Singularities of bihamiltonian systems and the multidimensional rigid body, страница 8

This means that kf indeed coincides withthe number of summands of type so(3, C), gC♦ .For a pair of pure imaginary roots there are three possibilities:1. νj ([fj , f−j ]) = 0 ⇒ g♦ .2. νj ([fj , f−j ]) < 0 ⇒ sl(2, R).3. νj ([fj , f−j ]) > 0 ⇒ so(3, R).Let us consider the second case and calculate the Killing form on the elementz = [fj , f−j ]. We have[z, fj ] = νj (z)f−j ,[z, f−j ] = −νj (z)fj .The value of the Killing form on the element z is equal totr (adz)2 = −2νi (z)2 < 0.Now let us consider the case of a pair of real roots.

There are two possibilities:591. λj ([ej , e−j ]) = 0 ⇒ gh♦ .2. λj ([ej , e−j ]) 6= 0 ⇒ sl(2, R).Consider the second case and calculate the Killing form on z = [ej , e−j ]. Wehave[z, ej ] = λj (z)ej ,[z, e−j ] = −λj (z)e−j .The value of the Killing form on z istr (adz)2 = 2λi (z)2 > 0.Therefore, the number ke is equal to the number of summands of typeg♦ + the number of summands of type so(3, R) + the number of summandsof type sl(2, R) with a negative value of Killing form on the intersectionsl(2, R) ∩ Ker A, while the number kh is equal to the number of summandsof type gh♦ + the number of summands of type sl(2, R) with a positive valueof Killing form on the intersection sl(2, R) ∩ Ker A.

The theorem is proved.5.5Proof of the non-degeneracy criterion forarbitrary pencilsBy Corollary 4.3.2, a singular point x is non-degenerate on a regular symplectic leaf of a bracket Pα if and only if the operators Df Pα , wheref ∈ F, df ∈ Ker Pα , generate a Cartan subalgebra in sp(L⊥ /L, Pα ). Thetype of the singular point coincides with the type of this subalgebra.Proposition 5.5.1. Suppose that a point x is non-degenerate.

Then thepencil Π is diagonalizable at the point x.60Proof. Indeed, since Df Pα generate a Cartan subalgebra in sp(L⊥ /L, Pα ),they should be diagonalizable. Moreover, we can find a linear combinationof these operators, which have distinct eigenvalues. The recursion operatormust commute with this linear combination (by Corollary 4.2.2). Therefore, the recursion operator must be diagonalizable. Now it suffices to applyCorollary 3.2.2.In the diagonalizable situation the space L⊥ /L is decomposed, togetherwith the form Pα , into the direct sum of the eigenspaces of the recursionoperator (Corollary 3.2.1).

These eigenspaces are invariant with respect tothe operators Df Pα (Corollary 4.2.3).Proposition 5.5.2. Let Π be a pencil diagonalizable at point x. Then thesingular point x is non-degenerate on a regular symplectic leaf of bracket Pαif and only if for each λ ∈ Λ(x) the set of operators Df Pα generate a Cartansubalgebra insp Re Ker Pλ |C⊗L⊥ /L , Pα .The type of x is the sum of types of these Cartan subalgebras.Proof.

By Corollary 3.2.1, the space L⊥ /L is decomposed into the sum ofreal parts of the eigenspaces of the recursion operator. In other words,L⊥ /L =MMKer Pλ |L⊥ /L ⊕Re Ker Pλ |C⊗L⊥ /Lλ∈R∩Λ(x)λ∈Λ(x),Im λ>0The summands of this decomposition are pairwise orthogonal with respectto all brackets of the pencil. Consequently,Msp(L⊥ /L,Pα ) =sp Ker Pλ |L⊥ /L , Pα ⊕λ∈R∩Λ(x)⊕Msp Re Ker Pλ |C⊗L⊥ /L , Pα .λ∈Λ(x),Im λ>061Now it suffices to apply Corollary 4.3.2 and Corollary 4.2.4.Proposition 5.5.3.

