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. 69
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on the well-known fact that the algebra AO(4) is decomposed into the direct sum of
two algebras AO(3). This is achieved by choosing the basis of AO(4) in the following
way:
1 1
±
Ja = ?abc Jbc ± Ja4 , (47)
2 2
?
where the indices a, b, c take the values 1, 2, 3. Due to (46) LVFs Ja , Ja ful?ll the
+

following commutation relations:
[Ja , Jb+ ] = ?abc Jc+ ,
+
(48)

[Ja , Jb? ] = 0,
+
(49)
On covariant realizations of the Euclid group 311

[Ja , Jb? ] = ?abc Jc? ,
?
(50)

which is the same as what was required. Now we are ready to formulate an assertion
giving an exhaustive description of LVFs (45) satisfying commutation relations (46)
or, equivalently, (48)–(50).
Theorem 3. Any realization of the Lie algebra AO(4) within the class of LVFs (45)
is given by the formulae (47) and by one of the formulae 1–6 presented below.

J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ,
+
1.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ,
+

J3 = ?u1 ,
+

?
J1 = ? sin u3 tan u4 ?u3 ? cos u3 ?u4 ,
?
J2 = ? cos u3 tan u4 ?u3 + sin u3 ?u4 ,
?
J3 = ?u3 ;
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ,
+
2.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ,
+

J3 = ?u1 ,
+

?
J1 = ? sin u3 tan u4 ?u3 ? cos u3 ?u4 ? sin u3 sec u4 ?u5 ,
?
J2 = ? cos u3 tan u4 ?u3 + sin u3 ?u4 ? cos u3 sec u4 ?u5 ,
?
J3 = ?u3 ;
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ? sin u1 sec u2 ?u3 ,
+
3.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ? cos u1 sec u2 ?u3 ,
+

J3 = ?u1 ,
+

?
J1 = sec u2 cos u3 ?u1 + sin u3 ?u2 ? tan u2 cos u3 ?u3 ,
?
J2 = ? sec u2 sin u3 ?u1 + cos u3 ?u2 + tan u2 sin u3 ?u3 ,
?
J3 = ?u3 ;
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ? sin u1 sec u2 ?u3 ,
+
4.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ? cos u1 sec u2 ?u3 ,
+

J3 = ?u1 ,
+

?
J1 = ? sin u4 tan u5 ?u4 ? cos u4 ?u5 ? sin u4 sec u5 ?u6 ,
?
J2 = ? cos u4 tan u5 ?u4 + sin u4 ?u5 ? cos u4 sec u5 ?u6 ,
?
J3 = ?u4 ;
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ? sin u1 sec u2 ?u3 ,
+
5.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ? cos u1 sec u2 ?u3 ,
+

J3 = ?u1 ,
+

?
J1 = k sin u4 sec u5 ?u3 ? sin u4 tan u5 ?u4 ? cos u4 ?u5 ,
?
J2 = k sin u4 sec u5 ?u3 ? cos u4 tan u5 ?u4 + sin u4 ?u5 ,
?
J3 = ?u4 ;
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ? sin u1 sec u2 ?u3 ,
+
6.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ? cos u1 sec u2 ?u3 ,
+
312 R.Z. Zhdanov, V.I. Lahno, W.I. Fushchych

J3 = ?u1 ,
+

?
J1 = u6 sin u4 sec u5 ?u3 ? sin u4 tan u5 ?u4 ? cos u4 ?u5 ,
?
J2 = u6 sin u4 sec u5 ?u3 ? cos u4 tan u5 ?u4 + sin u4 ?u5 ,
?
J3 = ?u4 ,
where k = const, k = 0.
Proof. We will give the principal steps of the proof omitting intermediate computa-
tions.
According to Theorem 1, there are two inequivalent realizations of the algebra
AO(3) with basis elements J1 , J2 , J3
+ + +


J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ,
+
1.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ,
+

J3 = ?u1 ;
+
(51)
J1 = ? sin u1 tan u2 ?u1 ? cos u1 ?u2 ? sin u1 sec u2 ?u3 ,
+
2.
J2 = ? cos u1 tan u2 ?u1 + sin u1 ?u2 ? cos u1 sec u2 ?u3 ,
+

