Steven Weinberg\'s book \"The Quantum Theory of Fields\", volume 3, page 46 give

Question

Steven Weinberg's book "The Quantum Theory of Fields", volume 3, page 46 gives the following argument against N = 3 supersymmetry:

"For global N = 4 supersymmetry there is just one supermultiplet ... This is equivalent to the global supersymmetry theory with N = 3, which has two supermultiplets: 1 supermultiplet... and the other the CPT conjugate supermultiplet... Adding the numbers of particles of each helicity in these two N = 3 supermultiplets gives the same particle content as for N = 4 global supersymmetry"

However, this doesn't directly imply (as far as I can tell) that there is no N = 3 QFT. Such a QFT would have the particle content of N = 4 super-Yang-Mills but it wouldn't have the same symmetry. Is such a QFT known? If not, is it possible to prove it doesn't exist? I guess it might be possible to examine all possible Lagrangians that would give this particle content and show none of them has N = 3 (but not N = 4) supersymmetry. However, is it possible to give a more fundumental argument, relying only on general principles such as Lorentz invariance, cluster decomposition etc. that would rule out such a model?

Explanation / Answer

Depending on what you mean by "exist", the answer to your question is Yes.

There is an N=3 Poincar

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