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What does the R-Symmetry do to scalars and fermions in N=4 U(N) SYM?

  1. Jul 1, 2011 #1

    I have forgotten all about my supersymmetry knowledge and all about my group theory knowledge. I am trying to understand what the R-symmetry in N=4 U(N) SYM does. Sadly I have never actually learnt anything about supersymmetry which is larger than N=1. I know the R-symmetry is SU(4) and I know that the gauge boson doesn't transform under R-symmetry. The 6 real scalars are "a 6" or SU(4) and the four Weyl fermions "a 4". So the question is:

    What does "a 6" and "a 4" mean? I know that means they are in different representations of SU(4). Does that mean the scalars transform into each other and the fermions transform into each other seperately and leave the particles in the other rep alone? If so, I thought R-symmetry came from supersymmetry, which basically transforms fermions into bosons and vice versa.

    Please, could you try and answer in not too technical term? Thanks so much!
  2. jcsd
  3. Jul 1, 2011 #2
    The R-symmetry is bosonic, so it transforms bosons into bosons and fermions into fermions. As far as the action on the scalars and fermions, su(4) (the algebra) is isomorphic to so(6). The 4 is the fundamental (spinor) rep of su(4), which takes the four (complex) spinors into each other. The 6 is the fundamental (vector) rep of so(6), which takes the six (real) scalars into each other. The specific numbers 4 and 6 just refer to the dimensionality of the representations. Hope this helps!
  4. Jul 1, 2011 #3
    This does really help, thank you!

    Do you also know, what the R-symmetry in this case has to do with the F-term in the SUSY Lagrangian? If I remember correctly, the F-term is some sort of mass term, isn't it? Or it could lead to a mass term, if we break SUYS?

    I have somehow this picture in my head that there is a F-term where scalars and fermions are multiplied. If we don't have R-symmetry for scalars and fermions than the F-term wouldn't be SUSY invariant. So is this how the R-symmetry manifests itself in the SUSY Lagrangian?
  5. Jul 1, 2011 #4
    Well, the F-terms also transform under R-symmetry, but I'm not entirely sure what you're referring to past that. If an F-term gets a VEV, then SUSY is spontaneously broken, but R-symmetry could remain intact. Likewise, you can spontaneously break R-symmetry without breaking SUSY. Those statements are at least true in N=1; I don't know if there's some stronger statement for higher-N SUSY theories.

    To answer your other question, if you have a supersymmetric mass term in the superpotential, W = m Phi^2, then by solving the F equations of motion, you generate masses for the scalars. Also, if you give an F a VEV, you spontaneously break SUSY and you can get different masses for the scalars as compared to the fermions that way.
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