ñòð. 105 |

+

called a splitting in the algebra ASch(n). The notion of a splitting subalgebra of the

?

algebra ASch(n) is defined in an analogous way. If every subalgebra F is a splitting,

we shall say that F has only splitting extensions in the algebra ASch(n) (resp. in the

algebra ASch(n)).

We shall find all subalgebras F , which possess only splitting extensions.

Let

J(a, b) = J2a?1,2a + · · · + J2b?1,2b (a ? b),

01

J(a) = J(a, a), J = ,

?1 0

X = S + T + ?1 J12 + · · · + ?t J2t?1,2t , 0 ? ?1 ? · · · ? ?t ,

Y2a?1 = G2a?1 + P2a , Y2a = G2a ? P2a?1 , Z2a?1 = G2a?1 ? P2a ,

Z2a = G2a + P2a?1 , La = Y2a?1 , Y2a , Na = Z2a?1 , Z2a .

Obviously, La + Na = M[2a ? 1, 2a]. If 1 ? a ? t, then

[X, Y2a?1 ] = ?(?a?1 )Y2a , [X, Y2a ] = (?a ? 1)Y2a?1 ,

(2)

[X, Y2a?1 ] = ?(?a + 1)Z2a , [X, Z2a ] = (?a + 1)Z2a?1 ,

Thus (?1 ? 1)J is the matrix of ad X in the basis Y2a?1 , Y2a of the space La , and

(?a + 1)J is the matrix of ad X in the basis Z2a?1 , Z2a of the space Na (1 ? a ? t).

If ?a = 0, we obtain a matrix corresponding to ad (S + T ).

Let ?a = 0, ?a = 1. The X module N is called an elementary module of the

first kind, and the X module Na is called an elementary module of the second kind.

A subdirect sum of elementary modules of the first kind is called a module of the

first kind, and a subdirect sum of elementary modules of the second kind is called a

module of the second kind.

Continuous subgroups of the generalized Schr?dinger groups

o 455

Lemma 4.1. Let C be a matrix obtained from the identily matrix of degree n as a

result of fulfilling a permutation over its columns

2k ? 1 2l ? 1

2k 2l

(k < l),

2l ? 1 2k ? 1

2l 2k

followed by the multiplication on (?1) columns which have number 2k and 2l. The

O(n) automorphism ? of the algebra ASch(n) which corresponds to the matrix C

has the following properties:

?(J2d?1,2d ) = J2d?1,2d , if d = k, d = l,

(1)

?(J2k?1,2k ) = J2l?1,2l , ?(J2l?1,2l ) = J2k?1,2k ;

?(G2k?1 ) = G2l , ?(G2k ) = ?G2l?1 ,

(2)

?(G2l?1 ) = G2k , ?(G2l ) = ?G2k?1 ;

?(Lk ) = Ll , ?(Ll ) = Ll ,

(3)

?(Nk ) = Nl , ?(Nl ) = Nk .

Proof. For simplicity we can take n = 4 and

?J

0

C= .

?J 0

Then

? ?? ?

?y4

y1

? y2 ? ? y3 ?

C ·? ?? ?.

C(?12 + ?J34 )C ?1 = ?J12 + ?J34 , ? y3 ? = ? ?y2 ?

y4 y1

Using the last equality we conclude that ?(G1 ) = G4 , ?(G2 ) = ?G3 , ?(G3 ) = G2 ,

and ?(G4 ) = ?G1 . The lemma is proved.

Lemma 4.2. Letting n > 4, 1 ? q ? [n/2] ? 1, and Ea be the identity matrix of

degree a,

? ?

1 ?

v v

? 1 + ?2 1 + ?2 ?

C1 (?) = ? ? ? E2 ,

? ?

?1

?

v v

1 + ?2 1 + ?2

? ?

1 ?

v v

0

? 1 + ?2 1 + ?2 ?

? ?

? ?

? ? E2 (k ? 2);

C1 (?) = ? 0 Ek?1 0

? ?

? ?

?1

?

v v

0

1 + ?2 1 + ?2

?(1, k; ?) = diag [Ck (?), En?2,(k+1) ], if 2(k + 1) < n,

?(1, k; ?) = Ck (?), if 2(k + 1) = n;

?(q, k; ?) = diag [E2q?2 , Ck (?), En?2(k+q) ], if q > 1, 2(k + q) < n,

?(q, k; ?) = diag [E2q?2 , Ck (?)], if q > 1, 2(k + q) = n;

456 L.F. Barannik, W.I. Fushchych

and ?(q, k; ?) is an O(n) automorphism of the algebra ASch(n) which corresponds

to a matrix ?(q, k; ?). Then

?(q, k; ?)(J(q, q + k)) = J(q, q + k),

v

?(q, k; ?)(G2q?1 + ?G2(q+k)?1 ) = 1 + ?2 · G2q?1 ,

v

?(q, k; ?)(G2q + ?G2(q+k) ) = 1 + ?2 G2q ,

Proof. We may restrict ourselves only to the case when n = 4, q = 1, k = 1. Since

X 0 X 0

C1 (?) · · C1 (?),

=

0 X 0 X

for every matrix X of degree 2, and

? ? ? ?

y1 y1

?y ? ? y2 ?

