ñòð. 38 |

? ?

Let ? be the projection of the algebra AP (1, n) onto AO(1, n), F a nonzero

? ? ? ?

subalgebra of AO(1, n), and F such subalgebra of AP (1, n) that ?(F ) = F . If

? ?

the algebra F is P (1, n) conjugated to the algebra W ? , where W is an F -

+F

?

invariant subspace of the space U , then we shall assume F to be splitting. If

? ? ?

every subalgebra F ? AP (1, n) satisfying ?(F ) = F is splitting, we shall say that

?

subalgebra F possesses only splitting extensions in the algebra AF (1, n). The spli-

ttability of subalgebras for other algebras of inhomogeneous transformations is defi-

ned by analogy. If nothing is reserved, then the investigation of subalgebras of given

algebra for conjugation is carried out with respect to the group of inner automorphi-

sms.

The affine group IGL(n, R) is defined as a group of matrices

B Y

(2.4)

,

0 1

where B ? GL(n, R), Y ? Rn . The Lie algebra AIGL(n, R) of this group consists of

matrices

XY

,

00

where X is a square matrix of degree n over R. Let 0a be the zero matrix of degree a,

Pa = Ea,n+1 . Let us identify X and diag [X, 01 ], then AIGL(n, R) = P1 , . . . , Pn ?

+

AGL(n, R). If m < n, then we shall assume that AGL(m, R) consists of the matrices

? ?

diag [X, 0n+1?m ], where deg X = m.

Lemma 2.1. Let F be a completely reducible subalgebra of the Lie algebra

AGL(m, R) (m < n), which is not semisimple. If Z is a nonzero central element

?

of the algebra F and F is the Lie algebra, which is obtained from F by replacing

?

Z by Z + Pm+1 , then the algebra F is nonsplitting in AIGL(n, R) with respect to

IGL(n, R) conjugation.

Proof. Let X0 be a square matrix of the degree m, T = diag [X0 , 0n?m ], Z =

diag [T, 01 ],

0n Ym+1

Pm+1 = .

0 01

?

If F is a splitting algebra, then there exists the matrix C of the form (2.4) such that

C(Z + Pm+1 )C ?1 = diag [T , 01 ]. It follows that ?BT B ?1 Y + BYm+1 = 0, which

implies that Ym+1 = T B ?1 Y . However,

X0 0 B1 B2 X0 B1 X0 B2

T B ?1 = · = ,

0 0n?m B3 B4 0 0n?m

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

and therefore

? ?

?1

?.?

?.?

?.?

?? ?

· Y = ? m ?.

T B ?1 ?0?

? ?

?.?

?. .?

0

This contradiction proves the lemma.

Proposition 2.1. Let F be a completely reducible Lie algebra of linear transfor-

mations of vector space V over the field R, W is an irreducible F submodule of

module V . If F W = 0, then algebra F possesses only splitting extensions in algebra

W? .

+F

Proof. Since F is a completely reducible subalgebra of the algebra gl(V ), then F =

Q ? Z(F ), where Q is Levy’s factor and Z(F ) is the center of F [23]. Using Jacobi

identity it is not difficult to conceive that F = F1 ? F2 , where F1 W = 0 and every

direct surnmand of algebra F2 annuls in W only zero subspace. Further we may

restrict ourselves only with the case when F = F2 .

?

Let Q = 0; F such a subalgebra of the algebra W ? that its projection onto F

+F

coincides with F . According to Whitehead’s theorem [23] H 1 (Q, W ) = 0. From this it

? ?

follows that the algebra F contains Q. Let J ? Z(F ), Y ? W , Y = 0, and J + Y ? F .

Since [Q, Y ] = 0, then there exists such an element X ? Q that [X, Y ] = 0. Let

Y1 = [X, Y ], W1 be the F submodule of module W , generated by Y1 . Because of the

?

fact that W1 = 0 and W is the irreducible F module we have W1 = W . Hence J ? F .

? ?

Therefore, if Q = 0 then F ? F , i.e., F is a splitting algebra.

Let Q = 0, J ? Z(F ). Since J annuls in W the only zero subspace is then

[J, W ] = W . Whence for every Y ? W there exists such element Y ? W that

? ?

[J, Y ] = Y . Consequently we may suppose that J ? F . If F contains J1 + Y1 , where

?

Y1 ? W and Y1 = 0, then [J, Y1 ] ? F and [J, Y1 ] = 0. Arguing as in the case Q = 0,

? ?

we get that J1 ? F , i.e., F is a splitting algebra. The proposition is proved.

Proposition 2.2. Let

?

AE(n ? 1) = G1 , . . . , Gn?1 ?

+(AO(n ? 1) ? J0n ),

where Ga = J0a ? Jan (a = 1, . . . , n ? 1). The subalgebra F ? AO(n ? 1) ? J0n

?

possesses only splittable extensions in AE(n ? 1) if and only if F is a semisimple

algebra or F not conjugated to a subalgebra of the algebra AO(n ? 2).

Proof. Let W = G1 , . . . , Gn?1 . Since every subalgebra of the algebra AO(n ? 1) is

completely reducible and [J0n , Ga ] = ?Ga , then every subalgebra F of the AO(n ?

