In field theory we associate to each Gauge theory a continuous group of local transformations (a Gauge group), and then we require\define fermion fields to be irreducible representations belonging to the fundamental representation of this Gauge group.
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What is the fundamental representation, and why do we require fermions to be in it?
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What does it mean for a field to belong to a certain representation? Is this just a way of stating that the fields are the "targets" of the Gauge transformations we introduced, meaning that they belong to the vector space where the representation of these transformations act?
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A Gauge transformation must by definition not change the physics. This means that given any field $\psi$ we have to identify as a single physical entity all the fields that can be obtained by $\psi$ by means of any element of the Gauge group. Is the formalization of this concept the reason we require fields to belong to irreducible representations? Is there any other justification for it?
Best Answer
The fundamental representation of a Lie group $G$, as commonly used in this context, is the smallest faithful (i.e. injective) representation of the group. We do not require fermions to belong to the fundamental rep, it is just the case that, in the Standard model, they always either belong to the fundamental or the trivial representation (as that is thoroughly indicated by experimental data), so there is rarely a need to look at other representations.
To belong to a certain representation $V_\rho$ of $G$ means that the field is a section of the associated vector bundle $P \times_G V_\rho$, where $P$ is the principal bundle belonging to our gauge theory. Equivalently, the field is a $G$-equivariant function $P \to V_\rho$ fulfilling $f(pg) = \rho(g^{-1})f(g)$ for all $p \in P$ and $g \in G$. If principal bundles are not talked about, the field is often just taken to be a function $\Sigma \to V_\rho$, though this is, strictly speaking, not the way to do it.
Every field must belong to a representation of the gauge group (even if it is the trivial one) since the gauge transformations must have a defined action upon everything in our theory.
It is not required that fields belong to irreducible representations (again, it is simply often just the case), but since every reducible representation can be split up into irreducible ones, it is enough to look at the behaviour of the irreducible representations. (Note though that there are fields transforming in reducible representations - the usual $\frac{1}{2}$-spinors (not Weyl spinors!) transform as members of the $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$-representation of the Lorentz group)