<< Ïðåäûäóùàÿ ñòð. 51(èç 70 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>
?1 + ? (1 + b ) 1 + ? (1 ? b )
2 2
2 2
t
?b
2? 2?
then

exp(??J0n ) exp(??1 G1 ? · · · ? ?n?1 Gn?1 ) = diag [1, ? ?1 , 1]C

from which it follows directly that

C = diag [1, B, 1] exp[(? ln ?)J0n ] exp(??1 G1 ? · · · ? ?n?1 Gn?1 )

and the lemma is proved.
The set of F of matrices of the form (5) with ? > 0 is a group under multiplication.
The mapping

?B ?v
C>
0 1
?
is an isomorphism of the group F onto the extended Euclidean group E(n ? 1). Thus
we shall mean the group F when talking of the extended Euclidean group, and the
?
connected identity component E1 (n ? 1) will be identi?ed with the group of matrices
of the form (5) with ? > 0 and B sin SO(n ? 1). From Lemma 2 it follows that the
Lie algebra AF of the group F is generated by the basis elements Jab , Ga , J0n (a < b;
a, b = 1, . . . , n ? 1).
Lemma 3. If C ? O1 (1, n) and C(P0 + Pn ) = ?(P0 + Pn ) then ? > 0 and B ?
SO(n ? 1) in (5).
Proof. Since
1 + ?2 (1 + v 2 )
> 0,
2?
then we have ? > 0. From Lemma 2, diag [1, B, 1] ? O1 (1, n), so that det B > 0. Thus
B ? SO(n ? 1) and the lemma is proved.
?
Lemma 4. If C ? O(1, n) and ±C ? E(n ? 1) then C = ±A1 C A2 where A1 , A2 ?
?
E(n ? 1) and C = diag [1, . . . , 1, ?1].
228 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

Proof. We can choose a matrix ? ? O(n?1) so that ?C(P0 +Pn ) = ?P0 +?P1 +?Pn
where ?2 ? ? 2 ? ? 2 = 0. If ? = 0 then ? ? ? = 0. Let ? = ?/(? ? ?). Then,
???
(P0 ? Pn )
exp(?G1 )(?P0 + ?P1 + ?Pn ) =
2
?
and so there exists a matrix ? ? E(n ? 1) such that ?C(P0 + Pn ) = ?(P0 + Pn ) or
?
?C(P0 + Pn ) = ?(P0 ? Pn ). In the ?rst case, ±?C ? E(n ? 1), so that then we have
?
±C ? E(n ? 1), which is impossible. In the second case, C ?C(P0 + Pn ) = ?(P0 + Pn ).
For ? > 0 we ?nd C ?C ? E(n ? 1). Put C ?C = A2 , ? = A?1 . Then C = A1 C A2 .
?
1
If ? < 0 then we put ?C ?C = A2 , in which case C = ?A1 C A2 , and the lemma is
proved.
?
Lemma 5. If C ? O1 (1, n) and C ? E1 (n ? 1), then C = D1 QD2 , where D1 , D2 ?
?
E1 (n ? 1), and Q = diag [1, ?1, 1, . . . , 1, ?1].
?
Proof. If ±C ? E(n ? 1), then C(P0 + Pn ) = ?(P0 + Pn ). By Lemma 3, ? > 0 and
? ?
C ? E1 (n?1), which contradicts the assumption. Thus, ±C ? E(n?1). By Lemma 4,
?
C = ±A1 C A2 . From this it follows that C = D1 ?D2 , where D1 , D2 ? E1 (n ? 1), and
?1 ?1
F is one of the matrices ±C , ±Q. However, ? ? O1 (1, n), since ? = D1 CD2 , ?nd
from this it follows that ? = Q. The Lemma is proved.
Direct calculation shows that the normalizer of the space P0 + Pn in AO(1, n)
is generated by the matrices Ga , Jab , J0n (a, b = 1, . . . , n ? 1), which satisfy the
commutation relations

[Ga , Jbc ] = gab Gc ? gac Gb , [Ga , Gb ] = 0, [Ga , J0n ] = Ga .

