A construction of a large family of commuting pairs of integrable symplectic birational four-dimensional maps

Abstract

We give a construction of completely integrable four-dimensional Hamiltonian systems with cubic Hamilton functions. Applying to the corresponding pairs of commuting quadratic Hamiltonian vector fields the so called Kahan–Hirota–Kimura discretization scheme, we arrive at pairs of birational four-dimensional maps. We show that these maps are symplectic with respect to a symplectic structure that is a perturbation of the standard symplectic structure on ${\mathbb{R}}^4$, and possess two independent integrals of motion, which are perturbations of the original Hamilton functions and which are in involution with respect to the perturbed symplectic structure. Thus, these maps are completely integrable in the Liouville–Arnold sense. Moreover, under a suitable normalization of the original pairs of vector fields, the pairs of maps commute and share the invariant symplectic structure and the two integrals of motion.

1. Introduction

The theory of integrable systems has a long record of fruitful interactions with various branches of mathe- matics, most prominently with algebraic geometry. A bright example is an in-depth study of the geometry of elliptic rational surfaces and their automorphisms by Duistermaat [1], following the discovery of a large Quispel–Roberts–Thompson (QRT) family of integrable birational two-dimensional maps by Quispel et al. [2].

The goal of this paper is to introduce a large family of integrable four-dimensional maps, along with their numerous remarkable properties. We hope that the study of the algebraic geometry of these maps will turn out to be as fruitful as the one in [1]. This family comes as a new instance in a long and still mysterious list of features of the so-called Kahan–Hirota–Kimura discretization method for quadratic vector fields.

This method was introduced in the geometric integration literature by Kahan in the unpublished notes [3] as a method applicable to any system of ordinary differential equations for $x:\mathbb{R}\to {\mathbb{R}}^n$ with a quadratic vector field:

$$\dot{x}=f(x)=Q(x)+Bx+c,$$ 1.1

where each component of $Q:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a quadratic form, while $B\in {\mathrm{Mat}}_{n\times n}(\mathbb{R})$ and $c\in {\mathbb{R}}^n$. Kahan’s discretization (with stepsize 2ε) reads as

$$\frac{\tilde{x}-x}{2\epsilon }=Q(x,\tilde{x})+\frac{1}{2}B(x+\tilde{x})+c,$$ 1.2

where

$$Q(x,\tilde{x})=\frac{1}{2}(Q(x+\tilde{x})-Q(x)-Q(\tilde{x}))$$

is the symmetric bilinear form corresponding to the quadratic form Q. Equation (1.2) is linear with respect to $\tilde{x}$ and therefore defines a rational map $\tilde{x}={\mathit{\Phi }}_f(x,\epsilon )$. Clearly, this map approximates the time 2ε shift along the solutions of the original differential system. Since equation (1.2) remains invariant under the interchange $x\leftrightarrow \tilde{x}$ with the simultaneous sign inversion ε↦−ε, one has the reversibility property

$${\mathit{\Phi }}_f^{-1}(x,\epsilon )={\mathit{\Phi }}_f(x,-\epsilon ).$$ 1.3

In particular, the map Φ_f is birational. Kahan applied this discretization scheme to the famous Lotka–Volterra system and showed that in this case it possesses a very remarkable non-spiralling property. This property was explained by Sanz-Serna [4] by demonstrating that in this case the numerical method preserves an invariant Poisson structure of the original system.

The next intriguing appearance of this discretization was in the two papers by Hirota and Kimura who (being apparently unaware of the work by Kahan) applied it to two famous integrable system of classical mechanics, the Euler top and the Lagrange top [5,6]. Surprisingly, the discretization scheme produced in both cases is integrable maps.

In [7–9], the authors undertook an extensive study of the properties of the Kahan’s method when applied to integrable systems (we proposed to use in the integrable context the term ‘Hirota–Kimura method’). It was demonstrated that, in an amazing number of cases, the method preserves integrability in the sense that the map Φ_f(x,ε) possesses as many independent integrals of motion as the original system $\dot{x}=f(x)$.

Further remarkable geometric properties of the Kahan’s method were discovered by Celledoni et al. in [10], see also [11,12]. They demonstrated that for an arbitrary Hamiltonian vector field f(x)=J∇H(x) with a constant Poisson tensor J and a cubic Hamilton function H(x), the map Φ_f(x,ε) possesses a rational integral of motion $\tilde{H}(x,\epsilon )$ such that $\tilde{H}(x,0)=H(x)$, and an invariant measure with a rational density, which is a small perturbation of the phase volume dx₁∧⋯∧dx_n as ε→0. It should be mentioned that, while for n=2 the existence of an invariant measure is equivalent to symplecticity of the map Φ_f(x,ε), the latter property was not established for any quadratic Hamiltonian system in dimension n≥4.

In this paper, we give a construction of a big family of completely integrable Hamiltonian systems in dimension n=4 for which the Kahan–Hirota–Kimura discretization possesses a whole series of novel features.

The main set of parameters of the construction is encoded in a 4×4 matrix

$$A=\left(\begin{array}{cccc}{a}_1& a_2& 0& -a_5\\ a_3& -a_1& a_5& 0\\ 0& a_6& a_1& a_3\\ -a_6& 0& a_2& -a_1\end{array}\right).$$

Such matrices form a five-dimensional vector space. To a generic matrix from this space, there corresponds an eight-dimensional vector space of homogeneous polynomials H(x) of degree 3 on ${\mathbb{R}}^4$, satisfying a certain system of second-order linear partial differential equations (PDEs), encoded in the matrix equation

$$A({\mathrm{\nabla }}^{2}H)=({\mathrm{\nabla }}^{2}H)A^{\mathrm{T}},$$ 1.4

where ∇²H is the Hesse matrix of the function H. To each such polynomial H(x) there corresponds a unique ‘dual’ polynomial K(x) from the same eight-dimensional vector space, characterized by

$$\mathrm{\nabla }K(x)=A\mathrm{\nabla }H(x).$$ 1.5

One can think of equation (1.4) as a matrix analog of the Laplace equation for harmonic functions, and of equation (1.5) as an analog of the Cauchy–Riemann equations relating conjugate pairs of harmonic functions. It turns out that the functions H(x) and K(x) are in involution with respect to the standard symplectic structure on ${\mathbb{R}}^4$, and for a generic A they are functionally independent. Therefore, each one defines a completely integrable Hamiltonian system. The flows of the Hamiltonian vector fields J∇H(x) and J∇K(x) commute. The following are the striking properties of the corresponding Kahan–Hirota–Kimura discretizations.

• — The map Φ_J∇H is symplectic with respect to a symplectic structure which is a perturbation of the canonical symplectic structure on ${\mathbb{R}}^4$, and possesses two functionally independent integrals. In other words, Φ_J∇H is completely integrable. Of course, the same holds true for Φ_J∇K.

• — The integrals $\tilde{H}(x,\epsilon )$ and $\tilde{K}(x,\epsilon )$ of Φ_J∇H are rational perturbations of the original polynomials H(x) and K(x), are related by the same equation as (1.5), that is, $\mathrm{\nabla }\tilde{K}=A\mathrm{\nabla }\tilde{H}$, and satisfy the same second-order differential equations (1.4) as H(x) and K(x) do. Moreover, they are in involution with respect to the invariant Poisson bracket of Φ_J∇H.

• — There exists a unique (up to sign) number α such that the maps Φ_J∇H and Φ_α⁻¹J∇K commute. For this value of α, the maps Φ_J∇H and Φ_α⁻¹J∇K share the invariant symplectic structure and the two functionally independent integrals.

We provide the reader with a quick reminder about the general properties of the Kahan–Hirota–Kimura discretization method in §2. Then we discuss the details of the general construction of the dual pairs H(x), K(x) in §3. The rich algebraic properties of the corresponding vector fields J∇H and J∇K are collected in §4. On the basis of these properties, we prove the main results in §5 (commutativity), §6 (two integrals of motion), §7 (invariant symplectic structure) and §8 (differential equations for the integrals of the maps).

