=

g3 ;

= (4.12)

(D ’ r1 ’ 2r2 )(gy1 ξ) (gy1 ) ξ + gy1 ξ ’ (r1 + 2r2 )gy1 ξ

=

’1

(r1 + r2 )gy1 ξ + gy1 (r2 ξ + g ’1 y1 ) ’ (r1 + 2r2 )gy1 ξ

=

= 1. (4.13)

Next we show that the second condition implies the third condition. Suppose R, S ∈ D

˜

are operators of order at most 1 such that L2 R = S L2 . Then R maps VL isomor-

˜

2

phically onto VL in KL . From Items 3 and 4 of Lemma 4.6.4, we see that R maps

2

VD’r1 ’2r2 = spanC g 2 y1 isomorphically onto itself; since this is an irreducible GL -

module, it follows that R|VD’r1 ’2r2 = c0 idVD’r1 ’2r2 for some c0 ∈ C \ {0} . This in

turn implies that R = f (D ’ r1 ’ 2r2 ) + c0 ∈ D for some f ∈ C(x). We have that

R maps the basis element gy1 ξ of VL onto some element of VL \ VD’2r2 . Using the

˜

2 2

basis of VL given in Item 4 of Lemma 4.6.4, we see that R(gy1 ξ) = ±g 2 y1 + βg 2 y2 for

2

some ±, β ∈ C, β = 0.

95

We now compute:

f (D ’ r1 ’ 2r2 ) + c0 (gy1 ξ)

R(gy1 ξ) =

f (D ’ r1 ’ 2r2 )(gy1 ξ) + c0 gy1 ξ

=

= f + c0 gy1 ξ by (4.13),

so that

f + c0 gy1 ξ = ±g 2 y1 + βg 2 y2 , ±, β ∈ C, β = 0.

Applying D ’ r1 ’ 2r2 to each side of this equation yields

(D ’ r1 ’ 2r2 )(f ) + c0 ±(D ’ r1 ’ 2r2 )(g 2 y1 ) + β(D ’ r1 ’ 2r2 )(g 2 y2 )

=

βg 3 by (4.11), (4.12) and (4.13).

=

It follows that y = β ’1 f ∈ C(x), c = ’β ’1 c0 ∈ C \ {0} satisfy

(D ’ r1 ’ 2r2 )(y) = g 3 + c,

so that the third condition holds.

Next we show that the third condition implies the second condition. Suppose (D ’

r1 ’2r2 )(f ) = g 3 +c for some f ∈ C(x), c ∈ C \{0} . De¬ne R = f (D ’r1 ’2r2 )’c ∈ D.

The following computations show that R maps VL isomorphically onto VL , so that

˜

2 2

2

the second condition holds. Consider the action of R on the basis g y1 , gy1 ξ of VL

˜

2

given in Item 4 of Lemma 4.6.4. First we compute

f (D ’ r1 ’ 2r2 ) ’ c (g 2 y1 )

R(g 2 y1 ) =

= ’cg 2 y1 by (4.11). (4.14)

Next, we claim that R(gy1 ξ) is a solution of the equation (D ’ r1 ’ 2r2 )(y) = g 3 .

Indeed, we have

(D ’ r1 ’ 2r2 )(R(gy1 ξ)) (D ’ r1 ’ 2r2 ) —¦ R (gy1 ξ)

=

(D ’ r1 ’ 2r2 ) —¦ f (D ’ r1 ’ 2r2 ) ’ c (gy1 ξ)

=

(D ’ r1 ’ 2r2 ) —¦ f (D ’ r1 ’ 2r2 ) (gy1 ξ) ’

=

c(D ’ r1 ’ 2r2 )(gy1 ξ)

(D ’ r1 ’ 2r2 )(f ) ’ c by (4.13)

=

(g 3 + c) ’ c by hypothesis

=

g3 .

=

96

This result, together with (4.12), implies that g 2 y2 and R(gy1 ξ) di¬er by a member

of VD’r1 ’2r2 . By (4.11), we have that VD’r1 ’2r2 = spanC g 2 y1 . Thus, we have

R(gy1 ξ) = g 2 y2 + cg 2 y1 for some c ∈ C.

˜ ˜ (4.15)

By Item 3 of Lemma 4.6.4, we have that g 2 y1 , g 2 y2 is a basis of VL . In light of this

2

2

fact, we conclude from formulas (4.14) and (4.15) that R maps g y1 , gy1 ξ onto a

basis of VL ; the desired result follows.

2

To prove that the third condition implies the fourth condition, simply di¬erentiate

both sides of the equation given in the third condition and divide by 3.

