LECTURE III. BiHamiltonian chains and it projections. Maciej B laszak. Poznań University, Poland


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1 LECTURE III BiHamiltonian chains and it projections Maciej B laszak Poznań University, Poland Maciej B laszak (Poznań University, Poland) LECTURE III 1 / 18
2 BiHamiltonian chains Let (M, Π) be a Poisson manifold with Π of constant rank (almost everywhere), with Poisson algebra given by the bracket {F 1, F 2 } Π = Π(dF 1, df 2 ) = df 1, ΠdF 2. Remark. In local Darboux coordinates (q 1,..., q n, p 1,..., p n, c 1,..., c m ) Π takes the form 0 I 0 Π = I where coordinates c i are Casimir functions. Definition A linear combination Π λ = π 1 λπ 0 (λ R) of two Poisson tensors Π 0, Π 1 is called a Poisson pencil if Π λ is Poisson for any value of λ. Then Π 0 and Π 1 are called compatible. Maciej B laszak (Poznań University, Poland) LECTURE III 2 / 18
3 BiHamiltonian chains Given a Π λ we can often construct a sequence of vector fields X i that have two Hamiltonian representations (bihamiltonian chains) X i = Π 1 dh i = Π 0 dh i+1, where H i C (M) are called Hamiltonians of a chain. Consider a bipoisson manifold (M, Π 0, Π 1 ) of dim M = 2n + m, where Π 0, Π 1 is a pair of compatible Poisson tensors of rank 2n. We further assume that Π λ admits m Casimir functions which are polynomial in λ H (k) (λ) = n k H (k) i λ n k i, k = 1,..., m i=0 so, that n n m = n and H (k) i are functionally independent. Maciej B laszak (Poznań University, Poland) LECTURE III 3 / 18
4 BiHamiltonian chains The collection of n bihamiltonian vector fields Π λ dh (k) (λ) = 0 Π 0 dh (k) 0 = 0 Π 0 dh (k) 1 = X (k) 1 = Π 1 dh (k) 0. Π 0 dh n (k) k = X n (k) k = Π 1 dh (k) n k 1 0 = Π 1 dh n (k) k (3.1) Lemma where k = 1,..., m, define bihamiltonian systems of Gel fandzakharevich type. Notice that each chain starts from a Casimir of Π 0 and terminates with a Casimir of Π 1. All H (k) i pairwise commute wit respect to Π 0 and Π 1. Maciej B laszak (Poznań University, Poland) LECTURE III 4 / 18
5 BiHamiltonian chains GZ bihamiltonian chain Liouville integrable system Having GZ bihamiltonian system the crucial toward construction of separation coordinates is the projection of second Poisson tensor Π 1 onto the symplectic foliation of the first Poisson tensor Π 0. Maciej B laszak (Poznań University, Poland) LECTURE III 5 / 18
6 Poisson projections onto submanifolds Consider manifold M of dim M = m and foliation S consisting of leaves S ν parametrized by ν R r, so r is codimension of every leave. Let Z be a distribution transversal to S, i.e. T x M = T x S ν Z x where S ν is a leave that passes through x. It defines a decomposition ( ) TM X = X + X, X x T xs ν, (X ) x Z x and induces splitting of dual space T x M = T x S ν Z x, where T x S ν annihilates Z x and Z x annihilates T x S ν. Maciej B laszak (Poznań University, Poland) LECTURE III 6 / 18
7 Poisson projections onto submanifolds Thus, any oneform α on M has a unique decomposition T ( ) M α = α + α, α x T x S ν, (α ) x Zx. Definition A function F C (M) is called Zinvariant if L Z F = Z (F ) = 0 for any Z Z. Obviously df T S. Definition The Poisson tensor Π is said to be Zinvariant if L Z {F, G } Π = 0 for any pair of Zinvariant functions F, G and any Z Z. Maciej B laszak (Poznań University, Poland) LECTURE III 7 / 18
8 Poisson projections onto submanifolds For any Poisson tensor Π on M define the following bivector Π D Π D  deformation of Π. Π D (α, β) := Π(α, β ), α, β T M (3.2) Lemma The image of Π D is tangent to the foliation S. Indeed, α, Π D β = α, Πβ = 0 for β Z = Z ker Π D. So, Π D can be naturally restricted to any leave S ν : π = Π D S. Theorem If Π is Zinvariant, then Π D is Poisson and so π. Maciej B laszak (Poznań University, Poland) LECTURE III 8 / 18
9 Poisson projections onto submanifolds Proof. From Zinvariant of Π : L Z df, ΠdG = 0, so d df, ΠdG T S = ( d df, ΠdG ) = d df, ΠdG, hence, the Jacobi identity {{F, G } ΠD + c.p. = ( d df, ΠdG ), ΠdH + c.p. = d df, ΠdG, Π is fulfilled. Observation. Annihilator Z of TS is defined as soon as S is determined. Definition Distribution D = Π(Z ), associated with the foliation S, is called Dirac distribution. Maciej B laszak (Poznań University, Poland) LECTURE III 9 / 18
