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’1
5. Suppose G = U(1,0) · Pσ T(d1 ,d2 ) Pσ , where d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, and σ ∈
2


{id, (1 2), (1 2 3)} . This group has a faithful parameterization
® 
dI1
w x
y
 
 
(w, x, y) ’  0 0 ,
y dI2
° »
dI3
0 0 y

where d3 = ’d1 ’d2 and σ(Ij ) = j for 1 ¤ j ¤ 3. Under this parameterization, we have
(Int y)(w, x) = (y dI1 ’dI2 w, y dI1 ’dI3 x). The G-invariant subspaces are a1 , a1 , a2 and
a1 , a3 , so that n1 = 1, n2 = 2.

’1
6. Suppose G = U(0,1) · Pσ T(d1 ,d2 ) Pσ , where d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, and σ ∈
2


{id, (2 3), (1 3 2)} . Then G has a faithful parameterization
® 
dI1
0 w
y
 
 
(w, x, y) ’  0 x ,
dI2
y
° »
dI3
0 0 y

where d3 = ’d1 ’ d2 and σ(Ij ) = j for 1 ¤ j ¤ 3. Under this parameterization, we
have (Int y)(w, x) = (y dI1 ’dI3 w, y dI2 ’dI3 x). The G-invariant subspaces are a1 , a2
and a1 , a2 , so that n1 = 2, n2 = 1.

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7. Suppose G = U(1,1) · T(1,0) . Then G has a faithful parameterization
2

® 
12
2w
y w x+
 
 
(w, x, y) ’  0 .
1 w
° »
y ’1
0 0

Under this parameterization, we have (Int y)(w, x) = (yw, y 2 x). The G-invariant sub-
spaces are a1 and a1 , a2 , so that n1 = n2 = 1.

’1
8. Suppose G = U(1,0) · P(1
2
3) T(1,1) P(1 3) . Then G has a faithful parameterization
® 
’2
y w x
 
 
(w, x, y) ’  0 .
0 y
° »
0 0 y

Under this parameterization, we have (Int y)(w, x) = (y ’3 w, y ’3 x). The G-invariant
subspaces are a1 and a1 , ±a2 + βa3 , where (± : β) ∈ P1 . Thus, we have n1 = 1, n2 =
∞.

9. Suppose G = U(0,1) · T(1,1) . Then G has a faithful parameterization
2

® 
y 0 w
 
 
(w, x, y) ’  0 .
y x
° »
y ’2
0 0

Under this parameterization, we have (Int y)(w, x) = (y 3 w, y 3 x). The G-invariant sub-
spaces are ±a1 + βa2 , where (± : β) ∈ P1 ; and a1 , a2 . Thus, we have n1 = ∞, n2 =
1.

The above listed subgroups are pairwise nonconjugate; in particular, they are distinct.

Proof. The statements in this lemma are veri¬ed by straightforward computations, as is
the fact that the above listed sets are subgroups. Nonconjugacy of the various groups
follows after studying their actions on invariant subspaces. In Item 1, the stipulation that
’1
σ ∈ {id, (1 2), (2 3)} if (d1 , d2 ) = (1, 0) is necessary because P(1 3) T(1,0) P(1 3) = T(1,0) .




C — , where
Lemma 4.5.3 Suppose G = [G]E0 is a subgroup of SL3 that is isomorphic to C r
C — . Let
r is either 1 or 2. Let G have Levi decomposition G = Ru T, where Ru C r and T

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F (resp., B) be a basis of V such that [T]F (resp., [Ru ]B ) is one of the subgroups listed in
Lemma 4.2.3 (resp., Lemmas 4.3.1 and 4.3.4).

1. Suppose [T]F = T(d1 ,d2 ) for some d1 , d2 ∈ Z with d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1.

’1
(a) If [Ru ]B = U(0,0) , then there exists a basis A such that [G]A = U(0,0) ·Pσ T(d1 ,d2 ) Pσ
1 1


for some σ ∈ S3 . If (d1 , d2 ) = (1, 0), then σ can be taken to be one of the
permutations {id, (1 2), (2 3)} .

