2.1. SO(3) distance metrics and Euler’s theorem

2.1.1. Upper bound from trace inequality

As will be shown in Section 3, (2) holds for skew-symmetric matrices with some caveats. This is relevant to the topic of $SO(3)$ matrices. It is well known that by Euler’s theorem, every $3\times 3$ rotation matrix can be written as

\begin{equation*} R = \exp (\theta \hat {\textbf {n}}) \end{equation*}

where $\textbf{n}$ is the unit vector in the direction of the rotation axis, $\hat{\textbf{n}}$ is the unique skew-symmetric matrix such that

\begin{equation*} \hat {\textbf {n}} \, \textbf {v} = \textbf {n} \,\times \, \textbf {v} \end{equation*}

for any $\textbf{v} \in \mathbb{R}^3$ , $\times$ is the cross product, and $\theta$ is the angle of the rotation. Letting $\textbf{n}$ roam the whole sphere and restricting $\theta \in [0,\pi ]$ covers all rotations, with redundancy at a set of measure zero. Since

\begin{equation*} \textrm {trace}(R) = 1 + 2 \cos \theta,\end{equation*}

from this equation, $\theta$ can be extracted from $R$ as

\begin{equation*} \theta (R) \,=\, \cos ^{-1} \left (\frac {\textrm {trace}(R) - 1}{2}\right ) \,.\end{equation*}

It can be shown that given two rotation matrices, then a valid distance metric is [Reference Park11]

\begin{equation*} d(R_1, R_2) \,=\, \theta (R_1^T R_2) \,. \end{equation*}

It is not difficult to show that this satisfies symmetry and positive definiteness. However, proving the triangle inequality is more challenging. But if the Golden-Thompson inequality (2) can be extended to the case of skew-symmetric matrices, it would provide a proof of the triangle inequality of the above $\theta (R_1^T R_2)$ distance metric. In order to see this, assume that

\begin{equation*} R_1^T R_2 = e^{\theta _1 \hat {\textbf {n}}_1} \,\,\,\textrm {and}\,\,\, R_2^T R_3 = e^{\theta _2 \hat {\textbf {n}}_2}, \end{equation*}

with $\theta _1 \in [0,\pi ]$ and $\theta _2 \in [0,\pi ]$ . It is not difficult to see that

\begin{equation*} \theta _1 + \theta _2 \,\geq \, \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \,\end{equation*}

since

\begin{equation*} \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \|^2 = \theta _1^2 + \theta _2^2 + 2\theta _1 \theta _2 \textbf {n}_1 \cdot \textbf {n}_2 \end{equation*}

and $\textbf{n}_1 \cdot \textbf{n}_2 \in [-1,1]$ . On one hand, if $\|\theta _1 \textbf{n}_1 +\theta _2 \textbf{n}_2 \| \leq \pi$ and (2) does hold for skew-symmetric matrices, then computing

\begin{equation*} f_1 \doteq \textrm {trace}\left (e^{\theta _1 \hat {\textbf {n}}_1 +\theta _2 \hat {\textbf {n}}_2}\right ) = 1 + 2 \cos \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \end{equation*}

and

\begin{equation*} f_2 \doteq \textrm {trace}\left (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}\right ) = 1 + 2 \cos \theta \left (e^{\theta _1 \hat {\textbf {n}}_1} e^{\theta _2 \hat {\textbf {n}}_2}\right ) \end{equation*}

and observing (2) would give

\begin{equation*} f_1 \leq f_2 \,.\end{equation*}

But the function $f(\theta ) = 1 + 2 \cos \theta$ is monotonically nonincreasing when $\theta \in [0, \pi ]$ , so $f(\theta ) \leq f(\phi )$ implies $\theta \geq \phi$ . Therefore, if (2) holds, then

\begin{equation*} \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \,\geq \, \theta (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}) \,. \end{equation*}

Then

\begin{equation*} d(R_1,R_2) + d(R_2,R_3) = \theta _1 + \theta _2 \geq \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \geq \theta (R_1^T R_2 R_2^T R_3)= \theta (R_1^T R_3) = d(R_1,R_3). \end{equation*}

On the other hand, if $\|\theta _1 \textbf{n}_1 +\theta _2 \textbf{n}_2 \| \gt \pi$ , then

\begin{equation*} d(R_1,R_2) + d(R_2,R_3) = \theta _1 + \theta _2 \geq \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \gt \pi \geq \theta (R_1^T R_3) = d(R_1,R_3). \end{equation*}

Therefore, if the Golden-Thompson inequality can be generalized to the case of skew-symmetric matrices, the result will be stronger than the triangle inequality for the $SO(3)$ metric $\theta (R_1^T R_2)$ since the latter follows from the former.

2.1.2. Lower bound from quaternion sphere

Alternatively, unit quaternions provide a simple way to encode the axis-angle representation, that is,

\begin{equation*} \textbf {q} (\textbf {n}, \theta )= \left (\textbf {n} \sin {\frac {\theta }{2}}, \cos {\frac {\theta }{2}}\right ), \ \text {where} \ \|\textbf {q} (\textbf {n}, \theta ) \| = 1, \end{equation*}

and we have the quaternion composition formula [Reference Rodrigues12]

(3) \begin{equation} \cos{\left (\frac{\theta (e^{\theta _1 \hat{\textbf{n}}_1}e^{\theta _2 \hat{\textbf{n}}_2})}{2}\right )} = \cos{\frac{\theta _1}{2}}\cos{\frac{\theta _2}{2}} - \textbf{n}_1 \cdot \textbf{n}_2 \sin{\frac{\theta _1}{2}} \sin{\frac{\theta _2}{2}}. \end{equation}

We can show that $\theta (e^{\theta _1 \hat{\textbf{n}}_1}e^{\theta _2 \hat{\textbf{n}}_2})$ is bounded from below such that

\begin{equation*} \theta (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}) \geq 2\|\textbf {q}(\textbf {n}_1, \theta _1) - \textbf { q}(-\textbf {n}_2, \theta _2)\|, \end{equation*}

provided that

\begin{equation*} \cos {\frac {\theta _1}{2}}\cos {\frac {\theta _2}{2}} - \textbf {n}_1 \cdot \textbf {n}_2 \sin {\frac {\theta _1}{2}} \sin {\frac {\theta _2}{2}} \geq 0. \end{equation*}

To see this, let

\begin{equation*} W = \cos {\frac {\theta _1}{2}}\cos {\frac {\theta _2}{2}} - \textbf {n}_1 \cdot \textbf {n}_2 \sin {\frac {\theta _1}{2}} \sin {\frac {\theta _2}{2}} \in [-1, 1], \end{equation*}

and

\begin{equation*} \mathcal {Q} = 2\|\textbf {q}(\textbf {n}_1, \theta _1) - \textbf {q}(-\textbf {n}_2, \theta _2)\| = 2 \sqrt {2 - 2\left ( \cos {\frac {\theta _1}{2}}\cos {\frac {\theta _2}{2}} - \textbf {n}_1 \cdot \textbf {n}_2 \sin {\frac {\theta _1}{2}} \sin {\frac {\theta _2}{2}} \right )} = 2 \sqrt {2-2W}. \end{equation*}

Let

\begin{equation*}f(h) = h^2 + 2\cos {h} - 2, h \in [0, +\infty ).\end{equation*}

It is easy to compute the derivative

\begin{equation*}f^{'}(h) =2h-2\sin h \geq 0.\end{equation*}

Thus,

\begin{equation*}f(h_*) = h_*^2 + 2\cos {h_*} - 2\geq f(0) = 0,\end{equation*}

that is, $\cos{h_*} \geq \frac{2-h_*^2}{2}$ for any $h_* \geq 0$ . Substituting $h_*$ with $\sqrt{2-2W} \in [0, 2]$ gives

\begin{equation*}\cos {\sqrt {2-2W}} \geq W.\end{equation*}

Let $\beta \in [0, \pi ]$ such that $\cos \beta = W$ , we have $\sqrt{2-2W} \leq \beta$ , that is, $\mathcal{Q} = 2\sqrt{2-2W} \leq 2\beta$ . If $W \in [0,1]$ , then $\beta \in [0, \frac{\pi }{2}]$ and $2\beta \in [0, \pi ]$ . So by (3),

\begin{equation*} \theta (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}) = 2\beta \geq 2\|\textbf {q}(\textbf {n}_1, \theta _1) - \textbf {q}(-\textbf {n}_2, \theta _2)\|. \end{equation*}

