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Lemma 4.3.5 Let G ⊆ SL3 be one of the two-dimensional unipotent subgroups listed in
Lemma 4.3.4.

2
1. If G = U(1,0) , then
Nor(G) = {(tij ) ∈ GL3 : t21 = t31 = 0}
GL3


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and
(tij ) ∈ Nor(G) : t11 (t22 t33 ’ t23 t32 ) = 1 .
Nor(G) =
SL3 GL3

2
2. If G = U(0,1) , then
Nor(G) = {(tij ) ∈ GL3 : t31 = t32 = 0}
GL3

and
(tij ) ∈ Nor(G) : (t11 t22 ’ t12 t21 )t33 = 1 .
Nor(G) =
SL3 GL3

3. If G = U(1,1) , then NorGL3 (G) = T3 , NorSL3 (G) = T3 © SL3 .
2



Proof. These statements are proved by straightforward calculations.




Semisimple subgroups of SL3
4.4
The author would like to thank Mohan Putcha for his help in proving the following lemma.

Lemma 4.4.1 The only semisimple algebraic groups that can be embedded in SL3 are SL2 ,
PSL2 and SL3 .

Proof. This proof relies on results from [FH91] and [Hum81]. The relevant results in
[FH91] are stated in [FH91] for the case in which C = C; these results are known to extend
to arbitrary algebraically closed ¬elds of characteristic zero. We will use basic terminology
from Lie theory without detailed explanation here, since it is only needed in this section.
Let G be a semisimple group and T a maximal torus of G. Then, the Weyl group WG is
de¬ned in [FH91], Sec. 14.1 to be the group generated by certain vector space involutions
on the dual of the Lie algebra of T. G also has a unique root system associated to it, and
this root system uniquely determines the Weyl group (ibid., Sec. 21.1). It is a fact (ibid,
Sec. 23.1) that WG is isomorphic to NG (T )/T. If B is a Borel subgroup of G and S is a full
set of coset representatives of T in NG (T ), then G has a Bruhat decomposition ([Hum81],
Sec. 28.3) given by G = BsB (disjoint union).
s∈S
The group SL3 has maximal torus T = D3 © SL3 . Its root system is referred to as (A2 )
([FH91], Sec. 21.1 and 23.1). The reader can verify that NSL3 (T ) is the group of matrices
in SL3 with exactly one nonzero entry in each row (resp., column); it is then easy to show
that
WSL3 NSL3 (T )/T S3 . (4.3)

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Now, let G0 ⊆ SL3 be semisimple, and let T0 be a maximal torus of G0 . Then T0 has
C — or T0 C — — C — . In the former case, it follows from
rank either 1 or 2, i.e., either T0
(ibid., Sec. 21.1 and 23.1) that G0 has root system (A1 ) and therefore is isomorphic to
either SL2 or PSL2 .
Suppose T0 has rank 2; we wish to show that G0 = SL3 . We may assume that T0 is
included in, hence equal to, D3 © SL3 . Also, by (ibid., Sec. 21.1), the root system of G0 is
one of (A1 — A1 ), (A2 ), (B2 ), (G2 ). By inspection of the diagrams given in (ibid.), the Weyl
groups associated to these root systems are dihedral groups of order 4, 6, 8, 12, respectively.
In our case, we have that T0 = T = D3 © SL3 is the maximal torus of both G0 and
SL3 , so that WG0 NG0 (T )/T is isomorphic to a subgroup of NSL3 (T )/T. Of the dihedral
groups listed above, the only one whose order divides 6 is the one whose order is exactly 6,
i.e., S3 . It now follows from (4.3) that G has root system (A2 ), that WG0 WSL3 , and that
NG0 (T ) = NSL3 (T ).
Since G0 has the same maximal torus and root system as SL3 , it follows from [Hum81],
Sec. 28.1, that G0 and SL3 have the same maximal unipotent subgroup U3 and thus the
same Borel subgroup B = T3 . Finally, we see that G0 and SL3 have identical Bruhat
decompositions, hence are identical.




Admissible subgroups of SL3
4.5
We begin with a technical lemma.

Lemma 4.5.1 Given:

• M is a nontrivial matrix of the form (4.1)

• v ∈ C — is a nonroot of unity

• d1 , d2 , d3 are integers such that d1 ≥ d2 ≥ 0, d1 > 0, GCD(d1 , d2 ) = 1, d1 +d2 +d3 = 0

• A1 , A2 , A3 are three distinct integers such that 1 ¤ Ai ¤ 3 for 1 ¤ i ¤ 3

• The matrix Q is given by

Q = diag v dA1 , v dA2 , v dA3 . (4.4)

1. Suppose QM Q’1 ∈ clos(M ). Then, the following statements hold:

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(a) If a = 0, b = 0, and c = 0, then d1 = d2 = 1 and A3 = 3.

