Predicting the transmission trends of COVID-19: an interpretable machine learning approach based on daily, death, and imported cases

Hyeonjeong Ahn; Hyojung Lee; Hyeonjeong Ahn; Hyojung Lee

doi:10.3934/mbe.2024270

Mathematical Biosciences and Engineering

2024, Volume 21, Issue 5: 6150-6166. doi: 10.3934/mbe.2024270

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Research article

Predicting the transmission trends of COVID-19: an interpretable machine learning approach based on daily, death, and imported cases

Hyeonjeong Ahn ,
Hyojung Lee ^,

Department of Statistics, Kyungpook National University, Daegu 41566, Republic of Korea

Academic Editor: Xi Huo

Received: 14 January 2024 Revised: 21 May 2024 Accepted: 29 May 2024 Published: 03 June 2024

COVID-19 is caused by the SARS-CoV-2 virus, which has produced variants and increasing concerns about a potential resurgence since the pandemic outbreak in 2019. Predicting infectious disease outbreaks is crucial for effective prevention and control. This study aims to predict the transmission patterns of COVID-19 using machine learning, such as support vector machine, random forest, and XGBoost, using confirmed cases, death cases, and imported cases, respectively. The study categorizes the transmission trends into the three groups: L0 (decrease), L1 (maintain), and L2 (increase). We develop the risk index function to quantify changes in the transmission trends, which is applied to the classification of machine learning. A high accuracy is achieved when estimating the transmission trends for the confirmed cases (91.5–95.5%), death cases (85.6–91.8%), and imported cases (77.7–89.4%). Notably, the confirmed cases exhibit a higher level of accuracy compared to the data on the deaths and imported cases. L2 predictions outperformed L0 and L1 in all cases. Predicting L2 is important because it can lead to new outbreaks. Thus, this robust L2 prediction is crucial for the timely implementation of control policies for the management of transmission dynamics.

Keywords:

Citation: Hyeonjeong Ahn, Hyojung Lee. Predicting the transmission trends of COVID-19: an interpretable machine learning approach based on daily, death, and imported cases[J]. Mathematical Biosciences and Engineering, 2024, 21(5): 6150-6166. doi: 10.3934/mbe.2024270

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Abstract

1. Introduction

In the study of perturbations of the three degree of freedom Kepler Hamiltonian pulling back, the regularized Hamiltonian by the Kustaanheimo-Stiefel (KS) map, gives a perturbation of the four degree of freedom harmonic oscillator Hamiltonian, when restricted to the zero level set of the KS symmetry. We use the formulation of the KS transformation in ^[6] which allows us to reduce the KS symmetry using invariant theory for the first time. As an illustration, we apply this procedure to the regularized Stark Hamiltonian, which is normalized after applying the KS transformation. We do not expect this Hamiltonian to be completely integrable (see Lagrange ^[4] and also ^[5]). Our treatment follows that of ^[3] and gives the full details of obtaining the second order normal form ^[1]. We use the notation of ^[2] and note that our procedure of regularization, pull back by the KS map, normalization and reduction may be used to study three degree of freedom perturbed Keplerian systems.

2. The basic set up

On $T_0{\mathbb{R }}^3 = ({\mathbb{R }}^3\setminus \{ 0 \}) \times {\mathbb{R }}^3$ with coordinates $(x, y)$ and standard symplectic form ${\omega }_3 = \sum^3_{i = 1}\mathrm{d} x_i \wedge \mathrm{d} y_i$ consider the Stark Hamiltonian

$\begin{equation} K(x,y) = \tfrac{{ 1}}{{ 2}} \langle y, y \rangle -\frac{1}{|x|} + fx_3. \end{equation}$

(2.1)

Here, $\langle \, \, , \, \, \rangle$ is the Euclidean inner product on ${\mathbb{R} }^3$ with associated norm $|\, \, |$ . On the negative energy level $-\tfrac{ 1}{ 2} k^2$ with $k > 0$ rescaling time $\mathrm{d} t \mapsto \frac{|x|}{k}\mathrm{d} s$ , we obtain

$\begin{equation} 0 = \tfrac{1}{2k}\big( |x| \langle y,y \rangle +k^2|x|\big) -\frac{1}{k} + fx_3 \frac{|x|}{k}. \end{equation}$

(2.2)

In other words, $(x, y)$ lies in the $\tfrac{1}{k}$ level set of

$\begin{equation} \widehat{K}(x,y) = \tfrac{1}{2k}|x|\big( \langle y, y \rangle +k^2 |x| \big) + f x_3\frac{|x|}{k}. \end{equation}$

(2.3)

We assume that $f$ is small, namely, $f = \varepsilon \beta$ . After the symplectic coordinate change $(x, y) \mapsto (\frac{1}{k}x, ky)$ the Hamiltonian $\widehat{K}$ becomes the preregularized Hamiltonian

$\begin{equation} \mathcal{K}(x,y) = \tfrac{ 1}{ 2} |x|(\langle y, y \rangle +|x|) +\varepsilon \beta x_3|x| \end{equation}$

(2.4)

on the level set ${\mathcal{K}}^{-1}(1)$ .

Let $T_0{\mathbb{R} }^4 = ({\mathbb{R} }^4\setminus \{ 0 \}) \times {\mathbb{R} }^4$ have coordinates $(q, p)$ and a symplectic form ${\omega }_4 = \sum^4_{i = 1}\mathrm{d} q_i \wedge \mathrm{d} p_i$ . Pull back $\mathcal{K}$ by the Kustaanheimo-Stiefel mapping

$\mathcal{KS}: T_0{\mathbb{R} }^4 \rightarrow T_0{\mathbb{R} }^3: (q,p) \mapsto (x,y),$

where

$\begin{align*} x_1 & = 2(q_1q_3+q_2q_4) = U_2-K_1 \\ x_2 & = 2(q_1q_4-q_2q_3) = U_3 -K_2 \\ x_3 & = q^2_1+q^2_2-q^2_3-q^2_4 = U_4 - K_3 \\ y_1 & = (\langle q,q\rangle )^{-1}(q_1p_3+q_2p_4 +q_3p_1 +q_4p_2) = (H_2 + V_1)^{-1}V_2 \\ y_2 & = (\langle q,q\rangle )^{-1}(q_1p_4-q_2p_3 -q_3p_2 +q_4p_1) = (H_2 + V_1)^{-1}V_3 \\ y_3 & = (\langle q,q\rangle )^{-1}(q_1p_1+q_2p_2 -q_3p_3 -q_4p_4) = (H_2 + V_1)^{-1}V_4 \end{align*}$

and

$\begin{align*} H_2 & = \tfrac{ 1}{ 2} (p^2_1+p^2_2+p^2_3+p^2_4 +q^2_1+q^2_2 +q^2_3+q^2_4) \\ \Xi & = q_1p_2-q_2p_1 +q_3p_4 -q_4p_3, \end{align*}$

to get the regularized Stark Hamiltonian

$\begin{equation} \mathcal{H} = H_2 +\varepsilon \beta \big( U_4V_1+H_2U_4 -K_3V_1 -H_2K_3 \big) \end{equation}$

