Bull. Kyushu lnst. Tech.
(M. & N. S.) No. 18. 1971
CONTRACTION OF LIE ALGEBRA OF THE MOTION GROUP OF THREE DIMENSIONAL EUCLIDEAN SPACE
By
Fumitake MiMuRA
(Receivid Oct. 26, 1970)
1. Introduction. The concept of contraction of Lie groups and Lie algebras was first introduced by segal [3] and was later discussed by In6nU and Wigner
[1] and Saletan [2]. The interesting examples of contraction in physics are that the Poincare algebra is contracted to the algebra of the inhogeneous Galilei group and the rotation algebra in three dimesions to the algebra of the Euclidean mo- tions in the plane. The contracting procedure is the following.
Consider a linear transformation Ut in a Lie algebra which depends on a pal- ameter t (O i.ltS..1) and define a new product called contracted Lie product, if exist, such as
(1.1) (X, Y).--lim U,-i(U,X, U,Y),
t-o
where U, is assumed to be nonsingular astlO. So long as U, is nonsingular the contracted Lie algebra is isomorphic to the original one. But if U, is singular we may obtain a new Lie algebra which is not isomorphic to the original one. In
[2], Saletan considered the contraction of Lie algebra G with respect to the linear transformation U, such as
(1.2) U,-tE+(1-t)u,
where E is a unit matrix and u is an arbitrary singular matrix, and he obtained the result that the necessary and suflELcient condition for (1.1) to exist is
(1.3) u2(.X, ]Y).-u(uX, IY).-u(X, u]Y).+(uX, u]Y).-=O,
and that the contracted Lie product is
(1.4) (X, IY).-=u-i(uX, ulY)R-u(X, Y)iv-l-(uX, Y)N+(X, UIY)Ar.
Here the subscription R and N of Lie product mean respectively the projections
into GR and G. defined by u"G-G. and u"G.--O (n-dim G). Let f(t) and g(t) be
arbitrary functions such as f(O) =O and g(O)==1, we can easily see by the same
way as Saletan that even if we take U,-=f(t)E+g(t)u in place of (1.2), the neces-
sary and suMcient condition for (1.1) to exist is condition (1.3), and then the
contracted Lie product is given by (1.4). And if A is a nonsingular matrix the
contracted algebras with respsct to U, and U,A are isomorphic, because
. (1.5) 1im(U,A)-i((U,A)X, (U,A) Y) =-A-ilimU,'i(U,(AX), U,(AY)).
t-o t-o Hence the contracted algebra with respect to U, -=f(t)AÅÄ• g(t)uA is isomorphic to the contracted algebra with respect to U, of the form (1.2).
The purpose of this paper is to find all types of singular matrices u of the form (1.2) which can contract the Lie algebra of the motion group of 3-dimen- sional Euclidean space, and to determine the contracted algebras whith respect
to u.
2. Lie subalgebras of Lie algebra of motion group. In this section we shall con- sider the Lie algebra G of motion group of 3-dimensional Euctidean space E,, and its Lie subalgebras. In [4], Stoka has been obtained the canonical forms of all Lie subalgebras of the Lie algebra G, i. e., he has been proved that with a suitable coordinate transformation y'=a;x`+bj (i, 1'=1, 2, 3) of E, the Lie subalgebra of G is transformed into one of the forms, i. e., the canonical forms:
Gl-[X,], Gi2-- [X,], Gia-[X, {-•X,],
(2,1) Gl === [X,, X,], G,2 =- [X,, X,,],
Gg--- [X,, X,, X,], Ge.-=- [X,, X,,, X,], Gi'::-- [X,, X,, X,],
G,,= [X,, X,,, X,, X,],
where X,=-aa-x-.-, (t-1, 2, 3), xs,i,-J,c't Da
x- ,. -Jxk aa
x, (h, k=-1, 2, 3) ; (XiL,-Xi, XL}:;==Xr, X:ii ==
X6). Xi, X2,..., X6 is a basis of Gsatisfymg the followmg relations of Liepreducts : (Xi, X2)-(Xi, X3)-=(X,, X,)[-: (.\i, X,)-- (X,, X6)-(X3, X4) :=::O,
(2.2) (Xi, X4)==-(X3, Xnr)=X2, (XL•, Xr,)=-(Xi, X6)==Xs', (X3, X6)=== ny (X2, X4)==Xi, (X4, X,.) = - X,, (X,, X6) = - X, , (X,, X,) =- - Xs.
LEMMA 2.1. The subalgebra containing GE is only G,, thesubalgebrascontainingGZ, are [-Ii, X2, pX3--l-qX4] or [Xi, X2, Xz, pX4-t--qXs+rX,,], the subalgebra containig G,l does not exist, the subalgebra containing G/7. is only [X,, X,, X,, X,] and the subalgebras containing Gg are [Xi, X2, X3, pX4+qXs+r"X'6].
PRooF. From (2.1) we may consider the tieubalgebra whose dimension is at most 4. Let H be any subalgebra of dimension 3 which contains GE.Wemay take X3, X,., and X-aXi+bX,+cl\s+dX, as a bas•is of H, Since ff contains (X,, X) =-diY,
-cX2 and (X4, X)----cXi-dX,, (c2+dL')X,EH and (c2+dZ)X,EH. If c2+d2 7kO, then Xi, X2 EH and hence dim H}.i..4; we have c===d==O i. e., aJ\,-i-bl\, E U. And since
(X4, aXi+b IY2)=bXi-aX2 E I]t/; (a2+bL)X, E H and (a2+b2)X, E H, vLTe have a--:b-.=.:O.
Thus we see that there is no subalgebra of dimen. sion 3 which contains G,i, Let Hbe any subalgebra of dimenson 4 which contains Ge, and let aiXi+biX2
+ciXs+diX6 and a2Xi+b2X2+c2Xs+d2X6 be two independent vectors of H, If cZ+d3 760 or cZ+dS 7E O, by the same way as the above, we have Xi, X2 E U. If ci--di :--
c,==d,=O, Hhas two independent vectors a,X,+btX2 and a2J\,+b,Xz,, and hen' ce
Contraciton of Lie Algebra of the Motion Group 19
X. ,, X, E H. Thus the subalgebra of dimension 4 which contains GS is only G,.
Let Hbe any .q.ubalgebra of dimension 3 which contains GZ. Let X-aX3+bX, +cX,,+clX, be any vector of H, since (Xi, X)=bX,-dX3 and (X,, X) ==-bXi+
cX,3, we have bX2-dX3 EH and bX,-cX3 E H. If c2+d2;O, X,, E H; i'f c2+d2== O, aX,
•"bX, E H. Thus the subalgebras of dimension 3 which contain GS are [Xi, X2, pX3
-I, • qX4 ] •
Let H be any ,q.ubalgebra of dimension 4 which contains GZ, and let a,X3+bi Xi+ciXf,+diXs and a2X3+b2X,i + d2X6 be two independent vectors of H. If c?+de ---i-4 o or cz+d3 # O, by the same way as the above, we have X3 E H. If ci-=di=--=-c2=- d2
==o, H has two independent vectors a,X3+b,X4 and a2X3+b2X4, and hence X3, X4 E H. Thus the subalgebras of dimension 4 which contain G3 are [Xi, X2, X3, pX4+
qXs + rXcj ] .
