Singularities of bihamiltonian systems and the multidimensional rigid body, страница 12

Rotations around the longest andthe shortest axes are elliptic singular points (see parabolic diagrams on figures6.2, 6.4). Rotation around the middle axis is a hyperbolic singularity (seeparabolic diagram on figure 6.3).Example 6.11.2 (Four-dimensional body). The first case from example 6.5.2corresponds to a center-center singular point (see figure 6.5).The second case corresponds to a center-saddle singular point (see figure6.6).91In the third case everything depends on the ratio of angular velocities. Aswe change the angular velocity of rotation in the “inner” plane, the followingbifurcations occurs:Center-center (figure 6.7) → degenerate (figure 6.10) → focus-focus (figure 6.8) → degenerate (figure 6.11) → saddle-saddle (figure 6.9).6.12Non-degeneracy: scheme of the proofAccording to the general scheme stated in section 2.4 to prove non-degeneracyand find type of singular point M we should do the following:• Find those λ, for which the rank of Pλ (M ) drops down, i.e.

describethe set Λ(M ). This is done in section 6.13.• Check that the pencil is diagonalizable at M . This is done in section6.14.• Linearize the pencil, check that linearisations are non-degenerate andfind their type. This is done in section 6.15.• Collect all this together. This is done in section 6.16.6.13Description of Λ(M )Let M be a zero-rank singular point. Let us find a basis such that J isdiagonal and M is block-diagonal. Let us introduce the following subspaces:• K ⊂ so(n) is generated by E2i−1,2i − E2i,2i−1 where i = 1, .

. . , m andall Eij − Eji for 2m < i < j ≤ n.92• Vij ⊂ so(n) is a subspace generated by E2i−1,2j−1 −E2j−1,2i−1 , E2i−1,2j −E2j,2i−1 , E2i,2j−1 − E2j−1,2i , E2i,2j − E2j,2i .• Wij ⊂ so(n) is a subspace generated by E2i−1,j − Ej,2i−1 , E2i,j − Ej,2i .We have a vector space decompositionso(n) = K ⊕M1≤i<j≤mVij ⊕MWij .1≤i≤m,2m<j≤nProposition 6.13.1. The space K belongs to the common kernel of all brackets of the pencil at the point M .

All spaces Vij , Wi are pairwise orthogonalwith respect to all brackets of the pencil at the point M .Proof. This is a simple straightforward computation.Therefore, the rank of a bracket of the pencil drops if and only if thisbracket is degenerate on one of Vij or Wij . Let us calculate our brackets onthese spaces.Identify Vij with the space of two-by-two matrices and Wij with R2 . LetM1 , . . . Mm be two-by-two diagonal blocks of M . Let A = J 2 − λE if λ 6= ∞and identity matrix otherwise. Write down A asA 1...AmA=a2m+1...an,where Ai are two-by-two diagonal matrices and ai are numbers.93Proposition 6.13.2.

The form Pλ restricted on Vij has the formPλ (X, Y ) = 2Tr (Mi XAj Y t + Mj X t Ai Y ).The form Pλ restricted on Wij has the formPλ (v, w) = −2aj Mi (v, w).Proof. This is a straightforward computation.Let us now calculate Pλ on Vij in coordinates. Leta00ms. , As =  2s−1Ms = 0a2s−ms 0Let alsoX=a bc d,Y = e fg h.Explicit calculation shows thatPλ (X, Y ) = 2(mi a2j−1 c + mj a2i−1 b)e + 2(mj a2i d − mi a2j−1 a)g++ 2(mi a2j d − mj a2i−1 a)f − 2(mi a2j b + mj a2i c)h.Consequently X ∈ Ker Pλ if and only ifmi a2j−1 c + mj a2i−1 b = 0,mj a2i d − mi a2j−1 a = 0,mi a2j d − mj a2i−1 a = 0,m a b + m a c = 0.i 2jj 2iThis system can be splitted onto two two-by-two systems and the determinantof both of them equalsdet = m2j a2i−1 a2i − m2i a2j−1 a2j .Consequently, we have proved the following94Proposition 6.13.3. Pλ is degenerate on Vij if and only ifm2j a2i−1 a2i − m2i a2j−1 a2j = 0,(6.5)where as = λ2s − λ if λ 6= ∞ and as = 1 if λ = ∞.The kernel in this case is given bya = αmj a2i ,b = βmi a2j−1 ,c = −βmj a2i−1 ,d = αm ai 2j−1 ,where α and β are arbitrary numbers.Now we shall study Pλ on Wij .

