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C2 :
iii. Subgroups satisfying Ru
(C — — C — ) with
A. U(1,0) · (D3 © SL3 ). In this case, we have G C2
2


’1
(Int(y1 , y2 ))(w, x) = (y1 y2 w, y1 y2 x) for y1 , y2 ∈ C — , w, x ∈ C;
2



we also have n1 = 1, n2 = 2.
(C — — C — ) with
C2
B. In this case, we have G

(Int(y1 , y2 ))(w, x) = (y1 y2 w, y1 y2 x) for y1 , y2 ∈ C — , w, x ∈ C;
2 2



we also have n1 = 2, n2 = 1.
(C — — C — )
U3 : T3 © SL3 . Here, we have G U3
iv. Subgroups satisfying Ru
«®  ® 
with
’1 2
1 v w y1 y2 v y1 y2 w
1
¬ ·  
¬ ·  
(Int(y1 , y2 )) ¬ 0 x · =  0 
2
1 1 y 1 y2 x
° » ° »
0 0 1 0 0 1

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for ® 
1 v w
 
 
y1 , y 2 ∈ C — ,  0 x  ∈ U3 ;
1
° »
0 0 1
we also have n1 = n2 = 1.

SL2 .
2. Subgroups satisfying H

(a) Subgroups satisfying T 1.

i. Subgroups satisfying Ru 0:

(tij ) ∈ SL3 : t13 = t23 = t31 = t32 = 0, t33 = 1.

SL2 and n1 = n2 = 1.
We have G
C2 :
ii. Subgroups satisfying Ru
A. {(tij ) ∈ SL3 : t31 = t32 = 0, t33 = 1} . In this case, we have G C2 SL2
with conjugation given by the unique irreducible representation of SL2 on
C 2 ; we also have n1 = 0, n2 = 1.
B. {(tij ) ∈ SL3 : t21 = t31 = 0, t11 = 1} . In this case, we have G C2 SL2
with conjugation given by the unique irreducible representation of SL2 on
C 2 ; we also have n1 = 1, n2 = 0.

C—.
(b) Subgroups satisfying T

i. Subgroups satisfying Ru 0:

{(tij ) ∈ SL3 : t13 = t23 = t31 = t32 = 0} .

GL2 and n1 = n2 = 1.
We have G
C2 :
ii. Subgroups satisfying Ru
A. {(tij ) ∈ SL3 : t31 = t32 = 0} . We have G C2 GL2 with conjugation
given by
M.v = (det M )M v for M ∈ GL2 , v ∈ C 2 ;

we also have n1 = 0, n2 = 1.
B. {(tij ) ∈ SL3 : t21 = t31 = 0} . We have G C2 GL2 with conjugation
given by

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

we also have n1 = 1, n2 = 0.

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PSL2 :
3. Subgroups satisfying H


{(tij ) ∈ SL3 : t2 = t11 t13 , t2 = 4t11 t31 , t2 = 4t13 t33 ,
12 21 23

t2 = t31 t33 , (t22 + 1)2 = 4t11 t33 , (t22 ’ 1)2 = 4t13 t31 }.
32



PSL2 and n1 = n2 = 0.
We have G


SL3 : In this case we have G = SL3 and n1 = n2 = 0.
4. Subgroups satisfying H




n2 0 1 2 3

PSL2 , C2 SL2 ,
n1 0
SL3 C2 GL2
C—
C
C—
C2
C— C—
C2 SL2 , U3 C2
C—
C2
1
C —2
C2 GL2 T3 © SL3 C2
SL2
GL2
C— C—
C2 C
C—
C
2
C —2 C —2
C2 C
C—
3
C —2
C—
— —
∞ C C C C
2
{1}


Table 4.1: Admissible subgroups of SL3 . “C —2 ” stands for C — — C — .



Below, in Section 4.2 (resp., Section 4.3; Section 4.4), we enumerate the ways in which
tori (resp., unipotent groups; semisimple groups) can be embedded in SL3 , up to conjugation.
Then, in Section 4.5, we prove Theorem 4.1.5, primarily by considering the ways in which an
admissible subgroup can be built from a torus, a semisimple group and a unipotent group.
In Section 4.6, we give an algorithm to compute the group of D3 + aD + b, a, b ∈ C[x]. The
main step of this algorithm relies on Theorem 4.1.5.

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Tori embedded in SL3
4.2
Lemma 4.2.1 The only (algebraic group) endomorphisms of C — are the maps x ’ xd , d ∈
Z. The only automorphisms of C — are the identity and the map x ’ x’1 .

Proof. The ¬rst statement is an easy exercise in the theory of rational functions on P1 . The
second statement follows easily from the ¬rst statement.




Throughout the remainder of this section, V = C 3 is a three-dimensional vector space
with ¬xed basis E0 = {e1 , e2 , e3 } ; this yields a bijection between GL3 and GL(V ).

Lemma 4.2.2 Assume G ⊆ SL(V ) and F is a basis of V. Let ni be the number of i-
dimensional G-invariant subspaces of V for i = 1, 2.

