Geometry of the Kahan discretizations of planar quadratic Hamiltonian systems

Abstract

Kahan discretization is applicable to any quadratic vector field and produces a birational map which approximates the shift along the phase flow. For a planar quadratic canonical Hamiltonian vector field, this map is known to be integrable and to preserve a pencil of cubic curves. Generically, the nine base points of this pencil include three points at infinity (corresponding to the asymptotic directions of cubic curves) and six finite points lying on a conic. We show that the Kahan discretization map can be represented in six different ways as a composition of two Manin involutions, corresponding to an infinite base point and to a finite base point. As a consequence, the finite base points can be ordered so that the resulting hexagon has three pairs of parallel sides which pass through the three base points at infinity. Moreover, this geometric condition on the base points turns out to be characteristic: if it is satisfied, then the cubic curves of the corresponding pencil are invariant under the Kahan discretization of a planar quadratic canonical Hamiltonian vector field.

1. Introduction

The Kahan discretization was introduced in the unpub- lished lecture notes [1] as a method applicable to any system of ordinary differential equations on ${\mathbb{R}}^n$ 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 discretiz- ation reads as

$$\frac{\tilde{x}-x}{\epsilon }=Q(x,\tilde{x})+\frac{1}{2}B(x+\tilde{x})+c,$$ 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 $\tilde{x}={\mathrm{\Phi }}_f(x,\epsilon )$. Explicitly, one has

$$\tilde{x}={\mathrm{\Phi }}_f(x,\epsilon )=x+\epsilon {(I-\frac{\epsilon }{2}{f}^{\prime }(x))}^{-1}f(x),$$ 1.3

where f′(x) denotes the Jacobi matrix of f(x).

Clearly, this map approximates the time ε 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

$${\mathrm{\Phi }}_f^{-1}(x,\epsilon )={\mathrm{\Phi }}_f(x,-\epsilon ).$$ 1.4

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 [2], who demonstrated 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 [3,4]. Surprisingly, the discretization scheme produced in both cases integrable maps.

In [5–7], the authors undertook an extensive study of the properties of Kahan's method when applied to integrable systems (we proposed that in the integrable context the term ‘Hirota–Kimura method’ should be used). 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 Kahan's method were discovered by Celledoni, McLachlan, Owren and Quispel.

Theorem 1.1 [8]. —

Consider a Hamiltonian vector field f(x) = J∇H(x), where J is a constant skew-symmetric n × n matrix, and the Hamilton function $H:{\mathbb{R}}^n\to \mathbb{R}$ is a polynomial of degree 3. Then the map Φ_f(x, ε) possesses the following rational integral of motion:

$$\tilde{H}(x,\epsilon )=H(x)+\frac{\epsilon }{3}{(\mathrm{\nabla }H(x))}^{\mathrm{T}}{(I-\frac{\epsilon }{2}{f}^{\prime }(x))}^{-1}f(x),$$ 1.5

as well as an invariant measure

$$\frac{\mathrm{d}{x}_1\wedge \dots \wedge \mathrm{d}{x}_n}{{\displaystyle det(I-(\epsilon /2)f^{\prime }(x))}}.$$ 1.6

The degree of the denominator $det(I-(\epsilon /2)f^{\prime }(x))$ of the function $\tilde{H}(x,\epsilon )$ is n, while the degree of the numerator of $\tilde{H}(x,\epsilon )$ is n + 1.

In the present paper, we study the case n = 2. Here, Φ_f is a birational planar map with an invariant measure and an integral of motion, thus it is completely integrable. Integral (1.5) in this case is given by

$$\tilde{H}(x,y,\epsilon )=\frac{C(x,y,\epsilon )}{D(x,y,\epsilon )},$$ 1.7

where C(x, y, ε) is a polynomial in x, y of degree 3 and D(x, y, ε) is a polynomial of degree 2. Thus, the level sets of the integral are cubic curves

$${\mathcal{E}}_{\lambda }=\{(x,y):C(x,y,\epsilon )-\lambda D(x,y,\epsilon )=0\},$$

which form a linear system (a pencil). Such a pencil is characterized by its base points (common points of all curves of the pencil). On each invariant curve, the map Φ_f induces a shift (respective to the addition law on this curve). In [9,10], it was shown that in several cases Φ_f can be represented as a Manin transformation, i.e. as a composition of two Manin involutions, as defined, for example, in [[11], p. 35; [12], sect. 4.2].

In this paper, we prove that, actually, in the generic situation Φ_f admits six different representations as a Manin transformation. Analysing these representations, we arrive at an amazing geometric characterization:

A pencil of elliptic curves consists of invariant curves for the Kahan discretization of a planar quadratic canonical Hamiltonian vector field if and only if the six finite base points can be ordered so that the resulting hexagon has three pairs of parallel sides which pass through the three base points at infinity.

The structure of the paper is as follows. In §2, we discuss the generalities about the pencils of cubic curves for Kahan discretizations and recall the definitions of Manin involutions and Manin maps, as well as formulae for computing those maps. In §3, we prove the sixfold representation of the Kahan discretization of a generic planar quadratic Hamiltonian vector field as a Manin map. Finally, in §4, we derive from the latter fact the remarkable geometric characterization given above.

2. The geometry of the invariant cubic curves of the Kahan map

The geometric properties of the birational map Φ_f become more uniform if we consider it in the complex domain, i.e. as a map of the complex plane ${\mathbb{C}}^2$, and actually as a map of the projective complex plane $\mathbb{C}{\mathbb{P}}^2$. In particular, we assume that all the coefficients a_i of the Hamilton function H(x, y) are complex numbers, as well as its arguments x, y.