10 Poisson projections onto submanifolds Two limited cases are possible: 1 TM = D TS Dirac case 2 D TS tangent case. Let S be parametrized by ϕ i (x) C (M) : S ν = {x M : ϕ i (x) = ν i, i = 1,..., r}, so {d ϕ i } basis in Z. Denote by {Z i } a basis dual to {d ϕ i }, so Z i (ϕ j ) = δ ij. Then, X = X r i=1 X (ϕ i )Z i, α = α r i=1 α(z i )d ϕ i. Obviously X (ϕ i ) = 0 and α (Z i ) = 0. Maciej B laszak (Poznań University, Poland) LECTURE III 10 / 18
11 Poisson projections onto submanifolds Then, from Π D (α, β) = Π(α, β ) we get r Π D = Π X i Z i + 1 i=1 2 where X i = Πd ϕ i, ϕ ij = {ϕ i, ϕ j } Π. r ϕ ij Z i Z j (3.3) i,j=1 In the Dirac case we have a canonical choice of Z = D, as all X i are transversal to S and are linearly independent (det(ϕ ij ) = 0). Moreover Π is naturally Zinvariant since L Xi Π = 0. ϕ i (x) are then second class constraints in Dirac terminology. The basis {Z i } dual to {d ϕ i } can be expressed by X i : Z i = r ( ϕ 1 ) ji X j, Z j (ϕ i ) = δ ij. j=1 Maciej B laszak (Poznań University, Poland) LECTURE III 11 / 18
12 Poisson projections onto submanifolds Π D (3.3) attains the form Π D = Π 1 2 r ( ϕ 1 ) ij X j X i. i,j=1 This Poisson tensor defines the following bracket on C (M) : {F, G } ΠD = {F, G } Π r i,j=1 known as a Dirac bracket. In tangent case all X i are tangent to S, i.e. X i (ϕ j ) = Π(d ϕ i, d ϕ j ) = 0, so Π D = Π {F, ϕ i } Π ( ϕ 1 ) ij {ϕ j, G } Π r i=1 X i Z i and Z i must be find separately from case to case. Maciej B laszak (Poznań University, Poland) LECTURE III 12 / 18
13 From bihamiltonian to quasibihamiltonian chains Our foliation is symplectic foliation of Π 0, so Z = Ker Π 0 = Sp{dc i }. For GZbiHamiltonian chains Π 1 (dc i, dc j ) = 0, so we are in tangent case when project Π 1 onto S. Assume we found Z transversal to S and such that Π 1 is Z  invariant. Then Π D is Poisson with Ker Π 0 Ker Π D, so both tensors can be restricted to S. Let π 0 = Π 0 S, π 1 = Π D S. Maciej B laszak (Poznań University, Poland) LECTURE III 13 / 18
14 From bihamiltonian to quasibihamiltonian chains Theorem BiHamiltonian chains (3.1) restricted to S take the form of quasibihamiltonian chains ( ) π 1 dh (k) i = π 0 dh (k) m i+1 α (k) ij dh (j) 1 j=1, (3.4) where ( ) α (k) ij = Z j (H (k) i ), S h(k) i = (H (k) i ) S. It follows from the fact that Π 1 = Π D + m k=1 X (k) 1 Z k. Maciej B laszak (Poznań University, Poland) LECTURE III 14 / 18
15 From bihamiltonian to quasibihamiltonian chains Lemma {H (k) i, H (r) j } ΠD = 0. Theorem Necessary and sufficient condition for compatibility of Π 0 and Π D (hence π 0 and π 1 ) is L Zi Π 0 = 0. Z i are symmetries of Π 0 (so Π 0 is Z  invariant), hence Z is integrable distribution and [Z i, Z j ] = 0. Maciej B laszak (Poznań University, Poland) LECTURE III 15 / 18
16 Nmanifolds Definition ωn manofold is a bipoisson manifold (π 0, π 1 ) with two compatible Poisson tensors, where at least one (say π 0 ) is nondegenerated. So M is endowed with a symplectic form ω = ω 0 = π 1 and the (1, 1) tensor field N = π 1 ω 0. From compatibility of π 0 and π 1 follows that Nijenhuis torsion of N vanishes T (N)(X, Y ) = [NX, NY ] N[NX, y] N[X, NY ] + N 2 [X, Y ] = 0 and hence, the second twoform is closed. L NX N = NL X N ω 1 = ω 0 π 1 ω 0 Maciej B laszak (Poznań University, Poland) LECTURE III 16 / 18
17 Nmanifolds We further restrict to generic case, when N has at every point n distinct eigenvalues which are functionally independent. It means that π 1 is also nondegenerated and ω 1 is symplectic. Notice that Moreover, π 1 = Nπ 0, ω 1 = N ω 0, N = ω 0 π 1. {f, g} πi = ω i (X f, X g ), X h = π 0 dh, i = 1, 2. Let us come back to our quasibihamiltonian chain (3.4). The distribution tangent to the foliation defined by (h (1) 1,..., h(m) n m ) is bilagrangian: ω 0 (X h (k) i, X h (r) j ) = ω 1 (X h (k) i, X (r) h ) = 0. j Maciej B laszak (Poznań University, Poland) LECTURE III 17 / 18
18 Nmanifolds Moreover, quasibihamiltonian equations can be put into one matrix equation N m dh i = F ij dh j N dh = Fdh j=1 where (h 1,..., h n ) = (h (1) 1,..., h(m) n m ) and F control matrix (recursion matrix). Maciej B laszak (Poznań University, Poland) LECTURE III 18 / 18
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