ñòð. 40 |

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

ñòð. 40 |