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˜
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.

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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
=

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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

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