On the other hand, if $W \in [-1,0)$ , then $\beta \in [\frac{\pi }{2}, \pi ]$ and $2\beta \in [\pi,2\pi ]$ . According to our definition of distance metric,

\begin{equation*} \theta (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}) = 2\pi - 2\beta, \end{equation*}

which does not guarantee to be larger or equal than $2\|\textbf{q}(\textbf{n}_1, \theta _1) - \textbf{q}(-\textbf{n}_2, \theta _2)\|$ . Geometrically speaking, $\theta (e^{\theta _1 \hat{\textbf{n}}_1}e^{\theta _2 \hat{\textbf{n}}_2})$ can be regarded as distance between two rotations $e^{\theta _1 \hat{\textbf{n}}_1}$ and $e^{-\theta _2 \hat{\textbf{n}}_2}$ , which equals to the arc length between $\textbf{q}(\textbf{n}_1, \theta _1)$ and $ \textbf{q}(-\textbf{n}_2, \theta _2)$ of the quaternion sphere. Furthermore, the arc length $\boldsymbol{s}$ is just the radian angle between $\textbf{q}(\textbf{n}_1, \theta _1)$ and $ \textbf{q}(-\textbf{n}_2, \theta _2)$ , that is,

\begin{equation*} \cos {\boldsymbol {s}} = \textbf {q}(\textbf {n}_1, \theta _1) \cdot \textbf {q}(-\textbf {n}_2, \theta _2) = \cos {\frac {\theta _1}{2}}\cos {\frac {\theta _2}{2}} - \textbf {n}_1 \cdot \textbf {n}_2 \sin {\frac {\theta _1}{2}} \sin {\frac {\theta _2}{2}}. \end{equation*}

But the arc length will always be larger than or equal to the Euclidean distance between $\textbf{q}(\textbf{n}_1, \theta _1)$ and $ \textbf{q}(-\textbf{n}_2, \theta _2)$ , that is,

\begin{equation*} \boldsymbol {s} \geq \|\textbf {q}(\textbf {n}_1, \theta _1) - \textbf {q}(-\textbf {n}_2, \theta _2)\|, \end{equation*}

which is equivalent to the lower bound discussed above (Fig. 1).

Figure 1.

Geometric interpretation of the lower bound inequality, where $\boldsymbol{s}$ is the arc length between $\textbf{q}(\textbf{n}_1, \theta _1)$ and $\textbf{q}(-\textbf{n}_2, \theta _2)$ on the quaternion sphere, as well as the angle between $O\textbf{q}(\textbf{n}_1, \theta _1)$ and $O\textbf{q}(-\textbf{n}_2, \theta _2)$ .

2.2. SO(4) distance metrics as an approximation for SE(3) using stereographic projection

It has been known in kinematics for decades that rigid-body motions in $\mathbb{R}^n$ can be approximated as rotations in $\mathbb{R}^{n+1}$ by identifying Euclidean space locally as the tangent plane to a sphere [Reference McCarthy13]. This has been used to generate approximately bi-invariant metrics for $SE(3)$ [Reference Etzel and McCarthy14]. Related to this are approaches that use the singular value decomposition [Reference Larochelle, Murray and Angeles15]. As with the $SO(3)$ case discussed above, if the Golden-Thompson inequality can be shown to hold for $4\times 4$ skew-symmetric matrices, then a sharper version of the triangle inequality would hold for $SO(4)$ metrics.

This is the subject of Section 3, which is the main contribution of this paper. In that section, it is shown that the Golden-Thompson inequality can be extended from Hermitian matrices to $4\times 4$ skew-symmetric matrices and therefore to the $3\times 3$ case as a special case. But before progressing to the main topic, some trace inequalities that arise naturally from other kinematic metrics are discussed. For example, the distance metric

\begin{equation*} d(R_1, R_2) \,\doteq \, \|R_1 - R_2\|_F \end{equation*}

is a valid metric where the Frobenius norm of an arbitrary real matrix is

\begin{equation*} \|A\|_F \,\doteq \, \sqrt {\textrm {trace}(A A^T)} \,. \end{equation*}

The triangle inequality for matrix norms then gives

\begin{equation*} \|R_1 - R_2\|_F + \|R_2 - R_3\|_F \,\geq \, \|R_1 - R_3\|_F \,. \end{equation*}

Since the trace is invariant under similarity transformations, the above is equivalent to

\begin{equation*} \|\mathbb {I} - R_1^T R_2\|_F + \|\mathbb {I} - R_2^T R_3\|_F \,\geq \, \|\mathbb {I} - R_1^T R_3\|_F \,. \end{equation*}

This is true in any dimension. But in the 3D case, we can go further using the same notation as in the previous section to get

(4) \begin{equation} \sqrt{3 - \textrm{trace}(e^{\theta _1 \hat{\textbf{n}}_1})} \,+\, \sqrt{3 - \textrm{trace}(e^{\theta _2 \hat{\textbf{n}}_2})} \,\geq \, \sqrt{3 - \textrm{trace}(e^{\theta _1 \hat{\textbf{n}}_1} e^{\theta _2 \hat{\textbf{n}}_2})} \,. \end{equation}

This trace inequality is equivalently written as

\begin{equation*} \sqrt {1 - \cos \theta _1} + \sqrt {1 - \cos \theta _2} \,\geq \, \sqrt {1 - \cos \theta (e^{\theta _1 \hat {\textbf {n}}_1} e^{\theta _2 \hat {\textbf {n}}_2}) \,. } \end{equation*}

2.3. SE(3) metrics as matrix norms and resulting trace inequalities

Multiple metrics for $SE(3)$ have been proposed over the past decades, as summarized recently in ref. [Reference Di Gregorio7]. The purpose of this section is to review in more detail a specific metric that has been studied in refs. [Reference Fanghella and Galletti16–Reference Martinez and Duffy18]. The concept of this metric for $SE(3)$ is to induce from the metric properties of the vector 2-norm

\begin{equation*} \|\textbf {x} \|_2 \,\doteq \, \sqrt {\textbf {x}^T \textbf {x}} \end{equation*}

in $\mathbb{R}^3$ . Since Euclidean distance is by definition invariant to Euclidean transformations, given the pair $g = (R, \textbf{t})$ , which contains the same information as a homogeneous transformation, and given the group action $g \cdot \textbf{x} \doteq R\textbf{x} + \textbf{t}$ , then

\begin{equation*} \|g \cdot \textbf {x} - g \cdot \textbf {y} \|_2 = \|\textbf {x} - \textbf {y} \|_2\,. \end{equation*}

Then, if a body with mass density $\rho (\textbf{x})$ is moved from its original position and orientation, the total amount of motion can be quantified as

(5) \begin{equation} d(g,e) \,\doteq \, \sqrt{\int _{\mathbb{R}^3}\|g\cdot \textbf{x}-\textbf{x}\|^2_2\,\rho (\textbf{x})\,d\textbf{x}} \,. \end{equation}