This means that the normalizer of the space P0 + Pn in the algebra AO(1, n) is the
extended Euclidean algebra
?
AE(n ? 1) = G1 , . . . , Gn?1 (AO(n ? 1) ? J0n )

in an (n?1)-dimensional space, where the generators of translations are G1 , . . . , Gn?1
and the generator of dilatations is the matrix J0n .
Let K be a subalgebra of AP (1, n) such that its projection onto AO(1, n) has an
invariant isotropic subspace in Minkowski space R1,n . The subalgebra K is conjugate
under Ad AP (1, n) with a subalgebra of the algebra A = AG1 (n ? 1) J0n where
AG1 (n ? 1) is the usual Galilei algebra with basis M , T , Pa , Ga , Jab (a, b = 1, . . . , n ?
1), and M = P0 + Pn , T = 1 (P0 ? Pn ).
2
Proposition 3. Let L1 and L2 be subalgebras of A, with L1 not conjugate under
Ad A to any subalgebra having zero projection onto G1 , . . . , Gn?1 . If ?(L1 ) = L2
for some ? ? Ad AP (1, n), then there exists an inner automorphism ? of the algebra
A with ?(L1 ) = L2 .
Proof. Since Ad A contains automorphisms which correspond to matrices of the form
n
(6)
exp a? P?
?=

and since P (1, n) is a semidirect product of the group of matrices of the form (6)
and the group O(1, n) of matrices of the form diag [?, 1], then we may assume that
On the classi?cation of subalgebras of the conformal algebra 229

?
? = ?C with C ? O1 (1, n). If C ? E1 (n ? 1), then by Lemma 5, C = D1 QD2 . In that
case we ?nd that
?1 ?1
(D1 QD2 )?(L1 )(D2 QD1 ) = ?(L2 ),
? ?

whence
?1 ?1
(7)
Q(D2 ?(L1 )D2 )Q = D1 ?(L2 )D1 .
? ?

However,

when a = 1,
J0a + Jan ,
QGa Q = Q(J0a ? Jan )Q =
?(J01 + J1n ), when a = 1.

This means that QGa Q ? A. Because of this, the left-hand side of (7) does not belong
to A, whereas the right-hand side of (7) is a subalgebra of A. This then implies that
?
we must have C ? E1 (n ? 1) and thus we have ?(L1 ) = L2 for some ? ? Ad A.
?
Proposition 4. Let A be a Lie algebra with basis P0 , Pa , Pn , Jab , J0n (a, b =
?
1, . . . , n ? 1) and let L1 , L2 be subalgebras of A such that at least one of them has a
nonzero projection onto J0n . If ?(L1 ) = L2 for some ? ? Ad AP (1, n), then there
?
exists an inner automorphism ? ? A so that either ?(L1 ) = L2 or ?(L1 ) = ?Q (L2 )
where Q = diag [1, ?1, 1, . . . , 1, ?1].
Proof. As in the proof of Proposition 3, we may assume that ? = ?C where C ?
O1 (1, n). We shall also assume that the projection of L1 onto J0n is nonzero. If
? ?
C ? E1 (n ? 1) and C ? O1 (n ? 1) then the projection of the algebra ?(L1 ) onto
G1 , . . . , Gn?1 is nonzero, and hence the projection of L2 onto G1 , . . . , Gn?1 is
?
nonzero, which contradicts the assumptions of the proposition. Thus, if C ? E1 (n ? 1)
?
? ?
Let C ? E1 (n ? 1). By Lemma 5, C = D1 QD2 where D1 , D2 ? E1 (n ? 1). Then
?(L1 ) = L2 can be written as

?Q (?D2 (L1 )] = ?D?1 (L2 ).
1

?
If D2 ? O1 (n ? 1) then the projection of ?D2 (L1 ) onto G1 , . . . , Gn?1 is nonzero and
hence ?Q [?D2 (L1 )] does not belong to A. But then ?D?1 (L2 ) is also not in A. This is
1
?
a contradiction. Thus D1 , D2 ? O1 (n ? 1). From this it follows that ?Q (?(L1 )) = L2
?
where ? = ?D is an inner automorphism of the algebra A. This proves the proposition.
Proposition 5. Suppose 2 ? m ? n ? 1. Let F be a subalgebra of the algebra AO(m)
which is not conjugate under Ad AO(m) to a subalgebra of AO(m ? 1), and let L be
a subalgebra of P0 , P1 , . . . , Pn F such that ?(L) = F . Then L is conjugate to an
?
algebra W K, where W is a subalgebra of P1 , . . . , Pm and K is a subalgebra of F ?
P0 , Pm+1 , . . . , Pn . Two subalgebras W1 K1 and W2 K2 of this type are conjugate
under Ad AP (1, n) if and only if there exists an automorphism ? ? Ad AO(m) ?
Ad AO(1, n ? m) such that ?(W1 K1 ) = W2 K2 or ?(W1 K1 ) = Q(W2 K2 )Q
where

