11Projective line (non-examinable)

IA Groups

11 Projective line (non-examinable)
We have seen in matrix groups that
GL
2
(
C
) acts on
C
2
, the column vectors.
Instead, we can also have GL
2
(C) acting on the set of 1-dimensional subspaces
(i.e. lines) of C
2
.
For any
v C
2
, write the line generated by
v
as
hvi
. Then clearly
hvi
=
{λv
:
λ C}
. Now for any
A GL
2
(
C
), define the action as
Ahvi
=
hAvi
. Check
that this is well-defined: for any
hvi
=
hwi
, we want to show that
hAvi
=
hAwi
.
This is true because
hvi
=
hwi
if and only if
w
=
λv
for some
λ C \ {
0
}
, and
then hAwi = hvi = hλ(Av)i = hAvi.
What is the kernel of this action? By definition the kernel has to fix all lines.
In particular, it has to fix our magic lines generated by
1
0
,
0
1
and
1
1
. Since
we want
Ah
1
0
i
=
h
1
0
i
, so we must have
A
1
0
=
λ
0
for some
λ
. Similarly,
A
0
1
=
0
µ
. So we can write
A
=
λ 0
0 µ
. However, also need
Ah
1
1
i
=
h
1
1
i
.
Since
A
is a linear function, we know that
A
1
1
=
A
1
0
+
A
0
1
=
λ
µ
. For the final
vector to be parallel to
1
1
, we must have
λ
=
µ
. So
A
=
λI
for some
I
. Clearly
any matrix of this form fixes any line. So the kernel Z = {λI : λ C \ {0}}.
Note that every line is uniquely determined by its slope. For any
v
=
(
v
1
, v
2
)
, w
= (
w
1
, w
2
), we have
hvi
=
hwi
iff
z
1
/z
2
=
w
1
/w
2
. So we have a
one-to-one correspondence from our lines to C
, that maps h
z
1
z
2
i z
1
/z
2
.
Finally, for each A GL
2
(C), given any line h
z
1
i, we have
a b
c d

z
1

=

az + b
cz + d

az + b
cz + d
So
GL
2
(
C
) acting on the lines is just “the same” as the obius groups acting
on points.