REPRESENTATION THEORY
For the groupS 3 of permutations on three objects, which has group multiplication ta-
ble 28.8 on p. 1055, with(in cycle notation)
I= (1)(2)(3),A=(123),B=(132
C= (1)(2 3),D= (3)(1 2),E= (2)(1 3),
use as the components of a basis vector the ordered letter triplets
u 1 ={PQR},u 2 ={QRP},u 3 ={RPQ},
u 4 ={PRQ},u 5 ={QPR},u 6 ={RQP}.
Generate a six-dimensional representationD={D(X)}of the group and confirm that the
representative matrices multiply according to table 28.8, e.g.
D(C)D(B)=D(E).It is immediate that the identity permutationI= (1)(2)(3) leaves alluiunchanged, i.e.
u′i=uifor alli. The representative matrixD(I) is thusI 6 ,the6×6 unit matrix.
We next takeXas the permutationA= (1 2 3) and, using (29.1), let it act on each of
the components of the basis vector:
u′ 1 =Au 1 =(123){PQR}={QRP}=u 2
u′ 2 =Au 2 =(123){QRP}={RPQ}=u 3
..
...
.
u′ 6 =Au 6 =(123){RQP}={QPR}=u 5.The matrixM(A) has to be such thatu′=M(A)u(here dots replace zeros to aid readability):
u′=
u 2
u 3
u 1
u 6
u 4
u 5
=
· 1 ····
·· 1 ···
1 ·····
····· 1
··· 1 ··
···· 1 ·
u 1
u 2
u 3
u 4
u 5
u 6
≡M(A)u.D(A)isthenequaltoMT(A).
The otherD(X) are determined in a similar way. In general, if
Xui=uj,
then[M(X)]ij= 1, leading to[D(X)]ji=1and[D(X)]jk=0fork=i. For example,
Cu 3 = (1)(23){RPQ}={RQP}=u 6implies that[D(C)] 63 =1and[D(C)] 6 k=0fork=1, 2 , 4 , 5 ,6. When calculated in full