Singularities of bihamiltonian systems and the multidimensional rigid body, страница 5

Consider so(3)∗ with the following bracketP0 = (x2 + y 2 + z 2 )Pso(3) ,where Pso(3) is the standard Lie-Poisson bracket on so(3).Consider also any constant bracket of rank 2 and denote it by P∞ . It iseasy to check that P0 and P∞ are compatible.The Casimir function of P0 is x2 + y 2 + z 2 . The restriction of this functionto the symplectic leaf of P∞ , passing through the origin, defines an integrablesystem. The origin is a non-degenerate elliptic point of this system. However,the linearization of our pencil at the origin is zero, therefore the conditionsof Theorem 8 do not hold.But if we take a Casimir function of a regular bracket, it will look like(x2 + y 2 + z 2 )2 + linear tems,and its restriction to the symplectic leaf of P∞ is degenerate, as it is predictedby Theorem 8.The problem is that if we consider the set of the Casimir functions of allbrackets of the pencil, the function x2 + y 2 + z 2 will be an “isolated point” inthis set.

But if the set of all Casimir functions formed a smooth family, thenTheorem 8 could be applied even if F contains Casimir functions of a bracket27singular at a given point (for example, we could take as F the set of Casimirfunctions of all brackets).

This can be proved by continuity arguments.Theorem 9 (Type theorem). Assume that the conditions of Theorem 8 hold.Then the type of the singular point x is (ke , kh , kf ), whereXke =ke (λ),λ∈Λ(x)∩Rkh =Xkh (λ),λ∈Λ(x)∩Rkf =Xkf (λ) +λ∈Λ(x)∩R1 X(dimC Ker Pλ − corank Π),2λ∈Λ(x),Im λ>0and (ke (λ), kh (λ), kf (λ)) is the type of Sing(dλ Π(x)).In other words, the type of a non-degenerate singular point x is the “sum”of types of Sing(dλ Π(λ)) for all λ ∈ Λ(x). The second summand in the formula for kf appears, because Sing(dλ Π(λ)) is always a focus-focus singularityif λ is not real.The proof of Theorem 9 is given in Section 5.5.Taking into account the Eliasson theorem (Theorem 3), Theorem 9 canbe reformulated as follows:Theorem 10 (Bihamiltonian linearization theorem).

Assume that the conditions of Theorem 8 hold. Then the Liouville foliation of the system F |O(α,x)is locally symplectomorphic to Y × (Rk × Rk ),Sing(dΠ(x))λλ∈Λ(x),Im λ≥0where Rk × Rk is a trivial Lagrangian foliation andX1k = rank Π −(dim Ker Pλ − corank Π)2λ∈Λ(x)28is the rank of x.A question arises: is it possible to generalise this theorem to the case ofdegenerate linearizations?Conjecture 1 (Bihamiltonian linearization conjecture). Assume that1.

A pencil Π is diagonalizable at a point x.2. For each singular λ the linearization dλ Π(x) is “nice” (for example,gλ (x) is reductive and Pα is exact on gλ (x)).Then for each regular α the statement of Theorem 10 holds.If the conjecture is true, then any bihamiltonian system is locally equivalent to a system of “argument shift type”, assuming that the linearizationof the initial system is “nice”.29Chapter 3Zero order theoryIn this chapter we study properties of two compatible Poisson brackets ata point, i.e. properties of two skew-symmetric bilinear forms on a vectorspace.

Note that all of these properties are known and can be deduced fromthe Jordan-Kronecker theorem.3.1Geometry of a pair of skew-symmetricforms on a vector spaceConsider a pencil Π and a point x. DefineΛ = {λ ∈ C : rank Pλ (x) < rank Π(x)}.XL=Ker Pλ (x) ⊂ T∗x M.λ∈R\ΛProposition 3.1.1 (Lagrange interpolation formula). Let Π be a Poissonpencil, α 6= β. ThenPγ (ξ, ψ)(x) =β−γγ−αPα (ξ, ψ)(x) +Pβ (ξ, ψ)(x)β−αβ−αfor any ξ, ψ ∈ T∗x M .30Proof. By definition we havePα = P0 + αP∞ ,Pβ = P0 + βP∞ .Expressing P0 and P∞ and substituting the result into the formulaPγ = P0 + γP∞ ,we obtain the desired statement.Proposition 3.1.2.1.