2. General properties of the Kahan–Hirota–Kimura discretization

Here we recall the main properties of the Kahan–Hirota–Kimura discretization, following mainly [8–10].

The explicit form of the map Φ_f defined by (1.2) is

$$\tilde{x}={\mathit{\Phi }}_f(x,\epsilon )=x+2\epsilon {(I-\epsilon f^{\prime }(x))}^{-1}f(x),$$ 2.1

where f′(x) denotes the Jacobi matrix of f(x). Moreover, if the vector field f(x) is homogeneous (of degree 2), then (2.1) can be equivalently rewritten as

$$\tilde{x}={\mathit{\Phi }}_f(x,\epsilon )={(I-\epsilon f^{\prime }(x))}^{-1}x.$$ 2.2

Owing to (1.3), in the latter case we also have

$$x={\mathit{\Phi }}_f(\tilde{x},-\epsilon )={(I+\epsilon f^{\prime }(\tilde{x}))}^{-1}\tilde{x}\hspace{0.17em}\iff \hspace{0.17em}\tilde{x}=(I+\epsilon f^{\prime }(\tilde{x}))x.$$ 2.3

One has the following expression for the Jacobi matrix of the map Φ_f:

$$\mathrm{d}{\mathit{\Phi }}_f(x)=\frac{\mathrm{\partial }\tilde{x}}{\mathrm{\partial }x}={(I-\epsilon f^{\prime }(x))}^{-1}(I+\epsilon f^{\prime }(\tilde{x})).$$ 2.4

Let the vector field f(x) be Hamiltonian, f(x)=J∇H(x), where $H:{\mathbb{R}}^n\to \mathbb{R}$ is a cubic polynomial and J is a non-degenerate skew-symmetric n×n matrix, so that H(x) is an integral of motion for $\dot{x}=f(x)$. Then the map Φ_f(x,ε) possesses the following rational integral of motion:

$$\tilde{H}(x,\epsilon )=\frac{1}{6\epsilon }{x}^{\mathrm{T}}{J}^{-1}\tilde{x}=H(x)+\frac{2\epsilon }{3}{(\mathrm{\nabla }H(x))}^{\mathrm{T}}{(I-\epsilon f^{\prime }(x))}^{-1}f(x),$$ 2.5

as well as an invariant measure

$$\frac{\mathrm{d}{x}_1\wedge \cdots \wedge \mathrm{d}{x}_n}{det(I-\epsilon f^{\prime }(x))}.$$ 2.6

These remarkable results from [10] hold true for all quadratic Hamiltonian vector fields with a constant Poisson tensor and therefore are not related to integrability.

3. A family of integrable four-dimensional Hamiltonian systems

Consider the canonical phase space ${\mathbb{R}}^4$ with coordinates (x₁,x₂,x₃,x₄), equipped with the standard symplectic structure in which the Poisson brackets of the coordinate functions are {x₁,x₃}={x₂,x₄}=1 (all other brackets being either obtained from these ones by skew-symmetry or otherwise vanish). Let H(x)=H(x₁,x₂,x₃,x₄) be a nonlinear Hamilton function on ${\mathbb{R}}^4$ (so that ∇H≠const.). The corresponding Hamiltonian system is governed by the equations of motion

$$\dot{x}=J\mathrm{\nabla }H(x),J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right).$$ 3.1

Proposition 3.1 —

Consider a constant non-degenerate non-scalar 4×4 matrix A, and suppose that for a function K(x)=K(x₁,x₂,x₃,x₄) the following relations are satisfied:

$$\mathrm{\nabla }K=A\mathrm{\nabla }H.$$ 3.2

If the matrix A satisfies

$$A^{\mathrm{T}}J=JA,$$ 3.3

then the functions H, K are in involution and functionally independent, so that the Hamiltonian system (3.1) is completely integrable.

Proof. —

We have

$$\{H,K\}={(\mathrm{\nabla }H)}^{\mathrm{T}}J\mathrm{\nabla }K={(\mathrm{\nabla }H)}^{\mathrm{T}}JA\mathrm{\nabla }H$$

and this vanishes if the matrix JA is skew-symmetric, which gives condition (3.3). Equation (3.2) with a non-scalar matrix A also ensures that H, K are functionally independent (generically). ▪

If A is written in the block form as

$$A=\left(\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right)$$

with 2×2 blocks A_i, then condition (3.3) reads

$$A_1^{\mathrm{T}}=A_4,A_2^{\mathrm{T}}=-A_2\mathrm{and}{A}_3^{\mathrm{T}}=-A_3.$$ 3.4

Such matrices form a six-dimensional vector space:

$$A=\left(\begin{array}{cccc}{a}_1& a_2& 0& -a_5\\ a_3& a_4& a_5& 0\\ 0& a_6& a_1& a_3\\ -a_6& 0& a_2& a_4\end{array}\right).$$ 3.5

We now discuss applicability of this construction. Let A be a generic matrix from family (3.5). For a given function H, differential equations (3.2) for K are solvable if and only if H satisfies the following condition:

$$A({\mathrm{\nabla }}^{2}H)=({\mathrm{\nabla }}^{2}H)A^{\mathrm{T}},$$ 3.6

where ∇²H is the Hesse matrix of the function H. Written down explicitly, one has the following five second-order PDEs for H:

$$\begin{array}{rl}{C}_1& =-a_3\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1^2}+a_2\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2^2}+(a_1-a_4)\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_2}-a_5(\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_3}+\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_4})=0,\end{array}$$ 3.7

$$\begin{array}{rl}{C}_2& =-a_2\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_3^2}+a_3\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_4^2}+(a_1-a_4)\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_3\mathrm{\partial }{x}_4}+a_6(\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_4}+\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_3})=0,\end{array}$$ 3.8

$$\begin{array}{rl}{C}_3& =a_6\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1^2}-a_5\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_4^2}+(a_1-a_4)\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_4}+a_2(\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_4}-\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_3})=0,\end{array}$$ 3.9

$$\begin{array}{rl}{C}_4& =a_5\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_3^2}-a_6\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2^2}-(a_1-a_4)\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_3}+a_3(\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_3}-\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_4})=0\end{array}$$ 3.10

$$\begin{array}{rl}\text{and}{C}_5& =-a_6\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_2}-a_5\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_3\mathrm{\partial }{x}_4}+a_2\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_2\mathrm{\partial }{x}_3}-a_3\frac{{\mathrm{\partial }}^{2}H}{\mathrm{\partial }{x}_1\mathrm{\partial }{x}_4}=0.\end{array}$$ 3.11

Only four of these PDEs are linearly independent, due to the linear relation

$$a_6{C}_1+a_5{C}_2+a_3{C}_3+a_2{C}_4+(a_1-a_4)C_5=0.$$

Proposition 3.2 —

For a generic matrix A from (3.5), the linear space of homogeneous polynomials of degree 3 satisfying the system of second-order PDEs (3.7)–(3.11), has dimension 8.

