QUANTIZATION OF ANTI DE SITTER AND SYMMETRIC SPACES 17

w' and m' are complicated functions of

(/3, ')', J)

and

l

is given by

(4.5)

l(/3

J)

=

-J((3b

+

')'C

+

Jd)

'')', 1 _

e-f3b-"fc-lid '

which is not surjective except when b

=

c

=

d

=

0. Taking the inverse a general

element of

RuH

reads [

e-ww

-mM -lM

eJ

1 h

u

J,

where the range of

l

is not the whole

JR.

Since the action of

R'

is

simply

transitive,

R

is not surjective on

RuH.

When b

=

c

=

d

=

0, the conditions for (4.3) to be an algebra are easy to solve,

leaving only two

a priori

possible two-parameter families of algebras:

Algebra 1.

1

A=J1+2.h+sV

B=W

C=M

D=L+rV

[A,B] =

B+sC

[A,C]

=

~C

1

[A, D]

=

2sB

+

2. D

[B,D]

=

-rC.

with

r =1-

0. The general symplectic form on that algebra is given by

(4.7)

-(3

0

0

0

-')')

2~r

0 '

0

Since det w

= (

2 ~r)

2

we must have

(3

=1-

0,

r

=1-

0. That algebra will be denoted by

t

1

.

The analytic subgroup of R whose Lie algebra is

t

1

is denoted by R

1

.

Algebra 2.

A=

J1

+rJ2 +sV

B=W

C=M

D=L.

[A,B]

=

B+sC

[A,C]=(r+1)C

[A,D]

=

2sB

+

(1- r)D

There is no way to get a non-degenerate symplectic form on that algebra.

REMARK

4.4. One can eliminate the two parameters in algebra

t 1

by the iso-

morphism

(4.8)

(

1 0

0 1

¢

=

0

2sr

0 0

0 0 )

0 4s

1/r

4s2 /r

0 1

which fixes

s

=

0 and

r

=

1 and transforms

t 1

into the algebra defined by [A', B']

=

B'

[A' C']

=

'lC'

[A' D']

= l

D'

[B' D']

=

-C'

' ' 2 ' ' 2 ' ' .

It is now easy to prove that

PROPOSITION

4.5.

The group

R

1

of algebra

t 1

acts transitively on U, i.e.

RuH

=

R1uH.

w' and m' are complicated functions of

(/3, ')', J)

and

l

is given by

(4.5)

l(/3

J)

=

-J((3b

+

')'C

+

Jd)

'')', 1 _

e-f3b-"fc-lid '

which is not surjective except when b

=

c

=

d

=

0. Taking the inverse a general

element of

RuH

reads [

e-ww

-mM -lM

eJ

1 h

u

J,

where the range of

l

is not the whole

JR.

Since the action of

R'

is

simply

transitive,

R

is not surjective on

RuH.

When b

=

c

=

d

=

0, the conditions for (4.3) to be an algebra are easy to solve,

leaving only two

a priori

possible two-parameter families of algebras:

Algebra 1.

1

A=J1+2.h+sV

B=W

C=M

D=L+rV

[A,B] =

B+sC

[A,C]

=

~C

1

[A, D]

=

2sB

+

2. D

[B,D]

=

-rC.

with

r =1-

0. The general symplectic form on that algebra is given by

(4.7)

-(3

0

0

0

-')')

2~r

0 '

0

Since det w

= (

2 ~r)

2

we must have

(3

=1-

0,

r

=1-

0. That algebra will be denoted by

t

1

.

The analytic subgroup of R whose Lie algebra is

t

1

is denoted by R

1

.

Algebra 2.

A=

J1

+rJ2 +sV

B=W

C=M

D=L.

[A,B]

=

B+sC

[A,C]=(r+1)C

[A,D]

=

2sB

+

(1- r)D

There is no way to get a non-degenerate symplectic form on that algebra.

REMARK

4.4. One can eliminate the two parameters in algebra

t 1

by the iso-

morphism

(4.8)

(

1 0

0 1

¢

=

0

2sr

0 0

0 0 )

0 4s

1/r

4s2 /r

0 1

which fixes

s

=

0 and

r

=

1 and transforms

t 1

into the algebra defined by [A', B']

=

B'

[A' C']

=

'lC'

[A' D']

= l

D'

[B' D']

=

-C'

' ' 2 ' ' 2 ' ' .

It is now easy to prove that

PROPOSITION

4.5.

The group

R

1

of algebra

t 1

acts transitively on U, i.e.

RuH

=

R1uH.