The Euclidean subgroup

• The affine subgroup can be further specialized by the requirement they leave a special conic invariant. The conic ${\bf\Omega_\infty}$ is intersection of the quadric of equation:
  $$\sum_{i=1}^{n+1} x_i^2 = x_{n+1} = 0$$    with $${\bf\pi_\infty}$$
  In a metric frame ${\bf\pi_\infty} = (0,0,0,1)^T$ , and points on ${\bf\Omega_\infty}$ satisfy
  $$\left. \begin{array}{r} X_1^2 + X_2^2 + X_3^3 \\ X_4 \end{array}\right\} = 0$$

• ${\bf\Omega_\infty}$ is called the absolute conic. In ${\bf\pi_\infty}$ , it can be interpreted as a circle of radius $i = \sqrt{-1}$ .

• For directions on ${\bf\pi_\infty}$ (with $X_4 = 0$ ), the absolute conic ${\bf\Omega_\infty}$ can be expressed as
  $$(X_1,X_2,X_3){\bf I}(X_1,X_2,X_3)^T = 0$$

• The absolute conic, ${\bf\Omega_\infty}$ , is fixed under a projective transformation ${\bf H}$ if and only if ${\bf H}$ is an Euclidean transformation.
  Proof. Since the absolute conic lies on ${\bf\pi_\infty}$ , a transformation fixing it must also fix ${\bf\pi_\infty}$ , hence it must be affine. Such a transformation is of the form

  $${\bf H}_A = \left[ \begin{array}{ll} {\bf A} & {\bf t} \\ {\bf0}^T & 1 \end{array}\right]$$

  Restricting to ${\bf\pi_\infty}$ , the absolute conic is represented by the matrix ${\bf I}_{3 \times 3}$ , and since it is fixed by ${\bf H}_A$ , one has (up to scale)

  $${\bf A}^{-T} {\bf I} {\bf A}^{-1} = I$$

  which yields ${\bf A}{\bf A}^T = {\bf I}$ . This means that ${\bf A}$ is orthogonal, hence a scaled rotation. $\qedsymbol$