Proof. —

A general homogeneous polynomial of degree 3 in four variables x₁, x₂, x₃, x₄ has 20 coefficients. Each of the four linearly independent expressions C_i is a linear polynomial in these four variables, so that each equation C_i=0 results in four linear equations for the coefficients of H. Altogether we get 16 linear homogeneous equations for 20 coefficients of H. A computation with a symbolic manipulation package (we used Maple) tells us that the rank of the matrix of the resulting linear system is equal to 12, so that the dimension of the space of solutions is equal to 20−12=8. ▪

Note that the solutions K(x) of the first-order PDEs (3.2) satisfy that the same compatibility conditions (3.7)–(3.11). To see this, observe that one has ∇H=A⁻¹∇K and

$$A^{-1}=\frac{1}{D}\left(\begin{array}{cccc}-a_4& a_2& 0& -a_5\\ a_3& -a_1& a_5& 0\\ 0& a_6& -a_4& a_3\\ -a_6& 0& a_2& -a_1\end{array}\right),\mathrm{where}\hspace{0.17em}D=-a_1{a}_4+a_2{a}_3+a_5{a}_6\ne 0.$$ 3.12

Clearly, the set of PDEs (3.7)–(3.11) generated by the latter matrix coincides with the original one. Thus, to any solution H of the system (3.7)–(3.11) there corresponds, via (3.2), another solution K (unique up to an additive constant). Changing A to A+βI would lead to changing K to K+βH, and would not change the set of all linear combinations of H and K. We will use this freedom to ensure that

$$\mathrm{tr}\hspace{0.17em}A=0\hspace{0.17em}\iff \hspace{0.17em}{a}_4=-a_1.$$ 3.13

Thus, from now on we assume that

$$A=\left(\begin{array}{cccc}{a}_1& a_2& 0& -a_5\\ a_3& -a_1& a_5& 0\\ 0& a_6& a_1& a_3\\ -a_6& 0& a_2& -a_1\end{array}\right).$$ 3.14

We mention also the following properties of the matrices A as in (3.14):

$$A^{-1}=\frac{1}{D}A\hspace{0.17em}\iff \hspace{0.17em}{A}^2=DI,\mathrm{where}\hspace{0.17em}D=a_1^2+a_2{a}_3+a_5{a}_6.$$ 3.15

This follows immediately from (3.12). Furthermore, a direct computation shows that

$$det(A-\lambda I)={({\lambda }^2-D)}^2.$$ 3.16

An (more or less randomly chosen) example of our construction is given by

$$\begin{array}{rl}A& =\left(\begin{array}{cccc}1& 1& 0& -1\\ 0& -1& 1& 0\\ 0& 1& 1& 0\\ -1& 0& 1& -1\end{array}\right),D=4,\\ H& =-3x_1^3+3x_1^2{x}_2-3x_1^2{x}_3-6x_1^2{x}_4-2x_1{x}_2^2-4x_1{x}_2{x}_3+14x_1{x}_2{x}_4-6x_1{x}_3^2\\ & -10x_1{x}_3{x}_4-11x_1{x}_4^2-2x_2^2{x}_4+4x_2{x}_3{x}_4+17x_2{x}_4^2+2x_3^2{x}_4-5x_3{x}_4^2-2x_4^3\\ \text{and}K& =-6x_1^2{x}_2+12x_1^2{x}_4-8x_1{x}_2{x}_3-24x_1{x}_2{x}_4-8x_1{x}_3^2+4x_1{x}_3{x}_4+12x_1{x}_4^2\\ & +4x_2^2{x}_4-22x_2{x}_4^2+4x_3^2{x}_4+12x_3{x}_4^2+4x_4^3.\end{array}$$

For this example, preliminary studies indicate algebraic complete integrability (see §9 for more comments on this point).

4. General algebraic properties of the vector fields f, g

From now on we will assume that A is a matrix as in (3.14). Define D is as in (3.15), and let α be a real or a purely imaginary number (depending on whether D>0 or D<0) satisfying

$${\alpha }^2=D.$$ 4.1

Assume that H(x) and K(x) are homogeneous polynomials of degree 3 satisfying (3.2). Set

$$f(x)=J\mathrm{\nabla }H(x)\mathrm{and}g(x)={\alpha }^{-1}J\mathrm{\nabla }K(x)={\alpha }^{-1}JA\mathrm{\nabla }H(x).$$ 4.2

Owing to (3.3), we have:

$$g(x)={\alpha }^{-1}{A}^{\mathrm{T}}f(x)\hspace{0.17em}\iff \hspace{0.17em}f(x)={\alpha }^{-1}{A}^{\mathrm{T}}g(x).$$ 4.3

This means that the roles of the vector fields f, g in all our constructions are absolutely symmetric. Furthermore,

$$f^{\prime }(x)=J{\mathrm{\nabla }}^{2}H(x)\mathrm{and}{g}^{\prime }(x)={\alpha }^{-1}J{\mathrm{\nabla }}^{2}K(x)={\alpha }^{-1}JA{\mathrm{\nabla }}^{2}H(x).$$ 4.4

Again, due to (3.3), we have

$$g^{\prime }(x)={\alpha }^{-1}{A}^{\mathrm{T}}{f}^{\prime }(x)\hspace{0.17em}\iff \hspace{0.17em}{f}^{\prime }(x)={\alpha }^{-1}{A}^{\mathrm{T}}{g}^{\prime }(x).$$ 4.5

Lemma 4.1 —

The following identities hold true

$$\begin{array}{rl}{(f^{\prime }(x))}^{\mathrm{T}}J& =-J{f}^{\prime }(x),{(g^{\prime }(x))}^{\mathrm{T}}J=-J{g}^{\prime }(x),\end{array}$$ 4.6

$$\begin{array}{rl}{A}^{\mathrm{T}}{f}^{\prime }(x)& =f^{\prime }(x)A^{\mathrm{T}},A^{\mathrm{T}}{g}^{\prime }(x)=g^{\prime }(x)A^{\mathrm{T}},\end{array}$$ 4.7

$$\begin{array}{rl}{f}^{\prime }(x)g(x)& =g^{\prime }(x)f(x),\end{array}$$ 4.8

$$\begin{array}{rl}{f}^{\prime }(x)f(x)& =g^{\prime }(x)g(x),\end{array}$$ 4.9

$$\begin{array}{rl}{f}^{\prime }(x)g^{\prime }(x)& =g^{\prime }(x)f^{\prime }(x)\end{array}$$ 4.10

$$\begin{array}{rl}\text{and}{(f^{\prime }(x))}^2& ={(g^{\prime }(x))}^2.\end{array}$$ 4.11

Proof. —

Equation (4.6) is the characteristic property of Jacobi matrices of Hamiltonian vector fields. Equation (4.7) is equivalent to (3.6), due to (3.3). To prove (4.8), (4.9), we compute with the help of (4.3), (4.7):

$$f^{\prime }(x)g(x)={\alpha }^{-1}{A}^{\mathrm{T}}{g}^{\prime }(x)g(x)={\alpha }^{-1}{g}^{\prime }(x)A^{\mathrm{T}}g(x)=g^{\prime }(x)f(x)$$

and similarly,

$$f^{\prime }(x)f(x)={\alpha }^{-1}{A}^{\mathrm{T}}{g}^{\prime }(x)f(x)={\alpha }^{-1}{g}^{\prime }(x)A^{\mathrm{T}}f(x)=g^{\prime }(x)g(x).$$

Observe that (4.8) expresses commutativity of the vector fields f(x) and g(x). Identities (4.10), (4.11) are proved along the same lines, with the help of (4.5), (4.7):

$$f^{\prime }(x)g^{\prime }(x)={\alpha }^{-1}{A}^{\mathrm{T}}{g}^{\prime }(x)g^{\prime }(x)={\alpha }^{-1}{g}^{\prime }(x)A^{\mathrm{T}}{g}^{\prime }(x)=g^{\prime }(x)f^{\prime }(x)$$

and

$${(f^{\prime }(x))}^2={\alpha }^{-1}{A}^{\mathrm{T}}{g}^{\prime }(x)f^{\prime }(x)={\alpha }^{-1}{g}^{\prime }(x)A^{\mathrm{T}}{f}^{\prime }(x)={(g^{\prime }(x))}^2.$$

 ▪

Corollary 4.2 —

There holds

$$det(I-\epsilon f^{\prime }(x))=det(I-\epsilon g^{\prime }(x)).$$ 4.12

Proof. —

It is enough to prove that tr (f′(x))^k=tr (g′(x))^k for k=1,2,3,4. Since vector fields f,g are Hamiltonian, we have tr (f′(x))^k= 0 and tr (g′(x))^k=0 for k=1,3. The equalities for k=2,4 follow from (4.11). ▪

Lemma 4.3 —

The following identities hold true:

$$det(f^{\prime }(x)+g^{\prime }(x))=0\mathrm{and}det(f^{\prime }(x)-g^{\prime }(x))=0.$$ 4.13

Proof. —

Indeed, these determinants are equal to

$$det((I\pm {\alpha }^{-1}{A}^{\mathrm{T}})f^{\prime }(x))$$

and they both vanish due to $det(\alpha I\pm A)=0$, which is a direct consequence of (3.16), (4.1). ▪

