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.

66

Proof. Item 1can 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

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.