This phase space is foliated by the one-parameter family (pencil) of invariant curves

$${\mathcal{E}}_{\lambda }=\{(x,y)\in {\mathbb{C}}^2\hspace{0.17em}:\hspace{0.17em}C(x,y,\epsilon )-\lambda D(x,y,\epsilon )=0\}.$$

Indeed, for any initial point $(x_0,y_0)\in {\mathbb{C}}^2$ the orbit of Φ lies on the cubic curve ${\mathcal{E}}_{\lambda }$ with $\lambda =\tilde{H}(x_0,y_0,\epsilon )=C(x_0,y_0,\epsilon )/D(x_0,y_0,\epsilon )$.

We consider ${\mathbb{C}}^2$ as an affine part of $\mathbb{C}{\mathbb{P}}^2$ consisting of the points $[x:y:z]\in \mathbb{C}{\mathbb{P}}^2$ with z≠0. We define the projective curves ${\overline{\mathcal{E}}}_{\lambda }$ as projective completion of ${\mathcal{E}}_{\lambda }$,

$${\overline{\mathcal{E}}}_{\lambda }=\{[x:y:z]\in \mathbb{C}{\mathbb{P}}^2\hspace{0.17em}:\hspace{0.17em}\overline{C}(x,y,z,\epsilon )-\lambda z\overline{D}(x,y,z,\epsilon )=0\},$$

where we set

$$\overline{C}(x,y,z,\epsilon )=z^{3}C(\frac{x}{z},\frac{y}{z},\epsilon )\mathrm{and}\overline{D}(x,y,z,\epsilon )=z^{2}D(\frac{x}{z},\frac{y}{z},\epsilon ).$$

We assume that the curve ${\overline{\mathcal{E}}}_0=\{[x:y:z]\in \mathbb{C}{\mathbb{P}}^2\hspace{0.17em}:\hspace{0.17em}\overline{C}(x,y,z,\epsilon )=0\}$ is non-singular. Note that the second basis curve of the pencil, ${\overline{\mathcal{E}}}_{\mathrm{\infty }}=\{[x:y:z]\in \mathbb{C}{\mathbb{P}}^2\hspace{0.17em}:\hspace{0.17em}z\overline{D}(x,y,z,\epsilon )=0\}$, is reducible, and consists of the conic $\{\overline{D}(x,y,z,\epsilon )=0\}$ and of the line at infinity {z = 0}.

All curves ${\overline{\mathcal{E}}}_{\lambda }$ pass through the set of base points which is defined as ${\overline{\mathcal{E}}}_0\cap {\overline{\mathcal{E}}}_{\mathrm{\infty }}$. According to the Bezout theorem, there are nine base points, counted with multiplicities. In our specific context, there are three base points at infinity,

$$\{F_1,F_2,F_3\}={\overline{\mathcal{E}}}_0\cap \{z=0\},$$

and six further base points $\{B_1,\dots ,B_6\}={\overline{\mathcal{E}}}_0\cap \{\overline{D}=0\}$ (figure 1). For any point of $P\in \mathbb{C}{\mathbb{P}}^2$ different from the base points, there is a unique curve ${\overline{\mathcal{E}}}_{\lambda }$ of the pencil such that $P\in {\overline{\mathcal{E}}}_{\lambda }$.

Figure 1.

The red curve is the cubic C = 0 passing through the origin; the blue curve is the conic D = 0. The black lines are the tangent lines to the red cubic at the points F₁, F₂, F₃ at infinity. The finite base points are B₁, …, B₆. (Online version in colour.)

Now we recall the definitions of Manin involutions and Manin transformations, following [[12, sect. 4.2]].

Definition 2.1 —

• (1)Consider a non-singular cubic curve $\overline{\mathcal{E}}$ in $\mathbb{C}{\mathbb{P}}^2$, and a point $P_0\in \overline{\mathcal{E}}$. The Manin involution on $\overline{\mathcal{E}}$ with respect to P₀ is the map $I_{\overline{\mathcal{E}},P_0}:\overline{\mathcal{E}}\to \overline{\mathcal{E}}$ defined as follows:
  • —For P₁≠P₀, the point $P_2=I_{\overline{\mathcal{E}},P_0}(P_1)$ is the unique third intersection point of $\overline{\mathcal{E}}$ with the line (P₀P₁);
  • —For P₁ = P₀, the point $P_2=I_{\overline{\mathcal{E}},P_0}(P_1)$ is the unique second intersection point of $\overline{\mathcal{E}}$ with the tangent line to $\overline{\mathcal{E}}$ at P₀ = P₁.
• (2)
  Let P₀, P₁ be two distinct points of the curve $\overline{\mathcal{E}}$. The Manin transformation $M_{\overline{\mathcal{E}},P_0,P_1}:\overline{\mathcal{E}}\to \overline{\mathcal{E}}$ is defined as

  $$M_{\overline{\mathcal{E}},P_0,P_1}=I_{\overline{\mathcal{E}},P_1}\circ I_{\overline{\mathcal{E}},P_0}.$$ 2.1

It is easy to see that $M_{\overline{\mathcal{E}},P_0,P_1}(P_1)=P_0$. Indeed, if P₂ is the third intersection point of $\overline{\mathcal{E}}$ with the line (P₀P₁) then $I_{\overline{\mathcal{E}},P_0}(P_1)=P_2$ and $I_{\overline{\mathcal{E}},P_1}(P_2)=P_0$. With a natural addition law on the elliptic curve $\overline{\mathcal{E}}$ (with a flex O as a neutral element, so that P₀ + P₁ + P₂ = O if and only if the points P₀, P₁, P₂ are collinear), we can write $I_{\overline{\mathcal{E}},P_0}(P)=-(P_0+P)$, and

$$\begin{array}{rl}{M}_{\overline{\mathcal{E}},P_0,P_1}(P)& =I_{\overline{\mathcal{E}},P_1}(I_{\overline{\mathcal{E}},P_0}(P))=-(P_1+I_{\overline{\mathcal{E}},P_0}(P))\\ & =-(P_1-(P_0+P))=P+P_0-P_1,\end{array}$$

so that $M_{\overline{\mathcal{E}},P_0,P_1}$ is the translation by P₀ − P₁ on the elliptic curve $\overline{\mathcal{E}}$.