Lemma 4.4 —

The following identities hold true:

$${(f^{\prime }(x))}^2={(g^{\prime }(x))}^2=p(x)I+q(x)A^{\mathrm{T}},$$ 4.14

where

$$p(x)=\frac{1}{4}\hspace{0.17em}\mathrm{tr}{(f^{\prime }(x))}^2$$ 4.15

and

$$q(x)=\frac{1}{4D}\hspace{0.17em}\mathrm{tr}(A^{\mathrm{T}}{(f^{\prime }(x))}^2)=\frac{1}{4\alpha }\hspace{0.17em}\mathrm{tr}(f^{\prime }(x)g^{\prime }(x)).$$ 4.16

Proof. —

We have, due to (4.13):

$$det(\lambda I-(f^{\prime }-g^{\prime }))={\lambda }^4-\frac{1}{2}{\lambda }^2\hspace{0.17em}\mathrm{tr}{(f^{\prime }-g^{\prime })}^2$$ 4.17

and

$$det(\lambda I-(f^{\prime }+g^{\prime }))={\lambda }^4-\frac{1}{2}{\lambda }^2\hspace{0.17em}\mathrm{tr}{(f^{\prime }+g^{\prime })}^2.$$ 4.18

By the theorem of Cayley–Hamilton, we have

$${(f^{\prime }-g^{\prime })}^4-\frac{1}{2}{(f^{\prime }-g^{\prime })}^2\hspace{0.17em}\mathrm{tr}{(f^{\prime }-g^{\prime })}^2=0$$ 4.19

and

$${(f^{\prime }+g^{\prime })}^4-\frac{1}{2}{(f^{\prime }+g^{\prime })}^2\hspace{0.17em}\mathrm{tr}{(f^{\prime }+g^{\prime })}^2=0.$$ 4.20

Add the latter two identities, taking into account (4.10) and (4.11). The result reads

$$16{(f^{\prime })}^4-4{(f^{\prime })}^2\hspace{0.17em}\mathrm{tr}{(f^{\prime })}^2-4f^{\prime }{g}^{\prime }\hspace{0.17em}\mathrm{tr}(f^{\prime }{g}^{\prime })=0$$

or, equivalently,

$${(f^{\prime })}^4-\frac{1}{4}{(f^{\prime })}^2\hspace{0.17em}\mathrm{tr}{(f^{\prime })}^2-\frac{1}{4D}{A}^{\mathrm{T}}{(f^{\prime })}^2\hspace{0.17em}\mathrm{tr}(A^{\mathrm{T}}{(f^{\prime })}^2)=0.$$

Upon dividing by a generically non-degenerate matrix (f′)² (its determinant is a non-vanishing homogeneous polynomial of degree 8), we arrive at the desired statement. ▪

Lemma 4.5 —

The following identity holds true:

$$p^2(x)-D{q}^2(x)=\frac{1}{16}{(\mathrm{tr}{(f^{\prime }(x))}^2)}^2-\frac{1}{16}{(\mathrm{tr}(f^{\prime }(x)g^{\prime }(x)))}^2=det{f}^{\prime }(x).$$ 4.21

Proof. —

We transform the left-hand side of this identity as follows, using tr(f′)²=tr(g′)²:

$$\begin{array}{rl}& (\mathrm{tr}{(f^{\prime })}^2-\mathrm{tr}(f^{\prime }{g}^{\prime }))(\mathrm{tr}{(f^{\prime })}^2+\mathrm{tr}(f^{\prime }{g}^{\prime }))\\ & =(\frac{1}{2}\hspace{0.17em}\mathrm{tr}{(f^{\prime })}^2+\frac{1}{2}\hspace{0.17em}\mathrm{tr}{(g^{\prime })}^2-\mathrm{tr}(f^{\prime }{g}^{\prime }))(\frac{1}{2}\hspace{0.17em}\mathrm{tr}{(f^{\prime })}^2+\frac{1}{2}\hspace{0.17em}\mathrm{tr}{(g^{\prime })}^2+\mathrm{tr}(f^{\prime }{g}^{\prime }))\\ & =\frac{1}{4}\hspace{0.17em}\mathrm{tr}{(f^{\prime }-g^{\prime })}^2\hspace{0.17em}\mathrm{tr}{(f^{\prime }+g^{\prime })}^2.\end{array}$$ 4.22

Owing to (4.17), (4.18), we find that the right-hand side of (4.22) is equal to the coefficient by λ⁴ in

$$det(\lambda I-(f^{\prime }-g^{\prime }))(\lambda I-(f^{\prime }+g^{\prime }))=det({(\lambda I-f^{\prime })}^2-{(g^{\prime })}^2)={\lambda }^{4}det(\lambda I-2f^{\prime })$$

(at the last step we used (4.11)). The latter coefficient is equal to $det(2f^{\prime })=16det{f}^{\prime }$, which finishes the proof. ▪

Lemma 4.6 —

The following identity holds true:

$$\mathrm{\nabla }p(x)=A\mathrm{\nabla }q(x).$$ 4.23

As a consequence, both quadratic polynomials p(x) and q(x) satisfy the second-order differential equations (3.6):

$$A({\mathrm{\nabla }}^{2}p)=({\mathrm{\nabla }}^{2}p)A^{\mathrm{T}}\mathrm{and}A({\mathrm{\nabla }}^{2}q)=({\mathrm{\nabla }}^{2}q)A^{\mathrm{T}}.$$ 4.24

Proof. —

We use the characteristic property (4.7) of the matrix f′ which in components reads

$$\sum _j{(A^{\mathrm{T}})}_{ij}\frac{\mathrm{\partial }{f}_j}{\mathrm{\partial }{x}_m}=\sum _j\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}{(A^{\mathrm{T}})}_{jm}\mathrm{\forall }\hspace{0.17em}i,m,$$ 4.25

as well as its derivative with respect to x_ℓ:

$$\sum _j{(A^{\mathrm{T}})}_{ij}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_m\mathrm{\partial }{x}_{\ell }}=\sum _j\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_{\ell }}{(A^{\mathrm{T}})}_{jm}\mathrm{\forall }\hspace{0.17em}i,m,\ell .$$ 4.26

We compute the components of ∇q:

$$\frac{\mathrm{\partial }q}{\mathrm{\partial }{x}_{\ell }}=\frac{1}{4D}\frac{\mathrm{\partial }\hspace{0.17em}\mathrm{tr}(A^{\mathrm{T}}{(f^{\prime })}^2)}{\mathrm{\partial }{x}_{\ell }}=\frac{1}{4D}\sum _{i,j,m}{(A^{\mathrm{T}})}_{ij}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_m\mathrm{\partial }{x}_{\ell }}\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}+\frac{1}{4D}\sum _{i,j,m}{(A^{\mathrm{T}})}_{ij}\frac{\mathrm{\partial }{f}_j}{\mathrm{\partial }{x}_m}\frac{{\mathrm{\partial }}^2{f}_m}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_{\ell }}.$$ 4.27

The contribution to

$${(A\mathrm{\nabla }q)}_k=\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\frac{\mathrm{\partial }q}{\mathrm{\partial }{x}_{\ell }}$$ 4.28

of the first sum in (4.27) is

$$\begin{array}{rl}& \frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,m}\left(\sum _j{(A^{\mathrm{T}})}_{ij}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_m\mathrm{\partial }{x}_{\ell }}\right)\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}(\mathrm{use}\hspace{0.17em}(4.26))\\ & =\frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,m}\left(\sum _j\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_m}{(A^{\mathrm{T}})}_{j\ell }\right)\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}\\ & =\frac{1}{4D}\sum _{i,j,m}\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_m}\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}{(A^{\mathrm{T}})}_{j\ell }(\mathrm{use}\hspace{0.17em}{(A^{\mathrm{T}})}^2=DI)\\ & =\frac{1}{4}\sum _{i,j,m}\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_m}\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}{\delta }_{jk}=\frac{1}{4}\sum _{i,m}\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_k\mathrm{\partial }{x}_m}\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}.\end{array}$$ 4.29

