REPRESENTATION THEORY
conjugacy classes, since any two elements ofGin the same class have the same
character set and therefore generate the same equation. For a unique solution
to simultaneous equations inNunknowns, exactlyNindependent equations are
needed. ThusNis also the number of classes, establishing the stated result.
Determine the irreps contained in the representation of the group 3 min the vector space
spanned by the functionsx^2 ,y^2 ,xy.We first note that although these functions are not orthogonal they form a basis set for a
representation, since they are linearly independent quadratic forms inxandyand any other
quadratic form can be written (uniquely) in terms of them. We must establish how they
transform under the symmetry operations of group 3m.Weneedtodosoonlyforarepre-
sentative element of each conjugacy class, and naturally we take the simplest in each case.
ThefirstclasscontainsonlyI(as always) and clearlyD(I)isthe3×3 unit matrix.
The second class contains the rotations,AandB, and we choose to findD(A). Since,
underA,
x→−1
2
x+√
3
2
y and y→−√
3
2
x−1
2
y,it follows that
x^2 →^14 x^2 −√
3
2 xy+3
4 y(^2) ,y (^2) → 3
4 x
(^2) +
√
3
2 xy+
1
4 y
(^2) (29.19)
and
xy →
√ 3
4 x
(^2) − 1
2 xy−
√ 3
4 y
(^2). (29.20)
HenceD(A) can be deduced and is given below.
The third and final class contains the reflections,C,DandE;oftheseCis much the
easiest to deal with. UnderC,x→−xandy→y,causingxyto change sign but leaving
x^2 andy^2 unaltered. The three matrices needed are thus
D(I)=I 3 , D(C)=
10 0
01 0
00 − 1
, D(A)=
1
43
4 −√
3
2
3
41
4√
3
√^2
3
4 −√
3
4 −1
2
;
their traces are respectively 3, 1 and 0.
It should be noticed that much more work has been done here than is necessary, since
the traces can be computed immediately from the effects of the symmetry operations on the
basis functions. All that is needed is the weight of each basis function in the transformed
expression for that function; these are clearly 1, 1, 1 forI,and^14 ,^14 ,−^12 forA, from (29.19)
and (29.20), and 1, 1,−1forC, from the observations made just above the displayed
matrices. The traces are then the sums of these weights. The off-diagonal elements of the
matrices need not be found, nor need the matrices be written out.
From (29.17) we now need to find a superposition of the characters of the irreps that
gives representationDin the bottom line of table 29.2.
By inspection it is obvious thatD=A 1 ⊕E, but we can use (29.18) formally:
mA 1 =^16 [1(1)(3) + 2(1)(0) + 3(1)(1)] = 1,
mA 2 =^16 [1(1)(3) + 2(1)(0) + 3(−1)(1)] = 0,
mE=^16 [1(2)(3) + 2(−1)(0) + 3(0)(1)] = 1.
Thus A 1 and E appear once each in the reduction ofD,andA 2 not at all. Table 29.1
gives the further information, not needed here, that it is the combinationx^2 +y^2 that
transforms as a one-dimensional irrep and the pair (x^2 −y^2 , 2 xy)thatformsabasisof
the two-dimensional irrep, E.