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. 38
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+
? ?
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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. 38
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