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Completeness of irreducible representations

  1. Mar 24, 2008 #1
    "Completeness" of irreducible representations

    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,

    p.s. The argument is given, for example, on page 41 of Group Theory in Physics by Tung
  2. jcsd
  3. Mar 24, 2008 #2


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    Because D(g) is invertible.
  4. Mar 24, 2008 #3
    Thanks for the quick reply. Would you mind explaining in a little more detail?
  5. Mar 24, 2008 #4


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    What part don't you understand? Why it's invertible, or how that answers your question? Both should be straightforward applications of your definitions and basic computational facts.
  6. Mar 24, 2008 #5
    The inverse of the matrix D(g) is D(g^-1) since a representation is a homomorphism from the group to the space of matrices:

    D(g)*D(g^-1) = D(g*g^-1) = D(e) = 1

    Knowing this, though, how can you show it's impossible that each D(g) have a 0 in the same place?

  7. Mar 24, 2008 #6
    I've got it. The orthogonality theorem says that each "vector" is normalized. There's no way to normalize a zero vector so none of the vectors can be a zero vector.

    Out of curiosity, how were you thinking of showing this by saying the representation matrices are invertible?
  8. Mar 25, 2008 #7


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    That sounds right!

    Characterizations of invertibility -- e.g. a square matrix is invertible if and only if it has full rank if and only if it has nonzero determinant. Either of those two conditions are easy to show if one of the rows or columns is zero; do you see how?
  9. Mar 25, 2008 #8
    I think. But the matrix that has the zero row would be the matrix built out of the vectors, not the actual matrices of the irrep.

    i.e. if the (I,J) element of each D(g) is 0 then build a new matrix whose (i,j) element is:
    This matrix is unitary (therefore invertible) by the orthogonality relation, because each row of the matrix is one of the "vectors" I described in the first post and these vectors are orthonormal.

    But the I'th row of the matrix will be all zeros since the elements in this row are [D(g_j)]_IJ, which is zero for every g.

    I guess you could imagine an irreducible representation as a 3-dimensional matrix whose third "axis" corresponds to the different group elements. Then a "cross-section" which included that third axis would have a row of zeros.

    edit: actually the matrix I described above would not be square. but still, a rectangular matrix with a zero row is not invertible.
    Last edited: Mar 25, 2008
  10. Mar 25, 2008 #9
    This is certainly not enough. You need to know that the representation is irreducible because for the trivial representation, we can just have all the entries mapping into a 2x2 identity matrix, and then the 01 entry of each matrix is 0.

    I don't know the answer yet, but I'll think about it and post it if I figure it out.
  11. Mar 25, 2008 #10


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    Oh! I misread the original question. I thought it was fixed i, fixed g, variable j... not fixed i fixed j variable g. :frown: I feel silly now.
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