Let H be any subalgebra of dimension 4 which contains Gg. Let X=aX,+bX, +cX, be any vector of H. Since (X,, X)==bX,-aX2, (Xs, X)L-cX,-bX3, (X,, X)=- cX,+aX3, (X,, (X,, X)) -= -- aX,-bX2, (Xs, (Xs, X))--bX2-cX3 and (X,, (X,, X)) -= - aX, -cX3, we have (a2+b2)X, E H, (a2+c2)Xi E H, (a2+b2)X, E H, (b2+c2)X, E H, (b2 +c)X, EHand (a2+c2)X, E H. If a2+b2+c2fO, then Xi E H, H2EH and X3 EH and hence dim ff--6; we have a--b-c=-O. Thus the subalgebra which contains Gg
doe.g not exist.
Let H be any subalgebra of dimension 4 which contains GZ. Let X=aX,+ bX, +cX, be any vector of H, since (X,, X)=bX2 and (X3, X) -==-cX,, we have bX,,c X, E H. If b2+c2 =7k O, X, E1 ;if b--c-O, aX, EH i. e., X,EH. Thus the the subalgebra which contains GZ is only [Xi, X2, X3, X6]•
The last assertion of Lemma 2.1 is obvious, because Xi, X2, Xp, and pX4+qXs+
rX, form a basis• of a subspace of dimension 4 which contains Gg and then [X,, X2, X3, pX,i+qXr.+rX6] becomes a subgebra.
3. Some general properties of contractien.
PRoposiTioN 3.1, Let u be a singulalar matrix such as uGDu2GD... =) uPG===uP'iG ik [O]. Then for arbitrary choice of basis of G. there exists a basis of GA,, in terms of which
u can be written as u=== [6t8] where rank A=dim GR, rank C=dim GN, det A fO and cp==o.
PRooF. Let Xi, X2 ..., X. be a basis of GR=uP6 and X.+i, X.+2, ..., X. be vectors of G .Q.uch that X,, X,,,.., X. form a basis of G. Since u is nonsinglar on G., we have
uPXj-- cjiX,=uP(cji(u-i)PX,) ; i--1, ,.. , m, J'--1, ,.., n.
If we put Y,-X,-c,i(u-')PX,, Xi,...X., IY..i,..., Y. form a basis ofG andthen also
u can be written as u == [# 8], where rank A=dim GR, rank C==dim GN and det ASO, with respcet to this basis. Moreover uPY,--O, from which it follows that
CP==- O and D--BAP-i+CBAP'2+...+CP-2BA+CP-iB-O, Then we have
.
,
CD === CBAP-i + C2BAP-2 + . . . + CP-2BA2 + CP-iBA = O,
and hence
CDA-i == CBAP-2+ CZBAP-3+ ... + CP-2BA + CP-iB=O.
Therefore we have D= B•AP-i==O i. e., B== O. This completes the proof.
PRoposi'rioN 3.2. Let u be a matrix as in Proposition 3.1 and assume p=1. Then u
satisfies (1.3).
PRooF. From the Proposition 3.1 in this case we have u-- [6t 8], IIence uG.=
O. We know that uG becomes a subalgebra of G ([2], p. 5), and sinceuG-=G. we have (uG, uG)N--O. Hence we see that-(1.3) is satisfied.
We shall find the form of singular matrices u satisfying i(1.3) for each of following cases:
I. uG- u2G=/- [O] , II. uGDu2GD... =) uPG-= uP+iG /: [O] (p J/ 1), III. uPG- [O] .
Throughout the following sections, let X, be the basis (2.2) of the Lie algebra G of the motion group and Yj,, Yj,,... Yj, be the basis of GN chosen by Proposi- tion3.1 with respect to the basis Xi,, Xi,,..., Xi, of GR. And A and C denote the matrices in Proposition 3.1. It can be seen from the proof of Proposition 3,1 that . Yj, is con- gruent to .)krj. modulo G. (q-=1,,.., k).
4. Gase I. uG==u2G-/--[O]. In this case from the Proposition 3.2 u satisfies (1.3), and from (2.1) with a suitable linear transformation of variables in E, u
can be transformed into the following forms.: '
ooo a, oo ooo a, oo ooo a, oo
Ui == ooo a, oo, where a, =-/ O, u,=
ooo a, oo ooo a, oo
OO ai ai OO OO a2 a2 OO l OO a3 a3 OO
OO a, a, oo , Where a3+a,T74o, OOa,, a.. OO
OO a, a, OO
U3=
a, ooooo a, ooooo a, ooooo
a, ooooo,where a,=/-O, u,=
a, ooooo a, ooooo
oo a, b, oo oo a, b, oo oo a, b, oo , where oo a, b, oo OO as b.r OO oo a, b, oo
a3 b3 a4 b,
yrO,
U5==
a, b, Oooo a, b,oooo a, b, oooo , where a, b, Oooe a, b, oooo a, b, ooo oJ
ai bi
a2 b2 IO, U6=
OOO a, b, c, OOO a, b, c, OOO a, b, c, OOO a, b, c, OOO a, b, c,
O O O a, b, c,laJ
, where
a4 b4 c4 as bs cs a6 b6 c6
lo,
Contraction of Lie Algebra of the Motion GJoup 21
U7 ===
a, Ob, OO c, a, O b, OO c, a, O b, OO c,
, where a, O b, OO c, a, O b, OO cs a, O b, OO c,
U8="
ai bi ci a3 b3 c3 a6 b6 c6
iO, U8==
a, b, c, d, ooo a, b, c, d, O O o a, b, c, d, o o o a, b, c, d, o o o as b.r C.r ds OOO a, b, c, d, o o o
a, b, c, OOO a, b, c, O O O a, b, c, O O O a, b, c, O O O a, b, c, O O O a, b, c, O O O
, where
a, bi ci d, a2 b, c2 d, a3 b3 c3 d3 a, b, c, d,
, where
;-o.
ai bi ci a2 b2 c2 a3 b3 c3
to,
5. Case ll. uGDu3G),.. )u"G-=uP'iG=7i= [O] (pll). Since uPG-G. is a subal- gebra of G ([2], p. 5), G. is transformed into one of the canonical forms (2.1) with a suitable linear transformation of variables in E3. Hence throughout in this section we shall assume that GR is one of the forms (2.1).
(a). Let dim G.-1, then 2LÅq.=p=ÅqL-4 because the dimension of a subalgebra of G is at most 4. If X is a nonzero vector ef GR, uX==aX (alO). From (1.3) we have U(U (X, Y) .- (X, U Y) ,.) =- a(U (X, Y) N- (X, U Y) N) .
If we put Z==u(X, Y)A,-(X, uY)N, uZ==aZ, Since u`ZE G. and u`ZE G., u`Z-= a`Z-rO i. e., Z==O. Hence we have
(5.1) u(X, Y).-(X, uY).===O, for XEG. and any YEG.