(2.5)

on ${\Xi}^{-1}(0)$ , since $|x| = \langle q, q \rangle = H_2+V_1$ . Here

$\begin{align} K_1 & = -(q_1q_3+q_2q_4+p_1p_3 +p_2p_4) \\ K_2 & = -(q_1q_4-q_2q_3+p_1p_4-p_2p_3) \\ K_3 & = \tfrac{ 1}{ 2} (q^2_3+q^2_4 +p^2_3+p^2_4 -q^2_1 -q^2_2 -p^2_1-p^2_2) \\ L_1 & = q_4p_1-q_3p_2+q_2p_3-q_1p_4 \\ L_2 & = q_1p_3+q_2p_4-q_3p_1-q_4p_2 \\ L_3 & = q_3p_4-q_4p_3+q_2p_1-q_1p_2 \\ U_1 & = -(q_1p_1+q_2p_2+ q_3p_3+ q_4p_4) \\ U_2 & = q_1q_3+q_2q_4-p_1p_3-p_2p_4 \\ U_3 & = q_1q_4 - q_2q_3 +p_2p_3 - p_1p_4 \\ U_4 & = \tfrac{ 1}{ 2} (q^2_1+q^2_2 -q^2_3 -q^2_4 +p^2_3 +p^2_4-p^2_1-p^2_2) \\ V_1 & = \tfrac{ 1}{ 2} (q^2_1+q^2_2 +q^2_3+q^2_4-p^2_1 -p^2_2-p^2_3-p^2_4) \\ V_2 & = q_1p_3 +q_2p_4 +q_3p_1 + q_4p_2 \\ V_3 & = q_1p_4 - q_2p_3 -q_3p_2 +q_4p_1 \\ V_4 & = q_1p_1 + q_2p_2 -q_3p_3 -q_4p_4. \end{align}$

generate the algebra of polynomials invariant under the $S^1$ action ${\varphi }^{\Xi}_s$ given by the flow of $X_{\Xi}$ on $(T{\mathbb{R} }^4 = {\mathbb{R} }^8, {\omega }_4)$ . The Hamiltonian $\mathcal{H}$ (2.5) is invariant under this $S^1$ action and thus is a smooth function on the orbit space ${\Xi}^{-1}(0)/S^1 \subseteq {\mathbb{R} }^{16}$ with coordinates $(K, L, H, {\Xi}; U, V)$ .

3. The first order normal form on ${\Xi}^{-1}(0)/S^1$

The harmonic oscillator vector field $X_{H_2}$ on $(T{\mathbb{R} }^4, {\omega }_4)$ induces the vector field $Y_{H_2} = \sum^4_{i = 1}\big(2V_i \frac{\partial }{\partial U_i} - 2U_i\frac{\partial }{\partial V_i }\big)$ on the orbit space ${\mathbb{R} }^8/S^1 \subseteq {\mathbb{R} }^{16}$ , which leaves ${\Xi}^{-1}(0)/S^1$ invariant.

We now compute the first order normal form of the Hamiltonian $\mathcal{H}$ (2.5) on the reduced space ${\Xi }^{-1}(0)/S^1 \subseteq {\mathbb{R} }^8/S^1$ .

The average of $H_2U_4-K_3V_1$ over the flow

${\varphi }^{Y_{H_2}}_t(K,L,H_2, \Xi ; U,V) = \big( K,L, H_2, \Xi ; U \cos 2t + V\sin 2t, -U\sin 2t + V \cos 2t \big)$

of $Y_{H_2}$ is

$\begin{align*} \overline{H_2U_4-K_3V_1} & = \tfrac{ 1}{ \pi} \int^{\pi }_0 ({\varphi }^{Y_{H_2}}_t)^{\ast }(H_2U_4 - K_3V_1 )\, \mathrm{d} t \\ & = \tfrac{ 1}{ \pi} \int^{\pi }_0 (U_4\cos 2t +V_4 \sin 2t) \mathrm{d} t \, H_2 - \tfrac{ 1}{ \pi} \int^{\pi }_0 (-U_1\sin 2t +V_1 \cos 2t) \mathrm{d} t \, K_3 = 0. \end{align*}$

The second equality above follows because $L_{X_{H_2}}K_3 = 0$ and the third because $\overline{\cos 2t} = \overline{\sin 2t } = 0$ . The average of $U_4V_1$ over the flow of $Y_{H_2}$ on ${\Xi }^{-1}(0)/S^1$ is

$\begin{align*} \overline{U_4V_1} & = \tfrac{ 1}{ \pi} \int^{\pi }_0 ({\varphi }^{Y_{H_2}}_t)^{\ast }(U_4V_1 )\, \mathrm{d} t \\ & = -\tfrac{ 1}{ 2} U_1U_4 \, \overline{\sin 4t} + U_4V_1\, \overline{{\cos }^2 \, 2t} - U_1V_4\, \overline{{\sin }^2 \, 2t} +\tfrac{ 1}{ 2} V_1V_4 \, \overline{\sin 4t} \\ & = \tfrac{ 1}{ 2} (U_4V_1 - U_1V_4) = -\tfrac{ 1}{ 2}H_2K_3, \end{align*}$

since $\overline{{\cos }^2 \, 2t} = \overline{{\sin }^2 \, 2t} = \tfrac{ 1}{ 2}$ and $\overline{\sin 4t} = 0$ . The last equality above follows from the explicit description of the orbit space ${\mathbb{R} }^8/S^1$ as the semialgebraic variety in ${\mathbb{R} }^{16}$ with coodinates $(K, L, U, V;H_2, \Xi)$ given by

$\begin{align} \langle U,U \rangle = U^2_1+U^2_2+U^2_3 +U^2_4 & = H^2_2 - {\Xi }^2 \ge 0 \, \, \, H_2 \ge 0 \\ \langle V,V \rangle = V^2_1+V^2_2+V^2_3 +V^2_4 & = H^2_2 - {\Xi }^2 \ge 0 \\ \langle U,V \rangle = U_1V_1 +U_2V_2+U_3V_3 + U_4V_4 & = 0 \\ U_2V_1-U_1V_2 & = L_1\Xi - K_1H_2 \\ U_3V_1-U_1V_3 & = L_2 \Xi - K_2H_2 \\ U_4V_1 - U_1V_4 & = L_3 \Xi -K_3H_2 \\ U_4V_3 -U_3V_4 & = K_1 \Xi -L_1H_2 \\ U_2V_4 -U_4V_2 & = K_2 \Xi -L_2H_2 \\ U_3V_2 -U_2V_3 & = K_3 \Xi -L_3H_2 . \end{align}$

(3.1)

So the average of $U_4V_1+ H_2U_4 - K_3V_1 - H_2K_3$ over the flow of $Y_{H_2}$ is $-\tfrac{ 3}{ 2}H_2K_3$ on ${\Xi }^{-1}(0)/S^1$ . Thus the first order normal form of the regularized Stark Hamiltonian $\mathcal{H}$ (2.5) on ${\Xi }^{-1}(0)/S^1$ is

$\begin{equation} {\mathcal{H}}^{(1)}_{\mathrm{nf}} = H_2 - \tfrac{ 3}{ 2} \beta \varepsilon H_2K_3. \end{equation}$

(3.2)

4. The second order normal form on ${\Xi }^{-1}(0)/S^1$

In order to compute the second order normal form of the Hamiltonian $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$ , we need to find a function $F$ on ${\mathbb{R} }^{16}$ such that changing coordinates by the time $\varepsilon$ value of the flow of the Hamiltonian vector field $Y_F$ brings the regularized Hamiltonian $\mathcal{H}$ (2.5) into first order normal form. Choose $F$ so that

$\begin{equation} L_{Y_F}H_2 = \beta \big( -U_4V_1 - \tfrac{ 1}{ 2}H_2K_3 -H_2U_4 +K_3V_1 \big) . \end{equation}$

(4.1)

The following calculation shows that this choice does the job.