(b) If a = 0, b = 0, and c = 0, then d1 = d2 = 1 and A1 = 3.

(c) If d1 > d2 and either a = 0 or c = 0, then two of {a, b, c} are zero.
1
(d) If a = 0 and c = 0, then d1 = 1, d2 = 0, A2 = 2, and b = ac.
2
2. Suppose QM Q’1 ∈ clos(M ) · U(0,0) and a = 0, c = 0. Then d1 = 1, d2 = 0, and
1


A2 = 2.

Proof. We compute
® 
dA1 ’dA2 dA1 ’dA3
a v b
1v
 
 
’1 dA2 ’dA3
= 0 c .
QM Q 1 v
° »
0 0 1

Let us ¬rst assume that this matrix is a member of clos(M ).
Suppose a = 0. Then, by Lemma 4.3.2, there exists x ∈ C such that v dA1 ’dA3 b = xb
and v dA2 ’dA3 c = xc. If both b and c are nonzero, then v dA1 ’dA3 = x = v dA2 ’dA3 . It quickly
follows that dA1 = dA2 , which (due to hypotheses on the dj ) implies d1 = d2 = 1. Also, A1
(resp., A2 ) is either 1 or 2, so that A3 = 3. This proves Item (1a) of the lemma. Item (1b)
is proved by similar logic.
Next, suppose a = 0 (resp., c = 0) and the other two parameters are nonzero. Then
Item (1a) (resp., Item (1b)) implies that d1 = d2 . This proves Item (1c) by contrapositive.
Now suppose a = 0 and c = 0. In this case, our computation of QM Q’1 , together with
Lemma 4.3.2, implies that

v dA1 ’dA2 = x = v dA2 ’dA3 for some x ∈ C. (4.5)

This yields dA1 ’ 2dA2 + dA3 = 0, which together with dA1 + dA2 + dA3 = 0 implies dA2 = 0.
The hypotheses on the dj now easily yield d1 = 1, d2 = 0. We also have dA2 = 0, dA1 =
, dA3 = ’ , where = ±1; this implies x = v . Lemma 4.3.2 also implies that the (1, 3)
coordinate of QM Q’1 is
x(x ’ 1)
v dA1 ’dA3 b = xb + ac.
2
We may rewrite this identity as
v2 ’ v
2
v b=v b+ ac.
2
Item (1d) of the lemma now follows after subtracting v b from both sides and dividing by
v2 ’ v .

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To prove Item Two, it follows from Lemma 4.3.3 that
±® 

 1 xa y 
 
 
  
 
clos(M ) · U(0,0) =  0 1 xc  : x, y ∈ C .
1
° 
»
 
 
00 
1

Therefore, (4.5) holds in this case, and the listed conclusions follow as in the previous case.




The following seven lemmas establish conjugacy classes for subgroups having certain Levi
decompositions. Lemma 4.5.2 (resp., Lemma 4.5.4; Lemma 4.5.6) enumerates a list of sub-
groups having trivial semisimple part and describes some of their properties. Lemma 4.5.3
(resp., Lemma 4.5.5; Lemma 4.5.7) shows that these subgroups form a set of conjugacy class
representatives. Lemma 4.5.8 addresses the case in which the reductive part is either SL2
or GL2 and the unipotent radical is C 2 . In each lemma, ni is the number of i-dimensional
G-invariant subspaces for i = 1, 2.

Lemma 4.5.2 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 = U(0,0) · Pσ T(d1 ,d2 ) Pσ , d1 > d2 ≥ 0, GCD(d1 , d2 ) = 1, σ ∈ S3 , σ ∈
1


{id, (1 2), (2 3)} if (d1 , d2 ) = (1, 0). Then G has a faithful parameterization
® 
dI1
0 x
y
 
 
(x, y) ’  0 0 ,
dI2
y
° »
dI3
0 0 y

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

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

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

Under this parameterization, we have (Int y)(x) = yx. The G-invariant subspaces are
a1 and a1 , a2 , so that n1 = n2 = 1.

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

® 
y0 x
 
 
(x, y) ’  0 y 0
° »
’2
00y

under which (Int y)(x) = y 3 x. The G-invariant subspaces are ±a1 + βa2 , where (± :
β) ∈ P1 ; a1 , a2 ; and a1 , a3 . Thus, we have n1 = ∞, n2 = 2.

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

under which (Int y)(x) = y ’3 x. The G-invariant subspaces are a1 , a2 , and

a1 , ±a2 + βa3 , (± : β) ∈ P1 .

Thus, we have n1 = 2, n2 = ∞.

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