Definition 2.2 —

Consider a pencil $\mathfrak{E}=\{{\overline{\mathcal{E}}}_{\lambda }\}$ of cubic curves in $\mathbb{C}{\mathbb{P}}^2$ with at least one non-singular member.

• (1)Let B be a base point of the pencil. The Manin involution $I_{\mathfrak{E},B}:\mathbb{C}{\mathbb{P}}^2⇢\mathbb{C}{\mathbb{P}}^2$ is a birational map defined as follows. For any $P\in \mathbb{C}{\mathbb{P}}^2$ which is not a base point, $I_{\mathfrak{E},B}(P)=I_{{\overline{\mathcal{E}}}_{\lambda },B}(P)$, where ${\overline{\mathcal{E}}}_{\lambda }$ is the unique curve of the pencil containing the point P.
• (2)
  Let B₁, B₂ be two distinct base points of the pencil. The Manin transformation $M_{\mathfrak{E},B_1,B_2}:\mathbb{C}{\mathbb{P}}^2⇢\mathbb{C}{\mathbb{P}}^2$ is a birational map defined as

  $$M_{\mathfrak{E},B_1,B_2}=I_{\mathfrak{E},B_2}\circ I_{\mathfrak{E},B_1}.$$ 2.2

We now provide formulae which can be used to compute Manin involutions.

Lemma 2.3 —

Let $\overline{\mathcal{E}}$ be a non-singular cubic curve in $\mathbb{C}{\mathbb{P}}^2$ given in the non-homogeneous coordinates by the equation

$$\mathcal{E}:u_1{x}^3+u_2{x}^{2}y+u_{3}x{y}^2+u_4{y}^3+u_5{x}^2+u_{6}xy+u_7{y}^2+u_{8}x+u_{9}y+u_{10}=0.$$

Let P₀ = (x₀, y₀) be a finite point on $\mathcal{E}$, and let F = [1:μ:0] be a point of $\overline{\mathcal{E}}$ at infinity. If $P_1=I_{\overline{\mathcal{E}},P_0}(F)$ is finite, then it is given by

$$P_1=(-(x_0+\alpha ),-\mu (x_0+\alpha )+\beta ),$$

where

$$\beta =y_0-\mu x_0$$

and

$$\alpha =\frac{(3u_4\mu +u_3){\beta }^2+(2u_7\mu +u_6)\beta +u_9\mu +u_8}{(3u_4{\mu }^2+2u_3\mu +u_2)\beta +u_7{\mu }^2+u_6\mu +u_5}.$$

Proof. —

The equation of the line (P₀F) is y = μx + β. To find the second (finite) intersection point P₁ = (x₁, y₁) of this line with $\mathcal{E}$, we substitute this equation into equation U(x, y) = 0 of the cubic curve. We have

$$U(x,\mu x+\beta )=A_3{x}^3+A_2{x}^2+A_{1}x+A_0,$$ 2.3

where

$$\begin{array}{rl}{A}_3& =u_4{\mu }^3+u_3{\mu }^2+u_2\mu +u_1,\end{array}$$ 2.4

$$\begin{array}{r}\\ A_2& =(3u_4{\mu }^2+2u_3\mu +u_2)\beta +u_7{\mu }^2+u_6\mu +u_5,\end{array}$$ 2.5

$$\begin{array}{rl}{A}_1& =(3u_4\mu +u_3){\beta }^2+(2u_7\mu +u_6)\beta +u_9\mu +u_8\end{array}$$ 2.6

$$\begin{array}{r}\\ \mathrm{and}{A}_0& =u_4{\beta }^3+u_7{\beta }^2+u_9\beta +u_{10}.\end{array}$$ 2.7

Moreover, A₃ = 0 since $F\in \overline{\mathcal{E}}$. Thus, for x₁ we get a quadratic equation. By the Vieta formula, we have: x₀ + x₁ = − α, where α = A₁/A₂, which is the desired formula. ▪

Lemma 2.4 —

Let $\overline{\mathcal{E}}$ be a non-singular cubic curve in $\mathbb{C}{\mathbb{P}}^2$ given in the non-homogeneous coordinates by the equation

$$\mathcal{E}:u_1{x}^3+u_2{x}^{2}y+u_{3}x{y}^2+u_4{y}^3+u_5{x}^2+u_{6}xy+u_7{y}^2+u_{8}x+u_{9}y+u_{10}=0.$$

Let P₁ = (x₁, y₁), P₂ = (x₂, y₂) be two finite points on $\mathcal{E}$ with x₁≠x₂. If the point $P_3=I_{\overline{\mathcal{E}},P_1}(P_2)=I_{\overline{\mathcal{E}},P_2}(P_1)$ is finite, then it is given by

$$P_3=(-(x_1+x_2+\gamma ),-(y_1+y_2+\delta )),$$

where

$$\nu =\frac{y_2-y_1}{x_2-x_1},\beta =y_1-\nu x_1=y_2-\nu x_2$$

and

$$\begin{array}{rl}\gamma & =\frac{(3u_4\beta +u_7){\nu }^2+(2u_3\beta +u_6)\nu +u_2\beta +u_5}{u_4{\nu }^3+u_3{\nu }^2+u_2\nu +u_1},\end{array}$$ 2.8

$$\begin{array}{r}\\ \delta & =\frac{u_7{\nu }^3-(u_3\beta -u_6){\nu }^2-(2u_2\beta -u_5)\nu -3u_1\beta }{u_4{\nu }^3+u_3{\nu }^2+u_2\nu +u_1}.\end{array}$$ 2.9

