ñòð. 52 |

If ?1 = ?4 , then also ?2 = ?3 . It is easy to verify that

F (?)?C(Q1 + Qn+3 ) = (?1 cos ? + ?2 sin ?)(Q1 + Qn+3 ) +

+ (?2 cos ? ? ?1 sin ?)(Q2 + Qn+2 ).

If ?1 = 0 then we put ? = (?/2), when ?2 > 0 and ? = ?(?/2), when ?2 < 0. If

?1 = 0 then we let ?2 cos ? ? ?1 sin ? = 0. In that case,

?2

?1 cos ? + ?2 sin ? = ?1 cos ?(1 + tan2 ?).

tan ? = ,

?1

We choose the value of ? so that ?1 cos ? > 0. With this choice of ? we have

F (?)?C(Q1 + Qn+3 ) = ?(Q1 + Qn+3 ),

??

where ? > 0. But then, as a result of Lemma 6 and Lemma 7, F (?)?C = ?, ? ?

P (1, n), and so C = A1 F (??)A2 , where A1 = ??1 , A2 = ?. The result is proved.

? ?

?

Lemma 10. Let L1 and L2 be subalgebras of AP (1, n) which are not conjugate under

? ?

AP (1, n) to subalgebras of AO(1, n) = AO(1, n) ? D . Then L1 , L2 are conjugate

?

under Ad AC(1, n) if and only if they are conjugate under Ad AP (1, n) or if one of

the following conditions holds:

?

(1) n is an odd number and there exists an automorphism ? ? Ad AP (1, n) with

?1

?(L1 ) = C2 L2 C2 (see Eq. (3) for notation);

?

(2) there exist automorphisms ?1 , ?2 ? AP (1, n) with

?1 (L1 ) = F (?)[?2 (L2 )]F (??).

Proof. Let CL1 C ?1 = L2 for some C ? O1 (2, n + 1). By Lemma 9, we may assume

?

that ±C ? P (1, n) or that C is one of the matrices ±A1 QA2 , A1 F (?)A2 (we use the

?

notation of Lemma 9) . If C ? P (1, n) then, by Lemma 8, C belongs to the identity

?

component of the group P (1, n) and thus ?C is an inner automorphism of the algebra

? ?

AP (1, n). Now suppose ?C ? P (1, n). Then by Lemma 7, C = ?diag [1, B, 1], where

?

B ? O(1, n) and ? ? P1 (1, n). Thus we may assume that C = ?diag [1, B, 1]. From

this it follows that B ? O1 (1, n) for odd n and we have

diag [1, 1, ?1, 1, . . . , 1, 1]B ? O1 (1, n)

232 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

For even n this means that the algebras L1 , L2 are conjugate to each other under

? ?

Ad AP (1, n) or that there exists an automorphism ? ? Ad AP (1, n) such that ?(L1 ) =

?1

C2 L2 C2 .

?

Let C = ±A1 QA2 . Then C = ?1 ??2 with ?1 , ?2 ? P (1, n) and ? = ±diag [1, ?1 , 1,

. . . , 1, ?2 , ?1] with ?1 , ?2 ? {?1, 1}. Clearly, ? ? O1 (2, n + 1). When C = A1 Q2 we

have ?1 = 1, ?2 = ?1 and when C = ?A1 QA2 , ?1 = 1, ?2 = (?1)n . Since

?Pn ??1 = ±Kn , ?P? ??1 = ±K?

with ? < n, then from ??1 L2 ?1 = ?(?2 L1 ??1 )??1 it follows that the algebra

1 2

??1 L2 ?1 has a nonzero projection onto K0 , K1 , . . . , Kn , which is impossible. Thus

1

the matrix C is di?erent from ±A1 QA2 .

Now let C = A1 F (?)A2 . If ? is one of the matrices (4), then ?F (?)??1 = F (±?),

so that

C = A1 F (?)A2 ?,

?

where A1 , A2 ? P (1, n) and ? = E or ? is one of the matrices (4). Since ? can be

represented as a product of matrices in O1 (2, n), then the last case is impossible, and

we have proved the Lemma.

?

Theorem 1. Let L1 and L2 be subalgebras of AP (1, n) which are not conjugate under

? ?

AP (1, n) to subalgebras of AO(1, n) and such that their projections onto AO(1, n)

have no invariant isotropic subspace in R1,n . The subalgebras L1 and L2 are conjugate

?

under AdAC(1, n) if and only if they are conjugate under Ad AP (1, n) or when there

?1

?

exists an automorphism ? ? Ad AP (1, n) such that ?(L1 ) = C2 L2 C2 , where C2 =

diag [1, 1, ?1, 1, . . . , 1].

Proof. By Lemma 10 we may assume that ?1 (L1 ) = F (?)[?2 (L2 )]F (??) for some

?

