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, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ 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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on field 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, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ 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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

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, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ 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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

I would like to prove and understand the statement

"Gauge transformations and Covariant derivatives commute on field 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, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ 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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on field 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, $\mathcal{D}_{\mu}=\partial_{\mu}-B_{\mu}{}^{A}T_A$ 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 $$\delta(\epsilon)\mathcal{D}_{\mu}\phi-\mathcal{D}_{\mu}\delta(\epsilon)\phi=0~?$$