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The super-Poincare algebra contains supersymmetry generators $Q^I$ which satisfy fermionic anticommutation relations. By the higher-dimensional analogue of the spin-statistics theorem, they must transform in spinor representations of the Lorentz group. In the familiar case of $d = 3+1$, they are usually taken to be Weyl spinors, or equivalently Majorana spinors.

It seems that in general, supersymmetry generators are always taken to transform in the minimal spinor representation, which are Dirac, Weyl or Majorana, or Majorana-Weyl, depending on the dimension. We then usually allow for $\mathcal{N}$ supersymmetry generators, all transforming in the same way.

However, this doesn't seem to me to be the most general possibility. Why not take the SUSY generators to transform in non-minimal spinor representations? For example, in $d = 3+1$ one could take the SUSY generator to have spin $3/2$. This would change the structure of the SUSY multiplets, e.g. I would imagine they are spaced by spin $3/2$ instead of $1/2$. It could be the case that even $\mathcal{N} = 1$ SUSY with a spin $3/2$ SUSY generator would generate particles with too high helicity, but perhaps the situation is less constrained in $d < 4$?

Has this kind of SUSY been investigated, or if not, what is the issue with it?

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    $\begingroup$ The answer is that SUSY generators can only be spin 1/2. I gave an answer in this thread: https://physics.stackexchange.com/questions/417989/is-there-a-3-2-supersymmetry/418025#418025 $\endgroup$
    – otillaf
    Commented Jan 20, 2019 at 20:48
  • $\begingroup$ @otillaf Does this work similarly for $d \neq 4$? $\endgroup$
    – knzhou
    Commented Jan 20, 2019 at 21:27
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    $\begingroup$ I don't know. One shoud see if Coleman-Mandula theorem works in a similar way in higher dimensions (and I think it should work, at least in d>1+1, considering only flat spacetimes). Then you have a strong constrain on the anticommutators of SUSY generators which together SO(N) clebsch-gordan decomposition could give us the right answer. Maybe I'll try to work it later. $\endgroup$
    – otillaf
    Commented Jan 20, 2019 at 22:22
  • $\begingroup$ This is addressed by Weinberg (cf. Vol.III, page 383). $\endgroup$ Commented Jan 22, 2019 at 21:25

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