The linear symmetries of Hill’s lunar problem

A symmetry of a Hamiltonian system is a symplectic or anti-symplectic involution which leaves the Hamiltonian invariant. For the planar and spatial Hill lunar problem, four resp. eight linear symmetries are well-known. Algebraically, the planar ones form a Klein four-group \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$\end{document}Z2×Z2 and the spatial ones form the group \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$\end{document}Z2×Z2×Z2. We prove that there are no other linear symmetries. Remarkably, in Hill’s system the spatial linear symmetries determine already the planar linear symmetries.


Introduction and result.
Hill's lunar problem is a limit case of the restricted three-body problem. In the restricted three-body problem, one considers two massive primaries and a massless object which does not influence the two masses and is attracted by them according to Newton's law of gravitation. In Hill's original set-up from 1877 [7], the primaries are the sun and the earth, and the massless body is the moon. The goal is to understand the dynamics of the moon. In Hill's system, the mass of the sun is infinitely much heavier than the mass of the earth, while the moon gets very close to the earth. Therefore, one shifts the earth to the origin and zooms in a region around the earth by pushing the huge sun off to infinity (see Figure 1).  There are two versions of Hill's lunar problem, the planar and the spatial one. The Hamiltonian describing the motion of the moon in the planar problem is and in the spatial problem, where each phase space is endowed with the canonical symplectic form ω = dq i ∧ dp i . For the derivation of the Hamiltonians, we refer to [4, pp. 77-78] for the planar problem, and to [2, pp. 3-4] for the spatial problem.
A symmetry ρ is, by definition, a symplectic or anti-symplectic involution of the phase space which leaves the Hamiltonian invariant, i.e., Symmetries of Hill's lunar problem play an important role since a natural class of periodic orbits are those that are invariant with respect to such involutions. In fact, linear symmetries are used traditionally for finding and studying invariant orbits; analytically by Birkhoff's "shooting method" [3] (1915) as well as numerically by Hill [7] in 1877, by Hénon [5,6], and by Michalodimitrakis [9].
Planar linear symmetries. The planar Hamiltonian (1) is invariant under the double-symmetry given by the two commuting linear anti-symplectic involutions Their product ρ 1 • ρ 2 = ρ 2 • ρ 1 = −id is symplectic. Geometrically, in the configuration space, the Hamiltonian H p is invariant under the reflections about the q 1 -and q 2 -axes, i.e., it is not possible to say whether we are going to the sun or away from it. The symplectic involutions ±id correspond to a rotation by 0 and π, respectively. Algebraically, ρ 1 , ρ 2 , and ±id form a Klein four-group, i.e., For the planar problem, these symmetries are already all linear symmetries.
Spatial linear symmetries. The spatial Hamiltonian (2) is invariant under the symplectic involution The linear symmetries of Hill's lunar problem 323 which arises from the reflection at the ecliptic {q 3 = 0}. Moreover, the planar problem can be viewed as the restriction of this system to the fixed point set Other linear symplectic symmetries are −σ and ±id, where −σ corresponds to a rotation around the q 3 -axis by π. Its fixed point set is Fix(−σ) = {(0, 0, q 3 , 0, 0, p 3 )}, thus the q 3 -axis is invariant under −σ. Four anti-symplectic symmetries are listed in Table 1.
Reflection at the q 2 q 3 -plane Rotation around the q 2 -axis by π Notice that all the maps σ, ρ 1 , ρ 2 , ρ 1 , and ρ 2 leave Fix(σ) invariant. Therefore, the two maps are exactly ρ 1 and ρ 2 from the planar problem. Furthermore, these eight linear symmetries form a group, which we denote by Σ 3 . The group structure is given by Table 2. Table 2. Group structure of Σ 3 This group is generated by {ρ 1 , ρ 2 , σ}, and By considering the four symplectic involutions ±id, ±σ, we see that, like in the planar case, a Klein four-group arises, namely as a sub-group of Σ 3 . It is generated by {±σ}, and we denote it by Σ ω In the spatial problem, Σ 3 provides already all linear symmetries.
The results of this paper are motivated by our work [1], in which we related Hill's variational orbit from 1878 [8] to the Babylonian lunar periods by using Floquet multipliers and Conley-Zehnder indices. Therein the linear symmetries of Hill's system and the invariance of Hill's orbit were significant.

