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Continuous subgroups of the generalized Schr?dinger groups

o 459

Proposition 4.3 is proved along with Proposition 2.4 and Theorem 3.4 in [13].

Proposition 4.4. Let

?b (a) = P1 + ?1 Pa+1 , . . . , Pb + ?b Pa+b ,

?

where 0 < ?1 ? · · · ? ?b , b ? a, a + b ? n. A subalgebra F of the algebra

? ?

ASch(n) such that ?(F ) = 0, D ? ? (F ), is Sch(n) conjugated to L ? U , where +

L ? ASL(2, R) and U is a subspace of the space M[1, n]. Let U = 0. If L =

ASL(2, R), then U is conjugated to M[1, d] (1 ? d ? n). If L = D, T , then U is

conjugated to one following spaces W [1, d], M[1, d], (1 ? d ? n); V [1, d] + W [1, t]

(1 ? d ? n ? 1, d + 1 ? t ? n). If L = D , then U is conjugated to one of the

following spaces:

W [1, d], M[1, d] (1 ? d ? n); V [1, d]+W [d+1, d+t] (1 ? d ? [n/2]; d ? t ? n?d);

V [1, d] + W [1, c] + W [d + 1, d + t] (1 ? d ? n ? 1; 1 ? c ? d; d ? c ? t ? n ? d, if

c = d; 1 ? t ? n ? d, if c = d);

V [1, d] + ?d (d) (1 ? d[n/2]); V [1, d] + ?t (d) + W [t + 1, d] (2 ? d ? n ? 1; 1 ? t ?

min{d ? 1, n ? d});

V [1, d] + ?t (d) + W [d + t + 1, b + t + s] + W [d + t + 1, d + t + s] (1 ? d ? [n/2];

1 ? t ? min{d, n ? d ? 1}; 1 ? s ? n ? d ? t; s + t ? d);

V [1, d] + ?t (d) + W [t + 1, b] + W [d + t + 1, d + t + s] (2 ? d ? n ? 2; 1 ? t ?

min{d ? 1, n ? d ? 1}; t + 1 ? b ? d; 1 ? s ? n ? d ? t; b + s ? d).

The proof of Proposition 4.4 is similar to the proof of Theorem 3.3 [13].

Using Theorem 4.2 to investigte splitting subalgebras wilh nonsplitting factor

? ?

algebra, it is enough to consider the algebras F ? ASch(n) for which ? (F ) = S + T

? ? ?

and ? (F ) ? F . In this case ?(F ) = F ? X , where F is a subalgebra of AO(n)

and X = S + T + ?i J12 + · · · + ?k J2k?1,2k . We may suppose that 0 < ?1 ? · · · ? ?k .

Henceforth we shall discuss subspaces of the space M[1, n] that are invariainl un-

der X.

Lemma 4.3. Let 1 ? a, b ? k. Then La ? Lb if and only if ?a = ?b or ?a + ?b = 2;

=

? Nb if and only if ?a = ?b ; La ? Nb if and only if ?a = 2 + ?b (a = b). Modules

Na = =

La and Na are not isomorphic.

Proof. The matrices ?J, µJ are similar if and only if ?2 = µ2 . It follows thal La ? Lb=

if and only if (?a ? 1) = (?b ? 1) . In the case ?a ? ?b = 0, ?a + ?b = 2.

2 2

If Na ? Nb then (?a + 1)2 = (?b + 1)2 , whence 2(?a ? ?b ) = ?(?a ? ?b )(?a + ?b ).

=

In the case ?a ? ?b = 0, 2 = ?(?a + ?b ). But this contradicts the fact that ?a , ?b > 0.

Let La ? Nb . Then (?a ? 1)2 = (?b + 1)2 , whence 2(?a + ?b ) = (?a ? ?b )(?a + ?b ).

=

Thus if ? = 0, then ?a ? ?b = 2. The lemma is proved.

Let us remark that if ?a = 1, then the X modules La and Na are irreducible,

and any X submodule of the module M[1, n] is completely reducible.