(b) If [Ru ]B = U(1,1) , then d1 = 1, d2 = 0, and there exists a basis A such that
1


[G]A = U(1,1) · T(1,0) .
1


(c) If [Ru ]B = U(1,0) , then there exists a permutation σ ∈ {id, (1 3), (1 2 3)} such that
2

’1
[G]Fσ = U(1,0) · Pσ T(d1 ,d2 ) Pσ .
2


(d) If [Ru ]B = U(0,1) , then there exists a permutation σ ∈ {id, (2 3), (1 3 2)} such that
2

’1
[G]Fσ = U(0,1) · Pσ T(d1 ,d2 ) Pσ .
2


(e) If [Ru ]B = U(1,1) , then d1 = 1, d2 = 0, and there exists a basis A such that
2


[G]A = U(1,1) · T(1,0) .
2


1 2 2
2. Suppose [T]F = T(1,1) . Then [Ru ]B is one of U(0,0) , U(1,0) , U(0,1) . Moreover, the
following statements hold:

(a) If [Ru ]B = U(0,0) , then there exist a basis A and a permutation σ ∈ {id, (1 3)}
1

’1
such that [G]A = U(0,0) · Pσ T(1,1) Pσ .
1


’1
= U(1,0) · P(1
2 2
(b) If [Ru ]B = U(1,0) , then [G]F(1 3) T(1,1) P(1 3) .
3)


(c) If [Ru ]B = U(0,1) , then [G]F = U(0,1) · T(1,1) .
2 2


ˆ ˆ ˆ
In particular, there exists a basis A and a unique subgroup G such that [G]A = G and G is
one of the subgroups listed in Lemma 4.5.2.

Proof. By the Lie-Kolchin theorem, there exists a ¬‚ag

V2 V
0 V1

that is preserved by the action of G.

1. Suppose [T]F = T(d1 ,d2 ) for some d1 , d2 ∈ Z with d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1. Let
d3 = ’d1 ’ d2 . Now, V1 ⊆ V is a G-invariant subspace and in particular a T-invariant
subspace, so that by Lemma 4.2.2 we have V1 = fA1 for some index A1 . An analogous

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argument shows that V2 = fA1 , fA2 for some index A2 . Let ω ∈ S3 be the permutation
such that ω(Aj ) = j for j = 1, 2; de¬ne A3 so that ω(A3 ) = 3. Following the notational
convention established in Section 2.1, we consider the ordered basis

Fω = {fA1 , fA2 , fA3 } .

Fix a nonroot of unity v ∈ C — . From the parameterization given in Lemma 4.2.2, we
see that [T]Fω is generated by the matrix Q, where Q is given by (4.4).

We will next consider the subgroup [Ru ]Fω . Since V1 and V2 are G-invariant, we see
that [G]Fω ⊆ T3 . This fact, together with the identity

[G]Fω = [Ru ]Fω [T]Fω , (4.6)

implies that [Ru ]Fω ⊆ U3 .

• Assume Ru C. Let φ ∈ SL(V ) be a generator of Ru . Then [Ru ]Fω is generated
by a nontrivial matrix M = [φ]Fω of the form (4.1). Since (4.6) is a Levi decom-
position, we have QM Q’1 ∈ [Ru ]Fω , where Q is the generator of [T]Fω speci¬ed
above.
Suppose [Ru ]B = U(0,0) . Then [Ru ]Fω is conjugate to U(0,0) . Let q : U3 ’ C 2 be
1 1


the mapping de¬ned in Lemma 4.3.3. Then q(M ) = (a, c) ∈ C 2 . Since U(0,0) is
1

1
nonconjugate to U(1,1) by Lemma 4.3.1, it follows from Item 3 of Lemma 4.3.4
that either a = 0 or c = 0. Lemma 4.5.1 now implies that two of {a, b, c} are zero.
De¬ne „ ∈ S3 to be (2 3) if a = 0, id if b = 0, (1 2) if c = 0. De¬ne σ = „ —¦ ω.
It is now an exercise involving the de¬nitions and Lemma 4.1.4 to check that
’1 ’1
[G]Fσ = U(0,0) · Pσ T(d1 ,d2 ) Pσ . If (d1 , d2 ) = (1, 0), then Pσ T(1,0) Pσ is one of the
1


following three groups:

’1 ’1
t, 1, t’1 = T(1,0) , 1, t, t’1 = P(1 t, t’1 , 1 = P(2
2) T(1,0) P(1 2) , 3) T(1,0) P(2 3) .


Item 1(a) of the conclusion now follows easily.
1
Now suppose [Ru ]B = U(1,1) , i.e., a = 0 and c = 0. Here, Lemma 4.5.1 implies that
1
d1 = 1, d2 = 0, A2 = 2, and b = ac. De¬ne A = {a1 , a2 , a3 } by a1 = afA1 , a2 =
® 
2
1 1 1/2
 
 
’1
f2 , a3 = c fA3 . It is easy to check that [φ]A =  0 1 1  , from which we
° »
00 1

77
1
obtain [Ru ]A = U(1,1) . Next, from A2 = 2 we see that ω is either id or (1 3). In
either case, we claim that [T]Fω = T(1,0) . Indeed, this statement is trivial in case
ω = id; if ω = (1 3), one checks that

’1
P(1 3) T(1,0) P(1 3) = T(1,0) ,

and the claim follows easily in this case as well. From [T]Fω = T(1,0) we obtain

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( 33 .)



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