Its commonly said that gauge bosons transform according to the adjoint representation of the corresponding gauge group. For example, for $SU(2)$ the gauge bosons live in the adjoint $3$ dimensional representation and the gluons in the $8$ dimensional adjoint of $SU(3)$.

Nevertheless, they transform according to

$$ A_μ→A′_μ=UA_μU^\dagger−ig(∂_μU)U^\dagger ,$$

which is not the transformation law for some object in the adjoint representation. For example the $W$ bosons transform according to

$$ (W_μ)_i=(W_μ)_i+∂_μa_i(x)+\epsilon_{ijk}a_j(x)(W_μ)_k.$$


1 Answer 1


A gauge field transforms in the adjoint of the gauge group, but not in the adjoint (or any other) representation of the group of gauge transformations.

In detail:

Let $G$ be the gauge group, and $\mathcal{G} = \{g : \mathcal{M} \to G \vert g \text{ smooth}\}$ the group of all gauge transformations.

A gauge field $A$ is a connection form on a $G$-principal bundle over the spacetime $\mathcal{M}$, which transforms as $$ A \mapsto g^{-1}Ag + g^{-1}\mathrm{d}g$$ for any smooth $g : \mathcal{M} \to G$. If $g$ is constant, i.e. not only an element of $\mathcal{G}$, but of $G$ itself, this obviously reduces to the adjoint action, so $A$ does transform in the adjoint of $G$, but not in the adjoint of $\mathcal{G}$. With respect to $\mathcal{G}$, it does not transform in any proper linear (or projective) representation in the usual sense, but like an element of a Jet bundle.


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