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## Main Question or Discussion Point

**"Completeness" of irreducible representations**

Hi,

For a finite group of order n each irreducible representation consists of n matrices [D(g)], one for each element in the group. For a given row and column (e.g. i,j) you can form an n-dimensional vector by taking the ij element of D(g) for each g. i.e. the vector:

{ [D(g1)]_ij, [D(g2)]_ij, ... , [D(gn)]_ij }

The orthogonality relations say that the vectors corresponding to different pairs ij are orthonormal and vectors formed from two different irreps are orthogonal.

Textbooks give an argument like:

Count the number of orthogonal vectors: If the dimension of the irrep is k then there are k^2 orthogonal vectors formed by all the combinations of i and j. Each different irrep adds k^2 vectors all orthogonal to each other and orthogonal to the vectors from the other irreps (k can be different for the different irreps). So we have in total k1^2 + k2^2 + ... + kp^2 different orthogonal vectors (where k1, k2, ... kp are the dimensions of the different irreps.) But there can only be at most n orthogonal n-dimensional vectors. Therefore, k1^2 + k2^2 + ... + kp^2 <= n.

My question is:

What happens if, in an irrep, all the entries are 0 for some particular row and column? i.e. for some i and j, [D(g)]_ij = 0 for every g. Then you should not count this vector in your list of orthogonal vectors.

That is, in your list of k^2 orthogonal vectors you might have some 0 vectors. This seems to break the proof.

How do you show that there is never such a situation where all the matrices of an irrep have a 0 in the ith row and jth column??

Thanks a lot,

Alex

p.s. The argument is given, for example, on page 41 of Group Theory in Physics by Tung