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How many independent components has a four-dimensional fully antisymmetric tensor?

  1. Sep 30, 2012 #1
    As the title suggests I am working on some general relativity and combinatorics seems to be my ever-returning Achilles heel. I have a four dimensional tensor, denoted by g_abcd with a,b,c,d ranging between 0 and 3, which is fully antisymmetric, i.e.: it is zero if any of the two (or more) indices are equal. Intuitively, I know that this tensor has only one independent component, but I would like to prove it using combinatorics.

    My idea is as follows: the tensor has 4^4 = 256 components --> calculate all zero components N --> (256 - N)/2^6 should be one, as 4 unequal indices can be arranged in 6 ways, so that every arrangements cuts the number of independent components in half.

    I believe this is correct, but correct me if I am wrong. The tricky part is calculating the number N. My idea: N = #(2 indices equal) + #(3 indices equal) + #(4 equal).
    Obviously: #(4 equal) = 4 and #(3 equal) = 4*4*3 = 48. Then:
    #(2 indices equal) = #(2 equal, other two equal) + #(2 equal, other not equal) = 36 + 12*4*3*2*0.5 = 180.
    This would yield N = 180 + 4 + 36 = 232, which is obviously not correct.

    Can anyone help me? Thanks in advance!
     
  2. jcsd
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