1) ? J0n algebra is also a completely reducible algebra of linear transformations of

space W .

Let W = W1 ? · · · ? Ws be the decomposition of W into the direct sum of

irreducible F modules. If projection F onto J0n is nonzero, then [F, Wi ] = Wi

for every i = 1, . . . , s. Whence according to Proposition 2.1 F has only splittable

?

extensions in AE(n ? 1). Let us assume that projection of F onto J0n is equal to

0. If F is a semisimple algebra then by Whitehead’s theorem every extension of F

?

On subalgebras of the Lie algebra of the extended Poincar? group P (1, n)

e 155

?

in AE(n ? 1) is splitting. Let F not be a semisimple algebra. When dim Wi ? 2 for

every i = 1, . . . , s we have [F, Wi ] = 0 and in view of Proposition 2.1 F possesses

?

only splitting extensions in AE(n ? 1). When dim Wi = 1 (1 ? i ? s), the module Wi

is annuled by the algebra F and the algebra F is conjugated to a subalgebra of the

algebra AO(n ? 2). If Z(F ) is the center of F and X is a nonzero element of Z(F )

? ?

then for every nonzero Y ? Wi there exists a subalgebra F of the algebra AE(n ? 1),

?

which is obtained from F by replacing X by X + Y . By Lemma 2.1 F is not splitting.

The proposition is proved.

From Theorem 2.1 and properties of solvable subalgebras of algebra AO(n) it

follows that if n is odd then AO(1, n) possesses with respect to O(1, n) conjugation

only one maximal solvable subalgebra

G1 , . . . , Gn?1 , J12 , J34 , . . . , Jn?2,n?1 , J0n .

If n is even then AO(1, n) possesses two maximal solvable subalgebras

J12 , J34 , . . . , Jn?1,n , G1 , . . . , Gn?1 , J12 , J34 , . . . , Jn?3,n?2 , J0n .

Since an extension of an Abelian algebra with the help of a solvable algebra is a

solvable algebra itself then maximal solvable subalgebras of the algebra AP (1, n)

are of the form U ? , where F is the maximal solvable subalgebra of the algebra

+F

?

AO(1, n). Maximal solvable subalgebras of the AP (1, n) are exhausted by algebras

U? +(F ? D ).

Proposition 2.3. Let AH(t) be the Cartan subalgebra of the algebra AO(t). The

?

maximal Abelian subalgebras of the algebra AO(1, n) are exhausted with respect to

?

O(1, n) conjugation by the following algebras: AH(n?1)? J0n , D ; AH(n)? D [n ?

0 (mod 2)]; G1 , . . . , Gn?1 , D ; AH(2a) ? G2a+1 , . . . , Gn?1 , D (a = 1, . . . , [n ? 2/2]).

The written algebras are pairwise nonconjugated.

?

Proof. If F is a maximal Abelian subalgebra of the algebra AO(1, n) then from

Proposition 2.2 F = ? ? L ? D , where L is a subalgebra of the algebra AO(l) ?

J0n or the algebra AO(n) and ? is a subalgebra of the algebra G1 , . . . , Gn?1 . If

projection L onto J0n is different from 0 then ? = 0. Let projection L onto J0n

be equal to 0. If L = AH(n), then ? = 0. If L = AH(2a), 1 ? a ? [n ? 2/2], then

? = G2a+1 , . . . , Gn?1 . The proposition is proved.

?

3. Completely reducible subalgebras of the algebra AO(1, n)

In this section we shall prove a number of general results on completely reducible

?

subalgebras of the algebra AO(1, n) and shall indicate how to search invariant subspa-

ces of space U for these algebras. The main results of this section are Proposition 3.3

and Theorem 3.1.

Proposition 3.1. If n ? 2 then any irreducible subalgebra of the algebra AO(1, n) is

semisimple and noncompact.

Proof. Let F be an irreducible subalgebra of the algebra AO(1, n), Z(F ) the center

of F . If Z(F ) = 0 then Z(F ) = J , where J 2 = ?En+1 . Let X be an arbitrary

element of the form (2.2) of the algebra AO(1, n). If X 2 = ?En+1 , then ?01 + ?02 +

2 2

· · · + ?0n = ?1. This contradiction proves that Z(F ) = 0.

2

If F is a compact algebra then there exists such symmetric matrix C that

?1

C F C ? AO(n + 1) [24]. Since

exp C ?1 F C = C ?1 · exp F · C

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

then in O(n + 1) there exists an irreducible subgroup conserving simultaneously

x2 + x2 + · · · + x2 and ?2 x2 ? ?2 x2 ? · · · ? ?2 x2

0 1 n 00 11 nn

(?0 , ?1 , . . . , ?n are nonzero real numbers). This contradiction proves the second part

of the proposition.

?

Proposition 3.2. A reducible subalgebra of the algebra AO(1, n) is completely redu-

cible if and only if it is conjugated to L1 ? L2 or a subalgebra of algebra L ? D ,

where L1 is an irreducible subalgebra of the algebra AO(1, k) (k ? 2), L2 is a

subalgebra of AO (n?k)? D and L is one of the algebras, AO(n), AO(n?1)? J0n .

