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In SUSY, why do fermions and gauge bosons in the same multiplet both transform in the adjoint representation of the gauge group?

+ 3 like - 0 dislike
11537 views

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.

This post imported from StackExchange Physics at 2014-10-18 12:47 (UTC), posted by SE-user Lior
asked Oct 18, 2014 in Theoretical Physics by Lior (15 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

Well, I assume you are talking about \(d=4\) \(\mathcal{N}=1\) pure SYM theory. In a pure gauge theory, forget for a moment the supersymmetry, I guess you know that the gauge fields transform indeed in the adjoint representation*. Ok, if we agree on this, a quick way to see that indeed the Weyl fermion, the gaugino \(\lambda_{\alpha}\), transforms in the adjoint is to consider that you can obtain one by making a supersymmetric transformation to the gauge boson. But a supersymmetric transformation cannot change the representation under which the multiplet transforms. You can see this by taking any kind of combinations of supersymmetry variations to your gauge boson and you will see that you won't be able to obtain a field from a chiral multiplet. If this is not convincing let me know and maybe myself or somebody else can help further. 

*Again, the basic point is to understand why in gauge theories the gauge bosons transform in the adjoint. The adjoint representation induces a linear mapping which is exactly what we want and the reason we introduce the covariant derivative and end up with interaction terms like in the SM. The gauge field must transform just like the covariant derivative such that the non-linearities cancel. This is a quick way to see that the gauge field transforms in the adjoint. 

answered Oct 18, 2014 by conformal_gk (3,535 points) [ no revision ]

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