Finally we show that the fourth condition implies the third condition. Suppose

¯

(D —¦ (D ’ r1 ’ 2r2 ))(f ) = r2 g 3

¯ ¯

for some f ∈ C(x). This condition can be rewritten (after de¬ning f = 3f ) as

(D ’ r1 ’ 2r2 )(f ) = [g 3 ] ,

or (D ’ r1 ’ 2r2 )(f ) ’ g 3 = 0, which in turn implies that (D ’ r1 ’ 2r2 )(f ) = g 3 + c

for some c ∈ C. We must show that c = 0. Suppose instead that c = 0. This yields

(D ’ r1 ’ 2r2 )(f ) = g 3 . One then checks that f ∈ VL . Recall that g ∈ C(x) and

2

(by Item 2 of the lemma) y1 , y2 ∈ C(x). It follows from Item 3 of Lemma 4.6.4 that

/

VL © C(x) = {0} , so that f = 0. This yields

2

g 3 = (D ’ r1 ’ 2r2 )(f ) = 0

and therefore g = 0, contradicting hypothesis on g. This completes the proof.

4. First of all, we compute a matrix representation of the action of GL on VL with

ˆ

respect to a certain basis; this will be useful in proving that the ¬rst two conditions

are equivalent. Let A = {·1 , ·2 , ·3 } be as in Item 1 of the conclusion of the lemma.

1 ’1 2

Let ·3 = ·1 ·2 . One checks that (D ’ r1 )(·1 ) = 0 and (D ’ r1 )(·2 ) = ±0 g for

ˆ

2

some ±0 ∈ C \ {0} ; using these facts, a straightforward set of computations shows that

ˆ

A = {·1 , ·2 , ·3 } is an (ordered) basis of VL . Let GL = Ru T be a Levi decomposition

ˆ ˆ

as above. Let σ ∈ T and let tσ ∈ C — be as in (4.9). Thus σ(·1 ) = tσ ·1 and σ(·2 ) =

·2 ; it follows that σ(ˆ3 ) = t’1 ·3 , so that [σ]A = diag(tσ , 1, t’1 ). We conclude that

· σˆ ˆ σ

[T ]A = T(1,0) . Next let „ ∈ Ru and let a„ , b„ , c„ ∈ C be as in (4.10). Thus „ (·1 ) = ·1

ˆ

97

and „ (·2 ) = ·2 + a„ ·1 . We compute

1 ’1

·1 (·2 + a„ ·1 )2

„ (·3 ) =

2

1 ’1 2

· (·2 + 2a„ ·1 ·2 + a2 ·1 )

2

= „

21

1

= ·3 + a„ ·2 + a2 ·1 .

ˆ

2„

®

We now see that

12

2 a„

1 a„

[„ ]A = 0 1

a„

° »

00 1

1 1

and therefore that [Ru ]A = U(1,1) . We conclude that [GL ]A = U(1,1) T(1,0) .

ˆ ˆ

Now, suppose the ¬rst condition holds. Then, it is clear that the map from VL to VL

ˆ

de¬ned on basis elements by

·i ’ ·i for i = 1, 2, ·3 ’ ·3 ,

ˆ

gives an isomorphism of GL -modules, so that the second condition holds.

Suppose now that the ¬rst condition fails. Then, Item 1 of the conclusion of the

lemma implies that the unipotent radical of GL has dimension 2 or 3. By contrast, we

1 1

have [GL ]A = U(1,1) T(1,0) , so that GL U(1,1) T(1,0) has a one-dimensional unipotent

ˆ ˆ

radical; we now see that the second condition fails. By contrapositive, we conclude

that the second condition implies the ¬rst condition.

Next we show that the second condition implies the third condition. Let ¦, Ψ ∈ D

ˆ

be operators of order at most 2 such that L¦ = ΨL. We have that VL2 is the unique

2-dimensional GL -invariant subspace of VL (resp., of VL ) by hypothesis (resp., by

ˆ

examining [GL ]A ). It follows that ¦|VL2 is a GL -invariant automorphism of VL2 . Thus,

ˆ

we have that [¦]A2 commutes with [GL ]A2 = U2 T2 , where T2 = {diag(t, 1) : t ∈ C — } .

A straightforward set of computations shows that

±®

±β

° » : ± ∈ C ,β ∈ C ,

—

Cen(U2 ) =

0±

GL2

= D2 .

Cen(T2 )

GL2

Since [¦]A2 is contained in the intersection of these two subgroups of GL2 , we conclude

that ¦|VL2 = ± idVL2 for some ± ∈ C \ {0} . This implies that ¦ = hL2 + ± for some

98

h ∈ C(x). Next we consider the basis {y1 , y2 , y3 } of VL and the basis {y1 , y2 , y3 } of VL

ˆ ˆ

as de¬ned in Lemma 4.6.4. Since ¦ is an isomorphism that restricts to ± id on VL2 ,

we see that

¦(y3 ) = β y3 + y0 for some β ∈ C — , y0 ∈ VL2 .

ˆ

Before proceeding further, remark that

’1

L2 (y3 ) = g ’1 y1 (4.16)

and

’1

L2 (ˆ3 ) = g 2 y1 ;

y (4.17)

these equalities can be veri¬ed by direct computation. Putting facts together, we

obtain