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The generators of the algebra AG4 (n ? 1) satisfy the following commutation rela-
tions:
[Jab , Jcd ] = gad Jbc + gbc Jad ? gac Jbd ? gbd Jac , [Ga , Jbc ] = gab Gc ? gac Gb ,
[Pa , Jbc ] = gab Pc ? gac Pb , [Ga , Gb ] = 0, [Pa , Gb ] = ?ab M, [Ga , M ] = 0,
[Pa , M ] = 0, [Jab , M ] = 0, [R, S] = 2S, [R, T ] = ?2T, [T, S] = R,
[Z, R] = [Z, S] = [Z, T ] = [Z, Jab ] = 0, [R, Ga ] = Ga , [R, Pa ] = ?Pa ,
[R, M ] = 0, [R, Jab ] = 0, [S, Ga ] = 0, [S, Pa ] = ?Ga , [S, M ] = 0,
[S, Jab ] = 0, [T, Ga ] = Pa , [T, Pa ] = 0, [T, M ] = 0, [T, Jab ] = 0,
[Z, Ga ] = ?Ga , [Z, Pa ] = ?Pa , [Z, M ] = ?2M,
with a, b, c, d = 1, . . . , n ? 1.
From these commutation relations we ?nd that

R, S, T = ASL(2, R), R, S, T ? Z = AGL(2, R),

where R denotes the ?eld of real numbers.
Let F be a reducible subalgebra of AO(2, n + 1). That is, there exists in R2,n+1
a nontrivial subspace W which is invariant under F . If W is isotropic, then there
exists a totally isotropic subspace W0 ? W which is invariant under F . Since dim W0
is 1 or 2, then, by Witt’s theorem [14] there exists an isometry C ? O(2, n + 1) such
that CW0 is either Q1 + Qn+3 or Q1 + Qn+3 , Q2 + Qn+2 . Taking into account
that the matrices (3) do not change these subspaces and represent all the components
of the group O(2, n + 1) di?erent from the identity component O1 (2, n + 1), then
we may assume that the above C lies in O1 (2, n + 1), the identity component. Thus
there exists an inner automorphism ? of the algebra AO(2, n + 1) such that either
?
?(F ) ? AP (1, n) or ?(F ) ? AG4 (n ? 1).
If W is a nondegenerate subspace, then, by Witt’s theorem, it is isometric with
one of the following subspaces: R1,k (k ? 2), R2,k (k ? 1), Rk (k ? 1). Each of the
isometrics (3) leaves invariant each of these subspaces, so that we may assume that the
224 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

isometry which maps W onto one of these subspaces belongs to O1 (2, n+1). From this,
it follows that a subalgebra F is conjugate under the group of inner automorphisms
of the algebra AO(2, n + 1) to a subalgebra of one of the following algebras:

AO (1, k) ? AO (1, n ? k + 1),
(1)
where AO (1, k) = ?ab : a, b = 1, 3, . . . , k + 2 and
AO (1, n ? k + 1) = ?ab : a, b = 2, k + 3, . . . , n + 3 with n ? 3
and k = 2, . . . , [(n + 1)/2];
AO(2, k) ? AO(n ? k + 1), where
(2)
AO(n ? k + 1) = ?ab : a, b = k + 3, . . . , n + 3 with k = 0, 1, . . . , n.

In order to classify the subalgebras of these direct sums it is necessary to know
the irreducible subalgebras of algebras of the type AO(1, m) (m ? 2) and AO(2, m)
(m ? 3). It has been shown in Ref. [15] that AO(1, m) has no irreducible subalgebras
di?erent from AO(1, m). In Refs. [16] and [17] it has been shown that every semisimple
irreducible subalgebra of AO(2, m) (m ? 3) can be mapped by an automorphism of
this algebra onto one of the following algebras:

(1) AO(2, m);
ASU (1, (m/2)] when m is even;
(2)
v v
?12 + 3?13 + ?25 , ??15 + ?24 ? 3?23 , ?12 ? 2?45 when m = 3.
(3)

