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Emergent behaviors of rotation matrix flocks

Razvan C. Fetecau
Department of Mathematics
Simon Fraser University, 8888 University, Burnaby, BC V5A 1S6, Canada.
razvan_fetecau@sfu.ca
Seung-Yeal Ha
Department of Mathematical Sciences
Seoul National University, Seoul 08826, and
Korea Institute for Advanced Study, Hoegiro 85, Seoul 02455, Republic of Korea
syha@snu.ac.kr
 and  Hansol Park
Department of Mathematical Sciences
Seoul National University, Seoul 08826, Republic of Korea
hansol960612@snu.ac.kr
Abstract.

We derive an explicit form for the Cucker-Smale (CS) model on the special orthogonal group SO(3)\mathrm{SO}(3) by identifying closed form expressions for geometric quantities such as covariant derivative and parallel transport in exponential coordinates. We study the emergent dynamics of the model by using a Lyapunov functional approach and La Salle’s invariance principle. Specifically, we show that velocity alignment emerges from some admissible class of initial data, under suitable assumptions on the communication weight function. We characterize the ω\omega-limit set of the dynamical system and identify a dichotomy in the asymptotic behavior of solutions. Several numerical examples are provided to support the analytical results.

Key words and phrases:
Cucker-Smale model, special orthogonal group, emergence, flocking, velocity alignment, rotation group.
2020 Mathematics Subject Classification:
70G60, 82C10
Acknowledgment. R.F. acknowledges support from NSERC Discovery Grant PIN-341834 during this research. The work of S.-Y. Ha was supported by National Research Foundation of Korea (NRF-2020R1A2C3A01003881).

1. Introduction

Collective behaviors of complex biological systems are ubiquitous in nature, e.g., aggregation of bacteria [36, 58], synchronous flashing of fireflies [9], synchronization of rhythms in pacemaker cells [47], flocking of migratory birds [7, 52] and swarming of fish [5, 16, 57]. We refer to [4, 46, 48, 62] for a brief survey of collective dynamics and engineering applications in the decentralized control of multi-agent systems. Among them, our main interest lies in flocking behaviors in which particles organize themselves from a disordered state to an ordered motion using simple rules based on environmental information. Despite its ubiquitous presence, systematic mathematical studies have begun only several decades ago by Vicsek et al. [61], a group of statistical physicists who were mainly motivated by Reynolds’s simulations [52] of bird flocking in computer graphics. Since then, several mechanical and phenomenological models have been used to study flocking behavior.

In this paper, we are interested in the Cucker-Smale (CS) model [13] for flocking of a matrix ensemble on the special orthogonal matrix group SO(3)\mathrm{SO}(3), given by:

(1.1) SO(3):={R3×3:RR=I and detR=1},\mathrm{SO}(3):=\{R\in\mathbb{R}^{3\times 3}:R^{\top}R=I\text{ and }~{}\mathrm{det}~{}R=1\},

where 3×3\mathbb{R}^{3\times 3} is the matrix group consisting of all 3×33\times 3 real matrices and II denotes the 3×33\times 3 identity matrix. Note that SO(3)\mathrm{SO}(3), also referred to as the rotation group, is a Lie group, having a group and a manifold structure at the same time. Elements in SO(3)\mathrm{SO}(3) are called rotation matrices; they are characterized by an axis and an angle of rotation. The rotation group is the configuration space of a rigid body in 3\mathbb{R}^{3} that undergoes rotations only (no translations). There has been recent growing interest in studying collective behavior on SO(3)\mathrm{SO}(3) [15, 14, 31], motivated by body attitude coordination of interacting agents in a variety of applications. For example, engineering applications include synchronization of satellite attitudes [8, 37, 39] and camera pose averaging [60]. In biology, it is very common for animals such as birds and fish to coordinate their body attitudes – see [15] for some images of collective motions in birds and dolphins.

Before we present the CS model on SO(3)\mathrm{SO}(3), we introduce the CS model on the Euclidean space d\mathbb{R}^{d}. Consider an ensemble of NN identical particles with unit mass in n\mathbb{R}^{n}, and let 𝐱i\mathbf{x}_{i} and 𝐯i\mathbf{v}_{i} be the position and velocity of the ii-th particle, respectively. The CS model reads:

(1.2) {𝐱˙i=𝐯i,t>0,i=1,,N,𝐯˙i=κNk=1Nϕik(𝐯k𝐯i),\begin{cases}\displaystyle{\dot{\mathbf{x}}}_{i}=\mathbf{v}_{i},\quad t>0,\quad i=1,\cdots,N,\\ \displaystyle{\dot{\mathbf{v}}}_{i}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}(\mathbf{v}_{k}-\mathbf{v}_{i}),\end{cases}

where κ\kappa is a nonnegative coupling strength and ϕik=ϕ(|𝐱k𝐱i|)\phi_{ik}=\phi(|\mathbf{x}_{k}-\mathbf{x}_{i}|) are communication weights given in terms of a bounded and Lipschitz continuous communication function ϕ\phi. In this setting, the global well-posedness of (1.2) follows directly from the Cauchy-Lipschitz theory together with uniform boundedness of the kinetic energy. The CS model (1.2) exhibits interesting flocking dynamics depending on the initial configuration, system parameters and communication functions (see [13, 32, 34]), and it has been studied extensively in applied mathematics and control communities in the last decade. The model has been investigated in diverse directions and areas of applications, e.g., network topology [11, 12, 20, 21, 41, 42], global and local flocking dynamics [13, 32, 34, 45], interaction with nonlinear damping [22], time-delayed interactions [17], kinetic and hydrodynamic CS equations [18, 40, 50, 51, 55, 56], coupling with thermodynamics [26, 27, 30, 33], complete predictability [28], extension to manifolds [1, 2, 3, 31] and relativistic CS model [29]. We refer to recent survey articles [4, 10, 44] for overview and further references.

In the present paper we address the“Cucker-Smale flocking realizability problem” on SO(3)\mathrm{SO}(3):

  • (Q1): Is there a CS type flocking model on SO(3)\mathrm{SO}(3)?

  • (Q2):  If so, under what conditions on system parameters and initial data, can we guarantee emergent dynamics?

To answer question (Q1), we need to replace the R.H.S. of (1.2)2\eqref{A-0}_{2} by a suitable internal force so that particles’ positions stay on SO(3)\mathrm{SO}(3) through the time evolution. This is not obvious at first glance, as we need to compare velocities belonging to different tangent spaces. In fact, question (Q1) was already answered affirmatively in an abstract setting in which the underlying manifold (,g)({\mathcal{M}},g) is a connected, complete and smooth Riemannian manifold with a metric gg (see [31]), whereas question (Q2) was answered only for the sphere and the hyperboloid (see [2, 3]).

Specifically, let (,g){\mathcal{M}},g) be a connected, complete and smooth Riemannian manifold with metric gg. Denote by Dvdt\frac{Dv}{dt} the covariant derivative of a tangent vector vv. For 𝐱k,𝐱i\mathbf{x}_{k},\mathbf{x}_{i}\in\mathcal{M}, let PkiP_{ki} be the parallel transport along the length minimizing geodesic from 𝐱k\mathbf{x}_{k} to 𝐱i\mathbf{x}_{i}. In this abstract Riemannian setting, the CS model on (M,g)(M,g) takes the following form:

(1.3) {𝐱˙i=𝐯i,t>0,i=1,,N,D𝐯idt=κNk=1Nϕik(Pki𝐯k𝐯i),\displaystyle\begin{cases}\displaystyle\dot{\mathbf{x}}_{i}=\mathbf{v}_{i},\quad t>0,\quad i=1,\cdots,N,\\ \displaystyle\frac{D\mathbf{v}_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}(P_{ki}\mathbf{v}_{k}-\mathbf{v}_{i}),\end{cases}

where κ\kappa is a nonnegative coupling strength and ϕik=ϕ(d(𝐱i,𝐱k))\phi_{ik}=\phi(d(\mathbf{x}_{i},\mathbf{x}_{k})) are communication weights. Here, d(,)d(\cdot,\cdot) denotes the geodesic distance on \mathcal{M}, and the communication weight function ϕ\phi is assumed to be a nonnegative, bounded, and smooth function on 0\mathbb{R}_{\geq 0}. Note that the parallel transport in model (1.3) is well-defined only if the length minimizing geodesic between 𝐱k\mathbf{x}_{k} and 𝐱i\mathbf{x}_{i} is unique, i.e., when the two points are not in the cut locus of each other. To alleviate this issue, if two points are in the cut locus of each other then we set ϕ\phi to be zero for such pairs of points111In the paper we will use the term pair of cut points to refer to two points on a manifold that are in the cut locus of each other., so that system (1.3) is well-defined and the local well-posedness of (1.3) is guaranteed. For certain special manifolds, one can explicitly calculate D𝐯idt\frac{D\mathbf{v}_{i}}{dt} and Pki𝐯kP_{ki}\mathbf{v}_{k}, and write the abstract CS model (1.3) in an explicit form. Examples of such manifolds are the Poincaré Half plane [31], unit sphere [2, 31] and hyperboloid [3]. For related first-order aggregation models on Riemannian manifolds, we also refer to [6, 23, 24, 25, 43, 53, 54].

We adopt the following definition of velocity alignment.

Definition 1.1.

[31] Let {(𝐱i,𝐯i)}i=1N\{(\mathbf{x}_{i},\mathbf{v}_{i})\}_{i=1}^{N} be a global smooth solution to (1.3). We say that the configuration exhibits asymptotic velocity alignment if the following relation holds:

limtmax1i,kNPki𝐯k(t)𝐯i(t)𝐱i=0,\lim_{t\to\infty}\max_{1\leq i,k\leq N}\|P_{ki}\mathbf{v}_{k}(t)-\mathbf{v}_{i}(t)\|_{\mathbf{x}_{i}}=0,

where Pki𝐯k𝐯i𝐱i2=g𝐱i(Pki𝐯k𝐯i,Pki𝐯k𝐯i)\|P_{ki}\mathbf{v}_{k}-\mathbf{v}_{i}\|^{2}_{\mathbf{x}_{i}}=g_{\mathbf{x}_{i}}(P_{ki}\mathbf{v}_{k}-\mathbf{v}_{i},P_{ki}\mathbf{v}_{k}-\mathbf{v}_{i}).

Next, we discuss briefly our main results. To simulate the particle system on a specific manifold, the abstract model (1.3) cannot be used directly, unless we find an explicit form for equation (1.3)2\eqref{A-1}_{2}. Our first main result is concerned with the explicit derivation of the CS model on SO(3)\mathrm{SO}(3) by finding closed-form representations for Dvidt\frac{Dv_{i}}{dt} and Pki𝐯kP_{ki}\mathbf{v}_{k}. For the model on SO(3)\mathrm{SO}(3) we will use Ri(t)R_{i}(t) and Vi(t)=dRi(t)dtV_{i}(t)=\frac{dR_{i}(t)}{dt} instead of 𝐱i(t)\mathbf{x}_{i}(t) and 𝐯i(t)\mathbf{v}_{i}(t), to denote the position and velocity of the ii-th particle at t>0t>0. Since Vi(t)V_{i}(t) lies in the tangent space of SO(3)\mathrm{SO}(3) at Ri(t)R_{i}(t), there exists a unique skew-symmetric 3×33\times 3 matrix Ai(t)A_{i}(t) (see Section 2) such that

Vi(t)=Ri(t)Ai(t).V_{i}(t)=R_{i}(t)A_{i}(t).

To close the dynamical system we need to find the dynamics for ViV_{i} or AiA_{i} from (1.3)2\eqref{A-1}_{2}. In order to present this explicit form, we need some preparations. First note that 𝔰𝔬(3)\mathfrak{so}(3), the Lie algebra of SO(3)\mathrm{SO}(3) consisting of all 3×33\times 3 skew-symmetric matrices, is isomorphic to 3\mathbb{R}^{3} via the hat map:

(1.4) 𝐱=(x1,x2,x3)3𝐱^=[0x3x2x30x1x2x10]𝔰𝔬(3).\displaystyle\mathbf{x}=(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}\quad\Longleftrightarrow\quad\widehat{\mathbf{x}}=\begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix}\in\mathfrak{so}(3).

For given i,k{1,,N}i,k\in\{1,\cdots,N\}, we set

𝐮^ki:=log(RkRi)=θki2sinθki(RkRiRiRk),θki:=𝐮ki=arccos(tr(RkRi)12),𝐧ki:=𝐮kiθki,\displaystyle\begin{aligned} \widehat{\mathbf{u}}_{ki}&:=\log(R_{k}^{\top}R_{i})=\frac{\theta_{ki}}{2\sin\theta_{ki}}\big{(}R_{k}^{\top}R_{i}-R_{i}^{\top}R_{k}\big{)},\\[5.0pt] \theta_{ki}&:=\|\mathbf{u}_{ki}\|=\arccos\left(\frac{\mathrm{tr}(R_{k}^{\top}R_{i})-1}{2}\right),\quad\mathbf{n}_{ki}:=\frac{\mathbf{u}_{ki}}{\theta_{ki}},\end{aligned}

where \|\cdot\| denotes the Euclidean norm in 3\mathbb{R}^{3}. The precise meanings of these variables will be given in Section 2. Equipped with these notations, the CS model on SO(3)\mathrm{SO}(3) is given by:

(1.5) {dRidt=RiAi,t>0,i=1,,N,d𝐚idt=κNk=1Nϕik[(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i],\displaystyle\hskip 28.45274pt\begin{cases}\displaystyle\frac{dR_{i}}{dt}=R_{i}A_{i},\quad t>0,\quad i=1,\cdots,N,\\[5.0pt] \displaystyle\frac{d\mathbf{a}_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right],\end{cases}

where 𝐚i3\mathbf{a}_{i}\in\mathbb{R}^{3} satisfies 𝐚^i=Ai\widehat{\mathbf{a}}_{i}=A_{i}.

As noted above, parallel transport between a pair of cut locus points is not well-defined, and in such cases we require that the communication weighs between the points is zero. On SO(3)\mathrm{SO}(3), the injectivity radius is π\pi and two rotation matrices RR and QQ are in the cut locus of each other if and only if d(R,Q)=πd(R,Q)=\pi. For system (1.5) to be well-defined and also locally well-posed, one of the following conditions will be used later.

  • (A)({\mathcal{H}}_{A}): ϕ(R,Q)\phi(R,Q) is a continuous function of d(R,Q)d(R,Q) that vanishes at pairs of cut locus points:

    ϕ(R,Q)=ϕ~(d(R,Q)),andϕ~(π)=0.\phi(R,Q)=\tilde{\phi}(d(R,Q)),\qquad\mbox{and}\qquad\tilde{\phi}(\pi)=0.
  • (B)({\mathcal{H}}_{B}): for any 1i,kN1\leq i,k\leq N and t0t\geq 0, the distance between Ri(t)R_{i}(t) and Rk(t)R_{k}(t) is strictly less than π\pi.

Our second main results deal with the emergent dynamics of (1.5). For this, we introduce a Lyapunov functional \mathcal{E} given by the total kinetic energy:

(t)=i=1NVi(t)Ri(t)2.\mathcal{E}(t)=\sum_{i=1}^{N}{\|V_{i}(t)\|}_{R_{i}(t)}^{2}.

Then, by straightforward calculations, one can derive a dissipation estimate (see Section 4.3) :

(1.6) ddt=κNi,k=1Nϕ(Ri,Rk)PkiVkVixi20.\frac{d\mathcal{E}}{dt}=-\frac{\kappa}{N}\sum_{i,k=1}^{N}\phi(R_{i},R_{k})\|P_{ki}V_{k}-V_{i}\|_{x_{i}}^{2}\leq 0.

We apply LaSalle’s invariance principle [38] and use (1.6) to obtain

limtddt=0,or equivalentlylimtϕ(Ri,Rk)PkiVkViRi2=0i,k{1,,N}.\lim_{t\to\infty}\frac{d\mathcal{E}}{dt}=0,\quad\mbox{or equivalently}\quad\lim_{t\to\infty}\phi(R_{i},R_{k})\|P_{ki}V_{k}-V_{i}\|_{R_{i}}^{2}=0\quad\forall~{}i,k\in\{1,\dots,N\}.

Under the assumptions (A)({\mathcal{H}}_{A}) or (B)({\mathcal{H}}_{B}), one can then derive velocity alignment (Theorem 4.1).

Finally, we characterize the ω\omega-limit set of model (1.5), and show that a certain dichotomy holds in the asymptotic behavior of the solutions (Theorem 4.2). This result is confirmed by numerical simulations.

The paper is organized as follows. In Section 2, we review briefly the geometry of the rotation group, e.g., the exponential map, the explicit form of the metric tensor, geodesic equations, as needed for later sections. In Section 3, we provide equations for geodesics and provide a closed form representation for parallel transport and its geometric interpretation. In Section 4, we explicitly derive the CS model on SO(3)\mathrm{SO}(3) from the abstract model (1.3) and study its emergent dynamics. In Section 5, we provide several numerical simulations and compare them with the analytical results in previous sections. Finally Section 6 is devoted to a brief summary of our main results and some remaining issues for future work. In Appendix we provide proofs for several lemmas and details to various calculations used in the paper.

Gallery of Notation: For notational simplicity, we use the following handy notation:

maxi:=max1iN,maxi,j:=max1i,jN,i:=i=1N,i,j:=i=1Nj=1N,\max_{i}:=\max_{1\leq i\leq N},\qquad\max_{i,j}:=\max_{1\leq i,j\leq N},\qquad\sum_{i}:=\sum_{i=1}^{N},\qquad\sum_{i,j}:=\sum_{i=1}^{N}\sum_{j=1}^{N},

and if there is no confusion, we also use Einstein’s notation on repeated indices from time to time. Let 3×3\mathbb{R}^{3\times 3} be the collection of real 3×33\times 3 matrices and for A1,A23×3A_{1},A_{2}\in\mathbb{R}^{3\times 3}, we introduce the Frobenius inner product and its associated norm:

A1,A2F:=Tr(A1A2),A1F:=A1,A1F.\langle A_{1},A_{2}\rangle_{\mathrm{F}}:=\mbox{Tr}(A_{1}^{\top}A_{2}),\quad\|A_{1}\|_{\mathrm{F}}:=\sqrt{\langle A_{1},A_{1}\rangle_{\mathrm{F}}}.

2. Geometry of the special orthogonal group SO(3)\mathrm{SO}(3)

In this section, we review minimal background on the geometry of the special orthogonal group SO(3)\mathrm{SO}(3) which will be necessary to follow discussions in later sections. In particular, we present SO(3)\mathrm{SO}(3) as a Riemannian manifold and derive explicit calculations using the parametrization by exponential coordinates.

2.1. General considerations

The Lie algebra 𝔰𝔬(3)\mathfrak{so}(3) of the Lie group SO(3)\mathrm{SO}(3) in (1.1), representing the tangent space of SO(3)\mathrm{SO}(3) at the identity II, is given by:

𝔰𝔬(3):={A3×3:A+A=O},\mathfrak{so}(3):=\{A\in\mathbb{R}^{3\times 3}:~{}A^{\top}+A=O\},

where OO is the zero matrix. In general, the tangent space at a generic point RSO(3)R\in\mathrm{SO}(3) is given by

(2.1) TRSO(3):={RA:A𝔰𝔬(3)}.T_{R}\mathrm{SO}(3):=\{RA:A\in\mathfrak{so}(3)\}.

The Riemannian metric gg on TRSO(3)T_{R}\mathrm{SO}(3) is defined as:

(2.2) g(RA1,RA2):=12RA1,RA2F=12A1,A2F,g(RA_{1},RA_{2}):=\frac{1}{2}\langle RA_{1},RA_{2}\rangle_{\mathrm{F}}=\frac{1}{2}\langle A_{1},A_{2}\rangle_{\mathrm{F}},

for any RA1,RA2TRSO(3)RA_{1},RA_{2}\in T_{R}\mathrm{SO}(3). Note that the norm induced by the metric satisfies

(2.3) RATRSO(3)=12RAF=12AF.\|RA\|_{T_{R}\mathrm{SO}(3)}=\frac{1}{\sqrt{2}}{\|RA\|}_{\mathrm{F}}=\frac{1}{\sqrt{2}}{\|A\|}_{\mathrm{F}}.

Throughout the paper, we use \|\cdot\| for the usual vector norm in 3\mathbb{R}^{3}, F\|\cdot\|_{\mathrm{F}} for the Frobenius norm, and R\|\cdot\|_{R} for the Riemannian norm at RSO(3)R\in\mathrm{SO}(3). We also denote the Frobenius and the Riemannian inner products by ,F{\langle\cdot,\cdot\rangle}_{\mathrm{F}} and ,R{\langle\cdot,\cdot\rangle}_{R}, respectively. In summary, we have the following relations between the Frobenius and Riemannian norms/inner-products as follows:

V1,V2R=12V1,V2F,VR=12VF,{\langle V_{1},V_{2}\rangle}_{R}=\frac{1}{2}{\langle V_{1},V_{2}\rangle}_{\mathrm{F}},\qquad{\|V\|}_{R}=\frac{1}{\sqrt{2}}{\|V\|}_{\mathrm{F}},

where VV, V1V_{1}, and V2V_{2} are tangent vectors in TRSO(3)T_{R}\mathrm{SO}(3).

2.1.1. Isomorphism between 𝔰𝔬(3)\mathfrak{so}(3) and 3\mathbb{R}^{3}

To any vector 𝐱=(x1,x2,x3)3\mathbf{x}=(x_{1},x_{2},x_{3})\in\mathbb{R}^{3} we associate its corresponding skew-symmetric matrix 𝐱^𝔰𝔬(3)\widehat{\mathbf{x}}\in\mathfrak{so}(3) given by (1.4). By introducing the canonical basis of 𝔰𝔬(3)\mathfrak{so}(3):

(2.4) E1=[000001010],E2=[001000100],E3=[010100000],E_{1}=\begin{bmatrix}0&0&0\\ 0&0&-1\\ 0&1&0\end{bmatrix},\quad E_{2}=\begin{bmatrix}0&0&1\\ 0&0&0\\ -1&0&0\end{bmatrix},\quad E_{3}=\begin{bmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{bmatrix},

the hat operation can be expressed as:

(2.5) ^:3𝔰𝔬(3),𝐱𝐱^=xαEα,\widehat{}:\mathbb{R}^{3}\to\mathfrak{so}(3),\qquad\mathbf{x}\mapsto\widehat{\mathbf{x}}=x_{\alpha}E_{\alpha},

where we used the Einstein convention for summation over repeated indices. Note that the components of EαE_{\alpha}, α=1,2,3\alpha=1,2,3 can be also expressed as

[Eα]βγ=ϵαβγ,[E_{\alpha}]_{\beta\gamma}=-\epsilon_{\alpha\beta\gamma},

where ϵ\epsilon denotes the Levi-Civita symbol.

The inverse of the hat map, denoted by ˇ\,\widecheck{}\,, is given by:

ˇ:𝔰𝔬(3)3,xαEα𝐱=(x1,x2,x3).\widecheck{}:\mathfrak{so}(3)\to\mathbb{R}^{3},\qquad x_{\alpha}E_{\alpha}\mapsto\mathbf{x}=(x_{1},x_{2},x_{3}).

The hat operator (and its inverse) establish an isomorphism between 3\mathbb{R}^{3} and 𝔰𝔬(3)\mathfrak{so}(3). In particular, 𝐱=𝐱^I\|\mathbf{x}\|={\|\widehat{\mathbf{x}}\|}_{I}, for all 𝐱3\mathbf{x}\in\mathbb{R}^{3}.

2.1.2. Angle-axis representation

As mentioned in the Introduction, an element in SO(3)\mathrm{SO}(3) is called a rotation matrix. Hence, any rotation RSO(3)R\in\mathrm{SO}(3) can be identified with a pair (θ,𝐯)[0,π]×𝕊2(\theta,\mathbf{v})\in[0,\pi]\times\mathbb{S}^{2}, where 𝕊2\mathbb{S}^{2} denotes the unit sphere in 3\mathbb{R}^{3}. The unit vector 𝐯\mathbf{v} indicates the axis of rotation and θ\theta represents the angle of rotation (by the right-hand rule) about the axis 𝐯\mathbf{v}. The pair (θ,𝐯)(\theta,\mathbf{v}) is also referred to as the angle-axis representation of the rotation RR. The expression of RR in terms of (θ,𝐯)(\theta,\mathbf{v}) is given by the exponential map via Rodrigues’s formula:

(2.6) R=exp(θ𝐯^)=I+sinθ𝐯^+(1cosθ)𝐯^2.\displaystyle R=\exp(\theta\widehat{\mathbf{v}})=I+\sin\theta\widehat{\mathbf{v}}+(1-\cos\theta)\widehat{\mathbf{v}}^{2}.

