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I'm trying to understand a certain point about supersymmetry.

We are dealing with a N=1 (i.e, one supersymmetric flavour), massless, four dimensional theory. Then the vector multiplet consists of a Weyl fermion and a gauge boson, and they both transform in the adjoint representation of the gauge group. I want to understand why and what this means.

I know that usually the number of gauge bosons is the number of generators in the gauge algebra, and that they transform in the adjoint. but each gauge boson has also a 4-vector index, so there is a total of four-times-the-number-of-gauge-generators operators that create these bosons.

Why do the Weyl fermions also need to transform in the adjoint? If I'm not wrong, this is not a feature of gauge theories, so I assume it comes from SUSY. Maybe the reason is that in a multiplet, the number of boson degrees of freedom(DOFs) is the same as the number of fermion DOFs? and then, since a Weyl fermion has two complex components, which means 4 DOFs, then to have the same number of total fermion and boson DOFs, we need the same number of fermions? but that only means that there is the same number, not necessarily the same representation of the gauge group.

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    $\begingroup$ See here for a rather intuitive explanation. $\endgroup$ – Dilaton Oct 18 '14 at 21:54
  • $\begingroup$ Have you ever written down the transformation laws under SUSY of a U(1), QED type gauge theory? Shouldn't it be obvious how to generalize them for a generic gauge group, which commutes (has nothing to do) with SUSY? This is not a group theory question, I suspect. $\endgroup$ – Cosmas Zachos Sep 3 '17 at 20:16

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