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
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