argument shows that two of {a, b, c} are zero. Thus, we may reorder the vectors of F if

1

necessary so that Ru = U(0,0) , and the result follows.

Lemma 4.5.6 Given G = [G]A ⊆ SL3 , where G is a subgroup of SL3 (V ) and A = {a1 , a2 , a3 }

is a basis of V.

’1

1. Suppose G = 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). Then G has a faithful parameterization

«®

®

dI1

v w

1vw y

¬

·

¬

·

¬ 0 1 x , y · ’ 0 x ,

dI2

y

° »

» °

dI3

0 0 y

001

where d3 = ’d1 ’ d2 and σ(Ij ) = j for 1 ¤ j ¤ 3. Under this parameterization, we

«®

®

have

dI1 ’dI2 dI1 ’dI3

v y w

1 v w 1 y

¬

·

¬

·

y dI2 ’dI3 x .

(Int y) ¬ 0 x · = 0 1

1

° »

» °

0 0 1

0 0 1

’1

2. Suppose G = U3 · P(2 3) T(1,1) P(2 3) . This subgroup has a faithful parameterization

«® ®

1 v w y v w

¬ ·

¬ ·

x , y· ’ 0 y ’2

¬ 0 x .

1

° » ° »

0 0 1 0 0 y

Under this parameterization, we have

«® ®

3

1 yv w

1 v w

¬ ·

¬ · ’3

x .

(Int y) ¬ 0 x · = 0 1 y

1

»

° » °

00 1

0 0 1

In either case, the G-invariant subspaces are a1 and a1 , a2 , so that n1 = n2 = 1. These

subgroups are pairwise nonconjugate; in particular, they are distinct.

82

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 — . Let G

Lemma 4.5.7 Suppose G = [G]E0 is a subgroup of SL3 that is isomorphic to U3

C — . Let F (resp., B) be a basis

have Levi decomposition G = Ru T, where Ru U3 and T

of V such that [T]F (resp., [Ru ]B ) is one of the subgroups listed in Lemma 4.2.3 (resp., U3 ).

1. If [T]F = T(d1 ,d2 ) , d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, then there exists a basis A such

’1

that [G]A = U3 · Pσ T(d1 ,d2 ) Pσ for some σ ∈ S3 . If (d1 , d2 ) = (1, 0), then we may take

σ to be one of {id, (1 2), (2 3)} .

’1

2. If [T]F = T(1,1) , then there exists a basis A such that [G]A = U3 · P(2 3) T(1,1) P(2 3) .

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 4.5.6.

Proof. By the Lie-Kolchin theorem, there exists a ¬‚ag 0 V1 V2 V that is preserved

by the action of G.

Let F be as in Lemma 4.2.3 and let [T]F = T(d1 ,d2 ) for some d1 , d2 , d1 ≥ d2 ≥ 0. If

d1 > d2 , then there exists a permutation ω such that [G]Fω ⊆ T3 . It follows that [G]Fω

’1

is one of the groups listed in Item 1. If (d1 , d2 ) = (1, 0), then Pω T(1,0) Pω is one of the

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 of the conclusion now follows easily.

If d1 = d2 = 1, then we may alter our de¬nition of the basis F if necessary so that V1 is

generated by either f1 or f3 . Therefore, there exists a permutation ω such that [G]Fω ⊆ T3

in this case as well. Now, a defect argument implies that ω = (2 3), and we conclude that

[G]Fω is the group listed in Item 2.

Lemma 4.5.8 Suppose G = [G]E0 is a subgroup of SL3 that is isomorphic to C 2 P, where

P = [P]E0 is isomorphic to either SL2 or GL2 . Then there exist a basis A and a unique

ˆ ˆ

subgroup G ⊆ SL3 such that G = [G]A is one of the following subgroups:

83

SL2 :

1. If P

(a) {(tij ) ∈ SL3 : t31 = t32 = 0, t33 = 1} C2 SL2 , in which conjugation is given

by the unique irreducible representation of SL2 on C 2 . In this case, the only G-

invariant subspace is a1 , a2 , so that n1 = 0, n2 = 1.

(b) {(tij ) ∈ SL3 : t21 = t31 = 0, t11 = 1} C2 SL2 , in which conjugation is given

by the unique irreducible representation of SL2 on C 2 . In this case, the only G-

invariant subspace is a1 , so that n1 = 1, n2 = 0.

GL2 :

2. If P

(a) {(tij ) ∈ SL3 : t31 = t32 = 0} C2 GL2 , in which conjugation is given by

M.v = (det M )M v for M ∈ GL2 , v ∈ C 2 .

In this case, the only G-invariant subspace is a1 , a2 , so that n1 = 0, n2 = 1.

(b) {(tij ) ∈ SL3 : t21 = t31 = 0} C2 GL2 , in which conjugation is given by

M.v = (det M )’1 (M ’1 )T v for M ∈ GL2 , v ∈ C 2 .