Similarly, the contribution of the second sum in (4.27) to (4.28) is

$$\begin{array}{rl}& \frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,m}\left(\sum _j{(A^{\mathrm{T}})}_{ij}\frac{\mathrm{\partial }{f}_j}{\mathrm{\partial }{x}_m}\right)\frac{{\mathrm{\partial }}^2{f}_m}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_{\ell }}(\mathrm{use}\hspace{0.17em}(4.25))\\ & =\frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,m}\left(\sum _j\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}{(A^{\mathrm{T}})}_{jm}\right)\frac{{\mathrm{\partial }}^2{f}_m}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_{\ell }}\\ & =\frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,j}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\left(\sum _m{(A^{\mathrm{T}})}_{jm}\frac{{\mathrm{\partial }}^2{f}_m}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_{\ell }}\right)(\mathrm{use}\hspace{0.17em}(4.26))\\ & =\frac{1}{4D}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}\sum _{i,j}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\left(\sum _m\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_m}{(A^{\mathrm{T}})}_{m\ell }\right)\\ & =\frac{1}{4D}\sum _{i,j,m}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_m}\sum _{\ell }{(A^{\mathrm{T}})}_{\ell k}{(A^{\mathrm{T}})}_{m\ell }(\mathrm{use}\hspace{0.17em}{(A^{\mathrm{T}})}^2=DI)\\ & =\frac{1}{4}\sum _{i,j,m}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_m}{\delta }_{km}=\frac{1}{4}\sum _{i,j}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_k}.\end{array}$$ 4.30

Collecting all the results, we find:

$${(A\mathrm{\nabla }q)}_k=\frac{1}{4}\sum _{i,m}\frac{{\mathrm{\partial }}^2{f}_i}{\mathrm{\partial }{x}_k\mathrm{\partial }{x}_m}\frac{\mathrm{\partial }{f}_m}{\mathrm{\partial }{x}_i}+\frac{1}{4}\sum _{i,j}\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}\frac{{\mathrm{\partial }}^2{f}_j}{\mathrm{\partial }{x}_i\mathrm{\partial }{x}_k}=\frac{1}{4}\frac{\mathrm{\partial }\hspace{0.17em}\mathrm{tr}({(f^{\prime })}^2)}{\mathrm{\partial }{x}_k}={(\mathrm{\nabla }p)}_k,$$

which finishes the proof. ▪

5. Commutativity of maps

Consider the Kahan–Hirota–Kimura discretizations Φ_f, Φ_g of the respective differential equations $\dot{x}=f(x)$ and $\dot{x}=g(x)$. Recall that in the case D<0, the vector field g(x) is purely imaginary, so that the rational map Φ_g has complex coefficients.

Theorem 5.1 —

The maps

$${\mathit{\Phi }}_f:x\mapsto \tilde{x}={(I-\epsilon f^{\prime }(x))}^{-1}x=(I+\epsilon f^{\prime }(\tilde{x}))x$$ 5.1

and

$${\mathit{\Phi }}_g:x\mapsto \hat{x}={(I-\epsilon g^{\prime }(x))}^{-1}x=(I+\epsilon g^{\prime }(\hat{x}))x,$$ 5.2

commute: Φ_f°Φ_g=Φ_g°Φ_f.

Proof. —

We have

$$({\mathit{\Phi }}_g\circ {\mathit{\Phi }}_f)(x)={(I-\epsilon g^{\prime }(\tilde{x}))}^{-1}(I+\epsilon f^{\prime }(\tilde{x}))x$$ 5.3

and

$$({\mathit{\Phi }}_f\circ {\mathit{\Phi }}_g)(x)={(I-\epsilon f^{\prime }(\hat{x}))}^{-1}(I+\epsilon g^{\prime }(\hat{x}))x.$$ 5.4

We prove the following matrix equation:

$${(I-\epsilon g^{\prime }(\tilde{x}))}^{-1}(I+\epsilon f^{\prime }(\tilde{x}))={(I-\epsilon f^{\prime }(\hat{x}))}^{-1}(I+\epsilon g^{\prime }(\hat{x})),$$ 5.5

which is stronger than the vector equation (Φ_f°Φ_g)(x)=(Φ_g°Φ_f)(x) expressing commutativity. Equation (5.5) is equivalent to

$$(I-\epsilon f^{\prime }(\hat{x})){(I-\epsilon g^{\prime }(\tilde{x}))}^{-1}=(I+\epsilon g^{\prime }(\hat{x})){(I+\epsilon f^{\prime }(\tilde{x}))}^{-1}.$$ 5.6

From (4.11), we find

$$\begin{array}{rl}{(I-\epsilon g^{\prime }(\tilde{x}))}^{-1}& =(I+\epsilon g^{\prime }(\tilde{x})){(I-{\epsilon }^2{(f^{\prime }(\tilde{x}))}^2)}^{-1}\\ \mathrm{and}{(I+\epsilon f^{\prime }(\tilde{x}))}^{-1}& =(I-\epsilon f^{\prime }(\tilde{x})){(I-{\epsilon }^2{(f^{\prime }(\tilde{x}))}^2)}^{-1}.\end{array}$$

With this at hand, equation (5.6) is equivalent to

$$(I-\epsilon f^{\prime }(\hat{x}))(I+\epsilon g^{\prime }(\tilde{x}))=(I+\epsilon g^{\prime }(\hat{x}))(I-\epsilon f^{\prime }(\tilde{x})).$$

Here the quadratic in ε terms cancel by virtue of (4.5) and (4.7):

$$f^{\prime }(\hat{x})g^{\prime }(\tilde{x})={\alpha }^{-1}{f}^{\prime }(\hat{x})A^{\mathrm{T}}{f}^{\prime }(\tilde{x})={\alpha }^{-1}{A}^{\mathrm{T}}{f}^{\prime }(\hat{x})f^{\prime }(\tilde{x})=g^{\prime }(\hat{x})f^{\prime }(\tilde{x}),$$

so that we are left with the terms linear in ε:

$$-f^{\prime }(\hat{x})+g^{\prime }(\tilde{x})=g^{\prime }(\hat{x})-f^{\prime }(\tilde{x}).$$ 5.7

Since the tensors f′′, g′′ are constant, we have

$$\begin{array}{rl}{f}^{\prime }(\hat{x})& =f^{\prime }(x)+f^{″}(\hat{x}-x)=f^{\prime }(x)+2\epsilon f^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x),\\ g^{\prime }(\hat{x})& =g^{\prime }(x)+g^{″}(\hat{x}-x)=g^{\prime }(x)+2\epsilon g^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x),\\ f^{\prime }(\tilde{x})& =f^{\prime }(x)+f^{″}(\tilde{x}-x)=f^{\prime }(x)+2\epsilon f^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x),\\ g^{\prime }(\tilde{x})& =g^{\prime }(x)+g^{″}(\tilde{x}-x)=g^{\prime }(x)+2\epsilon g^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x).\end{array}$$

Thus, equation (5.7) is equivalent to

$$\begin{array}{rl}& f^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x)+g^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x)\\ & =f^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x)+g^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x).\end{array}$$ 5.8

At this point, we use the following statement.

