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Xa ? Xb = 0 if Xa , Xb ? ?(n ? 1) and have no common summand.
Lemma 4.2. Let T = ?1 X1 + · · · + ?k Xk + Z, Z = ?J0n + ?D + ?P0 , where Xi ?
?(n ? 1), ?i = 0, ?i = ?j , Xi = Xj when i = j (i, j = 1, . . . , k). If W is a
2 2

?
subspace of the space M and [T, W ] ? W then W = W1 ? · · · ? Wk ? W , where
?
Wi = [Xi , W ] = [Xi , Wi ], [Z, Wi ] ? Wi , [Xj , Wi ] = 0 when j = i, [Xi , W ] = 0,
? ?
[Z, W ] ? W .
?
Proof. Let X = T ? Z, M = [X, M], M = {Y ? M|[X, Y ] = 0}, W be a projection
? ?
?
of W onto M , and W a projection of W onto M. Evidently, M = M ? M (as
162 L.F. Barannik, W.I. Fushchych

spaces). Since composition factors of the Z module M are one dimensional, then the
?
composition factors of the Z module W are one dimensional, too. Let M(P ) = {Pa ?
M|[X, Pa ] = 0}. It is easy to see that M(P ) and M / M(P ) can be represented as
direct sums of two-dimensional irreducible T submodules. Whence the dimensions
of composition factors of the T module W are equal to 2, too. When we now apply
?
Lemma 3.1 we conclude that W = W ? W .
Let Mi = [Xi , M] and Wi be a projection of W onto Mi . Clearly M = M1 ?
· · · ? Mk . At first let us establish that [Z, Wi ] ? Wi . Since for any Yi ? Wi we have
[J0n ? D, Yi ] = ?Yi , then we may assume that ? = 0. Obviously
[T, [T, Yi ]] = ??i Yi + 2?i [Xi , [Z, Yi ]] + ?[Z, Yi ].
2

Let
Yi = 2?i [Xi , [Z, Yi ]] + ?[Z, Yi ],
Yi = 2?i [Xi , [Z, Yi ]] + ?[Z, Yi ].
The space Wi contains Yi , Yi . It is easy to check that
Yi = 4?i ? 2 [Xi , [Z, Yi ]] + ?(? 2 ? 4?i )[Z, Yi ].
2

The determinant constructed by the coefficients of [Xi , [Z, Yi ]], [Z, Yi ] in Yi , Yi is
equal to ?2?i ?(? 2 + 4?i ). If ? = 0 then [Z, Yi ] ? Wi . If ? = 0 then Wi contains
2

Yi = [Xi , [?P0 , Yi ]] and Yi = [T, Yi ] = ??i [?P0 , Yi ].
In the composition factors of the T module Mi one can choose the basis so that
the matrix of the operator T is one of the matrices
??i ?? ??i
?
, .
??
?i ? ?i
If for i = j the modules Mi and Mj are possessed by isomorphic composition factors
then one of the following conditions is satisfied: ?i = ?j ; 2? = ?2?, ? 2 + ?i =
2 2 2

? 2 + ?j . Since it is impossible then on the basis of Lemma 3.1 we conclude that
2

W = W1 ? · · · ? Wk . The lemma is proved.
Proposition 4.3. Let L1 be a subalgebra of the AO(n ? 1), L2 = ?J0n + ?D + ?P0 ,
and F a subdirect sum of L1 and L2 . If W is a subspace of M and [F, W ] ? W then
[Lj , W ] ? W (j = 1, 2).
This is proved by virtue of Lemma 4.2.
Theorem 4.1. Let V1 = G1 , . . . , Gn?1 , V2 = [P0 , V1 ], V1,a be a subspace of V1 ,
V2,a = [P0 , V1,a ]; K1 , K2 , . . . , Kq be primary parts of nonzero subalgebra L1 of the
algebra AO(n ? 1); R be the maximal subalgebra of algebra M, annulled by L1 ; and
L2 be a subalgebra of the algebra R ? J0n , D . If F is the subdirect sum of L1 and
+
?
L2 , and W is a subspace of M invariant under F , then W = W1 ? · · · ? Wq ? W ,
?
where Wi = [Ki , W ] = [Ki , Wi ], [L2 , Wi ] ? Wi , [Kj , Wi ] = 0 when j = i, [Ki , W ] = 0,
? ?
[L2 , W ] ? W (i, j = 1, . . . , q).
If a primary algebra K is a subdirect sum of irreducible subalgebras of the
algebras AO(V1,1 , . . . , AO(V1,r ), respectively, then nonzero subspaces W of the
space M with the property [K, W ] = W are conjugated to
a a
V1,i , V2,i (a = 1, . . . , r)
i=1 i=1
?
On subalgebras of the Lie algebra of the extended Poincar? group P (1, n)
e 163

