# Projective Geometry/Classic/Projective Transformations/Transformations of the projective line

Projective transformation

Let X be a point on the x-axis. A projective transformation can be defined geometrically for this line by picking a pair of points P, Q, and a line m, all within the same x-y plane which contains the x-axis upon which the transformation will be performed.

Points P and Q represent two different observers, or points of view. Point R is the position of some object they are observing. Line m is the objective world which they are observing, and the x-axis is the subjective perception of m.

Draw line l through points P and X. Line l crosses line m at point R. Then draw line t through points Q and R: line t will cross the x-axis at point T. Point T is the transform of point X [Paiva].

## Analysis

The above is a synthetic description of a one-dimensional projective transformation. It is now desired to convert it to an analytical (Cartesian) description.

Let point X have coordinates (x0,0). Let point P have coordinates $(P_x,P_y)$. Let point Q have coordinates $(Q_x,Q_y)$. Let line m have slope m (m is being overloaded in meaning).

The slope of line l is

$P_y \over P_x - x_0,$

so an arbitrary point (x,y) on line l is given by the equation

${y \over x - x_0} = {P_y \over P_x - x_0}$,
$y = {P_y \over P_x - x_0} (x - x_0). \qquad \qquad (1)$

On the other hand, any point (x,y) on line m is described by

$y = m x + b. \qquad \qquad (2)$

The intersection of lines l and m is point R, and it is obtained by combining equations (1) and (2):

$m x + b = {P_y x \over P_x - x_0} - {P_y x_0 \over P_x - x_0}.$

Joining the x terms yields

$\left( {P_y \over P_x - x_0} - m \right) x = b + {P_y x_0 \over P_x - x_0}$

and solving for x we obtain

$x_1 = {b (P_x - x_0) + P_y x_0 \over P_y - m (P_x - x_0)}.$

x1 is the abscissa of R. The ordinate of R is

$y_1 = m \left[ {b (P_x - x_0) + P_y x_0 \over P_y - m (P_x - x_0)} \right] + b.$

Now, knowing both Q and R, the slope of line n is

${y_1 - Q_y \over x_1 - Q_x} .$

We want to find the intersection of line n and the x-axis, so let

$(Q_x, Q_y) + \lambda (x_1 - Q_x, y_1 - Q_y) = (x,0) \qquad \qquad (3)$

The value of λ must be adjusted so that both sides of vector equation (3) are equal. Equation (3) is actually two equations, one for abscissas and one for ordinates. The one for ordinates is

$Q_y + \lambda (y_1 - Q_y) = 0$

Solve for lambda,

$\lambda = {-Q_y \over y_1 - Q_y} \qquad \qquad (4)$

The equation for abscissas is

$x = Q_x + \lambda (x_1 - Q_x)$

which together with equation (4) yields

$x = Q_x - Q_y \left( {x_1 - Q_x \over y_1 - Q_y} \right) \qquad \qquad (5)$

which is the abscissa of T.

Substitute the values of x1 and y1 into equation (5),

$x = Q_x - Q_y \left[ { {b (P_x - x_0) + P_y x_0 \over P_y - m (P_x - x_0)} - Q_x \over {m b (P_x - x_0) + m P_y x_0 \over P_y - m (P_x - x_0)} + b - Q_y} \right].$

Dissolve the fractions in both numerator and denominator:

$x = Q_x - Q_y \left[ {b (P_x - x_0) + P_y x_0 - Q_x P_y + m Q_x (P_x - x_0) \over m b (P_x - x_0) + m P_y x_0 + b P_y - m b (P_x - x_0) - Q_y P_y + m Q_y (P_x - x_0) } \right].$

Simplify and relabel x as t(x):

$t(x) = Q_x - Q_y \left[ { (P_x - x_0) (b + m Q_x) + P_y (x_0 - Q_x) \over (P_x - x_0) m Q_y + P_y (m x_0 + b - Q_y) } \right].$

t(x) is the projective transformation.

Transformation t(x) can be simplified further. First, add its two terms to form a fraction:

$t(x) = { (m Q_x P_y - Q_y P_y + b Q_y) x_0 + (b Q_x P_y - b Q_y P_x) \over m (P_y - Q_y) x_0 + (m P_x Q_y + P_y (b - Q_y)) } \qquad \qquad (6)$

Then, define the coefficients α, β, γ and δ to be the following

$\alpha = m Q_x P_y - Q_y P_y + b Q_y,$
$\beta = b Q_x P_y - b Q_y P_x,$
$\gamma = m (P_y - Q_y),$
$\delta = m P_x Q_y + P_y (b - Q_y).$

Substitute these coefficients into equation (6), in order to produce

$t(x) = { \alpha x + \beta \over \gamma x + \delta }$

This is the Möbius transformation or linear fractional transformation.