?1 , ?2 ? AP (1, n). Under the given assumptions, the projection of ?2 (L2 ) onto

AO(1, n) contains an element of the form

n?1 n?1

X= (?b J0b + ?b Jbn ) + ?bc Jbc ,

b=1 b,c=1

where ?q = ??q for some q (1 ? q ? n ? 1). Since

1

F (?)J0q F (??) = J0q cos ? + (Kq + Pq ) sin ?

2

and

1

F (?)Jqn F (??) = Jnq cos ? + (Kq ? Pq ) sin ?

2

we have that F (?)XF (??) contains the term

F (?)[?q J0q + ?q Jqn ]F (??) = (?q J0q + ?q Jqn ) cos ? +

1

+ [?q (Kq + Pq ) + ?q (Kq ? Pq )] sin ?

2

and from this it follows that (?q + ?q ) sin ? = 0 so that sin ? = 0. But then ? = m?.

When m = 2d we have F (?) = En+3 . When m = 2d + 1 then F (?) = diag [?1, ?1,

En?1 , ?1, ?1]. However,

F (?)[?2 (L2 )]F (??) = (?F (?))[?2 (L2 )](?F (??))

On the classi?cation of subalgebras of the conformal algebra 233

from which it follows that we may assume that ?1 (L1 ) = C[?2 (L2 )]C ?1 where C =

?

diag [1, 1, ?En?1 , 1, 1]. If n is odd, then ?C is an inner automorphism of AP (1, n).

?

If n is even, then ?C2 ?C is an inner automorphism of the algebra AP (1, n). In the

?rst case, ?3 (L1 ) = L2 where ?3 = ?2 ??1 ?1 is an inner automorphism of the

?1

C

? (1, n). In the second case, ?(L1 ) = ?C (L2 ) for some ? ? Ad AP (1, n).

?

algebra AP 2

The theorem is proved.

?

Theorem 2. Let L1 and L2 be subalgebras of AO(1, n) having no invariant isotropic

subspaces in R1,n . The subalgebras L1 , L2 are conjugate under Ad AC(1, n) if and

?

only if they are conjugate under Ad AO(1, n) or when there exists an automorphism

? ? Ad AO(1, n) such that ?(L1 ) = CL2 C ?1 where C is one of the (n + 3) ? (n + 3)

?

matrices

diag [1, 1, ?1, 1, . . . , 1], diag [1, . . . , 1, ?1], diag [1, . . . , 1, ?1, ?1].

?

We note that AO(1, n) ? AO(2, n + 1) and that the matrix C is (n + 3) ? (n + 3).

5 Subalgebras of the full Galilei algebra

Lemma 11. Let C ? O(2, n + 1) and W = Q1 + Qn+3 , Q2 + Qn+2 . If CW = W ,

then

C = exp[?(S + T )] diag [1, ?, K, ?, 1] exp(?R + ?Z) ?

n?1 n?1

(9)

? exp ?i Gi ?M + ?T + µi Pi ,

i=1 i=1

where ? = ±1, K ? O(n ? 1).

Proof. We have

C(Q1 + Qn+3 ) = ?1 (Q1 + Qn+3 ) + ?2 (Q2 + Qn+2 )

and so

F (??)C(Q1 + Qn+3 ) = (?1 cos ? ? ?2 sin ?)(Q1 + Qn+3 ) +

+ (?2 cos ? + ?1 sin ?)(Q2 + Qn+2 ).

If ?1 = 0 then we put ? = (3?/2) when ?2 > 0 and ? = (?/2) when ?2 < 0. If ?1 = 0

then we put ?1 sin ? + ?2 cos ? = 0 and then tan ? = ??2 /?1 and ?1 cos ? ? ?2 sin ? =

?1 cos ?(1 + tan2 ?). We choose ? so that ?1 cos ? > 0. For this choice of ? we have

F (??)C(Q1 + Qn+3 ) = ?(Q1 + Qn+3 ), where ? > 0. Using Lemma 7, we obtain

n

?

F (??)C = A = diag [1, B, 1] exp([ln ?]D) exp ? ? P (1, n),

?i Pi

i=0

where B ? O(1, n). Then C = F (?)A. The matrix A has the form (8). Direct calcula-

tion gives

n+1

A(Q2 + Qn+2 ) = ?(Q1 + Qn+3 ) + ?Q2 + ?Qn+2 + ?i Qi .

i=3

234 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

From this it follows that

F (?)A(Q2 + Qn+2 ) = (? cos ? + ? sin ?)Q1 + (?? sin ? + ? cos ?)Q2 +

n+1

+ (? cos ? ? ? sin ?)Qn+2 + (? sin ? + ? cos ?)Qn+3 + ?i Qi .

i=3

Now we have F (?)A(Q2 + Qn+2 ) ? W , from which we have

?? sin ? + ? cos ? = ? cos ? ? ? sin ?

? cos ? + ? sin ? = ? sin ? + ? cos ?,

and so we conclude that ? = ? and ?j = 0, j = 3, . . . , n + 1. But in that case we have

diag [1, B, 1](Q2 + Qn+2 ) = ?(Q2 + Qn+2 ).

