lie-group

刚体运动的指数表示

旋转运动

根据欧拉定理，任意三维空间旋转运动可以表示为绕某一单位轴$w\in \mathcal{R}^3$转动角度$\theta$,则旋转矩阵可以表示为矩阵指数的形式： $R=e^{\hat{w}\theta} \in \mathcal{R}^{3\times 3} \\ \hat{w}=\begin{bmatrix}0 &-w_3 & w_2 \\w_3 & 0 & -w_1 \\-w_2 & w_1 & 0 \end{bmatrix}$ $so(3)与SO(3)的指数映射\text{(Exponential map)}关系 \{\begin{array}{lr} \hat{w}\in so(3)\in \mathcal{R}^{3\times 3} \\ \quad \\ \theta \in \mathcal{R} \end{array} \Rightarrow e^{\hat{w}\theta}\in SO(3)$ angles: $\theta=cos^{-1}(\frac{tr(R)-1}{2})$

rotation angle: $[n]\times=\frac{R-R^T}{2sin(\theta)}$, $n=[-[n]\times(2,3),[n]\times(1,3),-[n]\times(1,2)]$

In unit quaternions(四元数): q=$(cos(\theta/2),\omega sin(\theta/2))$

刚体运动

与旋转相似，根据Chasels定理，任意刚体运动均可以通过绕一轴的运动加上平行于该轴的移动实现: $T=e^{\hat{\xi}\theta} \in \mathcal{R}^{4\times 4} \\ \hat{\xi}=\begin{bmatrix} \hat{w} & \upsilon \\ 0& 0 \end{bmatrix} \\ e^{\hat{\xi}\theta}=\begin{bmatrix} e^{\hat{w}\theta}& (I-e^{\hat{w}\theta})(w\times \theta)+ww^T\upsilon\theta\\0 &1 \end{bmatrix}$

$se(3)与SE(3)的指数映射(Exponential map)关系 \{\begin{array}{lr} \hat{\xi}\in se(3)\in \mathcal{R}^{3\times 3} \\ \quad \\ \theta \in \mathcal{R} \end{array} \Rightarrow e^{\hat{\xi}\theta}\in SE(3)$

正运动学

for revolute joint: $\theta_i \in S^1$,unit circle in the plane
for prismatic joint: $\theta_i \in \mathcal{R}$
$T^p$ to denote the p-torus, defined to be the Cartesian product of p copies of $S^1$: $T^p=S^1\times S^1\times \cdots \times S^1$
joint space(configuration space) of a manipulator with p revolute joints and r prismatic joints: $Q=T^p \times R^ r$
the forward kinematics map: $g_{st}$, t-tool frame and s-base frame: $g_{st}(\theta)=g_{sl_1}(\theta_1)g_{l_1l_2}(\theta_2)\cdots g_{l_nt}$

指数积公式

If $\xi$ is a twist, then the rigid motion associated with rotating and translating along the axis of the twist is given by: $g_{ab}(\theta)=e^{\hat{\xi}\theta}g_{ab}(0)$ g: is a pose and also a transformation

The product of exponential formula(POE): $g_{st}(\theta)=e^{\hat{\xi}_1\theta_1}e^{\hat{\xi}_2\theta_2}\dots e^{\hat{\xi}_n\theta_n}g_{ab}(0)$

Generalize this procedure:

Define the reference configuration(初始位置) of the manipulator to be the configuration corresponding to $\theta=0$.
For each joint, construct a twist $\xi_i$ at $g_{st}(0)$
- $\omega_i \in \mathcal{R}^3$,unit vector in the direction of th twist axis
- $q_i \in \mathcal{R}^3$,any point on the axis(determine the direction)
- $v_i \in \mathcal{R}^3$,unit vector pointing in the direction of translation
- all vectors are specified relative to the base coordinate frame $\begin{align} \text{for revolute } \xi_i&=\begin{bmatrix}-\omega_i\times q_i \\ \omega_i \end{bmatrix} or \begin{bmatrix}q_i\times \omega_i \\ \omega_i \end{bmatrix} \\ \text{for prismatic } \xi_i&=\begin{bmatrix}v_i \\ 0 \end{bmatrix} \end{align}$
- 指数积公式只能描述末端关节对基坐标系的变换，而不能得知每个关节相对于坐标系的变换。

only math

matrix exponential(矩阵指数):

$e^A = \sum_{k=0} \frac{1}{k!}A^k$

for a skew-symmetric matrix:

$C=\left[\begin{array}{ccc} 0 & -a_{3} & a_{2} \\ a_{3} & 0 & -a_{1} \\ -a_{2} & a_{1} & 0 \end{array}\right]$

Let $x=\sqrt{a_1^2 + a_2^2 + a_3^2 }$， then $C^3 = -x^2C$, Hence:

$C^{2m+1}=(-1)^mx^{2m} \\ C^{2m}=(-1)^{m-1}x^{2m-2}C^2$

Therefore:

$\begin{aligned} &e^{C}=\sum_{n=0}^{\infty} \frac{1}{n !} C^{n}=I+\sum_{m=0}^{\infty} \frac{1}{(2 m+1) !} C^{2 m+1}+\sum_{m=1}^{\infty} \frac{1}{(2 m) !} C^{2 m} \\ &=I+\sum_{m=0}^{\infty} \frac{(-1)^{m} x^{2 m}}{(2 m+1) !} C+\sum_{m=1}^{\infty} \frac{(-1)^{m-1} x^{2 m-2}}{(2 m) !} C^{2} \\ &=I+\frac{1}{x} \sum_{m=0}^{\infty} \frac{(-1)^{m} x^{2 m+1}}{(2 m+1) !} C-\frac{1}{x^{2}} \sum_{m=1}^{\infty} \frac{(-1)^{m} x^{2 m}}{(2 m) !} C^{2} \\ &=I+\frac{\sin x}{x} C+\frac{1-\cos x}{x^{2}} C^{2} \end{aligned}$

SE2

基于李群的路径插值

Optimal Transport on the Lie Group of Roto-translations