## Inverse transformation

It is clear from the synthetic definition that the inverse transformation is obtained by exchanging points P and Q. This can also be shown analytically. If PQ, then αα′, ββ′, γγ′, and δδ′, where

$\alpha' = m P_x Q_y - P_y Q_y + b P_y = \delta,$
$\beta' = b P_x Q_y - b P_y Q_x = - \beta,$
$\gamma' = m (Q_y - P_y) = - \gamma,$
$\delta' = m Q_x P_y + b Q_y - Q_y P_y = \alpha.$

Therefore if the forwards transformation is

$t(x) = {\alpha x + \beta \over \gamma x + \delta}$

then the transformation t′ obtained by exchanging P and Q (PQ) is:

$t'(x) = {\delta x - \beta \over - \gamma x + \alpha }.$

Then

$t'(t(x)) = {\delta \left( {\alpha x + \beta \over \gamma x + \delta} \right) - \beta \over - \gamma \left( {\alpha x + \beta \over \gamma x + \delta} \right) + \alpha}$.

Dissolve the fractions in both numerator and denominator of the right side of this last equation:

$t'(t(x)) = {\alpha \delta x + \beta \delta - \beta \gamma x - \beta \delta \over - \alpha \gamma x - \beta \gamma + \alpha \gamma x + \alpha \delta}$
$= {\alpha \delta x - \beta \gamma x \over \alpha \delta - \beta \gamma} = x$.

Therefore t′(x) = t−1(x): the inverse projective transformation is obtained by exchanging observers P and Q, or by letting α ↔ δ, β → −β, and γ → −γ. This is, by the way, analogous to the procedure for obtaining the inverse of a two-dimensional matrix:

$\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \begin{bmatrix} \delta & - \beta \\ - \gamma & \alpha \end{bmatrix} = \Delta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

where Δ = α δ − β γ is the determinant.

## Identity transformation

Also analogous with matrices is the identity transformation, which is obtained by letting α = 1, β = 0, γ = 0, and δ = 1, so that

$t_I(x) = x.$

## Composition of transformations

It remains to show that there is closure in the composition of transformations. One transformation operating on another transformation produces a third transformation. Let the first transformation be t1 and the second one be t2:

$t_1(x) = {\alpha_1 x + \beta_1 \over \gamma_1 x + \delta_1 },$
$t_2(x) = {\alpha_2 x + \beta_2 \over \gamma_2 x + \delta_2 }.$

The composition of these two transformations is

$t_2(t_1(x)) = {\alpha_2 \left( {\alpha_1 x + \beta_1 \over \gamma_1 x + \delta_1} \right) + \beta_2 \over \gamma_2 \left( {\alpha_1 x + \beta_1 \over \gamma_1 x + \delta_1 } \right) + \delta_2 }$
$= {\alpha_2 \alpha_1 x + \alpha_2 \beta_1 + \beta_2 \gamma_1 x + \beta_2 \delta_1 \over \gamma_2 \alpha_1 x + \gamma_2 \beta_1 + \delta_2 \gamma_1 x + \delta_2 \delta_1 }$
$= {(\alpha_2 \alpha_1 + \beta_2 \gamma_1) x + (\alpha_2 \beta_1 + \beta_2 \delta_1) \over (\gamma_2 \alpha_1 + \delta_2 \gamma_1) x + (\gamma_2 \beta_1 + \delta_2 \delta_1)}.$

Define the coefficients α3, β3, γ3 and δ3 to be equal to

$\alpha_3 = \alpha_2 \alpha_1 + \beta_2 \gamma_1,$
$\beta_3 = \alpha_2 \beta_1 + \beta_2 \delta_1,$
$\gamma_3 = \gamma_2 \alpha_1 + \delta_2 \gamma_1,$
$\delta_3 = \gamma_2 \beta_1 + \delta_2 \delta_1.$

Substitute these coefficients into $t_2(t_1(x))$ to obtain

$t_2(t_1(x)) = { \alpha_3 x + \beta_3 \over \gamma_3 x + \delta_3}.$

Projections operate in a way analogous to matrices. In fact, the composition of transformations can be obtained by multiplying matrices:

$\begin{bmatrix} \alpha_2 & \beta_2 \\ \gamma_2 & \delta_2 \end{bmatrix} \begin{bmatrix} \alpha_1 & \beta_1 \\ \gamma_1 & \delta_1 \end{bmatrix} = \begin{bmatrix} \alpha_2 \alpha_1 + \beta_2 \gamma_1 & \alpha_2 \beta_1 + \beta_2 \delta_1 \\ \gamma_2 \alpha_1 + \delta_2 \gamma_1 & \gamma_2 \beta_1 + \delta_2 \delta_1 \end{bmatrix} = \begin{bmatrix} \alpha_3 & \beta_3 \\ \gamma_3 & \delta_3 \end{bmatrix}.$

Since matrices multiply associatively, it follows that composition of projections is also associative.