If α 6= β, then Ker Pα is orthogonal to Ker Pβ with respect to any bracketof the pencil.2. If α ∈/ Λ, then Ker Pα is isotropic with respect to any bracket of thepencil.Proof. Let ξ ∈ Ker Pα , ψ ∈ Ker Pβ . We havePα (ξ, ψ) = 0,Pβ (ξ, ψ) = 0If α 6= β then, taking into account Proposition 3.1.1, we have Pγ (ξ, ψ) = 0for any γ.Further if α is regular thenKer Pα = lim Ker Px .x→αPassing to the limit in the equality Pγ (Ker Px , Ker Pα ) = 0 we obtain thesecond statement.31Proposition 3.1.3.

The space L is isotropic with respect to any bracket ofthe pencil.Proof. By definition L is the direct sum of the kernels of the regular brackets.By Proposition 3.1.2, all these kernels are isotropic and orthogonal to eachother with respect to any bracket of the pencil. Therefore L is isotropic withrespect to all these brackets.Proposition 3.1.4. Orthogonal complement to L does not depend on thechoice of a bracket of the pencil.Proof. Assume that Pα (ξ, L) = 0. We need to show that Pβ (ξ, L) = 0 forany β which means that Pβ (ξ, Ker Pγ ) = 0 for any regular γ.

We havePγ (ξ, Ker Pγ ) = 0,Pα (ξ, Ker Pγ ) = 0, because Pα (ξ, L) = 0, Ker Pγ ⊂ L.If γ 6= α these two equalities imply Pβ (ξ, Ker Pγ ) = 0 (taking into accountProposition 3.1.2). And if γ = α then α is regular, and our statement canbe proven by passing to the limit.Proposition 3.1.5. Let b(x, y) be a skew-symmetric bilinear form on avector space V and let W be a vector subspace in V . Then dim W ⊥ =dim V − dim W + dim(W ∩ Ker b), where W ⊥ is the orthogonal complementto W with respect to the form b.Proof. Consider the operator b : V → V ∗ given by the formulab(x)(y) = b(x, y).By definition Ker b = Ker b, thereforeKer (b |W ) = Ker b ∩ W.32Consequently,dim b(W ) = dim W − dim(Ker b ∩ W ).ButW ⊥ = Ann b(W ),which gives us the desired formula.Proposition 3.1.6. Any regular bracket of the pencil is non-degenerate onL⊥ /L.Proof.

First of all we should say that that the forms Pα are well-defined onL⊥ /L, because L is isotropic with respect to all these forms. Since α isregular, there is an inclusion Ker Pα ⊂ L. Therefore, taking into accountProposition 3.1.5,dim L⊥ = n − dim L + dim Ker Pα ,where n is the dimension of the whole space. Furthermoredim(L⊥ )⊥ = n − dim L⊥ + dim Ker Pα = dim L.But this means that (L⊥ )⊥ = L, which implies non-degeneracy of Pα onL⊥ /L.Lemma 3.1.1. Denote W =kPKer Pαi and let α and β be different fromi=1all αk .

Thendim (W ∩ Ker Pα ) = dim (W ∩ Ker Pβ ) .Proof. Decompose each Ker Pαi into a sumKer Pαi =Ker Pαi ∩i−1Xj=133!!Ker Pαj⊕ Vi .Obviously, we haveL=kMVki=1and for each ξ ∈ (Ker Pα ∩ W ) there exists a unique decompostionξ=Xξj , ξj ∈ Vj .Defineπαβ (ξ) =X αj − αξj .αj − βWe will haveX Xαj − ααj − αξj , ψ =Pβ (ξj , ψ) == Pβαj − βαj − βX αj − α αj − ββ−α=Pα (ξj , ψ) +Pα (ξj , ψ) =αj − β αj − ααj − α jX=Pα (ξj , ψ) = Pα (ξ, ψ) = 0.Pβ (παβ (ξ), ψ)Consequently, παβ (Ker Pλ ∩ W ) = Ker Pβ ∩ W . Also note that the map παβ isinvertible: the inverse is given by the formulaπβα (ξ) =X αj − βξj .αj − αTherefore παβ is an isomorphism between Ker Pα ∩ W and Ker Pβ ∩ W , andthe dimensions of these spaces are equal.Lemma 3.1.2. Let k = dim L. Then for any regular different α1 , . .