The following is straightforwardProposition 6.13.4. Pλ is degenerate on Wij if and only if aj = 0 whereaj = λ2j − λ if λ 6= ∞ and aj = 1 otherwise.Proposition 6.13.5. The intersection of kernels of all brackets of the pencilis exactly K. For almost all brackets the kernel is exactly K.Proof. Indeed, only finite number of brackets are degenerate on each Vij andWij .Now we are able to describe the Bad set.Proof of Proposition 6.8.1. Indeed, for almost all brackets the kernel is exactly K, which means that all brackets are degenerate if and only ifdim K >hni2.This is equivalent to the condition dim Ker M > 2, q.e.d.95Now we can prove the followingProposition 6.13.6. Let M ∈/ Bad. Then Λ(M ) is the set of x coordinatesof the intersections on the parabolic diagram of M .Proof. Pλ is degenerate on Vi j if and only if m2j a2i−1 a2i − m2i a2j−1 a2j = 0.This can be rewritten asχi (λ) = χj (λ).But this means that λ is the x coordinate of the intersection point of twoparabolas.Further, Pλ is degenerate on Wij if and only if aj = 0 or, which is thesame, λ2j −λ = 0.

But this means that λ is the x coordinate of the intersectionpoint of the vertical line with any parabola.6.14When is the pencil diagonalizable?As a next step, we should check that the pencil is diagonalizable at point M .Proposition 6.14.1. The pencil is diagonalizable at point M ∈/ Bad if andonly if any two parabolas on the parabolic diagram of M intersect at twodifferent points.Proof. Proposition 3.2.2 implies that a pencil is diagonalizable if and only ifML⊥ /L =Ker Pλ |L⊥ /L .λ∈Λ(M )In our case L is the common kernel K and L⊥ /L can be naturally identifiedwithM1≤i<j≤mMVij ⊕1≤i≤m,2m<j≤n96Wij ,therefore the following relation must be satisfied:MMVij ⊕1≤i<j≤mWij =1≤i≤m,2m<j≤nMKer Pλ |L⊥ /L .λ∈Λ(M )This is satisfied if and only ifVij =MVij ∩ Ker Pλ |L⊥ /L(6.6)λ∈Λ(M )for 1 ≤ i < j ≤ m andWij =MWij ∩ Ker Pλ |L⊥ /L(6.7)λ∈Λ(M )for 1 ≤ i ≤ m, 2m < j ≤ n.For each Wij there is a unique λ such thatWij = Wij ∩ Ker Pλ |L⊥ /L .This follows from Proposition 6.13.4.

Therefore, relation (6.7) is alwayssatisfied.Relation (6.6) is satisfied if and only if the equation (6.5) has two distinctroots, i.e. if two corresponding two parabolas are not tangent to each other,q.e.d.6.15LinearizationThe last step is to linearise the pencil and to check whether the linearizationsare non-degenerate. We haveso(n) = K ⊕M1≤i<j≤m97Vij ⊕M1≤i≤m,2m<j≤nWij .The kernel of each Pλ (M ) can be decomposed in the following wayKer Pλ = K ⊕M1≤i<j≤mMVeij ⊕fij ,W1≤i≤m,2m<j≤nfij ⊂ Wij .where Veij ⊂ Vij , Wfij are invariant with respect to adjointProposition 6.15.1.