Proof of Theorems A and B.
First we show that Theorem B implies Theorem A, meaning that the spatial linear symmetries determine already the planar linear symmetries. This phenomenon is a remarkable property of Hill's lunar system.
Proof of Theorem A. Consider the projection map given by the restriction to Fix(σ), If ρ ∈ Σ 3 is symplectic or anti-symplectic, then ρ| Fix(σ) is a linear symplectic or anti-symplectic involution on Fix(σ) as well, respectively. Moreover, ρ| Fix(σ) leaves the planar Hamiltonian H p (1) invariant. Therefore the map π is welldefined.
While the map π is not injective (since π(ρ 1 ) = π(ρ 1 )), it is surjective. To see this, let ρ ∈ Σ 2 . If ρ is symplectic, then a symplectic extension is given by q 3 → q 3 and p 3 → p 3 . If ρ is anti-symplectic, then an anti-symplectic extension is given by q 3 → −q 3 and p 3 → p 3 . Theorem A thus follows from Theorem B.
In order to prove Theorem B, we first recall some basic prerequisites from linear symplectic geometry. Consider the standard symplectic vector space (R 2n , ω 0 ) with with respect to the splitting R 2n = R n × R n . Note that J 2 = −I 2n and The set of symplectic matrices in R 2n is denoted by It is easy to show that if Ψ, Φ ∈ Sp(n), then ΨΦ, Ψ −1 , Ψ T ∈ Sp(n) and also J ∈ Sp(n). In particular, Sp(n) is a group under matrix multiplication. Moreover, a 2n × 2n matrix, which is written as Vol. 120 (2023) The linear symmetries of Hill's lunar problem 325 with respect to the splitting R 2n = R n × R n , is symplectic if and only if Its inverse is given by We denote the set of anti-symplectic matrices in R 2n by which is not a group since for Ψ, Φ ∈ Sp − (n), the multiplication ΨΦ is symplectic. Nevertheless Ψ −1 , Ψ T ∈ Sp − (n) and −J ∈ Sp − (n). A 2n × 2n matrix given in the block form (4) is anti-symplectic if and only if The inverse matrix is given by Proof of Theorem B. Let ρ be a linear symmetry. We prove the theorem in three steps where the first one is obvious.
Step 1. The Hamiltonian (2) is the sum of where H 2 is homogeneous of degree 2 and H −1 is homogeneous of degree −1. Hence Step 2. The matrix form of ρ with respect to the splitting R 6 = R 3 × R 3 and to the coordinates (q 1 , q 2 , q 3 , p 1 , To see that, we write ρ in matrix form For fixed p, we take q with |q| very small and find Bp = 0. Hence Next, the ρ-invariance of H 2 yields for q = 0, |Dp| = |p| ∀p, whence also D ∈ O(3). Since ρ is an involution, we obtain and with AA T = DD T = I 3 , we obtain If ρ is symplectic, then (7) and the linear symplectic relations (5) imply With A 2 = I 3 , we have and therefore Since det(A) = ±1, the matrix C is skew-symmetric and this proves the first assertion of the second step.
If ρ is anti-symplectic, then by (7) and the linear anti-symplectic conditions (6), we obtain Therefore the matrix C is symmetric and the second assertion follows.
Step 3. In both cases, the matrix C is the zero matrix and

In both cases, A is of the form
Since A 2 = I 3 and A ∈ O(3), we have If ρ is symplectic, then ρ(q, p) is of the form ⎛ and therefore In view of the ρ-invariance of H 2 , we compare with H 2 ρ(q, p) , which is By the coefficients of p i q i for i = 1, 2, 3, we immediately have that the diagonal entries of AC in (9) are all zero. To see that the other entries of AC are also all zero, we set equal the coefficients of p 1 q 2 with p 2 q 1 , of p 1 q 3 with p 3 q 1 , of p 2 q 3 with p 3 q 2 , and use (10), which imply Arch. Math. dc 2 + bc 3 = ec 1 − cc 3 = 0.
Hence AC = 0 and thus C = 0. By the coefficients of p 1 q 2 and p 2 q 1 , which means that a = 0 and b = 0. In view of A 2 = I 3 in (8), the two equations ad + bd + ef = 0 and ae + df + ce = 0 together with the coefficients of q 1 q 2 and q 1 q 3 imply ad = ae = 0. Since a = 0, we obtain Furthermore, by the coefficient of p 1 q 3 , we have af = de = 0, hence f = 0. Together with the coefficients from the second until the fourth lines, we obtain ab = 1, a 2 = b 2 = c 2 = 1, which correspond to ±σ and ± id.
In a similar way to the symplectic case, we find that C = 0. In view of A 2 = I 3 in (8), the three equations ad+bd+ef = 0, ae+df +ce = 0, and de+bf +cf = 0 together with the coefficients of q 1 q 2 , q 1 q 3 , q 2 q 3 imply ad = ae = de = 0. Since the coefficients of p 1 q 3 and p 3 q 1 yield de = af , we have ad = ae = af = 0.
Suppose that a = 0, then in view of the coefficients of q 2 1 , we see that d 2 + e 2 = −2 which is a contradiction. Hence a = 0, d = e = f = 0.