Proof. —

Proceeding as before, we observe that in this case the equation of the line (P₁P₂) is y = νx + β. Thus, for the third intersection point P₃ = (x₃, y₃) of this line with $\mathcal{E}$ we get a cubic equation (2.3), where in the coefficients A_i one should change notation μ to ν. We compute x₃ from the Vieta formula x₁ + x₂ + x₃ = − γ, where γ = A₂/A₃ is found as in (2.8). To compute y₃, we observe that

$$y_3=\nu x_3+\beta =-\nu (x_1+x_2+\gamma )+\beta =-y_1-y_2+3\beta -\nu \gamma ,$$

and it remains to compute δ = νγ − 3β, which leads to (2.9). ▪

3. Kahan discretization as Manin transformation

For computations described in this section, we will need concrete formulae for the map Φ_f. The Hamiltonian vector field f = J∇H with $J=\left(\begin{array}{cc}\phantom{-}0& 1\\ -1& 0\end{array}\right)$ for the Hamilton function

$$H(x,y)=\frac{1}{3}{a}_1{x}^3+a_2{x}^{2}y+a_{3}x{y}^2+\frac{1}{3}{a}_4{y}^3+a_5{x}^2+2a_{6}xy+a_7{y}^2+a_{8}x+a_{9}y$$ 3.1

is given by

$$\dot{x}=H_y=a_2{x}^2+2a_{3}xy+a_4{y}^2+2a_{6}x+2a_{7}y+a_9$$ 3.2

and

$$\dot{y}=-H_x=-a_1{x}^2-2a_{2}xy-a_3{y}^2-2a_{5}x-2a_{6}y-a_8.$$ 3.3

(The normalization of the coefficients in the Hamilton function (3.1) is dictated by the desire to get rid of unnecessary factors in the equations of motion.) The Kahan discretization of this system is the map $(\tilde{x},\tilde{y})={\mathrm{\Phi }}_f(x,y,\epsilon )$ defined by the equations of motion

$$\frac{\tilde{x}-x}{\epsilon }=a_2\tilde{x}x+a_3(\tilde{x}y+x\tilde{y})+a_4\tilde{y}y+a_6(\tilde{x}+x)+a_7(\tilde{y}+y)+a_9$$ 3.4

and

$$\frac{\tilde{y}-y}{\epsilon }=-a_1\tilde{x}x-a_2(\tilde{x}y+x\tilde{y})-a_3\tilde{y}y-a_5(\tilde{x}+x)-a_6(\tilde{y}+y)-a_8,$$ 3.5

which can be solved for $\tilde{x},\tilde{y}$ according to

$$\begin{array}{rl}\left(\begin{array}{c}\tilde{x}\\ \tilde{y}\end{array}\right)& ={\left(\begin{array}{cc}1-\epsilon a_{2}x-\epsilon a_{3}y-\epsilon a_6& -\epsilon a_{3}x-\epsilon a_{4}y-\epsilon a_7\\ \epsilon a_{1}x+\epsilon a_{2}y+\epsilon a_5& 1+\epsilon a_{2}x+\epsilon a_{3}y+\epsilon a_6\end{array}\right)}^{-1}\\ & \times \left(\begin{array}{c}x+\epsilon a_{6}x+\epsilon a_{7}y+\epsilon a_9\\ y-\epsilon a_{5}x-\epsilon a_{6}y-\epsilon a_8\end{array}\right).\end{array}$$ 3.6

As a result,

$$\tilde{x}=\frac{R(x,y,ϵ)}{D(x,y,ϵ)}\mathrm{and}\tilde{y}=\frac{S(x,y,ϵ)}{D(x,y,ϵ)},$$ 3.7

where R, S and D are polynomials of degree 2. They can be found in appendix A. The integral (1.5) in the case n = 2 is given by

$$\tilde{H}(x,y,\epsilon )=\frac{C(x,y,\epsilon )}{D(x,y,\epsilon )},$$ 3.8

where C(x, y, ε) is a polynomial in x, y of degree 3, also given in appendix A.

Consider the pencil of cubic curves $\mathfrak{E}=\{{\overline{\mathcal{E}}}_{\lambda }\}$, which are level sets of the integral $\tilde{H}$. Recall that this pencil has three base points F₁, F₂, F₃ at infinity, as well as six further base points B₁, …, B₆, which are intersection points of the cubic curve $\overline{C}=0$ with the conic $\overline{D}=0$.

Theorem 3.1 —

To every base point F at infinity, there correspond two base points B, B′, such that the Kahan map Φ is a Manin transformation in two different ways,

$$\mathrm{\Phi }=I_{\mathfrak{E},B}\circ I_{\mathfrak{E},F}=I_{\mathfrak{E},F}\circ I_{\mathfrak{E},B^{\prime }}.$$ 3.9

Altogether, the Kahan map Φ is a Manin transformation in six different ways:

$$\begin{array}{rl}\mathrm{\Phi }& =I_{\mathfrak{E},B_1}\circ I_{\mathfrak{E},F_1}=I_{\mathfrak{E},F_1}\circ I_{\mathfrak{E},B_4}\end{array}$$ 3.10

$$\begin{array}{rl}& =I_{\mathfrak{E},B_5}\circ I_{\mathfrak{E},F_2}=I_{\mathfrak{E},F_2}\circ I_{\mathfrak{E},B_2}\end{array}$$ 3.11

$$\begin{array}{rl}& =I_{\mathfrak{E},B_3}\circ I_{\mathfrak{E},F_3}=I_{\mathfrak{E},F_3}\circ I_{\mathfrak{E},B_6}.\end{array}$$ 3.12

Proof. —

Both statements in (3.9) are proved similarly, therefore we restrict ourselves to the first one. We start with the following reformulation:

$$\begin{array}{rl}& \mathrm{\Phi }(P)=I_{\mathfrak{E},B}\circ I_{\mathfrak{E},F}(P)\\ & \iff \hspace{0.17em}{I}_{\mathfrak{E},B}(\mathrm{\Phi }(P))=I_{\mathfrak{E},F}(P)\\ & \iff \hspace{0.17em}{I}_{\mathfrak{E},\mathrm{\Phi }(P)}(B)=I_{\mathfrak{E},P}(F)\\ & \iff \hspace{0.17em}B=I_{\mathfrak{E},\mathrm{\Phi }(P)}\circ I_{\mathfrak{E},P}(F).\end{array}$$ 3.13

