The little orthogonality theorem consists of two relations that are reduced forms of the mathematical expression for the great orthogonality theorem.
1st little orthogonality relation
The 1st relation is derived from eq21 by letting and
, resulting in
Since , we have
and
Question
Show that .
Answer
Since the traces of a matrices of the same class are the same, eq22 is equivalent to
where
-
is the number of classes.
is the number of elements in the
-th class.
, called the character of a class, is the trace of a matrix belonging to the
-th class of the
-th irreducible representation.
Eq23 is known as the 1st little orthogonality relation.
Question
Determine whether the representations and
of the
point group are reducible or irreducible.
Answer
Since the 1st little orthogonality relation is derived from the great orthogonality theorem, which pertains only to irreducible representations, characters of a representation that do not satisfy the relation belong to a reducible representation. Using eq23,
Therefore, is an irreducible representation of the
point group, while
is a reducible representation of the
point group.
Let’s rewrite eq23 as
Eq24 has the form of the inner product of two vectors and
in a
-dimensional vector space, with components
and
respectively. The components are functions of
and the two vectors are orthogonal to each other when
. A
-dimensional vector space is spanned by
orthogonal vectors, each with
components. Since the number of vectors and the number of components of each vector are denoted by
and
respectively, the number of irreducible representations of a group is equal to the number of classes of that group. We say that the irreducible representations of a group form a complete set of basis vectors in the
-dimensional vector space, with the components of each basis vector being
.
2nd little orthogonality relation
Consider the matrices and
with entries
and
respectively.
Both are square matrices because the number of irreducible representations of a group is equal to the number of classes of that group. Comparing with eq24, the entries of the matrix product are given by:
Since and
are square matrices and
, then
, i.e.
Eq24b is the 2nd little orthogonality relation.
Question
Show that if and
are square matrices and
, then
.
Answer
Using the determinant identities of and
, we have
, which implies that
and
are not zero and are therefore non-singular. So,