In this case, the only G-invariant subspace is a1 , so that n1 = 1, n2 = 0.

Proof. By Lemma 4.3.4, there exists a basis B such that the subgroup [Ru ]B is one of

2 2 2

U(1,0) , U(0,1) , U(1,1) . Let P = [P]B . Then P is included in NorSL3 ([Ru ]B ). The desired

2

result now follows easily from Lemma 4.3.5; note that [Ru ]B cannot be U(1,1) because the

normalizer of the latter group is included in T3 while P cannot be upper-triangularized.

The statements about invariant subspaces are veri¬ed by calculations.

We give one more technical lemma before proceeding.

1. For d ≥ 0, de¬ne Xd C d+1 to be the space of homogeneous polyno-

Lemma 4.5.9

mials of degree d in C[x, y], where x and y are indeterminates. De¬ne

®

ab

° » .(xi y d’i ) = (ax + cy)i (bx + dy)d’i

cd

®

ab

for a typical element ° » ∈ SL2 and a typical basis vector xi y d’i of Xd , 0 ¤ i ¤

cd

d. Then, up to isomorphism, the induced action of SL2 on Xd is the unique irreducible

84

(d + 1)-dimensional representation of SL2 . If d is odd, then this representation is

faithful. If d is even and greater than zero, then the representation has kernel {±I2 }

and thus induces a faithful (d + 1)-dimensional representation of PSL2 . Moreover, the

only nontrivial irreducible representations of PSL2 are the ones which arise in this

way.

2. If G0 is an algebraic group and the vector space W is an irreducible G0 -module, then

any G0 -module endomorphism of W is a scalar multiple of the identity.

3. Given a faithful representation SL2 ’ SL(W ) with W C 3 . Then there exists a

decomposition

W = W1 + W2 (direct sum of SL2 -invariant subspaces),

C r , Xr’1 as described in Item 1, for r = 1, 2. Moreover, W1 and

where Wr Xr’1

W2 are the only nonzero proper SL2 -invariant subspaces of W.

4. Let G0 be one of {SL2 , PSL2 } and let A be one of {C, C — } . Then it is impossible to

embed G0 — A in SL3 .

Proof. The ¬rst statement is proved in Section 23.1 of [FH91]. The second statement is a

special case of Lemma 3.4.1.

To prove the third statement, notice that SL2 is reductive and therefore W is completely

reducible over SL2 . Since the representation is faithful, we can rule out W X2 as well as

X0 • X0 • X0 . Thus, we must have W X0 • X1 ; this yields the ¬rst part of the

W

statement. To prove the second part, let w ∈ W \ (W1 ∪ W2 ); we will show that the smallest

SL2 -invariant subspace of W containing w is W itself. Write w = w1 + w2 ∈ W, w1 ∈

W1 , w2 ∈ W2 , and assume w1 , w2 = 0. Then, it is easy to show that there exist φ1 , φ2 ∈ SL2

such that the set

{w, φ1 (w) = w1 + φ1 (w2 ), φ2 (w) = w1 + φ2 (w2 )}

is linearly independent. This proves the desired result.

We prove the fourth statement as follows: In this situation we may identify SL3 with

SL(W ), where W C 3 is a vector space. Suppose ¦ : G0 — A ’ SL(W ) is an embedding,

where G0 (resp., A) is one of the groups listed in the statement. Then ¦({1} — A) is the

group closure of some operator ψ ∈ SL(W ). Also, G0 G0 — {id} acts faithfully on W via

¦.

85

Suppose G0 = SL2 , so that W = W1 +W2 as in Item 3 of the lemma. Since ¦(SL2 —{id})

commutes with ψ, we see that ψ(W2 ) is a SL2 -invariant subspace of W, which by Item 3

must then be W2 itself. Moreover, ψ|W2 is a SL2 -module endomorphism. Item 2 of the

lemma then implies that ψ|W2 = ± · id |W2 for some ± ∈ C — . It quickly follows that, with

respect to some suitable basis B of W,

±®

c’2

0 0

c 0 :c=0 .

[¦(A)]B = 0

°

»

0 0c

Setting c = ’1 in the above formula, we see that the matrix diag(1, ’1, ’1) is contained

in ¦(A) © ¦(SL2 ), contradicting the hypothesis that the product of G0 and A is a direct

product.

Now suppose G0 = PSL2 . By Item 1 of the lemma, we see that W is isomorphic to the

C 3 , viewed as a PSL2 -module. In particular, W is irreducible over PSL2 .

vector space X2

We also see that ψ is a PSL2 -module endomorphism of W. As in the previous case, Item 2

implies that ψ is a scalar multiple of the identity, so that ¦(A) = D3 is the group of scalar

matrices. Here, we see that the matrix diag(ζ3 , ζ3 , ζ3 ), where ζ3 is a primitive cube root of

unity, is contained in ¦(A) © ¦(PSL2 ), contradicting hypothesis as in previous case.