Thus, it is sufficient to prove that there exists a base point B such that (3.13) is satisfied for all P (for which its right-hand side is defined). Since the expression $I_{\mathfrak{E},\mathrm{\Phi }(P)}\circ I_{\mathfrak{E},P}(F)$ is a continuous function of P on its definition domain, which is a connected set, and since the set of base points is finite, it is sufficient to prove that, for any point P, its image $I_{\mathfrak{E},\mathrm{\Phi }(P)}\circ I_{\mathfrak{E},P}(F)$ is a base point. Since translations of the argument by vectors in ${\mathbb{C}}^2$ act transitively on the set of cubic Hamilton functions, it is sufficient to prove the statement just for one point P₀ (and for all Hamilton functions). We will choose $P_0=(0,0)\in {\mathbb{C}}^2$. Thus, the proof of the theorem boils down to the following statement.

Lemma 3.2 —

The point $I_{\mathfrak{E},\mathrm{\Phi }(0,0)}\circ I_{\mathfrak{E},(0,0)}(F)$ is a base point of the pencil $\mathfrak{E}$. More precisely, the polynomial D(x, y, ε) vanishes at this point.

This lemma is proved by a direct computation, which is outlined as follows.

• (1)Determine the pencil parameter of the level curve containing the point (0, 0), that is, ${\lambda }_0=\tilde{H}(0,0,\epsilon )=c_{10}/d_6$.
• (2)Compute the coefficients of the level curve ${\overline{\mathcal{E}}}_{{\lambda }_0}$, which are u_i = c_i for i = 1, …, 4 and u_i = c_i − λ₀d_i−4 for i = 5, …, 10. Of course, u₁₀ = 0.
• (3)
  Compute, according to lemma 2.3,

  $$P_1=I_{{\overline{\mathcal{E}}}_{{\lambda }_0},(0,0)}(F)=(x_1,y_1),$$ 3.14

  where

  $$x_1=-\frac{u_9\mu +u_8}{u_7{\mu }^2+u_6\mu +u_5}\mathrm{and}{y}_1=-\frac{u_9{\mu }^2+u_8\mu }{u_7{\mu }^2+u_6\mu +u_5}.$$ 3.15
• (4)
  Compute

  $$P_2=\mathrm{\Phi }(0,0)=(x_2,y_2),$$ 3.16

  where

  $$x_2=\frac{\epsilon a_9+{\epsilon }^2(a_6{a}_9-a_7{a}_8)}{1+{\epsilon }^2(a_5{a}_7-a_6^2)}\mathrm{and}{y}_2=\frac{-\epsilon a_8+{\epsilon }^2(a_6{a}_8-a_5{a}_9)}{1+{\epsilon }^2(a_5{a}_7-a_6^2)}.$$ 3.17
• (5)Compute $I_{\mathfrak{E},\mathrm{\Phi }(0,0)}\circ I_{\mathfrak{E},(0,0)}(F)$ by applying lemma 2.4 with the point P₁ = (x₁, y₁) given in (3.15) and with the point P₂ = (x₂, y₂) given in (3.17). The coordinates of the resulting point $P_3=I_{{\overline{\mathcal{E}}}_{{\lambda }_0},P_1}(P_2)$ are rational functions of μ with numerators and denominators of degree 8.
• (6)Compute the value of the quadratic polynomial D(x, y, ε) at the point P₃. This value is a rational function of μ. Its numerator is a polynomial of μ of degree 16. Maple computation shows that this polynomial is divisible by u₄μ³ + u₃μ² + u₂μ + u₁. The following tricks can be used to lower the complexity of computations. First, for homogeneity reasons, it is sufficient to take ε = 1. Second, we can use affine transformations of the plane (x, y) (under which the Kahan discretization is covariant) to achieve vanishing of two of the coefficients of the polynomial H, for instance a₇ = 0 and a₉ = 0.

As a consequence of this remarkable factorization, the numerator of D(P₃, ε) vanishes, as soon as F is a base point at infinity, that is, as soon as u₄μ³ + u₃μ² + u₂μ + u₁ = 0. ▪

4. Geometry of the pencil of elliptic curves for the Kahan discretization

We now study the geometric consequences of the sixfold representation of the Kahan discretization as Manin maps. We restrict ourselves to the generic case, when the base points B₁, …, B₆ are finite and pairwise distinct.

Theorem 4.1 —

The lines (B₁B₂) and (B₄B₅) are parallel and pass through the point F₃ at infinity. Similarly, the lines (B₂B₃) and (B₅B₆) are parallel and pass through the point F₁ at infinity, and the lines (B₃B₄) and (B₆B₁) are parallel and pass through the point F₂ at infinity.

Proof. —

Consider an arbitrary invariant curve of the pencil, along with the addition law on this curve (assuming that the neutral element O is a flex, so that P₁ + P₂ + P₃ = O is equivalent to collinearity of the points P₁, P₂, P₃). In particular, since all three points F₁, F₂, F₃ lie on the line at infinity, we have

$$F_1+F_2+F_3=O.$$ 4.1

Now write down equations (3.10)–(3.12) in terms of this addition law. We have

$$F_1-B_1=B_2-F_2=F_3-B_3=B_4-F_1=F_2-B_5=B_6-F_3.$$ 4.2

Consider, for instance, the first of these equations. We have

$$B_1+B_2=F_1+F_2=-F_3\iff B_1+B_2+F_3=O.$$

Thus, the straight line (B₁B₂) passes through F₃ (figure 2). ▪

Figure 2.

The opposite side lines of the hexagon B₁B₂B₃B₄B₅B₆ are parallel. (Online version in colour.)