The inverse of Rodrigues’s formula, θ𝐯^=log(R)\theta\widehat{\mathbf{v}}=\log(R), is also given by

(2.7) θ=arccos(trR12),𝐯^=12sinθ(RR).\theta=\arccos\left(\frac{\mathrm{tr}R-1}{2}\right),\quad\widehat{\mathbf{v}}=\frac{1}{2\sin\theta}(R-R^{\top}).

2.1.3. Geodesic distance, exponential and logarithm maps

Given two rotation matrices RR, QSO(3)Q\in\mathrm{SO}(3), the shortest path between RR and QQ is the geodesic curve :[0,1]SO(3)\mathcal{R}:[0,1]\to\mathrm{SO}(3) given by

(t)=Rexp(tlog(RQ)).\mathcal{R}(t)=R\exp(t\log(R^{\top}Q)).

Note that

(t)=(t)log(RQ)T(t)SO(3),\mathcal{R}^{\prime}(t)=\mathcal{R}(t)\log(R^{\top}Q)\in T_{\mathcal{R}(t)}\mathrm{SO}(3),

and the Riemannian distance between RR and QQ on on SO(3)\mathrm{SO}(3) is

d(R,Q)=dRQ,d(R,Q)=d_{RQ},

where dRQd_{RQ} satisfies

dRQ𝐯^RQ=log(RQ).d_{RQ}\widehat{\mathbf{v}}_{{}_{RQ}}=\log(R^{\top}Q).

Then by (2.7), we also have

(2.8) d(R,Q)=arccos(tr(RQ)12),d(R,Q)=\operatorname{arccos}\left(\frac{\operatorname{tr}(R^{\top}Q)-1}{2}\right),

and in particular, d(I,R)=arccos(trR12)d(I,R)=\operatorname{arccos}\left(\frac{\operatorname{tr}R-1}{2}\right).

On the other hand, the Riemannian exponential and logarithm maps at RR are given as follows:

(2.9) {expR:TRSO(3)SO(3),expR(RA)=Rexp(A),logR:SO(3)TRSO(3),logR(Q)=Rlog(RQ).\begin{cases}\displaystyle\exp_{R}:T_{R}\mathrm{SO}(3)\to\mathrm{SO}(3),\qquad\exp_{R}(RA)=R\exp(A),\\[5.0pt] \displaystyle\log_{R}:\mathrm{SO}(3)\to T_{R}\mathrm{SO}(3),\qquad\log_{R}(Q)=R\log(R^{\top}Q).\end{cases}

Rodrigues’s formulas (2.6) and (2.7) can be expressed using the exponential map at the identity II, as

R=expI(θ𝐯^),θ𝐯^=logIR.R=\exp_{I}(\theta\widehat{\mathbf{v}}),\qquad\theta\widehat{\mathbf{v}}=\log_{I}R.

2.2. Exponential coordinates

For fixed R0SO(3)R_{0}\in\mathrm{SO}(3), we introduce exponential coordinates (also commonly referred to as Riemannian normal coordinates [49]) on SO(3)\mathrm{SO}(3) centerd at R0R_{0}. With this aim we consider the map R:3SO(3)R:\mathbb{R}^{3}\to\mathrm{SO}(3) (see (2.9)) given by

(2.10) 𝐱=(x1,x2,x3)R(𝐱)=R0exp(xαEα),\displaystyle\mathbf{x}=(x_{1},x_{2},x_{3})\longrightarrow R(\mathbf{x})=R_{0}\exp\left(x_{\alpha}E_{\alpha}\right),

which is an injective map on 𝐱<π\|\mathbf{x}\|<\pi. The chart covers all but the cut locus of R0R_{0}. Throughout the paper, we will use Greek letters (α,β,γ\alpha,\beta,\gamma, etc) for coordinates on SO(3)\mathrm{SO}(3). Consequently, indices in Greek letters take values in {1,2,3}\{1,2,3\}. We make this distinction to avoid possible confusion with indices for particles in the Cucker-Smale model, for which we use Roman letters (i,j,ki,j,k, etc); such indices take values in {1,2,,N}\{1,2,\dots,N\}. First, we compute the basis {1R(𝐱),2R(𝐱),3R(𝐱)}\{\partial_{1}R(\mathbf{x}),\partial_{2}R(\mathbf{x}),\partial_{3}R(\mathbf{x})\} of the tangent space TR(𝐱)SO(3)T_{R(\mathbf{x})}\mathrm{SO}(3) at R(𝐱)R(\mathbf{x}) induced by the coordinate chart (2.10). By differentiating (2.10), we have

(2.11) αR(𝐱)=R0α(exp(xβEβ)),whereα=xα.\partial_{\alpha}R(\mathbf{x})=R_{0}\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)},\quad\mbox{where}~{}\partial_{\alpha}=\partial_{x_{\alpha}}.

We set

(2.12) θ=𝐱,𝐯=𝐱θ.\theta=\|\mathbf{x}\|,\quad\mathbf{v}=\frac{\mathbf{x}}{\theta}.

Then, we use (2.6) to find

exp(xβEβ)=exp(θ𝐯^)=I+sinθθ𝐱^+1cosθθ2𝐱^2.\exp(x_{\beta}E_{\beta})=\exp(\theta\widehat{\mathbf{v}})=I+\frac{\sin\theta}{\theta}\widehat{\mathbf{x}}+\frac{1-\cos\theta}{\theta^{2}}\widehat{\mathbf{x}}^{2}.

This yields,

(2.13) α(exp(xβEβ))=(θcosθsinθθ2)(αθ)𝐱^+sinθθα𝐱^+(θsinθ2(1cosθ)θ3)(αθ)𝐱^2+1cosθθ2α𝐱^2.\displaystyle\begin{aligned} \partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)}&=\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{2}}\right)(\partial_{\alpha}\theta)\widehat{\mathbf{x}}+\frac{\sin\theta}{\theta}\partial_{\alpha}\widehat{\mathbf{x}}\\ &\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{3}}\right)(\partial_{\alpha}\theta)\widehat{\mathbf{x}}^{2}+\frac{1-\cos\theta}{\theta^{2}}\partial_{\alpha}\widehat{\mathbf{x}}^{2}.\end{aligned}
Lemma 2.1.

For any 𝐱3\mathbf{x}\in\mathbb{R}^{3}, one has

(2.14) α𝐱^=Eα,α𝐱^2={Eα,𝐱^},\partial_{\alpha}\widehat{\mathbf{x}}=E_{\alpha},\qquad\partial_{\alpha}\widehat{\mathbf{x}}^{2}=\{E_{\alpha},\widehat{\mathbf{x}}\},

where {,}\{\cdot,\cdot\} denotes the anticommutator {A,B}:=AB+BA\{A,B\}:=AB+BA.

Proof.

The first identity follows from (1.4) and (2.4) directly, and the second identity can be derived as follows.

α𝐱^2=α(xβxγEβEγ)=xβEβEα+xγEαEγ=xβ{Eα,Eβ}={Eα,xβEβ}={Eα,𝐱^}.\partial_{\alpha}\widehat{\mathbf{x}}^{2}=\partial_{\alpha}\left(x_{\beta}x_{\gamma}E_{\beta}E_{\gamma}\right)=x_{\beta}E_{\beta}E_{\alpha}+x_{\gamma}E_{\alpha}E_{\gamma}=x_{\beta}\{E_{\alpha},E_{\beta}\}=\{E_{\alpha},x_{\beta}E_{\beta}\}=\{E_{\alpha},\widehat{\mathbf{x}}\}.

Remark 2.1.

As an application of Lemma 2.1, we can check that αR(𝐱)\partial_{\alpha}R(\mathbf{x}) given by (2.11) lies in the tangent space of SO(3)\mathrm{SO}(3) at R(𝐱)R(\mathbf{x}). Indeed, use (2.14) and αθ=xαθ\partial_{\alpha}\theta=\frac{x_{\alpha}}{\theta} in (2.12) to rewrite (2.13) as:

(2.15) α(exp(xβEβ))=(θcosθsinθθ3)xα𝐱^+sinθθEα+(θsinθ2(1cosθ)θ4)xα𝐱^2+1cosθθ2{Eα,𝐱^}.\displaystyle\begin{aligned} \partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)}&=\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{3}}\right)x_{\alpha}\widehat{\mathbf{x}}+\frac{\sin\theta}{\theta}E_{\alpha}\\ &\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)x_{\alpha}\widehat{\mathbf{x}}^{2}+\frac{1-\cos\theta}{\theta^{2}}\{E_{\alpha},\widehat{\mathbf{x}}\}.\end{aligned}

By a straightforward but tedious calculation, one can show that (exp(xγEγ))α(exp(xβEβ))\big{(}\exp(x_{\gamma}E_{\gamma})\big{)}^{\top}\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)} is a skew-symmetric matrix. This implies

αR(𝐱)TR(𝐱)SO(3).\partial_{\alpha}R(\mathbf{x})\in T_{R(\mathbf{x})}\mathrm{SO}(3).

In the following lemma, we list several other elementary results to be used later.

Lemma 2.2.

For 𝐱3\mathbf{x}\in\mathbb{R}^{3}, let 𝐱^\widehat{\mathbf{x}} be given by (2.5) and θ=𝐱\theta=\|\mathbf{x}\|. Then the following identities hold:

𝐱^3=θ2𝐱^andxα𝐱^2+𝐱^2Eα𝐱^=0, for all α=1,2,3.\widehat{\mathbf{x}}^{3}=-\theta^{2}\widehat{\mathbf{x}}\qquad\mbox{and}\qquad x_{\alpha}\widehat{\mathbf{x}}^{2}+\widehat{\mathbf{x}}^{2}E_{\alpha}\widehat{\mathbf{x}}=0,\qquad\text{ for all }\alpha=1,2,3.
Proof.

See Appendix A.1. ∎

2.3. The metric tensor

In this subsection, we derive explicit representations for the metric tensor gαβ(𝐱)g_{\alpha\beta}(\mathbf{x}) and its inverse gαβ(𝐱)g^{\alpha\beta}(\mathbf{x}). First, we study a set of elementary estimates in the following lemma.

Lemma 2.3.

For 𝐱3\mathbf{x}\in\mathbb{R}^{3}, let 𝐱^\widehat{\mathbf{x}} be given by (2.5) and θ=𝐱\theta=\|\mathbf{x}\|. Then, the following identities hold

(i)Eα,EβF=2δαβ,𝐱^,𝐱^F=2θ2,𝐱^,EαF=2xα,\displaystyle(i)~{}\langle E_{\alpha},E_{\beta}\rangle_{\mathrm{F}}=2\delta_{\alpha\beta},\qquad\langle\widehat{\mathbf{x}},\widehat{\mathbf{x}}\rangle_{\mathrm{F}}=2\theta^{2},\qquad\langle\widehat{\mathbf{x}},E_{\alpha}\rangle_{\mathrm{F}}=2x_{\alpha},
(ii)𝐱^2,𝐱^2F=2θ4,𝐱^2,{Eα,𝐱^}F=4xαθ2,{Eα,𝐱^},{Eβ,𝐱^}F=6xαxβ+2δαβθ2.\displaystyle(ii)~{}\langle\widehat{\mathbf{x}}^{2},\widehat{\mathbf{x}}^{2}\rangle_{\mathrm{F}}=2\theta^{4},\quad\langle\widehat{\mathbf{x}}^{2},\{E_{\alpha},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}=4x_{\alpha}\theta^{2},\quad\langle\{E_{\alpha},\widehat{\mathbf{x}}\},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}=6x_{\alpha}x_{\beta}+2\delta_{\alpha\beta}\theta^{2}.
Proof.

(i) By direct calculation, one can find

Eα,EβF=2δαβ.\langle E_{\alpha},E_{\beta}\rangle_{\mathrm{F}}=2\delta_{\alpha\beta}.

Then, one has

𝐱^,𝐱^F=xαxβEα,EβF=2xαxα=2θ2,\langle\widehat{\mathbf{x}},\widehat{\mathbf{x}}\rangle_{\mathrm{F}}=x_{\alpha}x_{\beta}\langle E_{\alpha},E_{\beta}\rangle_{\mathrm{F}}=2x_{\alpha}x_{\alpha}=2\theta^{2},

and

𝐱^,EαF=xβEβ,EαF=xβEβ,EαF=2xα.\langle\widehat{\mathbf{x}},E_{\alpha}\rangle_{\mathrm{F}}=\langle x_{\beta}E_{\beta},E_{\alpha}\rangle_{\mathrm{F}}=x_{\beta}\langle E_{\beta},E_{\alpha}\rangle_{\mathrm{F}}=2x_{\alpha}.

(ii) Since the derivations for identities in (ii) are rather lengthy, we present them in Appendix A.2. ∎

Proposition 2.1.

For 𝐱3\mathbf{x}\in\mathbb{R}^{3}, let 𝐱^\widehat{\mathbf{x}} be given by (2.5) and θ=𝐱\theta=\|\mathbf{x}\|. Then, the metric tensor and its inverse are given as follows.

(i)gαβ(𝐱)=(2cosθ2+θ2θ4)xαxβ+2(1cosθ)θ2δαβ,(ii)gαβ(𝐱)=(2cosθ2+θ22θ2(cosθ1))xαxβ+θ22(1cosθ)δαβ.\displaystyle\begin{aligned} &(i)~{}g_{\alpha\beta}(\mathbf{x})=\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)x_{\alpha}x_{\beta}+\frac{2(1-\cos\theta)}{\theta^{2}}\delta_{\alpha\beta},\\ &(ii)~{}g^{\alpha\beta}(\mathbf{x})=\left(\frac{2\cos\theta-2+\theta^{2}}{2\theta^{2}(\cos\theta-1)}\right)x_{\alpha}x_{\beta}+\frac{\theta^{2}}{2(1-\cos\theta)}\delta^{\alpha\beta}.\end{aligned}
Proof.

(i) It follows from (2.2) and

αR(𝐱)=R0α(exp(xβEβ)),whereα=xα,\partial_{\alpha}R(\mathbf{x})=R_{0}\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)},\quad\mbox{where}~{}\partial_{\alpha}=\partial_{x_{\alpha}},

that

(2.16) gαβ(𝐱)=g(αR(𝐱),βR(𝐱))=12αR(𝐱),βR(𝐱)F=12α(exp(xβEβ)),α(exp(xβEβ))F\displaystyle\begin{aligned} g_{\alpha\beta}(\mathbf{x})&=g(\partial_{\alpha}R(\mathbf{x}),\partial_{\beta}R(\mathbf{x}))=\frac{1}{2}\Big{\langle}\partial_{\alpha}R(\mathbf{x}),\partial_{\beta}R(\mathbf{x})\Big{\rangle}_{\mathrm{F}}\\ &=\frac{1}{2}\Big{\langle}\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)},\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)}\Big{\rangle}_{\mathrm{F}}\end{aligned}

On the other hand, since EαE_{\alpha} and 𝐱^\widehat{\mathbf{x}} are skew-symmetric, {Eα,𝐱^}\{E_{\alpha},\widehat{\mathbf{x}}\} is also symmetric. Now, we decompose (2.15) in skew-symmetric and symmetric parts:

(2.17) α(exp(xβEβ))=Aα+Sα,\partial_{\alpha}\big{(}\exp(x_{\beta}E_{\beta})\big{)}=A_{\alpha}+S_{\alpha},

where AαA_{\alpha} and SαS_{\alpha} are skew-symmetric and symmetric matrices, respectively. One finds:

Aα\displaystyle A_{\alpha} =(θcosθsinθθ3)xα𝐱^+sinθθEα,\displaystyle=\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{3}}\right)x_{\alpha}\widehat{\mathbf{x}}+\frac{\sin\theta}{\theta}E_{\alpha},
Sα\displaystyle S_{\alpha} =(θsinθ2(1cosθ)θ4)xα𝐱^2+1cosθθ2{Eα,𝐱^}.\displaystyle=\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)x_{\alpha}\widehat{\mathbf{x}}^{2}+\frac{1-\cos\theta}{\theta^{2}}\{E_{\alpha},\widehat{\mathbf{x}}\}.

Next, we combine (2.16) and (2.17) to find

(2.18) gαβ(𝐱)=12αR(𝐱),βR(𝐱)F=12Aα,AβF+12Sα,SβF,g_{\alpha\beta}(\mathbf{x})=\frac{1}{2}\langle\partial_{\alpha}R(\mathbf{x}),\partial_{\beta}R(\mathbf{x})\rangle_{\mathrm{F}}=\frac{1}{2}\langle A_{\alpha},A_{\beta}\rangle_{\mathrm{F}}+\frac{1}{2}\langle S_{\alpha},S_{\beta}\rangle_{\mathrm{F}},

where we used that A,SF=0\langle A,S\rangle_{\mathrm{F}}=0 for any skew-symmetric matrix AA and symmetric matrix SS. For the estimation of the two terms in the R.H.S. of (2.18), we use Lemma 2.3.

\bullet Estimate of Aα,AβF\langle A_{\alpha},A_{\beta}\rangle_{\mathrm{F}}: We use Lemma 2.3 (i) to find:

(2.19) Aα,AβF=(θcosθsinθθ3)2xαxβ𝐱^,𝐱^F+(θcosθsinθθ3)sinθθ(xα𝐱^,EβF+xβ𝐱^,EαF)+(sinθθ)2Eα,EβF=2(θcosθsinθθ2)2xαxβ+4(θcosθsinθθ3)sinθθxαxβ+2(sinθθ)2δαβ=2(θ2cos2θsin2θθ4)xαxβ+2(sinθθ)2δαβ.\displaystyle\begin{aligned} &\langle A_{\alpha},A_{\beta}\rangle_{\mathrm{F}}\\ &=\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{3}}\right)^{2}x_{\alpha}x_{\beta}\langle\widehat{\mathbf{x}},\widehat{\mathbf{x}}\rangle_{\mathrm{F}}\\ &\quad+\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{3}}\right)\frac{\sin\theta}{\theta}\big{(}x_{\alpha}\langle\widehat{\mathbf{x}},E_{\beta}\rangle_{\mathrm{F}}+x_{\beta}\langle\widehat{\mathbf{x}},E_{\alpha}\rangle_{\mathrm{F}}\big{)}+\left(\frac{\sin\theta}{\theta}\right)^{2}\langle E_{\alpha},E_{\beta}\rangle_{\mathrm{F}}\\ &=2\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{2}}\right)^{2}x_{\alpha}x_{\beta}+4\left(\frac{\theta\cos\theta-\sin\theta}{\theta^{3}}\right)\frac{\sin\theta}{\theta}x_{\alpha}x_{\beta}+2\left(\frac{\sin\theta}{\theta}\right)^{2}\delta_{\alpha\beta}\\ &=2\left(\frac{\theta^{2}\cos^{2}\theta-\sin^{2}\theta}{\theta^{4}}\right)x_{\alpha}x_{\beta}+2\left(\frac{\sin\theta}{\theta}\right)^{2}\delta_{\alpha\beta}.\end{aligned}

\bullet Estimate of Sα,SβF\langle S_{\alpha},S_{\beta}\rangle_{\mathrm{F}}: We use Lemma 2.3 (ii) to find:

Sα,SβF=(θsinθ2(1cosθ)θ4)2xαxβ𝐱^2,𝐱^2F+(1cosθθ2)2{Eα,𝐱^},{Eβ,𝐱^}F+(θsinθ2(1cosθ)θ4)1cosθθ2(xα𝐱^2,{Eβ,𝐱^}F+xβ𝐱^2,{Eα,𝐱^}F)=2(θsinθ2(1cosθ)θ2)2xαxβ+(1cosθθ2)2(6xαxβ+2δαβθ2)+(θsinθ2(1cosθ)θ2)1cosθθ2(8xαxβ).\displaystyle\begin{aligned} &\langle S_{\alpha},S_{\beta}\rangle_{\mathrm{F}}\\ &=\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)^{2}x_{\alpha}x_{\beta}\langle\widehat{\mathbf{x}}^{2},\widehat{\mathbf{x}}^{2}\rangle_{\mathrm{F}}+\left(\frac{1-\cos\theta}{\theta^{2}}\right)^{2}\langle\{E_{\alpha},\widehat{\mathbf{x}}\},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}\\ &\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)\frac{1-\cos\theta}{\theta^{2}}(x_{\alpha}\langle\widehat{\mathbf{x}}^{2},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}+x_{\beta}\langle\widehat{\mathbf{x}}^{2},\{E_{\alpha},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}})\\ &=2\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{2}}\right)^{2}x_{\alpha}x_{\beta}+\left(\frac{1-\cos\theta}{\theta^{2}}\right)^{2}(6x_{\alpha}x_{\beta}+2\delta_{\alpha\beta}\theta^{2})\\ &\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{2}}\right)\frac{1-\cos\theta}{\theta^{2}}(8x_{\alpha}x_{\beta}).\end{aligned}

By grouping similar terms, one has

(2.20) Sα,SβF=2(θ2sin2θ(1cosθ)2θ4)xαxβ+2(1cosθ)2θ2δαβ\langle S_{\alpha},S_{\beta}\rangle_{\mathrm{F}}=2\left(\frac{\theta^{2}\sin^{2}\theta-(1-\cos\theta)^{2}}{\theta^{4}}\right)x_{\alpha}x_{\beta}+2\,\frac{(1-\cos\theta)^{2}}{\theta^{2}}\delta_{\alpha\beta}

Finally, we combine (2.18), (LABEL:eqn:A-prod) and (2.20) to obtain

(2.21) gαβ(𝐱)=(2cosθ2+θ2θ4)xαxβ+2(1cosθ)θ2δαβ.g_{\alpha\beta}(\mathbf{x})=\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)x_{\alpha}x_{\beta}+\frac{2(1-\cos\theta)}{\theta^{2}}\delta_{\alpha\beta}.

(ii) We postpone the calculation of the inverse of (gαβ(𝐱))(g_{\alpha\beta}(\mathbf{x})) to Appendix A.3. ∎

2.4. Christoffel symbols and geodesic equations.

A direct calculation in Appendix A.4 and formula for the Christoffel symbols

Γαβγ=12gγκ(βgκα+αgκβκgαβ),\Gamma^{\gamma}_{\alpha\beta}=\frac{1}{2}g^{\gamma\kappa}(\partial_{\beta}g_{\kappa\alpha}+\partial_{\alpha}g_{\kappa\beta}-\partial_{\kappa}g_{\alpha\beta}),

yield

(2.22) Γαβγ(𝐱)=(sinθ+θ)(cosθ1)+θ2sinθθ5(cosθ1)xαxβxγ+2cosθ2+θsinθ2θ2(1cosθ)(xαδβγ+xβδαγ)sinθθθ3δαβxγ.\displaystyle\begin{aligned} \Gamma^{\gamma}_{\alpha\beta}(\mathbf{x})&=\frac{(\sin\theta+\theta)(\cos\theta-1)+\theta^{2}\sin\theta}{\theta^{5}(\cos\theta-1)}x_{\alpha}x_{\beta}x_{\gamma}\\[3.0pt] &\quad+\frac{2\cos\theta-2+\theta\sin\theta}{2\theta^{2}(1-\cos\theta)}(x_{\alpha}\delta_{\beta}^{\gamma}+x_{\beta}\delta_{\alpha}^{\gamma})-\frac{\sin\theta-\theta}{\theta^{3}}\delta_{\alpha\beta}x_{\gamma}.\end{aligned}

One advantage of working with exponential coordinates is that geodesics are images of straight lines under the parametrization map. To check the expressions (2.22), it can be shown that for a fixed vector 𝐮3\mathbf{u}\in\mathbb{R}^{3}, the straight line 𝐱(t)=t𝐮\mathbf{x}(t)=t\mathbf{u} satisfies the geodesic equations

d2xγdt2+Γαβγ(𝐱(t))dxαdtdxβdt=0,\frac{d^{2}x_{\gamma}}{dt^{2}}+\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}(t))\frac{dx_{\alpha}}{dt}\frac{dx_{\beta}}{dt}=0,

for all γ=1,2,3\gamma=1,2,3. The second derivative is zero, so we only need to check

(2.23) Γαβγ(𝐱(t))uαuβ=0.\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}(t))u_{\alpha}u_{\beta}=0.

Detailed arguments will be given in Appendix A.5.