Proposition 4.5. Let

s

X =S+T + ?i J(ki?1 + 1, k),

i=1

where s ? 2, k0 = 0, ?i > 0, ?i = 1, ?i = ?j if i = j. There exists an

indecomposable X submodule with nonzero projections onto M[1, 2k1 ], M[2k1 +

1, 2k2 ], . . ., M[2ks?1 + 1, 2ks ] of the X module M[1, 2ks ] if and only if s = 2 and

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

one of the following conditions is satisfied: (i) ?1 = 2 + ?2 ; (ii) ?2 = 2 + ?1 ; (iii)

?1 + ?2 = 2. If U is ademanded indecomposable X module and U , is the projection

of U onto M[2ki?1 + 1, 2ki ] (i = 1, 2), then in case (i) U1 is a module of the first

kind and U2 is a module of the second kind; in the case (ii) U1 is a module of the

second kind; and U2 is a module of the first kind; and in cas (iii) U1 and U2 are

modules of the first kind.

Proof. By Lemma 3.1 of [9], in the X module M[1, 2ks ], there exists an indecompo-

sable submodule demanded if and only if the X modules M[1, 2k1 ], M[2k1 + 1, 2k2 ],

. . ., M[2ks?1 , +1, 2ks ], have isomorphic composition factors. If Lki ? Lkj , and Lkj ?

= =

Lkr then by Lemma 4.3 ?i + ?j = 2 and ?j + ?r = 2. From this it follows that

?i = ?r and that is why i = r. If Lki ? Nkj and Nkj ? Nkr , then ?i = 2 + ?j and

= =

?r = 2 + ?j , whence i = r. Thus s ? 2 and one of the following conditions is satisfied:

(1) ?1 = 2 + ?2 ; (2) ?2 = 2 + ?1 ; (3) 0 < ?1 < 2, ?2 = 2 ? ?1 . Statements about the

kinds of projections follow from Lemma 4.3. The proposition is proved.

Proposition 4.6. Let X = S + T + ?J(1, k) (? > 0). In the X module M[1, n] there

exists an indecomposable X submodule with nonzero projections onto M[1, 2k]

and M[2k + 1, n] if and only if ? = 2. If U is such a submodule and U1 is the

projection U onto M[1, 2k], then U1 is a module of the first kind.

5. Abelian subalgebras of the extended Schr?dinger algebra

o

The main results of this section are Theorem 5.1 and its two corollaries.

Let us use the following notation:

Xt = ?1 J12 + ?2 J34 + · · · + ?t J2t?1,2t ,

where ?1 = 1, 0 < ?2 ? · · · ? ?t ? 1 if t ? 2;

AH(0) = 0, AH(2d) = AH(2d + 1) = J12 , J34 , . . . , J2d?1,2d ;

?0 [r, t] = Gr + ?r Pr , . . . , Gt + ?t Pt , ?[r, t] = ?0 [r, t] + M ,

where r ? t, ?r ? · · · ? ?t , ?r = 0, and ?t = 1 if ?t = 0;

(a ? b).

?(a, b) = Y2a?1 , Y2a+1 , . . . , Y2b?1

We recall that Y2c?1 = G2c?1 + P2c and Y2c = G2c ? P2c?1 .

The algebra AH(n) is a maximal Abelian subalgebra of the algebra AO(n). It is

well known that any maximal Abelian subalgebra of the algebra AO(n) is conjugated

AH(n) with respect to inner automorphisms of the algebra AO(n). Henceforth when

speaking about Abelian subalgebras of the algebra AO(n) we shall mean subalgebras

of the algebra AH(n).

Lemma 5.1. Let L be an Abelian subalgebra of the algebra J(a, b) + S + T ?

+ M[2a ? 1, 2b] such that its projection onto J(a, b) + S + T is nonzero and its

projection onto M is equal to 0. Then L is conjugated to one of the following

algebras:

J(a, b) + S + T + ?Y2b?1 (? ? 0);

?(a, c) ? J(a, b) + S + T + ?Y2b?1 (? ? 0, c ? b).

