A group is an algebraic structure of a set plus an operation. (群是一种集合加上一种运算的代数结构)
A rotation group is a set of operator which can rotate a vector $$ r:\mathbb{R}^3\rightarrow \mathbb{R}^3; \mathbf{v} \rightarrow r(\mathbf{v})\qquad \forall\mathbf{v}\in\mathbb{R}^3 $$ They have the following properties
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Rotation preserves the vector norm $$ \Vert r(\mathbf{v})\Vert=\sqrt{\langle r(\mathbf{v}),r(\mathbf{v})\rangle}=\sqrt{\langle \mathbf{v},\mathbf{v}\rangle}=\Vert\mathbf{v}\Vert $$
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Rotation preserves angles between vectors $$ \langle r(\mathbf{v}),r(\mathbf{w})\rangle=\langle \mathbf{v},\mathbf{w}\rangle=\Vert\mathbf{v}\Vert\Vert\mathbf{w}\Vert\cos\alpha \qquad \forall \mathbf{v},\mathbf{w}\in\mathbb{R}^3 $$
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Rotation preserves the relative orientations of vectors $$ \mathbf{u}\times\mathbf{v}=\mathbf{w} \space \Longleftrightarrow \space r(\mathbf{u})\times r(\mathbf{v})=r(\mathbf{w}) $$
We can thus define the rotation group
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definition $$ Q = q_w + q_xi+q_yj+q_zk $$ a quaternion can be posed as a sum scalar + vector $$ Q = q_w + \mathbf{q}_v \ where \quad \mathbf{q}_v = [q_x,q_y,q_z]=q_xi+q_yj+q_zk $$ We should note that although a vector can be write as a matrix, it is still the weighted sum of basic vector
we mostly represent a quaternion
$Q$ as a 4-vector$\mathbf{q}$ $$ \mathbf{q}=\begin{bmatrix} q_w \ \mathbf{q}_v \end{bmatrix}\begin{bmatrix} q_w \ q_x \ q_y \ q_z \end{bmatrix} $$
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pure quaternion (where define
$\mathbf{u,v}$ ) $$ \mathbf{v} = [0,\mathbf{q}_v]^T = v_xi+v_yj+v_zk $$ let $$ \mathbf{u} = \frac{\mathbf{v}}{\Vert\mathbf{v}\Vert},\theta=\Vert\mathbf{v}\Vert \quad \rightarrow \quad \mathbf{v}=e^{\mathbf{u}\theta} $$ -
unit quaternion or normalized quaternion
Unit quaternions can be always be written in the form $$ \mathbf{q}=q_w+\mathbf{q}_v=\begin{bmatrix} \cos\theta \ \mathbf{u}\sin\theta \end{bmatrix} \quad where \quad \begin{cases} \mathbf{u}=u_xi+u_yj+u_zk & is\space a\space unit\space vector\ \theta = \arctan(q_w,\Vert \mathbf{q}_v\Vert) \end{cases} $$ this formula can represent any unit quaternion. since it is unit quaternion,
$0\leqslant q_w\leqslant 1$ .$\cos\theta$ can cover this range. also because$\mathbf{u}$ can represent any 3-vector while it independent with$\theta$ ,$\mathbf{u}\theta$ can represent any 3-vector. to sum up, $\begin{bmatrix} \cos\theta \ \mathbf{u}\sin\theta \end{bmatrix}$ can represent any 4-vector while maintain its norm$\Vert \mathbf{q}\Vert=1$ -
Exponential of pure quaternions
using Taylor expansions to derivate it. please refer to page 10 for details $$ \begin{split} e^{\mathbf{v}} &=e^{\mathbf{u}\theta}=\cos\theta+\mathbf{u}\sin\theta\ &= \cos\theta+\sin\theta(u_xi+u_yj+u_zk)\ &= \begin{bmatrix} \cos\theta \ \mathbf{u}\sin\theta \end{bmatrix} \end{split} $$ so, the exponential of a pure quaternion is a unit quaternion
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Exponential of general quaternions $$ e^{\mathbf{q}}=e^{q_w+\mathbf{q}_v}=e^{q_w}e^{\mathbf{q}_v}=e^{q_w}\begin{bmatrix} \cos\Vert \mathbf{q}_v\Vert\ \frac{\mathbf{q}_v}{\Vert \mathbf{q}_v \Vert}\sin\mathbf{q}_v \end{bmatrix} $$
