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Then, the following statements hold:

1
1. q is an algebraic group homomorphism with kernel U(0,0) .


2. U(0,0) is identical to (U3 , U3 ), the commutator subgroup of U3 .
1




3. Two matrices M1 , M2 ∈ U3 commute if and only if there is a one-dimensional subspace
of C 2 containing both q(M1 ) and q(M2 ), i.e., q(M1 ) and q(M2 ) di¬er by a scalar
multiple.


4. U(0,0) is identical to the center of U3 .
1




5. Given a nontrivial matrix M of the form (4.1). Then the centralizer of M in U3 is


Cen(M ) = clos(M ) · U(0,0) .
1
U3


Moreover, this subgroup is isomorphic to C r , where r is equal to 1 if clos(M ) = U(0,0) , 2
1


otherwise.

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Proof. Item 1can be veri¬ed by direct computation. The reader can verify that if M1 =
® ® 
1 a 1 b1 1 a2 b2
   
   
 0 1 c1  and M2 =  0 1 c2  , then
° » ° »
001 001
® 
0 a1 c2 ’ c1 a2
1
 
 
’1 ’1
= 0 ;
M 1 M2 M 1 M 2 (4.2)
1 0
° »
0 0 1

Items 2 and 3 follow easily from this identity. Item 4 follows easily from Item 3. To prove
˜
˜
Item 5, let M ∈ U3 . Suppose M commutes with M. Then, by Item 3, we have
® 
˜
1 xa b
 
˜ =  0 1 xc 
M 
° »
00 1

x(x ’ 1)
ac + ˆ ˆ ∈ C. It is easy
for some x, ˜ ∈ C. Write ˜ = xb + b, b to check that
b b
2
® ® 
x(x’1)
10ˆ
1 xa xb + 2 ac b
  
  
˜
M = 0  0 1 0 .
1 xc
° »° »
0 0 1 001

Consider the right-hand side of this identity. By Lemma 4.3.2, the ¬rst matrix is a member
˜
of clos(M ); by Lemma 4.3.1, the second matrix is a member of U(0,0) . Thus, M ∈ clos(M ) ·
1

1
U(0,0) ; this yields one inclusion. To prove the other inclusion, notice that elements of the
group closure clos(M ) commute with M ; meanwhile, Item 4 of the lemma implies that
U(0,0) ⊆ CenU3 (M ). Moreover, by Lemma 4.3.2 (resp., Lemma 4.3.1), the subgroup clos(M )
1


(resp., the subgroup U(0,0) ) is isomorphic to C. The desired result now follows easily.
1




Lemma 4.3.4 Let G = [G]E0 ⊆ SL3 be a unipotent group. Let E = {e1 , e2 , e3 } be such that
G = [G]E ⊆ U3 . Let q : U3 ’ C 2 be the map de¬ned in Lemma 4.3.3. Then, exactly one of
the following cases holds:

1. If q(G ) = C 2 , then G = U3 .

2. Suppose q(G ) = (a, c) is a one-dimensional vector subgroup of C 2 for some (a : c) ∈
P1 , and U(0,0) ⊆ G .
1



67
2
(a) If a = 0, then G = U(0,1) .
2
(b) If c = 0, then G = U(1,0) .

(c) If a and c are nonzero, then there exists a basis B such that [G]B = U(1,1) .
2



3. Suppose q(G ) = (a, c) is a one-dimensional vector subgroup of C 2 for some (a : c) ∈
P1 , and U(0,0) is not included in G .
1



(a) If a = 0 or c = 0, then there exists a basis B such that [G]B = U(0,0) .
1


(b) If a and c are both nonzero, then there exists a basis B such that [G]B = U(1,1) .
1


1
4. Suppose q(G ) = 0. Then either G = U(0,0) or G = 1.

ˆ
In particular, if G is nontrivial, then there exists a basis B and a unique subgroup G such
ˆ ˆ
that [G]B = G and G is one of the subgroups listed in Lemma 4.3.1.

Proof.

1. Suppose q(G ) = C 2 . Let M1 , M2 ∈ G be such that q(M1 ) = (1, 0), q(M2 ) = (0, 1). It
’1 ’1
follows from (4.2) that M1 M2 M1 M2 generates U(0,0) and therefore that U(0,0) ⊆ G .
1 1


We now see that the closed subgroup G is three-dimensional, hence identical to U3 .

To prove Items 2 and 3 of the lemma, suppose q(G ) = (a, c) as stated, and let
® 
1ab
 
 
M =  0 1 c  = [φ]E ∈ G ,
° »
001

where b ∈ C and φ ∈ G. It follows from Item 3 of Lemma 4.3.3 that G is abelian; it then
follows from Item 5 of that lemma that G ⊆ clos(M )·U(0,0) . Since clos(M )·U(0,0) C2
1 1


and M ∈ clos(M ) ⊆ G , we see that G is either clos(M ) or clos(M ) · U(0,0) .
1



2. Suppose U(0,0) ⊆ G . Here, we see that G = clos(M ) · U(0,0) C2.
1 1



(a) If a = 0, then it™s clear that G ⊆ U(0,1) ; a dimension count yields equality.
2


(b) If c = 0, then it™s clear that G ⊆ U(1,0) ; a dimension count yields equality.
2


(c) Assume a and c are nonzero. Write
® 
1 0 1
 
 
0  = [ψ0 ]E , ψ0 ∈ G,
0 1
° »
0 0 1

68
so in particular φ, ψ0 generate G. We may assume without loss of generality that
a = 1. Let B = {b®, b2 , b3 } , where bi = ei for i = 1, 2 and b3 = c’1 e3 . Then, one

1
’1
1 1 bc
 
 
1  and [ψ0 ]B ∈ U(0,0) . This implies [G]B ⊆ U(1,1) ,
calculates [φ]B =  0 1 1 2
° »
00 1
and a dimension count yields equality.

C and that G
1
3. Suppose U(0,0) is not included in G . Here, we see that G = clos(M )
(resp., G) is generated by M (resp., φ).

(a) If a = 0, then let B = {b1 , b2 , b3 } , where b1 = be1 + ce2 , b2 = e1 , and b3 = e3
for i = 1, 3. Then, one calculates [φ]B ∈ U(0,0) , and it follows that [G]B = U(0,0) .
1 1


If c = 0, then let B = {b1 , b2 , b3 } , where b1 = e1 , b2 = a’1 be2 ’ e3 , b3 = e2 . In
this case also, one calculates [φ]B ∈ U(0,0) and obtains [G]B = U(0,0) .
1 1


(b) If both a and c are nonzero, then we may assume without loss of generality
where bi = ei for i = 1, 2 and b3 = c’1 e3 .
˜˜˜ ˜ ˜
˜
that a = 1. Let B = b1 , b2 , b3 ,
® 
’1
1 1 bc
 
 
1  . De¬ne B = {b1 , b2 , b3 } , where
Then one checks that [φ]B =  0 1
˜
° »
0 0 1

b2 = b2 + (bc’1 ’ )b1 and bi = bi
˜ ˜ for i = 1, 3. One calculates
2
® 
1
11
 
2
 
[φ]B =  0 1 1 .
° »
00 1

1
By Lemmas 4.3.1 and 4.3.2, we see that U(1,1) is the group closure of the above
1
matrix; it follows that [G]B = U(1,1) .

4. Item 4 of the lemma follows easily from Item 1 of Lemma 4.3.3.


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