The written algebras are pairwise nonconjugated.

Continuous subgroups of the generalized Schr?dinger groups

o 461

Proof. The maximal subspace of the space M[2a ? 1, 2b] anulled by J(a, b) + S + T

and having zero projection onto M coincides with

b

Lc .

c=a

Let U = L ? M[2a ? 1, 2b]. By the same arguments as in the proof of Proposition 4.1

we can establish that if U = 0, then U contains Y2a?1 . As [Y2a?1 , Y2a ] = ?2M , so

U = Y2a?1 + U 1 , where U 1 is a subspace of the space

b

Lc .

c=a+1

Continuing these arguments we obtain that U = ?(a, c) (c ? b) and L contains

J(a, b) + S + T + ?Y2b?1 (? ? 0). The lemma is proved.

Theorem 5.1. Let L be a nonzero Abelian subalgebra of algebra ASch(n). If ? (L) =

D , then L is conjugated to the subdirect sum L1 + L2 + L3 of algebras L1 , L2 ,

? ?

L3 , where L1 ? AH(2d), L2 = D , L3 ? M (0 ? d ? [n/2]. If ? (L) = T ,

then L is conjugated to L1 + L2 + L3 + L4 , where L1 ? AH(2d), L2 = T + ?G2d+1

? ? ?

(? ? 0, 1), L3 = 0 or L3 = W [r, t], L4 ? M (0 ? d ? [n/2]; r = 2d + 1 if ? = 0,

2d + 1 ? n; r = 2d + 2 if ? = 1, 2d + 2 ? n); if ? (L) = S + T , then L is conjugated

to L1 + L2 + L3 + L4 , where L1 ? M , L2 ? AH(2d) (0 ? d ? [n/2]), and the

? ? ?

algebras L3 and L4 satisfy one of the following conditions:

(1) L3 = S + T , L4 = 0;

(2) L3 = J(d + 1, t) + S + T + ?Y2t?1 , L4 = 0 (? > 0);

L4 = ?(d + 1, s) (s ? t; ? ? 0).

(3) L3 = J(d + 1, t) + S + T + ?Y2t?1 ,

?

If L ? AG0 (n), then L is conjugated to L1 + L2 + L3 + L4 , where L1 ? AH(2d),

? ? ?

L2 = 0 or L2 = ?0 [2d + 1, s], L3 = 0 or L3 = W [k, l], L4 = 0 or L4 = M

(0 ? d ? [n/2]; k = s + 1 if L2 = 0; k = 2d + 1 if L2 = 0; l ? n).

Proof. lf ? (L) = D , then by Theorem 4.1 the algebra L is conjugated to the algebra

U + F , where U ? M[1, n] and F ? AH(n) ? D, M . Since D annuls only M in

M[1, n] and by Theorem 4.2 [D, U ] ? U , then U ? M . Thus L is conjugated to

some subalgebra of the algebra AH(2d) ? D, M (0 ? d ? [n/2]).

If ? (L) = T , then by Theorem 4.1 the algebra L is conjugated to the algebra

U +F satisfying one of the following conditions: U ? M[1, n] and F is a subalgebra of

AH(n)+ M, T ; or U ? M[1, 2d] and F is a subalgebra of AH(2d)+M[2d+1, n]+ T

(d ? 1). Let us consider the last case. Let us suppose that the projection K of the

algebra F onto AO(n) is not conjugated to any subalgebra of the algebra AH(2d ? 2).