Let G.--Gl. From (5.1) we haneu(Y,, YDN-(X,,uY,)N-O, from which it follows that
j-1; (a?+ab) Y,-(al-aZ) Y,+a: Y3+ (ag+a:) Ys- (af-aS) Y6 --O, J'-2; (al-ag) Y, + (a?+a}) Y,+a? Y,+ (a{-ag) Ys+ (ag+aS) Y,-O, j--3; ag Y,-ag Y,+ag Y,-ag Y, =-O,
J' -- 5 ; (aZ+ a}) Y, - (ag- ag) Y, + ag Y, + (ag+ ag) Y, - (ag-a2) Y, -- O, j-6 ; (ak-aZ) Y,+ (ag+ag) Y, +ag Y3 + (ag-a8) Y,+ (ag+ag) Y, ===O,
where uY,=a}•Yi+a?•Y,+a?•Y3+a?•Y,+a?•Y6. Hence we see that thematrix Cis of the form
A,1 O A,,
A,, O A,2.
Here the matrices A,, are of the form [mZ 2]. They form a field isomorphic to the
complex mumber field, Since rank C3$l.1 and C' -O, by computing the powers. of
matrix C we have
A,,3+A,,A,,(2A,,+A,,)=A,,3+A,,A,,(A,,+2A,,)=O
A,,(A. ,,2+A,,A,,+A,,A,,+A,,2) === A,, (A,,2+A,,A,,+A,,A,,+A,,2) =- O, (A,,2+,4. ,,A.,,,)2+A,,A,,(A. ,,+A,,)2== (A,,+A,,A, ,,)2+A,,A,,(A,,+A,,)2=O, A. ,, (A,, +A,,) (A,,2 + 2A,,A,, +A,,2) =A,, (A.,, + A,,) (A,,2 + 2A,,A,, + A,,2) ===. O.
Hence we have A,i+A,,,==Aii2+A,2A2t:=--O. And since uY:}-7=:ai,Y3,, we have a;',==O.
First assume that /!,, :=-:- O, then A,2A,,, =-,= O i. e., Ai2 --O or AL,, =- O. If A,ii =--O ; since u Yp, ==O
and C2-O, from (1.3) we have u(uY,, Y,,)N=O i. e., 2mnY,+ (m2--n2) Y, --O, where Ai2== [nynm Mn]. Hence we we have mn=m2-n2==O, from which it follows that C-.==.-O.
This contradicts to pJ-/•=1. If A,,===O; since uY,=uY2-uYr, =O and C2=O, we have
u( IYr, u Y6)rv +U(U Ys, ]Y6)Ar =m- (p2+t72) (ai IYi +a2 IY2) == (U IXr,, U Y6)N,
where A2i== [-:S] and Y,=X,+a,X,. (J'-1, 2, 3, 5, 6).We can easilyseethat (1.3) is satisfied by the other pairs of the basis X,, Y, (J'--1, 2, 3, 5, 6). IIence in this case u satisfies (1.3), and it is written as the following form with respect to the
basis X,,...,X, of G:
ooo a, o o"' ooo a, ooi ooo a, oo
UiO"= ooo a, o o i Where a4 -/ O and biny+c2/o.
J bcO a, OOf -cbO a, OO1
ttt
Next assume that A,,i/-O, Since uG.E [Y,, Y,, }l.., Y,], from (1.3) the coeflicient Y3 in (uYj,uY,). i.q. zero. Hence we have
an+bm-nq-mp==ap+bq==a2+b2-np-mq-O,
where Aii==[inba ba], Ai2== [mZi iZ] and A2i==-=[-: pq]. And since Ai]2+Ai,A2i--O, we have
a2-b2-np-mq-2ab+np-l-me-O.
From these equations we have a: b-=:-m: n--q: p and a2-mq-b2-np--O. Since A, iii740, we have m=-ka, n==kb, p--sb,e===-sa and ks--1. Since from (1.3) the coef- ficients of Ys and Y, in u(Y3, uY,). arezero, putting Y,=X,+a,X, we have k2ab==-k2 (a2-b2)=O. Hence we have k=O or a:T-b--O, but this contradicts to A,ifO or ks--1.
Thus in this case u does not exist.
Let GR:=-G?. From (5.1) we have
]' --1 ; (a?+af+aS) Y, + (al-ai'-a;) Y, +ag Y, + (ag+a;7) Y,- (a?-aS) Y, =- O,
1'-2; (al-a3-aS) Yi + (ai+aS+a2,) Y2 +a? Y3 + (ar-aS) Y, + (ag+aEJ) Y, -= O,
Contraction of Lie Algebra of the Motion Group 23
j-3; (aZ+al) Y,-(ag+air) Y,+ag Y,-ag' Y, =- O, .
J'-5; (a5+aft+ag+ae) Y,+(a3-ag-a;+a3) Y,
+ (a;+ a:) Y,, + (a: + ag -F ae7) Y., + (a:- a: + a2) Y,, =:--- O, 7-6 ; (al + a2- ag- ag) Y, + (a [' + ag + ag + ag) Y,
+ (ai+aS/) Y, + (a{+aP.-ag) Y,+ (ag+ag+aE) Y, -= O,
where uYj=al•Y,+a?•Y2+ai•Y3+a?•Y,+ag•Y6. From the first three equations of the ab- ove we have al-aZ-ar+aE--a?-a;=al.=ag::-aZ=a[j-O. Hence from the other three
equat ions of the above we have ai2, - a8 == ag + aar, :--=- aE =- ag' -- O. Th er efore the matrix C
is also of the form (5.2), and hence .q.imilarly we have Aii+A22===Aii2+Ai2A2i--O.
By the same way as the above; if Aii--A2i=:-O or Aii7- (--=O we have a contradiction, and if Aii=Ai2==O, u satisfies (1.3). It is- written a.T the following form with respect to the basis Xi,.,., X6 of G:
OO a, ai OO1 OO a2 a2 OO OO a, a3 OO
Uii= o o a, a, o o F Where a3+a4 I---O and b2+c27.!-o.
bc a, a, OO -cb a, a, OO
Let G.-G?. From (5.1) i. e., u(X,, Y,).-(X,, ulY,)N--O, it follows that 1'-2, 3, 5; a2•Y,-aSY3--O,
J'-4; (aZ-a2) Y2+ (a:+a2) Y,3+ag Y4+aE' Yj+afi2 Y6-O, J'-6; (ag+ag) IY,+(ai+ag) Y,+ag Y,,+ag/ Y.r+aS Y6-O,
wh ere u Y, -- a?• Y, + ...+a?• Y,. Hence we have aa=aE'=ag == af,=a2.=aS == ag. === aE =O and aZ- a2 =- ag+ aS -- aZ + ag =- a,, -ag• --- O. Th en since u Y, -- a?• \, +a?• Y, (i-2, 3) and u Y, -- aZ. Y, + ag' Y,
+agYs, we have u"Y.r=kY2+sY3+(aE)PY.. ; and since uPGN==O, we have aE---O. Moreover since uP Y, =- O and uY, == a?• Y2+a?• Y3 (j=2, 3), we have u2 Y,=u2 Y, -=O. Then since u2(Y,, IY,)A, -u(IY. uY..)-(uY,, uIY,)-:=O (J'--2, 3), from (1. 3) we'1 haveu(uIY,, IY,).-=O (j--2, 3)
i. e,,
a: (a; - ag) Y, + ( (a;• )2- a:aZ) Y, -O and (a:a3', - (aZ) 2) Y, + ag (a: - aZ) Y.. =- O.
