1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    arivero

    User Avatar
    Gold Member

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

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

    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
    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?
     
  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

    Haelfix

    User Avatar
    Science Advisor

    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

    arivero

    User Avatar
    Gold Member

    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

    samalkhaiat

    User Avatar
    Science Advisor

     
  9. Sep 3, 2008 #8

    arivero

    User Avatar
    Gold Member

    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

    samalkhaiat

    User Avatar
    Science Advisor

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Decomposition of direct product into symmetric/antisymmetric parts
Loading...