. , αkkXKer Pαi = L.i=1Proof. LetLm =mXKer Pαi .i=1By Lemma 3.1.1 for any α different from αi , where i ∈ {1, . . . , m}, we havedim (Lm ∩ Ker Pα ) = dim Lm ∩ Ker Pαm+1 .34If Ker Pαm+1 ⊂ Lm thendim (Lm ∩ Ker Pα ) = dim Ker Pαm+1 = corank Π.But if α is regular then dim Ker Pα = corank Π and therefore Ker Pα ⊂ Lm .This implies L ⊂ Lm , which means that Lm = L.Consequently, either there exists m ≤ dim L such that Lm = L, and inthis case everything is proven, or there exists a sequence of strict inclusionsL1 ⊂ L2 ⊂ · · · ⊂ Lk , where k = dim L.Since L1 is non-empty and all inculsions are strict, we have dim Lk ≥ k =dim L, therefore Lk = L, q.e.d.Corollary 3.1.1.

Let F be a system of functions defined in Section 1.5.Then dF = L.Proof. F is generated by local Casimir functions of infinite number of regularbrackets of the pencil. Differentials of local Casimir functions of a regularbracket generate kernel of this bracket. Therefore dF is a sum of kernels ofinfinite number of regular brackets of the pencil. But it is enough to takek = dim L of them to generate L.Lemma 3.1.3. dim(Ker Pλ ∩ L) = corank Π for all λ.Proof. By Lemma 3.1.2 there exists a decompositionL=kXKer Pαi .i=1By Lemma 3.1.1dim (Ker Pλ (x) ∩ L) = dim (Ker Pα (x) ∩ L)for any α different from all αi .

But if α is regular we havedim (Ker Pα (x) ∩ L) = dim Ker Pα = corank Π,which proves our Lemma.353.2The recursion operatorSince Pβ is non-degenerate on L⊥ /L for any regular β, the recursion operatorRαβ = Pβ−1 Pα for such β is well-defined.Proposition 3.2.1. For any α, γ and regular β, δ there exist constantsa, b, c, d such thatRαβ = (cRγδ + dE)−1 (aRγδ + bE).Consequently, the operators Rαβ and Rγδ commute.Proof. By definition we haveRαβ = Pβ−1 Pα ,Rγδ = Pδ−1 Pγ .Using interpolation formula we may writePα = aPγ + bPδ , Pβ = cPγ + dPδ ,which impliesRαβ = (cPγ + dPδ )−1 (aPγ + bPδ ) == (Pδ (cRγδ + dE))−1 Pδ (aRγδ + bE) = (cRγδ + dE)−1 (aRγδ + bE),q.e.d.Remark 3.2.1.

Since Rαβ is parametrized by α and β, we have a twodimensional family of recursion operators. But usually it does not matterwhich of them to consider. In this case we will simply say “the recursionoperator”.Lemma 3.2.1. The eigenspaces of the recursion operator are pairwise orthogonal with respect to all brackets of the pencil.36Remark 3.2.2. We say “the recursion operator” here, cause all recursion operators have common eigenspaces.Proof of Lemma 3.2.1.

Without loss of generality we may assume that thebracket P∞ is regular and consider the operator R0∞ .It is easy to see thatKer (R0∞ − λE) = Ker P−λ |L⊥ /L ,where E is the identity operator.But the kernels of the brackets of the pencil are pairwise orthogonal,which proves the desired statement.Proposition 3.2.2.