This metric has the left-invariance property

\begin{equation*} d(h \circ g_1, h \circ g_2) = d(g_1,g_2)\end{equation*}

where $h, g_1, g_2 \in SE(3)$ . This is because if $h = (R,\textbf{t}) \in SE(3)$ , then

\begin{equation*} (h \circ g_i) \cdot \textbf {x} = h \cdot (g_i \cdot \textbf {x}) \end{equation*}

and

\begin{align*} &\| h \cdot (g_1 \cdot \textbf {x}) - h \cdot (g_2 \cdot \textbf {x}) \|_2 = \\[6pt] &\|R [g_1\cdot \textbf {x}] + \textbf {t} - R [g_2\cdot \textbf {x}] - \textbf {t}\|_2 \\[6pt] &\quad= \|g_1\cdot \textbf {x}-g_2\cdot \textbf {x}\|_2. \end{align*}

It is also interesting to note that there is a relationship between this kind of metric for $SE(3)$ and the Frobenius matrix norm. That is, for $g = (R, \textbf{t})$ , and the corresponding homogeneous transformation $H(g) \in SE(3)$ , the integral in (5) can be computed in closed form, resulting in a weighted norm

\begin{equation*}d(g,e) = \|H(g)-\mathbb {I}_4\|_{W} \end{equation*}

where the weighted Frobenius norm is defined as

\begin{equation*} \|A\|_{W} \,\doteq \, \sqrt {\textrm {tr}(A^T W A)}\,.\end{equation*}

Here, with weighting matrix $W=W^T\in \mathbb{R}^{4\times 4}$ is $W=\left ( \begin{array}{cc} J & \textbf{0} \\ \textbf{0}^T & M \end{array} \right )$ . $M = \int _{\mathbb{R}^3} \rho (\textbf{x})\,d\textbf{x}$ is the mass, and $ J=\int _{\mathbb{R}^3} \textbf{x} \textbf{x}^{T} \rho (\textbf{x}) \,d\textbf{x}$ has a simple relationship with the moment of inertia matrix of the rigid body:

\begin{equation*} I = \int _{\mathbb {R}^3} \left ((\textbf {x}^T\textbf {x})\mathbb {I}_3-\textbf {x}\textbf {x}^T\right )\rho (\textbf {x})\,d\textbf {x}=\textrm {tr}(J)\mathbb {I}_3-J. \end{equation*}

The metric in (5) can also be written as

(6) \begin{equation} d(g,e) = \sqrt{2\textrm{tr}[(\mathbb{I}_3-R)J]+\textbf{t}\cdot \textbf{t}M} \,. \end{equation}

Furthermore, for $g_1, g_2 \in SE(3)$

(7) \begin{equation} d(g_1,g_2) = \|H(g_1)-H(g_2)\|_{W}\,, \end{equation}

as explained in detail in ref. [Reference Chirikjian and Zhou5]. When evaluating the triangle inequality for this metric,

\begin{equation*} d(g_1, g_2) + d(g_2, g_3) \,\geq \, d(g_1, g_3) \end{equation*}

gives another kind of trace inequality.

3.1. 4D case

Let $A$ be a $4 \times 4$ skew-symmetric matrix with its eigenvalues being $\{\pm \theta _1 i, \pm \theta _2 i\}$ . Without loss of generality, we assume that $\theta _1 \geq \theta _2 \geq 0$ . For every $A$ , there exists an orthogonal matrix $P$ such that [Reference Gallier and Xu19]:

\begin{equation*} \Omega _A = P^{\intercal }AP, \end{equation*}

where

\begin{equation*} \Omega _A = \left [\begin {array}{cccc} 0 & \theta _1 & 0 & 0\\ -\theta _1 & 0 & 0 & 0\\ 0 & 0 & 0 & \theta _2\\ 0 & 0 & -\theta _2 & 0\\ \end {array}\right ]. \end{equation*}

Let $\Omega _A = \theta _1 \Omega _1 + \theta _2 \Omega _2$ , where

\begin{equation*} \Omega _1 = \left [\begin {array}{cccc} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end {array}\right ] \ \text {and} \ \Omega _2 = \left [\begin {array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ \end {array}\right ]. \end{equation*}

Then

\begin{equation*} A = P\Omega _A P^{\intercal } = \sum _{i=1}^{2} \theta _i P\Omega _i P^{\intercal } = \sum _{i=1}^{2} \theta _i A_i, \end{equation*}

where $A_i = P\Omega _i P^{\intercal }$ . Notice that $\Omega _j^3 + \Omega _j = 0$ and $\Omega _1 \times \Omega _2 = 0 = \Omega _2 \times \Omega _1$ . So,

\begin{equation*} A_j^3 + A_j = \left (P\Omega _i P^{\intercal }\right )^3 + P\Omega _i P^{\intercal } = P\left (\Omega _j^3 + \Omega _j\right ) P^{\intercal } = 0, \end{equation*}

and

\begin{equation*} A_1 \times A_2 = P\Omega _1 P^{\intercal } \times P\Omega _2 P^{\intercal } = 0 = P\Omega _2 P^{\intercal } \times P\Omega _1 P^{\intercal } = A_2 \times A_1. \end{equation*}

In other words, $A_1 \ \text{and} \ A_2$ commute. Thus, we can expand the exponential of $A$ as follows:

(8) \begin{equation} \exp{A}= \exp{\left (\sum _{i=1}^{2} \theta _i A_i\right )} = \prod _{i=1}^2 \exp{\left (\theta _i A_i\right )} = \prod _{i=1}^2 \left (I+\sin{\theta _i}A_i + (1-\cos{\theta _i})A_i^2\right ). \end{equation}

The last equality comes from the fact that $A_j^3 + A_j = 0$ . Expanding the above equation gives

(9) \begin{equation} \exp{A} = I + \sum _{i=1}^2 \left (\sin{\theta _i}A_i + (1-\cos{\theta _i})A_i^2 \right ). \end{equation}

Given another $4 \times 4$ skew-symmetric matrix $B$ whose eigenvalues are $\{\pm \phi _1 i, \pm \phi _2 i\}$ and $\phi _1 \geq \phi _2 \geq 0$ , we have

(10) \begin{equation} \begin{split} & \textrm{trace} \left (\exp{A}\exp{B} \right ) = \textrm{trace} \left (P^{\intercal }\exp{(A)}\exp{(B)} P\right ) = \textrm{trace} \left (P^{\intercal }\exp{(A)}P P^{\intercal }\exp{(B)} P \right ) \\ & \hspace{2.8cm} = \textrm{trace} \left (\exp{\left (P^{\intercal } A P\right )}\exp{\left (P^{\intercal } B P\right )} \right ) = \textrm{trace} \left (\exp{\Omega _A}\exp{C} \right ), \end{split} \end{equation}

where $C=P^{\intercal } B P$ . Notice

\begin{equation*}C^{\intercal } = P^{\intercal } B^{\intercal } P = - P^{\intercal } B P = -C,\end{equation*}

so $C$ is a skew-symmetric matrix as well, and $C$ has exactly the same eigenvalues as $B$ since conjugation does not change the eigenvalues of a matrix. A similar conclusion can be drawn:

\begin{equation*} C = Q\Omega _C Q^{\intercal } = \sum _{i=1}^{2} \phi _i Q\Omega _i Q^{\intercal } = \sum _{i=1}^{2} \phi _i C_i \ \text {and} \ QQ^{\intercal } = \mathbb {I}_4. \end{equation*}

Expanding (10) via (9) gives

(11) \begin{equation} \begin{split} & \textrm{trace} \left (\exp{A}\exp{B} \right ) = \textrm{trace} \left (\exp{\Omega _A}\exp{C} \right ) = \\ & =\textrm{trace}\left (\left (I + \sum _{i=1}^2 \left (\sin{\theta _i}\Omega _{i} + (1-\cos{\theta _i})\Omega _{i}^2 \right )\right ) \left (I + \sum _{i=1}^2 \left (\sin{\phi _i}C_i + (1-\cos{\phi _i})C_i^2 \right )\right )\right ) \\ & = \textrm{trace} \left (I+\sum _{i=1}^2 \left (\sin{\theta _i}\Omega _{i} + (1-\cos{\theta _i})\Omega _{i}^2 \right )+\sum _{i=1}^2 \left (\sin{\phi _i}C_i + (1-\cos{\phi _i})C_i^2 \right )\right ) \\ & + \textrm{trace}\left (\sum _{i=1}^2 \sum _{j=1}^2 \left (\sin{\theta _i}\Omega _{i} + (1-\cos{\theta _i})\Omega _{i}^2 \right ) \left (\sin{\phi _j}C_j + (1-\cos{\phi _j})C_j^2 \right )\right ). \end{split} \end{equation}

