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The title leaves it quite clear, why must superpartners have the same gauge quantum numbers?

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  • $\begingroup$ How would you be able to add the thing SUSY produces to its superpartner if they transformed in different representations (and hence lay in different vector spaces)? $\endgroup$ – ACuriousMind Jul 5 '15 at 17:01
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    $\begingroup$ @ACuriousMind not sure what you are getting at. not all symmetries have to commute with SUSY (those that don't are called R-symmetries) $\endgroup$ – innisfree Jul 5 '15 at 17:02
  • $\begingroup$ gauge symmetries commute with SUSY. not sure whether there is any deeper answer than that $\endgroup$ – innisfree Jul 5 '15 at 17:08
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    $\begingroup$ @innisfree: While your answer is correct I think it needs to be expanded on to be clear to someone without experience with supersymmetry... $\endgroup$ – JeffDror Jul 5 '15 at 17:55
  • $\begingroup$ @JeffDror sure, yes, it needs to be expanded. tbh, i don't know much about gauging R-symmetries - i think it's problematic $\endgroup$ – innisfree Jul 5 '15 at 18:58
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A bosonic symmetry that acts differently on the different components of a supermultiplet is an R-symmetry. Such a symmetry does not commute with the supercharges. Since the commutator between an R-symmetry and a supercharge gives something Grassmann odd, it has to given another supercharge. Schematically $$ [R,Q] = Q $$ or in terms of variations acting on some field $$ [ \delta_R , \delta_Q ] = \delta_Q $$ If we gauge an R-symmetry then the variation $\delta_R$ is allowed to depend on the coordinates $x$. But then from the algebra above so is $\delta_Q$ $$ [ \delta_R(x) , \delta_Q(x) ] = \delta_Q(x) $$ So gauging an R-symmetry leads to local supersymmetry. To see where this takes us we can consider the supersymmetry algebra $$ \{ Q , Q \} = P $$ or $$ \{ \delta_Q , \delta_Q \} = \delta_P $$ where $\delta_P$ is a translation. If $\delta_Q$ depends on $x$ we get $$ \{ \delta_Q(x) , \delta_Q(x) \} = \delta_P(x) $$ so local supersymmetry automatically leads to invariance under local translations or in other words under diffeomorphisms. Hence, the only way to gauge an R-symmetry is in supergravity. In particular in a gauged supergravity the gravitino is charged with respect to some gauge field while the graviton is not, so this gives an example where superpartners don't carry the same gauge quantum numbers.

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  • $\begingroup$ How well motivated is it that that SM gauge group is chosen to commute with SUSY? Could I make the SM gauge groups an R-symmetry? $\endgroup$ – innisfree Jul 6 '15 at 16:21
  • $\begingroup$ @innisfree: to make the $SU(3) \times SU(2) \times U(1)$ gauge group of the standard model a gauged R-symmetry you would (obviously) need an R-symmetry group that contains this as a subalgebra. Unfortunately I don't think even the $SO(8)$ R-symmetry of maximally supersymmetric $\mathcal{N}=8$ sugra in 4d is enough to fit this (in normal unified theories need an $SU(5)$ or $SO(10)$) $\endgroup$ – Olof Jul 6 '15 at 16:40
  • $\begingroup$ does the R-symmetry group have to be simple or semi-simple? am i free to pick the R-symmetry group? why can't I pick $G_{SM}$ and build a theory? $\endgroup$ – innisfree Jul 6 '15 at 16:42
  • $\begingroup$ no, the supersymmetry algebra strongly restricts what R-symmetry is possible, at least as long as you don't want to break Lorentz invariance. In 4d you can organise the supercharges into eg Weyl spinors and the R-symmetry has to rotate among themselves. If you have $\mathcal{N}$ supercharges the 4d R-symmetry will in general be someting like $SU(\mathcal{N})$ or $SO(\mathcal{N})$ depending on what reality conditions you can impose. $\endgroup$ – Olof Jul 6 '15 at 16:55

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