, Hence we have a;=aZ. From this and u2 Y, =-= u2 Y3 -O we have u Y2 -== u Y3 === O, and th en a2 -a2=ag--ag--O i. e., uY, -= a2Y, + alY3+a[' Ys and uY6-aZY2+a:' Y3+agYs. Since uPY4-uPY6 ==o, by thesameway as the above we have a2=ag'=O. Thus we haveuY2===uY3=-O anduY,E[Y2, Y3] (1'--4, 5, 6). In this caseusatisfies (1.3), and u is written as
the following form with respect to the basis Xi,.,., X6 of G:
U12 ===
a, ooooo a, ooooo a, ooooo a, beOOO a, c fooo
a, d gO O Ov.
, where a-!,--O and b2+c2+...+g2740,
(b). Let dim 6R-=2. And let G.--GS, then from the Lemma 2,1 we have uG ==
G,== [Xi, X2, X3, X4] and u2G===GR. Hence we have u2G.=-=O and u6. c [Y,, }',]. The- refore from (1.3) we have u(uX3, Yj)N-(uX3, uYDN-:O i. e.,
J'-1; ag(a:+a}) Y,-ag(al-a:) Y,--O, J'-2; ag(al-a;) Y,+ag(a2+ai) Y, --O,
J'-5 ; (aSag+a4,ag. +agaL,) Y, + (agab3-atal. +ag,a?,) Y,-O, J'-6 ; (alag + agag- aEa:) Y, + (a?al + agal, + a3,a:i,) Y, -7- O,
where uX,-=a?•X3+af•X, (i-3, 4) and uY,--a}•Y,+a?Y, (j-1, 2, 5, 6). If ag:;O, from the first three equations of the above we have al-aZ==a?+aE-==O; then u2Y,==((al)2- (ar)2)Y,+2alaiY2-=O i. e,, al=- a?--aS-ag-O. Hence from the other equations of the above we have ag-ag===ag+ag===O. If a3• ==O, .q.ince u is nonsingular on G. we have ag .fO. Hence from the last three equations of the above we have al== a?==aS==aZ--O.
Then since u(X,, uY,),T==O, from (1.3) we have u(uX,, Y.r,)N-u(uX,, ul,)N =O i. e., aZ(ag+a}) Y,-a2(al.-ag) Y,-O, Since u is nonsingular on G., we have a2i7(=O; and then ag-ag-aZ.+aE=O. Therefore the matrix u must be uYi==uY2--O, uYs=pYi+qY2 and uY6==-qYi+pY2. This matrix u satisfies (1.3) ; because we have
U(U Ys, IYG)N+U(Yr,, UIY6)N= (P2+t7?) (bi IYi +b2 Y2) =" (U Ys ,U Y6)Ar, u(uXi, ]Ys) iv + u (Xi, u IYs)N === a: (q IYi -pIY2) ---- (uX,, u Y,.) ,, ; i-- 3, 4,
u(uXi, IY6)iv+u(Xi, uYc,)N=-aS•(plYi +qY2)=:- (uXi, u]Y6),v ; i-3, 4,
where Y,==Nj+a,X3+bjX, (J'--1, 2, 5, 6), and can easily see that (1.3) is satisfied by the other pairs of the basis X,, Y,. The matrix u is written as the following form with respect to the basis Xi,..., X6 of G:
U13=
oo a, b, oo oo a, b, oo oo a, b,, oo a3 b:3
O O a, b, o o , Where a, b, rr" O and c2+d2-fo.
cda, b, OO -dc a, b, OO
Let G.=GZ, then from the Lemma 2.1 we have uG== [Xi, X2, pX3+qX4] or uG==
[Xi, X2, X,,pX,+pXs+rX,]. First assume that uG== [X,, X,, pX3+qX,], then we have uY,--k,(pY3+gY,) (1'--3, 4, 5, 6). Since u2G==G., we have u2GN==O. And since (uG, uG).
.,.u(x,, uG).=O (i=1, 2), from (1.3) we have u(uXi, IYs)N-=u(uli, Y6)N ==O i. e., a}uY3
==a?•uY, ==O where uXi=al•Xi+a?X2 (i=1, 2). Since u is nonsingular on G., there exist
alXO ; hence we have uY, --O. Hence u2 Y,-- (k,)2q(pY,+qY,) ==O i. e., k,e==O. Therefore
we may put uY,=k,Y,. In this case we can easily see that u satisfies (1.3), and it
is written as the following form with respect to the basis Xi,..., X6 of G:
Contraction of Lie Algebra of the Motion Group 25
U14 ===
a, b, oooo a, b,oooo a, b, oooo
a, b, cO o O
a, b, dOoo
a6 b6 eOO Q..
, where
ai bi a2 b2
t-O and c2+d2+e21-O.
Next assume uG=- [X,, X2, X3, pX,+qX,+rX6], then we haveuY,=kjY3+s,(pY,+
qY,+rY6) (j=3, 4, 5, 6). Since (X,, Y,) - (uX,, Y,) =- O (i-1, 2), from (1.3) we have u(Xi, uIY3)N-(uXi, uIY3)N==O 1. e.,
i-- 1 ; s3 (k,r+ aiq-alr) Y3 + sgprY, + s3qr Y, + sgr2 Y, =O, i-2 ; s, (k3q-aZq+aSr) Y3 + s3pq Y, + sgq2 Y, + sgqr Y, === O,
where uX,==a:•Xi+a?•X2 (i--1, 2). If s3:if!O we have q==r==O. Hence we may put uyj=
k,Y,+s,Y,. Then since u(X,, uG)N==u(uXi, uG).-O (i--1, 2), from (1.3) we have u(uX,, IY,).,=u(uX,, IY..).=O,
u(uX,, IY,).-u2(X,, IY,).-nvO, u(uX,, IY,)N-u2(X,, Y,)N=O,
i. e., a?uY,=aÅruY, ===O and u2Y, == aluY3 -= a:uY3. Since s3V-O i. e., uY,:/:O; we have ai == a? == O, and from u'Y, == (al)'uY, ==(a:)PuY,=-O (p is some integer) we have al==
ag==O. This is a contradiction because u is nonsingular on GR. If s,=O we have
u Y3 == k, Y3, and hence uP Y3 - (k3)' Y3 =O i. e., k3 == O. Then since uP }ij -rm (s,)P Y, + (s,) P-ik, y,
=O, we have s,-O i. e. uY,=k, Y, (j=4, 5, 6). Hence we have uG c[Xi, X,, X,], but this contradicts to dinll uG=4.
(c). Let dim G.=-3. Int his case we see that dim uG-4, and that• G.--G; or G.==Gg because from the Lemma 2.1 there is no subalgebra containing Gl. Let G.
=-GZ,, then from the Lemma 2.1 we have uG== [Xi, X2, X3, X,]. Since u2G-=G., we have u2G.=-O. And since (uG, uG)N- (Xi, uG)N=-O (i--1, 3, 6), from (1.3) we have
u(uX,,IY,)N==u(Xi, Ys),v--•O i. e.,
a}•u Y2 -a9•u Y, -= a7,u Y2 -ae•u Y, ==- O, i-- 1, 3, 6 ;
where uX,-= a}•X,+a?X3+ag•X6 (i-1, 3, 6). Let Af• be the cofactor of the matrix
A-
[z"i':.i':ai•]
Then since u is nonsingular on G. i. e., detA:7LO, there exist Al•=7Åq:O and al• iftO. Hence from Al•;O we have uY, ==uY, =O and a9•uY, == O, and from a}:7iO we have uY, =O.