Lemma 5.2 —

For any vector $v\in {\mathbb{C}}^4$ we have:

$$g^{″}(x)v={\alpha }^{-1}{f}^{″}(x)(A^{\mathrm{T}}v)\mathrm{and}{f}^{″}(x)v={\alpha }^{-1}{g}^{″}(x)(A^{\mathrm{T}}v).$$ 5.9

We compute the matrices on the left-hand side of (5.8) with the help of (5.9), (4.3), (4.5):

$$\begin{array}{rl}{f}^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x)& =f^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}(g(x)+\epsilon g^{\prime }(x)g(x)),\\ g^{″}{(I-\epsilon g^{\prime }(x))}^{-1}g(x)& ={\alpha }^{-1}{f}^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}{A}^{\mathrm{T}}(g(x)+\epsilon g^{\prime }(x)g(x))\\ & =f^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}(f(x)+\epsilon f^{\prime }(x)g(x))\end{array}$$

and similarly

$$\begin{array}{rl}{f}^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x)& =f^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}(f(x)+\epsilon f^{\prime }(x)f(x))\\ g^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x)& ={\alpha }^{-1}{f}^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}{A}^{\mathrm{T}}(f(x)+\epsilon f^{\prime }(x)f(x))\\ & =f^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}(g(x)+\epsilon g^{\prime }(x)f(x)).\end{array}$$

Collecting all the results and using (4.8) and (4.9), we see that the proof is complete. ▪

Proof of lemma 5.2 —

The identities in question are equivalent to

$$A^{\mathrm{T}}(f^{″}(x)v)=f^{″}(x)(A^{\mathrm{T}}v)\mathrm{and}{A}^{\mathrm{T}}(g^{″}(x)v)=g^{″}(x)(A^{\mathrm{T}}v).$$ 5.10

(Actually, both tensors f′′ and g′′ are constant, i.e. do not depend on x.) To prove the latter identities, we start with equation (4.7) written in components:

$$\sum _k{(A^{\mathrm{T}})}_{ik}\frac{\mathrm{\partial }{f}_k}{\mathrm{\partial }{x}_{\ell }}=\sum _k\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_k}{(A^{\mathrm{T}})}_{k\ell }.$$

Differentiating with respect to x_j, we get

$$\sum _k{(A^{\mathrm{T}})}_{ik}\frac{{\mathrm{\partial }}^2{f}_k}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_{\ell }}=\sum _k\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_k}{(A^{\mathrm{T}})}_{k\ell }.$$

Hence,

$$\sum _{k,\ell }{(A^{\mathrm{T}})}_{ik}\frac{{\mathrm{\partial }}^2{f}_k}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_{\ell }}{v}_{\ell }=\sum _{k,\ell }\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_k}{(A^{\mathrm{T}})}_{k\ell }{v}_{\ell },$$

which is nothing but the (i,j) entry of the matrix identity (5.10). ▪

6. Integrals of motion

Theorem 6.1 —

The maps Φ_f and Φ_g share two functionally independent conserved quantities

$$\tilde{H}(x,\epsilon )={\epsilon }^{-1}{x}^{\mathrm{T}}J\tilde{x}={\epsilon }^{-1}{x}^{\mathrm{T}}J{(I-\epsilon f^{\prime }(x))}^{-1}x$$ 6.1

and

$$\tilde{K}(x,\epsilon )={\epsilon }^{-1}\alpha x^{\mathrm{T}}J\hat{x}={\epsilon }^{-1}\alpha x^{\mathrm{T}}J{(I-\epsilon g^{\prime }(x))}^{-1}x.$$ 6.2

Before proving this theorem, we observe different expressions for these functions. Expanding (6.1) in power series with respect to ε, we see

$$\tilde{H}(x,\epsilon )=\sum _{k=0}^{\mathrm{\infty }}{\epsilon }^{k-1}{x}^{\mathrm{T}}J{(f^{\prime }(x))}^{k}x=-\sum _{k=0}^{\mathrm{\infty }}{\epsilon }^{k-1}{x}^{\mathrm{T}}({\mathrm{\nabla }}^{2}H(x)J{\mathrm{\nabla }}^{2}H(x)\cdots J{\mathrm{\nabla }}^{2}H(x))x.$$

The matrix in the parentheses (involving k times ∇²H(x) and k−1 times J) is symmetric if k is odd, and skew-symmetric if k is even. Therefore, all terms with even k vanish, and we arrive at

$$\tilde{H}(x,\epsilon )=\sum _{k=0}^{\mathrm{\infty }}{\epsilon }^{2k}{x}^{\mathrm{T}}J{(f^{\prime }(x))}^{2k+1}x,$$

or, finally,

$$\tilde{H}(x,\epsilon )=x^{\mathrm{T}}J{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}{f}^{\prime }(x)x=2x^{\mathrm{T}}J{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x).$$ 6.3

At the last step we used that f(x) is homogeneous of degree 2, so that f′(x)x=2f(x). Of course, analogous expressions hold true for the function $\tilde{K}(x,\epsilon )$:

$$\tilde{K}(x,\epsilon )=\alpha x^{\mathrm{T}}J{(I-{\epsilon }^2{(g^{\prime }(x))}^2)}^{-1}{g}^{\prime }(x)x=2\alpha x^{\mathrm{T}}J{(I-{\epsilon }^2{(g^{\prime }(x))}^2)}^{-1}g(x).$$ 6.4

Formulae (6.3), (6.4) also clearly display the asymptotics

$$\tilde{H}(x,\epsilon )=2x^{\mathrm{T}}Jf(x)+O({\epsilon }^2)=-2x^{\mathrm{T}}\mathrm{\nabla }H(x)+O({\epsilon }^2)=-6H(x)+O({\epsilon }^2),$$

and analogously

$$\tilde{K}(x,\epsilon )=2\alpha x^{\mathrm{T}}Jg(x)+O({\epsilon }^2)=-2x^{\mathrm{T}}\mathrm{\nabla }K(x)+O({\epsilon }^2)=-6K(x)+O({\epsilon }^2).$$

Proof of theorem 6.1 —

We first show that $\tilde{H}(x,\epsilon )$ is an integral of motion of the map Φ_f (this is a result from [10], which holds true for arbitrary Hamiltonian vector fields with a constant Poisson tensor). For this goal, we compute with the help of (5.1):

$$\begin{array}{rl}\tilde{H}(\tilde{x},\epsilon )& ={\epsilon }^{-1}{\tilde{x}}^{\mathrm{T}}J{(I-\epsilon f^{\prime }(\tilde{x}))}^{-1}\tilde{x}\\ & ={\epsilon }^{-1}{x}^{\mathrm{T}}{(I+\epsilon f^{\prime }(\tilde{x}))}^{\mathrm{T}}J{(I-\epsilon f^{\prime }(\tilde{x}))}^{-1}{(I-\epsilon f^{\prime }(x))}^{-1}x.\end{array}$$

Taking into account (4.6) in the form ${(f^{\prime }(\tilde{x}))}^{\mathrm{T}}J=-J{f}^{\prime }(\tilde{x})$, we arrive at

$$\tilde{H}(\tilde{x},\epsilon )={\epsilon }^{-1}{x}^{\mathrm{T}}J(I-\epsilon f^{\prime }(\tilde{x})){(I-\epsilon f^{\prime }(\tilde{x}))}^{-1}{(I-\epsilon f^{\prime }(x))}^{-1}x=\tilde{H}(x,\epsilon ).$$ 6.5

Next, we show that $\tilde{K}(x,\epsilon )$ also is an integral of motion of the map Φ_f. For this goal, we first compute, based on (6.4):

$$\begin{array}{rl}\tilde{K}(\tilde{x},\epsilon )& =\alpha {\tilde{x}}^{\mathrm{T}}J{(I-{\epsilon }^2{(g^{\prime }(\tilde{x}))}^2)}^{-1}{g}^{\prime }(\tilde{x})\tilde{x}\\ & =\alpha x^{\mathrm{T}}{(I+\epsilon f^{\prime }(\tilde{x}))}^{\mathrm{T}}J{(I-{\epsilon }^2{(g^{\prime }(\tilde{x}))}^2)}^{-1}{g}^{\prime }(\tilde{x})(I+\epsilon f^{\prime }(\tilde{x}))x\\ & =\alpha x^{\mathrm{T}}J(I-\epsilon f^{\prime }(\tilde{x})){(I-{\epsilon }^2{(g^{\prime }(\tilde{x}))}^2)}^{-1}{g}^{\prime }(\tilde{x})(I+\epsilon f^{\prime }(\tilde{x}))x.\end{array}$$

By virtue of (4.10) and (4.11) we arrive at

$$\tilde{K}(\tilde{x},\epsilon )=\alpha x^{\mathrm{T}}J{g}^{\prime }(\tilde{x})x.$$ 6.6