Divide $Q$ into $2 \times 2$ blocks as follows:

(12) \begin{equation} Q = \left [\begin{array}{cc|cc} q_{11} & q_{12} & q_{13} & q_{14} \\ q_{21} & q_{22} & q_{23} & q_{24} \\ \hline q_{31} & q_{32} & q_{33} & q_{34} \\ q_{41} & q_{42} & q_{43} & q_{44} \\ \end{array}\right ] = \left [\begin{array}{cc} \boldsymbol{Q}_{11} & \boldsymbol{Q}_{12} \\ \boldsymbol{Q}_{21} & \boldsymbol{Q}_{22} \end{array} \right ], \end{equation}

where

\begin{equation*} \boldsymbol {Q}_{ij} = \left [\begin {array}{cc} q_{2i-1,2j-1} & q_{2i-1,2j} \\ q_{2i,2j-1} & q_{2i,2j} \end {array} \right ]. \end{equation*}

Denoting $\omega _{ij}=\det{\boldsymbol{Q}_{ij}}$ and $\varepsilon _{ij}=\|\boldsymbol{Q}_{ij}\|^2_{F}=q_{2i-1,2j-1}^2 + q_{2i-1,2j}^2 + q_{2i,2j-1}^2 + q_{2i,2j}^2$ , we have the following equities:

(13) \begin{equation} \begin{split} & \textrm{trace}\left (\Omega _i C_j\right ) = \textrm{trace}\left (\Omega _i Q\Omega _i Q^{\intercal } \right ) = -2\omega _{ij} \\ & \textrm{trace}\left (\Omega _i^2 C_j^2\right ) = \textrm{trace}\left (\Omega _i^2 Q\Omega _i^2 Q^{\intercal } \right ) = \varepsilon _{ij}\\ & \textrm{trace}(C_i) = \textrm{trace}(Q\Omega _i Q^{\intercal }) = \textrm{trace}(\Omega _{i})=0 \\ & \textrm{trace}(C_i^2)= \textrm{trace}(Q\Omega _i^2 Q^{\intercal }) =\textrm{trace}(\Omega _{i}^2)=-2 \\ & \textrm{trace}\left (\Omega _{i}^2 C_j\right ) = \textrm{trace}\left (\Omega _{i} C_j^2\right ) = 0. \end{split} \end{equation}

Combining (11) with (13) gives

(14) \begin{equation} \begin{split} & \textrm{trace} \left (\exp{A}\exp{B}\right ) = 4+ 2\sum _{i=1}^2\left (\cos{\theta _i}-1\right ) + 2\sum _{i=1}^2\left (\cos{\phi _i}-1\right ) \\ & \hspace{2.8cm} + \sum _{i=1}^2\sum _{j=1}^2 \left (-2\sin{\theta _i}\sin{\phi _j}\omega _{ij} + (1-\cos{\theta _i})(1-\cos{\phi _j})\varepsilon _{ij} \right ). \end{split} \end{equation}

Using the fact $\varepsilon _{11}+\varepsilon _{12}=\varepsilon _{11}+\varepsilon _{21}=\varepsilon _{22}+\varepsilon _{12}=\varepsilon _{22}+\varepsilon _{21}=2$ as $Q$ is an orthogonal matrix, we can reduce (14) to

(15) \begin{equation} \textrm{trace} \left (\exp{A}\exp{B}\right ) = \sum _{i=1}^2 \sum _{j=1}^2 \left ( \cos{\theta _i} \cos{\phi _j}\varepsilon _{ij} - 2\sin{\theta _i} \sin{\phi _j}\omega _{ij} \right ). \end{equation}

For convenience, in the following, we will use $L_1$ to denote $\textrm{trace} \left (\exp{A}\exp{B}\right )$ and $L_2$ to denote $\textrm{trace} \left (\exp{(A+B)}\right )$ .

Lemma 3.1. Let $\omega _{11} = \det{(\boldsymbol{Q_{11}})}, \omega _{12} = \det{(\boldsymbol{Q_{12}})}, \ \text{and} \ \varepsilon _{11}=\|\boldsymbol{Q}_{11}\|^2_{F}$ , where $\boldsymbol{Q_{11}}$ and $\boldsymbol{Q_{12}}$ are defined as in (12), then the following equity holds

\begin{equation*}\varepsilon _{11} = 1 + \omega _{11}^2 - \omega _{12}^2.\end{equation*}

Proof. Since $q_{11}q_{21}+q_{12}q_{22}+q_{13}q_{23}+q_{14}q_{24}=0$ (by orthogonality), we have

\begin{equation*} \begin {aligned} & (q_{11}q_{21}+q_{12}q_{22})^2 = q_{11}^2q_{21}^2 + q_{12}^2q_{22}^2 + 2q_{11}q_{12}q_{21}q_{22} \\ & \hspace {-0.4cm}=(q_{13}q_{23}+q_{14}q_{24})^2 = q_{13}^2q_{23}^2 + q_{14}^2q_{24}^2 + 2q_{13}q_{14}q_{23}q_{24}, \end {aligned} \end{equation*}

that is,

\begin{equation*} q_{11}^2q_{21}^2 + q_{12}^2q_{22}^2 - q_{13}^2q_{23}^2 - q_{14}^2q_{24}^2 = -2q_{11}q_{12}q_{21}q_{22} + 2q_{13}q_{14}q_{23}q_{24}. \end{equation*}

Then,

\begin{equation*} \begin {aligned} & \hspace {-0.7cm} RHS = 1 + \omega _{11}^2 - \omega _{12}^2 = 1 + (q_{11}q_{22}-q_{12}q_{21})^2 -(q_{13}q_{24}-q_{14}q_{23})^2 \\ & = 1 + q_{11}^2q_{22}^2 + q_{12}^2q_{21}^2 - q_{13}^2q_{24}^2 - q_{14}^2q_{23}^2 - 2q_{11}q_{12}q_{21}q_{22} + 2q_{13}q_{14}q_{23}q_{24} \\ & = 1 + q_{11}^2q_{22}^2+q_{12}^2q_{21}^2 - q_{13}^2q_{24}^2-q_{14}^2q_{23}^2 + q_{11}^2q_{21}^2 + q_{12}^2q_{22}^2 - q_{13}^2q_{23}^2 - q_{14}^2q_{24}^2\\ & = 1 + (q_{11}^2+q_{12}^2)(q_{21}^2+q_{22}^2) - (q_{13}^2+q_{14}^2)(q_{23}^2+q_{24}^2) \\ & = 1 + (q_{11}^2+q_{12}^2)(q_{21}^2+q_{22}^2) - (1-q_{11}^2-q_{12}^2)(1-q_{21}^2-q_{22}^2) \\ & = q_{11}^2+q_{12}^2 + q_{21}^2+q_{22}^2 = \varepsilon _{11} = LHS, \end {aligned} \end{equation*}

that is, $\varepsilon _{11} = 1 + \omega _{11}^2 - \omega _{12}^2$ .

