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Next: Laguerre formula Up: The Projective Plane Previous: Absolute points

Collineations

A collineation of \ensuremath{{\cal P}^2} is defined as a mapping from the plane to itself such that the collinearity of any set of points is preserved. Such a mapping can be achieved with matrix multiplication by a $3 \times 3$ matrix T. Each point $\ensuremath{{\bf p}} $ is transformed into a point $\ensuremath{{\bf p}} '$:

\begin{displaymath}\ensuremath{{\bf p}} ' = T \ensuremath{{\bf p}} .
\end{displaymath}

We will use the terms transformation and collineation interchangeably. Since scaling is unimportant, only eight elements of T are independent. Therefore, since each point contains two independent values, four pairs of corresponding points are necessary to determine T.

To transform a line $\ensuremath{{\bf u}} $ into a line $\ensuremath{{\bf u}} '$, we note that collinearity must be preserved, that is, if a point $\ensuremath{{\bf p}} $ lies on the line $\ensuremath{{\bf u}} $, then the corresponding point $\ensuremath{{\bf p}} '$ must lie on the corresponding line $\ensuremath{{\bf u}} '$. Therefore,

\begin{displaymath}\ensuremath{{\bf p}} ^T\ensuremath{{\bf u}} = 0 = (T^{-1}\ens...
...u}} = (\ensuremath{{\bf p}} ')^T(T^{-T}\ensuremath{{\bf u}} ),
\end{displaymath}

which indicates that

\begin{displaymath}\ensuremath{{\bf u}} ' = T^{-T}\ensuremath{{\bf u}} .
\end{displaymath}

From these results, it is not hard to show that a point conic C transforms to T-TCT-1, and a line conic |C|C-1 transforms to T|C|C-1TT.

Regarding transformations, recall that projective $\supset$ affine $\supset$ similarity $\supset$ Euclidean. Let's study the matrix T to uncover the relationships between these various geometries. First we will write out the elements of T, for reference:

\begin{displaymath}T_{projective} = \left[\matrix{t_{11} & t_{12} & t_{13} \cr
t_{21} & t_{22} & t_{23} \cr
t_{31} & t_{32} & t_{33}}\right].
\end{displaymath}

The affine plane is just the projective plane minus the ideal line. Therefore, affine transformations must preserve the ideal line and the ideal points, that is, any point [X, Y, 0]T must be transformed into $[\ensuremath{\alpha} X, \ensuremath{\alpha} Y, 0]^T$ for some arbitrary scaling $\ensuremath{\alpha} $:

\begin{displaymath}\ensuremath{\alpha}\left[\matrix{X \cr Y \cr 0}\right] = T
\left[\matrix{X \cr Y \cr 0}\right],
\end{displaymath}

which implies that t31 = t32 = 0. The matrix for affine transformation, then, is

\begin{displaymath}T_{af\!fine} = \left[\matrix{t_{11} & t_{12} & t_{13} \cr
t_{21} & t_{22} & t_{23} \cr
0 & 0 & t_{33}}\right],
\end{displaymath}

where once again only six of these parameters are independent, since scale is unimportant.

Unlike affine transformations, similarity transformations preserve angles and ratios of lengths. Delaying the derivation for a moment, we simply state the result:

 \begin{displaymath}T_{similarity} = \left[\matrix{\cos \theta & \sin \theta & t_...
...sin \theta & \cos \theta & t_{23} \cr
0 & 0 & t_{33}}\right],
\end{displaymath} (3)

where $\theta$ is an arbitrary angle.

Under Euclidean transformation, scale is important, and therefore the point $\ensuremath{{\bf p}} $ must first be converted to Euclidean coordinates by dividing by its third element. The transformation then is

\begin{displaymath}\left[\matrix{x' \cr y'}\right] =
\left[\matrix{\cos \theta ...
...t[\matrix{x \cr y}\right] +
\left[\matrix{t_x \cr t_y}\right].
\end{displaymath}

In closing this section, we offer one final proposition, along with its proof:

Proposition 1   A transformation is a similarity transformation if and only if it preserves the absolute points, $[1, \pm i, 0]$.

The ``only if'' is rather easy to see: The absolute point $[1, \pm i, 0]^T$is transformed through equation (3) to the point $e^{\pm i\theta}[1, \pm i, 0]^T$, which is equivalent because the scale factor is ignored.

The ``if'' is a little more complicated, but still rather straightforward. Starting with the unrestricted equation for T,

\begin{displaymath}T = \left[\matrix{t_{11} & t_{12} & t_{13} \cr
t_{21} & t_{22} & t_{23} \cr
t_{31} & t_{32} & t_{33}}\right],
\end{displaymath}

the fact that [1, i, 0]T is preserved yields the following two equations:

\begin{eqnarray*}\frac{t_{11} + it_{12}}{t_{21} + it_{22}} & = & \frac{1}{i} \\ \\
t_{31} + it_{32} & = & 0.
\end{eqnarray*}


Since the elements of T are constrained to be real, this leads to the following three constraints on the elements of T:

\begin{eqnarray*}t_{11} & = & t_{22} \\
t_{12} & = & -t_{21} \\
t_{31} & = & t_{32} \ \ = \ \ 0.
\end{eqnarray*}


So the matrix of T looks like this:

\begin{displaymath}T = \left[\matrix{t_{11} & t_{12} & t_{13} \cr
-t_{12} & t_{11} & t_{23} \cr
0 & 0 & t_{33}}\right].
\end{displaymath}

Given two arbitrary numbers t11 and t12, we can always reparameterize them as $t_{11} = k\cos \theta$ and $t_{12} = k\sin \theta$, where $\theta$ is an angle and k is a scalar. Multiplying the previous equation by 1/k (which is legal because we are working in homogeneous coordinates), we then get

\begin{displaymath}T = \left[\matrix{\cos \theta & \sin \theta & t_{13}/k \cr
-...
...\theta & \cos \theta & t_{23}/k \cr
0 & 0 & t_{33}/k}\right],
\end{displaymath}

which is seen to be the equation of a similarity transformation when compared with equation (3). (NOTE: Using the point [1,-i,0]T yields the same result.)


next up previous
Next: Laguerre formula Up: The Projective Plane Previous: Absolute points
Stanley Birchfield
1998-04-23