3. Parallel transport on SO(3)\mathrm{SO}(3)

In this section we derive an explicit formula for the parallel transport on SO(3)\mathrm{SO}(3) along a geodesic, and provide its geometric interpretation.

3.1. Parallel transport

Consider a geodesic curve on SO(3)\mathrm{SO}(3) given in exponential coordinates by xα(t)=tuαx_{\alpha}(t)=tu_{\alpha}, for a fixed 𝐮=(u1,u2,u3)3\mathbf{u}=(u_{1},u_{2},u_{3})\in\mathbb{R}^{3}, and the parallel transport along this geodesic of a vector of components vα(t)v_{\alpha}(t). Then, the equations for parallel transport are given by

dvγdt+Γαβγ(𝐱(t))x˙αvβ=0,γ=1,2,3.\frac{dv_{\gamma}}{dt}+\Gamma_{\alpha\beta}^{\gamma}(\mathbf{x}(t))\dot{x}_{\alpha}v_{\beta}=0,\qquad\gamma=1,2,3.

Since x˙α=uα\dot{x}_{\alpha}=u_{\alpha} along the geodesic, we can write the equations as

(3.1) v˙γ+Γαβγ(𝐱(t))uαvβ=0,γ=1,2,3.\dot{v}_{\gamma}+\Gamma_{\alpha\beta}^{\gamma}(\mathbf{x}(t))u_{\alpha}v_{\beta}=0,\qquad\gamma=1,2,3.

Next, we calculate Γαβγ(𝐱(t))uα\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}(t))u_{\alpha} with the Christoffel symbols given by (2.22). By direct substitution of 𝐱(t)=t𝐮\mathbf{x}(t)=t\mathbf{u} in (2.22) (in particular, use 𝐱(t)=𝐮t\|\mathbf{x}(t)\|=\|\mathbf{u}\|t), we obtain

(3.2) Γαβγ(𝐱(t))uα=(sin(𝐮t)+t𝐮)(cos(𝐮t)1)+t2𝐮2sin(𝐮t)t5𝐮5(cos(𝐮t)1)t3uαuβuγuα+2cos(𝐮t)2+t𝐮sin(𝐮t)2t2𝐮2(1cos(𝐮t))(tuαδβγ+tuβδαγ)uαsin(𝐮t)t𝐮t3𝐮3δαβtuγuα=2cos(𝐮t)2+t𝐮sin(𝐮t)2t(1cos(𝐮t))(δβγuβuγ𝐮2),\displaystyle\begin{aligned} &\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}(t))u_{\alpha}\\ &\hskip 5.69046pt=\frac{(\sin(\|\mathbf{u}\|t)+t\|\mathbf{u}\|)(\cos(\|\mathbf{u}\|t)-1)+t^{2}\|\mathbf{u}\|^{2}\sin(\|\mathbf{u}\|t)}{t^{5}\|\mathbf{u}\|^{5}(\cos(\|\mathbf{u}\|t)-1)}t^{3}u_{\alpha}u_{\beta}u_{\gamma}u_{\alpha}\\[3.0pt] &\hskip 5.69046pt+\frac{2\cos(\|\mathbf{u}\|t)-2+t\|\mathbf{u}\|\sin(\|\mathbf{u}\|t)}{2t^{2}\|\mathbf{u}\|^{2}(1-\cos(\|\mathbf{u}\|t))}(tu_{\alpha}\delta_{\beta}^{\gamma}+tu_{\beta}\delta_{\alpha}^{\gamma})u_{\alpha}-\frac{\sin(\|\mathbf{u}\|t)-t\|\mathbf{u}\|}{t^{3}\|\mathbf{u}\|^{3}}\delta_{\alpha\beta}tu_{\gamma}u_{\alpha}\\[3.0pt] &\hskip 5.69046pt=\frac{2\cos(\|\mathbf{u}\|t)-2+t\|\mathbf{u}\|\sin(\|\mathbf{u}\|t)}{2t(1-\cos(\|\mathbf{u}\|t))}\left(\delta_{\beta}^{\gamma}-\frac{u_{\beta}u_{\gamma}}{\|\mathbf{u}\|^{2}}\right),\end{aligned}

where the second equal sign follows from elementary algebra and uαuα=𝐮2u_{\alpha}u_{\alpha}=\|\mathbf{u}\|^{2}.

For calculations in this section we reset the notation for θ\theta used in Section 3, and denote:

θ:=𝐮,𝐧:=𝐮θ.\theta:=\|\mathbf{u}\|,\quad\mathbf{n}:=\frac{\mathbf{u}}{\theta}.

With this notation, rewrite (LABEL:AC-1) as

Γαβγ(𝐱(t))uα=2cos(θt)2+(θt)sin(θt)2t(1cos(θt))(δβγuβuγθ2).\Gamma_{\alpha\beta}^{\gamma}(\mathbf{x}(t))u_{\alpha}=\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t))}\left(\delta_{\beta}^{\gamma}-\frac{u_{\beta}u_{\gamma}}{\theta^{2}}\right).

Then, we use the above expression in the equations (3.1) for parallel transport to find

(3.3) 0=v˙γ+(2cos(θt)2+(θt)sin(θt)2t(1cos(θt)))(δβγuβuγθ2)vβ=v˙γ+(2cos(θt)2+(θt)sin(θt)2t(1cos(θt)))(vγuγθ2(uβvβ)).\displaystyle\begin{aligned} 0&=\dot{v}_{\gamma}+\left(\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t))}\right)\left(\delta_{\beta}^{\gamma}-\frac{u_{\beta}u_{\gamma}}{\theta^{2}}\right)v_{\beta}\\ &=\dot{v}_{\gamma}+\left(\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t))}\right)\left(v_{\gamma}-\frac{u_{\gamma}}{\theta^{2}}(u_{\beta}v_{\beta})\right).\end{aligned}

Multiply both sides of (3.3) by uγu_{\gamma} and sum up the resulting relation over γ\gamma to obtain

0\displaystyle 0 =v˙γuγ+(2cos(θt)2+(θt)sin(θt)2t(1cos(θt)))(vγuγuγuγθ2(uβvβ))=0.\displaystyle=\dot{v}_{\gamma}u_{\gamma}+\left(\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t))}\right)\underbrace{\left(v_{\gamma}u_{\gamma}-\frac{u_{\gamma}u_{\gamma}}{\theta^{2}}(u_{\beta}v_{\beta})\right)}_{=0}.

The second term in the R.H.S. above vanishes, and the vector 𝐮\mathbf{u} is constant. This leads to

ddt(vγuγ)=0,\frac{d}{dt}(v_{\gamma}u_{\gamma})=0,

and hence vγ(t)uγv_{\gamma}(t)u_{\gamma} is a conserved quantity:

(3.4) 𝒞=vγ(t)uγ=vγ(0)uγ,\mathcal{C}=v_{\gamma}(t)u_{\gamma}=v_{\gamma}(0)u_{\gamma},

for a constant 𝒞\mathcal{C}. The conservation property in (3.3) implies

(3.5) 0=v˙γ(t)+(2cos(θt)2+(θt)sin(θt)2t(1cos(θt)))(vγ(t)𝒞uγθ2).\displaystyle 0=\dot{v}_{\gamma}(t)+\left(\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t))}\right)\left(v_{\gamma}(t)-\frac{\mathcal{C}u_{\gamma}}{\theta^{2}}\right).

To find the exact solution of (3.5), we define the function ff by

(3.6) f(t):={1cos(θt)t2,ift>0,θ2,ift=0.f(t):=\begin{cases}\displaystyle\sqrt{\frac{1-\cos(\theta t)}{t^{2}}},\quad&\text{if}\quad t>0,\\[5.0pt] \displaystyle\frac{\theta}{\sqrt{2}},&\text{if}\quad t=0.\end{cases}

Here we can easily check that ff is a C1C^{1} function defined on [0,)[0,\infty), and ff satisfies

f(t)f(t)=2cos(θt)2+(θt)sin(θt)2t(1cos(θt),\frac{f^{\prime}(t)}{f(t)}=\frac{2\cos(\theta t)-2+(\theta t)\sin(\theta t)}{2t(1-\cos(\theta t)},

so one can write (3.5) as

0=f(t)v˙γ(t)+f(t)(vγ(t)𝒞uγθ2).0=f(t)\dot{v}_{\gamma}(t)+f^{\prime}(t)\left(v_{\gamma}(t)-\frac{\mathcal{C}u_{\gamma}}{\theta^{2}}\right).

Furthermore, we have

0=ddt(f(t)vγ(t)𝒞uγf(t)θ2).0=\frac{d}{dt}\left(f(t)v_{\gamma}(t)-\frac{\mathcal{C}u_{\gamma}f(t)}{\theta^{2}}\right).

This implies

f(t)vγ(t)𝒞uγf(t)θ2=limτ0(f(τ)vγ(τ)𝒞uγf(τ)θ2)=θ2vγ(0)𝒞uγ2θ,f(t)v_{\gamma}(t)-\frac{\mathcal{C}u_{\gamma}f(t)}{\theta^{2}}=\lim_{\tau\to 0}\left(f(\tau)v_{\gamma}(\tau)-\frac{\mathcal{C}u_{\gamma}f(\tau)}{\theta^{2}}\right)=\frac{\theta}{\sqrt{2}}v_{\gamma}(0)-\frac{\mathcal{C}u_{\gamma}}{\sqrt{2}\theta},

which yields the explicit solution of the parallel transport equations (3.1):

(3.7) vγ(t)=1f(t)(θ2vγ(0)𝒞uγ2θ)+𝒞uγθ2,v_{\gamma}(t)=\frac{1}{f(t)}\left(\frac{\theta}{\sqrt{2}}v_{\gamma}(0)-\frac{\mathcal{C}u_{\gamma}}{\sqrt{2}\theta}\right)+\frac{\mathcal{C}u_{\gamma}}{\theta^{2}},

with θ=𝐮\theta=\|\mathbf{u}\| and 𝒞\mathcal{C} given by (3.4).

3.2. Coordinate-free closed form expression

We use the explicit formula (3.7) to derive a coordinate free expression for parallel transport along a geodesic. We fix R0,R1SO(3)R_{0},R_{1}\in\mathrm{SO}(3), with d(R0,R1)<πd(R_{0},R_{1})<\pi, and consider the unique geodesic path between them. We set exponential coordinates centered at R0R_{0}, and also set 𝐮3\mathbf{u}\in\mathbb{R}^{3} by

𝐮^=log(R0R1).\widehat{\mathbf{u}}=\log(R_{0}^{\top}R_{1}).

Equivalently one has

logR0R1=R0𝐮^.\log_{R_{0}}R_{1}=R_{0}\widehat{\mathbf{u}}.

The geodesic curve between the two matrices is given by

:[0,1]SO(3),(t)=R0exp(t𝐮^).\mathcal{R}:[0,1]\to\mathrm{SO}(3),\quad\mathcal{R}(t)=R_{0}\exp(t\widehat{\mathbf{u}}).

In particular, θ=𝐮\theta=\|\mathbf{u}\| represents the geodesic distance between R0R_{0} and R1R_{1} (see (2.8)):

θ=d(R0,R1)=acos(tr(R0R1)12).\theta=d(R_{0},R_{1})=\operatorname{acos}\left(\frac{\operatorname{tr}(R_{0}^{\top}R_{1})-1}{2}\right).

Also, by (2.7), the unit vector 𝐧=𝐮θ\mathbf{n}=\frac{\mathbf{u}}{\theta} satisfies:

𝐧^=12sinθ(R0R1R1R0).\widehat{\mathbf{n}}=\frac{1}{2\sin\theta}(R_{0}^{\top}R_{1}-R_{1}^{\top}R_{0}).

For notational convenience, we adopt the following notation for the tangent vectors in exponential coordinates (see (2.11) and (2.15)):

Xα(R(𝐱)):=αR(𝐱).X_{\alpha}(R(\mathbf{x})):=\partial_{\alpha}R(\mathbf{x}).

Then, it follows from (2.11) and (2.15) that

(3.8) Xα(R0)=R0EαandXα(R1)=R0α(exp(xβEβ))|𝐱=𝐮=(sinθ+θcosθθ3)uαR0𝐮^+sinθθR0Eα+(θsinθ2(1cosθ)θ4)uαR0𝐮^2+1cosθθ2R0{Eα,𝐮^}.\displaystyle\begin{aligned} &X_{\alpha}(R_{0})=R_{0}E_{\alpha}\quad\mbox{and}\\[3.0pt] &X_{\alpha}(R_{1})=R_{0}\partial_{\alpha}\left(\exp\left(x_{\beta}E_{\beta}\right)\right)\big{|}_{\mathbf{x}=\mathbf{u}}\\[3.0pt] &\hskip 5.69046pt=\left(\frac{-\sin\theta+\theta\cos\theta}{\theta^{3}}\right)u_{\alpha}R_{0}\widehat{\mathbf{u}}+\frac{\sin\theta}{\theta}R_{0}E_{\alpha}+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)u_{\alpha}R_{0}\widehat{\mathbf{u}}^{2}\\ &\hskip 5.69046pt+\frac{1-\cos\theta}{\theta^{2}}R_{0}\{E_{\alpha},\widehat{\mathbf{u}}\}.\end{aligned}

Now take a tangent vector V(0)TR0SO(3)V(0)\in T_{R_{0}}\mathrm{SO}(3) given in components by

(3.9) V(0)=vα(0)Xα(R0).V(0)=v_{\alpha}(0)X_{\alpha}(R_{0}).

According to the calculations performed in Section 3.1 (see equation (3.7)), the parallel transport of V(0)V(0) from R0R_{0} to R1R_{1} along the geodesic curve is

(3.10) V(1)=vα(1)Xα(R1),V(1)=v_{\alpha}(1)X_{\alpha}(R_{1}),

where

vα(1)=1f(1)(θ2vα(0)𝒞uα2θ)+𝒞uαθ2.v_{\alpha}(1)=\frac{1}{f(1)}\left(\frac{\theta}{\sqrt{2}}v_{\alpha}(0)-\frac{\mathcal{C}u_{\alpha}}{\sqrt{2}\theta}\right)+\frac{\mathcal{C}u_{\alpha}}{\theta^{2}}.

Here, 𝒞\mathcal{C} is the constant of motion given by (3.4). Using (3.6) and some simple manipulations we write the components of V(1)V(1) as

(3.11) vα(1)=11cosθ(θ2vα(0)𝒞uα2θ)+𝒞uαθ2=θvα(0)2sinθ2+𝒞uα(1θ212θsinθ2).\displaystyle v_{\alpha}(1)=\sqrt{\frac{1}{1-\cos\theta}}\left(\frac{\theta}{\sqrt{2}}v_{\alpha}(0)-\frac{\mathcal{C}u_{\alpha}}{\sqrt{2}\theta}\right)+\frac{\mathcal{C}u_{\alpha}}{\theta^{2}}=\frac{\theta v_{\alpha}(0)}{2\sin\frac{\theta}{2}}+\mathcal{C}u_{\alpha}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right).

One can then use (LABEL:eqn:XalphaR1)2\eqref{eqn:XalphaR1}_{2} and (3.11) in (3.10) (see calculations in Appendix A.6) to find that the parallel-transported vector V(1)V(1) has the following coordinate-free expression:

(3.12) V(1)=𝒞R0𝐮^(cosθcosθ2θ2)+𝒞R0𝐮^2(sinθ2sinθ2θ3)+sinθ2θ(V(0)𝐮^+R0𝐮^R0V(0))+cosθ2V(0).\displaystyle\begin{aligned} V(1)&=\mathcal{C}R_{0}\widehat{\mathbf{u}}\left(\frac{\cos\theta-\cos\frac{\theta}{2}}{\theta^{2}}\right)+\mathcal{C}R_{0}\widehat{\mathbf{u}}^{2}\left(\frac{\sin\theta-2\sin\frac{\theta}{2}}{\theta^{3}}\right)\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(V(0)\widehat{\mathbf{u}}+R_{0}\widehat{\mathbf{u}}R_{0}^{\top}V(0)\right)+\cos\frac{\theta}{2}V(0).\end{aligned}

3.3. A geometric interpretation

By (2.1), the tangent vectors V(0)V(0) and V(1)V(1) take the following form:

V(0)=R0A0, and V(1)=R1A1,V(0)=R_{0}A_{0},\qquad\text{ and }\qquad V(1)=R_{1}A_{1},

where A0A_{0} and A1A_{1} are skew-symmetric matrices. To provide better insight, which leads to a nice geometric interpretation of parallel transport, we now express (3.12) in terms of these skew-symmetric matrices.

Therefore, we set

(3.13) A0=R0V(0),A1=R1V(1),A_{0}=R_{0}^{\top}V(0),\qquad A_{1}=R_{1}^{\top}V(1),

and also define 𝐚0\mathbf{a}_{0} and 𝐚1\mathbf{a}_{1} such that 𝐚^0=A0\widehat{\mathbf{a}}_{0}=A_{0} and 𝐚^1=A1\widehat{\mathbf{a}}_{1}=A_{1} For convenience of notation, denote

R=R0R1.R=R_{0}^{\top}R_{1}.

Then, it follows from (3.12) that

(3.14) A1=𝒞R𝐮^(cosθcosθ2θ2)+𝒞R𝐮^2(sinθ2sinθ2θ3)+sinθ2θR(A0𝐮^+𝐮^A0)+cosθ2RA0.\displaystyle\begin{aligned} A_{1}&=\mathcal{C}R^{\top}\widehat{\mathbf{u}}\left(\frac{\cos\theta-\cos\frac{\theta}{2}}{\theta^{2}}\right)+\mathcal{C}R^{\top}\widehat{\mathbf{u}}^{2}\left(\frac{\sin\theta-2\sin\frac{\theta}{2}}{\theta^{3}}\right)\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}R^{\top}\left(A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}A_{0}\right)+\cos\frac{\theta}{2}R^{\top}A_{0}.\end{aligned}

We take the transpose in Rodrigues’s formula (2.6) (applied to R=R0R1R=R_{0}^{\top}R_{1}) to get

(3.15) R=Isinθθ𝐮^+1cosθθ2𝐮^2.R^{\top}=I-\frac{\sin\theta}{\theta}\widehat{\mathbf{u}}+\frac{1-\cos\theta}{\theta^{2}}\widehat{\mathbf{u}}^{2}.

Hence, one has

(3.16) R𝐮^=cosθ𝐮^sinθθ𝐮^2, and R𝐮^2=cosθ𝐮^2+θsinθ𝐮^,R^{\top}\widehat{\mathbf{u}}=\cos\theta\widehat{\mathbf{u}}-\frac{\sin\theta}{\theta}\widehat{\mathbf{u}}^{2},\qquad\text{ and }\qquad R^{\top}\widehat{\mathbf{u}}^{2}=\cos\theta\widehat{\mathbf{u}}^{2}+\theta\sin\theta\widehat{\mathbf{u}},

where we used that 𝐮^3=θ2𝐮^\widehat{\mathbf{u}}^{3}=-\theta^{2}\widehat{\mathbf{u}} (see Lemma 2.1).

By using (3.15) and (3.16) in (3.14), after some manipulations (see Appendix A.7), we derive the following expression for A1A_{1}:

(3.17) A1=𝒞(1cosθcosθ22sinθsinθ2θ2)𝐮^+sinθ2θ(A0𝐮^𝐮^A0)2sin2θ2cosθ2θ2𝐮^A0𝐮^+cosθ2A0.\displaystyle\begin{aligned} A_{1}&=\mathcal{C}\left(\frac{1-\cos\theta\cos\frac{\theta}{2}-2\sin\theta\sin\frac{\theta}{2}}{\theta^{2}}\right)\widehat{\mathbf{u}}+\frac{\sin\frac{\theta}{2}}{\theta}\left(A_{0}\widehat{\mathbf{u}}-\widehat{\mathbf{u}}A_{0}\right)\\ &\quad-\frac{2\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}}{\theta^{2}}\widehat{\mathbf{u}}A_{0}\widehat{\mathbf{u}}+\cos\frac{\theta}{2}A_{0}.\end{aligned}

The expression for A1A_{1} can be further simplified using the following lemma.

Lemma 3.1.

Let 𝐱,𝐲3\mathbf{x},\mathbf{y}\in\mathbb{R}^{3} and their corresponding skew-symmetric matrices 𝐱^,𝐲^𝔰𝔬(3)\widehat{\mathbf{x}},\widehat{\mathbf{y}}\in\mathfrak{so}(3). The following identities hold:

(3.18) tr(𝐱^𝐲^)=2𝐱𝐲,𝐱^𝐲^𝐲^𝐱^=𝐱×𝐲^,𝐱^𝐲^𝐱^=(𝐱𝐲)𝐱^.\mathrm{tr}(\widehat{\mathbf{x}}\widehat{\mathbf{y}})=-2\,\mathbf{x}\cdot\mathbf{y},\qquad\widehat{\mathbf{x}}\widehat{\mathbf{y}}-\widehat{\mathbf{y}}\widehat{\mathbf{x}}=\widehat{\mathbf{x}\times\mathbf{y}},\qquad\widehat{\mathbf{x}}\widehat{\mathbf{y}}\widehat{\mathbf{x}}=-(\mathbf{x}\cdot\mathbf{y})\widehat{\mathbf{x}}.
Proof.

See Appendix A.8. ∎

First note that by (LABEL:eqn:XalphaR1)1\eqref{eqn:XalphaR1}_{1}, (3.9) and (3.13), we have:

A0=vα(0)R0Xα(R0)=vα(0)Eα,A_{0}=v_{\alpha}(0)R_{0}^{\top}X_{\alpha}(R_{0})=v_{\alpha}(0)E_{\alpha},

and hence, by (3.4),

𝒞=uαvα(0)=12𝐮^,A0F.\mathcal{C}=u_{\alpha}v_{\alpha}(0)=\frac{1}{2}\langle\widehat{\mathbf{u}},A_{0}\rangle_{\mathrm{F}}.

Then, it follows from (3.18)1\eqref{eqn:id1}_{1} and (3.18)3\eqref{eqn:id1}_{3} that

𝒞=12tr(𝐮^A0)=𝐮𝐚0, and 𝐮^A0𝐮^=(𝐮𝐚0)𝐮.\mathcal{C}=-\frac{1}{2}\mathrm{tr}(\widehat{\mathbf{u}}A_{0})=\mathbf{u}\cdot\mathbf{a}_{0},\qquad\text{ and }\qquad\widehat{\mathbf{u}}A_{0}\widehat{\mathbf{u}}=-(\mathbf{u}\cdot\mathbf{a}_{0})\mathbf{u}.

Using the equations above, combine the first and third terms in the R.H.S. of (3.17) to get

(3.19) A1=𝒞(1cosθcosθ2sinθsinθ2θ2)𝐮^+sinθ2θ(A0𝐮^𝐮^A0)+cosθ2A0=(𝐧𝐚0)(1cosθ2)𝐧^+sinθ2(A0𝐧^𝐧^A0)+cosθ2A0,\displaystyle\begin{aligned} A_{1}&=\mathcal{C}\left(\frac{1-\cos\theta\cos\frac{\theta}{2}-\sin\theta\sin\frac{\theta}{2}}{\theta^{2}}\right)\widehat{\mathbf{u}}+\frac{\sin\frac{\theta}{2}}{\theta}\left(A_{0}\widehat{\mathbf{u}}-\widehat{\mathbf{u}}A_{0}\right)+\cos\frac{\theta}{2}A_{0}\\ &=(\mathbf{n}\cdot\mathbf{a}_{0})\left(1-\cos\frac{\theta}{2}\right)\widehat{\mathbf{n}}+\sin\frac{\theta}{2}\left(A_{0}\widehat{\mathbf{n}}-\widehat{\mathbf{n}}A_{0}\right)+\cos\frac{\theta}{2}A_{0},\end{aligned}

where we also used

cosθcosθ2+sinθsinθ2=cosθ2, and 𝐧=𝐮θ.\cos\theta\cos\frac{\theta}{2}+\sin\theta\sin\frac{\theta}{2}=\cos\frac{\theta}{2},\quad\text{ and }\quad\mathbf{n}=\frac{\mathbf{u}}{\theta}.

Using (3.18)2\eqref{eqn:id1}_{2}, we can write (3.19) as:

(3.20) A1=(1cosθ2)(𝐧𝐚0)𝐧^+sinθ2𝐚0×𝐧^+cosθ2A0.A_{1}=\left(1-\cos\frac{\theta}{2}\right)(\mathbf{n}\cdot\mathbf{a}_{0})\,\widehat{\mathbf{n}}+\sin\frac{\theta}{2}\,\widehat{\mathbf{a}_{0}\times\mathbf{n}}+\cos\frac{\theta}{2}A_{0}.

