Video-based person re-identification with complementary local and global features using a graph transformer

Hai Lu; Enbo Luo; Yong Feng; Yifan Wang; Hai Lu; Enbo Luo; Yong Feng; Yifan Wang

doi:10.3934/mbe.2024293

Mathematical Biosciences and Engineering

2024, Volume 21, Issue 7: 6694-6709. doi: 10.3934/mbe.2024293

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Research article Special Issues

Video-based person re-identification with complementary local and global features using a graph transformer

Electric Power Research Institute of Yunnan Power Grid Co., Ltd., Kunming 650217, China

Received: 08 March 2024 Revised: 16 June 2024 Accepted: 05 July 2024 Published: 23 July 2024

In recent years, significant progress has been made in video-based person re-identification (Re-ID). The key challenge in video person Re-ID lies in effectively constructing discriminative and robust person feature representations. Methods based on local regions utilize spatial and temporal attention to extract representative local features. However, prior approaches often overlook the correlations between local regions. To leverage relationships among different local regions, we have proposed a novel video person Re-ID representation learning approach based on a graph transformer, which facilitates contextual interactions between relevant region features. Specifically, we construct a local relation graph to model intrinsic relationships between nodes representing local regions. This graph employs the architecture of a transformer for feature propagation, iteratively refining region features and considering information from adjacent nodes to obtain partial feature representations. To learn compact and discriminative representations, we have further proposed a global feature learning branch based on a vision transformer to capture the relationships between different frames in a sequence. Additionally, we designed a dual-branch interaction network based on multi-head fusion attention to integrate frame-level features extracted by both local and global branches. Finally, the concatenated global and local features, after interaction, are used for testing. We evaluated the proposed method on three datasets, namely iLIDS-VID, MARS, and DukeMTMC-VideoReID. Experimental results demonstrate competitive performance, validating the effectiveness of our proposed approach.

Keywords:

Citation: Hai Lu, Enbo Luo, Yong Feng, Yifan Wang. Video-based person re-identification with complementary local and global features using a graph transformer[J]. Mathematical Biosciences and Engineering, 2024, 21(7): 6694-6709. doi: 10.3934/mbe.2024293

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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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Mathematical Biosciences and Engineering

Video-based person re-identification with complementary local and global features using a graph transformer

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1. Introduction

2. The basic set up

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

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

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

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

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

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Video-based person re-identification with complementary local and global features using a graph transformer

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1. Introduction

2. The basic set up

3. The first order normal form on Ξ−1(0)/S1 {\Xi}^{-1}(0)/S^1

4. The second order normal form on Ξ−1(0)/S1 {\Xi }^{-1}(0)/S^1

5. The first order normal form of H(2)nf {\mathcal{H}}^{(2)}_{\mathrm{nf}} on ThS31 T_hS^3_1

6. The reduced Hamiltonian ˜H(1)nf {\widetilde{\mathcal{H}}}^{(1)}_{\mathrm{nf}} on S2h×S2h S^2_h\times S^2_h

7. The Hamiltonian system (ˆH,S2h×S2h,{,}) (\widehat{\mathcal{H}}, S^2_h \times S^2_h, \{ \, \, , \, \, \})

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

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

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

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

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