Now we compute as in the previous section:

$$\begin{array}{rl}{g}^{\prime }(\tilde{x})& =g^{\prime }(x)+g^{″}(\tilde{x}-x)=g^{\prime }(x)+2\epsilon g^{″}{(I-\epsilon f^{\prime }(x))}^{-1}f(x)\\ & =g^{\prime }(x)+2\epsilon g^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}(I+\epsilon f^{\prime }(x))f(x).\end{array}$$ 6.7

We will show that the contribution to (6.6) of the terms in (6.7) with odd powers of ε vanishes

$$\alpha x^{\mathrm{T}}J{g}^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x)x=0.$$ 6.8

For this, we use the fact that for an arbitrary vector $v\in {\mathbb{C}}^4$ we have

$$g^{″}vx=g^{\prime }(x)v.$$ 6.9

Indeed, due to homogeneity of g′(x),

$${(g^{″}vx)}_i=\sum _{j,k=1}^4\frac{{\mathrm{\partial }}^2{g}_i}{\mathrm{\partial }{x}_j\mathrm{\partial }{x}_k}{v}_j{x}_k=\sum _{j=1}^4\frac{\mathrm{\partial }{g}_i}{\mathrm{\partial }{x}_j}{v}_j=(g^{\prime }{(x)v)}_i.$$

Owing to (6.9), the left-hand side of (6.8) is equal to

$$\begin{array}{rl}& \alpha x^{\mathrm{T}}J{g}^{\prime }(x){(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x)\\ & =-x^{\mathrm{T}}{\mathrm{\nabla }}^{2}K(x){(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x)\\ & =-2{(\mathrm{\nabla }K(x))}^{\mathrm{T}}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x)\\ & =-2{(\mathrm{\nabla }H(x))}^{\mathrm{T}}(A^{\mathrm{T}}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}J)\mathrm{\nabla }H(x).\end{array}$$

One easily sees with the help of (4.6), (4.7) and (3.3) that the matrix A^T(I−ε²(f′(x))²)⁻¹J is skew-symmetric, which finishes the proof of (6.8). With this result, (6.6) turns into

$$\tilde{K}(\tilde{x},\epsilon )=\alpha x^{\mathrm{T}}J{g}^{\prime }(x)x+2\alpha {\epsilon }^2{g}^{″}{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}{f}^{\prime }(x)f(x)x.$$

By virtue of (4.11), (4.9) and (6.9), we put the latter formula as

$$\begin{array}{rl}\tilde{K}(\tilde{x},\epsilon )& =\alpha x^{\mathrm{T}}J{g}^{\prime }(x)x+2{\epsilon }^2\alpha x^{\mathrm{T}}J{g}^{″}{(I-{\epsilon }^2{(g^{\prime }(x))}^2)}^{-1}{g}^{\prime }(x)g(x)x\\ & =2\alpha x^{\mathrm{T}}Jg(x)+2{\epsilon }^2\alpha x^{\mathrm{T}}J{g}^{\prime }(x){(I-{\epsilon }^2{(g^{\prime }(x))}^2)}^{-1}{g}^{\prime }(x)g(x)\\ & =2\alpha x^{\mathrm{T}}J{(I-{\epsilon }^2{(g^{\prime }(x))}^2)}^{-1}g(x)=\tilde{K}(x,\epsilon ).\end{array}$$

This finishes the proof of the theorem. ▪

7. Invariant Poisson structure

Theorem 7.1 —

Both maps Φ_f and Φ_g are Poisson with respect to the brackets with the Poisson tensor Π(x) given by

$$\begin{array}{rl}\mathrm{\Pi }(x)& =J-{\epsilon }^2{(f^{\prime }(x))}^{2}J\end{array}$$ 7.1

$$\begin{array}{rl}& =(1-{\epsilon }^{2}p(x))J-{\epsilon }^{2}q(x)A^{\mathrm{T}}J,\end{array}$$ 7.2

where p(x) and q(x) are quadratic polynomials given in (4.15), (4.16).

Proof. —

First, we prove that

$$\mathrm{d}{\mathit{\Phi }}_f(x)\mathrm{\Pi }(x){(\mathrm{d}{\mathit{\Phi }}_f(x))}^{\mathrm{T}}=\mathrm{\Pi }(\tilde{x}).$$ 7.3

With the expression (2.4) for dΦ_f(x), (7.3) turns into

$$(I+\epsilon f^{\prime }(\tilde{x}))\mathrm{\Pi }(x){(I+\epsilon f^{\prime }(\tilde{x}))}^{\mathrm{T}}=(I-\epsilon f^{\prime }(x))\mathrm{\Pi }(\tilde{x}){(I-\epsilon f^{\prime }(x))}^{\mathrm{T}}.$$

Multiplying from the right by J and taking into account (4.6), we arrive at:

$$(I+\epsilon f^{\prime }(\tilde{x}))\mathrm{\Pi }(x)J(I-\epsilon f^{\prime }(\tilde{x}))=(I-\epsilon f^{\prime }(x))\mathrm{\Pi }(\tilde{x})J(I+\epsilon f^{\prime }(x)).$$

According to lemma 4.4, the matrix Π(x)J is a linear combination of I and A^T, therefore, by virtue of (4.7), it commutes with $f^{\prime }(\tilde{x})$ (actually, with f′ evaluated at any point). Thus, the latter equation is equivalent to

$$(I-{\epsilon }^2{(f^{\prime }(\tilde{x}))}^2)\mathrm{\Pi }(x)J=\mathrm{\Pi }(\tilde{x})J(I-{\epsilon }^2{(f^{\prime }(x))}^2),$$

which is obviously true due to (7.1).

It remains to prove that Π(x) is indeed a Poisson tensor. For this, one has to verify the Jacobi identity

$$\{x_i,\{x_j,x_k\}\}+\{x_j,\{x_k,x_i\}\}+\{x_k,\{x_i,x_j\}\}=0$$ 7.4

for the four different triples of indices {i,j,k} from {1,2,3,4}. A straightforward computation based on the expression (7.2) for Π(x) shows that the left-hand sides of the expressions (7.4) are polynomials of order 2 in ε², with the coefficients by ε² being the corresponding components of the vector ∇p(x)−A∇q(x), and the coefficients by ε⁴ being linear combinations of the latter. A reference to lemma 4.6 finishes the proof. ▪

8. Differential equations for the conserved quantities of maps Φ_f, Φ_g

Theorem 8.1 —

The rational functions $\tilde{H}(x,\epsilon )$, $\tilde{K}(x,\epsilon )$ are related by

$$\mathrm{\nabla }\tilde{K}(x,\epsilon )=A\mathrm{\nabla }\tilde{H}(x,\epsilon ).$$ 8.1

As a consequence, they satisfy the same second order differential equations (3.7)–(3.11) as the cubic polynomials H(x), K(x). Moreover, they are in involution with respect to the Poisson bracket with the Poisson tensor Π(x).

Proof. —

We start the proof with the derivation of the following formula for $\tilde{H}(x,\epsilon )$:

$$\tilde{H}(x,\epsilon )=-6\frac{(1-{\epsilon }^{2}p(x))H(x)+{\epsilon }^{2}q(x)K(x)}{{(1-{\epsilon }^{2}p(x))}^2-{\epsilon }^4{\alpha }^2{q}^2(x)}.$$ 8.2

For this aim, we observe the following formula:

$${(\mathrm{\Pi }(x))}^{-1}=-\frac{(1-{\epsilon }^{2}p(x))J+{\epsilon }^{2}q(x)AJ}{{(1-{\epsilon }^{2}p(x))}^2-{\epsilon }^4{\alpha }^2{q}^2(x)}.$$ 8.3

This is checked by a straightforward multiplication of expressions (7.1), (8.3). We note that, by lemma 4.5, the denominator in (8.3) is equal to

$$1-\frac{{\epsilon }^2}{2}\hspace{0.17em}\mathrm{tr}{(f^{\prime })}^2+{\epsilon }^{4}det{f}^{\prime }=det(I-\epsilon f^{\prime }).$$