Proof. Since $\omega _{11} = q_{11}q_{22}-q_{12}q_{21}$ , $\omega _{12} = q_{13}q_{24}-q_{14}q_{23}$ , $q_{11}^2+q_{12}^2+q_{13}^2+q_{14}^2 = 1$ , and $q_{21}^2+q_{22}^2+q_{23}^2+q_{24}^2 = 1$ , we have

\begin{equation*} \begin {aligned} & (q_{11}^2+q_{12}^2+q_{13}^2+q_{14}^2)(q_{21}^2+q_{22}^2+q_{23}^2+q_{24}^2) - (\omega _{11}+\omega _{12})^2 \\ & = (q_{11}q_{21}+q_{12}q_{22})^2 + (q_{11}q_{23}+q_{14}q_{22})^2 + (q_{11}q_{24}-q_{13}q_{22})^2 \\ & + (q_{12}q_{23}-q_{14}q_{21})^2 + (q_{12}q_{24}+q_{13}q_{21})^2 + (q_{13}q_{23}+q_{14}q_{24})^2 \geq 0, \end {aligned} \end{equation*}

that is, $(\omega _{11}+\omega _{12})^2 \leq 1$ . The same deduction gives $(\omega _{11}-\omega _{12})^2 \leq 1$ .

Let

\begin{equation*} m_1=\omega _{11}+\omega _{12} \ \text {and} \ m_2=\omega _{11}-\omega _{12}. \end{equation*}

Instantly by Lemma 3.2, we have $|m_1| \leq 1$ and $|m_2| \leq 1$ . Recall that $\varepsilon _{11} + \varepsilon _{12} = 2 = \varepsilon _{12} + \varepsilon _{22}$ , so $\varepsilon _{11}=\varepsilon _{22}$ and similarly $\varepsilon _{12}=\varepsilon _{21}$ . By Lemme 3.1, we have

\begin{equation*}\varepsilon _{11}=\varepsilon _{22}=1+m_1m_2\end{equation*}

and

\begin{equation*}\varepsilon _{12}=\varepsilon _{21}=2-\varepsilon _{11}=1-m_1m_2.\end{equation*}

Since $Q$ is an orthogonal matrix, $\det Q = \pm 1$ which is denoted as $\mu$ . In ref. [Reference Hudson20], the author has shown that

\begin{equation*}\det {\boldsymbol {Q}_{11}} = \det {\boldsymbol {Q}_{22}\det {Q}}\end{equation*}

and

\begin{equation*}\det {\boldsymbol {Q}_{12}} = \det {\boldsymbol {Q}_{21}\det {Q}}\end{equation*}

if $Q$ is an orthogonal matrix. Therefore, we have

\begin{equation*} \omega _{22} = \mu \omega _{11}= \frac {\mu (m1+m2)}{2}, \ \text {and} \ \omega _{21}=\mu \omega _{12}=\frac {\mu (m1-m2)}{2}. \end{equation*}

Substituting $\omega _{ij}$ and $\varepsilon _{ij}$ with $m_1$ and $m_2$ into (15) gives

(16) \begin{equation} \begin{split} & \hspace{-0.4cm} L_1 = (1+m_1m_2)\cos{\theta _1}\cos{\phi _1} + (1-m_1m_2)\cos{\theta _1}\cos{\phi _2} \\ & + (1-m_1m_2)\cos{\theta _2}\cos{\phi _1} + (1+m_1m_2)\cos{\theta _2}\cos{\phi _2} \\ & -(m_1+m_2)\sin{\theta _1}\sin{\phi _1} - (m_1-m_2)\sin{\theta _1}\sin{\phi _2} \\ & - \mu (m_1-m_2)\sin{\theta _2}\sin{\phi _1} - \mu (m_1+m_2)\sin{\theta _2}\sin{\phi _2}. \end{split} \end{equation}

On the other hand,

(17) \begin{equation} L_2 = \textrm{trace}\left (\exp{(A+B)} \right ) = \textrm{trace}\left (P^{\intercal }\exp{(A+B)} P\right ) = \textrm{trace}\left (\exp{(\Omega _A+C)} \right ). \end{equation}

Let $D =\Omega _A+C = \sum _{i=1}^2 \left ( \theta _i \Omega _i + \phi _i Q \Omega _i Q^{\intercal } \right )$ . The characteristic polynomial $\mathcal{P}(\lambda )$ of D is

\begin{equation*} \mathcal {P}(\lambda ) = \lambda ^4 + \mathcal {P}_1 \lambda ^2 + \mathcal {P}_2, \end{equation*}

where

\begin{equation*} \mathcal {P}_1 = \theta _1^2 + \theta _2^2 + \phi _1^2 + \phi _2^2 + 2\omega _{11}\theta _1 \phi _1 + 2\omega _{12}\theta _1 \phi _2 + 2\omega _{21}\theta _2 \phi _1 + 2\omega _{22}\theta _2 \phi _2, \end{equation*}

and

\begin{equation*} \mathcal {P}_2 = \left (\theta _1\theta _2 + \phi _1\phi _2 + \omega _{12}\theta _1\phi _1 + \omega _{11}\theta _1\phi _2 + \omega _{22}\theta _2\phi _1 + \omega _{21}\theta _2\phi _2\right )^2. \end{equation*}

Using the face that $\omega _{22}=\mu \omega _{11}$ and $\omega _{21}=\mu \omega _{12}$ , we can solve the above quartic equation:

(18) \begin{equation} \lambda _{1,2} = \pm \left (\frac{\sqrt{f_1}+\sqrt{f_2}}{2}\right )i = \pm \psi _1 i, \ \lambda _{3,4} = \pm \left (\frac{|\sqrt{f_1}-\sqrt{f_2}|}{2}\right )i=\pm \psi _2 i, \end{equation}

where

\begin{equation*} f_1 = (\theta _1+\mu \theta _2)^2 + (\phi _1+\phi _2)^2 + 2(\theta _1+\mu \theta _2)(\phi _1+\phi _2)(\omega _{11}+\omega _{12}), \end{equation*}

and

\begin{equation*} f_2 = (\theta _1-\mu \theta _2)^2 + (\phi _1-\phi _2)^2 + 2(\theta _1-\mu \theta _2)(\phi _1-\phi _2)(\omega _{11}-\omega _{12}). \end{equation*}

Both $f_1$ and $f_2$ are guaranteed to be greater than or equal to $0$ since $|\omega _{11}+\omega _{12}| \leq 1$ , $|\omega _{11}-\omega _{12}| \leq 1$ (Lemme 3.2), $\theta _1 \pm \mu \theta _2 \geq 0$ , and $\phi _1 \pm \phi _2 \geq 0$ . So, expanding (17) by (9) gives

(19) \begin{equation} \begin{split} & L_2 = \textrm{trace}\left (I+\sin{\psi _1}D_1+(1-\cos{\psi _1})D_1^2 + \sin{\psi _2}D_2+(1-\cos{\psi _2})D_2^2 \right ) \\ & \hspace{0.4cm} = 4 + 2(\cos{\psi _1}-1) + 2(\cos{\psi _2}-1) = 2\cos{\psi _1} + 2\cos{\psi _2} \\ & \hspace{0.4cm} = 2\cos{\left (\frac{\sqrt{f_1}+\sqrt{f_2}}{2}\right )} + 2\cos{\left (\frac{\sqrt{f_1}-\sqrt{f_2}}{2}\right )} = 4\cos{\frac{\sqrt{f_1}}{2}}\cos{\frac{\sqrt{f_2}}{2}}. \end{split} \end{equation}

Perform the following coordinate transformations:

\begin{equation*} \left [\begin {array}{c} x_1 \\ x_2 \\ y_1 \\ y_2 \\ \end {array} \right ] = \left [\begin {array}{cccc} 0.5 & 0.5\mu & 0.5 & 0.5 \\ 0.5 & -0.5\mu & 0.5 & -0.5 \\ -0.5 & -0.5\mu & 0.5 & 0.5 \\ -0.5 & 0.5\mu & 0.5 & -0.5 \end {array} \right ] \left [\begin {array}{c} \theta _1 \\ \theta _2 \\ \phi _1 \\ \phi _2 \\ \end {array} \right ]. \end{equation*}