or to subdirect sums of such spaces
?
a
? b a c
and and
V1,i V2,i ; V1,i V2,i ;
i=1 i=1 i=1 i=a+1
a b c
and
V1,i , V2,i , V2,i
i=1 i=1 i=a+1

(? = 1, . . . , r, ? = 1, . . . , a, a = 1, . . . , r ? 1, b = 1, . . . , a, c = a + 1, . . . , r).
a b ?
The subdirect sums of the spaces
a c
V1,i , V2,i
i=1 i=a+1

are exhausted with respect to O(n ? 1) conjugation by the following spaces:
a c
V1,i ? V2,j ;
i=1 j=a+1
b a c
(V1,i , V2,a+1 , ?i I(V1,i , V2,a+1 )) ? V1,j ? V2,k
i=1 j=b+1 k=a+b+1

(0 < ?1 ? · · · ? ?b , b = 1, . . . , min{a, c ? a}).
The written spaces are mutually nonconjugated.
Proof. Let Q = [L1 , W ], S be a projection of W onto R. It is easy to see that W is
the subdirect sum of Q and S. Since the composition factors of the L2 module R are
one dimensional and the composition factors of the L1 module [L1 , M] have dimension
not less than 2 then in view of Lemma 3.1 W = Q + S. In virtue of Proposition 4.3
[L2 , Q] ? Q. We can show, as in Theorem 3.1, that Q = W1 ? · · · ? Wq , where
Wi = [Ki , Q], Wi = [Ki , Wi ] (i = 1, . . . , q). The truthfulness of the further statements
is established earlier when considering the transformations of the coupling matrix of
elementary spaces in the space W . The theorem is proved.
Theorem 4.2. Let ?1 ? ?2 ? · · · ? ?s , ?1 = 0, and ?s ? {0, 1}, AH(0) = 0,
AH(2d) = J12 , J34 , . . . , J2d?1,2d , and L be a nonzero Abelian subalgebra of the
?
algebra AG(n ? 1). If the projection ?0 (L) of the algebra L onto P0 is equal to
0 then L is conjugated to the subdirect sum of the algebras L1 , L2 , L3 , and L4 ,
where L1 ? AH(2d) (0 ? d ? m), L2 = 0 or L2 = G2d+1 + ?1 P2d+1 , G2d+2 +
?2 P2d+2 , . . . , G2d+s + ?s P2d+s , L3 = 0 or L3 = P2d+s+1 , . . . , Pl , L4 = 0 or L4 =
M . If ?0 (L) = 0 then L is conjugated to the subdirect sum of the algebras L1 ,
L2 , L3 , and L4 , where L1 ? AH(2d), L2 = P0 + ?G2d+1 (? ? {0, 1}), L3 = 0
or L3 = Pr , . . . , Pt , L4 = 0 or L4 = M (0 ? d ? m; r = 2d + 1 when ? = 0;
r = 2d + 2 when ? = 1).
Proof. Let
2d+s
Xi = Gi + ?ji Pj , L = X2d+1 , . . . , X2d+s .
j=2d+1
164 L.F. Barannik, W.I. Fushchych

