SO(3)
SO(3) is the most common representation of rotation I’ve seen so far.
\[\begin{gather*} \exp(\hat{\omega}) = I + \frac{sin(\theta)}{\theta} \hat{\omega} + \frac{1 - cos(\theta)}{\theta} \hat{\omega}^2 \end{gather*}\]Common rotation matrices are:
\[\begin{gather*} R_x(\gamma) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \gamma & -\sin \gamma \\ 0 & \sin \gamma & \cos \gamma \end{bmatrix} R_y(\beta) = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{bmatrix} R_z(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{gather*}\]Skew Symmetric Matrices
Let $\phi \in R^3$ be a vector, and let $R \in SO(3)$ be a rotation matrix (i.e., $R^TR = I$ and $det(R)=1$). Define the “hat” (or wedge) operator $\phi^{\land}$ as the skew-symmetric matrix satisfying
\[\begin{gather*} \begin{aligned} & \phi^{\land} x = \phi \times x \end{aligned} \end{gather*}\]for any $x \in R^3$
Formalize Rotation into A Lie Group
What Are the Lie Group and Lie Algebras
This is an intuitive introduction of Lie Group and Lie algebras, inspired by this post.
A group is a set of algebraic structure (matrices, vectors) that is 1. globally connected to each other, 2. can transition between each other. Formally, a group has:
- Closure: if a and b are in the group,
a op bis also in the group - Associativity:
(a op b) op c = a op (b op c) - Identity: for any element a, identity element e has
e op a = a - Inverses:
a op a_inv = e
A manifold is a shape that’s locally euclidean and globally connected. Integers are not a Lie Group because they are discrete. But real numbers form a consecutive Lie Group. A Lie group has to be a smooth manifold, meaning it’s differentiable. - Elements in a Lie group can transition to each other through multiplication, inverse, and identity.
Lie Definitions
A Lie Algebra is a vector space that captures the infinitesimal structure of a Lie Group, e.g., the tangent space of the Lie Group. In the context of rotations,
- All 3D rotation matrices form a group called “Special Orthogonal Group, SO(3)”, whose definition is $RR^T = I, det(R) = I$
- Their corresponding skew symmetric matrices are their Lie Algebra so(3). The exponential $exp(\phi^{\land}) = R$ is called “exponential mapping”
- Within any Lie algebra, there is Lie Bracket for vectors A and B:
[A, B] = AB - BA
- Within any Lie algebra, there is Lie Bracket for vectors A and B:
From Skew Symmetric so(3) To SO(3)
We can see that R’R^T must be a skew matrix:
\[\begin{gather*} \begin{aligned} & RR^T = I \\ & \Rightarrow (RR^T)' = R'R^T + RR'^T = 0 \\ & \Rightarrow R'R^T = -(RR'^T) = -(R'R^T)^T = \phi(t)^{\land} \end{aligned} \end{gather*}\]Where $\phi(t)^{\land}$ the skew matrix of vector $\phi(t) = [\phi_1, \phi_2, \phi_1,]$
\[\begin{gather*} \phi(t)^{\land} = \begin{bmatrix} 0 & -\phi_3 & \phi_2 \\ \phi_3 & 0 & -\phi_1 \\ -\phi_2 & \phi_1 & 0 \end{bmatrix} \end{gather*}\]Then, one can see that taking the direvative of R is equivalent to multiplying the skew matrix $\phi(t)^{\land}$ with it.
\(\begin{gather*}
R'R^TR = R' = \phi(t)^{\land} R
\end{gather*}\)
The above is also known as the Poisson Formula. If we define
\[\begin{gather*} w^{\land} = R^{T}R' \end{gather*}\]We have another form of the Poisson Formula:
\[\begin{gather*} R' = Rw^{\land} \end{gather*}\]Finally, do matrix exponential:
\[\begin{gather*} R(t) = exp(\phi(t)^{\land}) \end{gather*}\]Rodrigues Formula
Let’s represent the skew matrix as angle multiplied by a unit vector, $a^{\land}$
\[\begin{gather*} -\phi(t) = [\phi_1, \phi_2, \phi_3] = \phi a^{\land} \end{gather*}\]By writing out $a^{\land} a^{\land}$ “Unit” skew matrices have the below two properties:
\[\begin{gather*} a^{\land} a^{\land} = aa^T - I \\ a^{\land} a^{\land} a^{\land} = -a^{\land} \end{gather*}\]- One trick for the second formula is: $a^{\land} (a^{\land} a^{\land}) = a \times (aa^T - I) = -a^{\land}$
So, \(\begin{gather*} \begin{aligned} & exp(\phi^{\land}) = exp(\theta a^{\land}) = \frac{1}{n!}(\theta a^{\land})^n \\ & = I + \theta a^{\land} + \frac{1}{2!} \theta^2 a^{\land} a^{\land} + \frac{1}{3!} \theta^3 a^{\land} a^{\land} a^{\land} + ... \\ & = aa^T - a^{\land} a^{\land} + \theta a^{\land} + \frac{1}{2!} \theta^2 a^{\land} a^{\land} - \frac{1}{3!} \theta^3 a^{\land}... \\ & = aa^T + (\theta - \frac{1}{3!} \theta^3 + ...)a^{\land} - (1 - \frac{1}{2!} \theta^2 + ... )a^{\land} a^{\land} \\ & = aa^T + sin(\theta) a^{\land} -cos(\theta) a^{\land} a^{\land} \\ & = cos(\theta) + (I-cos(\theta)) aa^T + sin(\theta)a^{\land} \\ Or \\ & = I + (1-cos \theta) a^{\land} a^{\land} + sin \theta a^{\land} \end{aligned} \end{gather*}\)
Properties
- $ \phi^{\land} R = R (R^T \phi)^{\land}$
Proof:
First, we know from $ (a \times b) \cdot c = a \times (b \cdot c)$
\[\begin{gather*} \begin{aligned} & (\phi^{\land} R)x = \phi^{\land}(Rx) \end{aligned} \end{gather*}\]One key property about rotation is commutation: a rotated cross product is the same as the rotated vectors’ dot product.
\[\begin{gather*} \begin{aligned} & R(a \times b) = (Ra) \times (Rb) \end{aligned} \end{gather*}\]So on the right,
\[\begin{gather*} \begin{aligned} & R (R^T \phi)^{\land} x= R((R^T \phi) \times x) \\ & = \phi \times (Rx) = (\phi^{\land} R)x \end{aligned} \end{gather*}\]