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I'm watching this intro to Supersymmetry lecture by Nathan Berkovits and, at roughly 33 minutes, he mentions he is going to extend the Poincaré group in an essentially unique manner, where "unique" means "so that the theory is unitary".

I'm new to both supersymmetry and the BMS group at asymptotic infinity, but I'm curious about to which extent that statement holds. For example, ain't BMS symmetries extensions of the Poincaré group that also lead to unitarity in a sense?

As I said, I'm new to both SUSY and the BMS group, so I believe I might be missing some trivial aspect.

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    $\begingroup$ Nitpick: I would call the question Coleman-Mandula theorem and the BMS group. $\endgroup$ Jul 18, 2023 at 3:09
  • $\begingroup$ The hidden assumption is that the extra symmetry generators preserve the vacuum state. The BMS symmetry does not do this (i.e. it is spontaneously broken). $\endgroup$
    – Prahar
    Jul 18, 2023 at 12:11
  • $\begingroup$ @ConnorBehan I don't really know what the Coleman-Mandula theorem is (perhaps that will be dealt with in more detail later in the course, but it is a really intro-level course for undergraduates with no knowledge of field theory), so perhaps that is already half an answer $\endgroup$ Jul 18, 2023 at 13:48
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    $\begingroup$ @Prahar If Connor's comment counts as half an answer, I'd say you just wrote the other half $\endgroup$ Jul 18, 2023 at 13:49

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The hidden assumption is that the extra symmetry generators preserve the vacuum state. The generators of the BMS symmetry does not do this (i.e. it is spontaneously broken). Note that the Poincare subgroup of the BMS group DOES preserve the vacuum.

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  • $\begingroup$ i'd say, a more important hidden assumption is that the symmetry group should have finitely many generators. There are infinitely many ways to extend Poincare using infinitely many generators, but only one way to do so using finitely many ones (namely, the SUSY algebra). Agree? $\endgroup$ Aug 5, 2023 at 19:02

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