# Why Is Abelian Gauge Theory So Special?

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 $$d$$?

## 2 Answers

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 $$d$$.

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.