Since K annuls only the zero subspace of V [1, 2d], then U ? M . Therefore we shall

assume that U = 0. As [T, Ga ] = ?Pa , so by Witt’s mapping theorem [14] (F ) = 0,

or (F ) = G2d+1 . Since

exp(?T )(T + ?G2d+1 + ?P2d+1 ) exp(??T ) = T + ?G2d+1 + (? ? ??)P2d+1

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

and

exp(?D)(T + ?G2d+1 ) exp(??D) = exp(?2?)(T + ? exp(3?) · G2d+1 ),

then if (F ) = 0, the projection of F onto T ? M[2d + 1, n] contains T + G2d+1 . In

+

this case, by Witt’s theorem ?(F ) coincides with 0 or W [2d + 2, t]. If (F ) = 0, then

?(F ) = 0 or ?(F ) = W [2d + 1, t].

Let ? (L) = S + T . If S + T ? L then (L) = 0 and ?(L) = 0. If S + T ? L, then

an algebra L contains

[n/2] n

Y =S+T + ?a J2a?1,2a + (?i Gi + ?i Pi ) + ?M.

a=1 i=1

We shall suppose that projections of the at rest basis elements of the algebra L onto

S +T are equal to zero, and ?a ? 0 for all a. If ?c = 1, then S +T +?c J2c?1,2c is a

completely reducible algebra of linear iransformalions of the vector space M[2c?1, 2c]

and annuls only the zero subspace of this space, whence by Proposition 2.1 of [9] we

conclude that the projection of L onto M[2c ? 1, 2c] is equal to zero. Therefore we

may assume that

2t

Y = J(d + 1, t) + S + T + (?i Gi + ?i Pi ).

i=2d+1

From Proposition 2.1 of [9] it also follows that

2t i

(?i Gi + ?i Pi ) ? Lj .

i=2d+1 j=d+1

Applying Theorem 4.1 and Lemma 5.1 we conclude that, with respect to the conjuga-

tion ?(L) ? AH(2d) + J(d + 1, t) ,

Y = J(d + 1, t) + S + T + ?Y2t?1 ,

and L ? M[1, n] = 0 or L ? M[1, n] = ?(d + 1, s) (? ? 0; s ? t).

?

Let us assume thet L ? AG0 (n). By Theorem 2 of [7] the algebra L is conjugated

to an algebra U + F , which statisfies one of the following conditions: U ? M[1, n]

and F is a subalgebra of AH(n) + M ; or U ? M[1, 2d] and F is a subalgebra of

AH(2d) + M[2d + 1, n] (1 ? d ? [n ? 1/2]). Let us restrict ourselves to the last

case. Let the projection K of the algebra AH(2d ? 2). Since K annuls only the zero

subspace of the space M[1, 2d], U ? M . Therefore we suppose that U = 0.

Let N be the projection of F onto M[2d + 1, n] and (N ) = V [2d + 1, 2d + q]. By

Witt’s mapping theorem the algebra N is a subdirect sum of the algebras N1 , N2 ,

N3 , where N1 ? M[2d + 1, 2d + q] (as a space), N2 = 0 or N2 = W [2d + q + 1, t], and

N3 ? M . Let

Zi = G1 + ?2d+1,i P2d+1 + · · · + ?2d+q,i P2d+q (i = (2d + 1), . . . , (2d + q)),

Nq = Z2d+1 , . . . , Z2d+q . Evidently [Zi , Zj ] = (?ij ? ?ji )M . Since N1 is an Abelian

algebra, ?ij = ?ji . Hence it foliows that the matrix B = (?ij ) is symmetric. Therefore

there exists a matrix Q ? O(q) such that QBQ?1 = diag [?1 , . . . , ?q ]. From this it

Continuous subgroups of the generalized Schr?dinger groups

o 463

follows that with respect to automorphisms from the group O(2d)?O(q)?O(n?2d?q)

we may assume that Z2d+j = G2d+j + ?j P2d+j (j = 1, . . . , q), where ?1 ? · · · ? ?q .

Applying the automorphism exp(?1 T ) we obtain the generators G2d+j + µj P2d+j

(j = 1, . . . , q), where µ1 = 0, 0 ? µ2 ? · · · ? µq . If µq > 0, then µq = exp(?2?).

Obviously

exp(?D)(G2d+j + µj P2d+j ) exp(??D) = exp ?(G2d+j + µj exp(?2?)P2d+j ).

