---
title: Homogeneous Coordinates
categories: lectures 3D mathematics
geometry: margin=2cm
fontsize: 12pt
---


# Homogenous Co-ordinates

## Six degrees of Freedom

+ Translation - add $T=[y_1,y_2,y_3]$
+ Rotation - multiply by $R=\exp(\hat{[\omega_1,\omega_2,\omega_3]})$
+ $x\mapsto xR+T$ is affine, not linear

This is the same for motion and for change of basis
(rotate and shift the origin).

## Points in Homogenous Co-ordinates

+ Point $\textbf{X}=[X_1,X_2,X_3]^\mathrm{T}\in\mathbb{R}^3$
+ Embed in $\mathbb{R}^4$ as
    $\mathbf{\tilde X}=[X_1,X_2,X_3,1]^\mathrm{T}\in\mathbb{R}^4$
+ Vector $\vec{pq}$ is represented as
    $$\mathbf{\tilde X}(q)-\mathbf{\tilde X}(p) =
    \begin{bmatrix} \mathbf{ X}(q) \\ 1 \end{bmatrix}
    - \begin{bmatrix} \mathbf{ X}(p) \\ 1 \end{bmatrix}
    =
    \begin{bmatrix} \mathbf{X}(q) - \mathbf{X}(p) \\ 0 \end{bmatrix}$$
+ In homogenous co-ordinates,
    - points have 1 in last position
    - vectors have 0 in last position
+ Arithmetics
    + Point + Point is undefined
    + Vector + Vector is a Vector
    + Point + Vector is a Point

## Rotation 

Let $R$ be a $3\times3$ rotation matrix.

$$
   R\cdot\vec{x}=
   R
\cdot
\begin{bmatrix} x\\y\\z \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z' \end{bmatrix}
$$

$$
\begin{bmatrix}
   R & 0 \\
   0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix} x\\y\\z\\1 \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z'\\1 \end{bmatrix}
$$

## Arbitrary motion

What happens if we change some of the zeroes?

$$
\begin{bmatrix}
   R & \vec{t} \\
   0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix} x\\y\\z\\1 \end{bmatrix}
=
\begin{bmatrix} x'\\y'\\z'\\0 \end{bmatrix}
+
\begin{bmatrix} \vec{t}\\1 \end{bmatrix}
=R\vec{x}+\vec{t}
$$

We have rotated and translated!

## The group structure

$$
g =
\begin{bmatrix}
   R & T\\
   0 & 1
\end{bmatrix}
\in \mathrm{SE}(3)
$$

+ This is a group 
    + motions can be applied sequentially

$$g_1\cdot g_2 =
\begin{bmatrix} R_1 & T_1\\ 0 & 1 \end{bmatrix}\cdot
\begin{bmatrix} R_2 & T_2\\ 0 & 1 \end{bmatrix}
=
\begin{bmatrix} R_1R_2 & R_1T_2+T_1\\ 0 & 1 \end{bmatrix}
\in \mathrm{SE}(3)
$$
    
$$g^{-1} =
\begin{bmatrix} R & T\\ 0 & 1 \end{bmatrix}^{-1}
=
\begin{bmatrix} R^T & -R^TT\\ 0 & 1 \end{bmatrix}
$$

# Motion as a function of time

1.  A homogeneous motion matrix as a function of time $t$
    $$ g(t) = \begin{bmatrix} R(t) & T(t)\\ 0 & 1 \end{bmatrix} $$
2.  We can differentiate, invert, and multiply to get
    $$
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \dot R(t) R^T(t) & \dot T(t)- \dot R(t)R^T(t)T(t) \\
    0 & 0 \end{bmatrix}
    \in\mathbb{R}^{4\times4}
    $$
3.  $\dot R(t) R^T(t)$ is skew-symmetric as we saw in
    [Representations of 3D Motion](), hence
    $$
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \hat\omega & \dot T(t)- \dot R(t)R^T(t)T(t) \\
    0 & 0 \end{bmatrix}
    $$
    for some $\omega\in\mathrm{so}(3)$
    + This can be seen by differentiation $R(t)R^TR(t)=I$
4.  Write 
    $$
    \begin{align}
    \dot v(t)= \dot T(t)- \hat\omega(t)T(t) \\
    \dot g(t)\cdot g^{-1}(t) =
    \begin{bmatrix}
    \hat\omega & \dot v \\
    0 & 0 \end{bmatrix}
    \end{align}
    $$
4.  Call this matrix $\hat\xi(t)$
    + i.e. $\hat\xi(t) =\dot g(t)\cdot g^{-1}(t) $
    + $\hat\xi$ is the tangent vector along $g(t)$
    + the $4\times4$ matrix corresponding $\hat\xi$ is called a *twist*
    + set of all twists is called $se(3)$; the tangent space or Lie group of $SE(3)$
5.  This give the differential equation
    $$\dot g(t) = \hat\xi\cdot g(t)$$
6.  If $\hat\xi$ is constant, we can integrate to solve the ODE
    $$g(t) = e^{\hat\xi t}g(0)$$
    + we typically assume $g(0)=I$ as the initial condition
  

+ $v(t)$ is linear velocity
+ $\omega(t)$ as angular velocity