Singularities of bihamiltonian systems and the multidimensional rigid body (792482), страница 13

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Considersome intersection point.One of the intersecting objects is necessary aparabola. Therefore R contains either some νeij = mi xi − mj xj or µei = mi xi .The coefficient in front of xi is non-zero in both cases. But there are no otherelements of R, containing the term mi xi .

Indeed, a parabola can’t intersectany other object at a point with the same x coordinate. Therefore, the element of R which is given by each pairwise intersection is independent with101other elements of R, which means that R is independent and the linearizationis indeed non-degenerate.Proposition 6.15.6. Let the λ-linearization of the pencil be non-degenerateat M , where λ is real.

Then the type of Sing(dλ Π(M )) is (ke , kh , 0), where• ke is the number of intersections on the parabolic diagram such thattheir x coordinate is λ and their y coordinate is positive;• kh is the number of intersections on the parabolic diagram such thattheir x coordinate is λ and their y coordinate is negative.Proof. Indeed, each intersection of parabolas correspond to the pair of roots±νij (X), whereνij (X) =p−χi (λ)(mj xj − mi xi ).The y coordinate of the intersection point isy = χi (λ).νij is real if and only if this number is negative.Intersection of a parabola with a vertical line corresponds to a pair±µi (X), whereµi (X) =p−χi (λ)mi xi .Again µi is real if and only if the y coordinate of the intersection given byy = χi (λ)is negative.We conclude that the number of pairs of real roots equals number ofintersections in the lower half-plane, while the number of pairs of imaginaryroots equals number of intersections in the upper half-plane.

Taking intoaccount Proposition 5.1.1, this proves our proposition.102Remark 6.15.1. Note that we didn’t calculate algebras gλ explicitly. Insteadof this we calculated their roots and used Proposition 5.1.1. However, it iseasy to show that• Intersection of two parabolas above the x axis corresponds to sl(2) ifthe intersection point belongs to the left branch of one parabola andto the right branch of another parabola.• Intersection of two parabolas above the x axis corresponds to so(3)if the intersection point belongs to either left or right branch of bothparabolas.

Intersection at infinity also corresponds to so(3).• Intersection of two parabolas below the x axis corresponds to sl(2).• Intersection of two parabolas at a complex point corresponds toso(3, C).• Intersection of a parabola with a vertical line above the x axis corresponds to e(2).• Intersection of a parabola with a vertical line below the x axis corresponds to e(1, 1).6.16Proof of non-degeneracy and type theoremsProof of Theorem 17.

M is non-degenerate if and only if the pencil is diagonalizable at M and all linearizations are non-degenerate (Theorem 8).Pencil is diagonalizable if and only if any two parabolas intersect exactly attwo points (Proposition 6.14.1). All linearizations are non-degenerate if and103only if there are no multiple intersections (Proposition 6.15.5). But thesetwo conditions together give exactly the condition for a parabolic diagramto be generic.Proof of Theorem 18. By Theorem 9, the type of a singular point M iske , kh , kf , whereke =Xke (λ),λ∈Λ(M )∩Rkh =Xkh (λ),λ∈Λ(M )∩Rkf =Xkf (λ) +1 X(dimC Ker Pλ − corank Π),2λ∈Λ(x),Im λ>0λ∈Λ(M )∩Rand (ke (λ), kh (λ), kf (λ)) is the type of Sing(dλ Π(M )).Let λ be real (or infinite). Then ke (λ) is the number of intersections withx = λ, y > 0, kh (λ) is the number of intersections with x = λ, y < 0, andkf (λ) = 0 (Proposition 6.15.6).

Therefore,Xke (λ) = the number of intersections in the upper half-plane,λ∈Λ(M )∩RXkh (λ) = the number of intersections in the lower half-plane,λ∈Λ(M )∩RXkf (λ) = 0.λ∈Λ(M )∩RIf λ is complex, then1(dimC Ker Pλ − corank Π)2is the number of intersections with x = λ. Since we count λ with Im λ ≥0, this sum is one half of the total number of complex intersections. Thetheorem is proved.1046.17Non-resonancyTheorem 19.

The system of Euler-Arnold equations is non-resonant, i.e.the trajectories of this system are dense on almost all Liouville tori.First we will prove several preliminary statements.Proposition 6.17.1. Let M be a regular equilibrium such that the parabolicdiagram of M is generic and all the eigenvalues of M are distinct. Thenthe eigenvalues of the linearization of the Euler-Arnold vector field at M are±σij1,2 , whereσij1,21=q−χi (x1,2ij )1,2x1,2xij+ λ2j−1 λ2jij + λ2i−1 λ2i−λ2i−1 + λ2iλ2j−1 + λ2j!,where 1 < i < j ≤ [n/2] and x1,2ij are two roots of the equation χi (x) = χj (x).If the dimension is odd, there are also eigenvalues ±τi , where 21λn + λ2i−1 λ2iτi = p− λn ,λ2i−1 + λ2i−χi (λ2n )where 1 < i ≤ [n/2].Proof.

Instead of the linearization of sgrad H, we can consider the operatorD = DH P∞ .Since Pλ dHλ = P∞ dH, we have D = DH P∞ = DHλ Pλ .Since the parabolic diagram of M is generic, the pencil is diagonalizableat M , and there is a decomposition∗TMso(n)∗ /K =MKer Pλ (M )/K,(6.10)λ∈Λ(M )where K is the common kernel of regular brackets of the pencil at point M .Since Ker Pλ is invariant with respect to DHλ Pλ , the decomposition (6.10)is invariant with respect to the operator D.