Spaces Veij , Woperators adX in gλ , where X ∈ K.Proof. Linearization gλ is simply a stabilizer of M with respect to the bracket[ , ]λ . Consequently, the commutator in gλ has the form[X, Y ]λ = XAY − Y AX,where A = J 2 − λE for finite λ and A = E for λ = ∞.Let X ∈ K, Y ∈ Vij . Then it is easy to see that [X, Y ] ∈ Vij , which meansthat adX (Vij ) ⊂ Vij . But Veij = Vij ∩ Ker Pλ , therefore Veij is invariant.

Thefij is the same.proof for WNow represent an element X ∈ K as0 x1 −x1 0X=0 x2−x2 0....Proposition 6.15.2. Consider the case when Veij is non-empty. Then theeigenvalues of adX restricted on Veij are ±νij (X), whereνij (X) =p−χi (λ)(mj xj − mi xi ).98Proof. Let X1 , . . . be the diagonal two-by-two blocks of X, and Y is a twoby-two matrix representing an element of Vij .

Let also A1 , . . . Am be thediagonal two-by-two blocks of A. Then[X, Y ]λ = Xi Ai Y − Y Aj Xj .Straightforward computation shows that adX sendsa bY =c dto the matrixadX Y = xi a2i c + xj a2j bxi a2i d − xj a2j−1 a−xi a2i−1 a + xj a2j d −xi a2i−1 b − xj a2j−1 c(6.8)Veij consists of matrices Y having formαmj a2iβmi a2j−1Y (α, β) = −βmj a2i−1 αmi a2j−1Substituting this into the formula (6.8), we get−β(mj a2i−1 a2i xi − mi a2j−1 a2j xj ) α(mi a2j−1 a2i xi − mj a2i a2j−1 xj )∗∗Therefore, in coordinates (α, β) the map adX restricted to Veij has the formmj0a2i (xi − mi xj )(6.9)m0−a2i−1 (xi − mji xj )The determinant of matrix (6.9) equalsdet = −a2i a2i−1 (xi −mj 2xj ) ,miwhich gives us the desired formula for the eigenvalues.99fij is non-empty.

Then theProposition 6.15.3. Consider the case when Wfij are ±µi (X), whereeigenvalues of adX restricted on Wµi (X) =p−χi (λ)mi xi .Proof. Represent an element of Wi by a vector v ∈ R2 . Then[X, v]λ = Xi Ai v,which means that adX restricted to Wi is given by the matrix0xi a2i,Xi Ai = −xi a2i−10and the eigenvalues have the desired form.Consider the following set R of linear functions on X:R = {νij (X), where i, j are such that Veij =6 0} ∪fij =∪ {µi (X), where i is such that W6 0 for some j}Now we see that the following is trueProposition 6.15.4. K is a diagonalizable subalgebra in gλ .

The set of rootsof gλ with respect to K is the set {±ξ, ξ ∈ R}.Proposition 6.15.5. λ-linearization of the pencil is non-degenerate at Mif and only if there are no three objects (which may be parabolas or verticallines) on the parabolic diagram which intersect at a point with x coordinateequal to λ.Proof. To prove our proposition it suffices to show that the linear functionsbelonging to R are independent if and only if there are on three objects on100the parabolic diagram which intersect at a point with x coordinate equal toλ.First note that up to multiplication on non-zero constant the elements ofR areνeij (X) = mi xi − mj xj ,µei (X) = mi xi .If there are three parabolas intersecting at one point (with x coordinate equalto λ), then the spaces Veij , Vejk , Veik are non-empty for some different i, j, k.

Butthis implies that R contains νeij , νejk , νeik and sinceνeij + νejk = νeik ,the elements of R are not independent and the linearization is degenerate(see Proposition 5.1.1).Now suppose that there is an intersection of two parabolas and one vertical line. Then R contains νeij , µei , µej and sinceνeij = µei − µej ,the elements of R are not independent and the linearization is again degenerate.Vice versa, suppose we do not have any triple intersections.