Corollary 4.2 —

For each cubic curve of the pencil, the second intersection points of the curve with the tangents at the points F₁, F₂, F₃ at infinity lie on the lines connecting the corresponding base points, (B₁B₄), (B₂B₅) and (B₃B₆), respectively.

Proof. —

This follows from equations such as

$$B_2-F_2=F_2-B_5\iff B_2+B_5+(-2F_2)=O,$$

which mean that the points B₂, B₅ and −2F₂ are collinear. Recall that −2F₂ is the second intersection point of the cubic curve with its tangent at F₂ (figure 3). ▪

Figure 3.

The (green) line (B₂B₅) passes through the point −2F₂, which is the second intersection point of the (black) tangent line to the curve $\overline{C}=0$ at point F₂ at infinity. (Online version in colour.)

A very remarkable statement is an inverse theorem to theorem 4.1.

Theorem 4.3 —

Let a hexagon B₁B₂B₃B₄B₅B₆ have three pairs of parallel side lines (B₁B₂)∥(B₄B₅), (B₂B₃)∥(B₅B₆) and (B₃B₄)∥(B₆B₁), passing through the points F₃, F₁ and F₂ at infinity, respectively. Consider the pencil $\mathfrak{E}$ of cubic curves with the base points {B₁, …, B₆, F₁, F₂, F₃}. Then the birational map Φ defined by the pencil $\mathfrak{E}$ according to equations (3.10)–(3.12) is the Kahan discretization of a planar quadratic canonical Hamiltonian vector field.

Proof. —

Observe that, according to Pascal's theorem, the points B₁, …, B₆ lie on a conic D. The pencil $\mathfrak{E}$ contains a reducible curve consisting of the conic D and the line at infinity. The fact that all six Manin maps in (3.10)–(3.12) coincide is an easy consequence of the condition of the theorem, if one reverses the arguments from the proof of theorem 4.1. It remains to prove the statement for one of the representations, say for $I_{\mathfrak{E},B_1}\circ I_{\mathfrak{E},F_1}$. This is again done with a direct Maple computation, which can be organized as follows.

• (1)
  Prescribe arbitrary nine coefficients of the side lines of the hexagon (three slopes μ₁, μ₂, μ₃ and six heights b₁, …, b₆), so that equations of these lines read

  $$(B_1{B}_2):\hspace{0.17em}y={\mu }_{3}x+b_1,(B_2{B}_3):\hspace{0.17em}y={\mu }_{1}x+b_2,(B_3{B}_4):\hspace{0.17em}y={\mu }_{2}x+b_3$$

  and

  $$(B_4{B}_5):\hspace{0.17em}y={\mu }_{3}x+b_4,(B_5{B}_6):\hspace{0.17em}y={\mu }_{1}x+b_5,(B_6{B}_1):\hspace{0.17em}y={\mu }_{2}x+b_6.$$
• (2)
  Compute the points B₁, …, B₆ as pairwise intersection points of these lines:

  $$B_1=(\frac{b_1-b_6}{{\mu }_2-{\mu }_3},\frac{{\mu }_2{b}_1-{\mu }_3{b}_6}{{\mu }_2-{\mu }_3}),\mathrm{etc}.$$
• (3)
  Compute the coefficients d₁, …, d₆ of the conic passing through these six points:

  $$D(x,y)=d_1{x}^2+d_{2}xy+d_3{y}^2+d_{4}x+d_{5}y+d_6=0.$$
• (4)
  Compute the coefficients c₅, …, c₁₀ of the pencil of cubic curves C(x, y) − λD(x, y) = 0 passing through B₁, …, B₆. Here

  $$C(x,y)=(y-{\mu }_{1}x)(y-{\mu }_{2}x)(y-{\mu }_{3}x)+c_5{x}^2+c_{6}xy+c_7{y}^2+c_{8}x+c_{9}y+c_{10}=0$$

  is the equation of an arbitrarily chosen non-singular curve of the pencil. For instance, setting c₁₀ = 0 defines c₅, …, c₉ uniquely.
• (5)
  Compute the involutions $I_{\overline{\mathcal{E}},F_1}(x,y)$ and $I_{\overline{\mathcal{E}},B_1}(x,y)$ according to formulae from lemmas 2.3 and 2.4 for the curve $\mathcal{E}:C-\lambda D=0$ of the pencil passing through a running point (x, y). Thus, we use the expressions

  $$u_1=-{\mu }_1{\mu }_2{\mu }_3,u_2={\mu }_1{\mu }_2+{\mu }_2{\mu }_3+{\mu }_3{\mu }_1,u_3=-({\mu }_1+{\mu }_2+{\mu }_3)\mathrm{and}{u}_4=1$$

  and

  $$u_i=c_i-\lambda d_{i-4},i=5,\dots ,10,\mathrm{where}\hspace{0.17em}\lambda =\frac{C(x,y)}{D(x,y)}.$$
• (6)
  Compute the map $(\tilde{x},\tilde{y})=\mathrm{\Phi }(x,y)=I_{\overline{\mathcal{E}},B_1}\circ I_{\overline{\mathcal{E}},F_1}(x,y)$. It turns out to be of the form

  $$\tilde{x}=\frac{R(x,y)}{D(x,y)}\mathrm{and}\tilde{y}=\frac{S(x,y)}{D(x,y)},$$

  where R and S are polynomials of degree 2.
• (7)
  Check whether these rational functions $\tilde{x}$, $\tilde{y}$ satisfy equations of motion of the Kahan discretization with ε = 1:

  $$\tilde{x}-x=a_2\tilde{x}x+a_3(\tilde{x}y+x\tilde{y})+a_4\tilde{y}y+a_6(\tilde{x}+x)+a_7(\tilde{y}+y)+a_9$$

  and

  $$\tilde{y}-y=-a_1\tilde{x}x-a_2(\tilde{x}y+x\tilde{y})-a_3\tilde{y}y-a_5(\tilde{x}+x)-a_6(\tilde{y}+y)-a_8,$$

