(ii) Let P and Q be two square matrices. Then, we have
Tr PQ
ð Þ ¼ Tr QP
ð Þ:
ð18:70Þ
This is because
X
i
PQ
ð Þ ii ¼
X
i
X
j
P ij Q ji ¼
X
j
X
i
Q ji P ij ¼
X
j
QP
ð Þ jj :
ð18:71Þ
Putting Q ¼ SP
À1 in (18.70), we get
Tr PSP
À1
À
Á ¼ Tr SP
À1 P
À
Á ¼ Tr S
ð Þ:
ð18:72Þ
Therefore, we have the following property:
(iii) Characters of group elements that are conjugate to each other are equal. If g i
and g j are conjugate, these elements are connected by means of a suitable
element g such that
gg i g
À1
¼ g j :
ð18:73Þ
Accordingly, a representation matrix is expressed as
D g
ð ÞD g i
ð ÞD g
À1
À Á ¼ D g
ð ÞD g i
ð Þ D g
ð Þ
½
Š
À1 ¼ D g j
À Á :
ð18:74Þ
Taking a trace of both sides of (18.74), we have
χ g i
ð Þ ¼ χ g j
À Á :
ð18:75Þ
(iv) Any two equivalent representations have the same trace. This immediately
follows from (18.30).
There are several orthogonality theorem about a trace. Among them, the
following theorem is well known.
Theorem 18.4 A trace of irreducible representations satisfies the following orthogonality relation:
X
g
χ
α
ð Þ g
ð Þ
à χ
β
ð Þ g
ð Þ ¼ nδ αβ ,
ð18:76Þ
where χ
(α) and χ
(β) are traces of irreducible representations D
(α) and D
(β) ,
respectively.
Proof In (18.51) putting i ¼ j and k ¼ l in both sides and summing over all i and k,
we have
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18 Representation Theory of Groups