Representation of a 3-dimensional moving scene

This post is a part of a multi-part series.
• Rigid motion is a linear transformation that preserves the distance between two points before and after transformation.
• They also preserve the cross-products between two vectors.
• They also preserve the inner-product, and the triple-product

i.e

$$\langle g(u) , g(v) \times g(w) \rangle = \langle u , v \times w \rangle$$

i.e, volume preserving.

Exponential coordinates of Rotation

Consider a family of rotation matrices $R(t)$ which continuously transform a point from its original location  $$R(0) = I$$ to a different one

$$X_t = R(t)*X_{original}$$ , where $R(t) \in SO(3)$

Since $R(t)R(t)^T = I$ , $\forall t$ we have

$$\frac {d(RR^T)}{dt} = \dot R R^T + R \dot R^T = 0 \implies \dot R R^T = -(\dot R R^T)^T \implies \dot R R^T$$ is a skew symmetric matrix.

(NOTE : $\frac {d(R)} {dt} = \dot R$ in case you were wondering what $\dot R$ meant. )

Now,

There exists a vector $w(t) \in \mathbb{R}^3$ such that

$$\dot R(t) R^T (t) = \hat w (t) \Leftrightarrow \dot R(t) = \hat w R(t)$$

This is a differential equation that links the rotation matrix with its derivative.

Since $R(0) = I$  ,  $R(0) = \hat w(0)$

Therefore skew-symmetric matrix $\hat w(0) = SO(3)$ gives the first order approximation of a rotation.

$\dot R$ is basically “how much a rotation changes” between two time intervals.

So, $$\color{red}{R(dt)} = \color{gold}{R(0)} + \color{blue}{dR} = \color{gold}{I} + \color{blue}{\hat w(0)dt} ——Observation(1)$$

This means “rotation matrix at time t ” is equal to  rotation matrix at time t-1 plus something.

So, why did we go through all this trouble?

Consider the “normal” scenario of Rotation matrices :

• we can have a new rotation by multiplying two other matrices , i.e

$$R_3 = R_2 \times R_1$$

• But rotations do not follow “linear” property, i.e , we cannot have a rotation matrix $R_3$ such that

$$R_3 = R_2 + R_2 ——Observation(2)$$

This is false! In fact, in this case $R^3$ is not even a rotation matrix!

(Mathematically, we say that such set of matrices do not belong to a linear “group”, since they don’t even follow simple rule of linear combination)

How nice would it be, if rotations could be represented in a sort of “linear” way……..

But that is exactly what we derived in $Observation(1)$!

Using magic of differential equations we’re able to represent rotations in a way that does not involve matrix-multiplication, hence simplifying the process intuitively. Not to mention it is also computationally cheaper.

The Exponential map

Given the formulation of rotation in terms of skew-symmetric matrix $\hat w$ , is it possible to determine a useful representation of $R(t)$ ?  Assuming $\hat w$ is constant in time, which helps us formulate a first-order differential equation given by :

$$\left\{\begin{matrix} \dot R(t) = \hat w R(t) \\ R(0) = I \end{matrix}\right.$$

That is, “change in R” at time t, isgiven by $\hat w$ times $R$. Also given initial condition is $R(0) = I$

Solving this equation uniquely determines these matrices $\because$ if I tell what the first order derivative is, it basically tells – from every time, to next how the rotation matrix is changes.

Now, the given differential equation has solution

$$R(t) = e^{\hat w t} ——Observation(3)$$

That’s right, we’re putting $e$ to the power of a matrix!

It is not actually different from computing exponents of regular numbers. Just doing Taylor expansion we get :

$$e^{\hat w t} = \sum_{n=0}^\infty \frac {(\hat w t)^n}{n!} = I + \hat w t + \frac {{(\hat w t)}^2} {2} + \cdots$$

What we see from $Observation(3)$ is that the function which maps skew-symmetric matrix $\hat w$ to a rotation matrix is just the exponential ($e$) function.

This is basically a rotation matrix that rotates around the axis “$w$” by angle “$t$” (only if $\left \| w \right \| =1$ )

(NOTE: we can also go back, i.e “inverse map” to $\hat w$ by applying logarithm. Again, logarithms of matrices can be found from their Taylor expansions)

Now, Taylor expansions are again computationally expensive because of matrix multiplications involved.

So we have another formula to get $w$ from $R$.

$$\left | w \right | = cos^{-1} \left ( \frac {trace(R) – 1}{2} \right )$$

And,

$$\frac {w}{\left | w \right |} = \frac {1}{2 sin(\left | w \right |)} \begin{pmatrix} r_{32} – r_{31} \\ r_{13} – r_{31} \\ r_{21} – r_{12} \end{pmatrix}$$

(we’re not proving this)