By Lemma 2, we have

n?1

±B = diag [1, K, 1] exp[(? ln |?|)J0n ] exp ?i Gi ,

i=1

where K ? O(n ? 1). We note that

1 1

K0 + P0 ? Kn ? Pn = 2(S + T ), (Z ? R), D = ? (Z + R),

J0n =

2 2

1 1

(M ? 2T ),

P0 = (M + 2T ), Pn = [D, Ga ] = 0, [D, J0n ] = 0.

2 2

The lemma is proved.

Lemma 12. Let C ? O1 (2, n + 1) and W = Q1 + Qn+3 , Q2 + Qn+2 . If CW = W

then the matrix C has the form (9) with ? = 1 and K ? SO(n ? 1).

Proof. From the conditions of Lemma 1 1 and the fact that we ask for C ? O1 (2, n+1),

it follows that diag [1, ?, K, ?, 1] ? O1 (2, n + 1). It follows now that ? > 0 and that

K 0

>0

0 ?

and thus we have ? = 1 and |K| > 0, whence K ? SO(n ? 1). This proves the lemma.

The matrices of the form (9) with ? = 1 and K ? SO(n ? 1) form a group under

multiplication, which we denote by G4 (n ? 1) since its Lie algebra is the full Galilei

algebra AG4 (n ? 1). It is easy to see that G4 (n ? 1) ? O1 (2, n + 1).

Lemma 13. If C ? O1 (2, n + 1) but C ? G4 (n ? 1), then C = A1 ?A2 , where

A1 , A2 ? G4 (n ? 1) and ? is one of the matrices

?1 = diag [1, . . . , 1, ?1], ?2 = diag [1, 1, ?1, 1, . . . , 1, ?1, 1]. (10)

Proof. Let

n+3

?1 + ?2 ? ?3 ? · · · ? ?n+3 = 0.

2 2 3 2

C(Q1 + Qn+3 ) = ?i Qi ,

i=1

There exists a matrix ? = diag [1, 1, ?, 1, 1] with ? ? SO(n ? 1) such that ?C(Q1 +

Qn+3 ) does not contain Q4 , . . . , Qn+1 . Hence we may assume ?1 +?2 ??n+2 ??n+3 =

2 2 2 2

0.

On the classi?cation of subalgebras of the conformal algebra 235

Since

1

(K0 + P0 + Kn ? Pn ) = ?12 + ?n+2,n+3 ,

S+T =

2

2 2

then, up to a factor exp[?(S + T )], we may suppose that ?1 = 0, ?2 = 0. If ?1 = ?n+3

then ?3 = 0, ?n+2 = 0. Assume ?1 = ?n+3 . As in the proof of Lemma 4, we ?nd that

exp(?1 P1 + ?2 P2 )(?1 Q1 + ?3 Q3 + ?n+2 Qn+2 + ?n+3 Qn+3 ) =

= ?1 Q1 + ?n+3 Qn+3 ,

where ?12 ? ?n+3 = 0. Thus there exists a matrix A1 ? G4 (n ? 1) such that

2

A?1 C(Q1 + Qn+3 ) = ?(Q1 ± Qn+3 ),

1

(11)

A?1 C(Q2 + Qn+2 ) = ?1 Q1 + ?2 Q2 + ?3 Q3 + ?4 Qn+2 + ?5 Qn+3 .

1

Since the pseudo-orthogonal transformations preserve the scalar product, it follows

that the right-hand sides in (11) are also orthogonal, which implies that ?(?1 ??5 ) = 0

so that ?5 = ±?1 . If ?2 = ?4 then multiplying the left- and right-hand sides in (11) by

exp(?G1 ) does not change the right-hand side of the ?rst equality, and allows us to

eliminate ?3 by transforming it into 0. If ?2 = ?4 , then one easily deduces that ?3 = 0.

Thus we may assume that ?3 = 0. But then we have ?4 = ±?2 because ?5 = ±?1 and

?1 + ?2 ? ?4 ? ?5 = 0.

2 2 2 2

Let W = Q1 + Qn+3 , Q2 + Qn+2 . The above reasoning implies that for some

matrix A1 ? G4 (n?1) we have ?A?1 CW = W where ? is one of the matrices (10). The

1

fact that ?A?1 C ? O1 (2, n + 1) implies, using Lemma 12, ?A?1 C = A2 ? G4 (n ? 1).

1 1

Thus C = A1 ?A2 and the lemma is proved.

Lemma 14. The subalgebras L1 and L2 of AG4 (n?1) are conjugate under AdAC(1, n)

if and only if they are conjugate under Ad AG(n ? 1) or if there exist automorphisms

?1 , ?2 in Ad AG4 (n ? 1) with ?1 (L1 ) = ?[?2 (L2 )]??1 , where ? is one of the matri-

ces (10).

Proof. The result follows immediately from Lemma 13.

In the following table we give the action on the full Galilei algebra AG4 (n ? 1) of

the automorphisms where

ñòð. 52 |