By applying the vee operator to (3.20) we then find:

(3.21) 𝐚1=(1cosθ2)(𝐧𝐚0)𝐧+sinθ2𝐚0×𝐧+cosθ2𝐚0.\mathbf{a}_{1}=\left(1-\cos\frac{\theta}{2}\right)(\mathbf{n}\cdot\mathbf{a}_{0})\,\mathbf{n}+\sin\frac{\theta}{2}\,\mathbf{a}_{0}\times\mathbf{n}+\cos\frac{\theta}{2}\,\mathbf{a}_{0}.
Remark 3.1.

Remarkably, (3.21) is in the form of Rodrigues’ rotation formula, with 𝐚1\mathbf{a}_{1} representing the rotation (according to the right-hand-rule) of 𝐚0\mathbf{a}_{0} about the axis 𝐧-\mathbf{n} by an angle θ2\frac{\theta}{2}. For the case in which 𝐚0\mathbf{a}_{0} is in the direction of 𝐧\mathbf{n}, we have 𝐚1=𝐚0,\mathbf{a}_{1}=\mathbf{a}_{0}, as expected.

Parallel transport by general theory on Lie groups. In [35, Chapter 2], listed as an exercise, one can find a general expression for parallel transport along geodesics on connected Lie groups. Specifically, consider a connected Lie group GG with Lie algebra 𝔤\mathfrak{g}, endowed with an affine connection that is invariant under left and right translations and under the map gg1g\mapsto g^{-1}. For gGg\in G, denote by LgL_{g} and RgR_{g} the left and right translations by gg, respectively. Take X,Y𝔤X,Y\in\mathfrak{g}. The parallel translate of XX along the geodesic exp(tY)\exp(tY), 0t10\leq t\leq 1, is given by:

DLexp(12Y)DRexp(12Y)X.DL_{\exp\left(\frac{1}{2}Y\right)}DR_{\exp\left(\frac{1}{2}Y\right)}X.

The rotation group and its connection satisfy the assumptions to apply the formula for parallel transport above. Adapting it to the setup of SO(3)\mathrm{SO}(3), we can find that the parallel transport of A0𝔰𝔬(3)=TISO(3)A_{0}\in\mathfrak{so(3)}=T_{I}\mathrm{SO}(3) from R0=IR_{0}=I to R1=exp(𝐮^)R_{1}=\exp(\widehat{\mathbf{u}}) is given by:

R1A1=exp(12𝐮^)A0exp(12𝐮^).R_{1}A_{1}=\exp\left(\frac{1}{2}\widehat{\mathbf{u}}\right)A_{0}\exp\left(\frac{1}{2}\widehat{\mathbf{u}}\right).

This is equivalent to

A1=exp(12𝐮^)A0exp(12𝐮^),A_{1}=\exp\left(-\frac{1}{2}\widehat{\mathbf{u}}\right)A_{0}\exp\left(\frac{1}{2}\widehat{\mathbf{u}}\right),

or using Rodrigues’ formula:

(3.22) A1=(Isin(θ2)𝐧^+(1cos(θ2))𝐧^2)A0(I+sin(θ2)𝐧^+(1cos(θ2))𝐧^2),A_{1}=\left(I-\sin\left(\frac{\theta}{2}\right)\widehat{\mathbf{n}}+\left(1-\cos\left(\frac{\theta}{2}\right)\right)\widehat{\mathbf{n}}^{2}\right)A_{0}\left(I+\sin\left(\frac{\theta}{2}\right)\widehat{\mathbf{n}}+\left(1-\cos\left(\frac{\theta}{2}\right)\right)\widehat{\mathbf{n}}^{2}\right),

where 𝐮^=θ𝐧^\widehat{\mathbf{u}}=\theta\widehat{\mathbf{n}}.

Next, we expand the R.H.S. of (3.22) and show that it is the same as that of (3.19). Indeed, we have

(3.23) (A0sin(θ2)𝐧^A0+(1cos(θ2))𝐧^2A0)(I+sin(θ2)𝐧^+(1cos(θ2))𝐧^2)=A0+sinθ2(A0𝐧^𝐧^A0)+(1cosθ2)(A0𝐧^2+𝐧^2A0)sin2(θ2)𝐧^A0𝐧^+(1cosθ2)2𝐧^2A0𝐧^2,\displaystyle\begin{aligned} &\left(A_{0}-\sin\left(\frac{\theta}{2}\right)\widehat{\mathbf{n}}A_{0}+\left(1-\cos\left(\frac{\theta}{2}\right)\right)\widehat{\mathbf{n}}^{2}A_{0}\right)\left(I+\sin\left(\frac{\theta}{2}\right)\widehat{\mathbf{n}}+\left(1-\cos\left(\frac{\theta}{2}\right)\right)\widehat{\mathbf{n}}^{2}\right)\\ &\hskip 28.45274pt=A_{0}+\sin\frac{\theta}{2}(A_{0}\widehat{\mathbf{n}}-\widehat{\mathbf{n}}A_{0})+\left(1-\cos\frac{\theta}{2}\right)(A_{0}\widehat{\mathbf{n}}^{2}+\widehat{\mathbf{n}}^{2}A_{0})\\ &\hskip 34.14322pt-\sin^{2}\left(\frac{\theta}{2}\right)\widehat{\mathbf{n}}A_{0}\widehat{\mathbf{n}}+\left(1-\cos\frac{\theta}{2}\right)^{2}\widehat{\mathbf{n}}^{2}A_{0}\widehat{\mathbf{n}}^{2},\end{aligned}

where the terms containing 𝐧^A0𝐧^2\widehat{\mathbf{n}}A_{0}\widehat{\mathbf{n}}^{2} and 𝐧^2A0𝐧^\widehat{\mathbf{n}}^{2}A_{0}\widehat{\mathbf{n}} have been cancelled out; this cancellation comes from the fact that 𝐧^A0𝐧^2=𝐧^2A0𝐧^\widehat{\mathbf{n}}A_{0}\widehat{\mathbf{n}}^{2}=\widehat{\mathbf{n}}^{2}A_{0}\widehat{\mathbf{n}}, since 𝐧^A0𝐧^=(𝐧𝐚𝟎)𝐧^\widehat{\mathbf{n}}A_{0}\widehat{\mathbf{n}}=-(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}} by (3.18)3\eqref{eqn:id1}_{3}.

By direct calculation, one can show

A0𝐧^2+𝐧^2A0=A0(𝐚𝟎𝐧)𝐧^,𝐧^2A0𝐧^2=𝐧^((𝐧𝐚𝟎)𝐧^)𝐧^=(𝐧𝐚𝟎)𝐧^3=(𝐧𝐚𝟎)𝐧^.\displaystyle\begin{aligned} &A_{0}\widehat{\mathbf{n}}^{2}+\widehat{\mathbf{n}}^{2}A_{0}=-A_{0}-(\mathbf{a_{0}}\cdot\mathbf{n})\widehat{\mathbf{n}},\\[2.0pt] &\widehat{\mathbf{n}}^{2}A_{0}\widehat{\mathbf{n}}^{2}=\widehat{\mathbf{n}}(-(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}})\widehat{\mathbf{n}}=-(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}}^{3}=(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}}.\end{aligned}

Using these identities in (3.23), (3.22) becomes:

A1\displaystyle A_{1} =A0+sinθ2(A0𝐧^𝐧^A0)(1cosθ2)(A0+(𝐚𝟎𝐧)𝐧^)\displaystyle=A_{0}+\sin\frac{\theta}{2}(A_{0}\widehat{\mathbf{n}}-\widehat{\mathbf{n}}A_{0})-\left(1-\cos\frac{\theta}{2}\right)(A_{0}+(\mathbf{a_{0}}\cdot\mathbf{n})\widehat{\mathbf{n}})
+sin2(θ2)(𝐧𝐚𝟎)𝐧^+(1cosθ2)2(𝐧𝐚𝟎)𝐧^.\displaystyle\hskip 5.69046pt+\sin^{2}\left(\frac{\theta}{2}\right)(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}}+\left(1-\cos\frac{\theta}{2}\right)^{2}(\mathbf{n}\cdot\mathbf{a_{0}})\widehat{\mathbf{n}}.

Finally, combine the terms containing (𝐚𝟎𝐧)𝐧^(\mathbf{a_{0}}\cdot\mathbf{n})\widehat{\mathbf{n}} to get (3.19).

While our expression for parallel transport is consistent with this general result on Lie groups, it presents an elementary and hands-on approach to its derivation. We are not aware of any other works presenting the explicit forms of the metric tensor, Christoffel symbols, geodesics and parallel transport in exponential coordinates, as done in Sections 2 and 3. These explicit calculations would be valuable tools for researchers on rigid-body dynamics.

4. The CS model on SO(3)\mathrm{SO}(3)

In this section, we provide a derivation of the CS model on SO(3)\mathrm{SO}(3) by gathering calculations from previous sections, and investigate its emergent dynamics. We also study the reduction of the model to 𝕊1\mathbb{S}^{1}.

4.1. Formulation of the CS model on SO(3)\mathrm{SO}(3)

Let Ri(t)SO(3)R_{i}(t)\in\mathrm{SO}(3) be the position of the ii-th CS particle, and we set its velocity as

(4.1) Vi(t):=dRi(t)dt.V_{i}(t):=\frac{dR_{i}(t)}{dt}.

Note that we require (Ri,Vi)(R_{i},V_{i}) to satisfy (see (1.3)):

(4.2) DVidt=κNk=1Nϕik(PkiVkVi).\frac{DV_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}(P_{ki}V_{k}-V_{i}).

Since Vi(t)V_{i}(t) lies in the tangent space of SO(3)\mathrm{SO}(3) at Ri(t)R_{i}(t), there exists a unique skew-symmetric matrix Ai(t)A_{i}(t) such that

Vi(t)=Ri(t)Ai(t).V_{i}(t)=R_{i}(t)A_{i}(t).

This yields

(4.3) ddtVi(t)=ddt(Ri(t)Ai(t))=(ddtRi(t))Ai(t)+Ri(t)(ddtAi(t))=Ri(t)Ai(t)2+Ri(t)(ddtAi(t)).\displaystyle\begin{aligned} \frac{d}{dt}V_{i}(t)&=\frac{d}{dt}(R_{i}(t)A_{i}(t))=\left(\frac{d}{dt}R_{i}(t)\right)A_{i}(t)+R_{i}(t)\left(\frac{d}{dt}A_{i}(t)\right)\\ &=R_{i}(t)A_{i}(t)^{2}+R_{i}(t)\left(\frac{d}{dt}A_{i}(t)\right).\end{aligned}

We will rewrite (4.2) in terms of AiA_{i}. By definition of the covariant derivative (using the embedding of SO(3)\mathrm{SO}(3) in 3×3\mathbb{R}^{3\times 3}), one has:

(4.4) DVi(t)dt=ProjTRi(t)SO(3)(ddtVi(t)).\frac{DV_{i}(t)}{dt}=\mathrm{Proj}_{T_{R_{i}(t)}\mathrm{SO}(3)}\left(\frac{d}{dt}V_{i}(t)\right).

Then, it follows from (4.3) and (4.4) that

DVi(t)dt\displaystyle\frac{DV_{i}(t)}{dt} =ProjTRi(t)SO(3)(Ri(t)Ai(t)2+Ri(t)(ddtAi(t)))\displaystyle=\mathrm{Proj}_{T_{R_{i}(t)}\mathrm{SO}(3)}\left(R_{i}(t)A_{i}(t)^{2}+R_{i}(t)\left(\frac{d}{dt}A_{i}(t)\right)\right)
=Ri(t)(ddtAi(t)),\displaystyle=R_{i}(t)\left(\frac{d}{dt}A_{i}(t)\right),

where we used the fact that Ai(t)2A_{i}(t)^{2} is symmetric and ddtAi(t)\frac{d}{dt}A_{i}(t) is skew-symmetric. This and (4.2) yield

(4.5) dAi(t)dt=κNk=1Nϕ(Ri,Rk)(Ri(t)PkiVk(t)Ri(t)Vi(t)).\frac{dA_{i}(t)}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi(R_{i},R_{k})\left(R_{i}(t)^{\top}P_{ki}V_{k}(t)-R_{i}(t)^{\top}V_{i}(t)\right).

We use now the calculations in Section 3.2. Here, RkR_{k} and VkV_{k} play the roles of R0R_{0} and V(0)V(0) from these calculations, while RiR_{i} and PkiVkP_{ki}V_{k} play the roles of R1R_{1} and V(1)V(1). In particular, Ak=RkVkA_{k}=R_{k}^{\top}V_{k} plays the role of A0A_{0}, and RiPkiVkR_{i}^{\top}P_{ki}V_{k} plays the role of A1A_{1}. Also, analogous to 𝐮\mathbf{u}, θ\theta and 𝐧\mathbf{n} from Section 3.2, we have 𝐮ki,θki\mathbf{u}_{ki},\theta_{ki} and 𝐧ki\mathbf{n}_{ki}, defined by:

(4.6) 𝐮^ki=log(RkRi)=θki2sinθki(RkRiRiRk),θki=𝐮ki=arccos(tr(RkRi)12),𝐧^ki=𝐮^kiθki.\displaystyle\begin{aligned} &\widehat{\mathbf{u}}_{ki}=\log(R_{k}^{\top}R_{i})=\frac{\theta_{ki}}{2\sin\theta_{ki}}\big{(}R_{k}^{\top}R_{i}-R_{i}^{\top}R_{k}\big{)},\\ &\theta_{ki}=\|\mathbf{u}_{ki}\|=\arccos\left(\frac{\mathrm{tr}(R_{k}^{\top}R_{i})-1}{2}\right),\qquad\widehat{\mathbf{n}}_{ki}=\frac{\widehat{\mathbf{u}}_{ki}}{\theta_{ki}}.\end{aligned}

From this framework, apply the parallel transport formula (3.20) to get

(4.7) Ri(t)(PkiVk(t))=(1cosθki2)(𝐧ki𝐚k)𝐧^ki+sinθki2𝐚k×𝐧ki^+cosθki2Ak,R_{i}(t)^{\top}(P_{ki}V_{k}(t))=\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\widehat{\mathbf{n}}_{ki}+\sin\frac{\theta_{ki}}{2}\,\widehat{\mathbf{a}_{k}\times\mathbf{n}_{ki}}+\cos\frac{\theta_{ki}}{2}A_{k},

where we denoted

(4.8) 𝐚k=Aˇk, for k=1,,N.\mathbf{a}_{k}=\widecheck{A}_{k},\qquad\text{ for }k=1,\dots,N.

Then, using (4.7) in (4.5) yields

dAidt=κNk=1Nϕik[(1cosθki2)(𝐧ki𝐚k)𝐧^ki+sinθki2𝐚k×𝐧ki^+cosθki2AkAi],\frac{dA_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\widehat{\mathbf{n}}_{ki}+\sin\frac{\theta_{ki}}{2}\widehat{\mathbf{a}_{k}\times\mathbf{n}_{ki}}+\cos\frac{\theta_{ki}}{2}A_{k}-A_{i}\right],

or equivalently, by removing the hats,

(4.9) d𝐚idt=κNk=1Nϕik[(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i].\frac{d\mathbf{a}_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right].

Finally, we combine (4.1) and (4.9) to obtain the CS model on SO(3)\mathrm{SO}(3):

(4.10) dRidt=RiAi,d𝐚idt=κNk=1Nϕik[(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i],\displaystyle\begin{aligned} \frac{dR_{i}}{dt}&=R_{i}A_{i},\\ \frac{d\mathbf{a}_{i}}{dt}&=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right],\end{aligned}

for t>0t>0, 1iN1\leq i\leq N, where we used notations (4.6) and (4.8), and ϕik=ϕ(Ri,Rk)\phi_{ik}=\phi(R_{i},R_{k}).

Since parallel transport between two cut locus points cannot be well defined, we should impose some conditions on ϕ\phi or a priori conditions (see (A)({\mathcal{H}}_{A}) and (B)({\mathcal{H}}_{B})). Some examples of communication weight functions ϕ(R,Q)=ϕ~(d(R,Q))\phi(R,Q)=\tilde{\phi}(d(R,Q)) that satisfy condition (A)({\mathcal{H}}_{A}) are:

ϕ~(d(R,Q))=sin(d(R,Q)),cos(d(R,Q)2),cos(d(R,Q))+1.\tilde{\phi}(d(R,Q))=\sin(d(R,Q)),\qquad\cos\left(\frac{d(R,Q)}{2}\right),\qquad\cos(d(R,Q))+1.

The well-posedness of the CS model on SO(3)\mathrm{SO}(3) is given by the following proposition.

Proposition 4.1.

Suppose either (A)({\mathcal{H}}_{A}) or (B)({\mathcal{H}}_{B}) holds. Then, the Cauchy problem to system (4.10) has a unique global solution.

Proof.

\bullet Case A: Suppose the first condition (A)({\mathcal{H}}_{A}) holds. Then dRidt\frac{dR_{i}}{dt} and d𝐚idt\frac{d\mathbf{a}_{i}}{dt} are Lipschitz continuous functions in the whole domain. Therefore, standard Cauchy-Lipschitz theory and continuity argument based on uniform a priori bound yield the global well-posedness of a smooth solution.

\bullet  Case B: Suppose the second condition (B)({\mathcal{H}}_{B}) holds. Then dRidt\frac{dR_{i}}{dt} and d𝐚idt\frac{d\mathbf{a}_{i}}{dt} are Lipschitz continuous functions in the whole domain, except at cut locus pairs. Then, similar to Case A, one can guarantee the global unique solvability of the Cauchy problem to (4.10). ∎

4.2. Reduction from SO(3)\mathrm{SO}(3) to 𝕊1\mathbb{S}^{1}

In this subsection, we study the relation between the CS model on SO(3)\mathrm{SO}(3) and the CS model on the circle 𝕊1\mathbb{S}^{1}.

Suppose that the initial data are given as follows.

Ri0=[cosθi0sinθi00sinθi0cosθi00001],Vi0=Ri0[0νi00νi000000],Ai0=[0νi00νi000000].R_{i}^{0}=\begin{bmatrix}\cos\theta_{i}^{0}&-\sin\theta_{i}^{0}&0\\ \sin\theta_{i}^{0}&\cos\theta_{i}^{0}&0\\ 0&0&1\end{bmatrix},\quad V_{i}^{0}=R_{i}^{0}\begin{bmatrix}0&-\nu_{i}^{0}&0\\ \nu_{i}^{0}&0&0\\ 0&0&0\end{bmatrix},\quad A_{i}^{0}=\begin{bmatrix}0&-\nu_{i}^{0}&0\\ \nu_{i}^{0}&0&0\\ 0&0&0\end{bmatrix}.

Then, by the isomorphism between 3\mathbb{R}^{3} and 𝔰𝔬(3)\mathfrak{so}(3) in Section 2.1.1, one has

𝐚i0=(0,0,νi0).\mathbf{a}_{i}^{0}=(0,0,\nu_{i}^{0}).

Now take the following ansatz for the solution Ri(t)R_{i}(t) and 𝐚i(t)\mathbf{a}_{i}(t):

(4.11) Ri(t)=[cosθi(t)sinθi(t)0sinθi(t)cosθi(t)0001],𝐚i(t)=(0,0,νi(t)),Ai(t)=νi(t)E3,\displaystyle R_{i}(t)=\begin{bmatrix}\cos\theta_{i}(t)&-\sin\theta_{i}(t)&0\\ \sin\theta_{i}(t)&\cos\theta_{i}(t)&0\\ 0&0&1\end{bmatrix},\quad\mathbf{a}_{i}(t)=(0,0,\nu_{i}(t)),\quad A_{i}(t)=\nu_{i}(t)E_{3},

where

θi(0)=θi0andνi(0)=νi0.\theta_{i}(0)=\theta_{i}^{0}\qquad\mbox{and}\qquad\nu_{i}(0)=\nu_{i}^{0}.

Then we have

{θki=arccos(cos(θiθk)),𝐮^ki=θki2sinθki(RkRiRiRk)=θkisinθki[0sin(θiθk)0sin(θiθk)00000]=θki(sin(θiθk)sinθki)E3.\begin{cases}\displaystyle\theta_{ki}&=\arccos(\cos(\theta_{i}-\theta_{k})),\\ \displaystyle\widehat{\mathbf{u}}_{ki}&=\frac{\theta_{ki}}{2\sin\theta_{ki}}(R_{k}^{\top}R_{i}-R_{i}^{\top}R_{k})=\frac{\theta_{ki}}{\sin\theta_{ki}}\begin{bmatrix}0&-\sin(\theta_{i}-\theta_{k})&0\\ \sin(\theta_{i}-\theta_{k})&0&0\\ 0&0&0\end{bmatrix}\\ &\displaystyle=\theta_{ki}\left(\frac{\sin(\theta_{i}-\theta_{k})}{\sin\theta_{ki}}\right)E_{3}.\end{cases}

From here, one has

𝐧^ki=(sin(θiθk)sinθki)E3𝐧ki=(0,0,sin(θiθk)sinθki).\widehat{\mathbf{n}}_{ki}=\left(\frac{\sin(\theta_{i}-\theta_{k})}{\sin\theta_{ki}}\right)E_{3}\quad\Longrightarrow\quad\mathbf{n}_{ki}=\left(0,0,\frac{\sin(\theta_{i}-\theta_{k})}{\sin\theta_{ki}}\right).

By direct calculation, we have

R˙i(t)=θ˙i(t)[sinθi(t)cosθi(t)0cosθi(t)sinθi(t)0000]=θ˙i(t)Ri(t)E3.\displaystyle\dot{R}_{i}(t)=\dot{\theta}_{i}(t)\begin{bmatrix}-\sin\theta_{i}(t)&-\cos\theta_{i}(t)&0\\ \cos\theta_{i}(t)&-\sin\theta_{i}(t)&0\\ 0&0&0\end{bmatrix}=\dot{\theta}_{i}(t)R_{i}(t)E_{3}.

So we set

θ˙i(t)=νi(t)\dot{\theta}_{i}(t)=\nu_{i}(t)

to satisfy R˙i=RiAi\dot{R}_{i}=R_{i}A_{i}. Since 𝐚k\mathbf{a}_{k} and 𝐧ki\mathbf{n}_{ki} are parallel, and 𝐧ki\mathbf{n}_{ki} is a unit vector, it holds that

(𝐧ki𝐚k)𝐧ki=𝐚k,𝐚k×𝐧ki=0.(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}=\mathbf{a}_{k},\qquad\mathbf{a}_{k}\times\mathbf{n}_{ki}=0.

From these relations, we have

κNk=1Nϕik((1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i)=κNk=1Nϕik(𝐚k𝐚i).\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}\left(\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right)=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}(\mathbf{a}_{k}-\mathbf{a}_{i}).

Hence, if (θi,νi)(\theta_{i},\nu_{i}) satisfy the CS model on the circle, given by:

{dθidt=νi,dνidt=κNk=1Nϕik(νkνi),\displaystyle\begin{cases}\displaystyle\frac{d\theta_{i}}{dt}=\nu_{i},\\ \displaystyle\frac{d\nu_{i}}{dt}=\frac{\kappa}{N}\sum_{k=1}^{N}\phi_{ik}(\nu_{k}-\nu_{i}),\end{cases}

then (Ri,𝐚i)(R_{i},\mathbf{a}_{i}) defined by (4.11) satisfy (4.10). This shows that in this case, the CS model on SO(3)\mathrm{SO}(3) can be reduced to the CS model on the circle.

4.3. Emergent dynamics

In this subsection, we study the emergent behavior of the CS model (4.10) by using an explicit Lyapunov functional and LaSalle’s invariance principle. For a given ensemble {Ri,Vi=RiAi}i=1N\{R_{i},V_{i}=R_{i}A_{i}\}_{i=1}^{N}, we define a Lyapunov functional \mathcal{E} which corresponds to the total kinetic energy:

(t):=i=1NVi(t)Ri(t)2.\mathcal{E}(t):=\sum_{i=1}^{N}{\|V_{i}(t)\|}_{R_{i}(t)}^{2}.