$\begin{align} ({\varphi }^{Y_F}_{\varepsilon })^{\ast }\mathcal{H} & = \mathcal{H} + \varepsilon L_{Y_F}\mathcal{H} + \tfrac{ 1}{ 2} {\varepsilon}^2 L^2_{Y_F}\mathcal{H} + \mathrm{O}({\varepsilon}^3) \\ & = H_2 +\varepsilon \beta \big( U_4V_1+H_2U_4-K_3V_1 -H_2K_3 \big) + \varepsilon L_{Y_F}H_2 \\ & +{\varepsilon }^2\beta L_{Y_F}\big( U_4V_1+ H_2U_4 -K_3V_1 -H_2K_3 \big) + \tfrac{ 1}{ 2} {\varepsilon }^2 L^2_{Y_F}H_2 + \mathrm{O}({\varepsilon }^3) \\ & = H_2 +\varepsilon \beta \big( U_4V_1+H_2U_4-K_3V_1 -H_2K_3 \big) \\ & + \varepsilon \beta \big( -U_4V_1 - \tfrac{ 1}{ 2}H_2K_3 -H_2U_4 + K_3V_1 \big) \\ & + {\varepsilon }^2 \big[ L_{Y_F}\big( -L_{Y_F}H_2 -\tfrac{ 3}{ 2}\beta H_2K_3 \big) +\tfrac{ 1}{ 2} L^2_{Y_F}H_2 \big] + \mathrm{O}({\varepsilon }^3) \\ & = H_2 -\tfrac{ 3}{ 2}\varepsilon \beta H_2K_3 -\tfrac{ 1}{ 2} {\varepsilon }^2\big( L^2_{Y_F}H_2 +3\beta L_{Y_F}(H_2K_3) \big) + \mathrm{O}({\varepsilon}^3). \end{align}$

(4.2)

To determine the function $F$ , we solve equation (4.1). Write $F = F_1+F_2$ , where $L_{Y_{H_2}}F_1 = \beta (U_4V_1+ \tfrac{ 1}{ 2} H_2K_3)$ and $L_{Y_{H_2}}F_2 = \beta (H_2U_4-K_3V_1)$ . Then

$\begin{align} L_{Y_F}H_2 & = -L_{Y_{H_2}}F = -L_{Y_{H_2}}F_1 - L_{Y_{H_2}}F_2 \\ & = -\beta (U_4V_1+ \tfrac{ 1}{ 2} H_2K_3) - \beta (H_2U_4-K_3V_1). \end{align}$

(4.3)

Since $L_{Y_{H_2}}V_4 = -2 U_4$ and $L_{Y_{H_2}}U_1 = 2V_1$ , it follows that

$\begin{equation} F_2 = - \tfrac{ \beta}{ 2}(H_2V_4 +K_3U_1). \end{equation}$

(4.4a)

Now

$\begin{align*} F_1 & = \tfrac{ \beta }{ \pi} \int^{\pi}_0 t({\varphi }^{Y_{H_2}}_t)^{\ast }\big( U_4V_1 + \tfrac{ 1}{ 2} H_2K_3 \big) \, \mathrm{d} t = \tfrac{ \beta}{ \pi} \int^{\pi}_0 t({\varphi }^{Y_{H_2}}_t)^{\ast }(U_4V_1) \mathrm{d} t + \tfrac{ \pi \beta }{ 4} H_2K_3, \end{align*}$

see ^[1], and

$\begin{align*} \tfrac{ \beta }{ \pi} \int^{\pi}_0 t({\varphi }^{Y_{H_2}}_t)^{\ast }(U_4V_1) \mathrm{d} t & = \\ & = -\tfrac{ \beta}{ 2}\ (U_1U_4) \, \tfrac{ 1}{ \pi} \int^{\pi }_0 t \, \sin 4t \mathrm{d} t + \beta (U_4V_1)\, \tfrac{ 1}{ \pi} \int^{\pi }_0 t \, {\cos }^2\, 2t \mathrm{d} t \\ &-\beta (U_1V_4)\, \tfrac{ 1}{ \pi} \int^{\pi }_0 t \, {\sin }^2\, 2t \mathrm{d} t + \tfrac{ \beta}{ 2} (V_1V_4)\, \tfrac{ 1}{ \pi} \int^{\pi }_0 t \, \sin 4t \mathrm{d} t \\ & = \tfrac{ \beta }{ 8} (U_1U_4 - V_1V_4) + \tfrac{ \pi \beta}{ 4}(U_4V_1 -U_1V_4) , \end{align*}$

since $\tfrac{1}{\pi } \int^{\pi}_0 t \, \sin 4t \mathrm{d} t = -\tfrac{1}{4}$ and $\tfrac{1}{\pi } \int^{\pi}_0 t \, {\sin }^2\, 2 t \mathrm{d} t = \tfrac{1}{\pi } \int^{\pi}_0 t \, {\cos }^2\, 2 t \mathrm{d} t = \tfrac{\pi }{4}$ . Thus

$\begin{equation} F_1 = \tfrac{ \beta}{ 8}(U_1U_4-V_1V_4) +\tfrac{ \pi \beta }{ 4} (U_4V_1 -U_1V_4 + H_2K_3) \notag \\ = \tfrac{ \beta }{ 8}(U_1U_4-V_1V_4) \label{eq-eightb} \end{equation}$

on ${\Xi}^{-1}(0)/S^1$ , see (3.1). Hence on ${\Xi}^{-1}(0)/S^1$

$\begin{equation} F = F_1+F_2 = \tfrac{ \beta }{ 8}(U_1U_4-V_1V_4) -\tfrac{ \beta}{ 2}(H_2V_4 + K_3U_1). \end{equation}$

(4.5)

We now calculate the average over the flow of $Y_{H_2}$ of

$\begin{equation} -\tfrac{ 3}{ 2}\beta L_{Y_F}(H_2K_3) -\tfrac{ 1}{ 2} L^2_{Y_F}H_2, \end{equation}$

(4.6)

which is the ${\varepsilon }^2$ term in the transformed Hamiltonian $({\varphi }^{Y_F}_{\varepsilon})^{\ast }\mathcal{H}$ , see (4.2). This determines the second order normal form of $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$ . We begin with the term

