Singularities of bihamiltonian systems and the multidimensional rigid body, страница 6

A pencil is diagonalizable at x if and only if the corresponding recursion operator is diagonalizable (over C).Remark 3.2.3. Lemma 3.2.1 implies that if Rγδ is diagonalizable, then Rαβ isdiagonalizable as well.Proof of Proposition 3.2.2. Suppose that the recursion operator is diagonalizable. Then the (complexified) space L⊥ /L is decomposed into the sum ofthe eigenspaces of the recursion operator.

All summands are orthogonal toeach other, therefore any regular bracket of the pencil is non-degenerate onthese summands, which means thatKer Pα |Ker Pλ /(Ker Pλ ∩L) = 0.But this impliesKer (Pα |Ker Pλ ) = Ker Pλ ∩ L.Taking into account Lemma 3.1.3 we obtain the followingdim Ker Pα |Ker Pλ = corank Π,37q.e.d.Now let the pencil be diagonalizable at x. This immediately implies thatPα is non-degenerate on Ker Pλ /(Ker Pλ ∩ L) for any λ. Now suppose thatthe recursion operator R0∞ has a non-trivial Jordan block for some eigenvalueλ.

This means that(R0∞ − λE)y = xfor some non-zero x, y ∈ L⊥ /L.Then for any z ∈ L⊥ /L we haveP−λ (y, z) = P∞ (x, z).If z ∈ Ker P−λ /(Ker P−λ ∩ L), then P∞ (x, z) = 0. But this means that P∞has a kernel on Ker P−λ /(Ker P−λ ∩ L). Contradiction.Corollary 3.2.1. If a pencil is diagonalizable at point x, then the space L⊥ /Lis decomposed into the sum of the real parts of the eigenspaces of the recursionoperator.

The summands of this decomposition are pairwise orthogonal withrespect to all brackets of the pencil.38Chapter 4First order theory4.1Definition of the operator Df PIn this section we will define the operator Df P , which will be very usefullater. Let P be a Poisson bracket on M , x ∈ M , df (x) ∈ Ker P (x). DefineDf P (x) : T∗x M → T∗x M by the following formulaDf P (x)(ξ) = d{f, g}P ,where g is an arbitrary function such that dg(x) = ξ.LetCkij =∂P ij.∂xkProposition 4.1.1.(Df P (x)(ξ))k = Ckij2∂fij ∂ fξ+Pξjj∂xi∂xi ∂xkand therefore does not depend on the choice of g.Proof.2∂∂ij ∂fij ∂f ∂gij ∂ f{f,g}=(P)=Cξ+Pξj ,Pjk∂xk∂xk∂xi ∂xj∂xi∂xi ∂xkq.e.d.39Proposition 4.1.2. Df P (x) is dual to the linearisation Af of a vector fieldsgrad f = P df at a point x.

In other words,hDf P (x)(ξ), vi = hξ, Af (v)ifor any v ∈ Tx M .Proof. By definitionAf (v) =∂∂.∂v ∂(sgrad f )Let ξ = dg. Thenhξ, Af (v)i =∂∂g∂{f, g}== hDf P (x)(ξ), vi,∂v ∂(sgrad f )∂vq.e.d.Proposition 4.1.3. Df P (x) is skew-symmetric with respect to P , whichmeansP (Df P (x)(ξ), η) + P (ξ, Df P (x)(η)) = 0Proof. By definition we haveDf P (x)(ξ) = d{f, g}(x), where dg(x) = ξ,Df P (x)(η) = d{f, h}(x), where dh(x) = η.ThereforeP (Df P (x)(ξ), η) = {{f, g}, h}(x),P (ξ, Df P (x)(η)) = {{h, f }, g}(x),andP (Df P (x)(ξ), η) + P (ξ, Df P (x)(η)) == {{f, g}, h}(x) + {{h, f }, g}(x) = −{{g, h}, f }(x) = 0,because df (x) ∈ Ker P .40Proposition 4.1.4.

Let W ⊂ Tx∗ M be invariant under Df P (x). Then W ⊥is also invariant.Proof. Let ξ ∈ W ⊥ , η ∈ W . ThenP (Df P (x)(ξ), η) = −P (ξ, Df P (x)(η)) = 0,because W is invariant. Therefore Df P (x)(ξ) ∈ W ⊥ , q.e.d.Proposition 4.1.5. The kernel of P at a point x is invariant with respectto Df P (x).Proof.

Indeed, Ker P = (Tx∗ M )⊥ and, therefore, is invariant.Proposition 4.1.6. Let ξ ∈ Ker P . ThenDf P (x)(ξ) = [df (x), ξ],where [, ] is the commutator in the linearization of P .Proof. By definitionDf P (x)(ξ) = d{f, g}(x), where dg(x) = ξ.On the other hand,[df (x), ξ] = [df (x), dg(x)] = d{f, g}(x)by the definition of linearization. Proposition is proved.Proposition 4.1.7.

