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Irreducible representation of tensor field

  1. Jan 17, 2013 #1
    In Mark Srednicki's book "Quantum Field Theory"
    He says that a tensor field [itex] B^{αβ} [/itex] with no particular symmetry can be written as :-

    [itex] B^{αβ} = A^{αβ} + S^{αβ} + (1/4) g^{αβ} T(x) [/itex] Equn. 33.6

    where A - Antisymmetric, S = symmetric and T(x) = trace of [itex] B^{αβ} [/itex] .

    Is there any reason for explicit addition of trace term ?
    Coz generally we split things into symmetric and antisymmetric parts and trace is included in symmetric part.
    Last edited: Jan 17, 2013
  2. jcsd
  3. Jan 17, 2013 #2


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    If you're talking about the general linear group GL(n), the irreducible representations are the tensors whose indices have been symmetrized in a particular way. When you go to the orthogonal group, there are fewer transformations in the group, and some of these representations are no longer irreducible. The operation of contraction (forming a trace) commutes with the orthogonal transformations.
  4. Jan 17, 2013 #3


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    This is true only if the symmetric part is traceless. In general, for any rank-2 tensor, we write
    [tex]B^{ ab } = B^{ (ab) } + B^{ [ab] } .[/tex]
    Then we take the symmetric part and decompose it as
    B^{ (ab) } = \left( B^{ (ab) } - \frac{ 1 }{ 4 } g^{ ab } B \right) + \frac{ 1 }{ 4 } g^{ ab } B .
    The tensor (call it [itex]S^{ ab }[/itex]) in the bracket on the left-hand side is symmetric and traceless, because
    [tex]B = \mbox{ Tr } ( B^{ (ab) } ) = g_{ ab } B^{ (ab) }[/tex]
    So, our original tensor can now be written as
    B^{ ab } = A^{ ab } + S^{ ab } + \frac{ 1 }{ 4 } g^{ ab } B ,
    where [itex]A^{ ab } = - A^{ ba } \equiv B^{ [ab] }[/itex]

    See posts #24 and 25 in


  5. Jan 17, 2013 #4
    yea, i got just it, thank u :)
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