[T]A = [id]Fω ,A [T]Fω [id]A,Fω

diag(a’1 , 1, c)T(1,0) diag(a, 1, c’1 ) = T(1,0) .

=

We conclude that [G]A = [Ru ]A [T]A = U(1,1) · T(1,0) as desired.

1

• Now assume Ru C 2 . If [Ru ]B = U(1,0) , then Lemma 4.3.1 implies that V1 = b1

2

and Ru acts trivially on V /V1 . It follows that b1 and fA1 di¬er by a scalar multiple

and that φ(fAj ) ∈ fAj + V1 for φ ∈ Ru and j ∈ {2, 3} . This implies that [Ru ]Fω =

U(1,0) . Moreover, we may swap fA2 and fA3 if necessary so that A2 < A3 . Since

2

Aj = σ ’1 (j), this means that we only need consider the permutations id, (1 3)

and (1 2 3). Item 1(c) of the lemma is now clear.

2

If [Ru ]B = U(0,1) , then Lemma 4.3.1 implies that

V2 = fA1 , fA2 = b1 , b2

and Ru acts trivially on V2 and V /V2 . It follows that [Ru ]Fω = U(0,1) . Moreover,

2

in this case, we may swap fA1 and fA2 if necessary so that A1 < A2 . Thus, we

need only consider the permutations id, (2 3) and (1 3 2). Item 1(d) of the lemma

follows.

Next, suppose [Ru ]B = U(1,1) . Since [Ru ]Fω ⊆ U3 , the proof of Lemma 4.3.4 implies

2

1

that [Ru ]Fω = clos(M )·U(0,0) for some matrix M of the form (4.1) with a = 0, c =

0. Here, Lemma 4.5.1 yields d1 = 1, d2 = 0, A2 = 2. De¬ne A = {a1 , a2 , a3 } by

a1 = afA1 , a2 = f2 , a3 = c’1 fA3 . Then, a computation analogous to that used in

the case [Ru ]B = U(1,1) shows that [G]A = U(1,1) · T(1,0) , as stated in Item 1(e).

1 2

2. Suppose [T]F = T(1,1) , so that d1 = d2 = 1 and we de¬ne d3 = ’2. In this case,

[T]F is generated by a matrix of the form diag(v, v, v ’2 ), v ∈ C — a nonroot of unity.

Recall that V1 is a G-invariant subspace and in particular a T-invariant subspace.

Therefore V1 is either f3 or ±f1 + βf2 for some nonzero (±, β). In order to narrow

78

down our list of choices, note that our choice of basis F is arbitrary, so long as

[T]F = T(1,1) . Therefore, in case V1 = f3 , we may swap f1 and f2 in F or replace

f1 with f1 + γf2 (γ ∈ C) if necessary so that V1 = f1 ; Lemma 4.2.2 then implies

that either V2 = f1 , f2 or V2 = f1 , f3 . In case V1 = f3 , we may perform a similar

operation if necessary so that V2 = f2 , f3 . Thus, without loss of generality, we may

assume that one of the following holds:

• V1 = f1 , V2 = f1 , f2 (ω = id)

• V1 = f1 , V2 = f1 , f3 (ω = (2 3))

• V1 = f3 , V2 = f2 , f3 (ω = (1 3))

We now see that [G]Fω is included in T3 , where ω is the appropriate permutation

de¬ned above. Let Aj = ω ’1 (j) for 1 ¤ j ¤ 3, so that [T]Fω is generated by a matrix

Q of the form (4.4).

• 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.

1

Suppose [Ru ]B = U(0,0) , i.e., either a = 0 or c = 0. We distinguish three subcases:

“ If a = c = 0, then [Ru ]Fω = U(0,0) and we have [G]Fω = U(0,0) · T(1,1) .

1 1

“ If a = 0, b = 0, c = 0, then Lemma 4.5.1 implies that A3 = 3, so that ω = id

by hypothesis. If we now let a1 = bf1 +cf2 , a2 = f2 , a3 = f3 , then [T]A = T(1,1)

since T acts in the same way on all subspaces of the form ±f1 + βf2 , ±, β ∈ C.

Meanwhile, it is easy to check that [φ]A ∈ U(0,0) , so that [Ru ]A = U(0,0) ; we

1 1

conclude that [G]A = U(0,0) · T(1,1) .

1

“ If a = 0, b = 0, c = 0, then Lemma 4.5.1 implies that A1 = 3, so that

ω = (1 3) by hypothesis; if we now let a1 = f3 , a2 = a’1 bf2 ’ f1 , a3 = f2 , then

a sequence of calculations similar to the previous case shows that [G]A =

U(0,0) · P(1

1

3) T(1,1) P(1 3) .

Each subcase is consistent with Item (2a) of the lemma. Note that [Ru ]B cannot

1

be U(1,1) if [T]F = T(1,1) , by Item (1d) of Lemma 4.5.1.

