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. 105
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algebra F ? N, where N is an F -invariant subspace of the space M[1, n], then F is
+
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

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( 145 .)



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