AO(1, n ? m) = J?? : ?, ? = 0, m + 1, . . . , n

and Q = diag [1, ?1, 1, . . . , 1, ?1].
230 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

4 Conjugacy of subalgebras of the extended Poincar?
e
algebra AP (1, n) under Ad AC(1, n)
?
Lemma 6. If C ? O(2, n + 1) and C(Q1 + Qn+3 ) = ?(Q1 + Qn+3 ) then ? = 0 and
? ?
1 + ?2 (1 ? v 2 ) ?1 + ?2 (1 + v 2 )
??v t E1,n B
? ?
2? 2?
? ?
C=? ?,
v ?v
B (8)
? ?
? ?1 + ?2 (1 ? v 2 ) 1 + ?2 (1 + v 2 ) ?
??v E1,n B
t
2? 2?
where B ? O(1, n), E1,n = diag [1, ?1, . . . , ?1], v is an (n + 1) ? 1 matrix and v 2
is its scalar square in R1,n . Conversely, every matrix C of the form (8) satis?es the
condition C(Q1 + Qn+3 ) = ?(Q1 + Qn+3 ).
Proof. Direct calculation.
Lemma 7. Let C ? O(2, n + 1) have the form (8), with ? > 0. Then

C = diag [1, B, 1] exp[(ln ?)D] exp(??0 P0 ? ?1 P1 ? · · · ? ?n Pn ),

where
? ?
?0
? ?
?1
? ?
? = B ?1 v.
? .
? ?
.
.
?n

Proof. The proof of Lemma 7 is similar to that of Lemma 2.
The mapping

?B ?v
f :C>
0 1

is a homomorphism of the group of matrices (8) onto the extended Poincar? group
e
? (1, n). The kernel of this homomorphism is the group of order two, {?En+3 , En+3 }.
P
Let us denote by H the set of matrices of the form (8) with ? > 0. Then f is an
?
isomorphism of H onto P (1, n). For this reason we shall, in the remainder of this
?
article, mean the group H when referring to P (1, n). Its Lie algebra is the extended
?
Poincar? algebra AP (1, n) given in Section 2.
e
Lemma 8. Let C ? O1 (2, n + 1) and let it be of the form (8) with ? > 0. Then
B ? B1 (1, n).
Remark 1. Note that when ? < 0 it is possible that B does not belong to O1 (2, n+1).
?
Lemma 9. If C ? O1 (2, n + 1) and ±C ? P (1, n) then either C = ±A1 QA2 or C =
?
A1 F (?)A2 , where A1 , A2 ? P (1, n), Q = diag [1, . . . , 1?1] and F (?) = exp[(?/2)(K0 +
P0 + Kn ? Pn )].
?
Proof. There exists a matrix ?P (1, n) such that

?C(Q1 + Qn+3 ) = ?1 Q1 + ?2 Q2 + ?3 Qn+2 + ?4 Qn+3 ,
On the classi?cation of subalgebras of the conformal algebra 231

where ?1 + ?2 ? ?3 ? ?4 = 0 and ?2 ?3 ? 0. If ?1 = ?4 then, as in the proof of
2 2 2 2

Lemma 4, we obtain that

exp(?0 P0 + ?n Pn )?C(Q1 + Qn+3 ) = ?(Q1 ± Qn+3 )

for some real numbers ?0 , ?n , ?. From this it follows that

? exp(?0 P0 + ?n Pn )?C(Q1 + Qn+3 ) = ?(Q1 + Qn+3 ),

where ? > 0 and ? = ±En+3 or ? = ±Q. By Lemma 6 and Lemma 7, we obtain
? ? ?
? ? P (1, n).
? exp(?0 P0 + ?n Pn )?C = ?,

Since ±C ? P (1, n), then ? = ±Q, and so C = ±A1 QA2 , where A1 = ??1 exp(??0 P0
?
?
 << Ïðåäûäóùàÿ ñòð. 51(èç 70 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>