneracy works for any involutive system of integrals.87And if the condition of non-degeneracy is satisfied, then the system of integrals is automatically complete.We will give a simple criteria (see Section 6.11) which makes it possibleto check non-degeneracy of a zero rank singular point. Therefore, we have asimple sufficient condition for completeness on a given symplectic leaf.It is a kind of a combinatorial problem to prove that there is a nondegenerate point on each regular symplectic leaf (and, therefore, there iscomplete integrability on each orbit).

This problem is not discussed in thethesis.6.10Rank zero singular pointsTheorem 16 (Bolsinov, Oshemkov, see [8]). M ∈/ Bad is a zero-rank singular point if and only if there exists an orthonormal basis such that J isdiagonal and M is block-diagonal with two-by-two blocks on the diagonal (i.e.M is a regular equilibrium point).Proof. M is a zero-rank singular point if and only if Hamiltonian vectorfields generated by all the integrals vanish in this point, i.e. Pα dF = 0. Butwe know that dF = L, where L is the sum of kernels of regular brackets.Therefore,L ⊂ Ker Pα .Butdim L ∩ Ker Pα = corank Π,thereforedim L = corank Πand the kernels of all regular brackets at point M coincide.

On the other88hand, if the kernels coincide, then Pα dF = 0 and M is a zero-rank singularpoint.If M is block diagonal with two-by-two blocks on the diagonal and J isdiagonal, then the set of block-diagonal skew-symmetric matrices with twoby-two blocks on the diagonal lies in kernel of all brackets. By dimensionargument this set coincides with kernel for almost all brackets, which meansthat our point is a zero-rank singular point.Vice versa, let M be a point such that the kernels of all regular bracketscoincide at M .

First let us consider the case when the standard so(n) bracketis regular at M . This means that M is a matrix with pairwise distincteigenvalues.Now find an orthonormal basis such that M is block-diagonal with twoby-two blocks on the diagonal. The kernel of so(n) bracket coincides withthe centraliser of M .

Consequently, it is just the set of block-diagonal skewsymmetric matrices with two-by-two blocks on the diagonal. Since the kernelis common, we havehM, XJ 2 Y − Y J 2 Xi = 0for all X ∈ K and all Y .Suppose that B = J 2 is not diagonal, i.e. bij 6= 0 for some i 6= j. TakeY = Eij − Eji . Then(J 2 Y )ii = −bij ,(Y J 2 )ii = bij ,(J 2 Y )jj = bij ,(Y J 2 )jj = −bij ,and all other diagonal elements vanish.89Take X = M . ThenhM, XJ 2 Y − Y J 2 Xi = −Tr M 2 (J 2 Y − Y J 2 ) = 2bij ((M 2 )jj − (M 2 )ii ).Since M is regular, (M 2 )ii − (M 2 )jj can only be zero in the case when i =2k, j = 2k + 1 or vice versa. Consequently, J has block-diagonal form withtwo-by-two blocks on the diagonal.

But for such J we can find an orthogonaltransformation which preserves Ω and brings J to diagonal form.Now let us consider the case when the standard bracket is singular atpoint M . Since M ∈/ Bad, there is a bracket { , }λ in our pencil which is stillregular at M . Moreover, since almost all brackets of the pencil are regularat M , we can choose a regular bracket { , }λ such that J 2 − λE is positivedefinite. DenoteA=√J 2 − λEand consider transformation φ : so(n) → so(n) given byφ(X) = AXA.It is easy to see thatφ([X, Y ]) = φ(X)A−2 φ(Y ) − φ(Y )A−2 φ(X),φ([X, Y ]λ ) = [φ(X), φ(Y )].Consequently, our pencil is mapped to the pencil^[X,Y ]λ = X(A−2 − λE)Y − Y (A−2 − λE)X.The point φ∗ (M ) = A−1 M A−1 is going to be the point where all bracketsof the new pencil have coinciding kernels.

This point is regular with respectto the standard bracket, consequently there is an orthogonal transformationwhich brings A−2 to diagonal form and A−1 M A−1 to block-diagonal form.But this implies that J and M will have necessary form as well.906.11Non-degeneracy and type theoremsTheorem 17. Zero-rank singular point M ∈/ Bad is non-degenerate if andonly if the parabolic diagram of M is generic.Remark 6.11.1. A sufficient condition for non-degeneracy is given in [8]. Inour terms it means the following:• The parabolic diagram is generic.• For each λ there is no more than one intersection point on the parabolicdiagram with x coordinate equal to λ.Theorem 18. The type of a non-degenerate zero-rank singular point M ∈/Bad is (ke , kh , kf ), where• ke is the number of real intersections on the parabolic diagram in theupper half-plane plus the number of intersections at infinity,• kh is the number of real intersections in the lower half-plane,• kf is half the number of complex intersections.Sections 6.12-6.16 are dedicated to the proof of these theorems.Example 6.11.1 (Three-dimensional body).

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