Then, by direct calculations, one has

(4.12) d(t)dt=2i=1NDVidt,ViRi=2i=1NκNk=1Nϕ(Ri,Rk)(PkiVkVi),ViRi=2κNi,k=1Nϕ(Ri,Rk)PkiVkVi,ViRi=2κNi,k=1Nϕ(Ri,Rk)(PkiVk,ViRiViRi2)=κNi,k=1Nϕ(Ri,Rk)(2PkiVk,ViRiViRi2VkRk2)=κNi,k=1Nϕ(Ri,Rk)PkiVkViRi20.\displaystyle\begin{aligned} \frac{d\mathcal{E}(t)}{dt}&=2\sum_{i=1}^{N}\left\langle\frac{DV_{i}}{dt},V_{i}\right\rangle_{{R_{i}}}=2\sum_{i=1}^{N}\left\langle\frac{\kappa}{N}\sum_{k=1}^{N}\phi(R_{i},R_{k})(P_{ki}V_{k}-V_{i}),V_{i}\right\rangle_{{R_{i}}}\\ &=\frac{2\kappa}{N}\sum_{i,k=1}^{N}\phi(R_{i},R_{k}){\langle P_{ki}V_{k}-V_{i},V_{i}\rangle}_{R_{i}}\\ &=\frac{2\kappa}{N}\sum_{i,k=1}^{N}\phi(R_{i},R_{k})\left({\langle P_{ki}V_{k},V_{i}\rangle}_{R_{i}}-{\|V_{i}\|}_{R_{i}}^{2}\right)\\ &=\frac{\kappa}{N}\sum_{i,k=1}^{N}\phi(R_{i},R_{k})\left(2{\langle P_{ki}V_{k},V_{i}\rangle}_{R_{i}}-{\|V_{i}\|}_{R_{i}}^{2}-{\|V_{k}\|}_{R_{k}}^{2}\right)\\ &=-\frac{\kappa}{N}\sum_{i,k=1}^{N}\phi(R_{i},R_{k}){\|P_{ki}V_{k}-V_{i}\|}_{R_{i}}^{2}\leq 0.\end{aligned}

For the last two equalities in (4.12), we used the symmetry of the communication function, ϕ(Ri,Rk)=ϕ(Rk,Ri)\phi(R_{i},R_{k})=\phi(R_{k},R_{i}), and the fact that parallel transport preserves lengths, specifically,

VkRk2=PkiVkRi2.{\|V_{k}\|}^{2}_{R_{k}}={\|P_{ki}V_{k}\|}^{2}_{R_{i}}.

Let {(Ri,Ai)}i=1N(SO(3)×𝔰𝔬(3))N\{(R_{i},A_{i})\}_{i=1}^{N}\in(\mathrm{SO}(3)\times\mathfrak{so}(3))^{N} be a solution of system (4.10). From (4.12), we know that (t)\mathcal{E}(t) is decreasing, and hence,

supt[0,)Vi(t)Ri(t)2supt[0,)(t)(0),\sup_{t\in[0,\infty)}\|V_{i}(t)\|^{2}_{R_{i}(t)}\leq\sup_{t\in[0,\infty)}\mathcal{E}(t)\leq\mathcal{E}(0),

which implies the boundedness of 12Ai(t)F2=Vi(t)Ri(t)2\frac{1}{2}{\|A_{i}(t)\|}_{\mathrm{F}}^{2}={\|V_{i}(t)\|}_{R_{i}(t)}^{2} for all i=1,,Ni=1,\dots,N and t0t\geq 0. Then, by defining the subset of 𝔰𝔬(3)\mathfrak{so}(3),

𝒟:={A𝔰𝔬(3):AF2(0)},\mathcal{D}:=\{A\in\mathfrak{so}(3):{\|A\|}_{\mathrm{F}}\leq\sqrt{2\mathcal{E}(0)}\},

we can restrict the domain of states as follows:

{(Ri,Ai)}i=1N(SO(3)×𝒟)N.\{(R_{i},A_{i})\}_{i=1}^{N}\in(\mathrm{SO}(3)\times\mathcal{D})^{N}.

Since this domain is compact, we can apply LaSalle’s invariance principle to (4.12) to get

(4.13) limtd(t)dt=0,i.e.,limtϕ(Ri,Rk)PkiVkViRi2=0,i,k{1,,N}.\lim_{t\to\infty}\frac{d\mathcal{E}(t)}{dt}=0,\quad\mbox{i.e.,}\quad\lim_{t\to\infty}\phi(R_{i},R_{k}){\|P_{ki}V_{k}-V_{i}\|}_{R_{i}}^{2}=0,\qquad\forall~{}i,k\in\{1,\cdots,N\}.

On the other hand, using (3.20) and the fact that the hat operator is an isomorphism between 𝔰𝔬(3)\mathfrak{so}(3) and 3\mathbb{R}^{3}, one has:

(4.14) PkiVkViRi=(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i.{\|P_{ki}V_{k}-V_{i}\|}_{R_{i}}=\left\|\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right\|.

Finally, we combine (4.13) and (4.14) to derive the following theorem on emergent dynamics.

Theorem 4.1.

Suppose that either (A)({\mathcal{H}}_{A}) or (B)({\mathcal{H}}_{B}) in Section 1 holds, and let {(Ri,Ai)}i=1N\{(R_{i},A_{i})\}_{i=1}^{N} be a solution to (4.10). Then, one has the following assertions.

  1. (1)

    Suppose that the condition (A)({\mathcal{H}}_{A}) holds. Then, for each 1ikN1\leq i\neq k\leq N we have

    (4.15) limtϕ(Ri,Rk)PkiVkViRi2=0,\lim_{t\to\infty}\phi(R_{i},R_{k}){\|P_{ki}V_{k}-V_{i}\|}_{R_{i}}^{2}=0,

    or equivalently,

    limtϕ(Ri,Rk)[(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i]=0.\lim_{t\to\infty}\phi(R_{i},R_{k})\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right]=0.
  2. (2)

    Suppose that the condition (B)({\mathcal{H}}_{B}) holds. Then, for each 1ikN1\leq i\neq k\leq N we have

    limtPkiVkViRi2=0,\lim_{t\to\infty}{\|P_{ki}V_{k}-V_{i}\|}_{R_{i}}^{2}=0,

    or equivalently,

    limt[(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k𝐚i]=0.\lim_{t\to\infty}\left[\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\mathbf{a}_{k}-\mathbf{a}_{i}\right]=0.

4.4. The ω\omega-limit set

In this subsection, we investigate the structure of the ω\omega-limit set of the Cucker-Smale model on SO(3)\mathrm{SO}(3). By Theorem 4.1, each point in the ω\omega-limit set satisfies (4.15). Moreover, since the ω\omega-limit set is positively invariant, each trajectory that starts in the ω\omega-limit set satisfies (4.2)\eqref{D-0-0}. Therefore, each particle is either at rest or moves along a constant speed geodesic.

Geodesic loops in SO(3)\mathrm{SO}(3). Let us first specify what is a closed geodesic curve on SO(3)\mathrm{SO}(3). Consider the unit-speed geodesic curve starting at R0R_{0} in the direction R0𝐧^R_{0}\widehat{\mathbf{n}}, where 𝐧3\mathbf{n}\in\mathbb{R}^{3}, with 𝐧=1\|\mathbf{n}\|=1. By the explicit representation of the exponential map in (2.9), this curve is given as

(t)=R0exp(t𝐧^)=I+sint𝐧^+(1cost)𝐧^2,t[0,π].\mathcal{R}(t)=R_{0}\exp(t\widehat{\mathbf{n}})=I+\sin t\,\widehat{\mathbf{n}}+(1-\cos t)\widehat{\mathbf{n}}^{2},\qquad t\in[0,\pi].

At t=πt=\pi, (π)=R0(2𝐧𝐧I)\mathcal{R}(\pi)=R_{0}(2\mathbf{n}\mathbf{n}^{\top}-I) is a point in the cut locus of R0R_{0}, at distance π\pi from R0R_{0}.

Now consider the unit-speed geodesic curve 𝒬(t)\mathcal{Q}(t) starting from R0R_{0} in the opposite direction R0𝐧^-R_{0}\widehat{\mathbf{n}}, as given by

𝒬(t)=R0exp(t𝐧^)=Isint𝐧^+(1cost)𝐧^2,t[0,π].\mathcal{Q}(t)=R_{0}\exp(-t\widehat{\mathbf{n}})=I-\sin t\,\widehat{\mathbf{n}}+(1-\cos t)\widehat{\mathbf{n}}^{2},\qquad t\in[0,\pi].

This curve reaches the same point 𝒬(π)=R0(2𝐧𝐧I)=(π)\mathcal{Q}(\pi)=R_{0}(2\mathbf{n}\mathbf{n}^{\top}-I)=\mathcal{R}(\pi) at time π\pi. By construction,

(t)=(t)𝐧^,and𝒬(t)=𝒬(t)𝐧^,for all t[0,π].\mathcal{R}^{\prime}(t)=\mathcal{R}(t)\widehat{\mathbf{n}},\qquad\mbox{and}\qquad\mathcal{Q}^{\prime}(t)=-\mathcal{Q}(t)\widehat{\mathbf{n}},\quad\mbox{for all $t\in[0,\pi]$}.

In particular, the tangents to the two curves at t=πt=\pi are opposite to each other, that is,

(π)=(π)𝐧^=𝒬(π)𝐧^=𝒬(π).\mathcal{R}^{\prime}(\pi)=\mathcal{R}(\pi)\widehat{\mathbf{n}}=\mathcal{Q}(\pi)\widehat{\mathbf{n}}=-\mathcal{Q}^{\prime}(\pi).

To obtain a geodesic loop starting and ending at R0R_{0} we patch together the two geodesic arcs, and consider the curve ~(t)\tilde{\mathcal{R}}(t) on [0,2π][0,2\pi] defined by:

~(t)={(t),for t[0,π],𝒬(2πt),for t(π,2π].\tilde{\mathcal{R}}(t)=\begin{cases}\mathcal{R}(t),&\text{for }t\in[0,\pi],\\ \mathcal{Q}(2\pi-t),&\text{for }t\in(\pi,2\pi].\end{cases}

Then, we can extend the curve for all t>2πt>2\pi, with the image of the curve being this closed geodesic loop.

In the visualization of SO(3)\mathrm{SO}(3) as a ball of radius π\pi with center at R0R_{0} (see also Section 5), this geodesic loop consists of two parts. The first part is the curve (t)\mathcal{R}(t) starting at the center and reaching the boundary (the sphere of radius π\pi) at t=πt=\pi. The second part of the loop starts at the antipodal of the point where the first curve ended at t=πt=\pi, and ends at the center R0R_{0}. So visually, it seems like that a jump has occurred at t=πt=\pi, however, antipodal points on the boundary are identified, and the curve is indeed smooth.

Structure of the ω\omega-limit set. Consider NN constant speed geodesic loops Ri(t)R_{i}(t) on SO(3)\mathrm{SO}(3), as described above, initialized in the ω\omega-limit set of (4.10). The loops start at Ri0SO(3)R_{i}^{0}\in\mathrm{SO}(3), in the direction Ri0AiR_{i}^{0}A_{i}, where Ai=𝐚^i𝔰𝔬(3)A_{i}=\widehat{\mathbf{a}}_{i}\in\mathfrak{so}(3), i=1,,Ni=1,\dots,N, are constant matrices. Note that 𝐚i\mathbf{a}_{i} have not been assumed of unit length, so a generic loop Ri(t)R_{i}(t) closes after t=2π/𝐚it=2\pi/\|\mathbf{a}_{i}\| units of time. For all ii, we have

R˙i(t)=Ri(t)Ai.\dot{R}_{i}(t)=R_{i}(t)A_{i}.

For the moment, we ignore the issue of points reaching the cut locus of each other dynamically. To characterize the ω\omega-limit set of the CS model (4.10), we require

(4.16) PkiR˙k(t)=R˙i(t),for all i,k{1,2,,N} and t0,P_{ki}\dot{R}_{k}(t)=\dot{R}_{i}(t),\qquad\text{for all }i,k\in\{1,2,\cdots,N\}\quad\text{ and }t\geq 0,

such that θki(t)=d(Ri(t),Rk(t))<π\theta_{ki}(t)=d(R_{i}(t),R_{k}(t))<\pi. The goal is to show that the only possibility for this to happen is to have a single geodesic loop along which all particles move.

Fix two arbitrary indices ii and kk and suppose that θki(0)π\theta_{ki}(0)\neq\pi. Then there exists ϵ>0\epsilon>0 such that

θki(t)π,t[0,ϵ).\theta_{ki}(t)\neq\pi,\quad\forall~{}t\in[0,\epsilon).

Consider γki(t,)\gamma_{ki}(t,\cdot), the length-minimizing geodesic between Rk(t)R_{k}(t) and Ri(t)R_{i}(t), and the parallel transport of R˙k(t)\dot{R}_{k}(t) along γki\gamma_{ki}, from Rk(t)R_{k}(t) to Ri(t)R_{i}(t). Using the calculations detailed in Section 4.1 (see equation (4.7)), the parallel transport relationship (4.16) can be written as:

(4.17) Ai=(1cosθki2)(𝐧ki𝐚k)𝐧^ki+sinθki2𝐚k×𝐧ki^+cosθki2Ak,A_{i}=\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\,\widehat{\mathbf{n}}_{ki}+\sin\frac{\theta_{ki}}{2}\,\widehat{\mathbf{a}_{k}\times\mathbf{n}_{ki}}+\cos\frac{\theta_{ki}}{2}A_{k},

or without the hats,

(4.18) 𝐚i=(1cosθki2)(𝐧ki𝐚k)𝐧ki+sinθki2𝐚k×𝐧ki+cosθki2𝐚k,\mathbf{a}_{i}=\left(1-\cos\frac{\theta_{ki}}{2}\right)(\mathbf{n}_{ki}\cdot\mathbf{a}_{k})\,\mathbf{n}_{ki}+\sin\frac{\theta_{ki}}{2}\,\mathbf{a}_{k}\times\mathbf{n}_{ki}+\cos\frac{\theta_{ki}}{2}\,\mathbf{a}_{k},

where we used notations (4.6) and (4.8). Note that θki\theta_{ki} and 𝐧ki\mathbf{n}_{ki} depend on time, while 𝐚i\mathbf{a}_{i} and 𝐚k\mathbf{a}_{k} are fixed. Equations (4.17) (and (4.18)) need to hold for all times.

First, since parallel transport preserves lengths, we have

RiAiRi=RkAkRk,{\|R_{i}A_{i}\|}_{R_{i}}={\|R_{k}A_{k}\|}_{R_{k}},

or equivalently, 𝐚i=𝐚k\|\mathbf{a}_{i}\|=\|\mathbf{a}_{k}\|. We denote by θ\theta the common length, and hence we can write

(4.19) 𝐚i=θ𝐧iand𝐚k=θ𝐧k,\mathbf{a}_{i}=\theta\mathbf{n}_{i}\quad\text{and}\quad\mathbf{a}_{k}=\theta\mathbf{n}_{k},

where 𝐧i\mathbf{n}_{i} and 𝐧k\mathbf{n}_{k} are unit vectors in 3\mathbb{R}^{3}.

Second, parallel transport preserves angles, so the angle that R˙k(t)\dot{R}_{k}(t) makes with the geodesic γki\gamma_{ki} at Rk(t)R_{k}(t) is the same as the angle between R˙i(t)\dot{R}_{i}(t) and γki\gamma_{ki} at Ri(t)R_{i}(t). Consequently,

R˙klogRkRi=R˙ilogRiRk,\dot{R}_{k}\cdot\log_{R_{k}}R_{i}=-\dot{R}_{i}\cdot\log_{R_{i}}R_{k},

and hence, we have

(4.20) RkAkRklog(RkRi)=RiAiRilog(RiRk).R_{k}A_{k}\cdot R_{k}\log(R_{k}^{\top}R_{i})=-R_{i}A_{i}\cdot R_{i}\log(R_{i}^{\top}R_{k}).

By the expression of matrix logarithm (see (2.7)), we can write (4.20) as

θki2sinθkiAk,RkRiRiRkF=θki2sinθkiAi,RiRkRkRiF,\frac{\theta_{ki}}{2\sin\theta_{ki}}{\langle A_{k},R_{k}^{\top}R_{i}-R_{i}^{\top}R_{k}\rangle}_{\mathrm{F}}=-\frac{\theta_{ki}}{2\sin\theta_{ki}}{\langle A_{i},R_{i}^{\top}R_{k}-R_{k}^{\top}R_{i}\rangle}_{\mathrm{F}},

and furthermore, by restoring the dependence on tt,

θki(t)2sinθki(t)AkAi,Rk(t)Ri(t)Ri(t)Rk(t)F=0, for all t0.\frac{\theta_{ki}(t)}{2\sin\theta_{ki}(t)}{\langle A_{k}-A_{i},R_{k}(t)^{\top}R_{i}(t)-R_{i}(t)^{\top}R_{k}(t)\rangle}_{\mathrm{F}}=0,\qquad\text{ for all }t\geq 0.

The expression θki2sinθki\frac{\theta_{ki}}{2\sin\theta_{ki}} is well-defined when θki(0,π)\theta_{ki}\in(0,\pi). Since the function can be extended to θki[0,π)\theta_{ki}\in[0,\pi) continuously, we define the value at θki=0\theta_{ki}=0 as 12\frac{1}{2}.

Now we work out the following equation explicitly:

(4.21) AkAi,Rk(t)Ri(t)Ri(t)Rk(t)F=0.\langle A_{k}-A_{i},R_{k}(t)^{\top}R_{i}(t)-R_{i}(t)^{\top}R_{k}(t)\rangle_{\mathrm{F}}=0.

Note that for any skew-symmetric matrix AA and rotation matrix RR, one has

A,RF=tr(AR)=tr(RA)=tr(RA)=A,RF\langle A,R^{\top}\rangle_{\mathrm{F}}=\mathrm{tr}(AR)=\mathrm{tr}(RA)=-\mathrm{tr}(RA^{\top})=-\langle A,R\rangle_{\mathrm{F}}

Hence, since AkAiA_{k}-A_{i} is skew-symmetric and Rk(t)Ri(t)SO(3)R_{k}(t)^{\top}R_{i}(t)\in\mathrm{SO}(3), (4.21) can be written as

AkAi,Rk(t)Ri(t)F=0,{\langle A_{k}-A_{i},R_{k}(t)^{\top}R_{i}(t)\rangle}_{\mathrm{F}}=0,

or equivalently,

(4.22) Ai,Rk(t)Ri(t)F=Ak,Rk(t)Ri(t)F.\langle A_{i},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\langle A_{k},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}.

Since Ai=𝐚^i=θ𝐧^iA_{i}=\widehat{\mathbf{a}}_{i}=\theta\widehat{\mathbf{n}}_{i} and Ak=𝐚^k=θ𝐧^kA_{k}=\widehat{\mathbf{a}}_{k}=\theta\widehat{\mathbf{n}}_{k} (see (4.19)), we can cancel θ\theta and write (4.22) as:

(4.23) 𝐧^i,Rk(t)Ri(t)F=𝐧^k,Rk(t)Ri(t)F.\langle\widehat{\mathbf{n}}_{i},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\langle\widehat{\mathbf{n}}_{k},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}.

In the following lemma, we calculate 𝐧^i,Rk(t)Ri(t)F{\langle\widehat{\mathbf{n}}_{i},R_{k}(t)^{\top}R_{i}(t)\rangle}_{\mathrm{F}} and 𝐧^k,Rk(t)Ri(t)F{\langle\widehat{\mathbf{n}}_{k},R_{k}(t)^{\top}R_{i}(t)\rangle}_{\mathrm{F}}.

Lemma 4.1.

The following relations hold:

(i)𝐧^i,Rk(t)Ri(t)F=tr(12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k)+tr(𝐧^i2R0𝐧^i2R0𝐧^k2)sin(θt)+tr(𝐧^iR0+𝐧^iR0𝐧^k2)cos(θt)+tr(12𝐧^iR0𝐧^k+12𝐧^i2R0𝐧^k2)sin(2θt)+tr(12𝐧^iR0𝐧^k2+12𝐧^i2R0𝐧^k)cos(2θt),(ii)𝐧^k,Rk(t)Ri(t)F=tr(12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k)+tr(R0𝐧^k2+𝐧^i2R0𝐧^k2)sin(θt)+tr(R0𝐧^k+𝐧^i2R0𝐧^k)cos(θt)+tr(12𝐧^iR0𝐧^k12𝐧^i2R0𝐧^k2)sin(2θt)+tr(12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k)cos(2θt),\displaystyle\begin{aligned} &(i)~{}\langle\widehat{\mathbf{n}}_{i},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)\\[2.0pt] &\hskip 56.9055pt+\mathrm{tr}\left(-\widehat{\mathbf{n}}_{i}^{2}R^{0}-\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)\sin(\theta t)+\mathrm{tr}\left(\widehat{\mathbf{n}}_{i}R^{0}+\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)\cos(\theta t)\\[2.0pt] &\hskip 56.9055pt+\mathrm{tr}\left(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)\sin(2\theta t)+\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)\cos(2\theta t),\\[5.0pt] &(ii)~{}\langle\widehat{\mathbf{n}}_{k},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)\\[2.0pt] &\hskip 56.9055pt+\mathrm{tr}\left(R^{0}\widehat{\mathbf{n}}_{k}^{2}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)\sin(\theta t)+\mathrm{tr}\left(R^{0}\widehat{\mathbf{n}}_{k}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)\cos(\theta t)\\[2.0pt] &\hskip 56.9055pt+\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)\sin(2\theta t)+\mathrm{tr}\left(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)\cos(2\theta t),\end{aligned}

where we used the notation R0=Ri0Rk0R^{0}={R_{i}^{0}}^{\top}R_{k}^{0}.

Proof.

Since the proof of this lemma is rather lengthy, we present it in Appendix A.9. ∎

We substitute the result of Lemma 4.1 into (4.23) and equate the coefficients of the trigonometric functions sin(θt)\sin(\theta t), cos(θt)\cos(\theta t), sin(2θt)\sin(2\theta t), and cos(2θt)\cos(2\theta t) to obtain

(4.24) {tr(𝐧^i2R0𝐧^i2R0𝐧^k2)=tr(R0𝐧^k2+𝐧^i2R0𝐧^k2),tr(𝐧^iR0+𝐧^iR0𝐧^k2)=tr(R0𝐧^k+𝐧^i2R0𝐧^k),tr(12𝐧^iR0𝐧^k+12𝐧^i2R0𝐧^k2)=tr(12𝐧^iR0𝐧^k12𝐧^i2R0𝐧^k2),tr(12𝐧^iR0𝐧^k2+12𝐧^i2R0𝐧^k)=tr(12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k).\begin{cases}\displaystyle\mathrm{tr}(-\widehat{\mathbf{n}}_{i}^{2}R^{0}-\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})=\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{k}^{2}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}),\vspace{0.2cm}\\ \displaystyle\mathrm{tr}(\widehat{\mathbf{n}}_{i}R^{0}+\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2})=\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{k}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}),\vspace{0.2cm}\\ \displaystyle\mathrm{tr}\left(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right)=\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2}\right),\vspace{0.2cm}\\ \displaystyle\mathrm{tr}\left(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right)=\mathrm{tr}\left(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}\right).\end{cases}

From (4.24)4\eqref{J-1}_{4}, we get

(4.25) tr(𝐧^iR0𝐧^k2)=tr(𝐧^i2R0𝐧^k).\mathrm{tr}(\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2})=\mathrm{tr}(\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}).

Then, (4.25) and (4.24)2\eqref{J-1}_{2} yield

(4.26) tr(R0𝐧^i)=tr(R0𝐧^k).\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{i})=\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{k}).

Note that (4.24)3\eqref{J-1}_{3} can be rewritten as

(4.27) tr(𝐧^iR0𝐧^k+𝐧^i2R0𝐧^k2)=0.\mathrm{tr}(\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})=0.

Finally, (4.24)1\eqref{J-1}_{1} leads to

(4.28) tr(R0𝐧^k2+R0𝐧^i2+2𝐧^i2R0𝐧^k2)=0.\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{k}^{2}+R^{0}\widehat{\mathbf{n}}_{i}^{2}+2\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})=0.

Without any loss of generality, we assume that R0R^{0} is a rotation about the x1x_{1}-axis. Since R0=Ri0Rk0R^{0}={R_{i}^{0}}^{\top}R_{k}^{0}, the angle of rotation is θki0\theta_{ki}^{0}. For simplicity of notation, set

α=θki0.\alpha=\theta_{ki}^{0}.

Then R0R^{0} becomes

R0=[1000cosαsinα0sinαcosα].\displaystyle R^{0}=\begin{bmatrix}1&0&0\\ 0&\cos\alpha&-\sin\alpha\\ 0&\sin\alpha&\cos\alpha\end{bmatrix}.