  with some a₁, …, a₉. This is a system of linear equations for the coefficients a₁, …, a₉, which turns out to have a unique solution. This solution can be found in appendix B. It leads to the following compact formula for the Hamilton function H(x, y) in terms of the data μ_i, b_i of the pencil of cubic curves (which stay invariant under the map Φ_f(x, y;ε) with f = J∇H and ε = 1):

  $$\begin{array}{rl}H(x,y)& =\frac{2{\mu }_{12}}{b_{14}{\mu }_{23}{\mu }_{13}}(\frac{1}{3}{({\mu }_{3}x-y)}^3+\frac{1}{2}(b_1+b_4){({\mu }_{3}x-y)}^2+b_1{b}_4({\mu }_{3}x-y))\\ & -\frac{2{\mu }_{23}}{b_{25}{\mu }_{12}{\mu }_{13}}(\frac{1}{3}{({\mu }_{1}x-y)}^3+\frac{1}{2}(b_2+b_5){({\mu }_{1}x-y)}^2+b_2{b}_5({\mu }_{1}x-y))\\ & +\frac{2{\mu }_{13}}{b_{36}{\mu }_{12}{\mu }_{23}}(\frac{1}{3}{({\mu }_{2}x-y)}^3+\frac{1}{2}(b_3+b_6){({\mu }_{2}x-y)}^2+b_3{b}_6({\mu }_{2}x-y)),\end{array}$$

  where b_ij = b_i − b_j, μ_ij = μ_i − μ_j.

 ▪

5. Conclusion

We proved an amazing characterization of integrable maps arising as Kahan discretizations of quadratic planar Hamiltonian vector fields, in terms of the geometry of their set of invariant curves. Such a neat characterization supports our belief expressed in [[6,7]] that Kahan–Hirota–Kimura discretizations will serve as a rich source of novel results concerning algebraic geometry of integrable birational maps. It will be desirable to find similar characterizations for further classes of integrable Kahan discretizations, in dimensions n = 2 (which belongs to our agenda in the near future) and n > 2 (even more important but also more difficult). For systems in dimension n = 2 of the type $\dot{x}=\ell (x)J\mathrm{\nabla }H(x)$ with $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, a linear form ℓ(x) and a quadratic Hamilton function H(x), this task was accomplished in [[10,13]].

Appendix A. Appendix A. Formulae for the planar Kahan map and its integral

The numerators of the components of map (3.7) are

$$R(x,y,\epsilon )=r_1{x}^2+r_{2}xy+r_3{y}^2+r_{4}x+r_{5}y+r_6$$ A.1

and

$$S(x,y,\epsilon )=s_1{x}^2+s_{2}xy+s_3{y}^2+s_{4}x+s_{5}y+s_6,$$ A.2

with

$$\begin{array}{rl}& r_1=\epsilon a_2+{\epsilon }^2(a_2{a}_6-a_3{a}_5),\\ & r_2=2\epsilon a_3+{\epsilon }^2(a_2{a}_7-a_4{a}_5),\\ & r_3=\epsilon a_4+{\epsilon }^2(a_3{a}_7-a_4{a}_6),\\ & r_4=1+2\epsilon a_6+{\epsilon }^2(a_6^2-a_5{a}_7-a_3{a}_8+a_2{a}_9),\\ & r_5=2\epsilon a_7+{\epsilon }^2(a_3{a}_9-a_4{a}_8)\\ \mathrm{and}& r_6=\epsilon a_9+{\epsilon }^2(a_6{a}_9-a_7{a}_8)\end{array}$$

and

$$\begin{array}{rl}& s_1=-\epsilon a_1+{\epsilon }^2(a_2{a}_5-a_1{a}_6),\\ & s_2=-2\epsilon a_2+{\epsilon }^2(a_3{a}_5-a_1{a}_7),\\ & s_3=-\epsilon a_3+{\epsilon }^2(a_3{a}_6-a_2{a}_7),\\ & s_4=-2\epsilon a_5+{\epsilon }^2(a_2{a}_8-a_1{a}_9),\\ & s_5=1-2\epsilon a_6+{\epsilon }^2(a_6^2-a_5{a}_7+a_3{a}_8-a_2{a}_9)\\ \mathrm{and}& s_6=-\epsilon a_8+{\epsilon }^2(a_6{a}_8-a_5{a}_9).\end{array}$$

The common denominator is

$$D(x,y,\epsilon )=d_1{x}^2+d_{2}xy+d_3{y}^2+d_{4}x+d_{5}y+d_6,$$ A.3

where

$$\begin{array}{rl}& d_1={\epsilon }^2(a_1{a}_3-a_2^2),\\ & d_2={\epsilon }^2(a_1{a}_4-a_2{a}_3),\\ & d_3={\epsilon }^2(a_2{a}_4-a_3^2),\\ & d_4={\epsilon }^2(a_1{a}_7-2a_2{a}_6+a_3{a}_5),\\ & d_5={\epsilon }^2(a_4{a}_5-2a_3{a}_6+a_2{a}_7)\\ \mathrm{and}& d_6=1+{\epsilon }^2(a_5{a}_7-a_6^2).\end{array}$$