It follows then that when m > 3 is odd, the algebra AO(2, m) has no irreducible
subalgebras other than AO(2, m). If m = 2k and k ? 2, then, up to inner automor-
phisms, AO(2, m) has two nontrivial maximal irreducible subalgebras: ASU (l, k) ?
Y , and ASU (l, k) ? Y , where

Y = diag [J, ?J, J . . . , J]
Y = diag [J, . . . , J],

with

0 ?1
J= .
10

We note that a subalgebra L of AG4 (n?1) is conjugate under Ad AO(2, n+1) with
?
a subalgebra the algebra AP (1, n) if and only if the projection of L onto AGL(2, R) =
R, S, T ? Z is conjugate under Ad AGL(2, R) with a subalgebra of the algebra
R, T, Z .

Conjugacy under Ad AP (1, n) of subalgebras
3
of the Poincar? algebra AP (1, n)
e
The Poincar? group P (1, n) is the multiplicative group of matrices
e

? Y
,
0 1
On the classi?cation of subalgebras of the conformal algebra 225

where ? ? O(1, n) and Y ? Rn+1 . Let Iab , a, b = 0, 1, . . . , n + 1 be the (n + 2) ? (n + 2)
matrix whose entries are all zero except for the ab-entry, which is unity. Then a basis
for AP (1, n) is given by the matrices

J0a = ?I0a ? I0a , Jab = ?Iab + Iba , P0 = I0,n+1 , Pa = Ia,n+1 ,

with a < b; a, b = 1, . . . , n. These basis elements obey the commutation relations (1).
It is sometimes useful in calculations to identify elements of AO(1, n) with matrices
of the form
? ?
?02 · · · ?0n
0 ?01
? ?01 ?12 · · · ?1n ?
0
? ?
X = ? ?02 ??12 · · · ?2n ?
0
? ?
?· ·?
· · ·
?0n ??1n ??2n · · · 0

and elements of the space U = P0 , . . . , Pn are represented by n + 1-dimensional
columns Y . In this case, we take
?? ? ? ? ?
1 0 0
?0? ?1 ? ? ?
0
?? ? ? ? ?
P0 = ? . ? , P1 = ? . ?, . . . , Pn = ? ?
.
?.? ?. ? ? ?
.
. . .
0 0 1

and with this notation it is easy to see that [X, Y ] = XY . We endow the space U
with the metric of the pseudo-Euclidean space R1,n , so that the inner product of two
vectors
? ? ? ?
x0 y0
? x1 ? ? y1 ?
? ? ? ?
? . ?, ? . ?
?.? ?.?
. .
xn yn

is x0 y0 ? x1 y1 ? · · · ? xn yn . The projection of AP (1, n) onto AO(1, n) is denoted
by ?. We also note that AO(n), contained in AO(1, n), is generated by Jab (a < b;
?
a, b = 1, . . . , n).
Let B be a Lie subalgebra of the algebra AO(1, n) which has no invariant isotropic
subspaces in R1,n . Then B is conjugate under Ad AO(1, n) to a subalgebra of AO(n)
or to AO(1, k) ? C, where k ? 2 and C is a subalgebra of the orthogonal algebra
AO (n ? k) generated by the matrices Jab (a, b = k + 1, . . . , n). In the ?rst case, B is
not conjugate to any subalgebra of AO(n ? 1).
Proposition 1. Let B be a subalgebra of AO(n) which is not conjugate to a subalgebra
of AO(n ? 1). If L is a subalgebra of AP (1, n) and ?(L) = B, then L is conjugate to
?
an algebra W C, where W is a subalgebra of P1 , . . . , Pn , and C is a subalgebra of
B ? P0 . Two subalgebras W1 C1 and W2 C2 of this type are conjugate to each
other under Ad AP (1, n) if and only if they are conjugate under Ad AO(n).
Proof. The algebra B is a completely reducible algebra of linear transformations of
the space U and annuls only the subspace P0 (other than the null subspace itself).
Thus, by Theorem 1.5.3 [9], the algebra L is conjugate to an algebra of the form
226 L.F. Barannyk, P. Basarab-Horwath, W.I. Fushchych