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Logarithm of unit quaternions $$ \log\mathbf{q}=\log(\cos\theta+\mathbf{u}\sin\theta)=\log(e^{\mathbf{u}\theta})=\mathbf{u}\theta=\begin{bmatrix} 0 \ \mathbf{u}\theta \end{bmatrix}\ where \quad \begin{cases} \mathbf{u}=\mathbf{q}_v/\Vert \mathbf{q}_v \Vert\ \theta = \arctan(\Vert \mathbf{q}_v\Vert,q_w) \end{cases} $$
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Logarithm of general quaternions $$ \log \mathbf{q} = \log(\Vert \mathbf{q}\Vert \frac{\mathbf{q}}{\Vert \mathbf{q}\Vert})=\log \Vert \mathbf{q}\Vert+\log \frac{\mathbf{q}}{\Vert \mathbf{q}\Vert}=\log \Vert \mathbf{q}\Vert+\mathbf{u}\theta =\begin{bmatrix} \log\Vert \mathbf{q}\Vert \ \mathbf{u}\theta \end{bmatrix} $$
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rotation action using quaternion $$ r(\mathbf{v}) = \mathbf{q}\otimes \mathbf{v}\otimes \mathbf{q}^* \quad where \quad \mathbf{v}=\begin{bmatrix} 0 \ x \ y \ z \end{bmatrix}, \quad \Vert\mathbf{q}\Vert=1 $$
$(x,y,z)$ is the coordinate of the point -
The group of unit quaternion
$S^3$ The set of unit quaternions forms a group under the operation of multiplication. This group is topologically a 3-sphere, the 3-dimensional surface of the unit sphere of
$\mathbb{R}^4$ and is commonly noted as$S^3$ since the unit quaternion can be written as $$ \mathbf{q}=q_w+\mathbf{q}_v=\begin{bmatrix} \cos\theta \ \mathbf{u}\sin\theta \end{bmatrix}=\cos\theta+\mathbf{u}\sin\theta=e^{\mathbf{u}\theta}=e^{\mathbf{v}} $$ it is actually 3-dimension
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The exponential map (where define
$\mathbf{\Omega,V}$ ) $$ \mathbf{q}^\otimes \mathbf{q}=1 \quad \rightarrow \quad \frac{d(\mathbf{q}^\otimes \mathbf{q})}{dt}=\dot{\mathbf{q}}^\otimes \mathbf{q}+\mathbf{q}^\otimes \dot{\mathbf{q}}=0 $$ solute the above difference equation $$ \mathbf{q}^*\otimes \dot{\mathbf{q}}=\mathbf{\Omega} =\begin{bmatrix} 0 \ \mathbf{\Omega} \end{bmatrix} \in \mathbb{H}_p \quad \rightarrow \quad \mathbf{q}(t)=\mathbf{q}(0)\otimes e^{\mathbf{\Omega}t} $$ defining$\mathbf{V}=\mathbf{\Omega}\Delta t\space\in\space\mathbb{H}_p$ $$ \mathbf{q}=e^{\mathbf{V}} $$ the exp map $$ \mathrm{exp}: \mathbb{H}_p\rightarrow S^3;\quad \mathbf{V}\rightarrow \mathrm{exp}(\mathbf{V})=e^{\mathbf{V}} $$ -
The capitalized exponential map (where define
$\phi,\boldsymbol{\phi}$ )since
$\mathbf{V}$ is a pure quaternion, we can write$\mathbf{V}$ as a 3-vector $$ \mathbf{V}=\begin{bmatrix} 0 \ V_x \ V_y \ V_z \end{bmatrix} = V_xi+V_yj+V_zk =\theta\mathbf{u} =\theta\begin{bmatrix}u_x\u_y\u_z \end{bmatrix} \ here\quad \theta = \Vert\mathbf{V}\Vert,\space \mathbf{u}=\frac{\mathbf{V}}{\Vert\mathbf{V}\Vert} $$ this is the same as pure quaternion in section Quaternionand here we define $$ \theta = \frac{\phi}{2} \rightarrow \mathbf{V}=\frac{\phi\mathbf{u}}{2} \in \mathbb{R}^3\ \boldsymbol{\phi}=\phi\mathbf{u} \rightarrow \mathbf{V}=\frac{\boldsymbol{\phi}}{2}\in\mathbb{R}^3 $$ and the proof of
$\theta=\phi/2$ will be given as below $$ \mathbf{q}=e^{\mathbf{V}}=e^{\boldsymbol{\phi}/2} $$ the Exp map $$ \mathrm{Exp}: \mathbb{R}^3\rightarrow S^3; \space \boldsymbol{\phi}\rightarrow \mathrm{Exp}(\boldsymbol{\phi})=e^{\boldsymbol{\phi}/2}=\mathrm{exp}(\boldsymbol{\phi}/2) $$
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$\phi$ and$\theta$ 
