Product rule for exterior covariant derivative

[Physics_Stuff]

It is well known that the product rule for the exterior derivative reads
$$d(a\wedge b)=(da)\wedge b +(-1)^p a\wedge (db),$$where a is a p-form.
In gauge theory we then introduce the exterior covariant derivative $$D=d+A\wedge.$$ What is then D(a ∧ b) and how do you prove it?

I obtain
$$D(a\wedge b)=d(a\wedge b)+A\wedge a \wedge b=(da)\wedge b +(-1)^p a\wedge (db)+A\wedge a \wedge b,$$
which is neither (Da) ∧ b +(-1)^p a ∧ (Db) nor (Da)∧ b + a∧ (Db). I have, however, seen the latter been used without proof.

 

[martinbn]

May be the definition is for 1-forms and then you extend it to satisfy the product rule.

 

[Ben Niehoff]

$$D=d+A\wedge.$$

This statement is your error. The correct statement is

$$D = d + \rho(A) \wedge$$
where $\rho : G \to GL(n)$ is the homomorphism which defines the representation of the gauge group you need. If you have two 1-forms $\alpha$ and $\beta$, each in the fundamental representation, then the 2-form $\alpha \wedge \beta$ is no longer in the fundamental representation! It will be in the antisymmetrized product of two fundamental representations (which is usually the adjoint rep, I think). Higher-degree wedge products will give you antisymmetrized products of more fundamental reps.