If x is regular, then Df P (x) vanishes on Ker P .Proof. Indeed, the linearization of P is Abelian in this case.414.2Operators Df Pα for f ∈ FThe following two lemmas will allow us to rewrite the operator Df Pα , f ∈ Fas Dg Pλ for an appropriate function g.Lemma 4.2.1. Letf=kXf αi ,i=1where fαi is a Casimir function of Pαi .Let also df (x) ∈ Ker Pα (x).Let λ ∈ C and λ 6= αi for any i. Consider a functiong=kXα − αii=1λ − αif αi .Then1.

dg(x) ∈ Ker Pλ ,2. Df Pα (x) = Dg Pλ (x).Proof.1. We havePλ (dg, ξ) = Pλ=Xα − αiXλ − αidfαi , ξ=X α − αiλ − αiPλ (dfαi , ξ) =Pα (dfαi , ξ) = Pα (df, ξ) = 0.2. We haveX α − αihDg Pλ (x), dhi = dPλ (dg, dh) = dPλ (dfαi , dh) =λ − αiX=dPα (dfαi , dh) = dPα (df, dh) = hDf Pα (x), dhi.42Lemma 4.2.2. Letf=kXf αi ,i=1where fαi is a Casimir function of Pαi .Let also df (x) ∈ Ker Pα (x).Suppose that λ = αj and αj is regular at a point x. Then there exists afunction g ∈ F such that1. dg(x) ∈ Ker Pλ ,2. Df Pα (x) = Dg Pλ (x).Proof.

Consider a smooth family fν such that fν |ν=αj = fαj and fν is aCasimir function of Pν . Letf (ν) =kXf αi − f αj + f νi=1By the previous lemma, for each ν in a sufficiently small punctured neighbourhood of αj we haveDf (ν) Pα (x) = Dg(ν) Pλ (x)for some function g(ν) ∈ F. Since Df (ν) Pα (x) → Df Pα (x) as ν → αj ,the operator Df Pα (x) belongs to the closure of the set {Dg Pλ }g∈F ,dg∈Ker Pλ .But this latter set is a finite-dimensional vector space, therefore it containsDf Pα (x), q.e.d.Corollary 4.2.1.

Let f ∈ F, df (x) ∈ Ker Pα .Then Df Pα is skew-symmetric with respect to all brackets of the pencil.Proof. Taking into account Lemmas 4.2.1, 4.2.2, we see that for each λ thereexists a function g such thatDf Pα = Dg Pλ .But Dg Pλ is skew-symmetric with respect to Pλ .43Proposition 4.2.1.

Let f ∈ F, df (x) ∈ Ker Pα . Then Df Pα vanishes on L.Proof. It suffices to show that Df Pα vanishes on Ker Pγ for each regular γ.As it was shown, there exists a function g such that Df Pα = Dg Pγ . Due tothe regularity of Pγ the operator Dg Pγ vanishes on Ker Pγ (see Proposition4.1.7), q.e.d.Consequently, the operators Df Pα are well-defined on the space L⊥ /L.By Corollary 4.2.1 the following is trueProposition 4.2.2.

Let f ∈ F, df (x) ∈ Ker Pα . ThenDf Pα |L⊥ /L ∈ sp(L⊥ /L, Pβ )for any regular β, i.e. operators Df Pα |L⊥ /L are bi-symplectic.Corollary 4.2.2. Let f ∈ F, df (x) ∈ Ker Pα . Then Df Pα |L⊥ /L commuteswith the recursion operator.Proof. Let D = Df Pα |L⊥ /L . Due to the skew-symmetry we haveDPα + Pα DT = 0,DPβ + Pβ DT = 0,which impliesDR = DPα−1 P β = Pα−1 DT Pβ = Pα−1 Pβ D = RD,q.e.d.Corollary 4.2.3. Let f ∈ F, df (x) ∈ Ker Pα . Then Df Pα |L⊥ /L preservesthe eigenspaces of the recursion operator.The following Proposition allows us to calculate Df Pα on such aneigenspace.44Proposition 4.2.3.1.

Let f =kPfαi , where fαi is a Casimir function of a regular bracketi=1Pαi . Let also df (x) ∈ Ker Pα , λ ∈ Λ(x). Then" k#X α − αiDf Pα |Ker Pλ (ξ) =dfαi , ξ ,λ − αii=1where [ , ] is the commutator in gλ .2. The following sets of operators are equal{Df Pα |Ker Pλ }f ∈F ,df ∈Ker Pα = {adξ }ξ∈gλ ∩L ,where adξ is the adjoint operator in gλProof. This directly follows from Proposition 4.1.6 and Lemma 4.2.1.Now note that if the recursion operators are diagonalizable, we are ableto express Df Pα on the whole L⊥ /L via adjoint operators.