J3 = ?u1 .
+


To complete a classi?cation of inequivalent realization of AO(4) we have to ?nd all
? ? ?
triplets of operators J1 , J2 , J3 which together with the operators (51) satisfy (49),
(50).
Analyzing the commutation relations (49) we arrive at the following expressions
? ? ?
for operators J1 , J2 , J3 :
n
?
Ja
1. = fai (u3 , . . . , un )?ui ,
i=3
3 n
?
Ja
2. = fab (u4 , . . . , un )Qb + fai (u4 , . . . , un )?ui ,
i=4
b=1

where fij are arbitrary smooth functions and
Q1 = sec u2 cos u3 ?u1 + sin u3 ?u2 ? tan u2 cos u3 ?u3 ,
Q2 = ? sec u2 sin u3 ?u1 + cos u3 ?u2 + tan u2 sin u3 ?u3 ,
Q3 = ?u3 .
Note that the operators Qa ful?ll the commutation relations of the algebra AO(3).
?
Hence, we conclude that for the case 1 from (51) the operators Ja are given by
the formulae (51), where one should replace ui by ui+2 , correspondingly.
Let us turn now to the second realization of the algebra AO(3) from (51).
?
Case 1. fai = 0, a = 1, 2, 3, i = 4, . . . , n. In this case we can reduce J1 to the
form
?
J1 = r(u4 , . . . , n)Q1
?
with the help of equivalence transformation
3
X > X = VXV ?1 ,
? V = exp Fa Qa (52)
,
a=1
On covariant realizations of the Euclid group 313

where Fa are some functions of u4 , . . . , un . Note that transformation (52) does not
change the form of the operators Ja , since [Ja , Qb ] = 0, a, b = 1, 2, 3.
+ +
?
From commutation relations (50) it follows that r = 1 and furthermore J2 = Q2 ,
?
? ?
J3 = Q3 . Thus we get the following forms of the operators Ja :
?
J1 = sec u2 cos u3 ?u1 + sin u3 ?u2 ? tan u2 cos u3 ?u3 ,
?
J2 = ? sec u2 sin u3 ?u1 + cos u3 ?u2 + tan u2 sin u3 ?u3 ,
?
J3 = ?u3 .
? ? ?
Case 2. Not all fai vanish. Then the operators J1 , J2 , J3 can be transformed
to become
?
Ja = fa (u4 , . . . , un )Q1 + ga (u4 , . . . , un )Q2 + ha (u4 , . . . , un )Q3 + Za ,
where a = 1, 2, 3, and
Z1 = ? sin u4 tan u5 ?u4 ? cos u4 ?u5 ? ? sin u4 sec u5 ?u6 ,
Z2 = ? cos u4 tan u5 ?u4 + sin u4 ?u5 ? ? cos u4 sec u5 ?u6 ,
Z3 = ?u4 ,
and ? = 0, 1.
?
Now using the transformation (52) we reduce the operator J3 to the form Z3 =
?u4 . Next, from commutation relations
? ? ? ? ? ?
[J3 , J1 ] = J2 , [J3 , J2 ] = ?J1
we get
3
?
J1 (Ga cos u4 + Ha sin u4 )Qa + Z1 ,
=
a=1
3
?
J2 = (Ha cos u4 ? Ga sin u4 )Qa + Z2 ,
a=1

where Ga , Ha are arbitrary smooth functions of u5 , . . . , un .
Making use of the equivalence transformation (52) with Fa being functions of
? ?
u5 , . . . , un we can cancel coe?cients Ga . The remaining commutation relation [J1 , J2 ]
?
= J3 yields equations for H1 , H2 , H3
Hau5 ? tan u5 Ha = 0,
whence
?
Ha = Ha sec u5 , a = 1, 2, 3,
?
?
Ha being arbitrary functions of u6 , . . . , un . Consequently, the operators Ja read
3
? ?
J1 sin u4 sec u5 Ha Qa + Z1 ,
=
a=1
3
? ?
J2 = cos u4 sec u5 Ha Qa + Z2 ,
a=1
?
J3 = Z3 .
314 R.Z. Zhdanov, V.I. Lahno, W.I. Fushchych

If ? = 1, then using the transformation (52) with Fa being functions of u6 , . . . , un
?
?
we can cancel Ha , thus getting Ja = Za , a = 1, 2, 3. If ? = 0, then making use of the
? ?
transformation (52) with Fa being functions of u6 , . . . , un we can put H1 = H2 = 0.
?
Provided H3 = 0, we get the realization which is reduced to that given by the
formulae 2 from the statement of the theorem.
? ?
Provided H3 = const = 0, we get the formulae 5. At last, if H3 = const, then
performing a proper change of variables we arrive at the realization given by the
formulae 6 from the statement of the theorem. The theorem is proved.

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( 70 .)



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