Similarly,

$$C(x,y,\epsilon )=c_1{x}^3+c_2{x}^{2}y+c_{3}x{y}^2+c_4{y}^3+c_5{x}^2+c_{6}xy+c_7{y}^2+c_{8}x+c_{9}y+c_{10},$$ A.4

with the coefficients

$$\begin{array}{rl}{\tilde{c}}_1& =\frac{1}{3}{a}_1+\frac{1}{3}{\epsilon }^2(a_1{a}_3{a}_8-a_1{a}_6^2+2a_2{a}_5{a}_6-a_2^2{a}_8-a_3{a}_5^2),\\ {\tilde{c}}_2& =a_2+\frac{1}{3}{\epsilon }^2(a_1{a}_3{a}_9+a_1{a}_4{a}_8-2a_1{a}_6{a}_7+2a_2{a}_5{a}_7-a_4{a}_5^2-a_2^2{a}_9-a_2{a}_3{a}_8+a_2{a}_6^2),\\ {\tilde{c}}_3& =a_3+\frac{1}{3}{\epsilon }^2(a_2{a}_4{a}_8+a_1{a}_4{a}_9-2a_4{a}_5{a}_6+2a_3{a}_5{a}_7-a_1{a}_7^2-a_3^2{a}_8-a_2{a}_3{a}_9+a_3{a}_6^2),\\ {\tilde{c}}_4& =\frac{1}{3}{a}_4+\frac{1}{3}{\epsilon }^2(a_2{a}_4{a}_9-a_4{a}_6^2+2a_3{a}_6{a}_7-a_3^2{a}_9-a_2{a}_7^2),\\ {\tilde{c}}_5& =a_5+\frac{1}{3}{\epsilon }^2(a_1{a}_7{a}_8-2a_1{a}_6{a}_9+2a_2{a}_5{a}_9-a_3{a}_5{a}_8-a_5^2{a}_7+a_5{a}_6^2),\\ {\tilde{c}}_6& =2a_6+\frac{1}{3}{\epsilon }^2(-a_1{a}_7{a}_9-a_4{a}_5{a}_8-2a_2{a}_6{a}_9-2a_3{a}_6{a}_8+3a_2{a}_7{a}_8\\ & +3a_3{a}_5{a}_9-a_5{a}_6{a}_7+2a_6^3),\\ {\tilde{c}}_7& =a_7+\frac{1}{3}{\epsilon }^2(a_4{a}_5{a}_9-2a_4{a}_6{a}_8+2a_3{a}_7{a}_8-a_2{a}_7{a}_9-a_5{a}_7^2+a_6^2{a}_7),\\ {\tilde{c}}_8& =a_8+\frac{1}{3}{\epsilon }^2(2a_2{a}_8{a}_9-a_1{a}_9^2-a_3{a}_8^2-a_5{a}_7{a}_8+a_6^2{a}_8),\\ {\tilde{c}}_9& =a_9+\frac{1}{3}{\epsilon }^2(2a_3{a}_8{a}_9-a_4{a}_8^2-a_2{a}_9^2-a_5{a}_7{a}_9+a_6^2{a}_9)\\ \mathrm{and}{\tilde{c}}_{10}& =\frac{1}{3}{\epsilon }^2(2a_6{a}_8{a}_9-a_5{a}_9^2-a_7{a}_8^2).\end{array}$$

Appendix B. Coefficients of the Hamiltonian from the data of the pencil

Here are the formulae referred to at the end of the proof of theorem 4.3:

$$\begin{array}{rl}{a}_1& =\frac{b_{25}{b}_{36}{\mu }_{12}^2{\mu }_3^3-b_{14}{b}_{36}{\mu }_{23}^2{\mu }_1^3+b_{14}{b}_{25}{\mu }_{13}^2{\mu }_2^3}{D},\\ a_2& =\frac{-b_{25}{b}_{36}{\mu }_{12}^2{\mu }_3^2+b_{14}{b}_{36}{\mu }_{23}^2{\mu }_1^2-b_{14}{b}_{25}{\mu }_{13}^2{\mu }_2^2}{D},\\ a_3& =\frac{b_{25}{b}_{36}{\mu }_{12}^2{\mu }_3-b_{14}{b}_{36}{\mu }_{23}^2{\mu }_1+b_{14}{b}_{25}{\mu }_{13}^2{\mu }_2}{D},\\ a_4& =\frac{-b_{25}{b}_{36}{\mu }_{12}^2+b_{14}{b}_{36}{\mu }_{23}^2-b_{14}{b}_{25}{\mu }_{13}^2}{D},\\ a_5& =\frac{(b_1+b_4)b_{25}{b}_{36}{\mu }_3^2{\mu }_{12}^2-(b_2+b_5)b_{14}{b}_{36}{\mu }_1^2{\mu }_{23}^2+(b_3+b_6)b_{14}{b}_{25}{\mu }_2^2{\mu }_{13}^2}{2D},\\ a_6& =\frac{-(b_1+b_4)b_{25}{b}_{36}{\mu }_3{\mu }_{12}^2+(b_2+b_5)b_{14}{b}_{36}{\mu }_1{\mu }_{23}^2-(b_3+b_6)b_{14}{b}_{25}{\mu }_2{\mu }_{13}^2}{2D},\\ a_7& =\frac{(b_1+b_4)b_{25}{b}_{36}{\mu }_{12}^2-(b_2+b_5)b_{14}{b}_{36}{\mu }_{23}^2+(b_3+b_6)b_{14}{b}_{25}{\mu }_{13}^2}{2D},\\ a_8& =\frac{b_1{b}_4{b}_{25}{b}_{36}{\mu }_3{\mu }_{12}^2-b_2{b}_5{b}_{14}{b}_{36}{\mu }_1{\mu }_{23}^2+b_3{b}_6{b}_{14}{b}_{25}{\mu }_2{\mu }_{13}^2}{D},\\ \mathrm{and}{a}_9& =\frac{-b_1{b}_4{b}_{25}{b}_{36}{\mu }_{12}^2+b_2{b}_5{b}_{14}{b}_{36}{\mu }_{23}^2-b_3{b}_6{b}_{14}{b}_{25}{\mu }_{13}^2}{D},\end{array}$$

where

$$D=\frac{1}{2}{b}_{14}{b}_{25}{b}_{36}{\mu }_{12}{\mu }_{13}{\mu }_{23}$$

and b_ij = b_i − b_j, μ_ij = μ_i − μ_j.

Data accessibility

This paper does not have any experimental data.

Competing interests

We have no competing interests.

Author's contributions

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

Funding

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