Therefore if µq > 0, we may suppose that µq = 1. This proves that the algebra N1 is

conjugated to ?0 [2d + 1, 2d + 1, 2d + q]. The theorem is proved.

Corollary 1. The maximal Abelian subalgebras of the algebra ASch(n) are exhaus-

ted with respect to the Sch(n) conjugation by the following algebras:

AH(n) ? T, M (n ? 0 (mod 2)); AH(n) ? S + T, M ;

AH(n) ? D, M ; AH(n ? 1) ? Gn + T, M ; (n ? 1(mod 2));

AH(2d) ? ?[2d + 1, n] (d = 0, 1 . . . , [(n ? 1)/2]);

AH(2d) ? ?[2d + 1, t] ? W [t + 1, n] (d = 0, 1, . . . , [(n ? 2)/2]; t = 2d + 1, . . . , n ? 1);

AH(2d) ? T, M ? W [2d + 1, n] (d = 0, 1, . . . , [(n ? 1)/2]);

AH(2d) ? G2d+1 + T ? W [2d + 2, n] ? M (d = 0, 1, . . . , [(n ? 2)/2]);

AH(2d) ? J(d + 1, r) + S + T ? M ? ?(d + 1, r) (d = 0, 1, . . . , [(n ? 2)/2]); r =

d + 1, . . . , [n/2]).

Corollary 2. Let ?, ? ? R, ? > 0, ? > 0; t = 1, . . . , [(n ? 1)/2]; n ? 3. One-

dimensional subalgebras of the algebra ASch(n) are exhausted with respect to the

Sch(n) conjugation by the following algebras:

D ; T ; S + T ; M ; D + ?M ; T ± M ; S + T ± ?M ; P1 ;

G1 + P2 ; G1 + T ; Xt ; Xt + ?D ; Xt + ?D + ?M ; Xt + T ;

Xt + ?(S + T ) ; Xt + ?M ; Xt + ?(S + T ) ± ?M ; Xs + P2s+1 ;

Xr + G2r+1 + ?P2r+2 (r = 1, . . . , [(n ? 2)/2]); Xs + T + ?G2s+1 ;

Xt + S + T + ?(G1 + P2 ) .

Remark. One-dimensional subalgebras of the algebra ASch(n) are exhausted with

respect to the Sch(n) conjugation by one-dimensional subalgebras, of the algebra

ASch(n) whose generators do not contain ?M as an addend (? = 0.

Theorem 5.2. Let L be a nonzero Abelian subalgebra of the algebra ASch(n). If

? (L) = D , then L is conjugated to a subdirect sum of D and the subalgebra

of the algebra AH(2d) (0 ? d ? [n/2]). If ? (L) = T , then L is conjugated to

L1 + L2 + L3 , where L1 ? AH(2d), L2 = T + ?G2d+1 , and L3 is one of the

? ?

following algebras:

0; W [2d + 2, t]; P2d+1 + ?P2d+2 + ?W [2d + 2] + ?W [2d + 3, t] (0 ? d ? [n/2];

t ? n; ?, ?, ? ? {0, 1}; ? ? 0).

If ? (L) = S + T , then L is conjugated to L1 + L2 + L3 , where L1 ? AH(2d)

? ?

(0 ? d ? [n/2]) and the algebras L2 , L3 satisfy one of the following conditions:

(1) L2 = S + T and L3 = 0; or (2) L2 = J(d + 1, t) + S + T + ?Y2t?1 (? ? 0) and

L3 is a subalgebra of the algebra

t

La

a=d+1

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

If L ? AG0 (n), then L is conjugated to L1 + L2 , where L1 ? AH(2d) and L2 ?

?

M[2d + 1, n] (0 ? d ? [n/2]).

The theorem is proved along the same lines as Theorem 5.1.

Corollary. The maximal Abelian subalgebras of the algebra ASch(n) are exhausted

with respect to the Sch(n) conjugation by the following algebras:

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