Now note thatD |Ker Pλ = addHλ .105Now, applying Propositions 6.15.2, 6.15.3 and taking into account formula(6.4), we get the formula for the eigenvalues.Proposition 6.17.2. Let λ2i−1 λ2i 6= λ2j−1 λ2j for all i, j. Then σij1,2 and τiare linearly independent as functions of M .Proof. Suppose thatXX(a1ij σij1 + a2ij σij2 ) +1<i<j≤[n/2]bi τi = 0.1<i≤[n/2]Fix k and choose those members which do not depend on mk . Their sumdoes not depend on mk , therefore the sum of all other membersSk =X12(a1ik σik+ a2ik σik)+1<i<kX12(a1kj σkj+ a2kj σkj) + b k τk .(6.11)k<j≤[n/2]does not depend on mk as well.

For simplicity denoteσjk = −σkj , ajk = −akjand rewrite (6.11) asSk =X12(a1ik σik+ a2ik σik) + bk τ k .i6=kLet mk tend to zero. It is easy to see that1,2lim σik= 0,mk →0lim τk = 0.mk →0Consequently,lim Sk = 0.mk →0But Sk does not depend on mk which means that Sk = 0.106(6.12)Now fix l 6= k. Again, there are two summands in (6.12), which dependon ml .

Their sum21+ a2lk σlkSlk = a1lk σlkmust not depend on ml . But this sum tends to 0 as ml → 0, thereforeSlk = 0. But1lim σlk= Aml ,mk →∞2lim σlk= Bml ,mk →∞where1λ2l−1 + λ2ls1B=λ2l−1 + λ2lsA=−−1,(λ2k−1 + λ2l−1 )(λ2k−1 + λ2l )(λ2k − λ2l−1 )(λ2k − λ2l ).(λ2k + λ2l−1 )(λ2k + λ2l )Consequently,a1lk A + a2lk B = 0.On the other hand,1lim σlk= Cmk ,ml →∞2lim σlk= Dmk ,ml →∞1C=−λ2k−1 + λ2ks1D=−λ2k−1 + λ2ks−−(λ2l−1 − λ2k−1 )(λ2l−1 − λ2k ),(λ2l−1 + λ2k−1 )(λ2l−1 + λ2k )(λ2l − λ2k−1 )(λ2l − λ2k ).(λ2l + λ2k−1 )(λ2l + λ2k )Consequently,a1lk C + a2lk D = 0.107It is easy to see that AD − BC = 0 if and only if λ2l−1 λ2l = λ2k−1 λ2k .

Butwe assumed that this equality is not satisfied. Thereforea1lk = a2lk = 0.1,2Since k and l were arbitrary, all coefficients aijvanish. But this implies thatbi vanish as well and our functions are linearly independent, q.e.d.Remark 6.17.1. It is easy to see that if λ2i−1 λ2i = λ2j−1 λ2j for some i, j, thenσij1 = σij2 , which means that the eigenvalues are not linearly independent.Lemma 6.17.1. Let f1 , . .

. fn be continuous functions on a manifold N .Suppose that these functions are linearly independent on any open subsetV ⊂ N . Then their values are independent over Z almost everywhere on N .Proof. Assume that this is not the case. Then there is a closed ball V0 ⊂ Nsuch that the numbers f1 (x), . . . fn (x) are dependent over Z for any x ∈ V0 .Denote f (x) = (f1 (x), . . . fn (x)) ∈ Rn . LetΓ(x) = {v ∈ Zn such that hv, f (x)i = 0}.By our assumption Γ(x) is non-empty for all x ∈ V0 . Letl(x) = min ||v||v∈Γ(x)Let alsoKv = {x : hv, f (x)i = 0}.We claim that we can find a closed ball V1 ⊂ V0 such thatmin l(x) > 1.x∈V1Indeed,{x : l(x) ≤ 1} =[v:||v||≤1108Kv .The sets Kv are closed and nowhere dense by linear independency assumptionand there is only finite number of them in the union.

Therefore, this union isalso closed and nowhere dense and we can find V1 with the desired property.Analogously, we can find a sequenceV0 ⊃ V1 ⊃ V2 ⊃ . . .of closed balls such thatmin l(x) > i.x∈V1Since the balls Vi are closed, there is a point x which belongs of them. Thevalue of l(x) for this point must be bigger than any natural number. Contradiction.Proof of Theorem 19.

Let us consider an equilibrium M such that λ1 < λ2 <. . . and the parabolic diagram is generic. In the neighbourhood of M eigenvalues of sgrad H are linearly independent. Then, by the previous lemma,they are independent over Z almost everywhere. Take a point M1 such thatthey are independent. We claim that our system is non-resonant in the neighbourhood of M1 .

Indeed, M1 is an elliptic point and the Liouville foliationon any regular symplectic leaf passing through M1 is locally given by thefunctions2s1 = p21 + q12 , . . . , sm = p2m + qmin some symplectic coordinates p, q (see Theorem 3).The functions si are action variables and the rotation numbers are givenbyci (s1 , . .

. , sm ) =∂H.∂siIt is easy to see that ci (0) are exactly eigenfrequencies of sgrad H. Theyare independent over Z. But this implies that ci are independent on almost109all tori in the neighbourhood of M1 and our system is non-resonant.

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