Obviously, [Xi , Xk ] = (?ki ? ?ik )M . Since L is an Abelian algebra then ?ik = ?ki
and therefore B = (?ik ) (i, k = 2d + 1, . . . , 2d + s) is a symmetric matrix. Hence,
there exists a matrix C ? O(s) such that CBC ?1 = diag [?1 , . . . , ?s ]. Whence we
can assume up to conjugacy under O(n ? 1) that X2d+j = G2d+j + ?j P2d+j (j =
1, . . . , s). O(n ? 1) automorphisms permit us to change the numeration of generators
G2d+1 , . . . , G2d+s . That is why we can suppose that ?1 ? · · · ? ?s . Applying the
automorphism exp(??1 P0 ) we get generators G2d+j + µj P2d+j (j = 1, . . . , s), where
µ1 = 0, 0 ? µ2 ? · · · ? µs . If µs > 0 then µs = exp ? (? ? R). Evidently,

exp(??J0n )(G2d+j + µj P2d+j ) exp(?J0n ) = exp ? · (G2d+j + µj exp(??)P2d+j ).

Therefore when µs > 0 we can assume that µs = 1.
The rest of the assertion of the theorem follows from Proposition 4.1. The theorem
is proved.
Corollary 1. Let

A(r, t) = Gr + ?r Pr , Gr+1 + ?r+1 Pr+1 , . . . , Gt + ?t Pt , M ,

where ?r ? ?r+1 ? · · · ? ?t , ar = 0 and ?t = 1 when ?t = 0. The maximal Abelian
? ?
subalgebras of the algebra AG(n ? 1) are exhausted up to conjugacy under P (1, n)
by the following algebras:

U ; A(1, n ? 1); A(1, s) ? V2 (s + 1, n ? 1) (s = 1, . . . , n ? 2);

G1 + P0 , M ? V2 (2, n ? 1); AH(n ? 2) ? Gn?1 + P0 , M [n ? 0 (mod 2)];

AH(2d) ? P0 ? V2 (2d + 1, n) (d = 1, . . . , [(n ? 1)/2]);

AH(2d) ? A(2d + 1, n ? 1) (d = 1, . . . , [(n ? 2)/2]);

AH(2d) ? A(2d + 1, s) ? V2 (s + 1, n ? 1) (d = 1, . . . , [(n ? 3)/2]);

AH(2d) ? G2d+1 + P0 , M ? V2 (2d + 2, n ? 1) (d = 1, . . . , [(n ? 3)/2]).

The written algebras are not mutually conjugated.
Corollary 2. Let n ? 3, Xt = ?1 J12 + ?2 J34 + · · · + ?t X2t?1,2t ; ?1 = 1, 0 < ?2 ?
· · · ? ?t ? 1; t = 1, . . . , [(n ? 1)/2]; s = 1, . . . , [(n ? 2)/2].
?
The one-dimensional subalgebras of the algebra AG(n ? 1) are exhausted with
?
respect to P (1, n) conjugation by the following algebras: P0 ; M ; P1 ; G1 ;
G1 + P2 ; G1 + P0 ; Xt ; Xt + P0 ; Xt + M ; Xt + P2t+1 ; Xs + G2s+1 ;
Xs + G2s+1 + P0 ; Xr + G2r+1 + P2r+2 (r = 1, . . . , [(n ? 3)/2]).
The written algebras are not mutually conjugated.
Let
?(0) = M , ?(i) = M, P1 , . . . , Pi , ?(0) = M, P0 ,
?(i) = M, P0 , P1 , . . . , Pi , V2 (s, t) = Ps , . . . , Pt (s ? t), (4.2)
?r+1,k+1 (j) = Pr+d + ?d Pk+d |d = 1, 2, . . . , j ,

where 0 < ?1 ? · · · ? ?j (1 ? j ? k ? r).
?
On subalgebras of the Lie algebra of the extended Poincar? group P (1, n)
e 165