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$\phi$ is the actually rotation angle -
$\theta$ is the angle between$\mathbf{q}$ and the identity quaternion$\mathbf{q}_1$
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Quaternion and rotation vector
since
$\mathbf{u}$ is pure quaternion $$ \mathbf{q}=\mathrm{Exp}(\boldsymbol{\phi})=e^{\phi\mathbf{u}/2}=\cos\frac{\phi}{2}+\mathbf{u}\sin\frac{\phi}{2}=\begin{bmatrix} \cos(\phi/2) \ \mathbf{u}\sin(\phi/2) \end{bmatrix} $$ -
The logarithmic maps
we define the logarithmic map as the inverse of the exponential map $$ \mathrm{log}: S^3\rightarrow \mathbb{H}_p;\quad \mathbf{q}\rightarrow \mathrm{log}(\mathbf{q})=\mathbf{V}=\mathbf{u}\theta \ \mathrm{Log}: S^3\rightarrow \mathbb{R}^3;\quad \mathbf{q}\rightarrow \mathrm{Log}(\mathbf{q})=\mathbf{u}\phi\ $$
and $$ \mathrm{Log}(\mathbf{q})=2\mathrm{log}(\mathbf{q})\ \phi=2\arctan(\Vert\mathbf{q}_v\Vert,q_w)\ \mathbf{u}=\mathbf{q}_v/\Vert\mathbf{q}_v\Vert $$
although quaternion and rotation matrix both use
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The exponential map (where define
$\boldsymbol{\omega,\phi}$ ) $$ \begin{split} \mathbf{R}^T\mathbf{R}=\mathbf{I} &\rightarrow \frac{d}{dt}(\mathbf{R}^T\mathbf{R})=\dot{\mathbf{R}}^T\mathbf{R}+\mathbf{R}^T\dot{\mathbf{R}}=0\ &\rightarrow \dot{\mathbf{R}}^T\mathbf{R}=-\mathbf{R}^T\dot{\mathbf{R}} \end{split} $$ meaning that$\mathbf{R}^T\dot{\mathbf{R}}$ is skew-symmetric. So we can take a vector$\boldsymbol{\omega}=[\omega_x,\omega_y,\omega_z]\in\mathbb{R}^3$ and write $$ \mathbf{R}^T\dot{\mathbf{R}}=[\boldsymbol{\omega}]{\times} \qquad [\boldsymbol{\omega}]{\times}=\begin{bmatrix} 0 & -\omega_z & \omega_y\ \omega_y & 0 & -\omega_x\ -\omega_y & \omega_x & 0\ \end{bmatrix}\in\mathfrak{se}(3) $$ so we have $\dot{\mathbf{R}}=\mathbf{R}[\boldsymbol{\omega}]{\times}$ and solve this difference equation $$ \mathbf{R}(t)=\mathbf{R}(0)e^{[\boldsymbol{\omega}]{\times}t}=\mathbf{R}(0)e^{[\boldsymbol{\omega}t]{\times}} $$ we define the vector $$ \boldsymbol{\phi}=\boldsymbol{\omega}\Delta t = \phi\boldsymbol{u} \quad\rightarrow\quad \mathbf{R}=e^{[\boldsymbol{\phi}]{\times}} $$ this is known as the exponential map $$ \mathrm{exp}:\mathfrak{se}(3)\rightarrow SO(3); \quad [\boldsymbol{\phi}]{\times}\rightarrow \mathrm{exp}([\boldsymbol{\phi}]{\times})=e^{[\boldsymbol{\phi}]_{\times}} $$ -
The capitalized exponential map $$ \mathrm{Exp}:\mathbb{R}^3\rightarrow SO(3); \quad \boldsymbol{\phi}\rightarrow \mathrm{Exp}(\boldsymbol{\phi})=e^{[\boldsymbol{\phi}]{\times}}=\mathrm{exp}([\boldsymbol{\phi}]{\times}) $$
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The logarithmic maps $$ \mathrm{log}:SO(3)\rightarrow \mathfrak{se}(3); \quad \mathbf{R} \rightarrow \mathrm{log}(\mathbf{R})=[\boldsymbol{\phi}]{\times} $$ with $$ \begin{split} \phi &= \arccos(\frac{\mathrm{trace}(\mathbf{R})-1}{2})\ \mathbf{u} &= \frac{(\mathbf{R}-\mathbf{R}^T)^{\lor}}{2\sin\phi} \end{split} $$ where $(\cdot)^{\lor}$ is the inverse of $[\cdot]{\times}$, that is $([\mathbf{v}]{\times})^{\lor}=\mathbf{v}$ and $[\mathbf{V}^{\lor}]{\times}=\mathbf{V}$
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The capitalized logarithmic maps $$ \mathrm{Log}:SO(3)\rightarrow \mathbb{R}^3; \space \mathbf{R}\rightarrow\mathrm{Log}(\mathbf{R})=\boldsymbol{\phi}=\mathbf{u}\phi $$