C1 (?) · ? 2 ? = 1 + ?2 ? ?

? ?y1 ? ? 0?

?y2 0

then

?(1, 1; ?)(J(1, 2)) = J(1, 2),

v

?(1, 1; ?)(G1 + ?G3 ) = 1 + ?2 G1 ,

v

?(1, 1; ?)(G2 + ?G4 ) = 1 + ?2 G2 ,

The lemma is proved.

Proposition 4.1. Let X = S + T + ?J(k, l), where ? > 0, ? = 1. If U is an X

submodule of the first (the second) king of the module M[2k ? 1, 2l], then U is

conjugated to the module

t t

La Na (t ? l).

a=k a=k

Proof. Let us assume that U is a module of the first king. By Lemma 4.1 we shall

suppose that a projection of U onto Lk differs from 0. As

exp(?J2a?1,2a )(?a Y2a?1 + ?a Y2a ) exp(??J2a?1,2a ) =

= (?a cos ? + ?a sin ?)Y2a?1 + (?a cos ? ? ?a sin ?)Y2a ,

putting ?a cos ? ? ?a sin ? = 0, we may assume that if a projection of an element

Y ? U onto La is equal to ?a Y2a?1 + ?a Y2a , then ?a = 0. Hence it follows that U has

the element

Y = Y2k?1 + ?k+1 Y2k+1 + · · · + ?l Y2l?1 =

= (G2k?1 + ?k+1 G2k+1 + · · · + ?l G2l?1 ) + (P2k + ?k+1 P2k+2 + · · · + ?l P2l )

In view of Lemma 4.2, for some O(n) automorphism ? = ?(k, 1; µ1 ) · ?(k, 2; µ2 ) · · · · ·

?(k, l ? k; µl?k ) of the algebra ASch(n) the following equalities hold true: ?(X) = X,

?(Y ) = ?(G2k?1 + P2k ) (? ? R, ? = 0). Therefore we may assume that Y2k?1 ? U .

Then Y2k ? U , and thus Lk ? U . Using induction we conclude that U = La .

Continuous subgroups of the generalized Schr?dinger groups

o 457

The case when U is a module of the second king is treated similarly. The proposi-

tion is proved.

Theorem 4.1. Let F be a subalgebra of the algebra AO(n) ? ASL(2, R). Then

F has only splitting extensions in ASch(n) if and only if one of the following

conditions is satisfied: (1) D ? ? (F ); (2) ? (F ) = S + T and F is not conjugated to

J12 + S + T + K, where K is a subalgebra of the algebra Jab | a, b = 3, . . . , n; (3)

?

? (F ) ? T and ?(F ) is not conjugated to any subalgebra of the algebra AO(n ? 1);

or (4) ? (F ) = 0 and ?(F ) is a semisimple algebra.

Proof. Let D ? ? (F ). If ? (F ) = ASL(2, R), then by Theorem 3.1 F is a completely

reducible algebra. Since in this case F annuls only zero subspace in M[1, n], then

by Proposition 2.1 of [9] the algebra F has only splitting extensions in ASch(n).

If ? (F ) = D, T , then T ? F . Algebra F/ T acts completely reducible in M[1, n]

and annuls only zero subspace in this space. From this, using Proposition 2.1 and

Lemma 3.1, we conclude that F has only splitting extensions in ASch(n). At the

same time the case ? (F ) = D is considered.

Let ? (F ) = S + T . If S + T ? F , then F annuls only zero subspace in M[1, n].

Because of Theorem 3.1 the algebra F is completely reducible; then by Proposition 2.1

?

of [9] any algebra F in the algebra ASch(n), then F contains X = S + T + ?1 J12 +

· · · + ?t J2t?1,2t , where 0 < ?1 ? · · · ? ?t . We may assume that prolections of other

basis elements of the algebra F onto S + T are equal to 0. In view of Proposition

2.1 of [9] the algebra F annuls in M[1, n] a certain nonzero subspace U . It follows

from this and formula (2) that U ? Y1 , Y2 , . . . , Y2k and X = S + T + J(1, t) (k ? t)

or

X = S + T + J(1, k) + ?k+1 J(k + 1) + · · · + ?t J(t) (t > k),

where ?k+1 > 0, . . ., ?t > 0, ?k+1 = 1, . . ., ?t = 1. Arguing as in the proof

of Proposition 4.1 we obtain that Y1 ? U up to conjugacy. Hence it follows that

F = S + T + J12 + K, where K ? Jab | a, b = 3, . . . , n . By Lemma 2.1 of [9] the

?

?

algebra F which is obtained from F by replacing S + T + J12 by S + T + J12 + Y1 , is

nonsplitting.

?

