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??n Pn ), A2 = ?.
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

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