$-\tfrac{ 3}{ 2} \beta L_{Y_F}(H_2K_3) = -\tfrac{ 3}{ 2}\beta [K_3(L_{Y_F}H_2) -H_2(L_{Y_{K_3}}F)] .$

The average of

$-\tfrac{ 3}{ 2} beta \, K_3(L_{Y_F}H_2) = \tfrac{ 3}{ 2}{\beta }^2\, K_3(U_4V_1 +\tfrac{ 1}{ 2} H_2K_3 + H_2U_4 -K_3V_1)$

vanishes on ${\Xi}^{-1}(0)/S^1$ . The term

$\begin{align*} \tfrac{ 3}{ 2}\beta H_2(L_{Y_{K_3}}F) & = \tfrac{ 3}{ 2}{\beta }^2H_2\, L_{Y_{K_3}}\big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) -\tfrac{ 1}{ 2} (H_2V_4 +K_3U_1) \big) \\ & = \tfrac{ 3}{ 2}{\beta }^2 H_2 \big[ \big( -2L_2\frac{\partial }{\partial K_1} + 2L_1\frac{\partial }{\partial K_2} - 2K_2\frac{\partial }{\partial L_1} +2K_1\frac{\partial }{\partial L_2} \\ &-2U_4\frac{\partial }{\partial U_1} + 2U_1 \frac{\partial }{\partial U_4} -2V_4\frac{\partial }{\partial V_1} + 2V_1\frac{\partial }{\partial V_4 } \big) \big] \big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) \\ &-\tfrac{ 1}{ 2} (H_2V_4 +K_3U_1) \big) , \, \, \mbox{see [2, table 1] } \\ & = \tfrac{ 3}{ 2} {\beta }^2H_2 \big[ \tfrac{ 1}{ 4} (-U^2_4+U^2_1+V^2_4-V^2_1) -H_2V_1 +K_3U_4 \big] . \end{align*}$

Next we calculate $\frac{ 3}{ 2}\beta \overline{H_2(L_{Y_{K_3}}F)}$ . Since $\overline{H_2V_1} = 0 = \overline{K_3U_4}$ we need only calculate the average of $U^2_1$ , $U^2_4$ , $V^2_1$ and $V^2_4$ . We get $\overline{U^2_1} = \tfrac{ 1}{ 2} (U^2_1+V^2_1) = \overline{V^2_1}$ and $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_4+V^2_4) = \overline{V^2_4}$ . Thus $\tfrac{ 3}{ 2}\beta \overline{H_2(L_{Y_{K_3}}F)} = 0$ . So the average $-\tfrac{ 3}{ 2}\beta \overline{L_{Y_F}(H_2K_3)}$ of the first term in expression (4.6) vanishes on ${\Xi}^{-1}(0)/S^1$ .

Next we calculate the average of the term $L^2_{Y_F}H_2$ in expression (4.6) on ${\Xi }^{-1}(0)/S^1$ . We have

$\begin{align*} L^2_{Y_F}H_2 & = -L_{Y_F}(L_{Y_{H_2}}F) \\ & = -\beta L_{Y_F}\big( U_4V_1 + \tfrac{ 1}{ 2} H_2K_3 +H_2U_4 -K_3V_1) , \, \, \mbox{using (4.3)} \\ & = \beta \big[ \overbrace{(L_{Y_{U_4}}F)V_1}^{I} + \overbrace{U_4(L_{Y_{V_1}}F)}^{II} -\tfrac{ 1}{ 2} \overbrace{(L_{Y_F}H_2)K_3}^{III} + \tfrac{ 1}{ 2} \overbrace{H_2(L_{Y_{K_3}}F)}^{IV} \\ & -\overbrace{(L_{Y_F}H_2)U_4}^{V} + \overbrace{H_2(L_{Y_{U_4}}F)}^{VI} - \overbrace{(L_{Y_{K_3}}F)V_1}^{VII} -\overbrace{K_3(L_{Y_{V_1}}F)}^{VIII} \big] . \end{align*}$

We begin by finding

$\begin{align*} L_{Y_{H_2}}F & = \beta \big[ \big( 2V_1 \frac{\partial }{\partial U_1} + 2V_2 \frac{\partial }{\partial U_2} +2V_3 \frac{\partial }{\partial U_3} + 2V_4 \frac{\partial }{\partial U_4} -2U_1 \frac{\partial }{\partial V_1} - 2U_2 \frac{\partial }{\partial V_2} \\ & - 2U_3 \frac{\partial }{\partial V_3} -2U_4 \frac{\partial }{\partial V_4} \big) \big] \big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) -\tfrac{ 1}{ 2} (H_2V_4 + K_3U_1) \big) \\ & = \beta \big[ \tfrac{ 1}{ 2}(V_1U_4+U_1V_4) +H_2U_4 - K_3V_1 \big] ; \\ L_{Y_{K_3}}F & = \beta \big( \big( -2L_2 \frac{\partial }{\partial K_1} + 2L_1 \frac{\partial }{\partial K_2} -2K_2 \frac{\partial }{\partial L_1} + 2K_1 \frac{\partial }{\partial L_2} -2U_4 \frac{\partial }{\partial U_1} + 2U_1 \frac{\partial }{\partial U_4} \\ & - 2V_4 \frac{\partial }{\partial V_1} +2V_1 \frac{\partial }{\partial V_4} \big) \big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) -\tfrac{ 1}{ 2} (H_2V_4+K_3U_1) \big) \big) \\ & = \beta \big[ \tfrac{ 1}{ 4}(-U^2_4 +U^2_1 +V^2_4 - V^2_1) -H_2V_1 +K_3U_4 \big] ; \\ L_{Y_{U_4}}F & = \beta \big[ \big( -2U_1 \frac{\partial }{\partial K_3} - 2U_3 \frac{\partial }{\partial L_1} +2U_2 \frac{\partial }{\partial L_2} - 2V_4 \frac{\partial }{\partial H_2} -2K_3 \frac{\partial }{\partial U_1} + 2L_2 \frac{\partial }{\partial U_2} \\ & + 2V_3 \frac{\partial }{\partial U_3} - 2H_2 \frac{\partial }{\partial V_4} \big) \big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) -\tfrac{ 1}{ 2} (H_2V_4+K_3U_1) \big) \big] \\ & = \beta \big[ V^2_4 +K^2_3 -\tfrac{ 1}{ 4}(K_3U_4 - H_2V_1) +H^2_2 + U^2_1\big] ; \\ L_{Y_{V_1}}F & = \beta \big[ \big( 2V_2 \frac{\partial }{\partial K_1} + 2V_3 \frac{\partial }{\partial K_2} +2V_4 \frac{\partial }{\partial K_3} + 2U_1 \frac{\partial }{\partial H_2} +2H_2 \frac{\partial }{\partial U_1} + 2K_1 \frac{\partial }{\partial V_2} \\ & + 2K_2 \frac{\partial }{\partial V_3} + 2K_3 \frac{\partial }{\partial V_4} \big) \big( \tfrac{ 1}{ 8}(U_1U_4-V_1V_4) -\tfrac{ 1}{ 2} (H_2V_4+K_3U_1) \big) \big] \\ & = \beta [-2(U_1V_4+H_2K_3) + \tfrac{ 1}{ 4}(H_2U_4-K_3V_1)] . \end{align*}$

