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

58

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.

59

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.

60

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

61

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.

62

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