The cross-ratio of four collinear points is an invariant under the homography that is fundamental for the study of the homographies of the lines. Three distinct points , and on a projective line over a field form a projective frame of this line. There is therefore a unique homography of this line onto that maps to , to 0, and to 1. Given a fourth point on the same line, the
cross-ratio of the four points , , and , denoted , is the element of . In other words, if has
homogeneous coordinates over the projective frame , then . \frac{w-a}{w-c} / \frac{b-a}{b-c}. This transformation is composed of two elementary homographies: • The separating homography \begin{pmatrix}1 & 1 \\ -a & -c \end{pmatrix} that sends U(
a,1) to U(0,1) and U(
c,1) to ∞ = U(1,0), and • The normalizing homography \begin{pmatrix}e & 0 \\ 0 & 1 \end{pmatrix} where \scriptstyle e = \frac{b - c}{b - a}, which sends U(
b,1) to U(1,1) and has U(0,1) and U(1,0) as
fixed points. In the general
ring, the fraction
e may not be available. Then the construction of the cross-ratio homography may be approached as follows: Suppose
p,
q,
r ∈
A with :
t = (
r –
p)−1 and
v = (
t + (
q −
r)−1)−1. When these inverses
t and
v exist we say "
p,
q, and
r are separated sufficiently". Up to sufficient separation, the group of homographies is
3-transitive: : \begin{pmatrix} 1 & 0 \\ -r & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} \begin{pmatrix} v & 0 \\ 0 & 1 \end{pmatrix}. The first two factors put
r at
U(1, 0) = ∞ where it stays. The third factor moves
t, the image of
p under the first two factors, to
U(0, 1), or zero in the canonical embedding. Finally, the fourth factor has traced
q through the first three factors and formation of the rotation with
v places U(
q, 1) at
U(1, 1). Thus the composition displayed places the triple
p,
q,
r at the triple 0, 1, ∞. Evidently it is the
unique such homography considering the pivotal use of fixed points of generators to bring the triple to 0, 1, ∞.
Proposition: If the group of homographies is sharply 3-transitive, then there is a cross-ratio function that is
invariant under the permutation of the projective line by homographies. : proof: If
s and
t are two sufficiently separated triples, they correspond to homographies
g and
h respectively, which map each of
s and
t to (0, 1, ∞). Thus the homography
h−1 o
g maps
s to
t . : Denote by (
x,
p ,
q,
r) the image of
x under the homography determined by
p,
q,
r as above. This function f(
x) is the
cross-ratio determined by
p,
q,
r ∈
A. The uniqueness of this function (sharp transitivity) implies that when a single homography
g ∈
G(
A) is used to form another triple
g(
p),
g(
q),
g(
r) from the first one, then the new cross-ratio function
h must agree with
f ∘
g. Hence
h ∘
g−1 =
f so that : (
g(
x),
g(
p),
g(
q),
g(
r) ) = (
x,
p,
q,
r). As the sharpness does not hold in non-commutative rings like quaternions and biquaternions, there are limits to usage of cross-ratios. --> == Over a ring ==