• Assume Ru C2.

2

Note that [Ru ]B cannot be U(1,1) if [T]F = T(1,1) , by Item 2 of Lemma 4.5.1.

79

2

Suppose [Ru ]B = U(1,0) . Then, by Lemma 4.3.4, we have

V1 = b1 = fA1

and Ru acts trivially on V /V1 . We conclude that [Ru ]Fω ⊆ U(1,0) and hence

2

2

[Ru ]Fω = U(1,0) . A defect argument then implies that ω = (1 3), and we con-

’1

= U(1,0) · P(1

2

clude that [G]F(1 3) T(1,1) P(1 3) .

3)

Suppose [Ru ]B = U(0,1) . Here, Ru ¬xes V2 elementwise. We conclude that [Ru ]Fω =

2

U(0,1) . A defect argument implies ω = id . We conclude that [G]F = U(0,1) · T(1,1) .

2 2

Lemma 4.5.4 Given G = [G]A ⊆ SL3 , where G is a subgroup of SL3 (V ) and A = {a1 , a2 , a3 }

is a basis of V.

1. Suppose G = U(0,0) · (D3 © SL3 ). Then G admits a faithful

1

parameterization

®

y 0 x

1

(x, y1 , y2 ) ’ 0 y2 ,

0

° »

0 0 (y1 y2 )’1

under which (Int(y1 , y2 ))(x) = y1 y2 x. The G-invariant subspaces are a1 , a2 , a1 , a2

2

and a1 , a3 , so that n1 = n2 = 2.

2. Suppose G = U(1,0) · (D3 © SL3 ). Then G has

2

a faithful parameterization

®

y w x

1

(w, x, y1 , y2 ) ’ 0 ,

y2 0

° »

’1

0 0 (y1 y2 )

’1

under which (Int(y1 , y2 ))(w, x) = (y1 y2 w, y1 y2 x). The G-invariant subspaces are

2

a1 , a1 , a2 and a1 , a3 , so that n1 = 1, n2 = 2.

3. Suppose G = U(0,1) · (D3 © SL3 ). Then G has

2

a faithful parameterization

®

0 w

y

1

(w, x, y1 , y2 ) ’ 0 ,

y2 x

° »

’1

0 0 (y1 y2 )

under which (Int(y1 , y2 ))(w, x) = (y1 y2 w, y1 y2 x). The G-invariant subspaces are a1 ,

2 2

a2 and a1 , a2 , so that n1 = 2, n2 = 1.

80

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.

(C — — C — ),

Lemma 4.5.5 Suppose G = [G]E0 is a subgroup of SL3 that is isomorphic to C r

where r is either 1 or 2. Let G have Levi decomposition G = Ru T, where Ru C r and T

C — —C — . Let F (resp., B) be a basis of V such that [T]F (resp., [Ru ]B ) is D3 ©SL3 (resp., one of

1 2 2

the subgroups listed in Lemmas 4.3.1 and 4.3.4). Then [Ru ]B is one of U(0,0) , U(1,0) , U(0,1) .

Moreover, the following statements hold:

1. If [Ru ]B = U(0,0) , then there exists a permutation σ ∈ S3 such that [G]Fσ = U(0,0) ·

1 1

(D3 © SL3 ).

2. If [Ru ]B = U(1,0) , then there exists a permutation σ ∈ S3 such that [G]Fσ = U(1,0) ·

2 2

(D3 © SL3 ).

3. If [Ru ]B = U(0,1) , then there exists a permutation σ ∈ S3 such that [G]Fσ = U(0,1) ·

2 2

(D3 © SL3 ).

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 4.5.4.

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

V2 V

0 V1

that is preserved under the action of G.

It follows from Lemma 4.2.2 that (after reordering the basis vectors of F if necessary)

V1 = f1 , V2 = f1 , f2 . It follows that [Ru ]F ⊆ U3 . Let M be a member of [Ru ]F , where M

is as described in (4.1). If [Ru ]F = U(0,0) , then assume M ∈ U(0,0) .

1

/1

Let Q = diag v1 , v2 , (v1 v2 )’1 ∈ [T]F with v2 a nonroot of unity. We compute

®

’1 2

1 v1 v2 a v1 v2 b

QM Q’1 = 0 .

2

1 v1 v2 c

° »

0 0 1

81

If both a and c are nonzero, then the proof of Lemma 4.3.3 and the fact that QM Q’1 ∈ [Ru ]F

’1

imply that v1 v2 = x = v1 v2 for some x ∈ C, which in turn implies that v2 = 1. This

2 3

contradicts hypothesis on v2 , and we conclude that either a = 0 or c = 0.

C 2 , we may apply Lemma 4.3.3 and its proof to show that [Ru ]F = clos(M )·

In case Ru

C, an additional normalization

1