Applying the above transformation to (16) and (19) gives

(20) \begin{equation} L_1 = (\cos{x_1}+\cos{y_1}+m_1\cos{x_1}-m_1\cos{y_1})(\cos{x_2}+\cos{y_2}+m_2\cos{x_2}-m_2\cos{y_2}), \end{equation}

and

(21) \begin{equation} L_2 = 2\cos{\frac{K_1}{2}} \cdot 2\cos{\frac{K_2}{2}}, \end{equation}

where

\begin{equation*} K_1 = \sqrt {2(1+m_1)x_1^2 + 2(1-m_1)y_1^2}, \end{equation*}

and

\begin{equation*} K_2 = \sqrt {2(1+m_2)x_2^2 + 2(1-m_2)y_2^2}. \end{equation*}

Lemma 3.3. Let $a = \frac{K}{\sqrt{2(1+\zeta )}}$ and $b = \frac{K}{\sqrt{2(1-\zeta )}}$ , where $\zeta \in (-1,1)$ and $K \in (0,\pi ]$ . Then

\begin{equation*} (1+\zeta )\cos a + 1-\zeta \gt 2\cos \frac {K}{2} \ \operatorname {and} \ (1-\zeta )\cos b+1+\zeta \gt 2\cos \frac {K}{2}. \end{equation*}

Proof. Let

\begin{equation*} f(p) = -\frac {\sin ^2(p)}{p^2}, \ \operatorname {where} \ p \in (0, +\infty ), \end{equation*}

and the derivative of $f(p)$ is

\begin{equation*} \frac {df}{dp} = \frac {2\sin ^2(p) - 2\sin (p)\cos (x)p}{p^3}. \end{equation*}

It is not difficult to conclude that $\frac{df}{dp} \gt 0$ when $p \in (0,\pi )$ ; that is, $f(p)$ is strictly increasing as $p \in (0,\pi ]$ . Assume that there exists a $p_0$ such that $f(p_0) \lt f(\frac{\pi }{2})$ , then

\begin{equation*} -\frac {\sin ^2{p_0}}{p_0^2} \lt -\frac {1}{\frac {\pi ^2}{4}}, \ i.e., \ \frac {4p_0^2}{\pi ^2} \lt \sin ^2 p_0 \leq 1. \end{equation*}

So if such $p_0$ exists, it must satisfy $p_0 \lt \frac{\pi }{2}$ , which means for any $p\geq \frac{\pi }{2}$ , we have

\begin{equation*} f(p) \geq f(\frac {\pi }{2}) \gt f(\frac {\pi }{4})\geq f(\frac {K}{4}). \end{equation*}

Let $q=\sqrt{2(1+\zeta )} \in (0,2)$ , then $\frac{1}{q} \gt \frac{1}{2}$ and $\frac{K}{2q} \gt \frac{K}{4}$ . If $\frac{K}{2q} \lt \frac{\pi }{2}$ , then

\begin{equation*} f\left (\frac {\pi }{2}\right ) \gt f\left (\frac {K}{2q}\right ) \gt f\left (\frac {K}{4}\right ) \end{equation*}

since $f$ is strictly increasing within that range. Otherwise if $\frac{K}{2q} \geq \frac{\pi }{2}$ , we have $f\left (\frac{K}{2q}\right ) \gt f(\frac{\pi }{4})\geq f(\frac{K}{4})$ . In other words, the following inequality is always valid:

\begin{equation*} f\left (\frac {K}{2q}\right ) =\frac {4q^2(-\sin ^2\frac {K}{2q})}{K^2}=\frac {2q^2(\cos \frac {K}{q}-1)}{K^2} \gt f\left (\frac {K}{4}\right ) = \frac {8(\cos \frac {K}{2}-1)}{K^2}. \end{equation*}

Multiplying both sides by $\frac{K^2}{4}$ gives

\begin{equation*} \frac {q^2}{2}(\cos \frac {K}{q} -1) = (1+\zeta )\left (\cos {\frac {K}{\sqrt {2(1+\zeta )}}}-1 \right )\gt 2(\cos \frac {K}{2}-1), \end{equation*}

that is,

\begin{equation*} (1+\zeta )\cos {a} + 1 - \zeta \gt 2\cos \frac {K}{2}. \end{equation*}

By letting $\zeta = -\zeta$ , we have

\begin{equation*} (1-\zeta )\cos b+1+\zeta \gt 2\cos \frac {K}{2}. \end{equation*}

Lemma 3.4. If $x\gt 0$ , $y\gt 0$ , $\zeta \in (-1,1)$ , and $K \in (0,\pi ]$ , then the only solution to the following equation is $x=y=K/2$

\begin{equation*} \left \{\begin {array}{l} \frac {\sin x}{x} = \frac {\sin y }{y} \\ (1+\zeta ) x^2+(1-\zeta ) y^2 = \frac {K^2}{2} \end {array}\right .. \end{equation*}

Proof. Let $h(x)=\frac{\sin x}{x}$ . Assume that there exists $x_1 \gt x_2 \gt 0$ such that $h(x_1)=h(x_2)$ . The derivative of $h(x)$ is

\begin{equation*} h^\prime (x) = \frac {x\cos x - \sin x}{x^2}. \end{equation*}

When $x\in (0,\frac{\pi }{2}]$ , $x\cos x - \sin x$ will always be smaller than 0; that is, $h(x)$ is strictly decreasing. Therefore, to have $h(x_1)=h(x_2)$ , $x_1$ must be greater than $\frac{\pi }{2}$ . Assume $x_2 \leq \frac{\pi }{2}$ , then

\begin{equation*} h(x_1) = h(x_2) \geq h(\frac {\pi }{2})=\frac {2}{\pi }. \end{equation*}

Thus, we have

\begin{equation*} \frac {2}{\pi } \leq h(x_1) \leq \frac {\sin x_1}{x_1}, \ i.e. \ \frac {2x_1}{\pi } \leq \sin x_1 \leq 1, \end{equation*}

which leads to $x_1 \leq \frac{\pi }{2}$ , contradicting what we previously stated. Thus, both $x_1$ and $x_2$ need to be larger than $\frac{\pi }{2}$ . However,

\begin{equation*} \frac {K^2}{2} = (1+\zeta ) x_1^2+(1-\zeta ) x_2^2 \gt (1 + \zeta + 1 - \zeta ) \cdot \frac {\pi ^2}{4}= \frac {\pi ^2}{2}, \end{equation*}

which causes a contradiction since $K \in (0,\pi ]$ .

Theorem 3.5. If $(1+\zeta ) x^2+(1-\zeta ) y^2=\frac{K^2}{2}, \zeta \in [-1,1], \ \operatorname{and} \ K \in [0,\pi ]$ , then

\begin{equation*} \cos x + \cos y + \zeta \cos x - \zeta \cos y \geq 2\cos {\left (\frac {K}{2} \right )} \geq 0. \end{equation*}

Proof. If $\zeta = 1$ , then $x = \pm \frac{K}{2}$ . So, $LHS = 2\cos{\left (\frac{K}{2} \right )} = RHS$ . Same for $\zeta = -1$ . If $K = 0$ , then $x=y=0$ . So, $LHS = 0 = RHS$ . Now, let us restrict $\zeta \in (-1,1)$ and $K \in (0,\pi ]$ . Let

\begin{equation*} f(x,y) = \cos x + \cos y + \zeta \cos x - \zeta \cos y - 2\cos {\left (\frac {K}{2} \right )}, \end{equation*}

and

\begin{equation*} g(x,y) = (1+\zeta ) x^2+(1-\zeta ) y^2 - \frac {K^2}{2}. \end{equation*}