So the average of term I on ${\Xi}^{-1}(0)/S^1$ is

$\begin{align} \beta \overline{(L_{Y_{U_4}}F)V_1} & = {\beta }^2(\overline{V_1V^2_4} + \overline{K^2_3V_1} -\tfrac{ 1}{ 4}\overline{K_3U_4V_1} + \tfrac{ 1}{ 4}\overline{H_2V^2_1} +\overline{H^2_2V_1} + \overline{U^2_1V_1}) \\ & = {\beta }^2 \big( \tfrac{ 1}{ 8}H_2K^2_3 + \tfrac{ 1}{ 8}H_2(U^2_1+V^2_1) \big) , \end{align}$

(4.7a)

since the average of $V_1V^2_4$ , $K^2_3V_1$ , $H^2_2V_1$ , and $U^2_1V_1$ are each $0$ , $\overline{U_4V_1} = -\tfrac{ 1}{ 2} H_2K_3$ and $\overline{V^2_1} = \tfrac{ 1}{ 2} (U^2_1+V^2_1)$ .

Term II is

$\beta U_4(L_{Y_{V_1}}F) = {\beta }^2 \big( - 2U_1U_4V_4 - 2H_2K_3U_4 + \tfrac{ 1}{ 4} H_2U^2_4 - \tfrac{ 1}{ 4}K_3U_4V_1 \big) .$

$\begin{align} \beta \overline{U_4(L_{Y_{V_1}}F)} & = \tfrac{ 1}{ 4} {\beta }^2 H_2 \overline{U^2_4} - \tfrac{ 1}{ 4} {\beta }^2 \, K_3 \overline{U_4V_1} \\ & = \tfrac{ 1}{ 8} {\beta }^2 H_2(U^2_4+V^2_4) + \tfrac{ 1}{ 8} {\beta }^2 H_2K^2_3. \end{align}$

(4.7b)

For term III, we have already shown that

$\begin{equation} -\tfrac{ \beta}{ 2}\overline{(L_{Y_F}H_2)K_3} = 0. \end{equation}$

(4.7c)

and for term IV we have already shown that

$\begin{equation} \tfrac{ \beta}{ 2}\overline{H_2(L_{Y_{K_3}}F)} = 0. \end{equation}$

(4.7d)

Term V is

$-{\beta }(L_{Y_F}H_2)U_4 = {\beta }^2 \big( \tfrac{ 1}{ 2} U^2_4V_1 + \tfrac{ 1}{ 2} U_1U_4V_1 + H_2U^2_4 - K_3U_4V_1 \big) .$

$\begin{align} -{\beta }\overline{(L_{Y_F}H_2)U_4} & = {\beta }^2 \big( \tfrac{ 1}{ 2} \overline{U^2_4V_1} + \tfrac{ 1}{ 2} \overline{U_1U_4V_1} + \overline{H_2U^2_4} - \overline{K_3U_4V_1} \big) \\ & = \tfrac{ {\beta }^2}{ 2} H_2 (U^2_4 + V^2_4) - \tfrac{ {\beta }^2}{ 2}K_3(U_4V_1 - U_1V_4), \end{align}$

(4.7e)

since the average of $U^2_4V_1$ and $U_1U_4V_1$ vanish; while $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_4+V^2_4)$ and $\overline{U_4V_1} = \tfrac{ 1}{ 2} (U_4V_1 - U_1V_4)$ .

Term VI is

$\beta H_2(L_{Y_{U_4}}F) = {\beta }^2H_2\big( V^2_4 +K^2_3 - \tfrac{ 1}{ 4}K_3U_4 + \tfrac{ 1}{ 4} H_2V_1 +H^2_2 + U^2_1 \big) .$

$\begin{align} \beta \overline{H_2(L_{Y_{U_4}}F)} & = {\beta }^2 \, H_2\overline{V^2_4} +{\beta }^2\, H_2K^2_3 + {\beta }^2\, H^3_2 + {\beta }^2\, H_2 \overline{U^2_1} \\ & = \tfrac{ 1}{ 2} {\beta }^2\, H_2(U^2_4+V^2_4) + {\beta }^2H_2K^2_3 +{\beta }^2H^3_2 + \tfrac{ 1}{ 2} {\beta }^2\, H_2(U^2_1+V^2_1), \end{align}$

(4.7f)

since $\overline{K_2U_4} = 0 = \overline{H_2V_1}$ .

Term VII is

$-\beta (L_{Y_{K_3}}F)V_1 = {\beta }^2\big( \tfrac{ 1}{ 4}[U^2_4V_1-U^2_1V_1-V_1V^2_4 +V^3_1] +H_2V^2_1 - K_3U_4V_1 \big) .$

$\begin{align} -\beta \overline{(L_{Y_{K_3}}F)V_1} & = {\beta }^2 H_2\overline{V^2_1} - {\beta }^2K_3 \overline{U_4V_1} \\ & = \tfrac{ {\beta }^2}{ 2} H_2(U^2_1+V^2_1) - \tfrac{ {\beta }^2}{ 2} K_3(U_4V_1 - U_1V_4). \end{align}$

(4.7g)

Term VIII is

$-\beta K_3(L_{Y_{V_1}}F) = {\beta }^2\big( 2K_3 U_1V_4 + 2H_2K^2_3 - \tfrac{ 1}{ 4} H_2K_3 U_4 + \tfrac{ 1}{ 4} K^2_3V_1 \big) .$

$\begin{equation} -\beta \overline{K_3(L_{Y_{V_1}}F)} = 2{\beta }^2K_3\overline{U_1V_4} + 2{\beta }^2\, H_2K^2_3 = 3{\beta }^2\, H_2K^2_3, \end{equation}$

(4.7h)

since $\overline{U_1V_4} = \tfrac{ 1}{ 2} H_2K_3$ . Collecting together the results of all the above term calculations gives

$\begin{align*} \overline{L^2_{Y_F}H_2} & = \beta \overline{(L_{Y_{U_4}}F)V_1} + \beta \overline{U_4(L_{Y_{V_1}}F)} - \beta \overline{(L_{Y_F}H_2)U_4} +{\beta } \overline{H_2(L_{Y_{U_4}}F)} \\ & -{\beta }\overline{(L_{Y_{K_3}}F)V_1} - \beta \overline{K_3(L_{Y_{V_1}}F)} \\ & = {\beta }^2\big( [ \tfrac{ 1}{ 8}H_2K^2_3 + \tfrac{ 1}{ 8}H_2(U^2_1+V^2_1) ] + [\tfrac{ 1}{ 8} H_2(U^2_4+V^2_4) + \tfrac{ 1}{ 8} H_2K^2_3] \\ &+ [ \tfrac{ 1}{ 2} H_2(U^2_4+V^2_4) - \tfrac{ 1}{ 2} K_3(U_4V_1 -U_1V_4) ] + [ \tfrac{ 1}{ 2} H_2(U^2_4+V^2_4) + H_2K^2_3 + H^3_2 \\ & + \tfrac{ 1}{ 2} H_2(U^2_1+V^2_1)] + [ \tfrac{ 1}{ 2} H_2(U^2_1+V^2_1) - \tfrac{ 1}{ 2} K_3(U_4V_1-U_1V_4)] + 3 H_2 K^2_3 \big) \\ & = {\beta }^2 \big[ \tfrac{ 21}{ 4} H_2K^2_3 +H^3_2 + \tfrac{ 9}{ 8} H_2(U^2_1+V^2_1) + \tfrac{ 9}{ 8} H_2(U^2_4+V^2_4) \big] , \end{align*}$

using $U_4V_1-U_1V_4 = -K_3H_2$ . Thus the second order normal form of the regularized Stark Hamiltonian $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$ is

