Preserve length, preserve cross product $\Leftrightarrow$ rigid body motion

$\hat{u} \in so(3)$

• $g(v) = |v|$, length preserved
• $g(u) \times g(v) = g(u \times v)$, cross product preserved

$\left<u,v\right> = \frac{1}{4}(|u+v|^2-|u-v|^2)$

$\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.

$\exp: so(3) \rightarrow SO(3)\quad \hat{w} -> e^{\hat{w}}$