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U(0,0) ; the desired result then follows easily. In case Ru
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.

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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:

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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

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(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
¦.

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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.




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