# Who chooses the representation of SM particles?

I refer to this wikipedia table where the SM particles and their transformation behaviour are listed. When considering the field $W$ e.g., the 'representation' is listed as $(\mathbf {1} ,\mathbf {3} ,0)$.

So far my understanding of that is, that $W$ does not transform under $SU(3)$, transforms as a triplet under $SU(2)$ (means it transforms in the adjoint) and does not transform under $U(1)$.

The transformation in some representation of the gauge bosons make (at least a bit of) sense to me, since it has to keep the lagrangian invariant as the fermions transform.

However, the transformation of the fermions seem to have an arbitrary transformation behaviour to me. Is there any reason why we have a fermion field that transforms as $({\bar {\mathbf {3} }},\mathbf {1} ,\textstyle {\frac {2}{3}})$ (left handed downtype anti fermion) but no $({\bar {\mathbf {7} }},\mathbf {3} ,-\textstyle {\frac {11}{3}})$ fermion field?

If the answer is "because it describes what we see": Why does the number of fields and their transformation behaviour then not count as free parameters of the model?

• So, what would you do if someone actually did discover your notional crazy fermion? To be sure, the quantum numbers in each family obey anomaly constraints necessary for quantum preservation of gauge invariance, but a new fermion simply signals the beginning of the discovery of a new family. How did you get the idea the quantities you are mentioning are not dictated by what we see, but, instead, by some other principle? – Cosmas Zachos Mar 17 '17 at 15:47
• ok so there are anomaly constraints, thanks for that hint! But assumed i could write down some anomaly free theory with additional crazy fermions that would be ok? So any fermion can transform in any representation of any group? In this case why does the coupling strength then count as a free parameter but not the fermion transformation under a group? – Mr Puh Mar 17 '17 at 15:59
• Model builders do write oddball representations and particles incessantly and exuberantly, for several contexts, hidden sectors, technicolor, etc. Some will lose you asymptotic freedom. Some will violate anomaly constraints. For pity's sake, why are you focussing on the daffy, deprecated, jejune, pointless counting of the SM free parameters? Do they mean anything? Are they saturating a genie's limited number of wishes? Most people guffaw with drollery when they hear this type of drivel at seminars. What do you get out of it? Why are you attracted to it? – Cosmas Zachos Mar 17 '17 at 16:05
• @'Do they mean anything': Free parameters are a measure for the fundamentality of a theory. A theory without free parameters, describing what we see is definitely favored. A theory with too many free parameters has no deeper concept. So they do play a role (at least in my understanding). But the question was rather why these transformations do not count as such free parameters – Mr Puh Mar 17 '17 at 16:24
• No, there is no deeper formalism. People are pleased that they fit nicely into such of "small" groups such as SU(5), but the harsh voice of nature chose them, not a fundamental principle. Any attempt to "derive" our world, its dimensions, and SM parameters out of "fundamental" theories like stringery, etc.. have fallen so flat on their face it is not funny.... – Cosmas Zachos Mar 17 '17 at 16:31