Let ? (F ) = T , F1 = ?(F ), and F be a subalgebra of the algebra ASch(n) such

?

that ?(F ) = F . If F1 is not conjugated to a subalgebra of the algebra AO(n ? 1), then

?

by Proposition 2.1 and Lemma 3.1 of [9] an algebra F is splitting. If F1 is conjugated

to a subalgebra of the algebra AO(n ? 1), then F = X ? F2 , where X = 0, and X

and F2 are subalgebras of the algebra AO(n ? 1) ? T . An algebra F2 ? Pn X + Gn

+

is nonsplitting.

The case ? (F ) = 0 is considered in [5, 7]. The theorem is proved.

Proposition 4.2. A subalgebra F of the algebra AO(n) ? ASL(2, R) possesses only

splitable extensions in ASch(n) if and only if F is a semisimple algebra.

The proof of Proposition 4.2 is similar to the proof of Theorem 4.1.

Let ? : X > X be the trivial representation of a subalgebra F of the algebra

AO(n). Then ? is O(n) equivalent to diag [?1 , . . . , ?m ], where ?i is an irreducible

subrepresentation (i = 1, . . . , m). It is well known that if representations ? and ? of

Lie algebra L by skew-symmetric matrices are equivalent over R, then C?(X)C ?1 =

? (X) for some orthogonal matrix C (X ? L), hence we conclude that if ?i and ?j

458 L.F. Barannik, W.I. Fushchych

are equivalent representations, then we can assume that for every X ? F the equality

?i (X) = ?j (X) takes place. Uniting equivalent nonzero irreducible subrepresentations

we shall get nonzero disjunctive primary subrepresentations ?1 , . . . , ?q of the repre-

sentation ?. An algebra

Ki = {diag [0, . . . , ?i (X), . . . , 0] | X ? F } (1 ? i ? q)

is called a primary part of the algebra F . Evidently F is a subdirect sum of its

primary parts.

We shall say that the splitting subalgebra F of the algebra ASch(n) or of the

?

algebra ASch(n) has a splitting factor algebra in the case ?(F ) = F1 ? F2 , where

F1 ? AO(n), F2 ? ASL(2, R). If this condition does not hold, then the factor algebra

? ?

?(F ) of an algebra F is called nonsplitting.

Theorem 4.2. Let K1 , K2 , . . . , Kq be primary parts of the nonzero subalgebra L

of the algebra AO(n), L be a subalgebra of the algebra ASL(2, R) differing from

S + T , and L be a subdirect sum of L and L . If U is a subspace of M[1, n],

?

being invariant under L, then U = U1 ? · · · ? Uq ? U , where Ui = [Ki , U ] = [Ki , Ui ];

? ? ?

[L , Ui ] ? Ui ; [Kj , Ui ] = 0 in the case j = i; [Ki , U ] = 0, [L , U ] ? U (i, j = 1, . . . , q).

Proof. If L = ASL(2, R), then L ? L. Therefore from [L, U ] ? U it follows that

[L , U ] ? U . Since M[a] is invariant under ASL(2, R) for any a, 1 ? a ? n, then each

?

subspace Ui = [Ki , U ] is invariant under this algebra. Let U be a maximal subspace

of the space U annulled by L , U = [L , U ]. Since L is a completely reducible

?

algebra, U = U ? U and [L , U ] = U . Applying Lemma 3.1 of [9] we conclude that

U = U1 ? · · · ? Uq , where Ui = [Ki , U ] = [Ki , Ui ] (i = 1, . . . , q).

Let L = T, D . Since T, D is a non-Abelian solvable algebra and every subal-

gebra of the algebra AO(n) is reductive, then applying the Goursat twist method [11]

we obtain that T ? L. Therefore it is enough to consider the case L = D . By

? ?

Lemma 4.2 of [9], [D, U ] ? U , it follows that [D, Ui ] ? Ui , [D, U ] ? U (i = 1, . . . , q).

The case L = T is considered in [5, 7]. The theorem is proved.

?

Because of Theorem 4.2, the study of splitting subalgebras F of the algebra

? ?

ASch(n), for which ? (F ) = S +T , is reduced to the study of splitting subalgebras K

?

of the algebra ASch(n) having the splitting factor algebra ?(K) and zero or primary

projection onto AO(n). Such subalgebras have been described in [13].

Proposition 4.3. Nonzero subspaces of the space M[1, n] invariant under S + T

are exhausted with respect to O(n) conjugalion by the following spaces: M[1, d]

(d = 1, . . . , n); Uq (q = 1, . . . , [n/2]); Um + M[2m + 1, t] (m = 1, . . . , [(n ? 1)/2]; t =

2m + 1, . . . , n), where Uq is a subdirect sum of V [1, 2q] and W [1, 2q] having zero

interseclions with there spaces. If

{Gj + ?1j P1 + · · · + ?2q,j P2q | j = 1, . . . , 2q}

is a basis of Uq , then with respect to O(2q) conjugation a matrix (?kj ) (k, j =

1, . . . , 2q) coincides with diag [?(?1 ), . . . , ?(?q )], where 0 < ?1 ? · · · ? ?q ? 1 and

0 ?

?(?) = .

???1 0

ñòð. 105 |