$\begin{align} {\mathcal{H}}^{(2)}_{\mathrm{nf}} & = H_2 -\tfrac{ 3}{ 2} \varepsilon \beta H_2K_3 - \tfrac{ 1}{ 2} {\varepsilon }^2 \overline{L^2_{Y_F}H_2} = H_2 -\tfrac{ 3}{ 2}\varepsilon \beta H_2K_3 \\ & -\tfrac{ 1}{ 2} {\varepsilon }^2 {\beta }^2 H_2 \big[ \tfrac{ 21}{ 4} K^2_3 +H^2_2 + \tfrac{ 9}{ 8} (U^2_1+V^2_1) +\tfrac{ 9}{ 8} (U^2_4+V^2_4) \big] . \end{align}$

(4.8)

5. The first order normal form of ${\mathcal{H}}^{(2)}_{\mathrm{nf}}$ on $T_hS^3_1$

Since $L_{X_{H_2}}{\mathcal{H}}^{(2)}_{\mathrm{nf}} = 0$ by construction, the second order normal form ${\mathcal{H}}^{(2)}_{\mathrm{nf}}$ (4.8) is a smooth function on $(H^{-1}_2(h) \cap {\Xi}^{-1}(0))/S^1 = T_hS^3_1$ , the tangent $h$ -sphere bundle of the unit $3$ -sphere $S^3_1$ , given by

$\begin{align} \widetilde{\mathcal{H}} & = h - \tfrac{ 1}{ 2} {\varepsilon }^2{\beta }^2 h^3 - \tfrac{ 3}{ 2} \varepsilon \beta h K_3 \\ & - \tfrac{ 1}{ 2} {\varepsilon }^2{\beta }^2 h \big[ \tfrac{ 21}{ 4} K^2_3 + \tfrac{ 9}{ 8} (U^2_1+V^2_1) + \tfrac{ 9}{ 8} (U^2_4+V^2_4) \big] . \end{align}$

(5.1)

We now show that the Hamiltonian $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ can be normalized again. On $(T{\mathbb{R} }^4, {\omega }_4)$ the Hamiltonian

$K_3(q,p) = \tfrac{ 1}{ 2} (q^2_3+q^2_4+p^2_3+p^2_4-q^2_1-q^2_2-p^2_1-p^2_2)$

gives rise to the Hamiltonian vector field $X_{K_3}$ , whose flow ${\varphi }^{X_{K_3}}_t(q, p)$ is

$\begin{align*} &\big( q_1 \cos t -p_1\sin t, q_2\cos t -p_2 \sin t, q_3 \cos t + p_3 \sin t , q_4 \cos t +p_4 \sin t ,\\ &q_1 \sin t +p_1 \cos t, q_2 \sin t +p_2 \cos t, -q_3\sin t +p_3 \cos t, -q_4 \sin t + p_4\cos t \big) , \end{align*}$

which is periodic of period $2\pi$ .

The vector field $X_{K_3}$ on $T{\mathbb{R} }^4$ induces the vector field

$\begin{align*} Y_{K_3} & = -2L_2 \frac{\partial }{\partial K_1} +2L_1 \frac{\partial }{\partial K_2} -2K_2 \frac{\partial }{\partial L_1} +2K_1\frac{\partial }{\partial L_2 } \\ & - 2U_4\frac{\partial }{\partial U_1} + 2U_1 \frac{\partial }{\partial U_4} -2V_4\frac{\partial }{\partial V_1} + 2V_1 \frac{\partial }{\partial V_4} , \end{align*}$

on ${\Xi}^{-1}(0)/S^1 \subseteq {\mathbb{R} }^{16}$ with coordinates $(K, L, H_2, \Xi; U, V)$ , whose flow

$\begin{align*} {\varphi }^{Y_{K_3}}_s(K,L,H_2, \Xi ; U, V) & = \big( -L_2 \sin 2s + K_1 \cos 2s, L_1 \sin 2s +K_2 \cos 2s, K_3 \\ & -K_2\sin 2s +L_1\cos 2s , K_1 \sin 2s + L_2 \cos 2s, L_3, H_2, \Xi ; U_1 \cos 2s -U_4 \sin 2s, \\ &U_2, U_3, U_1 \sin 2s +U_4 \cos 2s, V_1 \cos 2s - V_4 \sin 2s, V_2, V_3, V_1\sin 2s + V_4 \cos 2s \big) \end{align*}$

is periodic of period $\pi$ . Since $L_{Y_{K_3}}$ maps the ideal of smooth functions which vanish identically on ${\Xi}^{-1}(0)/S^1$ into itself, $Y_{K_3}$ is a vector field on ${\Xi}^{-1}(0)/S^1$ . Since $L_{X_{K_3}}H_2 = 0$ , it follows that $Y_{K_3}$ induces a vector field on $T_hS^3_1$ with periodic flow. So we can normalize again.

To compute the normal form of the Hamiltonian $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ we need only calculate the average of the term

$T = \tfrac{ 21}{ 4} K^2_3 + \tfrac{ 9}{ 8} (U^2_1+V^2_1) +\tfrac{ 9}{ 8} (U^2_4+V^2_4)$

over the flow ${\varphi }^{Y_{K_3}}_s$ . Since $L_{Y_{K_3}}K_3 = 0$ , we need only calculate $\overline{U^2_1}$ , $\overline{U^2_4}$ , $\overline{V^2_1}$ , and $\overline{V^2_4}$ . Now

$\begin{align*} \overline{U^2_1} & = \tfrac{ 1}{ \pi} \int^{\pi }_0 (U_1 \cos 2s -U_4 \sin 2s)^2 \, \mathrm{d} s \\ & = \tfrac{ 1}{ \pi} \int^{\pi }_0 \big( U^2_1 {\cos }^2\, 2s - U_1U_4\sin 4s + U^2_4 {\sin }^2\, 2s \big) \mathrm{d} s = \tfrac{ 1}{ 2} (U^2_1+U^2_4). \end{align*}$

Similarly, $\overline{V^2_1} = \tfrac{ 1}{ 2} (V^2_1+V^2_4)$ , $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_1+U^2_4)$ , and $\overline{V^2_4} = \tfrac{ 1}{ 2} (V^2_1+V^2_4)$ . Thus