W C where W ? P1 , . . . Pn and C ? B ? P0 . Now let W1 C1 , and W2 C2 be of
this form, conjugate under Ad AP (1, n). Then there exists a matrix ? ? P1 (1, n) such
that ?? (W1 C1 ) = W2 C2 , and from this it follows that ?? (B1 ) = B2 for some
? ? O1 (1, n). Let V = P1 , . . . , Pn . Since [B1 , V ] = V , then [B2 , ?? (V )] = ?? (V )
and ?? (V ) = V . Thus we can assume that ? = diag [1, ?1 ] where ?1 ? SO(n), so
that the given algebras are conjugate under Ad AO(n). The converse is obvious.
Proposition 2. Let B = AO(1, k) ? C, where k ? 2 and C ? AO (n ? k). If L
is a subalgebra of AP (1, n) and ?(L) = B then L is conjugate to L1 ? L2 where
?
L1 = AO(1, k) or L1 = AP (1, k), and L2 is a subalgebra of the Euclidean algebra
AE (n ? k) with basis Pa , Jab (a, b = k + 1, . . . , n). Two subalgebras of this form,
L1 ? L2 and L1 ? L2 are conjugate under Ad AP (1, n) if and only if L1 = L1 and L2
is conjugate to L2 under the group of E (n ? k)-automorphisms.
Proof. The proof is as in the proof of Proposition 1.
Lemma 1. If C ? O(1, n) and C(P0 + Pn ) = ?(P0 + Pn ) then ? = 0 and
? ?
1 + ?2 (1 + v 2 ) ?1 + ?2 (1 ? v 2 )
t
?v B
? ?
2? 2?
? ?
? ?,
v ?v
B (5)
C=? ?
? ?1 + ?2 (1 + v 2 ) ?
1 + ?2 (1 ? v 2 )
?v t B
2? 2?
where B ? B(n?1), v is an (n?1)-dimensional column vector, v 2 is the scalar square
of v and v t is the transpose of v. Conversely, every matrix C of this form satis?es
C(P0 + Pn ) = ?(P0 + Pn ).
Proof. Proof is by direct calculation.
Lemma 2. Let C ? O(1, n) have the form (5), with ? > 0. Then

C = diag [1, B, 1] exp[(? ln ?)J0n ] exp(??1 G1 ? · · · ? ?n?1 Gn?1 ),

where Ga = J0a ? Jan and
? ?
?1
?.? ?1
? . ? = B v.
.
?n?1

Proof. Direct calculation gives us
? ?
cosh ? 0 sinh ?
exp(??J0n ) = ? ?
0 En?1 0
sinh ? 0 cosh ?

and
? ?
b2 b2
b t
1+
? ?
2 2
? ?
exp(??1 G1 ? · · · ? ?n?1 Gn?1 ) = ? ?,
b ?b
En?1
? ?
? ?
b b2
2
b 1?
t
2 2
On the classi?cation of subalgebras of the conformal algebra 227

where b = (?1 , . . . , ?n?1 )t . On putting ? exp ? we have

?2 ? 1
?2 + 1
cosh ? = , sinh ? = .
2? 2?
Since we have
?2 ?? ?
?2 ? 1 b2 b2
? +1
b t
0 1+
? 2? 2? ? ? ?
2 2
? ?? ?
? ?? ?=
b ?b
0 En?1 0 En?1
? ?? ?
? ?2 ? 1 ?2 + 1 ? ? b2 ?
b2
b 1?
t
0
2? 2? 2 ?2
?
1 + ? (1 + b ) ?1 + ? (1 ? b )
2 2
2 2
?bt
? ?
2? 2?
? ?
=? ?,
b ?b
En?1
? ?
? ?
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