Gauge transformations and Covariant derivatives commute

Question

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on fields on which the algebra is closed off-shell"

which was taken from section 11.2.1 (page 223) of Supergravity by Freedman and Van Proeyen. In this text the authors support this statement by proving the following $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\epsilon^A\mathcal{D}_{\mu}T_A\phi=0 \tag{1}$$ where $\epsilon^A(x)$ is the gauge parameter for the symmetry transformation generated by the operator $T_A$, $\delta(\epsilon)$ is a gauge transformation, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ is the covariant derivative and $B_{\mu}{}^{A}(x)$ is the gauge field corresponding to each gauged symmetry.

I have no problem in deriving the result $(1)$, but I don't understand why the result $(1)$ is equivalent to the statement in the yellow box. In particular, why is the statement in the yellow box not equivalent to $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

Answer 1

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_\mu$ that transforms to $D'_\mu$ and a field $\phi$ that transforms as $\phi' = U(\epsilon)\phi$ under a finite gauge transformation $U(\epsilon) = \mathrm{e}^{\epsilon^a T^a}$, we require that the transform of $D_\mu \phi$ is equal to $U(\epsilon) D_\mu \phi$. In formulae: $$ D'_\mu \phi' = UD_\mu\phi.$$

Infinitesimally, this means $$ \delta(\epsilon)(D_\mu \phi) = (\epsilon^a T^a) D_\mu \phi = \epsilon^a D_\mu(T^a\phi),$$ since the infinitesimal version of $D'_\mu \phi'$ is $\delta(\epsilon) D_\mu \phi$ by definition and the infinitesimal action of $U$ on $D_\mu\phi$ is $(\epsilon^a T^a)D_\mu \phi)$. This is precisely what Freedman and Van Proeyen show.