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 = hAλvi = 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 M¨obius groups acting

on points.