To find the minimum of $f(x,y)$ subjected to the equality constraint $g(x,y)=0$ , we form the following Lagrangian function:

\begin{equation*} \mathcal {L}(x,y,\lambda )=f(x,y) + \lambda g(x,y), \end{equation*}

where $\lambda$ is the Lagrange multiplier. Notice that $\mathcal{L}(\pm x,\pm y, \lambda )=f(\pm x,\pm y)+\lambda g(\pm x,\pm y)=f(x,y) + \lambda g(x,y) = \mathcal{L}(x,y,\lambda )$ . Thus, the Lagrangian function is symmetric about $x=0$ and $y=0$ . So, we only need to study how $\mathcal{L}(x,y,\lambda )$ behaves with $(x,y)\in [0,+\infty ) \times [0,+\infty )$ . To find stationary points of $\mathcal{L}$ , we have

\begin{equation*} \left \{\begin {array}{l} \frac {\partial \mathcal {L}}{\partial x}=[-(1+\zeta ) \sin x]+2 \lambda (1+\zeta ) x=0 \\ \frac {\partial \mathcal {L}}{\partial y}=[-(1-\zeta ) \sin y]+2 \lambda (1-\zeta ) y=0 \\ \frac {\partial \mathcal {L}}{\partial \lambda }=(1+\zeta ) x^2+(1-\zeta ) y^2-\frac {K^2}{2}=0 \end {array}\right .. \end{equation*}

We can readily obtain three sets of solutions to the above equation:

1. 1. $x=0$ , $y=\sqrt{\frac{K^2}{2(1-\zeta )}}$ and $\lambda =\frac{\sin y}{2y}$ ;

2. 2. $x=y$ , $x=y=\frac{K}{2}$ and $\lambda =\frac{\sin{\frac{K}{2}}}{K}$ ;

3. 3. $y=0$ , $x=\sqrt{\frac{K^2}{2(1+\zeta )}}$ and $\lambda =\frac{\sin x}{2x}$ .

To have a fourth solution, we need to satisfy

\begin{equation*} \frac {(1+\zeta ) \sin x}{2(1+\zeta )x} = \lambda = \frac {(1-\zeta ) \sin y}{2(1-\zeta )y} \end{equation*}

and

\begin{equation*} (1+\zeta ) x^2+(1-\zeta ) y^2-\frac {K^2}{2}=0. \end{equation*}

However, by Lemma 3.4, we conclude that there are no other solutions. Substituting those solutions back into $f(x,y)$ gives

\begin{equation*} \begin {aligned} &f\left (0, \sqrt {\frac {K^2}{2(1-\zeta )}}\right ) = (1-\zeta )\cos \left (\sqrt {\frac {K^2}{2(1-\zeta )}}\right ) + 1+\zeta - 2\cos \left (\frac {K}{2}\right ) \\ &f\left (\frac {K}{2},\frac {K}{2}\right ) = 2\cos \left (\frac {K}{2}\right ) - 2\cos \left (\frac {K}{2}\right ) = 0 \\ & f\left (\sqrt {\frac {K^2}{2(1+\zeta )}},0\right ) = (1+\zeta )\cos \left (\sqrt {\frac {K^2}{2(1+\zeta )}}\right ) + 1-\zeta - 2\cos \left (\frac {K}{2}\right ). \end {aligned} \end{equation*}

By Lemma 3.3, we have $f\left (0, \sqrt{\frac{K^2}{2(1-\zeta )}}\right )\gt 0$ and $f\left (\sqrt{\frac{K^2}{2(1+\zeta )}},0\right )\gt 0$ . Therefore, we can conclude that the global minimum for $f(x,y)$ subjected to $g(x,y)=0$ is zero, that is,

\begin{equation*} \cos x + \cos y + \zeta \cos x - \zeta \cos y \geq 2\cos {\left (\frac {K}{2} \right )}. \end{equation*}

Now recall that

\begin{equation*} L_1 = (\cos {x_1}+\cos {y_1}+m_1\cos {x_1}-m_1\cos {y_1})(\cos {x_2}+\cos {y_2}+m_2\cos {x_2}-m_2\cos {y_2}), \end{equation*}

and

\begin{equation*} L_2 = 2\cos {\frac {K_1}{2}} \cdot 2\cos {\frac {K_2}{2}}. \end{equation*}

where

\begin{equation*} K_1 = \sqrt {2(1+m_1)x_1^2 + 2(1-m_1)y_1^2}, \end{equation*}

and

\begin{equation*} K_2 = \sqrt {2(1+m_2)x_2^2 + 2(1-m_2)y_2^2}. \end{equation*}

By (18), we know

\begin{equation*} \psi _1 = \frac {K_1+K_2}{2} \geq 0 \ \text {and} \ \psi _2 = \frac {|K_1-K_2|}{2} \geq 0, \end{equation*}

where $\{\pm \psi _1i,\pm \psi _2i\}$ are eigenvalues of $A+B$ . With the condition $\psi _1 + \psi _2 \leq \pi$ , if $K_1 \geq K_2$ , then $\psi _1 + \psi _2=K_1 \leq \pi$ and so $K_2 \leq K_1 \leq \pi$ ; otherwise if $K_1 \lt K_2$ , then $\psi _1 + \psi _2=K_2 \leq \pi$ and $K_1 \lt K_2 \leq \pi$ . In both cases, we have $K_1 \leq \pi$ and $K_2 \leq \pi$ . By Theorem 3.5, we have

\begin{equation*} \cos {x_1}+\cos {y_1}+m_1\cos {x_1}-m_1\cos {y_1} \geq 2\cos {\frac {K_1}{2}} \geq 0, \end{equation*}

and

\begin{equation*} \cos {x_2}+\cos {y_2}+m_2\cos {x_2}-m_2\cos {y_2} \geq 2\cos {\frac {K_2}{2}} \geq 0. \end{equation*}

Therefore, we have $L_1 \geq L_2$ , that is,

\begin{equation*} \textrm {trace} \left (\exp {A} \exp {B}\right ) \geq \operatorname {tr}\left (\exp {(A+B)}\right ), \end{equation*}

subjected to

\begin{equation*} \psi _1 + \psi _2 \leq \pi . \end{equation*}

3.2. 3D case

Given two $3 \times 3$ skew-symmetric matrices $A$ and $B$ such that the eigenvalues of $A+B$ is $\{\pm \psi i, 0\}$ and $\psi \in [0,\pi ]$ , we can pad both $A$ and $B$ with zeros as follows:

\begin{equation*} \bar {A} = \left [\begin {array}{cc} A & \boldsymbol {0}_{3\times 1} \\ \boldsymbol {0}_{1\times 3} & 0 \end {array}\right ] \ \text {and} \ \bar {B} = \left [\begin {array}{cc} B & \boldsymbol {0}_{3\times 1} \\ \boldsymbol {0}_{1\times 3} & 0 \end {array}\right ]. \end{equation*}

Then,

\begin{equation*} \textrm {trace} \left (\exp {\bar {A}} \exp {\bar {B}}\right ) = \textrm {trace} \left (\exp {A} \exp {B} \right ) + 1, \end{equation*}

and

\begin{equation*} \textrm {trace} \left (\exp {(\bar {A}+\bar {B})}\right ) = \textrm {trace} \left (\exp {(A+B)}\right ) + 1. \end{equation*}

Notice that by padding zeros, the eigenvalues of $\bar{A}+\bar{B}$ become $\{\psi i,-\psi i,0,0\}$ . As $\psi + 0 = \psi \leq \pi$ , we have

\begin{equation*} \textrm {trace}\left (\exp {\bar {A}} \exp {\bar {B}}\right ) \geq \textrm {trace}\left (\exp {(\bar {A}+\bar {B})}\right ), \end{equation*}

that is,

\begin{equation*} \textrm {trace}\left (\exp {A} \exp {B} \right ) \geq \textrm {trace}\left (\exp {(A+B)}\right ). \end{equation*}