$\begin{align} \overline{T} & = \tfrac{ 21}{ 4} K^2_3 + \tfrac{ 9}{ 8} h(U^2_1+V^2_1 +U^2_4+V^2_4) , \end{align}$

(5.2)

which is no surprise since $L_{Y_{K_3}}T = 0$ . So the first order normal form of $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ is

$\begin{align} {\widetilde{\mathcal{H}}}^{(1)}_{\mathrm{nf}} & = h - \tfrac{ 1}{ 2} {\varepsilon}^2 {\beta }^2h^3 - \tfrac{ 3}{ 2} \varepsilon \beta h K_3 \\ & - {\varepsilon }^2 {\beta }^2 h [ \tfrac{ 21}{ 8} K^2_3 + \tfrac{ 9}{ 16} (U^2_1+V^2_1+U^2_4+V^2_4)] . \end{align}$

(5.3)

6. The reduced Hamiltonian ${\widetilde{\mathcal{H}}}^{(1)}_{\mathrm{nf}}$ on $S^2_h\times S^2_h$

The polynomial $U^2_1+V^2_1 +U^2_4+V^2_4$ is invariant under the flows ${\varphi }^{X_{H_2}}_t$ , ${\varphi }^{X_{\Xi }}_u$ , and thus is a polynomial on the orbit space $T_hS^3_1/S^1 = S^2_h \times S^2_h$ , defined by

$\begin{align*} K^2_1+K^2_2+K^2_3 + L^2_1+L^2_2+L^2_3 & = h^2 \\ K_1L_1 +K_2L_2 +K_3L_3 & = 0. \end{align*}$

We now find this polynomial. From the explicit description of ${\mathbb{R} }^8/S^1$ in (3.1) it follows that on $\big(H^{-1}_2(h) \cap {\Xi}^{-1}(0) \big) /S^1$

$\begin{align*} U_2V_1 - U_1V_2 & = -h K_1 \\ U_3V_1-U_1V_3 & = -hK_2 \\ U_4V_1 -U_2V_4 & = -h K_3. \end{align*}$

So on $S^2_h \times S^2_h$

$\begin{align*} h^2(K^2_1+K^2_2+K^2_3) & = (U_2V_1 - U_1V_2)^2 +(U_3V_1-U_1V_3)^2 +(U_4V_1-U_1V_4)^2 \\ & = (U^2_1+U^2_2+U^2_3+U^2_4)V^2_1 - U^2_1V^2_1 \\ & -2(U_1V_1)(U_1V_1+U_2V_2+U_3V_3+U_4V_4) + 2U^2_1V^2_1 \\ & +(V^2_1+V^2_2+V^2_3+V^2_4)U^2_1 - U^2_1V^2_1 \\ & = h^2(V^2_1+U^2_1), \end{align*}$

since $\langle U, U \rangle = h^2$ , $\langle V, V \rangle = h^2$ , and $\langle U, V \rangle = 0$ . Thus $U^2_1+V^2_1 = K^2_1+K^2_2+K^2_3$ . Again from the explicit description of $\big(H^{-1}_2(h) \cap {\Xi }^{-1}(0)\big) /S^1$ we have

$\begin{align*} U_4V_3 -U_3V_4 & = -hL_1 \\ U_4V_2 - U_2V_4 & = h L_2 \\ U_4V_1-U_1V_4 & = -hK_3. \end{align*}$

So on $S^2_h \times S^2_h$

$\begin{align*} h^2(L^2_1+L^2_2+K^2_3) & = (U_4V_3 - U_3V_4)^2 +(U_4V_2-U_2V_4)^2 +(U_4V_1-U_1V_4)^2 \\ & = (V^2_1+V^2_2+V^2_3+V^2_4)U^2_4 - U^2_4V^2_4 \\ & -2(U_4V_4)(U_1V_1+U_2V_2+U_3V_3+U_4V_4) + 2U^2_4V^2_4 \\ & +(U^2_1+U^2_2+U^2_3+U ^2_4)V^2_4 - U^2_4V^2_4 \\ & = h^2(U^2_4 +V^2_4). \end{align*}$

Thus $U^2_4+V^2_4 = L^2_1+L^2_2+K^2_3$ . Consequently

$\begin{align*} U^2_1+V^2_1+U^2_4+V^2_4 & = K^2_1+K^2_2 +2K^2_3 +L^2_1+L^2_2 \\ & = K^2_1+K^2_2+K^2_3 + L^2_1+L^2_2+L^2_3 + K^2_3 - L^2_3 \\ & = 2h^2 + K^2_3 - L^2_3. \end{align*}$

on $S^2_h \times S^2_h$ . Hence on $S^2_h \times S^2_h$

$\begin{align} \widehat{\mathcal{H}} = {\widetilde{\mathcal{H}}}^{(1)}_{\mathrm{nf}} & = h - \tfrac{ 13}{ 8} {\varepsilon}^2 {\beta }^2h^3 - \tfrac{ 3}{ 2} \varepsilon \beta h K_3 - \tfrac{ 51}{ 16} {\varepsilon}^2 {\beta }^2 h K^2_3 + \tfrac{ 9}{ 16} {\varepsilon }^2 {\beta }^2h L^2_3. \end{align}$

(6.1)

7. The Hamiltonian system $(\widehat{\mathcal{H}}, S^2_h \times S^2_h, \{ \, \, , \, \, \})$

Using the coordinates $(\xi, \eta) = \big((K+L)/2, (K-L)/2 \big)$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ , the space of smooth functions on the reduced space $S^2_h \times S^2_h$ , defined by

${\xi }^2_1+{\xi }^2_2+{\xi }^2_3 = h^2\, \, \, \mathrm{and} \, \, \, {\eta }^2_1+{\eta }^2_2 + {\eta }^2_3 = h^2,$

has a Poisson structure with bracket relations

$\{ {\xi }_i, {\xi }_j \} = \sum\limits^3_{k = 1}{\epsilon }_{ijk}{\xi }_k, \, \, \, \{ {\eta }_i, {\eta }_j \} = -\sum\limits^3_{k = 1}{\epsilon }_{ijk}{\eta }_k, \, \, \, \{ {\xi }_i, {\eta }_j \} = 0.$

Since $\{ K_3, L_3 \} = 0$ , it follows that $\{ K_3, \widehat{\mathcal{H}} \} = 0$ . Thus the flow ${\varphi }^{Z_{K_3}}_r$ of the Hamiltonian vector field $Z_{K_3}$ on $(S^2_h \times S^2_h, \{ \, \, , \, \, \})$ generates an $S^1$ symmetry of the Hamiltonian system $(\widehat{\mathcal{H}}, S^2_h \times S^2_h, \{ \, \, , \, \, \})$ . So this system is completely integrable.