Also for ease of notation, we set

𝐧i=(x1,x2,x3),and𝐧k=(y1,y2,y3).\mathbf{n}_{i}=(x_{1},x_{2},x_{3}),\qquad\mbox{and}\qquad\mathbf{n}_{k}=(y_{1},y_{2},y_{3}).

By direct calculation, one finds

tr(R0(𝐧^i𝐧^k))=2sinα(y1x1).\mathrm{tr}(R^{0}(\widehat{\mathbf{n}}_{i}-\widehat{\mathbf{n}}_{k}))=2\sin\alpha(y_{1}-x_{1}).

For α<π\alpha<\pi, (4.26) implies

y1=x1.y_{1}=x_{1}.

By direct calculations one finds

tr(𝐧^iR0𝐧^k)=((cosα+1)x2+sinαx3)y2+(sinαx2(cosα+1)x3)y32x1y1cosα,tr(R0𝐧^i2)=(1cosα)x121cosα.\displaystyle\begin{aligned} &\mathrm{tr}(\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k})=-((\cos\alpha+1)x_{2}+\sin\alpha\,x_{3})y_{2}+(\sin\alpha\,x_{2}-(\cos\alpha+1)x_{3})y_{3}-2x_{1}y_{1}\cos\alpha,\\[3.0pt] &\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{i}^{2})=(1-\cos\alpha)x_{1}^{2}-1-\cos\alpha.\end{aligned}

Note that since tr(R0𝐧^i2)\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{i}^{2}) depends only on x1x_{1}, one has

tr(R0𝐧^i2)=tr(R0𝐧^k2).\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{i}^{2})=\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{k}^{2}).

Hence, we can combine (4.27) and (4.28) to get

tr(R0𝐧^i2)=tr(𝐧^iR0𝐧^k),\mathrm{tr}(R^{0}\widehat{\mathbf{n}}_{i}^{2})=\mathrm{tr}(\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}),

and by using the coordinate expressions for the two traces, we find

(4.29) ((cosα+1)x2+sinαx3)y2+(sinαx2(cosα+1)x3)y3=(cosα+1)(x121).-((\cos\alpha+1)x_{2}+\sin\alpha\,x_{3})y_{2}+(\sin\alpha\,x_{2}-(\cos\alpha+1)x_{3})y_{3}=(\cos\alpha+1)(x_{1}^{2}-1).

By using y1=x1y_{1}=x_{1} one finds that equation (4.25) leads to an identity, with both sides being equal to

x1(sinαx2+(1cosα)x3)y2+x1((cosα1)x2+sinαx3)y3\displaystyle x_{1}(\sin\alpha\,x_{2}+(1-\cos\alpha)x_{3})y_{2}+x_{1}((\cos\alpha-1)x_{2}+\sin\alpha\,x_{3})y_{3}
+x1(1+x12)sinα+((x321)cosαx2x3sinα.\displaystyle\hskip 142.26378pt+x_{1}(1+x_{1}^{2})\sin\alpha+((x_{3}^{2}-1)\cos\alpha-x_{2}x_{3}\sin\alpha.

Finally, (4.27) yields

(4.30) ((x221)cosα+x2x3sinα)y22+((x321)cosαx2x3sinα)y32\displaystyle((x_{2}^{2}-1)\cos\alpha+x_{2}x_{3}\sin\alpha)y_{2}^{2}+((x_{3}^{2}-1)\cos\alpha-x_{2}x_{3}\sin\alpha)y_{3}^{2}
+(sinαx22+sinαx32+2cosαx2x3)y2y3=x12(x121)cosα.\displaystyle\qquad\qquad+(-\sin\alpha\,x_{2}^{2}+\sin\alpha\,x_{3}^{2}+2\cos\alpha\,x_{2}x_{3})y_{2}y_{3}=x_{1}^{2}(x_{1}^{2}-1)\cos\alpha.

We solve equations (4.29) and (4.30) for unknowns y2y_{2} and y3y_{3}. By direct (but tedious) calculations, the two sets of solutions turn out to be:

{y2=x2,y3=x3;andy2=cosαx2+sinαx3,y3=sinαx2+cosαx3,\begin{cases}\displaystyle y_{2}=x_{2},\quad y_{3}=x_{3};\quad\mbox{and}\\ \displaystyle y_{2}=\cos\alpha\,x_{2}+\sin\alpha\,x_{3},\quad y_{3}=-\sin\alpha x_{2}+\cos\alpha x_{3},\end{cases}

or in vector form,

{𝐧k=𝐧i;and𝐧k=R0𝐧i.\begin{cases}\mathbf{n}_{k}=\mathbf{n}_{i};\quad\mbox{and}\\ \mathbf{n}_{k}={R^{0}}^{\top}\mathbf{n}_{i}.\end{cases}

Note that the latter solution satisfies

Rk0𝐧k=Ri0𝐧i.R_{k}^{0}\mathbf{n}_{k}=R_{i}^{0}\mathbf{n}_{i}.

Next we study the two cases separately.

\bullet Case A  (𝐧i=𝐧k:=𝐧\mathbf{n}_{i}=\mathbf{n}_{k}:=\mathbf{n}; also, 𝐚i=𝐚k=θ𝐧\mathbf{a}_{i}=\mathbf{a}_{k}=\theta\mathbf{n}, and Ai=Ak=θ𝐧^:=AA_{i}=A_{k}=\theta\widehat{\mathbf{n}}:=A): Recall from the geometric interpretation of (4.18) that at all times, 𝐚i\mathbf{a}_{i} can be obtained as the rotation by right-hand-rule of 𝐚k\mathbf{a}_{k} about the axis 𝐧ki-\mathbf{n}_{ki}, by an angle θki2\frac{\theta_{ki}}{2}. In particular, this statement holds at t=0t=0. Provided θki0:=θki(0)0\theta_{ki}^{0}:=\theta_{ki}(0)\neq 0, a rotation keeps a vector unchanged only if the vector is parallel to the rotation axis. Hence, 𝐧𝐧ki0\mathbf{n}\parallel\mathbf{n}_{ki}^{0}, which implies that the particles move along the same geodesic loop, which includes the geodesic curve between Ri0R_{i}^{0} and Rk0R_{k}^{0}. In the trivial case θki0=0\theta_{ki}^{0}=0, one gets Ri0=Rk0R_{i}^{0}=R_{k}^{0} and Ri(t)=Rk(t)R_{i}(t)=R_{k}(t) for all tt. Note that moving along the same geodesic loop, the distance between the two particles remains constant in time, that is, θki(t)=θki0\theta_{ki}(t)=\theta_{ki}^{0} for all tt.

\bullet Case B  (𝐧i=Ri0Rk0𝐧k\mathbf{n}_{i}={R_{i}^{0}}^{\top}R_{k}^{0}\,\mathbf{n}_{k}, or equivalently 𝐚i=Ri0Rk0𝐚k\mathbf{a}_{i}={R_{i}^{0}}^{\top}R_{k}^{0}\mathbf{a}_{k}): Note that Ri0Rk0{R_{i}^{0}}^{\top}R_{k}^{0} is a rotation about 𝐧ki0-\mathbf{n}_{ki}^{0} by angle θki0\theta_{ki}^{0}. This can only be consistent with the geometric interpretation of (4.18) provided θki0=0\theta_{ki}^{0}=0 or equivalently Ri0=Rk0R_{i}^{0}=R_{k}^{0}, in which case we get Ri(t)=Rk(t)R_{i}(t)=R_{k}(t) for all tt.

The considerations above lead to the following proposition.

Proposition 4.2.

Let {(Ri,Ai)}i=1N\{(R_{i},A_{i})\}_{i=1}^{N} be a solution of (4.10) which lies in the ω\omega-limit set. Then there exists a constant matrix A=𝐚^𝔰𝔬(3)A=\widehat{\mathbf{a}}\in\mathfrak{so}(3) such that

Ai=A, for all i=1,2,,N.\displaystyle A_{i}=A,\qquad\text{ for all }i=1,2,\dots,N.

Furthermore, if θki0<π\theta_{ki}^{0}<\pi for all i,ki,k, then 𝐮ki\mathbf{u}_{ki} is parallel to 𝐚\mathbf{a} (see notations (4.6)), and this implies that all particles rotate along the same closed geodesic curve.

Proof.

Fix an index i{1,,N}i\in\{1,\dots,N\} and take another generic index kik\neq i. If θki0<π\theta_{ki}^{0}<\pi, the discussion above applies and we have

Ai=Ak:=A,𝐮ki𝐚,andθki(t)=θki0<πt0,\displaystyle A_{i}=A_{k}:=A,\qquad\mathbf{u}_{ki}\parallel\mathbf{a},\qquad\text{and}\quad\theta_{ki}(t)=\theta_{ki}^{0}<\pi\quad\forall t\geq 0,

where 𝐮^ki=log(RkRi)\widehat{\mathbf{u}}_{ki}=\log(R_{k}^{\top}R_{i}) and 𝐚^=A\widehat{\mathbf{a}}=A. Furthermore, RiR_{i} and RkR_{k} rotate on the same closed geodesic curve. Note that when θki0=0\theta_{ki}^{0}=0 this reduces to the trivial case Ri(t)=Rk(t)R_{i}(t)=R_{k}(t) and 𝐮^ki=0\widehat{\mathbf{u}}_{ki}=0, for all tt.

Consider now the case θki0=π\theta_{ki}^{0}=\pi. The matrices RiR_{i} and RkR_{k} are governed by the dynamics

R˙i=RiAi,R˙k=RkAk,\dot{R}_{i}=R_{i}A_{i},\quad\dot{R}_{k}=R_{k}A_{k},

for constant matrices Ai,Ak𝔰𝔬(3)A_{i},A_{k}\in\mathfrak{so}(3). If AiAkA_{i}\neq A_{k}, then 0<θki(t)<π0<\theta_{ki}(t)<\pi for 0<t<ϵ0<t<\epsilon, where ϵ\epsilon is a sufficiently small positive constant. Then we can restart the time and apply the result for the first case, to obtain that Ai=AkA_{i}=A_{k}, which gives a contradiction. Therefore, we have Ai=AkA_{i}=A_{k} in this case as well.

When particles satisfy θki0<π\theta_{ki}^{0}<\pi for all i,ki,k, the first case applies to the entire group and all particles rotate along the same closed geodesic curve. ∎

The main result of this section is the following dichotomy on the long time behavior of the CS model on SO(3).

Theorem 4.2.

Let {(Ri(t),Ai(t))}i=1N\{(R_{i}(t),A_{i}(t))\}_{i=1}^{N} be a solution of (4.10). Then, we have the following dichotomy for its asymptotic dynamics:

  1. (1)

    either the kinetic energy tends to zero:

    limt(t)=0,\lim_{t\to\infty}\mathcal{E}(t)=0,
  2. (2)

    or the kinetic energy converges to a nonzero positive value \mathcal{E}^{\infty}, and

    (4.31) limt(Ai(t)Ak(t))=0, for all 1i,kN,\lim_{t\to\infty}(A_{i}(t)-A_{k}(t))=0,\qquad\text{ for all }1\leq i,k\leq N,
    (4.32) limtAi(t)F=2N, for all 1iN.\lim_{t\to\infty}{\|A_{i}(t)\|}_{\mathrm{F}}=\frac{2}{N}\mathcal{E}^{\infty},\qquad\text{ for all }1\leq i\leq N.
Proof.

It follows from (4.12) that (t)\mathcal{E}(t) is non-increasing. Since (t)0\mathcal{E}(t)\geq 0, there exists a limit:

(4.33) limt(t)=.\displaystyle\lim_{t\to\infty}\mathcal{E}(t)=\mathcal{E}^{\infty}.

If =0\mathcal{E}^{\infty}=0, we can obtain the first case. Now we assume that >0\mathcal{E}^{\infty}>0.

By LaSalle’s invariance principle we infer that

(4.34) limtdist(𝒳(t),𝒮)=0,\displaystyle\lim_{t\to\infty}\mathrm{dist}(\mathcal{X}(t),\mathcal{S})=0,

where 𝒳(t)={(Ri(t),Ai(t))}i=1N\mathcal{X}(t)=\{(R_{i}(t),A_{i}(t))\}_{i=1}^{N} and 𝒮\mathcal{S} is the ω\omega-limit set of system (4.10). From Proposition 4.2, we know that any state {(Ri,Ai)}i=1N\{(R_{i}^{\ast},A_{i}^{\ast})\}_{i=1}^{N} in 𝒮\mathcal{S} satisfies

AiA𝔰𝔬(3), for all i=1,2,.N.\displaystyle{A}_{i}^{\ast}\equiv A^{\ast}\in\mathfrak{so}(3),\qquad\textrm{ for all }i=1,2,\dots.N.

If we combine this fact with (4.34), we can then obtain (4.31).

Note that the asymptotic behavior in (4.31) only states that the differences between Ai(t)A_{i}(t) and Ak(t)A_{k}(t) converge to zero for all i,ki,k. In particular, (4.31) does not imply that each Ai(t)A_{i}(t) converges to some fixed A𝔰𝔬(3)A\in\mathfrak{so}(3) as tt\to\infty. On the other hand, one can show that the norms Ai(t)F{\|A_{i}(t)\|}_{\mathrm{F}} have a common fixed limit as tt\to\infty. To show this, express the energy functional \mathcal{E} as:

(t)=k=1NVk(t)Rk(t)2=12k=1NAk(t)F2.\mathcal{E}(t)=\sum_{k=1}^{N}{\|V_{k}(t)\|}_{R_{k}(t)}^{2}=\frac{1}{2}\sum_{k=1}^{N}{\|A_{k}(t)\|}_{\mathrm{F}}^{2}.

By (4.33) we then have

(4.35) limtk=1NAk(t)F2=2.\displaystyle\lim_{t\to\infty}\sum_{k=1}^{N}{\|A_{k}(t)\|}_{\mathrm{F}}^{2}=2\mathcal{E}^{\infty}.

For a fixed index 1iN1\leq i\leq N, we have

k=1NAkF2\displaystyle\sum_{k=1}^{N}{\|A_{k}\|}_{\mathrm{F}}^{2} =k=1NAi+(AkAi)F2\displaystyle=\sum_{k=1}^{N}{\|A_{i}+(A_{k}-A_{i})\|}_{\mathrm{F}}^{2}
=k=1N(AiF2+2Ai,AkAiF+AiAkF2)\displaystyle=\sum_{k=1}^{N}\left({\|A_{i}\|}_{\mathrm{F}}^{2}+2\langle A_{i},A_{k}-A_{i}\rangle_{\mathrm{F}}+{\|A_{i}-A_{k}\|}_{\mathrm{F}}^{2}\right)
=NAiF2+2k=1NAi,AkAiF+k=1NAiAkF2.\displaystyle=N{\|A_{i}\|}_{\mathrm{F}}^{2}+2\sum_{k=1}^{N}\langle A_{i},A_{k}-A_{i}\rangle_{\mathrm{F}}+\sum_{k=1}^{N}{\|A_{i}-A_{k}\|}_{\mathrm{F}}^{2}.

Letting tt\to\infty in the equation above and using (4.35), we get

(4.36) 2=limt(NAiF2+2k=1NAi,AkAiF+k=1NAiAkF2).\displaystyle 2\mathcal{E}^{\infty}=\lim_{t\to\infty}\left(N{\|A_{i}\|}_{\mathrm{F}}^{2}+2\sum_{k=1}^{N}\langle A_{i},A_{k}-A_{i}\rangle_{\mathrm{F}}+\sum_{k=1}^{N}{\|A_{i}-A_{k}\|}_{\mathrm{F}}^{2}\right).

Now combine (4.31) and (4.36) to obtain

2=NlimtAi(t)F,2\mathcal{E}^{\infty}=N\lim_{t\to\infty}{\|A_{i}(t)\|}_{\mathrm{F}},

which yields (4.32).

Remark 4.1.

As noted above, (4.31) does not imply that each Ai(t)A_{i}(t) converges to a fixed skew-symmetric matrix as tt\to\infty. Consequently, the directions of motion of the particles are not guaranteed to have a limit as tt\to\infty. By (4.32) however, the particles’ speeds Vi(t)Ri(t)=12Ai(t)F\|V_{i}(t)\|_{R_{i}(t)}=\frac{1}{2}{\|A_{i}(t)\|}_{\mathrm{F}} become equal asymptotically. Numerical simulations suggest (see Section 5) that the directions of motion of the particles have in fact a common limit as tt\to\infty, meaning that asymptotically, particles approach and rotate with constant speed along a common geodesic.

5. Numerical Simulations

We solve numerically the CS model on SO(3)\mathrm{SO}(3) using the 4th order Runge-Kutta method. In the simulations presented below we initialize the rotation matrices RiR_{i} in the angle-axis representation (see Section 2.1.2). Specifically, the rotation angles θi\theta_{i} (i=1,,Ni=1,\dots,N) were initialized randomly in the interval (0,π/2)(0,\pi/2), while the unit vectors 𝐯i\mathbf{v}_{i} were generated in spherical coordinates, with the polar and azimuthal angles drawn randomly in the intervals (0,π)(0,\pi) and (0,2π)(0,2\pi), respectively. For the initial velocities RiAiR_{i}A_{i} (i=1,,Ni=1,\dots,N), we generate the components of 𝐚i\mathbf{a}_{i} (here, 𝐚i^=Ai\widehat{\mathbf{a}_{i}}=A_{i}) randomly in the interval (1,1)(-1,1).

For plotting purposes, SO(3)\mathrm{SO}(3) is identified with the ball in 3\mathbb{R}^{3} of radius π\pi, centered at the origin. The center of the ball corresponds to the identity matrix II. A generic point within the ball represents a rotation matrix, with rotation angle given by the distance from the point to the center, and axis given by the ray from the center to the point. Antipodal points on the surface of the ball are identified, as they represent the same rotation matrix (rotation by π\pi about a ray gives the same result as rotation by π\pi about the opposite ray).

The numerical results we present correspond to the communication function

ϕ(R,Q)=cos(d(R,Q)2).\phi(R,Q)=\cos\left(\frac{d(R,Q)}{2}\right).

Note that this function vanishes for pairs of cut points, i.e., for d(R,Q)=πd(R,Q)=\pi. Similar results were obtained with other communication functions as well.

Figure 1 corresponds to a simulation where all particles come to a stop asymptotically. In Figure 1(a) the initial and final locations of the particles are indicated by black dots and red diamonds, respectively. At t=10,000t=10,000 particles have reached a configuration with an energy \mathcal{E} of order 10910^{-9}. Also note the decay to 0 over time of the energy, as shown in Figure 1(b).

Refer to caption Refer to caption
(a) (b)
Figure 1. A group of N=10N=10 particles reach a stationary state where all particles are at rest. (a) The initial state is indicated by black dots. Red diamonds indicate the particle locations at t=10000t=10000, where the particle configuration has reached an energy of order 10910^{-9}. (b) Energy decaying to 0 over time.

Figure 2 shows a simulation in which particles reach asymptotically a flocking state. In Figure 2(a) the initial locations of the particles are indicated by black dots. At t=2,000t=2,000 the particles (red diamonds) have aligned along the closed geodesic path indicated by small blue dots, and we will rotate on this closed geodesic loop indefinitely. Note that the initial and end points of the geodesic path are antipodal of each other and hence, by the visualization of SO(3)\mathrm{SO}(3) that we use, they represent the same rotation matrix. Therefore, the geodesic path is in fact closed, though it does not appear so in the figure. Figure 2(b) shows the energy decay over time. As opposed to the previous simulation, the energy now approaches a non-zero value as tt\to\infty, as particles do not stop, but keep moving on the geodesic loop.

Refer to caption Refer to caption
(a) (b)
Figure 2. Asymptotic flocking of N=10N=10 particles along a closed geodesic path. (a) The black dots and the red diamonds indicate the particle locations at t=0t=0 and t=2000t=2000, respectively. The particles have aligned their velocities (in the parallel transport sense) along the closed geodesic path shown by small blue dots, and will move along this path indefinitely. (b) Energy decay over time. Note that the energy approaches a non-zero value as tt\to\infty.

6. Conclusion

In this paper, we have studied the Cucker-Smale model on SO(3)\mathrm{SO}(3) which exhibits collective behaviors of a rotation matrix ensemble. In earlier work [31], a general and abstract Cucker-Smale model was proposed on a connected, complete and smooth Riemannian manifold using the covariant derivative and parallel transport of tangent vectors along length minimizing geodesics. Thus, for the explicit dynamics and numerical simulations of particles lying on a given specific Riemannian manifold, one needs to find explicit forms for the covariant derivative and parallel transport in terms of state variables and the underlying geometric information. Fortunately, for the special orthogonal group, we can explicitly calculate the covariant derivative and parallel transport which results in the explicit representation of the Cucker-Smale model on SO(3)\mathrm{SO}(3). For the proposed model, we used a Lyapunov approach based on an energy functional. By deriving a dissipative estimate, we use La Salle’s invariance principle to conclude the velocity alignment without any a priori conditions.

Of course, there are still various issues that have not been addressed in the current work, for example, we do not have explicit information on the spatial structure of emerging asymptotic configurations and asymptotic velocity, due to the lack of enough conservation laws, etc. These interesting issues will be addressed in future work.

Appendix A Proofs of several lemmas

In this appendix, we present proofs of several lemmas and details to various calculations employed in the main body of the paper.

A.1. Proof of Lemma 2.2.

By direct calculation, one has

(A.1) 𝐱^2=[x22x32x1x2x1x3x1x2x12x32x2x3x1x3x2x3x12x22].\widehat{\mathbf{x}}^{2}=\begin{bmatrix}-x_{2}^{2}-x_{3}^{2}&x_{1}x_{2}&x_{1}x_{3}\\ x_{1}x_{2}&-x_{1}^{2}-x_{3}^{2}&x_{2}x_{3}\\ x_{1}x_{3}&x_{2}x_{3}&-x_{1}^{2}-x_{2}^{2}\end{bmatrix}.

This yields

𝐱^3\displaystyle\widehat{\mathbf{x}}^{3} =[x22x32x1x2x1x3x1x2x12x32x2x3x1x3x2x3x12x22][0x3x2x30x1x2x10]\displaystyle=\begin{bmatrix}-x_{2}^{2}-x_{3}^{2}&x_{1}x_{2}&x_{1}x_{3}\\ x_{1}x_{2}&-x_{1}^{2}-x_{3}^{2}&x_{2}x_{3}\\ x_{1}x_{3}&x_{2}x_{3}&-x_{1}^{2}-x_{2}^{2}\end{bmatrix}\begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix}
=[0x3(x12+x22+x32)x2(x12+x22+x32)x3(x12+x22+x32)0x1(x12+x22+x32)x2(x12+x22+x32)x1(x12+x22+x32)0]\displaystyle=\begin{bmatrix}0&x_{3}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})&-x_{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})\\ -x_{3}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})&0&x_{1}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})\\ x_{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})&-x_{1}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})&0\end{bmatrix}
=θ2𝐱^,\displaystyle=-\theta^{2}\widehat{\mathbf{x}},

which shows the first identity.

We verify the second identity for α=1\alpha=1. The other cases can be treated similarly.

E1𝐱^=[000001010][0x3x2x30x1x2x10]=[000x2x10x30x1].E_{1}\widehat{\mathbf{x}}=\begin{bmatrix}0&0&0\\ 0&0&-1\\ 0&1&0\end{bmatrix}\begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix}=\begin{bmatrix}0&0&0\\ x_{2}&-x_{1}&0\\ x_{3}&0&-x_{1}\end{bmatrix}.

Furthermore,

x1𝐱^2+𝐱^2E1𝐱^\displaystyle x_{1}\widehat{\mathbf{x}}^{2}+\widehat{\mathbf{x}}^{2}E_{1}\widehat{\mathbf{x}} =𝐱^2(x1I+E1𝐱^)=𝐱^2[x100x200x300]\displaystyle=\widehat{\mathbf{x}}^{2}(x_{1}I+E_{1}\widehat{\mathbf{x}})=\widehat{\mathbf{x}}^{2}\begin{bmatrix}x_{1}&0&0\\ x_{2}&0&0\\ x_{3}&0&0\end{bmatrix}
=[x22x32x1x2x1x3x1x2x12x32x2x3x1x3x2x3x12x22][x100x200x300]=[000000000].\displaystyle=\begin{bmatrix}-x_{2}^{2}-x_{3}^{2}&x_{1}x_{2}&x_{1}x_{3}\\ x_{1}x_{2}&-x_{1}^{2}-x_{3}^{2}&x_{2}x_{3}\\ x_{1}x_{3}&x_{2}x_{3}&-x_{1}^{2}-x_{2}^{2}\end{bmatrix}\begin{bmatrix}x_{1}&0&0\\ x_{2}&0&0\\ x_{3}&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}.