Indeed, L⊥ /L inthis case is going to be the direct sum of Ker Pλ |L⊥ /L , λ ∈ Λ(x).DenoteD = Df Pα |L⊥ /L f ∈F ,df ∈Ker Pα ⊂ sp(L⊥ /L).By Lemmas 4.2.1, 4.2.2, the subspace D does not depend on the choice of α.Lemma 4.2.3. D is invariant with respect to the recursion operator. Inother words, if D ∈ D, then RD ∈ D.Proof. Without loss of generality we may assume that D = Df Pα , where αis regular. Choose a Casimir function fα of Pα such that dfα (x) = df (x).ThenRD = Pβ Pα−1 D = Pβ Pα−1 Df −fα Pα = Pβ d2 (f − fα ) = Df −fα Pβ ∈ D.45on. SupposeCorollary 4.2.4.

Let Dλ = Df Pα |Re Ker (Pλ | ⊥ )LC⊗L /L f ∈F ,df ∈Ker Pαthat the pencil is diagonalizable. Then D =Dλ .4.3OperatorDf Pandlinearizationsofhamiltonian vector fieldsLet us consider the integrable system F |O(α,x) defined in Section 1.5. Weassume that the symplectic leaf O(α, x) is regular.The tangent space Tx O is equipped with a natural symplectic form ωαgiven by the formulaωα (Pα df, Pα dg) = {f, g}α (x).LetW = {sgrad α f = Pα df, where f ∈ F }.Let W ⊥ be the orthogonal complement to it (with respect to ωα ). Then thespace W ⊥ /W is symplectic with respect to ωα .Proposition 4.3.1. Consider the map Pα : T∗x M → Tx M . The following istrue:1.

ωα (Pα (ξ), Pα (ψ)) = Pα (ξ, ψ).2. Pα (L) = W .3. Pα (L⊥ ) = W ⊥ .4. Let Af be the linearization of sgrad f on W ⊥ /W , where df ∈ Ker Pα .46Then the following diagram is commutative:Df PαL⊥ /L −−−→ L⊥ /LPPy αy αAfW ⊥ /W −−−→ W ⊥ /WProof.1. Let ξ, ψ ∈ Tx∗ M . Choosing functions f, g such that df = ξ, dg = ψ, wewill haveωα (Pα (ξ), Pα (ψ)) = ωα (Pα (df ), Pα (dg)) == ωα (sgrad f, sgrad g) = {f, g}α (x) = Pα (ξ, ψ),q.e.d.2. Indeed, by definition W = Pα dF. On the other hand, dF = L.3. Let ξ ∈ L⊥ , v ∈ W .

Then v = Pα (ψ) for some ψ ∈ L. Thereforeωα (Pα (ξ), v) = ωα (Pα (ξ), Pα (ψ)) = Pα (ξ, ψ) = 0.Consequently, Pα (L⊥ ) ⊂ W ⊥ . Now let w ∈ W ⊥ . Since w lies in Tx Owe can find ξ ∈ T∗x M such that w = Pα (ξ). Let us show that ξ ∈ L⊥ .Let ψ ∈ L. ThenPα (ξ, ψ) = ωα (Pα (ξ), Pα (ψ)) = ωα (w, Pα (ψ)).Since Pα (ψ) ∈ W , while w ∈ W ⊥ , the latter expression vanishes, q.e.d.4.

We need to prove thatPα (Df Pα (ξ)) = Af (Pα (ξ)).47Let ξ = dg(x). Then we havePα (Df Pα (dg)) = Pα (d{f, g}α ) = sgrad {f, g}α == [sgrad f, sgrad g] = Af (sgrad g) = Af (Pα (ξ)),q.e.d.Corollary 4.3.1. Pα defines a symplectomorphism between L⊥ /L andW ⊥ /W . This symplectomorphism sends Df Pα to Af , which is the linearization of sgrad f .Corollary 4.3.2. A singular point x is non-degenerate on a regular symplectic leaf of Pα if and only if the set of operators Df Pα , where f ∈ F, df ∈Ker Pα , generate a Cartan subalgebra in sp(L⊥ /L, Pα ).

Type of the point xcoincides with the type of the Cartan subalgebra.Proof. By definition x is non-degenerate if and only if the linearizations ofthe hamiltonian vector fields sgrad f, f ∈ F generate a Cartan subalgebra insp(W ⊥ /W, ωα ). Now we need to apply the isomorphism constructed above.48Chapter 5Proof of the main theorems5.1Non-degeneracy of linear pencilsLet Πg,A be an integrable linear pencil. Construct the system F for thispencil and consider the singular point 0 on the orbit of the regular bracketA. Due to Corollary 4.3.1 there is a symplectomorphismA : L⊥ /L → W ⊥ /W,which sends Df A to the linearization of sgrad f for each f ∈ F, df ∈ Ker A.Consequently, to check non-degeneracy and find the type we need to calculateoperators Df A on L⊥ /L for f ∈ F, df ∈ Ker A.By Proposition 4.2.3 we have{Df A |Ker P0 }f ∈F ,df ∈Ker A = {adξ }ξ∈L .But Ker P0 = g∗ , while Ker A = L, therefore{Df A}f ∈F ,df ∈Ker A = {adξ }ξ∈Ker A .Since L⊥ = g∗ , we haveDf A |L⊥ /Lf ∈F ,df ∈Ker A49= {adξ |g/Ker A }ξ∈Ker A ,which proves the followingLemma 5.1.1.