Therefore we have uG==G., but this contradicts to dim uG==4.
Let G,==Gg, then frorn the Lemma 2,1 we have uG=[Xi, X2, X3, pX4+eXs+rX6].
' '
Hence we have uY,=k,(pY,+eY,+rY,) i. e., the matrix C is the following form k4p k4q k4r
C=-=-, ksp ks-q k.rr . .k6p k6q k6r
Since u2G,, =O i. e., C2 =r-O, by computing the powers; of matrix C we have k,p+k,q +k,,r-==O, Therefore since u2GN=:= (uY,, uY,)=O (.i, lc--4, 5, 6), from (1.3) we have u(I.YJ. uY,),,+u(IY,, uiYh)N=-O i. e.
J'-4, k-5 ; r(lcZ + kg. + lt-g) (p Y, +gYs +r Y6) -- O, f- 4, k=- 6 ; e(kZ + lcg + kg) (p Y, +qY, +rY,) =- O, 1' -ww 5, k-6; p(k2+ke,+k?,) (pY4 +qY.r +rY6) =- O•
Since pY4 +ql s+ rY6 =/- O, we have k, = ks = lc c, =O; but this contradicts to dim uG= 4.
Thus in this case the matrix which satisfies (1.3) does not exist.
6. Cfise III. uPG=[O]. Since u6 is a subalgebi'a of G, we shall also assume that uG is the canonical form (2.1) as in section 5. Put uX,==afX,;then since Giei- O and G. :-G, by straightforward computations of (1.3) we have the follovkTings:
( 1 ) atuX,+aSuX2- (ar+aS)uX3- (uXi, uX2) -: O, (2) a2uX,- (a['+aS)uX, +al,u,X?- (uX,,, uN,)- O, (3) -(aS+a3)uX,+aE'uX2+a5'uXo,-(uX2, uX,i)--O,
(4) -(ai-a,`,)uX,+(a:-aS.:)uN,+afuX,-a2uX,,-(uX,,uX,,)=--TO, ( 5 ) (a:-ag)uX,- (a}-ag.')uX,-aE'uX, + aEttX,, - (uX,, uN,,) -- O, ( 6 ) - (aZ - aE') uX, + (aS - af,) u, X,, - aguX, + ai/uX, - (uX,, , uX,) -=- ]-- O,
( 7 ) -u2' X, -aiuX, + (al + a2) uX,, -a2uX3 - a2uX,, + a?uX, - (uXi , uXt)=:O, (8) u2X, .+ aguX, - (ag+aP,)uX, + aZuX,,+ aSuX•,-a3.uX,- (uX3, uXs) =O, ( 9 ) -u2X, - ag.uX, - a:uX, + (a3 + ag.)uX, + aSuX, -aSuX6 - (uX2, uXs) ==- O, (10) u2X, + aiuX, + aguX, - (al+ag')uX3 - a15uX, +atuX,., - (uXi , uX6) -- O, (11) - u2X, + (ah? + ag) ul, - agruX, - aguX, -- abruX, + ag,uXr, - (uX3 , uX6) =O, (12) u2X, - (al + a2) uX, + aEttX, + a['uX,, -aSuX,. + aE'uX, - (uX,,., uX, ) =:-r O,
(13) u?X, + ae,u.X, - (a2 + a:) u, X. , + a?,uX,, + afiuX, + ag.uX.. - (a2 + a/)uX6 - (uX, , uXr,) =O, (14) u?X, - (aZ + aZ)uX, + aguXL, + aJ,uN, + ai'uX, - (a2 + aE)uX, + aEuX6 - (uX,,, uX,) =- O, (15) u2X, + aguX, + aguX, - (a:. + aZ)uX, - (ag' + a8)uX, -I ag.uX.. +ag•uX6 - (uXF,, uX6) -O.
(a). Let dim uG=-1. In this case we have u2G-(uG, uG)-O. Let uG-61; then putting uX,=a,X,, frem (1) and (15) we have aiuXi+a2uX2==O and asuXs+a6uX6==O
i. e. (a,) 2X, + (a,) 2X, =O and (a,) 2X, + (a,) 2X, =- O. Hen. ce we have a,=a2 -- a, == a, :== O.
Since u2X,--(a,t)2X,=O, we have a,==O. Therefore we have uX,--O (i-1, 2, 4, 5, 6).
In this case we can see that u satisfies (1),,,., (15), aiid it is written as the fol-
)owing form ;
Contraction of Lie Algebra of the Motion Group 27
ooooo o- oooooo oooaoo
Ui,rm oooooo,wllerea/-O.
oooooo oooooo
Let uG==GZ then putting uX,=---a,Xi, from (11) and (12) we have a2uX2=--as•uX3-- O i. e., (a,) 2X, =- (a,) ZXi == O. Hence we have a2 ==-:- a3 ==T- O. Since u2Xi =- (ai) 2X, =- O, we have a, -= O. Th erefore we have tt Xi :== Lt X2 =-=- uN3 ;=-:- O. In this case we can see that u J.atislies (1),..., (15), and it is written as the following form:
U16 ===
ooooo o'
oooooo oooooo aooooo bOOOOO cooooo
, where a2+b2+c3-/LO.
Let uG--Gi, then putting uX,-a,(X,+X,), from (1) and (15) we have a,uX,+
a,uX,---O and a,uX,+a,uX,+a,uX,+a,uX,-=O i. e. ((ai)2+(a,)2)(X,+X,)-O and (a,a,+
a,a, + (a,) 2 + (a,) 2) (X3 + X4) -=- O. Hence we have a! ==: a, == a, == aG== O. Since u2 (X, + X4) =-==--
(a,+a,)L(X,,+X,)=-=-O, we have. a,+a,,=-:O. Therefore we have uX,=uX,==uX,=uX,--.
O, t{X3rm-a3(X3+X4) and uX4--ats(X3+X4). In this case we can see that u satisfies (1),..., (15), and it is vvritten as the following form:
ft7oo o ooo oo o ooo OO a aOO
Ui7 =`- o o -a -a o o , where a-fO.
oo o ooo
.
.-o o o oo o.
(b). Let dim u6-=:2. Let uG--=GS. Then putting u2Y,=--atlX3+ag•X,;since (uG, uG)-- O, from (1) we have atuX, + aSuX, =--O i. e., (aiat+aÅr'aS)X, + ((af)2+ (aa)2)X, --- O. Hence we have at==a3--O. Then from (2) and (3) we have aguXi=aguX2--o i. e., aiag.X,- a:aSX3-O. If ag-fO; a:'-=aS--O i, e., uX,=-uX,-==O, and hence from (8) and (11) we have aguX6-ag.uX,--O i. e., uX,--uX,=O. Hence from (9) and (15) we have u2X,.-.u2X,==
O i. e.,
[".51, a.'l12-:-io, i. e. [Zil ll" =--o
Therefore uX3 and uX, are linearly dependefit,. "IAhen we have u6--- [uX,l, but th.is,,
contradicts to dim uG =- 2. If ag -- O ; since u"X3 = (a k') 'X3 -= O, we have ag -O i. e,, uX3 ==
O. Hence (7) and (12) we have a?a2X,--a;a2X,-O. Assume a2=7kO, then we have ai=- ai-O i. e,, uXi-uX2-O. Hence from (14) and (13) we have u2X,-aauX..-O and u2 X,-a2uX,===O. Then since uPX,=(a2)P-iuX,==O and uPX,= (a2)P-iuX,=O, we have uX,
=:uX,==O; and hence we have uG-= [uX,]. But this contradicts to dim uG=2. The- refore a2-=O; then since u2X,===O, from (15) we have aguXi + aguX, +aguX,,+aguX6=O.
