Let me first of all give a more precise definition of the curvature and exterior covariant derivative. To start with, lets fix the following data:
- A smooth manifold $\mathcal{M}$
- A principal $G$-bundle $P$, where $G$ is a (finite-dimensional) Lie group with Lie algebra $\mathfrak{g}$.
Now, as you have said, a connection $1$-form is a $\mathfrak{g}$-valued $1$-form, i.e. $A\in\Omega^{1}(P,\mathfrak{g})$ satisfying some extra properties. In order to define the corresponding curvature $2$-form $F^{A}\in\Omega^{2}(P,\mathfrak{g})$ there are several (equivalent) ways. One is to define it directly via the structure equation, which is
$$F^{A}:=\mathrm{d}A+\frac{1}{2}[A\wedge A]$$
where $\mathrm{d}$ is just the usual exterior covariant derivative of Lie-algebra valued forms, i.e.
$$\mathrm{d}A=\mathrm{d}(A^{a}T_{a}):=(\mathrm{d}A^{a})T_{a}$$
where $A^{a}\in\Omega^{1}(P)$ and $\{T_{a}\}_{a}$ is a basis of $\mathfrak{g}$ and where $[\cdot\wedge\cdot]$ is just the wedge product defined via the commutator, i.e.
$$[A\wedge A]:=\sum_{a,b}(A^{a}\wedge A^{b})[T_{a},T_{b}],$$
where $A^{a}\wedge A^{b}$ is just the standard wedge-product of real-valued forms. Starting from this, it is straight-forward to get the coordinate expression: Take a local section ("local gauge") $s\in\Gamma(U,P)$, where $U\subset\mathcal{M}$ open. Then, we define
$$A_{s}:=s^{\ast}A\in\Omega^{1}(U,\mathfrak{g})$$
as well as
$$F^{A}_{s}:=s^{\ast}F^{A}\in\Omega^{2}(U,\mathfrak{g}),$$
which are now forms defined on $U\subset\mathcal{M}$.
A straight-forward computation then yields
$$F_{\mu\nu}^{a}=\partial_{\mu}A^{a}_{\nu}-\partial_{\nu}A_{\mu}^{a}+c_{bd}^{a}A_{\mu}^{b}A_{\nu}^{d}$$
where $F_{\mu\nu}^{a}$ are defined via $F_{s}^{A}=F^{a}_{\mu\nu}T_{a}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}$ and $A_{\mu}^{a}$ via $A_{s}=A_{\mu}^{a}T_{a}\mathrm{d}x_{\mu}$. The constants $c^{a}_{bc}$ are the structure constants of the Lie algebra, i.e. $[T_{a},T_{b}]=c_{ab}^{d}T_{d}$.
Now, secondly, you can define the curvature via the covariant derivative $D_{A}$. Let $(\rho,V)$ be some (finite-dimensional) representation of $G$. Then, $D_{A}$ is a (family of) map(s) $$D_{A}:\Omega^{k}(P,V)\to\Omega^{k+1}(P,V)$$ defined via
$$(D_{A}\omega)_{p}(v_{1},\dots,v_{k}):=(\mathrm{d}\omega)_{p}(\mathrm{pr}(v_{1}),\dots,\mathrm{pr}(v_{k}))$$
for all $p\in P$, $v_{1},\dots,v_{k}\in T_{p}P$ and $\omega\in\Omega^{k}(P,V)$, where $\mathrm{d}$ on the right-hand side is the standard exterior derivative of $V$-valued forms (as explained above) and where $\mathrm{pr}:TP\to H$ is the projection onto the horizontal tangent space $H_{p}:=\mathrm{ker}(A_{p})\subset T_{p}P$ (the "Ehresmann connection corresponding to $A$").
Now, here is the most important point: You stated that $D_{A}$ can be
calculated via the formula
$$D_{A}\cdot=\mathrm{d}\cdot+\rho(A)\wedge\cdot.$$ That is in general not
true! In fact, it is only true for forms living in the subset
$$\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}\subset\Omega^{k}(P,V)$$
This set consists of all forms $\omega\in\Omega^{k}(P,V)$ satisfying the following two extra properties:
- "$\omega$" is horizontal": $\omega_{p}(v_{1},\dots,v_{k})=0$ whenever at least one of the tangent vectors $v_{i}$ is vertical (i.e. not contained in $H_{p}$).
- "$\omega$ is of type $\rho$": $(r_{g}^{\ast}\omega)_{p}=\rho(g^{-1})\circ\omega_{p}$ for all $g\in G$, where $r_{g}:P\to G,p\mapsto p\cdot g$ (the action $\cdot:P\times G\to P$ denotes the right-group action contained in the definition of a principal bundle)
In other words, your "definition" of the covariant derivative is actually only valid for this subset. Now, a connection $1$-form is in general not an element of $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$! (*)
To sum up, you cannot use your formula for a connection $1$-form. However, you can actually easily show that
$$D_{A}A\stackrel{!}{=}F^{A}=\mathrm{d}A+\frac{1}{2}[A\wedge A],$$
which then also leads to the correct coordinate expression as explained above.
* However, the curvature is an element of $\Omega^{2}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$ and also the difference of two connection $1$-forms is an element of $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$. The latter statement implies that the set of connection $1$-forms $\mathcal{C}(P)$ of a principal bundle is an infinite-dimensional affine space with vector space $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$.
Best Answer
OP is considering Yang-Mills theory over a curved base space $(M,g)$. If the base space connection is the Levi-Civita connection $\nabla^{LC}=\partial+\Gamma$, then it doesn't matter whether one uses the gauge-covariant derivative $D=\partial+A$ or the full covariant derivative $\nabla=D+\Gamma$ since the Christoffel symbols $\Gamma$ drops out of the Yang-Mills theory and OP's eqs. (3), (4) and (5). This is mainly due to the torsionfreeness of the Levi-Civita connection $\nabla^{LC}$.