16 P. BIELIAVSKY, L. CLAESSENS, D. STERNHEIMER, ANDY. VOGLAIRE

splits

a

into two parts:

a= a+

EB

a-

with J1

E

a+

= an

f) and J2

E

a-n

q.

Let

{31

,

{32

E a*

be the dual basis; we have W

E 9,6u

V

E 9,62

,

L

E 9,61 -,62

,

M

E 9,61 +!32

,

and, in terms of positive roots, the space n is given by

W

E 9a+i3

V

E 913,

L

E 9a,

ME

9a+2!3·

We are now able to compute the vectors X E

t

such that aB(X) =X.

Let us take X

E

t

=

a

EB n and apply aB:

X

=

XJ1

+

Xh

+

Xa

+

X13

+

Xa+/3

+

Xa+2i3,

aBX

=

-Xh

+

Xh

+

Z-(a+2i3)

+

Z13

+

Z-(a+/3)

+

Z_a

(4.2)

where

X'P

and

Z'P

denote elements of 9p· It is directly apparent that

X

=

J2

belongs to

te.

The only other component common to X and aBX is in

913,

but

it is a priori not clear that

X13

=

Z13.

The dimension of

U

is 4 and that of

R

is 6, hence Re is at least 2-dimensional; it is generated by

J

2

and

913

=

JRV, i.e.

te

=

Span { J2

,

V}. This proves that the orbit of uH is open.

The fact that R' acts freely on U

=

R/Re proves that U is locally of group

type and since, by definition, U is only one orbit of R, the space U is globally of

group type. From now on,

Mo

=

R/Re will be identified with U as homogeneous

space, so what we have to find is a group

R

which

• acts transitively on U, i.e. RuH

=

RuH,

• admits a symplectic structure.

It is immediate to see that the algebra

t'

fails to fulfil the symplectic condition.

The algebra t

=

Span{

A, B, C,

D} of a group which fulfils the first condition must

at least act transitively on a small neighborhood of uH and thus be of the form

(4.3a)

(4.3b)

(4.3c)

(4.3d)

A

=

J1

+

aJ2

+

a'V

B

=

W

+

bJ2

+

b'V

C

=

M

+

cJ2

+

c'V

D

=

L

+

dJ2

+

d'V.

The problem is now to fix the parameters

a, a',

b, b',

c, c',

d, d'

in such a way that

Span{A, B, C, D} is a Lie algebra (i.e. it is closed under the Lie bracket) which

admits a symplectic structure and whose group acts transitively on U. We will

begin by proving that the surjectivity condition imposes

b

=

c

=

d

=

0. Then the

remaining conditions fort to be an algebra are easy to solve by hand.

First, remark that A acts on the algebra Span{B, C, D} because

J

1

does not

appears in

[t, t].

We can write t

=

JRA EBad Span{B, C, D} and therefore a general

element of the group

R

reads

f(a,

{3, "(, 8)

=

eAei3B+yC+OD

because a subalgebra

of a solvable exponential Lie algebra is solvable exponential. Our strategy will be to

split this expression in order to get a product

SR'

(which is equivalent to a product

R'S).

As Lie algebras, Span{B,C,D}

C

JRJ2

EBad

{W,M,L,

V}. Hence there exist

functions

w,

m, l, v and x of (a, {3, 'Y, 8) such that

(4.4)

We are now going to determine l(a,{3,"(,8) and study the conditions needed in

order for l to be surjective on JR. Since

J

2

does not appear in any commutator,

the Campbell-Baker-Hausdorff formula yields

x

=

{3b

+

"(C

+

8d. From the fact that

[J2

,

L]

=

-L, we see that the coefficient of Lin the left hand side of (4.4) is -l(l-

e-x)jx. The V-component in the exponential can also get out without changing

the coefficient of

L.

We are left with

f(a,

{3, "(, 8)

=

eAexheyV ew'W+m' M+lL

where