(tij ) ∈ D3 © SL3 : td2 = td1 , d1 , d2 ∈ Z, d1 ≥ d2 ≥ 0, d1 > 0,
1. De¬ne T(d1 ,d2 ) = 11 22

GCD(d1 , d2 ) = 1, and assume [G]F = T(d1 ,d2 ) . Then, the following statements hold:

(a) T(d1 ,d2 ) has a faithful parameterization

y ’ diag(y d1 , y d2 , y ’d1 ’d2 ), y ∈ C — ,

C—.
so in particular T(d1 ,d2 )

(b) If d1 > d2 , then the only G-invariant subspaces of V are f1 , f2 , f3 , and
fi , fj , 1 ¤ i < j ¤ 3. In particular, we have n1 = n2 = 3.

(c) If d1 = d2 = 1, then the G-invariant subspaces are ±f1 + βf2 , where (± : β) ∈
P1 ; f3 ; f1 , f2 ; and ±f1 + βf2 , f3 , where (± : β) ∈ P1 . In particular, we have
n1 = n2 = ∞.

2. The group D3 © SL3 has a faithful parameterization

(y, z) ’ diag(y, z, (yz)’1 ), y, z ∈ C — ,

so in particular it is isomorphic to C — — C — . If [G]F = D3 © SL3 , then the G-invariant
subspaces are f1 , f2 , f3 , and fi , fj , 1 ¤ i < j ¤ 3. In particular, we have n1 =
n2 = 3.

Proof. A set of straightforward calculations veri¬es that each of the given parameterizations
is correct for the corresponding algebraic group. The statements about invariant subspaces

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follow from further calculations. We remark that in the case of T(d1 ,d2 ) , the three integers
d1 , d2 and ’d1 ’ d2 are distinct if d1 = d2 .




Lemma 4.2.3 If G = [G]E0 ⊆ SL3 is isomorphic to a torus, then there exist a basis F =
ˆ ˆ ˆ
{f1 , f2 , f3 } of V and a unique subgroup G ⊆ SL3 such that G = [G]F and G is one of the
following groups:

1. 1 = {diag(1, 1, 1)} .

C — , d1 , d2 ∈ Z, d1 ≥ d2 ≥ 0, d1 > 0, GCD(d1 , d2 ) = 1.
2. T(d1 ,d2 )

C— — C—.
3. D3 © SL3

Proof. It is known ([HK71]) that a commuting group of diagonalizable operators can be
simultaneously diagonalized. We may therefore assume that G ⊆ D3 . Since SL3 © D3
C — — C — , it follows that G is isomorphic to a torus of dimension at most 2.
C — . Let φ : C — ’ G be an isomorphism. It follows from Lemma 4.2.1
Suppose G
C — — C — — C — that there exist integers d1 , d2 , d3 such that φ(y) =
and the fact that D3
diag(y d1 , y d2 , y d3 ) for all y ∈ C — . Note that φ(ζ) = diag(1, 1, 1) if ζ is a dth root of unity,
where d is a common divisor of d1 , d2 , d3 . It follows that GCD(d1 , d2 , d3 ) = 1. Since G ⊆ SL3 ,
we also have d1 + d2 + d3 = 0. Let σ be the automorphism of C — given by σ(x) = x’1 . We
have
(φ —¦ σ)(x) = diag(x’d1 , x’d2 , x’d3 ), x ∈ C — .

After replacing φ with φ —¦ σ if necessary, we may assume that two of {d1 , d2 , d3 } are nonneg-
ative. After reordering indices if necessary, we may assume that d1 ≥ d2 ≥ 0 > d3 . If d2 = 0,
then G = T(1,0) . If d1 = d2 , then G = T(1,1) . Otherwise, G = T(d1 ,d2 ) with d1 > d2 > 0.
Next we show that the subgroups T(d1 ,d2 ) are distinct. Let T(d1 ,d2 ) , T(d1 ,d2) be two such
ˆˆ

subgroups. De¬ne φ, φ : C — ’ SL3 © D3 by
ˆ

ˆ ˆ ˆˆ
φ(y) = diag(y d1 , y d2 , y ’d1 ’d2 ), φ(y) = diag(y d1 , y d2 , y ’d1 ’d2 ),
ˆ

and suppose that φ(C — ) = φ(C — ). It follows that φ’1 —¦ φ is an automorphism of C — . If this
ˆ ˆ

map is the identity, then φ = φ. Otherwise, by Lemma 4.2.1, φ’1 —¦ φ is the inverse map σ,
ˆ ˆ
ˆ ˆˆ ˆ ˆ
i.e., φ = φ —¦ σ. This implies (d1 , d2 , ’d1 ’ d2 ) = ’(d1 , d2 , ’d1 ’ d2 ) ∈ Z3 , a contradiction
ˆ
since d1 , d1 > 0. We conclude that any two of these parameterizations have distinct images
in D3 © SL3 , and the T(d1 ,d2 ) are distinct.

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It remains to be shown that no two of these subgroups are conjugate. Suppose T(d1 ,d2 )

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