10Mobius group

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10.4 Cross-ratios
Finally, we’ll look at an important concept known as cross-ratios. Roughly
speaking, this is a quantity that is preserved by obius transforms.
Definition
(Cross-ratios)
.
Given four distinct points
z
1
, z
2
, z
3
, z
4
C
,
their
cross-ratio is [
z
1
, z
2
, z
3
, z
4
] =
g
(
z
4
), with
g
being the unique obius map that
maps z
1
7→ , z
2
7→ 0, z
3
7→ 1. So [, 0, 1, λ] = λ for any λ 6= , 0, 1. We have
[z
1
, z
2
, z
3
, z
4
] =
z
4
z
2
z
4
z
1
·
z
3
z
1
z
3
z
2
(with special cases as above).
We know that this exists and is uniquely defined because
M
acts sharply
three-transitively on C
.
Note that different authors use different permutations of 1
,
2
,
3
,
4, but they
all lead to the same result as long as you are consistent.
Lemma. For z
1
, z
2
, z
3
, z
4
C
all distinct, then
[z
1
, z
2
, z
3
, z
4
] = [z
2
, z
1
, z
4
, z
3
] = [z
3
, z
4
, z
1
, z
2
] = [z
4
, z
3
, z
2
, z
1
]
i.e. if we perform a double transposition on the entries, the cross-ratio is retained.
Proof. By inspection of the formula.
Proposition. If f M, then [z
1
, z
2
, z
3
, z
4
] = [f(z
1
), f(z
2
), f(z
3
), f(z
4
)].
Proof.
Use our original definition of the cross ratio (instead of the formula). Let
g be the unique obius map such that [z
1
, z
2
, z
3
, z
4
] = g(z
4
) = λ, i.e.
z
1
g
7−
z
2
7→ 0
z
3
7→ 1
z
4
7→ λ
We know that gf
1
sends
f(z
1
)
f
1
7− z
1
g
7−
f(z
2
)
f
1
7− z
2
g
7− 0
f(z
3
)
f
1
7− z
3
g
7− 1
f(z
4
)
f
1
7− z
4
g
7− λ
So [f (z
1
), f(z
2
), f(z
3
), f(z
4
)] = gf
1
f(z
4
) = g(z
4
) = λ.
In fact, we can see from this proof that: given
z
1
, z
2
, z
3
, z
4
all distinct and
w
1
, w
2
, w
3
, w
4
distinct in
C
, then
f M
with
f
(
z
i
) =
w
i
iff [
z
1
, z
2
, z
3
, z
4
] =
[w
1
, w
2
, w
3
, w
4
].
Corollary. z
1
, z
2
, z
3
, z
4
lie on some circle/straight line iff [z
1
, z
2
, z
3
, z
4
] R.
Proof.
Let
C
be the circle/line through
z
1
, z
2
, z
3
. Let
g
be the unique obius
map with
g
(
z
1
) =
,
g
(
z
2
) = 0,
g
(
z
3
) = 1. Then
g
(
z
4
) = [
z
1
, z
2
, z
3
, z
4
] by
definition.
Since we know that obius maps preserve circle/lines,
z
4
C g
(
z
4
) is on
the line through , 0, 1, i.e. g(z
4
) R.