Similarly one check the identity for α=2\alpha=2 and α=3\alpha=3.

A.2. Proof of Lemma 2.3.

We provide derivations for the following three identities:

𝐱^2,𝐱^2F=2θ4.𝐱^2,{Eα,𝐱^}F=4xαθ2.{Eα,𝐱^},{Eβ,𝐱^}F=6xαxβ+2δαβθ2.\displaystyle\begin{aligned} &~{}\langle\widehat{\mathbf{x}}^{2},\widehat{\mathbf{x}}^{2}\rangle_{\mathrm{F}}=2\theta^{4}.\\[2.0pt] &~{}\langle\widehat{\mathbf{x}}^{2},\{E_{\alpha},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}=4x_{\alpha}\theta^{2}.\\[2.0pt] &~{}\langle\{E_{\alpha},\widehat{\mathbf{x}}\},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}=6x_{\alpha}x_{\beta}+2\delta_{\alpha\beta}\theta^{2}.\end{aligned}

\bullet (Derivation of the first identity): we use (A.1) to calculate

𝐱^2,𝐱^2F\displaystyle\langle\widehat{\mathbf{x}}^{2},\widehat{\mathbf{x}}^{2}\rangle_{\mathrm{F}} =(x22x32)2+(x1x2)2+(x1x3)2+(x1x2)2+(x12x32)2\displaystyle=(-x_{2}^{2}-x_{3}^{2})^{2}+(x_{1}x_{2})^{2}+(x_{1}x_{3})^{2}+(x_{1}x_{2})^{2}+(-x_{1}^{2}-x_{3}^{2})^{2}
+(x2x3)2+(x1x3)2+(x2x3)2+(x12x22)2\displaystyle+(x_{2}x_{3})^{2}+(x_{1}x_{3})^{2}+(x_{2}x_{3})^{2}+(-x_{1}^{2}-x_{2}^{2})^{2}
=2(x12+x22+x32)2=2θ4.\displaystyle=2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2}=2\theta^{4}.

\bullet (Derivation of the second identity): we use the commutativity of the trace to get

(A.2) 𝐱^2,{Eα,𝐱^}F=tr(𝐱^2(Eα𝐱^+𝐱^Eα))=2tr(𝐱^3Eα).\langle\widehat{\mathbf{x}}^{2},\{E_{\alpha},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}=\mathrm{tr}(\widehat{\mathbf{x}}^{2}(E_{\alpha}\widehat{\mathbf{x}}+\widehat{\mathbf{x}}E_{\alpha}))=2\mathrm{tr}(\widehat{\mathbf{x}}^{3}E_{\alpha}).

Then use the first identity in Lemma 2.1 and the fact that 𝐱^\widehat{\mathbf{x}} is skew-symmetric to find

(A.3) 2tr(𝐱^3Eα)=2θ2tr(𝐱^Eα)=2θ2𝐱^,EαF.2\mathrm{tr}(\widehat{\mathbf{x}}^{3}E_{\alpha})=-2\theta^{2}\mathrm{tr}(\widehat{\mathbf{x}}E_{\alpha})=2\theta^{2}\langle\widehat{\mathbf{x}},E_{\alpha}\rangle_{\mathrm{F}}.

Finally, we combine (A.2), (A.3) and the third identity in Lemma 2.2 to obtain the desired second identity.

\bullet (Derivation of the third identity): Note that

(A.4) {Eα,𝐱^},{Eβ,𝐱^}F=xκxλ{Eα,Eκ},{Eβ,Eλ}F=xκxλEαEκ+EκEα,EβEλ+EλEβF=xκxλtr(EαEκEβEλ+EκEαEβEλ+EαEκEλEβ+EκEαEλEβ).\displaystyle\begin{aligned} &\langle\{E_{\alpha},\widehat{\mathbf{x}}\},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}\\ &\hskip 14.22636pt=x_{\kappa}x_{\lambda}\langle\{E_{\alpha},E_{\kappa}\},\{E_{\beta},E_{\lambda}\}\rangle_{\mathrm{F}}\\ &\hskip 14.22636pt=x_{\kappa}x_{\lambda}\langle E_{\alpha}E_{\kappa}+E_{\kappa}E_{\alpha},E_{\beta}E_{\lambda}+E_{\lambda}E_{\beta}\rangle_{\mathrm{F}}\\ &\hskip 14.22636pt=x_{\kappa}x_{\lambda}\mathrm{tr}(E_{\alpha}E_{\kappa}E_{\beta}E_{\lambda}+E_{\kappa}E_{\alpha}E_{\beta}E_{\lambda}+E_{\alpha}E_{\kappa}E_{\lambda}E_{\beta}+E_{\kappa}E_{\alpha}E_{\lambda}E_{\beta}).\end{aligned}

Next, we want to express

tr(EιEκEμEν)\mathrm{tr}(E_{\iota}E_{\kappa}E_{\mu}E_{\nu})

for general ι,κ,μ,ν{1,2,3}\iota,\kappa,\mu,\nu\in\{1,2,3\}. To simplify this term, we will use the notation introduced in Remark 2.1. We use Einstein’s convention (repeated index means summation) to find

(A.5) tr(EιEκEμEν)=[EιEκEμEν]αα=[Eι]αβ[Eκ]βγ[Eμ]γδ[Eν]δα=ϵιαβϵκβγϵμγδϵνδα=(ϵβιαϵβγκ)(ϵδμγϵδαν)=(διγδακδικδαγ)(δμαδγνδμνδγα)=διγδακδμαδγνδιγδακδμνδγαδικδαγδμαδγν+δικδαγδμνδγα=δινδκμδικδμνδικδμν+δικδμνδαα=δινδκμ+δικδμν,\displaystyle\begin{aligned} \mathrm{tr}(E_{\iota}E_{\kappa}E_{\mu}E_{\nu})&=[E_{\iota}E_{\kappa}E_{\mu}E_{\nu}]_{\alpha\alpha}=[E_{\iota}]_{\alpha\beta}[E_{\kappa}]_{\beta\gamma}[E_{\mu}]_{\gamma\delta}[E_{\nu}]_{\delta\alpha}\\ &=\epsilon_{\iota\alpha\beta}\epsilon_{\kappa\beta\gamma}\epsilon_{\mu\gamma\delta}\epsilon_{\nu\delta\alpha}=(\epsilon_{\beta\iota\alpha}\epsilon_{\beta\gamma\kappa})(\epsilon_{\delta\mu\gamma}\epsilon_{\delta\alpha\nu})\\ &=(\delta_{\iota\gamma}\delta_{\alpha\kappa}-\delta_{\iota\kappa}\delta_{\alpha\gamma})(\delta_{\mu\alpha}\delta_{\gamma\nu}-\delta_{\mu\nu}\delta_{\gamma\alpha})\\ &=\delta_{\iota\gamma}\delta_{\alpha\kappa}\delta_{\mu\alpha}\delta_{\gamma\nu}-\delta_{\iota\gamma}\delta_{\alpha\kappa}\delta_{\mu\nu}\delta_{\gamma\alpha}-\delta_{\iota\kappa}\delta_{\alpha\gamma}\delta_{\mu\alpha}\delta_{\gamma\nu}+\delta_{\iota\kappa}\delta_{\alpha\gamma}\delta_{\mu\nu}\delta_{\gamma\alpha}\\ &=\delta_{\iota\nu}\delta_{\kappa\mu}-\delta_{\iota\kappa}\delta_{\mu\nu}-\delta_{\iota\kappa}\delta_{\mu\nu}+\delta_{\iota\kappa}\delta_{\mu\nu}\delta_{\alpha\alpha}\\ &=\delta_{\iota\nu}\delta_{\kappa\mu}+\delta_{\iota\kappa}\delta_{\mu\nu},\end{aligned}

where we used δαα=3\delta_{\alpha\alpha}=3 in last equality. Now we combine (LABEL:Ap-3) and (A.5) to get

{Eα,𝐱^},{Eβ,𝐱^}F\displaystyle\langle\{E_{\alpha},\widehat{\mathbf{x}}\},\{E_{\beta},\widehat{\mathbf{x}}\}\rangle_{\mathrm{F}}
=xκxλ(δακδβλ+δκβδαλ+δκαδβλ+δαβδλκ+δακδλβ+δκλδαβ+δκαδλβ+δαλδκβ)\displaystyle\qquad=x_{\kappa}x_{\lambda}\big{(}\delta_{\alpha\kappa}\delta_{\beta\lambda}+\delta_{\kappa\beta}\delta_{\alpha\lambda}+\delta_{\kappa\alpha}\delta_{\beta\lambda}+\delta_{\alpha\beta}\delta_{\lambda\kappa}+\delta_{\alpha\kappa}\delta_{\lambda\beta}+\delta_{\kappa\lambda}\delta_{\alpha\beta}+\delta_{\kappa\alpha}\delta_{\lambda\beta}+\delta_{\alpha\lambda}\delta_{\kappa\beta}\big{)}
=xκxλ(4δακδβλ+2δβκδαλ+2δαβδκλ)\displaystyle\qquad=x_{\kappa}x_{\lambda}\big{(}4\delta_{\alpha\kappa}\delta_{\beta\lambda}+2\delta_{\beta\kappa}\delta_{\alpha\lambda}+2\delta_{\alpha\beta}\delta_{\kappa\lambda}\big{)}
=6xαxβ+2δαβθ2.\displaystyle\qquad=6x_{\alpha}x_{\beta}+2\delta_{\alpha\beta}\theta^{2}.

A.3. Inverse of the metric tensor

Let gαβg_{\alpha\beta} be the metric tensor given in Proposition 2.1(i). We look for the coefficients gβγg^{\beta\gamma} of the metric’s inverse in the following ansatz:

(A.6) gβγ=𝒜xβxγ+δβγ.g^{\beta\gamma}=\mathcal{A}x_{\beta}x_{\gamma}+\mathcal{B}\delta^{\beta\gamma}.

Now we calculate gαβgβγg_{\alpha\beta}g^{\beta\gamma} using (2.21) as follows:

gαβgβγ\displaystyle g_{\alpha\beta}g^{\beta\gamma} =((2cosθ2+θ2θ4)xαxβ+2(1cosθ)θ2δαβ)(𝒜xβxγ+δβγ)\displaystyle=\left(\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)x_{\alpha}x_{\beta}+\frac{2(1-\cos\theta)}{\theta^{2}}\delta_{\alpha\beta}\right)(\mathcal{A}x_{\beta}x_{\gamma}+\mathcal{B}\delta^{\beta\gamma})
=𝒜(2cosθ2+θ2θ2)xαxγ+𝒜(2(1cosθ)θ2)xαxγ\displaystyle=\mathcal{A}\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{2}}\right)x_{\alpha}x_{\gamma}+\mathcal{A}\left(\frac{2(1-\cos\theta)}{\theta^{2}}\right)x_{\alpha}x_{\gamma}
+(2cosθ2+θ2θ4)xαxγ+(2(1cosθ)θ2)δαγ,\displaystyle+\mathcal{B}\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)x_{\alpha}x_{\gamma}+\mathcal{B}\left(\frac{2(1-\cos\theta)}{\theta^{2}}\right)\delta_{\alpha}^{\gamma},

where we used xβxβ=θ2x_{\beta}x_{\beta}=\theta^{2} in the second equality.

Since we require to have gαβgβγ=δαγg_{\alpha\beta}g^{\beta\gamma}=\delta_{\alpha}^{\gamma}, we simply set

(2(1cosθ)θ2)=1,𝒜(2cosθ2+θ2θ2)+𝒜(2(1cosθ)θ2)+(2cosθ2+θ2θ4)=0.\mathcal{B}\left(\frac{2(1-\cos\theta)}{\theta^{2}}\right)=1,\quad\mathcal{A}\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{2}}\right)+\mathcal{A}\left(\frac{2(1-\cos\theta)}{\theta^{2}}\right)+\mathcal{B}\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)=0.

Then, this yields

(A.7) 𝒜=2cosθ2+θ22θ2(cosθ1),=θ22(1cosθ).\displaystyle\begin{aligned} \mathcal{A}=\frac{2\cos\theta-2+\theta^{2}}{2\theta^{2}(\cos\theta-1)},\quad\mathcal{B}=\frac{\theta^{2}}{2(1-\cos\theta)}.\end{aligned}

Finally, (A.6) and (A.7) imply the desired relation for gαβg^{\alpha\beta} in Proposition 2.1(ii).

A.4. Christoffel symbols (2.22)

Note that

gαβ=(2cosθ2+θ2θ4)xαxβ+2(1cosθ)θ2δαβ.g_{\alpha\beta}=\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)x_{\alpha}x_{\beta}+\frac{2(1-\cos\theta)}{\theta^{2}}\delta_{\alpha\beta}.

We differentiate (2.21) to find

γgαβ(𝐱)\displaystyle\partial_{\gamma}g_{\alpha\beta}(\mathbf{x}) =(2cosθ2+θ2θ4)(xαδβγ+xβδαγ)\displaystyle=\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)(x_{\alpha}\delta_{\beta\gamma}+x_{\beta}\delta_{\alpha\gamma})
+((88cosθ2θsinθ2θ2θ5)xαxβ+2(2+2cosθ+θsinθθ3)δαβ)xγθ\displaystyle\quad+\left(\left(\frac{8-8\cos\theta-2\theta\sin\theta-2\theta^{2}}{\theta^{5}}\right)x_{\alpha}x_{\beta}+2\left(\frac{-2+2\cos\theta+\theta\sin\theta}{\theta^{3}}\right)\delta_{\alpha\beta}\right)\frac{x_{\gamma}}{\theta}
=(2cosθ2+θ2θ4)(xαδβγ+xβδαγ+2xγδαβ)+2(θsinθθ2θ4)δαβxγ\displaystyle=\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)(x_{\alpha}\delta_{\beta\gamma}+x_{\beta}\delta_{\alpha\gamma}+2x_{\gamma}\delta_{\alpha\beta})+2\left(\frac{\theta\sin\theta-\theta^{2}}{\theta^{4}}\right)\delta_{\alpha\beta}x_{\gamma}
+(88cosθ2θsinθ2θ2θ6)xαxβxγ.\displaystyle\quad+\left(\frac{8-8\cos\theta-2\theta\sin\theta-2\theta^{2}}{\theta^{6}}\right)x_{\alpha}x_{\beta}x_{\gamma}.

This yields

(A.8) βgγα+αgγβγgαβ=2(2cosθ2+θ2θ4)(xβδαγ+xαδβγ)+2(θsinθθ2θ4)(δαγxβ+δβγxαδαβxγ)+(88cosθ2θsinθ2θ2θ6)xαxβxγ.\displaystyle\begin{aligned} &\partial_{\beta}g_{\gamma\alpha}+\partial_{\alpha}g_{\gamma\beta}-\partial_{\gamma}g_{\alpha\beta}=2\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)(x_{\beta}\delta_{\alpha\gamma}+x_{\alpha}\delta_{\beta\gamma})\\ &+2\left(\frac{\theta\sin\theta-\theta^{2}}{\theta^{4}}\right)(\delta_{\alpha\gamma}x_{\beta}+\delta_{\beta\gamma}x_{\alpha}-\delta_{\alpha\beta}x_{\gamma})+\left(\frac{8-8\cos\theta-2\theta\sin\theta-2\theta^{2}}{\theta^{6}}\right)x_{\alpha}x_{\beta}x_{\gamma}.\end{aligned}

Now, we use explicit relation of the inverse metric together with xκxκ=θ2x_{\kappa}x_{\kappa}=\theta^{2} to see

gλγxγ\displaystyle g^{\lambda\gamma}x_{\gamma} =(2cosθ2+θ22θ2(cosθ1))xλθ2+(θ22(1cosθ))xλ=xλ.\displaystyle=\left(\frac{2\cos\theta-2+\theta^{2}}{2\theta^{2}(\cos\theta-1)}\right)x_{\lambda}\theta^{2}+\left(\frac{\theta^{2}}{2(1-\cos\theta)}\right)x_{\lambda}=x_{\lambda}.

Now use (A.8) to find the Christoffel symbols:

(A.9) Γαβλ=12gλγ(βgγα+αgγβγgαβ)=gλγ(xβδαγ+xαδβγ)(2cosθ2+θ2θ4)+gλγ(δαγxβ+δβγxαδαβxγ)(θsinθθ2θ4)+gλγxαxβxγ(44cosθθsinθθ2θ6)=(gλαxβ+gλβxα)(2cosθ2+θ2θ4)+(gλαxβ+gλβxαδαβxλ)(θsinθθ2θ4)=(gλαxβ+gλβxα)(2cosθ2+θsinθθ4)δαβxλ(θsinθθ2θ4)+xαxβxλ(44cosθθsinθθ2θ6).\displaystyle\begin{aligned} \Gamma^{\lambda}_{\alpha\beta}&=\frac{1}{2}g^{\lambda\gamma}(\partial_{\beta}g_{\gamma\alpha}+\partial_{\alpha}g_{\gamma\beta}-\partial_{\gamma}g_{\alpha\beta})\\ &=g^{\lambda\gamma}(x_{\beta}\delta_{\alpha\gamma}+x_{\alpha}\delta_{\beta\gamma})\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)+g^{\lambda\gamma}(\delta_{\alpha\gamma}x_{\beta}+\delta_{\beta\gamma}x_{\alpha}-\delta_{\alpha\beta}x_{\gamma})\left(\frac{\theta\sin\theta-\theta^{2}}{\theta^{4}}\right)\\[2.0pt] &\quad+g^{\lambda\gamma}x_{\alpha}x_{\beta}x_{\gamma}\left(\frac{4-4\cos\theta-\theta\sin\theta-\theta^{2}}{\theta^{6}}\right)\\ &=\left(g^{\lambda\alpha}x_{\beta}+g^{\lambda\beta}x_{\alpha}\right)\left(\frac{2\cos\theta-2+\theta^{2}}{\theta^{4}}\right)+(g^{\lambda\alpha}x_{\beta}+g^{\lambda\beta}x_{\alpha}-\delta_{\alpha\beta}x_{\lambda})\left(\frac{\theta\sin\theta-\theta^{2}}{\theta^{4}}\right)\\[2.0pt] &=\left(g^{\lambda\alpha}x_{\beta}+g^{\lambda\beta}x_{\alpha}\right)\left(\frac{2\cos\theta-2+\theta\sin\theta}{\theta^{4}}\right)-\delta_{\alpha\beta}x_{\lambda}\left(\frac{\theta\sin\theta-\theta^{2}}{\theta^{4}}\right)\\ &\quad+x_{\alpha}x_{\beta}x_{\lambda}\left(\frac{4-4\cos\theta-\theta\sin\theta-\theta^{2}}{\theta^{6}}\right).\end{aligned}

By (A.6) and (A.7), write:

(A.10) gλαxβ+gλβxα=2𝒜xλxαxβ+(xαδβλ+xβδαλ)=2cosθ2+θ2θ2(cosθ1)xλxαxβ+θ22(1cosθ)(xαδβλ+xβδαλ).\displaystyle\begin{aligned} g^{\lambda\alpha}x_{\beta}+g^{\lambda\beta}x_{\alpha}&=2\mathcal{A}x_{\lambda}x_{\alpha}x_{\beta}+\mathcal{B}(x_{\alpha}\delta_{\beta}^{\lambda}+x_{\beta}\delta_{\alpha}^{\lambda})\\[3.0pt] &=\frac{2\cos\theta-2+\theta^{2}}{\theta^{2}(\cos\theta-1)}x_{\lambda}x_{\alpha}x_{\beta}+\frac{\theta^{2}}{2(1-\cos\theta)}(x_{\alpha}\delta_{\beta}^{\lambda}+x_{\beta}\delta_{\alpha}^{\lambda}).\end{aligned}

Finally, we combine (A.9) and (A.10) to derive (2.22) for the Christoffel coefficients.

A.5. Equation (2.23) for geodesics

Let xα(t)=tuα.x_{\alpha}(t)=tu_{\alpha}. Then, (2.23) is equivalent to show

Γαβγ(𝐱(t))xα(t)xβ(t)=0.\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}(t))x_{\alpha}(t)x_{\beta}(t)=0.

We multiply Γαβγ(𝐱)\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x}) from (2.22) by xαxβx_{\alpha}x_{\beta}, and then sum up the resulting relation over α\alpha and β\beta, and use xαxα=𝐱2=θ2x_{\alpha}x_{\alpha}=\|\mathbf{x}\|^{2}=\theta^{2} to obtain

Γαβγ(𝐱)xαxβ\displaystyle\Gamma^{\gamma}_{\alpha\beta}(\mathbf{x})x_{\alpha}x_{\beta} =(sinθ+θ)(cosθ1)+θ2sinθθ5(cosθ1)xαxβxγxαxβ\displaystyle=\frac{(\sin\theta+\theta)(\cos\theta-1)+\theta^{2}\sin\theta}{\theta^{5}(\cos\theta-1)}x_{\alpha}x_{\beta}x_{\gamma}x_{\alpha}x_{\beta}
+2cosθ2+θsinθ2θ2(1cosθ)(xαδβγ+xβδαγ)xαxβsinθθθ3δαβxγxαxβ\displaystyle\quad+\frac{2\cos\theta-2+\theta\sin\theta}{2\theta^{2}(1-\cos\theta)}(x_{\alpha}\delta_{\beta}^{\gamma}+x_{\beta}\delta_{\alpha}^{\gamma})x_{\alpha}x_{\beta}-\frac{\sin\theta-\theta}{\theta^{3}}\delta_{\alpha\beta}x_{\gamma}x_{\alpha}x_{\beta}
=(sinθ+θ)(cosθ1)+θ2sinθθ(cosθ1)xγ+2cosθ2+θsinθ1cosθxγsinθθθxγ.\displaystyle=\frac{(\sin\theta+\theta)(\cos\theta-1)+\theta^{2}\sin\theta}{\theta(\cos\theta-1)}x_{\gamma}+\frac{2\cos\theta-2+\theta\sin\theta}{1-\cos\theta}x_{\gamma}-\frac{\sin\theta-\theta}{\theta}x_{\gamma}.

Note that xγx_{\gamma} factors out in the R.H.S. above. By a direct calculation, one can see that sum of coefficients equal to zero:

(sinθ+θ)(cosθ1)+θ2sinθθ(cosθ1)+2cosθ2+θsinθ1cosθsinθθθ=0.\frac{(\sin\theta+\theta)(\cos\theta-1)+\theta^{2}\sin\theta}{\theta(\cos\theta-1)}+\frac{2\cos\theta-2+\theta\sin\theta}{1-\cos\theta}-\frac{\sin\theta-\theta}{\theta}=0.

This establishes the desired result.

A.6. Closed form expression (3.12) of parallel transport

We use (LABEL:eqn:XalphaR1), (3.11) and vα(1)uα=𝒞v_{\alpha}(1)u_{\alpha}=\mathcal{C} (see (3.4)) to express V(1)V(1) as follows:

V(1)\displaystyle V(1) =vα(1)Xα(R1)\displaystyle=v_{\alpha}(1)X_{\alpha}(R_{1})
=(sinθ+θcosθθ3)𝒞R0𝐮^+sinθθ(θvα(0)2sinθ2+𝒞uα(1θ212θsinθ2))R0Eα\displaystyle=\left(\frac{-\sin\theta+\theta\cos\theta}{\theta^{3}}\right)\mathcal{C}R_{0}\widehat{\mathbf{u}}+\frac{\sin\theta}{\theta}\left(\frac{\theta v_{\alpha}(0)}{2\sin\frac{\theta}{2}}+\mathcal{C}u_{\alpha}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)R_{0}E_{\alpha}
+(θsinθ2(1cosθ)θ4)𝒞R0𝐮^2\displaystyle\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)\mathcal{C}R_{0}\widehat{\mathbf{u}}^{2}
+1cosθθ2(θvα(0)2sinθ2+𝒞uα(1θ212θsinθ2))(R0Eα𝐮^+R0𝐮^Eα).\displaystyle\quad+\frac{1-\cos\theta}{\theta^{2}}\left(\frac{\theta v_{\alpha}(0)}{2\sin\frac{\theta}{2}}+\mathcal{C}u_{\alpha}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)(R_{0}E_{\alpha}\widehat{\mathbf{u}}+R_{0}\widehat{\mathbf{u}}E_{\alpha}).

