°

»

0 0 1

±®

1 0 x

y : x, y ∈ C

2

=

U(0,1) 0 1

°

»

0 0 1

±®

12

1 x 2x + y

: x, y ∈ C = U(0,0) U(1,1)

2 1 1

=

U(1,1) 0 1 x

°

»

0 0 1

Let S3 be the group of permutations of the ordered set {1, 2, 3} . Given σ ∈ S3 , then we

de¬ne the permutation matrix Pσ to be the matrix whose (i, j)th entry is 1 if i = σ(j), 0

otherwise.

Given an ordered basis E = {e1 , e2 , e3 } and a permutation σ ∈ S3 , then Eσ is the ordered

basis given by

Eσ = eσ’1 (1) , eσ’1 (2) , eσ’1 (3) .

That is, if σ(Ij ) = j for 1 ¤ j ¤ 3, then Eσ = {eI1 , eI2 , eI3 } . Otherwise stated, if Eσ =

{˜1 , ˜2 , ˜3 } , then ej = ˜σ(j) for 1 ¤ j ¤ 3. It is a fact that Pσ = [id]E,Eσ . For example, if

eee e

σ = (1 2 3), then Eσ = {e3 , e1 , e2 } and

®

0 0 1

= 1 0 .

Pσ = [id]E,Eσ 0

° »

0 1 0

The following lemma is easily veri¬ed.

’1

Lemma 4.1.4 1. If [φ]E = M, then [φ]Eσ = Pσ M Pσ .

2. If σ = ω —¦ ω, then Eσ = (Eω )ω and Pσ = Pω Pω .

˜ ˜

˜

In what follows, the groups C, C — , U3 , SL2 , GL2 , PSL2 , SL3 , T3 and D3 and the inner auto-

morphism Int y, de¬ned for a member y of a group G, are as de¬ned in Section 2.2.

The following result is the main theorem of this chapter.

55

Theorem 4.1.5 Let G ⊆ SL3 be an admissible subgroup. Then G is conjugate to exactly

one of the following subgroups of SL3 . The subgroups are classi¬ed according to Levi decom-

position and decomposition of a reductive group into product of torus and semisimple group;

they are also classi¬ed in Table 4.1 below by number of invariant subspaces of C 3 of dimen-

sion 1 (resp., 2), denoted n1 (resp., n2 ). Assume G = Ru P is a Levi decomposition with P

(resp., Ru ) a maximal reductive subgroup (resp., the unipotent radical) of G; H = (P, P ) is

semisimple and T = Z(P )—¦ is a torus.

1. Subgroups satisfying H 1.

1 : {diag(1, 1, 1)} . Here, we have G {1} and n1 =

(a) Subgroups satisfying T

n2 = ∞.

C—.

(b) Subgroups satisfying T

C — . If d1 > d2 ,

i. Subgroups satisfying Ru 0 : T(d1 ,d2 ) . Here, we have G

then n1 = n2 = 3; otherwise, n1 = n2 = ∞.

C:

ii. Subgroups satisfying Ru

’1

A. U(0,0) · Pσ T(d1 ,d2 ) Pσ , d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, σ ∈ S3 , σ ∈

1

C—

{id, (1 2), (2 3)} if (d1 , d2 ) = (1, 0). In this case, we have G C

with

(Int y)(x) = y dσ’1 (1) ’dσ’1 (3) x for y ∈ C — , x ∈ C;

we also have n1 = n2 = 2.

C — with

B. U(1,1) · T(1,0) . In this case, we have G C

1

(Int y)(x) = yx for y ∈ C — , x ∈ C;

we also have n1 = n2 = 1.

C — with

C. U(0,0) · T(1,1) . In this case, we have G C

1

(Int y)(x) = y 3 x for y ∈ C — , x ∈ C;

we also have n1 = ∞, n2 = 2.

’1

C — with

D. U(0,0) · P(1 C

1

3) T(1,1) P(1 3) . In this case, we have G

(Int y)(x) = y ’3 x for y ∈ C — , x ∈ C;

we also have n1 = 2, n2 = ∞.

56

C2 :

iii. Subgroups satisfying Ru

’1

A. U(1,0) · Pσ T(d1 ,d2 ) Pσ , where d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, and σ ∈

2

C — with

{id, (1 2), (1 2 3)} . In this case, we have G C2

(Int y)(w, x) = (y dσ’1 (1) ’dσ’1 (2) w, y dσ’1 (1) ’dσ’1 (3) x)

for y ∈ C — , (w, x) ∈ C 2 ; we also have n1 = 1, n2 = 2.

’1

B. U(0,1) · Pσ T(d1 ,d2 ) Pσ , where d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, and σ ∈

2

C — with

{id, (2 3), (1 3 2)} . In this case, we have G C2

(Int y)(w, x) = (y dσ’1 (1) ’dσ’1 (3) w, y dσ’1 (2) ’dσ’1 (3) x)

for y ∈ C — , (w, x) ∈ C 2 ; we also have n1 = 2, n2 = 1.

C — with

C. U(1,1) · T(1,0) . In this case, we have G C2

2

(Int y)(w, x) = (yw, y 2 x) for y ∈ C — , (w, x) ∈ C 2 ;

we also have n1 = n2 = 1.

’1

C — with

D. U(1,0) · P(1 C2

2

3) T(1,1) P(1 3) . In this case, we have G

(Int y)(w, x) = (y ’3 w, y ’3 x) for y ∈ C — , (w, x) ∈ C 2 ;

we also have n1 = 1, n2 = ∞.

C — with

E. U(0,1) · T(1,1) . In this case, we have G C2

2

(Int y)(w, x) = (y 3 w, y 3 x) for y ∈ C — , (w, x) ∈ C 2 ;

we also have n1 = ∞, n2 = 1.

U3 :

iv. Subgroups satisfying Ru

’1

A. U3 · Pσ T(d1 ,d2 ) Pσ , d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, σ ∈ S3 , σ ∈

{id, (1 2), (2 3)} if (d1 , d2 ) = (1, 0). We have G U3 C — with

«® ®

κ1 κ2

1vw 1yvyw

¬ ·

¬ ·

(Int y) ¬ 0 1 x · = 0 y κ3 x ,

1

° » ° »

001 0 0 1

where κ1 = dσ’1 (1) ’ dσ’1 (2) , κ2 = dσ’1 (1) ’ dσ’1 (3) ,

κ3 = dσ’1 (2) ’ dσ’1 (3) , for

®

1 v w

y ∈ C—, 0 x ∈ U3 ;

1

° »

0 0 1

57

we also have n1 = n2 = 1.

’1

C — with

B. U3 · P(2 U3

3) T(1,1) P(2 3) . We have G

«®

®

3

yv w

1 v w 1

¬

·

¬

· ’3

(Int y) ¬ 0 x

x · = 0 1 y

1

° »

» °

0 0 1

0 0 1

for ®

1 v w

y ∈ C—, 0 x ∈ U3 ;

1

° »

0 0 1

we also have n1 = n2 = 1.

C— — C—.

(c) Subgroups satisfying T

C — — C — and

0 : D3 © SL3 . Here, we have G

i. Subgroups satisfying Ru

n1 = n2 = 3.

C : U(0,0) · (D3 © SL3 ). Here, we have G C

1

ii. Subgroups satisfying Ru

(C — — C — ) with

(Int(y1 , y2 ))(x) = y1 y2 x for y1 , y2 ∈ C — , x ∈ C;

2

we also have n1 = n2 = 2.