We reduce this $S^1$ symmetry as follows. Consider the vector field $Z_{K_3}$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ corresponding to the Hamiltonian $K_3 = \tfrac{ 1}{ 2} ({\xi }_3 + {\eta }_3)$ . Its integral curves satisfy

$\begin{align*} {\dot{\xi}}_1 & = \{ {\xi }_1, K_3 \} = \tfrac{ 1}{ 2}\{ {\xi }_1, {\xi }_3 \} = - \tfrac{ 1}{ 2}{\xi }_2 \\ {\dot{\xi}}_2 & = \{ {\xi }_2, K_3 \} = \tfrac{ 1}{ 2} \{ {\xi }_2, {\xi }_3 \} = \tfrac{ 1}{ 2} {\xi }_1 \\ {\dot{\xi}}_3 & = \{ {\xi }_3, K_3 \} = 0 \\ {\dot{\eta}}_1 & = \{ {\eta }_1, K_3 \} = \tfrac{ 1}{ 2}\{ {\eta }_1, {\eta }_3 \} = \tfrac{ 1}{ 2}{\eta }_2 \\ {\dot{\eta}}_2 & = \{ {\eta }_2, K_3 \} = \tfrac{ 1}{ 2} \{ {\eta }_2, {\eta }_3 \} = -\tfrac{ 1}{ 2} {\eta }_1 \\ {\dot{\xi}}_3 & = \{ {\eta }_3, K_3 \} = 0. \end{align*}$

Thus the flow of $Z_{K_3}$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ is

$\begin{align*} {\varphi }^{Z_{K_3}}_t(\xi , \eta ) & = \big( {\xi }_1 \cos t/2 -{\xi}_2 \sin t/2, {\xi }_1 \sin t/2 +{\xi }_2 \cos t/2, {\xi }_3 , \\ & {\eta }_1 \cos t/2 +{\eta }_2 \sin t/2, {\eta }_1 \sin t/2 -{\eta }_2 \cos t/2, {\eta }_3 \big) , \end{align*}$

which preserves $S^2_h \times S^2_h$ and is periodic of period $4\pi$ .

We now determine the space $(S^2_h\times S^2_h)/S^1$ of orbits of the vector field $Z_{K_3}$ . We use invariant theory. The algebra of polynomials on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ , which are invariant under the $S^1$ action given by the flow ${\varphi }^{Z_{K_3}}_t$ , is generated by

$\begin{array}{lclcl} {\sigma }_1 = {\xi }^2_1 +{\xi }^2_2 & & {\sigma }_2 = {\eta }^2_1 +{\eta }^2_2 & & {\sigma }_3 = {\xi }_1{\eta }_2 - {\xi}_2{\eta }_1 \\ {\sigma }_4 = {\xi }_1{\eta }_1 +{\xi }_2{\eta }_2 & & {\sigma }_5 = \tfrac{ 1}{ 2} ({\xi }_3+{\eta }_3) & & {\sigma }_6 = \tfrac{ 1}{ 2} ({\xi }_3-{\eta }_3) , \end{array}$

which are subject to the relation

$\begin{align} {\sigma }^2_3 +{\sigma }^2_4 & = ({\xi }_1{\eta }_2-{\xi}_2{\eta }_1)^2 + ({\xi }_1{\eta }_1+{\xi}_2{\eta }_2)^2 \\ & = ({\xi }^2_1+{\xi }^2_2)({\eta }^2_1+{\eta }^2_2) = {\sigma }_1 {\sigma }_2, \, \, \, {\sigma }_1 \ge 0 \, \, & \, \, {\sigma }_2 \ge 0. \end{align}$

(7.1a)

In terms of invariants the defining equations of $S^2_h \times S^2_h$ become

$\begin{align} {\sigma }_1 +({\sigma }_5+{\sigma }_6)^2 & = {\xi }^2_1+{\xi }^2_2 +{\xi }^2_3 = h^2 \end{align}$

(7.1b)

$\begin{align} {\sigma }_2 +({\sigma }_5-{\sigma }_6)^2 & = {\eta }^2_1+{\eta }^2_2 +{\eta }^2_3 = h^2 . \end{align}$

(7.1c)

Eliminating ${\sigma }_1$ and ${\sigma }_2$ from (7.1a) using (7.1b) and (7.1c) gives

$\begin{equation} {\sigma }^2_3 +{\sigma }^2_4 = \big( h^2-({\sigma }_5+{\sigma }_6)^2 \big) \big( h^2 - ({\sigma }_5 - {\sigma }_6)^2 \big) , \, \, |{\sigma }_5 +{\sigma }_6| \le h \, \, \, \, |{\sigma }_5-{\sigma }_6| \le h, \end{equation}$

(7.2a)

which defines $(S^2_h \times S^2_h)/S^1$ as a semialgebraic variety in ${\mathbb{R} }^4$ with coordinates $({\sigma}_3, {\sigma }_4, {\sigma }_5, {\sigma }_6)$ . Thus the reduced space $\big(K^{-1}_3(2 k) \cap (S^2_h\times S^2_h) \big)/S^1$ is defined by (7.2a) and

$\begin{equation} {\sigma }_5 = \tfrac{ 1}{ 2} ({\xi }_3+{\eta }_3) = \tfrac{ 1}{ 2} K_3 = k . \end{equation}$

(7.2b)

Consequently, $\big(K^{-1}_3(2k) \cap (S^2_h \times S^2_h) \big) /S^1$ is the semialgebraic variety

$\begin{align} {\sigma }^2_3+{\sigma }^2_4 & = \big( h^2-(k +{\sigma }_6)^2 \big) \big( h^2-(k-{\sigma }_6)^2 \big) \\ & = \big( (h-k )^2-{\sigma }^2_6 \big) \big( (h+k )^2-{\sigma }^2_6\big) , \, \, |{\sigma }_6| \le h - |\ell | \end{align}$

(7.3)

in ${\mathbb{R} }^3$ with coordinates $({\sigma }_3, {\sigma }_4, {\sigma }_6)$ . When $0 < |k | < h$ the reduced space (7.3) is a smooth $2$ -sphere. When $|k| = h$ it is a point. When $k = 0$ it is a topological $2$ -sphere with conical singular points at $(0, 0, \pm h)$ . These singular points correspond to the fixed points $h(0, 0, \pm 1, 0, 0, \mp 1)$ of the $S^1$ action on $S^2_h \times S^2_h$ generated by the flow of the vector field $Z_{K_3}$ .

By (6.1) the reduced Hamiltonian on $\big(K^{-1}_3(2k) \cap (S^2_h \times S^2_h) \big)/S^1$ is

$\begin{equation} {\widehat{\mathcal{H}}}_{\mathrm{red}} = \tfrac{ 9}{ 4} {\varepsilon}^2 {\beta }^2 h\, {\sigma }^2_6, \end{equation}$

(7.4)

using $L_3 = {\xi }_3-{\eta }_3 = 2{\sigma }_6$ , having dropped the constant $h - \tfrac{13}{8} {\varepsilon}^2 {\beta }^2 h^3 - \tfrac{3}{2} \varepsilon \beta h k - \tfrac{51}{16}{\varepsilon }^2{\beta }^2 h k^2$ .

Use of AI tools declaration

The author declares that he has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author would like to thank the referees for their careful reading and comments on the manuscript. Especially he thanks the one who pointed out errors in his calculations.

The author receied no funding for the research in this article.

Conflict of interest

The author declares there is no conflict of interest.

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