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I have a perhaps stupid question about Maxwell equations.

Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+A\wedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as

$$D\star F=\star J \tag{1.1}$$

$$DF=0 \tag{1.2}$$

Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.

If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.

However, as we all know, the Maxwell equations are

$$d\star F=\star J$$

$$dF=ddA=0$$

Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?

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Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $\star F$ transform in the adjoint of the gauge group).

For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^{abc} = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.

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We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = \mathrm{d}F$, since the adjoint representation of $\mathrm{U}(1)$ is trivial.

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