Preserve length, preserve cross product $\Leftrightarrow$ rigid body motion
$\hat{u} \in so(3)$
$\left<u,v\right> = \frac{1}{4}(|u+v|^2-|u-v|^2)$
$\Rightarrow \left<g(u), g(v) \times g(w)\right> = \left<u, v\times w\right> $, volume preserving
Representation of Rigid-Body Motion
The RBM preserves length and cross product.
$r_i = g_t(e_i)$
$R = (r_1, r_2, r_3)$
$r_i^Tr_j = g_t(e_i)^T g_t(e_j) = e_i^T e_j = \delta_{ij}$
$\quad r_1 \times r_2 = r_3 \Rightarrow \det(R) =1$
$R^T R = RR^T = I$
$R$ is in the Special Orthogonal Group $SO(3)$
Rotation nasty to be represented
$R_1 + R_2 \notin SO(3)$, not a linear space!
$R_1 \cdot R_2 \in SO(3)$, not a linear space!
R has 9 elements, choice is not independent. You have to guarantee: orthogonal and det(R) = 1
Infinitesimal rotation
better representation of non-linear
Group which is a smooth Manifold, is a Lie group.
$x_{trans}(t) = R(t)x_{orig}$
$R(t)R(t)^T = I$, rotate forward and backward. You land where you started.
$\frac{d}{dt}(RR^T) = \dot{R}R^T + R\dot{R^T} = 0$
$\dot{R}R^T = - (R\dot{R^T})^T$, is skew symmetric!
$\dot{R}R^T = \hat{w} \Leftrightarrow \dot{R}(t)= \hat{w}R(t)$
$R(dr) = R(0) + dR = I + \hat{w}(0)dt$
Lie Algebra: $so(3) = \{\hat{w} | w \in \mathbb{R}^3\}$ of skew symmetric matrices
Lie Group: $SO(3)$, smooth manifold and a group
The Lie algebra $so(3)$ is the tangent space at the identity of the rotation group $SO(3)$
algebra over a field K is a vector space over K with multiplication on the space V. Not commutative.
Lie Brackets:
$\left[\hat{w},\hat{v}\right] \identical \hat{w}\hat{v}-\hat{v}\hat{w} \in so(3)$
Two Lie groups we will study: $so(3), se(3)$
$\left\{ \dot{R}(t)=\hat{w}R(t)\right. $
$R(0) = I$
How does the rotation matrix change from one time within a little time step?
$R(t) = e^{\hat{w}t} = \sum \ldots $ expansion. That solves the diff. eqn. above.
$\exp: so(3) \rightarrow SO(3)\quad \hat{w} -> e^{\hat{w}}$