Proposition 4.4. Let L = G1 , . . . , Gk . The subspaces of the space U = P0 , P1 , . . . ,
?
Pn , which are invariant under L, are exhausted with respect to O(1, n) conjugation
by the following spaces:
0, ?(i), ?(k), V2 (k + 1, t), ?(i) ? V2 (k + 1, t), ?(k) ? V2 (k + 1, t),
?(r) ? ?r+1,k+1 (j), ?(r) ? ?r+1,k+1 (j) ? V2 (k + j + 1, s),
where i = 0, 1, . . . , k, t = k + 1, . . . , n ? 1, r = 0, 1, . . . , k ? 1, j = 1, . . . , k ? r,
s = k + j + 1, . . . , n ? 1.
Proof. Let W be a subspace of the space ?(k) invariant under L. Since [Pa , Ga ] = M
then with W = 0 we have M ? W . The normalizer of the algebra L in O(n ? 1)
contains O(k). It follows from this and Witt’s theorem that if W = M and P0 ? W /
then W = ?(i) (1 ? i ? k). If P0 ? W then W = ?(k).
For a description of all subspaces of the space U which are invariant under L
we shall use the Goursat twist method [25]. Since by Witt’s theorem the nonzero
subspaces of the space V2 (k + 1, n ? 1) are exhausted with respect to O(n ? 1)
conjugation by the spaces V2 (k + 1, t) (t = k + 1, . . . , n ? 1) we need to classify the
subdirect sums of the following pairs of spaces ?(k), V2 (k + 1, t); ?(i), V2 (k + 1, t)
(i = 0, 1, . . . , k, t = k + 1, . . . , n ? 1).
Let N be the subdirect sum of ?(k) and V2 (k + 1, t). If P0 + ?Pk+1 ? N (? = 0)
then N contains P1 , P1 = ?[G1 , P0 + ?Pk+1 ], and whence it contains M , too. Let
N = exp(?Gk+1 ) · N · exp(??Gk+1 ).
The space N contains P0 + (? ? ?)Pk+1 + (?2 /2 ? ??)M . Since M ? N then P0 + (? ?
?)Pk+1 ? N . Putting ? = ? we get that P0 ? N and whence ?(k) ? N . Therefore
N = ? ? V2 (k + 1, t ).
Let N be the subdirect sum of ?(i) and V2 (k + 1, t). If i = 0, M + ?Pk+1 ? N
(? = 0) then N contains (1 ? ??)M + ?Pk+1 . Putting 1 ? ?? = 0 we get that
N = V2 (k + 1, t). If i = 0 then M ? N . Let us assume that N = ?(i) ? V2 (k + 1, t).
Then ?(i)/S1 ? V2 (k + 1, t)/S2 , where S1 = N ? ?(i), S2 = N ? V2 (k + 1, t). Let
=
dim (?(i)/S1 ) = i ? r = j. Within the conjugation we can assume that S1 = ?(r)
and S2 = 0 or S2 = V2 (k + j + 1, s) and that is why by means of Lemma 4.1 N is
conjugated to one of the spaces,
?(r) ? ?r+1,k+1 (j); ?(r) ? ?r+1,k+1 (j) ? V2 (k + j + 1, s).
The proposition is proved.

5. On subalgebras of the normalizer of isotropic space
?
In virtue of Theorem 2.1 the normalizer of the isotropic space P0 +Pn in AP (1, n)
?
coincides with the algebra K = AG(n?1) ? J0n , D . In this section we shall establish
+
a number of assertions on subalgebras of the algebra K possessing nonzero projection
onto J0n , D . On the grounds of these results in Theorem 5.1 we describe all Abelian
?
subalgebras of the algebra K that are nonconjugate to the subalgebras of AG(n ? 1).
As a corollary, we obtain the list of maximal Abelian subalgebras and one-dimensional
subalgebras of the algebra K as well as one-dimensional subalgebras of the algebra
?
AP (1, n).
Further ? denotes the projection of K onto J0n , D and ? denotes the projection
of K onto AO(n ? 1) ? J0n , D .
166 L.F. Barannik, W.I. Fushchych

Proposition 5.1. Let L = G1 , . . . , Gk (1 ? k ? n ? 1), and F be a subdirect sum
?
of L and D . The algebra F has only splitting extensions in AP (1, n).
? ? ?
Proof. Let F be a subalgebra of AP (1, n) such that ?(F ) = F . Up to an O(n ? 1)
?
automorphism one can assume that F contains the generator
n
X1 = G1 + ?? P? + ?D (? = 0).
?=0

Clearly,
n n
· X1 · exp ? = G1 + ?D + (?0 ? ?b0 + b1 )P0 +
exp bµ P µ bµ P µ
µ=0 µ=0
n?1
+(?1 + b0 ? bn ? ?b1 )P1 + (?n + b1 ? ?bn )Pn + (?i ? ?bi )Pi .
i=2

We put

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