Th en since uX, -a?• X, (i -- 1, 2) , we have ((a2.)2+ (ag) 2) X, -;- O i. e,, ag -- ag -- O. Hence
from (9) and (10) we have aiuX,=-a;uX,=Oi. e., (a?)2X,==(ai)2X,--O. Hence we have ai-a;=O i. e,, uXi==uX2==O. Threfore we have uG=== [X,], but this contradicts to dim uG=2. Thus in this case u does not exist.
Let uG===Gg. Then putting ulY,--a}•X,+a?•X, ; since (uG, uG) --;O, from (4) and (5)
we have aiuX3-aSuX,=O. If uX3=/O we have a?=aS=O. Then sinceuPX,=--L(al)PX,=-O and u'X, -=(a:)"X,=-O, we have al-a:=--O i. e., uX,-=uX,-O. Hence from (8) and (11)
we have a5uX3===aZuX,=O, but this contradict•s to uX,-/--O. Therefore we have uX, -O; then from (8) and (11) we have u2X,=-u2X,--O, and hence from (7) and (12)
we have aZuX, - aluX, -- O and aZuX, - abu X, -= O i. e., al (a? - a5) X, + ((ai) 2- alaZ) X, === O and
(ala:-(aS)2)X,+a:(a?-a5)X,--O. Hence we have ar-al. Then since uPX,=-O (i-1, 2),
we have al == aZ -= aS == a: == O i. e., uX, == uX, == O. In this case we can see that u satisfi es (1),..., (15), and it is written as the following form:
ooooo oL'
oooooo oooooo uis= adoooO beoooo cfoooo
2
1
, where rank
a d) b e'
cf
'
==- 2.
(c) Let dim uG--3. Let uG-=GS. Then since uG is isomorphic to the 3-dimen- sional rotation algebra, uG has no 2-dimensional subalgebras. Hence the dimen- sion of the subalgebra u2G of the algebra uG is at most 1. Therefore since uX, E u2G (i=4, 5, 6), we shall put uX,=kulY, anduX,--suX,. Then putting uXi===af•X4+
a?X,+agX, (i-1, 2, 3, 4);since (uX. uX,,)--O (J', k-=4, 5, 6), from (13), (14) and (15)
we have
. uLX, + a2uX, + kaSuX, - (a2 + ka2)uX, - O, u2X.. + a2uX, - (a2 + sa2) uX.. + sa2uX, == O, u2X, - (ka[' + sa2) uX, + ka2uX,. + sa2uX6 =- O,
i. e., a2(1+k2+s2)uX,===O, a2(1+k2+s2)uX,==O and a2(1+k2+s2)uX,=O. Hence we have uX,=O and then uX..==uX,==O. Therefore from (1),..., (9) we have the following:
(1)' (al) 2+ (aS) 2- (arr+aS) ag+ (a?aS-a2asr)-O,
(2)' alai+aSa:- (a?+aS) ag+ (a?aS-ataS) -- O,
(3)' atag+aSaS- (a2+ag) ag+ (afaL7-arraS) -= O,
(4)' a2al- (ai'+ag) aS+aSag+ (afiag-al"ag) -= O,
(5), afia?- (ar"+ag) a8+agag+ (aiaS-a?al,) -O,
Contraction of Lie Algebra of the Motion Group 29
.
(6)' (a?)2- (ar+ag) ag+(ag)2+ (a2ag-a2ag') =- O, (7)' -(aS+ag) a2+aaaS+agag. +(aS'ag-aSag) -= O, (8)' -(ag+ag.)a?+ (a8)2+ (ag)2+ (aSag-aaaS) -O, (9)' -(aS+ag) a2+aZa2+ab5afi,+ (aSag -- agag) -O.
Frorn the above we have
" ai aS -(al'+aS)' 'a2 al' al Ag A.S' Afi•l) a2 -(ar"+aS) ag aS a: aS --l-- A: A: Agl:=--O,
-(aS+ag) aE' 'agr ag ag aS. Ai AI' A?)
where A; is the cofactor of the coefficient af of the matrix A-- [a;] (i---1, 2, 3; i==-:--
det A
a2 aS -(ai5+aS) Ag Ab5 Ag a? -(a?+ag) ag + A3 AS Ag
-(aS+ag) a: ag A2 A? A?
A2 AS Ag A2 AS A5• -=O.
Af AS Ag
Since dim uG=3 we see det A-/--O. IIence from the above we have at=ag, aa--ag and
ag -- as". Th en from (2) ' we have afa:-aSa2 -- O i. e., at : a? =- aS : aS ; and hence putting aS=--kal and aS-ka?, from (2)' we have al(ag-kartr)-O. If aY--O we see that a?-ka?, and that aS: a:: aS =- aS: a?: aS === ka2: ka l': ka?. If k=O, uX, and uX, are linearly dep en- dent; and if k==O, aS=a:=aS:=-=O i. e., uX,==O. Hence we have uG=[uX,,uX3],but this contradicts to dim uG=3. Therefore we have al=ag-=O, and then from (2)', (3)' and (5)' we have agaS-=a{ag-a?ag-O. If aS-"O we have a2-a2-O, i. e., uX, =-O, but this contradicts to diin uG===3. Hence we have aS-=ag=O. Since uX37(=O and ag==ag-=O, we have ag7AO; and then we have ag ==a:==O. Then from (1)' and (6)' we have (ar +a:)ag-a?aS--O and (ar+ag)aS-al'ag-O, from which we have aSag--O. Since ag:/:O, we have aS==O i. e., uX3==O; but this contradicts to dim uG=3. Thus in this case u does not exist.
Let uG===Gg. Then putting uX,== al•X,+a?•X,+a9•X, ; from (2) we have aiuXi+a3uX3 +(uXi, uX3)=-O. Then since (uX,, uX,) E [X,, X,], from which we have ((a?)2+
(ag) 2) X, -= O i. e., ag-ag-O. From (1) we have aq,uX, + (uXi , uX2) === O i. e. aS (ai+ ak) Xi -
aS(al-ag)X3-O. If aS =- O, we have uX, E [X,, X,] (i-1, 2, 3), and then from (13) we have ((a2)2+ (ag)2 + (ag)2)X,-O i. e. a2--ag--ag. Hence we have uGc[Xi, X3], but this contradicts to dim uG--3. Therefore we have aSi;O, and then al-a;- a?+a5-O. Since uPX, -=O (i=1, 2), from which we have al == a? -al-ag -O i. e., uX, ==uX3===O. Then from (7) we have u2X,=O i. e., a2tuX,=O; and since a:=7LO we have uX,=O. Therefore u must be uX,==uX,,=uX,===O, and in this case we can see that u satisfies (1), (2),..., (8), (10), (11), (14) and (15). From (9), (12) and (13)
we have
a:uX, - aguX, + (uN, , uX,) = O, aEuX, -,aguX, - (uX, , uX, ) = = O, (a2+al,)uX, -a2uX, -afi.uX, + (uX,, uXs) =O,
i. e.,
.
