As n < m by supposition, this implies that as in (18.35), D would be decomposed
into representations whose dimension is smaller than m. It is in contradiction to
the supposition that D is irreducible. To avoid this contradiction, we must have
M ¼ 0. This makes (18.41) trivially hold.
(b) Next we consider a case of m < n. Taking a complex conjugate of (18.41) and
exchanging both sides, we get
e
D g
ð Þ
{ M
{
¼ M
{ D g
ð Þ
{ :
ð18:44Þ
Then, from (18.21) and (18.28) we have
D g
ð Þ
{ D g
ð Þ ¼ E and D g
ð ÞD g
À1
À Á ¼ D e
ð Þ:
ð18:45Þ
Therefore, we get
D g
À1
À Á ¼ D g
ð Þ
À1 ¼ D g
ð Þ
{ :
Similarly, we have
e
D g
À1
À Á ¼ e
D g
ð Þ
À1 ¼ e
D g
ð Þ
{ :
Thus, (18.44) is rewritten as
e
D g
À1
À Á
M
{
¼ M
{ D g
À1
À Á :
ð18:46Þ
The number of rows of M
{ is larger than that of columns. This time,
dimension n of e
D g
À1
ð Þ is larger than m of D(g
À1 ). A group element
g designates an arbitrary element of ℊ, so does g
À1 . Thus, (18.46) can be treated
in parallel with (18.41) where m > n with a (m, n) matrix M. Figure 18.2
graphically shows the magnitude relationship in representation dimensions
between representation matrices and M. Thus, in parallel with the above argument of (a), we exclude the case where m < n as well.
(c) We consider the third case of m ¼ n. Similarly as before we make a linear
combination of vectors contained in the basis set B ¼ ψ 1 , ψ 2 , Á Á Á, ψ m
f
g such that
ϕ ν ¼
X m
μ¼1
ψ μ M μν 1 ν m
ð
Þ ,
ð18:47Þ
where M is a (m, m) square matrix. If detM ¼ 0, then ϕ 1 , ϕ 2 , Á Á Á, and ϕ m are
linearly dependent (see Sect. 11.4). With the number of linearly independent
vectors p, we have p < m accordingly. As in (18.43) again, this implies that we
would have obtained a representation of a smaller dimension p for e
D , in
contradiction to the supposition that e
D is irreducible. To avoid this contradiction,
we must have detM 6 ¼ 0. These complete the proof.
∎
18.3 Schur’s Lemmas and Grand Orthogonality Theorem (GOT)
693