Graphics Reference
In-Depth Information
An inherent problem when using coordinates is that one has to show that different
coordinatizations give the same answer.
Let
A
,
B
,
C
, and
D
be four distinct points on a line
L
in
P
2
. We shall denote the
cross-ratio in which
B
and
C
divide
A
and
D
by (
AD
,
BC
). Here are two definitions for
it:
Definition 1.
Assume that
A
= [
a
] and
D
= [
d
] and express
B
and
C
in the form
[
a
+ k
d
] and [
a
+ k¢
d
], respectively, for some nonzero k and k¢. Define
k
k
(
)
=
AD BC
,
.
(3.19)
¢
Definition 2.
Let
P
= [
p
] and
Q
= [
q
] be any distinct points of
L
and express
A
,
B
,
C
, and
D
in the form [
p
+ a
q
], [
p
+ b
q
], [
p
+ c
q
], and [
p
+ d
q
], respectively.
Define
(
)
(
)
abdc
acdb
-
-
(
)
=
AD BC
,
.
(3.20)
(
)
(
)
-
-
3.4.1.6. Theorem.
The two definitions in equations (3.19) and (3.20) of the cross-
ratio (
AD
,
BC
) of four distinct points
A
,
B
,
C
, and
D
on a line in the projective plane
are well defined and agree.
Proof.
The proof consists of some straightforward computations.
To prove that Definition 1 is well defined, let
A
= [s
a
] and
D
= [t
d
] and let
B
= [(s
a
) + m(t
d
)] and
C
= [(s
a
) + m¢(t
d
)]. We must show that
m
m
k
k
=
¢
.
¢
Now, (s
a
) + m(t
d
) = e(
a
+ k
d
) and (s
a
) + m¢(t
d
) = f(
a
+ k¢
d
) for nonzero constants
e and f. It follows that
(
)
(
)
(
)
(
)
s
-
e
a
=-
ek
mt
d
and
s
-
f
a
=¢ -¢
fk
m t
d
.
Since the vectors
a
and
d
are linearly independent, s = e = f, ek = mt, and fk¢=m¢t,
which easily implies that our two ratios are the same.
To prove that Definition 2 is well defined, one needs to show that it depends
neither on the choice of
P
and
Q
nor on their representatives. The independence of
their representatives is easy to show. For the rest, first replace
P
by
P
¢=[
p
¢ ],
p
¢=
p
+ s
q
, express the points
A
,
B
,
C
, and
D
in terms of
P
¢ and
Q
as [
p
¢+a¢
q
], [
p
¢+b¢
q
],
[
p
¢+c¢
q
], and [
p
¢+d¢
q
], respectively, and compare the new expressions with the old
ones. One will find that a = s + a¢, b = s + b¢, etc., so that it is easy to check that
(
)
(
)
(
)
(
)
abdc
acdb
¢- ¢
¢- ¢
abdc
acdb
.
-
-
=
(
)
(
)
(
)
(
)
¢- ¢
¢- ¢
-
-
