and T(d1 ,d2) are conjugate. Then there exist a basis E = {˜1 , ˜2 , ˜3 } and a subgroup G ⊆

eee

˜˜

GL(V ) such that T(d1 ,d2 ) = [G]E0 , T(d1 ,d2) = [G]E . Let T(d1 ,d2 ) (resp., T(d1 ,d2) ) have the

˜

˜˜ ˜˜

˜

φ φ ˜ ˜ ˜ ˜

parameterization y ’ diag y d1 , y d2 , y ’d1 ’d2 (resp., z ’ diag z d1 , z d2 , z ’d1 ’d2 ). Let

P = [id]E0 ,E ; de¬ne ω : GL3 ’ GL3 by ω(M ) = P M P ’1 , so that ω(T(d1 ,d2 ) ) = T(d1 ,d2) .

˜ ˜˜

Then the map „ = φ’1 —¦ ω —¦ φ is an automorphism of C — . As above, Lemma 4.2.1 implies

˜

˜ ˜

that either „ = id or „ = σ. We may replace φ with φ —¦ σ if necessary so that „ = id . This

˜

implies that φ = ω —¦ φ, i.e.,

˜ ˜ ˜ ˜

diag(y d1 , y d2 , y ’d1 ’d2 ) = P diag(y d1 , y d2 , y ’d1 ’d2 )P ’1 for all y ∈ C — .

Now, if V includes in¬nitely many one-dimensional G-invariant subspaces, then it is easy

to see that T(d1 ,d2 ) = T(d1 ,d2) = T(1,1) . Therefore, we may assume that V includes exactly

˜˜

three one-dimensional G-invariant subspaces. It follows that

˜1 = c1 ej1 , ˜2 = c2 ej2 , ˜3 = c3 ej3

e e e

where the cj are nonzero constants and the ordered 3-tuple (j1 , j2 , j3 ) is a permutation of

(1, 2, 3). This implies that the change-of-coordinates matrix P is a product of a diagonal

matrix by a permutation matrix. From this, we see that

˜ ˜ ˜ ˜ ¯ ¯ ¯

diag(y d1 , y d2 , y ’d1 ’d2 ) = diag(y d1 , y d2 , y d3 )

for all y ∈ C — , where the ordered 3-tuple d1 , d2 , d3 is a permutation of the ordered 3-tuple

¯¯¯

¯ ˜

(d1 , d2 , d3 ), d3 = ’d1 ’ d2 . This yields di = di for 1 ¤ i ¤ 3. Since d1 > d2 > 0 and

˜ ˜ ˜˜

d1 > d2 > 0, we see that (d1 , d2 ) = (d1 , d2 ), so that the parameterizations and in particular

their images are identical.

C — — C — . Then G ⊆ D3 is a two-dimensional connected subgroup

Finally, suppose G

of the two-dimensional torus SL3 © D3 and therefore must be all of SL3 © D3 .

Unipotent subgroups of SL3

4.3

As in the previous section, V = C 3 is a three-dimensional vector space with ¬xed basis

E0 = {e1 , e2 , e3 } , yielding a bijection between GL3 and GL(V ).

Lemma 4.3.1 Assume G ⊆ SL(V ) and B is a basis of V.

63

1. De¬ne U(0,0) = {(tij ) ∈ U3 : t12 = 0, t23 = 0} .

1

1

(a) U(0,0) has a faithful parameterization

®

1 0x

x’ 0 0 , x ∈ C,

1

° »

0 0 1

C.

1

so in particular U(0,0)

(b) If [G]B = U(0,0) , then the G-invariant subspaces of V are:

1

• ±b1 + βb2 , where (± : β) ∈ P1 (C); ¬xed elementwise

• b1 , b2 ; ¬xed elementwise

• b1 , γb2 + b3 , where γ ∈ C.

2. De¬ne U(1,1) = (tij ) ∈ U3 : t12 = t23 , t13 = 1 t2 .

1

2 12

1

(a) U(1,1) has a faithful parameterization

®

12

2x

1x

x’ 0 1 : x ∈ C,

x

° »

00 1

C.

1

so in particular U(1,1)

(b) If [G]B = U(1,1) , then the G-invariant subspaces of V are b1 and b1 , b2 .

1

3. De¬ne U(1,0) = {(tij ) ∈ U3 : t23 = 0} .

2

2

(a) U(1,0) has a faithful parameterization

®

1 xy

(x, y) ’ 0 0 , x, y ∈ C,

1

° »

0 0 1

C2.

2

so in particular U(1,0)

(b) If [G]B = U(1,0) , then the G-invariant subspaces are b1 and b1 , ±b2 + βb3 ,

2

where (± : β) ∈ P1 (C).

4. De¬ne U(0,1) = {(tij ) ∈ U3 : t12 = 0} .

2

64

2

(a) U(0,1) has a faithful parameterization

®

10x

(x, y) ’ 0 1 y , x, y ∈ C,

° »

001

C2.

2

so in particular U(0,1)

(b) If [G]B = U(0,1) , then the G-invariant subspaces are ±b1 + βb2 , where (± : β) ∈

2

P1 (C), and b1 , b2 , all ¬xed elementwise.

5. De¬ne U(1,1) = {(tij ) ∈ U3 : t12 = t23 } .

2

2

(a) U(1,1) has a faithful parameterization

®

12

2x

1x y+

(x, y) ’ 0 1 , x, y ∈ C,

x

° »

00 1

C2.

2

so in particular U(1,1)

(b) If [G]B = U(1,1) , then the G-invariant subspaces are b1 and b1 , b2 .

2

6. If [G]B = U3 , then the G-invariant subspaces are b1 and b1 , b2 .

The above listed subgroups are pairwise nonconjugate; in particular, they are distinct.

Proof. These statements are veri¬ed by straightforward calculations.

We shall see that every nontrivial unipotent subgroup of SL3 is conjugate to one of the

groups listed above. First we need two technical lemmas.

Lemma 4.3.2 Given a nontrivial matrix

®

1 a b

c ∈ U3 ,

M = 0 (4.1)

1

° »

0 0 1

then the group closure of M in U3 is

±

0

ct12 ’ at23

=

clos(M ) = (tij ) ∈ U3 : 1 at12 t23 + bt12 ’ at13 ’ 1 a2 t23 .

= 0

2 2

0

2 ct12 t23 ’ 2 c t12 ’ ct13 + bt23

1 12

=

65

Moreover, clos(M ) has a faithful parameterization

®

x(x’1)

ac

1 xa xb +

2

x’ 0 , x ∈ C,

1 xc

° »

0 0 1

so in particular it is isomorphic to C.

Proof. A sequence of straightforward computations shows that the given set is a group con-

taining M, that it has the given parameterization, and that it is isomorphic to C. The desired

result then follows from the fact that C includes no nontrivial proper closed subgroup.

Lemma 4.3.3 De¬ne q : U3 ’ C 2 by

«®

1 x y

¬ ·

¬ ·

q ¬ 0 z · = (x, z).

1

° »

0 0 1