Projections have: an operation (composition), associativity, an identity, an inverse and closure, so they form a group.

## The cross-ratio defined by means of a projection

Let there be a transformation ts such that ts(A) = $\infty$, ts(B) = 0, ts(C) = 1. Then the value of ts(D) is called the cross-ratio of points A, B, C and D, and is denoted as [A, B, C, D]s:

$[A,B,C,D]_s = t_s(D).$

Let

$t_s(x) = {\alpha x + \beta \over \gamma x + \delta},$

then the three conditions for ts(x) are met when

$t_s(A) = {\alpha A + \beta \over \gamma A + \delta} = \infty, \qquad \qquad (7)$
$t_s(B) = {\alpha B + \beta \over \gamma B + \delta} = 0, \qquad \qquad (8)$
$t_s(C) = {\alpha C + \beta \over \gamma C + \delta} = 1. \qquad \qquad (9)$

Equation (7) implies that $\gamma A + \delta = 0$, therefore $\delta = - \gamma A$. Equation (8) implies that $\alpha B + \beta = 0$, so that $\beta = - \alpha B$. Equation (9) becomes

${\alpha C - \alpha B \over \gamma C - \gamma A} = 1,$

which implies

$\gamma = \alpha {C - B \over C - A}.$

Therefore

$t_s(D) = {\alpha D - \alpha B \over \alpha \left( {C - B \over C - A} \right) D - \gamma A} = {\alpha (D - B) \over \alpha \left( {C - B \over C - A} \right) D - \alpha \left( {C - B \over C - A} \right) A}$
$= {D - B \over C - B} \; {C - A \over D - A} = {A - C \over A - D}\; {B - D \over B - C}. \qquad \qquad (10)$

In equation (10), it is seen that ts(D) does not depend on the coefficients of the projection ts. It only depends on the positions of the points on the "subjective" projective line. This means that the cross-ratio depends only on the relative distances among four collinear points, and not on the projective transformation which was used to obtain (or define) the cross-ratio. The cross ratio is therefore

$[A,B,C,D] = {A - C \over A - D} \; {B - D \over B - C}. \qquad \qquad (11)$

## Conservation of cross-ratio

Transformations on the projective line preserve cross ratio. This will now be proven. Let there be four (collinear) points A, B, C, D. Their cross-ratio is given by equation (11). Let S(x) be a projective transformation:

$S(x) = {\alpha x + \beta \over \gamma x + \delta}$

where $\alpha \delta \ne \beta \gamma$. Then

$[S(A) S(B) S(C) S(D)] = {{\alpha A + \beta \over \gamma A + \delta} - {\alpha C + \beta \over \gamma C + \delta} \over {\alpha A + \beta \over \gamma A + \delta} - {\alpha D + \beta \over \gamma D + \delta}} \cdot {{\alpha B + \beta \over \gamma B + \delta} - {\alpha D + \beta \over \gamma D + \delta} \over {\alpha B + \beta \over \gamma B + \delta} - {\alpha C + \beta \over \gamma C + \delta}}$
$= { [(\alpha A + \beta) (\gamma C + \delta) - (\alpha C + \beta) (\gamma A + \delta)] [(\alpha B + \beta) (\gamma D + \delta) - (\alpha D + \beta) (\gamma B + \delta)] \over [(\alpha A + \beta) (\gamma D + \delta) - (\alpha D + \beta) (\gamma A + \delta)] [(\alpha B + \beta) (\gamma C + \delta) - (\alpha C + \beta) (\gamma B + \delta)] }$
$= { [\alpha A \delta + \beta \gamma C - \alpha C \delta - \beta \gamma A] [\alpha B \delta + \beta \gamma D - \alpha D \delta - \beta \gamma B] \over [\alpha A \delta + \beta \gamma D - \alpha D \delta - \beta \gamma A] [\alpha B \delta + \beta \gamma C - \alpha C \delta - \beta \gamma B]}$
$= { [\alpha \delta (A - C) + \beta \gamma (C - A)] [\alpha \delta (B - D) + \beta \gamma (D - B)] \over [\alpha \delta (A - D) + \beta \gamma (D - A)] [\alpha \delta (B - C) + \beta \gamma (C - B)]}$
$= {(\alpha \delta - \beta \gamma) (A - C) (\alpha \delta - \beta \gamma) (B - D) \over (\alpha \delta - \beta \gamma) (A - D) (\alpha \delta - \beta \gamma) (B - C)}$
$= {A - C \over A - D} \cdot {B - D \over B - C}$

Therefore [S(A) S(B) S(C) S(D)] = [A B C D], Q.E.D.