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U(1,0) 0 1
° 
»
 
 
 
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 = ∞.

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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.

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( 33 .)



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