Proposition 3.2 follows from Theorem 2.1, Propositions 2.2 and 3.1, and the fact

that Ga acts noncompletely reducible onto the space P0 + Pn , Pa .

Let L be a direct sum of the Lie algebras L1 , . . . , Ls , B a Lie subalgebra of L, and

?i the projection L onto Li . If ?i (B) = Li for i = 1, . . . , s then B is called a subdirect

sum of L1 , . . . , Ls .

?

Proposition 3.3. A completely reducible subalgebra F ? AO(1, n) has only splitting

?

extensions in AP (1, n) if and only if F is semisimple or F is nonconjugate to sub-

algebra of one of the algebras, AO(n) or AO(1, n ? 1).

The proof of Proposition 3.3 is analogous to that of Proposition 2.2.

Let Ai be a Lie algebra over R (i = 1, 2), f : A1 > A2 is an isomorphism,

B = {(X, f (X)|X ? A1 }. Here B is the Lie algebra over R with “componentwise”

operational rules,

[(X, f (X)), (X , f (X ))] = ([X, X ], f ([X, X ])),

(X, f (X)) + (X , f (X )) = (X + X , f (X + X )),

?(X, F (X)) = (?X, f (?X)),

where X, X ? A1 , ? ? R. Let us denote it as (A1 , A2 , ?). Evidently (A1 , A2 , ?) is the

subdirect sum of the algebras A1 and A2 .

Let Wi be a left Ai module (i = 1, 2). It is easy to see that Wi is the B module if

we put

(X, f (X)) · Y1 = X · Y1 , (X, f (X)) · Y2 = f (X) · Y2 ,

for every X ? A1 , Yi ? Wi (i = 1, 2). Let W be a B submodule of the module

W1 ? W2 . If W = W1 ? W2 , where Wi ? Wi (i = 1, 2) then W is called a splitting B

module. Otherwise the module W is called nonsplitting B module.

Lemma 3.1. Let B = (A1 , A2 , ?) and Vi be a left Ai module (i = 1, 2). In the B

module V1 ? V2 exists a nonsplitting B submodule if and only if the B modules V1

and V2 have isomorphic composition factors.

Proof. Let W be a nonsplitting B submodule of the module V1 ? V2 . Then W is the

subdirect sum of the modules W1 and W2 , where Wi ? Vi (i = 1, 2). Let Si = W ? Vi

(i = 1, 2). Evidently, Si is the B submodule of the module W . The module W/(S1 ?S2 )

is nonsplitting B submodule of the module V1 /S1 ? V2 /S2 . Whence we shall assume

that W ? Vi = 0 (i = 1, 2).

For every element Y1 ? W1 there exists only one such element Y2 ? W2 such that

(Y1 , Y2 ) ? W . We put ?(Y1 ) = Y2 . The mapping ? is the isomorphism of B modules

?

On subalgebras of the Lie algebra of the extended Poincar? group P (1, n)

e 157

W1 and W2 . In this case modules W1 and W2 have isomorphic composition factors.

The necessity is proved.

Let Wi be a left B submodule of the module Vi (i = 1, 2) and let the composition

factor W1 /N1 of the module W1 be isomorphic to the composition factor W2 /N2 of the

module W2 . We denote as W the vector space over the field R generated by the pairs

(Z1 , 0), (0, Z2 ), (Y1 , Y2 ), where Zi ? Ni , Yi ? Wi (i = 1, 2) and ?(Y1 + N1 ) = Y2 + N2

for the isomorphism ? : W1 /N1 > W2 /N2 . It is easy to see that W is a nonsplitting

B module. The sufficiency of the lemma is proved.

Let ? : X > X be the trivial representation of the completely reducible algebra

?

F ? AO(1, n), the projection of which onto AO(1, n) has not any invariant isotropic

subspaces in the space U or annuls the isotropic subspaces. Then ? is O(1, n) equiva-

lent to diag [?1 , . . . , ?m ], where ?i is an irreducible subrepresentation (i = 1, . . . , m).

One may suppose that algebra Fi = {diag [0, . . . , ?i (X), . . . , 0]|X ? F } is an irreducib-

?

le subalgebra AO(Wi ), where

(k0 = ?1, km = n, i = 1, . . . , m).

Wi = Pki?1 +1 , Pki?1 +2 , . . . , Pki

If Fi = 0 then we shall call algebra Fi an irreducible part of the algebra F . It is

well known that if representations ? and ? of the Lie algebra L by skew-symmetric

matrices are equivalent over R, then C · ?(X) · C ?1 = ? (X) for some orthogonal

matrix C (X ? L). Whence and from Proposition 3.1 we conclude that if ?i and

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

equality ?i (X) = ?j (X) takes place. Having united equivalent nonzero irreducible

subrepresentations we shall get a nonzero disjunctive primary subrepresentation of

?

the representation ?. Corresponding to those subalgebras of the algebra AO(1, n),

built by the same rule as the irreducible parts of Fi , we shall call them primary parts

of the algebra F . If F coincides with its primary part then F is called a primary

algebra.

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