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Hamiltonian flows and the exact splitting map preserve the complexified canonical form and the Poisson bracket

Statement

Let d≥1 and let (U,Uρ,A~,B~,M,K,δ,Ψh) be analytic splitting data as in Analytic splitting data with uniform holomorphic neighbourhood, and the exact symmetric splitting map Ψh, with J the standard matrix and ωC, { ⋅ , ⋅ } the complexified form and bracket of Complex phase domains, holomorphic maps on Cm, and the complexified canonical form and bracket. Let F∈{A~,B~} and let φFt be the flow of XF as in The flow of a bounded holomorphic field on a polydisc: existence in complex time, displacement bound, holomorphy, and the composition law, on the domain where it is defined.

  1. For every (t,Z) in the flow domain, φFt is holomorphic near Z, its derivative is invertible, and DφFt(Z)TJ DφFt(Z)=J.
  2. For every h and every point Z where Ψh is defined and holomorphic, DΨh(Z) is invertible and DΨh(Z)TJ DΨh(Z)=J.
  3. For all holomorphic f,g on an open set containing Ψh(Z), {f∘Ψh,  g∘Ψh}(Z)={f,g}(Ψh(Z)).

Facts & Assumptions

Given: The analytic splitting data of Analytic splitting data with uniform holomorphic neighbourhood, and the exact symmetric splitting map Ψh; a field F∈{A~,B~} with ∥XF∥≤M on Uρ; the flow φFt of XF; the exact splitting map Ψh=φA~h/2∘φB~h∘φA~h/2.

[L2]

The variational equation. Since t↦φFt(Z) solves ∂tφ=XF(φ) and both sides are C1 in (t,Z) by [L1], Clairaut's theorem permits exchanging the mixed partials, and the chain rule gives ∂t(DφFt(Z))=DXF(φFt(Z))∘DφFt(Z) (Clairaut--Schwarz theorem for continuous second partial derivatives, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder, Directional derivatives and partial derivatives of a map U⊆Rm→Rn).

[L3]

For holomorphic H:D→C, its gradient column ∇H has total derivative D(∇H)(Z)=Hess⁡H(Z)=(∂zk∂zlH(Z))k,l, a complex-symmetric matrix, and DXH=JHess⁡H because XH=J∇H (Calculus of holomorphic maps: chain rule, Wirtinger derivatives, coefficient extraction in one variable, and locally uniform limits clause 2, Clairaut--Schwarz theorem for continuous second partial derivatives, Complex phase domains, holomorphic maps on Cm, and the complexified canonical form and bracket).

[L4]

Matrix algebra: JT=−J, J2=−I2d, hence JTJ=I2d; for invertible A, the identity ATJA=J is equivalent to AJAT=J; and products of matrices satisfying the identity satisfy it again (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, The transpose AT of a matrix, Invertible matrices and the general linear group GL⁡n(F)).

[L5]

Bracket matrix identity: for holomorphic f,g, {f,g}=(∇f)TJ∇g (Calculus of holomorphic maps: chain rule, Wirtinger derivatives, coefficient extraction in one variable, and locally uniform limits clause 2).

Proof

technique · direct; differentiate the matrix identity along the flow with the variational equation, then compose and rewrite the bracket
1.1givenL1L2L3L4algebra

The derivative of the matrix identity along the flow. Fix (Z,t) in the flow domain and put A(τ):=DφFτ(Z). By [L2], A′(τ)=DXF(φFτ(Z))A(τ), and by [L3], DXF=JHess⁡F with Hess⁡F symmetric. Then ddτ(A(τ)TJA(τ))=AT[(JHess⁡F)TJ+J(JHess⁡F)]A=AT[Hess⁡FTJTJ+J2Hess⁡F]A=0, using JTJ=I, J2=−I and Hess⁡FT=Hess⁡F ([L4]). Since the expression is continuous in τ and has zero derivative on the connected flow interval, it equals its value at τ=0, namely ITJI=J. Thus DφFt(Z)TJDφFt(Z)=J.

2.1givenstep 1.1L4

Invertibility. Along the reverse continuation of the same solution branch, clause 3 of The flow of a bounded holomorphic field on a polydisc: existence in complex time, displacement bound, holomorphy, and the composition law supplies the local inverse φF−t near φFt(Z). The chain rule gives DφF−t(φFt(Z))∘DφFt(Z)=I: the derivative is invertible with inverse DφF−t(φFt(Z)), and the identity of step 1.1 transfers to AJAT=J by [L4].

3.1givenstep 1.1step 2.1L4algebra

Composition. The chain rule gives DΨh(Z)=A1A2A3 with A1=DφA~h/2 at φB~h(φA~h/2(Z)), A2=DφB~h at φA~h/2(Z) and A3=DφA~h/2 at Z. Each Ai satisfies AiTJAi=J by step 1.1 and is invertible by step 2.1; hence DΨh(Z)TJDΨh(Z)=A3T(A2T(A1TJA1)A2)A3=J by [L4], and DΨh(Z) is invertible as a product of invertible matrices.

4.1givenstep 3.1L4L5algebra

Bracket identity. Let f,g be holomorphic near Ψh(Z). By the chain rule, ∇(f∘Ψh)(Z)=DΨh(Z)T∇f(Ψh(Z)), and likewise for g; hence by [L5] {f∘Ψh,g∘Ψh}(Z)=∇f(Ψh(Z))T(DΨh(Z) J DΨh(Z)T)∇g(Ψh(Z)). By step 3.1 and the equivalence of [L4] (applicable because DΨh(Z) is invertible), DΨh J DΨhT=J, so the last display equals ∇f(Ψh(Z))TJ∇g(Ψh(Z))={f,g}(Ψh(Z)) by [L5] again.

5.1step 1.1step 3.1step 4.1∎

Conclusion. Step 1.1 gives clause 1 for the two fields, step 3.1 gives clause 2, and step 4.1 gives clause 3.

Remarks

Depends on

Dependency tree · 2 levels

Sources