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Decomposition of direct product into symmetric/antisymmetric parts

  1. Sep 2, 2008 #1
    Can anyone explain to me why

    the 3-rep of SU(3) gives

    [itex]3\otimes 3 = \overline{3}\oplus 6[/itex]

    whereas for the 5 of SU(5)

    [itex]5\otimes 5 = 10\oplus 15[/itex]?

    I thought the general pattern was

    [itex]N \otimes N = \overline{\frac{1}{2}N(N-1)}\oplus \frac{1}{2}N(N+1)[/itex]

    but this second example seems to contradict that.

    Edit: I wrote 15* instead of 15.
    Last edited: Sep 2, 2008
  2. jcsd
  3. Sep 2, 2008 #2


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    Slansly report is available in the net, scanned by KEK
    http://www.slac.stanford.edu/spires/find/hep/www?j=PRPLC,79,1 [Broken]

    SU(3) is here
    http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198102061+194+265 [Broken]
    and SU(5) here
    http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198102061+199+265 [Broken]

    Someone has different conventions somewhere.

    A related thing that intrigues me, btw. The authors of this
    http://www.slac.stanford.edu/spires/find/hep/www?j=PHLTA,B188,58 [Broken]
    are doing, classically, [itex]5\otimes 5 [/itex] or [itex]5\otimes \bar 5 [/itex] or [itex](5 \oplus \bar 5) \otimes (5 \oplus \bar 5) [/itex] but the point is that they get SO(32). How is it? Does SO(32) and SU(5)^2 share representations, or dimensionality of representations?
    Last edited by a moderator: May 3, 2017
  4. Sep 2, 2008 #3
    I don't think it's a difference in convention.

    Slansky gets [itex]5\otimes 5 = 10 \oplus 15[/itex] for SU(5) and [itex]\overline{3}\otimes \overline{3} = \overline{6}\oplus 3[/itex] for SU(3) which is equivalent to [itex]3\otimes 3 = 6\oplus\overline{3}[/itex].

    I understand how to derive 3 x 3 = 6 + 3* by contracting with the permutation symbol.

    Do you know how to derive 5 x 5 = 10 + 15?
  5. Sep 2, 2008 #4


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    Probably the easiest way is to do a Young Tableux calculation. Its not a very long calculation (about 4 - 5 lines) but completely impossible to show in this sort of forum

    You can check yourself with the other possibility
    5 * 5bar = 1 + 24
  6. Sep 2, 2008 #5


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    but note the subscript _s in Slansky to mark the symmetric. It differs from the bar.
  7. Sep 2, 2008 #6
    Here's my proposed proof. Work under the assumption that a tensor is irreducible if it a tensor of lower rank can not be formed by contraction with the isotropic tensors.

    Consider an arbitrary element of the tensor product space [itex]T \in 5\otimes 5[/itex].

    Contracting with the lower index permutation symbol gives

    [itex]U_{klm} = \varepsilon_{ijklm}T^{ij}[/itex].

    Contracting again with the upper index permutation symbol gives

    [itex]V^{no} = \varepsilon^{klmno}U_{klm}[/itex].

    V is irreducible, has two upper anti-symmetric indices. Therefore [itex]V\in 10[/itex].
  8. Sep 3, 2008 #7


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  9. Sep 3, 2008 #8


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    So, the original question is based in confusion between "conjugate" and "anti-symmetric", isn't it? Is this confusion a notational or a conceptual one?
  10. Sep 5, 2008 #9
    A lecturer told me.

    Yes, I believe that [itex]T^2(V) = S^2(V)\oplus \Lambda^2(V)[/itex]. I've had trouble reconciling this with physicists' notation for SU(3) where the anti-symmetric part is represented by [itex]\overline{3}[/itex]. For consistency let's use upper indices to denote tensors which transform according to the fundamental rep from now on.

    I think you're saying that there's an isomorphism between the group actions on [itex]\Lambda^2(\mathbb{C}^3)[/itex] (represented by [itex]T^{ij}=T^{[ij]}[/itex], say) and the representation on the dual space [itex](\mathbb{C}^{3})^\ast[/itex], which is defined (in component notation) by [itex]\mathrm{SU}(3) : v_i \mapsto {U_{i}}^j v_j[/itex] where [itex]{U_{i}}^j \equiv {U^{i}}_j^\ast[/itex] and the action on the fundamental representation is [itex]\mathrm{SU}(3) : v^i \mapsto {U^{i}}_j v^j[/itex].

    I suppose the isomorphism is [itex]Z_i \mapsto \varepsilon^{ijk}Z_k[/itex] but I don't see why this is an isomorphism. Ie it's not obvious why they have the same transformation properties.

    Once again, it's not obvious to me why for SU(4), [itex]T_{[ij]}[/itex] transforms in the same way as [itex]T^{[ij]}[/itex], and thus why [itex]\mathbf{6} \cong \overline{\mathbf{6}}[/itex].

    I understand the concept of irreducibility by the non-existence of proper non-trivial subspaces which transform only amongst themselves. What I fail to see is why this is equivalent to the inability for form lower rank tensors by contraction with the isotropic tensors.
  11. Sep 8, 2008 #10


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