(a!a;' - aSal + agag' - aSa'2L) X, + ( (aZ) 2 - aSa2 - aSag. + a2ag) X, + aS (a; -- a2') X, =- O,
( (a}) 2 - aSag - aia2 + aSa2) X, + (aSai - a6,ai. + ala2 - aSal) X, + a9, (aS - ag) X, --- O, (aS (a2 + ag) - ala2- al.aE + a2ag - a2ag) X, + (ag (a2 + al.) - aEa2- agag - aaag + a2ag.) X, + (aS (aP, + ag) - (a2) 2- (ag.) 2) X, - O.
Since ag/O, from which we see that aL,-agi=a;'-a9,--O, and that
(ab) 2 - (a:) 2 - aS (aS, -- a2) , (al) 2 + (a;) 2 -=-- aE, (al -ai.) , 2aEaE' -- aS (al + ait/,) ,
2aEai - ai (al + a?.) , 2aZag -- aS (al + ag) .
We can see that the above equations are equivalent to the first four equations of them, and then we have ail -- cbZ, al. -=-: ca2 and al+al]-;--2abc where al-a, aZ=:b and aq,=---1/c. In this case tt satisfieG' (1),..., (15), and it is written as the follow-
ing form:
Ul9=
oo o oo o'!
aO b OOI/c
oo o ooo
d O cb •?• OO b
ca re O 2abc-d O O a
oo o ooo
1
!
, where abc / d.
The condition abc;d follows from dim uG--3.
Let uG-Gg. Then putting uX,-al•X,+a?•X,+a?X,; since (uG, uG)-O, from (5) and (6) we have a/2uX,-a!uX,-=O and a:,uX,-ag,uX,=-O. Assume that uX,-/=O (i-1, 2, 3). Then if a5ag-/LO we can put uX,L---kuXi and uX,,=-;=-suX,. IIence from the above
we have
a;uX, - aluX, -- k(al - sal) uX, = O, aZuX, - aluX, -=- s(a? -- kal)uX, -- O,
and then since uX,-O, we have a?-sal and ai -kal, Moreover since u"X, = (al+ka?
+ su3,)P-'uX, =-= O, we have al + kai + sai -- O ; and th en since al + ka2, + sai -- al (1 + k2 + s2) = O,
we have al=O and a?-ai-O. This contradicts to uX,--rZO. Therfore we have aSak
==O. If a5-O, we have a:uX,=-O i. e., a:--O. Hence we have uX2-=aZX,; then since uPX,--O, from which we have uX,-•==O. This contradicts to teX,=/;-O. If ak-O; we have aZ---O i. e., uX,-=agX3, then by the same way as the above we have uX,-O.
This contradicts to uJY,=--i.O. Therefore we have uXi=O or uX2==O or wX3=O. Let uX,-==O; then from (7) and (10) we have u2X,= u2X,==O, and hence 'from (8) and (9) we have aguX,-a",uX,--O and aiuX2-ae.uX3=::O i. e.,
(aga2' - (ag) i' ) X, + ag (a;- ag) X, -- O and aÅr' (a; - aZ) X, + ( (ag) 2- a3al ) X, - O.
Contraction of Lie Algebra of the Motion Group 31
Hence we have a;-ag -O, and then since u'X,-O (i--2, 3), we have aZ--a:=-ag=ag- O. By the same way as this;if •uX2-O from (9), (12), (10) and (11) we have uX,
==uX,==O, and if uX,=-O frdm (8), (11), (7) and (12) we have uX,=uX,===O. Therefore we have uXi===uX2==uX3=O. In this casewecan see thatusatisfies (1),..., (15), and it is written as the following form :
l-o oooo ol OOOOOOI
ooooo ol i' ad g' )i
U2o == adgo o ol, where rank b eh -== 3.
i• behoo ol .c fi.,
L-c fiOO OJ
(d). Let dim uG-=4. Let uG-G,. Then putting uXi--a}•Xi+a?•X2+a?•X3+a3•X4 ; from (1) we have aluX,+aSuX,-(uX!, uX,)==O, and then since (uXi uX,) E [X,, X,] we have ((al)2+(aS)2)X,--O i. e., al-=aS=O. Hence from (3) and (2) we have aguX,+(uX,,uX,,) -O and aguX,+(uX,, uX,)-=O i. e.,
ag (al- a:) X, + ag (al + aS) X, + agaiX, -- O, ag (a? + aS) X, - ag (al - a;) X, + aga;X, == O.
If ag-O; since uXi E [X,,X,, X,] (i-1, 2, 3) and (uG, uG)c[X,, X,], from (15) we have ((a2)2+(ag)2+(ag)2)X,===O i. e., a2-ag===ag--O. Hence we haveuGc[X,, X2, X3], but this contradicts to dim uG=4. Therefore ag:SO; then we have ag--ag-O and al-a:-al+
aS == O. Since uPX, == O (i- 1, 2) , we have al -= al === al == ag == O i. e., uX, -uX2 - O. Therefore from (9) we have u2X,==O i. e.,
(agag + agal) X, + (agag + agak) X, + ( (ag) 2 + aga2) .2r, + ag (ag + a2) X, -= O.
Since ag#O we have ag+at=O, and then we see
alat-agal-O, aZa2-agak-=O and (a2)Z+agai===O.
From the first and the last of the above we have a2(ala2--ala2)=-O. If at===O, since ag74=O we have al=-aZ==a?-O i. e., uX,-O; and if a27iO, we have alat-agag-O. The- refore we have ag:ag: ag: ag--al: aZ:a2: at i. e., uX3 and uX, are linearly dependent, and then we have uGc[uX,, uX,, uX,]. This contradicts to dim uG==4. Thus in
" this caseu does not exist.
7. Relations in u,. If there exists a nonsingular matrix A such as uA=v and if u and v satisfy (1.3), we shall call u is equivalent to v. We see that if we drop off the second condition of uio,..., ui4, then ui,..., us become special cases of uio,..., u,, respectively. Since there exist nonsingular matrices A and B such as
ro oo ooo ooo ooo ooo aoo
ooo a, oo1
ooo a, oo
ooo a, oo
ui7A= OOO -a OO , uiiB== OOo a, Oo L bco a, OO
ooo ooo
[-c bo a, ool ooo ooo
.
u,, is equivalent to the spesial case of u,,, and if a, /=O in u,,, u,, is equivalent to the special case of u,,. If a, ==O in u,,, we shall denote u,,=u',,. We know that the con- tracted algebra with respect to u such as uG-=u2Gf[O] is isomorphic to the In6ntt- Wigner contracted algebra ([2], pp. 7-8). This result can be seen as following.
Let X, X'E G. and Y, Y'E GN;then since uGN ==O, uG== GR and (uGR, GR) == (GR, GR)c GR, from (1.4) we have
(X, X').==-u-i(uX, uX'), (X, Y).----(uX, Y). and (Y, Y').=-=O.
Consider the linear transformation Fsuch as F=u on GR and F==identity on GN;
then since u is nonsingular on 6., we can see that F transforms isomorphically the above contracted products to the followings :
(7.1) (X, X').--(X, X'), (X, Y).---(X, Y)., and (Y, Y').=-O.