Now we derive from (6.3), (7.1), (7.2):

$$\begin{array}{rl}\tilde{H}(x,\epsilon )& =2x^{\mathrm{T}}J{(I-{\epsilon }^2{(f^{\prime }(x))}^2)}^{-1}f(x)\\ & =-2x^{\mathrm{T}}{(\mathrm{\Pi }(x))}^{-1}f(x)\\ & =2\frac{(1-{\epsilon }^{2}p(x))x^{\mathrm{T}}Jf(x)+{\epsilon }^{2}q(x)x^{\mathrm{T}}AJf(x)}{det(I-\epsilon f^{\prime }(x))}\\ & =-2\frac{(1-{\epsilon }^{2}p(x))x^{\mathrm{T}}\mathrm{\nabla }H(x)+{\epsilon }^{2}q(x)x^{\mathrm{T}}\mathrm{\nabla }K(x)}{det(I-\epsilon f^{\prime }(x))}\\ & =-6\frac{(1-{\epsilon }^{2}p(x))H(x)+{\epsilon }^{2}q(x)K(x)}{det(I-\epsilon f^{\prime }(x))},\end{array}$$

which is formula (8.2). Interchanging the roles of H and K, that is, replacing (H,K) by (K,α²H), we find

$$\tilde{K}(x,\epsilon )=-6\frac{(1-{\epsilon }^{2}p(x))K(x)+{\epsilon }^2{\alpha }^{2}q(x)H(x)}{{(1-{\epsilon }^{2}p(x))}^2-{\epsilon }^4{\alpha }^2{q}^2(x)}.$$ 8.4

We write formulas (8.2), (8.4) as

$$\tilde{H}(x,\epsilon )=r(x)H(x)+s(x)K(x)$$ 8.5

and

$$\tilde{K}(x,\epsilon )=r(x)K(x)+{\alpha }^{2}s(x)H(x).$$ 8.6

We will prove the following relation for the coefficients r(x), s(x):

$$\mathrm{\nabla }r(x)=A\mathrm{\nabla }s(x).$$ 8.7

Then (8.1) will be an immediate corollary of (8.5), (8.6), combined with (3.2) and (8.7). Indeed, we have:

$$\begin{array}{rl}\mathrm{\nabla }\tilde{H}& =r\mathrm{\nabla }H+H\mathrm{\nabla }r+s\mathrm{\nabla }K+K\mathrm{\nabla }s\\ & =r\mathrm{\nabla }H+HA\mathrm{\nabla }s+sA\mathrm{\nabla }H+K\mathrm{\nabla }s,\\ \mathrm{\nabla }\tilde{K}& =r\mathrm{\nabla }K+K\mathrm{\nabla }r+{\alpha }^{2}s\mathrm{\nabla }H+{\alpha }^{2}H\mathrm{\nabla }s\\ & =rA\mathrm{\nabla }H+KA\mathrm{\nabla }s+{\alpha }^{2}s\mathrm{\nabla }H+{\alpha }^{2}H\mathrm{\nabla }s\end{array}$$

and (8.1) follows directly by virtue of A²=α²I.

Thus, it remains to prove (8.7). We have

$$\begin{array}{rl}-\frac{1}{3}r(x)& =\frac{2(1-{\epsilon }^{2}p(x))}{{(1-{\epsilon }^{2}p(x))}^2-{\epsilon }^4{\alpha }^2{q}^2(x)}\\ & =\frac{1}{1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x)}+\frac{1}{1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x)}\end{array}$$ 8.8

and

$$\begin{array}{rl}-\frac{1}{3}s(x)& =\frac{2{\epsilon }^{2}q(x)}{{(1-{\epsilon }^{2}p(x))}^2-{\epsilon }^4{\alpha }^2{q}^2(x)}\\ & =\frac{{\alpha }^{-1}}{1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x)}-\frac{{\alpha }^{-1}}{1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x)}.\end{array}$$ 8.9

Therefore, by virtue of (4.23)

$$\begin{array}{rl}-\frac{1}{3}{\epsilon }^{-2}\mathrm{\nabla }r(x)& =\frac{\mathrm{\nabla }p(x)+\alpha \mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x))}^2}+\frac{\mathrm{\nabla }p(x)-\alpha \mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x))}^2}\\ & =\frac{(A+\alpha I)\mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x))}^2}+\frac{(A-\alpha I)\mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x))}^2},\\ -\frac{1}{3}{\epsilon }^{-2}\mathrm{\nabla }s(x)& =\frac{{\alpha }^{-1}(\mathrm{\nabla }p(x)+\alpha \mathrm{\nabla }q(x))}{{(1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x))}^2}-\frac{{\alpha }^{-1}(\mathrm{\nabla }p(x)-\alpha \mathrm{\nabla }q(x))}{{(1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x))}^2}\\ & =\frac{{\alpha }^{-1}(A+\alpha I)\mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)-{\epsilon }^2\alpha q(x))}^2}-\frac{{\alpha }^{-1}(A-\alpha I)\mathrm{\nabla }q(x)}{{(1-{\epsilon }^{2}p(x)+{\epsilon }^2\alpha q(x))}^2}.\end{array}$$

Now (8.7) follows by virtue of

$$\begin{array}{rl}& {\alpha }^{-1}A(A+\alpha I)=A+{\alpha }^{-1}{A}^2=A+\alpha I,\\ & {\alpha }^{-1}A(A-\alpha I)=-A+{\alpha }^{-1}{A}^2=-(A-\alpha I).\end{array}$$

This finishes the proof of $\mathrm{\nabla }\tilde{K}=A\mathrm{\nabla }\tilde{H}$. As for the last statement of the theorem, we observe that

$${\{\tilde{H},\tilde{K}\}}_{\mathrm{\Pi }}(x)={(\mathrm{\nabla }\tilde{K}(x))}^{\mathrm{T}}\mathrm{\Pi }(x)\mathrm{\nabla }\tilde{H}(x)={(\mathrm{\nabla }\tilde{H}(x))}^{\mathrm{T}}{A}^{\mathrm{T}}\mathrm{\Pi }(x)\mathrm{\nabla }\tilde{H}(x)=0,$$

since, according to (7.2), the matrix A^TΠ(x) is a linear combination of A^TJ and (A^T)²J, and therefore it is skew-symmetric. ▪

9. Conclusion

Completely integrable Hamiltonian systems lying at the basis of our constructions, seem to be worth studying on their own. In particular, their invariant surfaces are intersections of two cubic hypersurfaces in the four-dimensional space. Algebraic geometry of such surfaces and of the corresponding foliations of the phase space does not seem to be elaborated very well in the existing literature. Preliminary studies (in collaboration with Pol Vanhaecke) indicate that the systems of our class are algebraically completely integrable, that is, their invariant surfaces are (affine parts of) Abelian surfaces and the vector fields are translation invariant with respect to the corresponding group law. Still more interesting and intriguing are the algebraic–geometric aspects of the commuting pairs of integrable maps introduced here.

After the first version of this paper has been submitted for publication, we found out that our constructions are generalizable into higher dimensions, as well as to a certain class of Lie–Poisson Hamiltonian systems [13,14]. We are confident that this line of research will lead us to still more exciting discoveries.

Note added in proof

After the final version of the paper went to print, we observed that any system of our class can be transformed by a linear symplectic change of variables to two uncoupled two-dimensional systems. Indeed, one can find a matrix $S\in S_p(4,\mathbb{R})$ such that $SA{S}^{-1}=\left(\begin{array}{cc}W& 0\\ 0& W^{\mathrm{T}}\end{array}\right)$, where W=diag(α,−α). Perform the change of variables x=s^Ty, and denote $H(x)=\overline{H}(y)$ and similarly $K(x)=\overline{K}(y)$. Then $\overline{H}(y)=h_1(y_1,y_3)+h_2(y_2,y_4)$, while $\overline{K}(y)=\alpha h_1(y_1,y_3)-\alpha h_2(y_2,y_4)$.

Authors' contributions

Both authors contributed to the theoretical results and to the drafting and checking of the manuscript, and gave final approval for the publication.

Competing interests

We have no competing interests.

Funding

This research is supported by the DFG Collaborative Research Center TRR 109 ‘Discretization in Geometry and Dynamics’.