4.1. BCH formula

The Baker-Campbell-Hausdorff (BCH) formula gives the value of $\boldsymbol{Z}$ that solves the following equation:

\begin{equation*} \boldsymbol {Z}(\boldsymbol {X},\boldsymbol {Y}) = \log \left (\exp (\boldsymbol {X})\exp (\boldsymbol {Y})\right ) = \boldsymbol {X} + \boldsymbol {Y} + \frac {1}{2}[\boldsymbol {X},\boldsymbol {Y}] + \frac {1}{12}\left ([\boldsymbol {X},[\boldsymbol {X},\boldsymbol {Y}]] - [\boldsymbol {Y},[\boldsymbol {X},\boldsymbol {Y}]]\right ) + \cdots, \end{equation*}

where $\boldsymbol{X},\boldsymbol{Y}, \text{and} \ \boldsymbol{Z}$ are in the Lie algebra of a Lie group, $[\boldsymbol{X},\boldsymbol{Y}] = \boldsymbol{X}\boldsymbol{Y}-\boldsymbol{Y}\boldsymbol{X}$ , and $\cdots$ indicates terms involving higher commutators of $\boldsymbol{X}$ and $\boldsymbol{Y}$ . The BCH formula is used for robot state estimation [Reference Barfoot21] and error propagation on the Euclidean motion group [Reference Wang and Chirikjian22]. Let us denote all the terms after $\boldsymbol{X} + \boldsymbol{Y}$ as $\boldsymbol{W}$ and so

\begin{equation*} \begin {aligned} & \exp (\boldsymbol {Z}) = \exp (\boldsymbol {X})\exp (\boldsymbol {Y}) \\ &\boldsymbol {Z} = \log \left (\exp (\boldsymbol {X})\exp (\boldsymbol {Y})\right ) = \boldsymbol {X} + \boldsymbol {Y} + \boldsymbol {W}. \end {aligned} \end{equation*}

Considering the case of $SO(3)$ , we can write

\begin{equation*} \boldsymbol {X} = \theta _1\hat {\textbf {n}}_1 \ \text {and} \ \boldsymbol {Y} = \theta _2\hat {\textbf {n}}_2, \end{equation*}

where $\textbf{n}_i$ is the unit vector in the direction of the rotation axis, $\hat{\textbf{n}}_i$ is the unique skew-symmetric matrix such that

\begin{equation*} \hat {\textbf {n}}_i \, \textbf {v} = \textbf {n}_i \,\times \, \textbf {v} \end{equation*}

for any $\textbf{v} \in \mathbb{R}^3$ , and $\boldsymbol\theta _{\textbf{\textit{i}}} \boldsymbol\in \boldsymbol[\boldsymbol 0,\boldsymbol +\boldsymbol\infty \boldsymbol )$ is the angle of the rotation. Then,

\begin{equation*} d(\exp (\boldsymbol {Z}),\mathbb {I}_3) = d(\exp (\boldsymbol {X}+\boldsymbol {Y}+\boldsymbol {W}),\mathbb {I}_3) \leq \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \|= d(\exp (\boldsymbol {X}+\boldsymbol {Y}), \mathbb {I}_3), \end{equation*}

provided that $\|\theta _1 \textbf{n}_1 +\theta _2 \textbf{n}_2 \| \leq \pi$ . So, we conclude that the existence of $\boldsymbol{W}$ will reduce the distance between $\exp (\boldsymbol{X}+\boldsymbol{Y})$ and the identity $\mathbb{I}_3$ if $\|\theta _1 \textbf{n}_1 +\theta _2 \textbf{n}_2 \| \leq \pi$ .

4.2. Rotation fine-tuning

Euler angles are a powerful approach to decomposing rotation matrices into three sequential rotation matrices. Let us assume that a manipulator can rotate around the $x$ , $y$ , and $z$ -axis, respectively. Therefore, to rotate the manipulator to a designated orientation $R_d$ , we can compute the corresponding Euler angles $\alpha _1$ , $\beta _1$ , and $\gamma _1$ such that

\begin{equation*} R_x(\alpha _1)R_y(\beta _1)R_z(\gamma _1) = R_d. \end{equation*}

Assuming that whenever rotated, the device will incur some random noise to the input angle, that is,

\begin{equation*} R_1 = R_x(\alpha _1+\delta \alpha _1)R_y(\beta _1+\delta \beta _1)R_z(\gamma _1+\delta \gamma _1) \neq R_d, \end{equation*}

leading to deviations of the final orientation. To reduce the error, one can measure the actual rotation $R_1$ and compute another set of Euler angles $\{\alpha _2,\beta _2,\gamma _2\}$ such that

\begin{equation*} R_x(\alpha _2)R_y(\beta _2)R_z(\gamma _2) = R_dR_1^T. \end{equation*}

But inevitably, noise will again be introduced, and the actual rotation will become

\begin{equation*} R_2R_1 = R_x(\alpha _2+\delta \alpha _2)R_y(\beta _2+\delta \beta _2)R_z(\gamma _2+\delta \gamma _2)R_1 \neq R_d. \end{equation*}

Therefore, one can repeat the above process until $d(\prod _{i=1}^N R_{N-i+1}, R_d)$ is within tolerance. Another approach to reducing the inaccuracy caused by the noise is applying the following inequality:

\begin{equation*} \|\theta _1 \textbf {n}_1 +\theta _2 \textbf {n}_2 \| \,\geq \, \theta (e^{\theta _1 \hat {\textbf {n}}_1}e^{\theta _2 \hat {\textbf {n}}_2}). \end{equation*}

To refine the current rotation by rotating the x-axis, that is, minimizing $d(R_x(\alpha )R_1, R_d) = \theta (R_x(\alpha )R_1R_d^T)$ , we let $R_s = R_1R_d^T = \exp ({\theta _s\hat{\textbf{n}}_s})$ , where $\theta _s \in [0, \pi ]$ . If $\alpha = \arg \min _{\alpha } \|\theta _s \textbf{n}_s +\alpha \textbf{e}_1 \|$ , that is, $\alpha = -(\textbf{n}_s \cdot \textbf{e}_1)\theta _s$ , then

\begin{equation*} \theta (R_x(\alpha )R_1R_d^T) = \theta (R_x(\alpha )R_s) = \theta (e^{\alpha \hat {\textbf {e}}_1}e^{\theta _s \hat {\textbf {n}}_s}) \leq \|\alpha \textbf {e}_1 +\theta _s \textbf {n}_s \| = \theta _s\sqrt {1-(\textbf {n}_s \cdot \textbf {e}_1)^2} \leq \theta _s. \end{equation*}

In other words, the inequality provides a simple way to reduce the angle of the resulting rotation by rotating around an axis with a specific angle. In practice, when given the $R_s$ , we compute $|\textbf{n}_s \cdot \textbf{e}_1|$ , $|\textbf{n}_s \cdot \textbf{e}_2|$ , and $|\textbf{n}_s \cdot \textbf{e}_3|$ and choose the axis that has the largest dot value to rotate. The above process is repeated until the tolerance requirement is met.

Figure 2.

Average radian distance between the target rotation and the actual rotation.

To prove the effectiveness, we conduct the following experiment. The target rotation is chosen as $R_d = R_x(\alpha _*)R_y(\beta _*)R_z(\gamma _*)$ , where $\alpha _*$ , $\beta _*$ , and $\gamma _*$ are all random numbers from $[0,2\pi ]$ . We assume that whenever the device is rotated, there will be a noise, which is uniformly distributed within the range $[-0.15,0.15]$ , added to the input angle. In the first step, the manipulator is rotated according to the Euler angles for both methods. Then in the subsequent steps, it is refined three times either by Euler angles or by angles calculated from the inequality. For each approach, we refine the orientation to 100 steps, and at each step, the distance between the current rotation and the target rotation is measured. We conduct the above experiments 500 times, and the average distance is computed at each step for both approaches. The results are shown in Fig. 2. Overall, the radian distance is smaller if we refine the rotation by the inequality. In other words, the inequality provides a simple yet effective way to fine-tune the rotation in the presence of noise.