Let Π be a pencil diagonalizable at x. Let K = R if λis real and C otherwise. Then the set of operators Df Pα generate a Cartansubalgebra in sp(Ker Pλ |K⊗L⊥ /L , Pα ) if and only if the pencil dλ Π(x) isnon-degenerate. The type of this subalgebra for real λ coincides with the typeof Sing(dλ Π(x)).Proof. By Proposition 4.2.3{Df Pα |Ker Pλ , f ∈ F, df ∈ Ker Pα } = {adξ , ξ ∈ gλ ∩ K ⊗ L}.In the diagonalizable case we have gλ ∩ K ⊗ L = Ker (Pα |gλ ).Buth = {adξ , ξ ∈ Ker (Pα |gλ )}is a Cartan subalgebra in sp(gλ /h) if and only if the pencil dλ Π(x) is nondegenerate (Proposition 5.1.1). The type of this subalgebra coincides withthe type of Sing(dλ Π(x)) by the same Proposition 5.1.1.Proposition 5.5.4. For a complex value of λ the set of operators Df Pαgenerate a Cartan subalgebra in sp Re Ker Pλ |C⊗L⊥ /L , Pα if and only ifthe pencil dλ Π(x) is non-degenerate.

The type of this subalgebra is (0, 0, kf ),where kf equals half of its dimension.Proof. Let the pencil dλ Π(x) be non-degenerate. Then the pencil dλ Π(x) isalso non-degenerate. But this means that the set of operators Df Pα generatea Cartan subalgebra insp Ker Pλ |C⊗L⊥ /L ⊕ Ker Pλ |C⊗L⊥ /L , Pα .Now note that(Ker Pλ |C⊗L⊥ /L ⊕ Ker Pλ |C⊗L⊥ /L ) ∩ (L⊥ /L) = Re Ker Pλ |C⊗L⊥ /L .62Thereforeoperatorssp Re Ker Pλ |C⊗L⊥ /LDf Pα generate a Cartan subalgebra in, Pα as well (taking into account that theseoperators are real). The inverse statement is proved analogously.To prove the statement about the type note that if Df Pα has pure imaginary or real eigenvalue on Re Ker Pλ |C⊗L⊥ /L , then adξ in gλ has the sameeigenvalue.

But a generic element in Cartan subalgebra of sp(2n, C) doesn’thave such eigenvalues.Proof of Theorem 8. We have already shown that it is a necessary for a pencilto be diagonalizable. Therefore it suffices to show that for a diagonalizablepencil x is non-degenerate if and only if for each λ ∈ Λ(x) the linear pencildλ Π(x) is non-degenerate. But taking into account Propositions 5.5.3, 5.5.4,our statement directly follows from Proposition 5.5.2.Proof of Theorem 9. Taking into account Propositions 5.5.3, 5.5.4, our statement directly follows from Proposition 5.5.2.63Chapter 6Multidimensional rigid body6.1IntroductionSpeaking informally, free multidimensional rigid body is simply a rigid bodyrotating in multidimensional space without action of any external forces (i.e.by inertia).Let us first discuss three-dimensional free rigid body (the so-called Eulercase in the rigid body dynamics).

A good model for such a body is a bookor a parallelepiped shaped box.Now throw the book in the air spinning it in arbitrary direction. If weneglect the gravity force, then what we get is exactly the Euler case.Note that a general trajectory of a body is not a rotation in a usual sense.At each moment of time our body is indeed rotating around some axis, butthis axis is changing as time goes.

What we are interested in, are the relativeequilibria of the system, i.e. such trajectories for which the axis of rotationremains fixed. Such rotations are also called stationary.It is well known that a generic three-dimensional rigid body (i.e.

a bodywith pairwise distinct principal moments of inertia) admits three stationary64Figure 6.1: Three-dimensional rigid body. Letters S,M,L stand for the small,middle and large axes of symmetry respectively.rotations: these are the rotations around three principal axes of inertia. If wedeal with a parallelepiped shaped body, then these axes coincide with threeaxes of symmetry (see fig.

6.1).Now spin our body around one of these axes. Of course, since our handsand eyes are not too precise, we will have a small mistake in the initial data.But this is not going to be fatal if we spin the body around the shortestor the longest axis. We will not have a stationary rotation, but somethingvery close to it. This is due to the fact that the rotations of a free threedimensional rigid body around the shortest and the longest principal axes ofinertia are (Lyapunov) stable.

But if we rotate around the middle axis, wewill se something essentially different: the axis of rotation will start changingrapidly and the body will start rotating in other direction. This is becausethe rotation of a free three-dimensional rigid body around the middle principalaxis of inertia is (Lyapunov) unstable.Basically, what we are interested in, is to generalise this result to the caseof multidimensional body. The equations of multidimensional free rigid bodyin modern literature were first written by Arnold (see his book [2], however,these equations were already known to Schottky and Frahm).65In matrix representation Arnold equations look the same as Euler equations for a three-dimensional body.