The great orthogonality theorem establishes the orthogonal relation between entries of matrices of irreducible representations of a group. Mathematically, it is expressed as

where

The proof of eq14 involves analysing two cases and then combining the results. Consider the matrix

where is an matrix associated with representation , is and matrix associated with representation and is an arbitrary matrix with rows and columns.

Multiplying eq15 on the left by some matrix associated with representation ,

Using the matrix identity ,

###### Question

Proof the matrix identity .

###### Answer

and so .

According to the closure property of a group, and and thus

__Case 1__: .

Eq15 and eq16 becomes and respectively (we have changed the dummy index from to in eq16). As every element of a representation can undergo a similarity transformation to a unitary matrix, we shall assume and its inverse are unitary matrices. According to Schur’s first lemma, eq16 implies that and eq15 becomes

is a similarity transformation, where is similar to some other arbitrary matrix. Since the traces of similar matrices are the same,

where is the identity matrix’s dimension, which is equal to the dimension of the matrix .

Substitute eq18 in eq17, we have , or in terms of matrix entries,

The RHS of the above equation is a finite summation of the product of three scalars and their order can be changed. So,

With ,

Since is an arbitrary matrix, the above equation must satisfy any. This is only possible if

Since is unitary,

__Case 2__: and is not equivalent to .

According to Schur’s second lemma, eq15 becomes , or in terms of matrix entries,

Since is an arbitrary matrix, the above equation must satisfy any . This is only possible if

Since is unitary,

Combining eq19 and eq20, we have the expression for the great orthogonality theorem:

which can also be expressed as

because the RHS vanishes if , which renders the subscript unnecessary for .

###### Question

What about the case where and is equivalent to ?

###### Answer

In this case, can undergo a similarity transformation to become , which in turn can undergo a similarity transformation to become elements of a unitary representation. This implies that both and can be expressed as the same unitary representation because if is similar to and is similar to , then is similar to . In other words, we have (where is unitary), which is case 1.

Let’s rewrite eq20b as

Eq21 has the form of the inner product of two vectors, where and are components of vectors and respectively in a -dimensional vector space. We can regard the components of the vectors as a function of three indices , and such that the two vectors are orthogonal to each other when . This orthogonal relation of matrix entries is why eq21 is called the great orthogonality theorem.

###### Question

Verify eq21 for

i) ,

ii) with , and

iii) , with .

###### Answer

i)

ii)

iii)