This is the contracted products with respect to v such as v=identity on G. and v
==O on G.. Therefore instead of u,, ..., ug we may take, v, , ..., v,, of the following forms respectively :
ooooool rm1 oooo os e'1ooooo
ooooool oooooo Ioioooo
OOOOOO OOIOOOi OOIOOO
I
Vt= OOO1OO,VL= OOOOOOI, Vi:= OOOOOO
i
OOOOIO OOOOOO OOOOOO
OOOOO1-, OOOOO 1-r.j [.O OOOO Or
'100000' iOIOOOO OOIOOO
, V4 =- oOO1OO
oooooo oooooo
There exists nonsingular matrix A such that u',,A is equal to
oo a, oooI ooa,ooo OO1OOO
V5='l
oooooo
[m2g".g'so,g
Then since v,X3==X3, vsYj--O (J'=1, 2, 4), and v,Y, E [Y,, Y,, X,] (k==5, 6) where Y,
==Xj-ajX3, Y4 =X4, Ys=Xs-(bai+ca2+as)X3 and Y6===X6+(cai-ba2-a6)X3,we can see that v, satisfies (1.3). Therefore ui, is equivalent to the special case of vs and u',,
is equivalent to v,. There exist nonsingular matrices Ai such as uioAi, ui3A2, ui2A3,
ui4A4 and uigAs are equal to the followings respectively:
Contraction of Lie Algebra of the Motion Group 33
i ooo a, oo 1 ooo a, oo ] ooo a, oo
Vf; --
ooo 1 OO
F
l' bco a, oo i -c bO a, OO
sV7=
j oo a, b,ool 'Ml oooo o'i
l oo a, b, ool
OOI O OOi
ooo 1 oor Vs -=
l cd a, b, OO L --dc a6 bG• OO,
a,oooooii a, oooool a, beOOO
a,cfOOO1 a,dgOOOI
V9L'
r'
1 O OOOO]
O1 OOOO
a, b, oooo a, b, cOoo a, b, dooo a6 b6 eOOO
L. ,"
, V]o=:=
[oo oooo loo oooi oo oooo 10 OOOO
OO -1 OOO
oo oooo
Since the above matrices are special cases of uie, ui3, ui2, ui4 and uig respectively, they satisfy (1.3). Therefore uio, ui3, ui2, ui4 and uig are equivalent to v6,..., vio re- spectively. Since even if we drop off the condition of ui6, uis and u2o, they satisfy
(1.3) ; u,6 and u,, become special case of u2e, i. e.,
Vl1M
oooooo oooooo oooooo adgOOO behooo cfiooo
8. Contracted Lie algebras. In this section.we shall compute the contracted Lie products with respect to each vi. By straightforward computations of (7.1), we can obtain the contracted products with respect to vi (i=1, ..., 4). We have that tvX3--X3,vY,-O (J'--1, 2, 4), (Yj-Xj-ajX3 (i-1, 2), IY,-X,,
if v==vs; {v Ys ==bYi+eY2, where SYs ==Xs-(bai+ca2+as)X3, Kv Y,= -cY, +b Y2, KY6 === X6+(cai-ba2-a6) X3, - tvX,-X,,vYj, ==O (j=1, 2, 3), tY,=-X,-a,X, (J'=1, 2, 3),
if v==v6; SvYs ==bY,+cY2, where SYs ==Xs-(bai+ca,+as)X,, Kv Y, === -cYi +bY2, XY6 =X6+(cai-ba2-a6)X4,
vXi=:X, (i-1, 2),vY,-O (j-1,2), `
if v=v7; {v Ys ==cYi +dY2, where
tv Y, == -dY, +cY,,
]Yj== Xj-ajX3-biX, (J'=1, 2),
Ys==Xs-(cai+da2+as)X3 -(cb,+dbz+bs)X4,
Y, === X6 + (da, - ca, -a6)x3
+ (db,-cb,-b,)X,,
if V=V8 ;
if v=r• vg;
vX, =-=X,, vIY,--O (1'-rm2, 3), v Y4 === b Y2 +eY3,
vYs -=cY, +fy,, where
vY6 ==dY2 +gY3,
Y,--X,-a,X, (i--2, 3), Y, = X4 - (ba2 + ea3+ a4)Xi , Ys =Xs -L (Ca2+fa3+as)Xi,
Y6=X6-(da2+ga3+a6)Xi, vxi-:Xi (i=1, 2),vl 3=:--O, IY3==X3-a3Xi-b3X2,
,V
yY l==SYylll where1Yy`,=m.Xx`,:((2ai,++a.1))Xx',1((Cdbb3,++bb`.?)Xx'l,
vY6 =- eY:,, Y6 =-= X6-(ea3+a6)X,-(eb3+b6) X2.
From the above, by straightforward computations of (1.4) we can obtain the con- tracted products vLrith respect to v, (i=5, ...,9). If v==v,o orv==vii; since GR=O and G,v--G, from (1.4) we have
(8.1) (X, Y).-(vX, Y)+(X, vY)-v(X, Y).
By straitforward computations of (8.1) we can obtain the contracted products with respect to vio and to v,,. Thus as the main result we have obtained the following theorem.
THEoREM. Mith a suitable linear transformation of variables in 3-dimensional Euclidean space Es•, the singular matrix u of the form (1.2) which can contract the Lie algebra ofmo- tion group of E3 is vi or v2 or... or v, (this is the I-PV cases) or is equivalent to themqtrix v.r or v6 or ... or vn whose contracted Lie algebra is the following:
1' vi
I.. "-Lrm the contracted products are given by (2.2)
(Xi ,X4)o i=::-(X3, Xs)o==X2, (Xl X6)e ==-X3, (X3, X6)e='Xi,
V2 (Xs, X6)e=:-'-X4, (X6, X4)e= keXs,
i other produets=o
V3
j V4
lt...--LL l
V5
V6
V7
all.of the products=:O
(Xl, X4)o==X2, (X2, X4)o=mXi, (X4, Xs)o=mX6, (X6, X4)t.,:==-Xs, other products=O
(X3, Ys).==-Y2, (X3, Y6).==Yi, (Ys, Y6).=-(bai+ca2)Yi+(ca1-ba2)Y2, other products=O
(Yi, Ys).==--(bai+ca2)Y2, (Y2, Y6).:=-(eai-ba2)Yl, (Yl, Y4).=Y2, (Y2, Ys).=(bai+ca2)Yi, (Yi, Y6),=:(cai-ba2)Y2,
(Ys, Y6).:==-2bY3-(baii-ca2)Ys+(cai-ba2)Y6,
1 other products=o
(Y,, Y,).=-(cb,+db,)Y,, (Y,, Y,). ==-(db,-cb,)Y,,
(Yi, Y4). =-(X3, Ys).--Y2, (Y2, Ys).==(Y2, Ys).=(cbi+db2)Yi, (Yi,Y6)e=(dbi-Cb2)Y2, (X3, Y6)o=-(Y2, X4)e==Yi, (X4, Ys).==-Y6, (Y6, X4).=-Ys,
(Ys, Y6). =-2cY3-(cai+da2)Yi+(dai-ca2)Y2-(cbi+db2)Ys-(dbi+cb2)Y6,
other products=O
Contraction of Lie Algebra of the Motion Group ' 35
. l