By (3.9) and (LABEL:eqn:XalphaR1), one can write

V(0)=vα(0)R0Eαand𝐮^=uαEα.V(0)=v_{\alpha}(0)R_{0}E_{\alpha}\quad\mbox{and}\quad\widehat{\mathbf{u}}=u_{\alpha}E_{\alpha}.

So one can continue the calculation above to get:

V(1)\displaystyle V(1) =(sinθ+θcosθθ3)𝒞R0𝐮^+sinθθ(θV(0)2sinθ2+𝒞R0𝐮^(1θ212θsinθ2))\displaystyle=\left(\frac{-\sin\theta+\theta\cos\theta}{\theta^{3}}\right)\mathcal{C}R_{0}\widehat{\mathbf{u}}+\frac{\sin\theta}{\theta}\left(\frac{\theta V(0)}{2\sin\frac{\theta}{2}}+\mathcal{C}R_{0}\widehat{\mathbf{u}}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)
+(θsinθ2(1cosθ)θ4)𝒞R0𝐮^2\displaystyle\quad+\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}\right)\mathcal{C}R_{0}\widehat{\mathbf{u}}^{2}
+1cosθθ2(θ2sinθ2(V(0)𝐮^+R0𝐮^R0V(0))+2R0𝐮^2𝒞(1θ212θsinθ2))\displaystyle\quad+\frac{1-\cos\theta}{\theta^{2}}\left(\frac{\theta}{2\sin\frac{\theta}{2}}(V(0)\widehat{\mathbf{u}}+R_{0}\widehat{\mathbf{u}}R_{0}^{\top}V(0))+2R_{0}\widehat{\mathbf{u}}^{2}\mathcal{C}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)
=𝒞R0𝐮^(sinθ+θcosθθ3+sinθθ(1θ212θsinθ2))\displaystyle=\mathcal{C}R_{0}\widehat{\mathbf{u}}\left(\frac{-\sin\theta+\theta\cos\theta}{\theta^{3}}+\frac{\sin\theta}{\theta}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)
+𝒞R0𝐮^2(θsinθ2(1cosθ)θ4+21cosθθ2(1θ212θsinθ2))\displaystyle\quad+\mathcal{C}R_{0}\widehat{\mathbf{u}}^{2}\left(\frac{\theta\sin\theta-2(1-\cos\theta)}{\theta^{4}}+2\frac{1-\cos\theta}{\theta^{2}}\left(\frac{1}{\theta^{2}}-\frac{1}{2\theta\sin\frac{\theta}{2}}\right)\right)
+1cosθθ2(θ2sinθ2(V(0)𝐮^+R0𝐮^R0V(0)))+sinθθ(θV(0)2sinθ2)\displaystyle\quad+\frac{1-\cos\theta}{\theta^{2}}\left(\frac{\theta}{2\sin\frac{\theta}{2}}(V(0)\widehat{\mathbf{u}}+R_{0}\widehat{\mathbf{u}}R_{0}^{\top}V(0))\right)+\frac{\sin\theta}{\theta}\left(\frac{\theta V(0)}{2\sin\frac{\theta}{2}}\right)
=𝒞R0𝐮^(cosθcosθ2θ2)+𝒞R0𝐮^2(sinθθ32sinθ2θ3)\displaystyle=\mathcal{C}R_{0}\widehat{\mathbf{u}}\left(\frac{\cos\theta-\cos\frac{\theta}{2}}{\theta^{2}}\right)+\mathcal{C}R_{0}\widehat{\mathbf{u}}^{2}\left(\frac{\sin\theta}{\theta^{3}}-\frac{2\sin\frac{\theta}{2}}{\theta^{3}}\right)
+sinθ2θ(V(0)𝐮^+R0𝐮^R0V(0))+cosθ2V(0).\displaystyle\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(V(0)\widehat{\mathbf{u}}+R_{0}\widehat{\mathbf{u}}R_{0}^{\top}V(0)\right)+\cos\frac{\theta}{2}V(0).

A.7. Expression (3.17) for A1A_{1}

We use (3.15), (3.16) in (3.14) and arrange terms to get

(A.11) A1=𝒞(cosθcosθ2θ2)(cosθ𝐮^sinθθ𝐮^2)+𝒞(sinθ2sinθ2θ3)(cosθ𝐮^2+θsinθ𝐮^)+sinθ2θ(Isinθθ𝐮^+1cosθθ2𝐮^2)(A0𝐮^+𝐮^A0)+cosθ2(Isinθθ𝐮^+1cosθθ2𝐮^2)A0=𝒞(1cosθcosθ22sinθsinθ2θ2)𝐮^+𝒞(cosθ2sinθ2sinθ2cosθθ3)𝐮^2+sinθ2θ(A0𝐮^+𝐮^A0)sinθ2sinθθ2(𝐮^A0𝐮^+𝐮^2A0)+(sinθ2(1cosθ)θ3)(𝐮^2A0𝐮^+𝐮^3A0)+cosθ2A0cosθ2sinθθ𝐮^A0+cosθ2(1cosθ)θ2𝐮^2A0.\displaystyle\begin{aligned} A_{1}&=\mathcal{C}\left(\frac{\cos\theta-\cos\frac{\theta}{2}}{\theta^{2}}\right)\left(\cos\theta\widehat{\mathbf{u}}-\frac{\sin\theta}{\theta}\widehat{\mathbf{u}}^{2}\right)+\mathcal{C}\left(\frac{\sin\theta-2\sin\frac{\theta}{2}}{\theta^{3}}\right)\left(\cos\theta\widehat{\mathbf{u}}^{2}+\theta\sin\theta\widehat{\mathbf{u}}\right)\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(I-\frac{\sin\theta}{\theta}\widehat{\mathbf{u}}+\frac{1-\cos\theta}{\theta^{2}}\widehat{\mathbf{u}}^{2}\right)\left(A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}A_{0}\right)+\cos\frac{\theta}{2}\left(I-\frac{\sin\theta}{\theta}\widehat{\mathbf{u}}+\frac{1-\cos\theta}{\theta^{2}}\widehat{\mathbf{u}}^{2}\right)A_{0}\\[3.0pt] &=\mathcal{C}\left(\frac{1-\cos\theta\cos\frac{\theta}{2}-2\sin\theta\sin\frac{\theta}{2}}{\theta^{2}}\right)\widehat{\mathbf{u}}+\mathcal{C}\left(\frac{\cos\frac{\theta}{2}\sin\theta-2\sin\frac{\theta}{2}\cos\theta}{\theta^{3}}\right)\widehat{\mathbf{u}}^{2}\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}A_{0}\right)-\frac{\sin\frac{\theta}{2}\sin\theta}{\theta^{2}}(\widehat{\mathbf{u}}A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}^{2}A_{0})+\left(\frac{\sin\frac{\theta}{2}(1-\cos\theta)}{\theta^{3}}\right)\left(\widehat{\mathbf{u}}^{2}A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}^{3}A_{0}\right)\\ &\quad+\cos\frac{\theta}{2}A_{0}-\frac{\cos\frac{\theta}{2}\sin\theta}{\theta}\widehat{\mathbf{u}}A_{0}+\frac{\cos\frac{\theta}{2}(1-\cos\theta)}{\theta^{2}}\widehat{\mathbf{u}}^{2}A_{0}.\end{aligned}

After some simple trigonometric manipulations, we can write (A.11) as

(A.12) A1=𝒞(1cosθcosθ22sinθsinθ2θ2)𝐮^+𝒞(2sin3θ2θ3)𝐮^2+sinθ2θ(A0𝐮^+𝐮^A0)2sin2θ2cosθ2θ2(𝐮^A0𝐮^+𝐮^2A0)+(2sin3θ2θ3)(𝐮^2A0𝐮^θ2𝐮^A0)+cosθ2A02cos2θ2sinθ2θ𝐮^A0+2sin2θ2cosθ2θ2𝐮^2A0=𝒞(1cosθcosθ22sinθsinθ2θ2)𝐮^+𝒞(2sin3θ2θ3)𝐮^2+sinθ2θ(A0𝐮^𝐮^A0)2sin2θ2cosθ2θ2𝐮^A0𝐮^+(2sin3θ2θ3)𝐮^2A0𝐮^+cosθ2A0.\displaystyle\begin{aligned} A_{1}&=\mathcal{C}\left(\frac{1-\cos\theta\cos\frac{\theta}{2}-2\sin\theta\sin\frac{\theta}{2}}{\theta^{2}}\right)\widehat{\mathbf{u}}+\mathcal{C}\left(\frac{2\sin^{3}\frac{\theta}{2}}{\theta^{3}}\right)\widehat{\mathbf{u}}^{2}\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}A_{0}\right)-\frac{2\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}}{\theta^{2}}(\widehat{\mathbf{u}}A_{0}\widehat{\mathbf{u}}+\widehat{\mathbf{u}}^{2}A_{0})+\left(\frac{2\sin^{3}\frac{\theta}{2}}{\theta^{3}}\right)\left(\widehat{\mathbf{u}}^{2}A_{0}\widehat{\mathbf{u}}-\theta^{2}\widehat{\mathbf{u}}A_{0}\right)\\ &\quad+\cos\frac{\theta}{2}A_{0}-\frac{2\cos^{2}\frac{\theta}{2}\sin\frac{\theta}{2}}{\theta}\widehat{\mathbf{u}}A_{0}+\frac{2\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}}{\theta^{2}}\widehat{\mathbf{u}}^{2}A_{0}\\[5.0pt] &=\mathcal{C}\left(\frac{1-\cos\theta\cos\frac{\theta}{2}-2\sin\theta\sin\frac{\theta}{2}}{\theta^{2}}\right)\widehat{\mathbf{u}}+\mathcal{C}\left(\frac{2\sin^{3}\frac{\theta}{2}}{\theta^{3}}\right)\widehat{\mathbf{u}}^{2}\\ &\quad+\frac{\sin\frac{\theta}{2}}{\theta}\left(A_{0}\widehat{\mathbf{u}}-\widehat{\mathbf{u}}A_{0}\right)-\frac{2\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}}{\theta^{2}}\widehat{\mathbf{u}}A_{0}\widehat{\mathbf{u}}+\left(\frac{2\sin^{3}\frac{\theta}{2}}{\theta^{3}}\right)\widehat{\mathbf{u}}^{2}A_{0}\widehat{\mathbf{u}}+\cos\frac{\theta}{2}A_{0}.\end{aligned}

To derive further simplification of (A.12), we present an elementary lemma as follows.

Lemma A.1.

The following identity holds:

𝒞𝐮^2+𝐮^2R0V(0)𝐮^=0,\mathcal{C}\widehat{\mathbf{u}}^{2}+\widehat{\mathbf{u}}^{2}R_{0}^{\top}V(0)\widehat{\mathbf{u}}=0,

where the constant 𝒞\mathcal{C} is given in (3.4).

Proof.

By the second identity in Lemma 2.1, we have

uα𝐮^2+𝐮^2Eα𝐮^=0.u_{\alpha}\widehat{\mathbf{u}}^{2}+\widehat{\mathbf{u}}^{2}E_{\alpha}\widehat{\mathbf{u}}=0.

We multiply by vα(0)v_{\alpha}(0) and sum up the resulting relation over α\alpha to get

vα(0)uα𝐮^2+𝐮^2vα(0)Eα𝐮^=0.v_{\alpha}(0)u_{\alpha}\widehat{\mathbf{u}}^{2}+\widehat{\mathbf{u}}^{2}v_{\alpha}(0)E_{\alpha}\widehat{\mathbf{u}}=0.

By vα(0)uα=𝒞v_{\alpha}(0)u_{\alpha}=\mathcal{C} and V(0)=vα(0)R1EαV(0)=v_{\alpha}(0)R_{1}E_{\alpha} (see (3.9) and (LABEL:eqn:XalphaR1)), one can obtain the desired identity. ∎

Now, we use Lemma A.1 to get

𝒞𝐮^2=𝐮^2R0V(0)𝐮^=𝐮^2A0𝐮^.\mathcal{C}\widehat{\mathbf{u}}^{2}=-\widehat{\mathbf{u}}^{2}R_{0}^{\top}V(0)\widehat{\mathbf{u}}=-\widehat{\mathbf{u}}^{2}A_{0}\widehat{\mathbf{u}}.

Finally, we use this identity in the second and the fifth terms in the R.H.S. of (A.12) to cancel, and we derive (3.17).

A.8. Proof of Lemma 3.1

Take 𝐱=(x1,x2,x3)\mathbf{x}=(x_{1},x_{2},x_{3}), 𝐲=(y1,y2,y3)\mathbf{y}=(y_{1},y_{2},y_{3}) in 3\mathbb{R}^{3}.

By direct calculation, we have

(A.13) 𝐱^𝐲^=[0x3x2x30x1x2x10][0y3y2y30y1y2y10]=[x3y3x2y2x2y1x3y1x1y2x3y3x1y1x3y2x1y3x2y3x2y2x1y1].\displaystyle\widehat{\mathbf{x}}\widehat{\mathbf{y}}=\begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix}\begin{bmatrix}0&-y_{3}&y_{2}\\ y_{3}&0&-y_{1}\\ -y_{2}&y_{1}&0\end{bmatrix}=\begin{bmatrix}-x_{3}y_{3}-x_{2}y_{2}&x_{2}y_{1}&x_{3}y_{1}\\ x_{1}y_{2}&-x_{3}y_{3}-x_{1}y_{1}&x_{3}y_{2}\\ x_{1}y_{3}&x_{2}y_{3}&-x_{2}y_{2}-x_{1}y_{1}\end{bmatrix}.

This implies

tr(𝐱^𝐲^)=2(x1y1+x2y2+x3y3)=2𝐱𝐲,\mathrm{tr}(\widehat{\mathbf{x}}\widehat{\mathbf{y}})=-2(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})=-2\,\mathbf{x}\cdot\mathbf{y},

which is the first identity in (3.18).

Then, from (A.13) we have

𝐱^𝐲^𝐲^𝐱^\displaystyle\widehat{\mathbf{x}}\widehat{\mathbf{y}}-\widehat{\mathbf{y}}\widehat{\mathbf{x}}
=[x3y3x2y2x2y1x3y1x1y2x3y3x1y1x3y2x1y3x2y3x2y2x1y1][x3y3x2y2x1y2x1y3x2y1x3y3x1y1x2y3x3y1x3y2x2y2x1y1]\displaystyle=\begin{bmatrix}-x_{3}y_{3}-x_{2}y_{2}&x_{2}y_{1}&x_{3}y_{1}\\ x_{1}y_{2}&-x_{3}y_{3}-x_{1}y_{1}&x_{3}y_{2}\\ x_{1}y_{3}&x_{2}y_{3}&-x_{2}y_{2}-x_{1}y_{1}\end{bmatrix}-\begin{bmatrix}-x_{3}y_{3}-x_{2}y_{2}&x_{1}y_{2}&x_{1}y_{3}\\ x_{2}y_{1}&-x_{3}y_{3}-x_{1}y_{1}&x_{2}y_{3}\\ x_{3}y_{1}&x_{3}y_{2}&-x_{2}y_{2}-x_{1}y_{1}\end{bmatrix}
=[0x2y1x1y2x3y1x1y3x1y2x2y10x3y2x2y3x1y3x3y1x2y3x3y30].\displaystyle=\begin{bmatrix}0&x_{2}y_{1}-x_{1}y_{2}&x_{3}y_{1}-x_{1}y_{3}\\ x_{1}y_{2}-x_{2}y_{1}&0&x_{3}y_{2}-x_{2}y_{3}\\ x_{1}y_{3}-x_{3}y_{1}&x_{2}y_{3}-x_{3}y_{3}&0\end{bmatrix}.

Since the matrix on the r.h.s. above is 𝐱×𝐲^\widehat{\mathbf{x}\times\mathbf{y}}, we get the second identity in (3.18).

Finally, also from (A.13), we get

𝐱^𝐲^𝐱^\displaystyle\widehat{\mathbf{x}}\widehat{\mathbf{y}}\widehat{\mathbf{x}} =[x3y3x2y2x2y1x3y1x1y2x3y3x1y1x3y2x1y3x2y3x2y2x1y1][0x3x2x30x1x2x10]\displaystyle=\begin{bmatrix}-x_{3}y_{3}-x_{2}y_{2}&x_{2}y_{1}&x_{3}y_{1}\\ x_{1}y_{2}&-x_{3}y_{3}-x_{1}y_{1}&x_{3}y_{2}\\ x_{1}y_{3}&x_{2}y_{3}&-x_{2}y_{2}-x_{1}y_{1}\end{bmatrix}\begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix}
=[0x3(x1y1+x2y2+x3y3)x2(x1y1+x2y2+x3y3)x3(x1y1+x2y2+x3y3)0x1(x1y1+x2y2+x3y3)x2(x1y1+x2y2+x3y3)x1(x1y1+x2y2+x3y3)0]\displaystyle=\begin{bmatrix}0&x_{3}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})&-x_{2}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})\\ -x_{3}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})&0&x_{1}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})\\ x_{2}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})&-x_{1}(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})&0\end{bmatrix}
=(𝐱𝐲)𝐱^.\displaystyle=-(\mathbf{x}\cdot\mathbf{y})\widehat{\mathbf{x}}.

A.9. Proof of Lemma 4.1.

(i) By the explicit formula of Ri(t)R_{i}(t) and Rk(t)R_{k}(t), we have

Ri(t)=Ri0exp(tAi),Rk(t)=Rk0exp(tAk).R_{i}(t)=R_{i}^{0}\exp(tA_{i}),\quad R_{k}(t)=R_{k}^{0}\exp(tA_{k}).

Recall that Ai=𝐚^i=θ𝐧^iA_{i}=\widehat{\mathbf{a}}_{i}=\theta\widehat{\mathbf{n}}_{i} and Ak=𝐚^k=θ𝐧^kA_{k}=\widehat{\mathbf{a}}_{k}=\theta\widehat{\mathbf{n}}_{k} (see (4.19)). Hence, we have:

(A.14) 𝐧^i,Rk(t)Ri(t)F=tr(𝐧^iRi(t)Rk(t))=tr(𝐧^iexp(t𝐚^i)Ri0Rk0exp(t𝐚^k))=tr(𝐧^i(Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2)),\displaystyle\begin{aligned} &\langle\widehat{\mathbf{n}}_{i},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\mathrm{tr}(\widehat{\mathbf{n}}_{i}R_{i}(t)^{\top}R_{k}(t))=\mathrm{tr}(\widehat{\mathbf{n}}_{i}\exp(-t\widehat{\mathbf{a}}_{i}){R_{i}^{0}}^{\top}R_{k}^{0}\exp(t\widehat{\mathbf{a}}_{k}))\\ &\hskip 14.22636pt=\mathrm{tr}\left(\widehat{\mathbf{n}}_{i}(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2})\right),\end{aligned}

where we used Rodrigues’s formula and the notation R0=Ri0Rk0R^{0}={R_{i}^{0}}^{\top}R_{k}^{0}.

Note that

(A.15) 𝐧^i(Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2)=(𝐧^isin(θt)𝐧^i2(1cos(θt))𝐧^i))R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2),=(cos(θt)𝐧^isin(θt)𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2),\displaystyle\begin{aligned} &\widehat{\mathbf{n}}_{i}(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2})\\ &\qquad=(\widehat{\mathbf{n}}_{i}-\sin(\theta t)\widehat{\mathbf{n}}_{i}^{2}-(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}))R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2}),\\ &\qquad=(\cos(\theta t)\widehat{\mathbf{n}}_{i}-\sin(\theta t)\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2}),\end{aligned}

where we used that 𝐧^i3=𝐧^i\widehat{\mathbf{n}}_{i}^{3}=-\widehat{\mathbf{n}}_{i} for the second equal sign. Now, by the trigonometric identities:

sin2(θt)=1cos(2θt)2, and cos2(θt)=1+cos(2θt)2,\sin^{2}(\theta t)=\frac{1-\cos(2\theta t)}{2},\quad\text{ and }\quad\cos^{2}(\theta t)=\frac{1+\cos(2\theta t)}{2},

we write the R.H.S. of (LABEL:New-3) as a linear combination of trigonometric functions:

(A.16) (cos(θt)𝐧^isin(θt)𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2)=12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k+(𝐧^i2R0𝐧^i2R0𝐧^k2)sin(θt)+(𝐧^iR0+𝐧^iR0𝐧^k2)cos(θt)+(12𝐧^iR0𝐧^k+12𝐧^i2R0𝐧^k2)sin(2θt)+(12𝐧^iR0𝐧^k2+12𝐧^i2R0𝐧^k)cos(2θt).\displaystyle\begin{aligned} &(\cos(\theta t)\widehat{\mathbf{n}}_{i}-\sin(\theta t)\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2})\\ &\hskip 5.69046pt=-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}+(-\widehat{\mathbf{n}}_{i}^{2}R^{0}-\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})\sin(\theta t)+(\widehat{\mathbf{n}}_{i}R^{0}+\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2})\cos(\theta t)\\ &\hskip 5.69046pt+(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})\sin(2\theta t)+(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}+\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k})\cos(2\theta t).\end{aligned}

Combining (LABEL:eqn:term1), (LABEL:New-3) and (LABEL:eqn:calc1) leads to identity (i) in Lemma 4.1.

(ii) Similarly, one has

(A.17) 𝐧^k,Rk(t)Ri(t)F=tr(𝐧^kRi(t)Rk(t))=tr(Ri(t)Rk(t)𝐧^k)=tr((Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2)𝐧^k).\displaystyle\begin{aligned} &\langle\widehat{\mathbf{n}}_{k},R_{k}(t)^{\top}R_{i}(t)\rangle_{\mathrm{F}}=\mathrm{tr}(\widehat{\mathbf{n}}_{k}R_{i}(t)^{\top}R_{k}(t))=\mathrm{tr}(R_{i}(t)^{\top}R_{k}(t)\widehat{\mathbf{n}}_{k})\\ &\hskip 14.22636pt=\mathrm{tr}((I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2})\widehat{\mathbf{n}}_{k}).\end{aligned}

Note that

(A.18) (Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(I+sin(θt)𝐧^k+(1cos(θt))𝐧^k2)𝐧^k=(Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(𝐧^k+sin(θt)𝐧^k2(1cos(θt))𝐧^k)=(Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(cos(θt))𝐧^k+sin(θt)𝐧^k2).\displaystyle\begin{aligned} &(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(I+\sin(\theta t)\widehat{\mathbf{n}}_{k}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{k}^{2})\widehat{\mathbf{n}}_{k}\\ &\qquad=(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(\widehat{\mathbf{n}}_{k}+\sin(\theta t)\widehat{\mathbf{n}}_{k}^{2}-(1-\cos(\theta t))\widehat{\mathbf{n}}_{k})\\ &\qquad=(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(\cos(\theta t))\widehat{\mathbf{n}}_{k}+\sin(\theta t)\widehat{\mathbf{n}}_{k}^{2}).\end{aligned}

By writing the R.H.S. as a linear combination of trigonometric functions we find:

(A.19) (Isin(θt)𝐧^i+(1cos(θt))𝐧^i2)R0(cos(θt))𝐧^k+sin(θt)𝐧^k2)=12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k+(R0𝐧^k2+𝐧^i2R0𝐧^k2)sin(θt)+(R0𝐧^k+𝐧^i2R0𝐧^k)cos(θt)+(12𝐧^iR0𝐧^k12𝐧^i2R0𝐧^k2)sin(2θt)+(12𝐧^iR0𝐧^k212𝐧^i2R0𝐧^k)cos(2θt).\displaystyle\begin{aligned} &(I-\sin(\theta t)\widehat{\mathbf{n}}_{i}+(1-\cos(\theta t))\widehat{\mathbf{n}}_{i}^{2})R^{0}(\cos(\theta t))\widehat{\mathbf{n}}_{k}+\sin(\theta t)\widehat{\mathbf{n}}_{k}^{2})\\ &=-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}+(R^{0}\widehat{\mathbf{n}}_{k}^{2}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})\sin(\theta t)+(R^{0}\widehat{\mathbf{n}}_{k}+\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k})\cos(\theta t)\\ &+(-\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k}^{2})\sin(2\theta t)+(\frac{1}{2}\widehat{\mathbf{n}}_{i}R^{0}\widehat{\mathbf{n}}_{k}^{2}-\frac{1}{2}\widehat{\mathbf{n}}_{i}^{2}R^{0}\widehat{\mathbf{n}}_{k})\cos(2\theta t).\end{aligned}

Combining (LABEL:eqn:term2), (LABEL:New-4) and (LABEL:eqn:calc2) leads to identity (ii). ∎

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