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
Forgive me but I’ll use $y$ to mean the other coordinate system. Your last two terms are \begin{align} &\quad\,\,V^\rho \omega_\alpha\frac{\partial}{\partial x^{\rho}} \left( \frac{\partial x^\alpha}{\partial y^\mu}\right) + \frac{\partial x^\gamma}{\partial y^\rho} \omega_\gamma V^\beta \frac{\partial}{\partial y^{\mu}} \left( \frac{\partial y^\rho}{\partial x^\beta}\right)\\ &=V^{\rho}\omega_{\alpha}\left[\frac{\partial}{\partial x^{\rho}}\left( \frac{\partial x^\alpha}{\partial y^\mu}\right)+ \frac{\partial x^{\alpha}}{\partial y^{\sigma}}\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right)\right]\tag{index juggling}\\ &= V^{\rho}\omega_{\alpha} \left[\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial}{\partial y^{\sigma}}\left(\frac{\partial x^\alpha}{\partial y^\mu}\right)+ \frac{\partial}{\partial y^{\mu}} \underbrace{\left(\frac{\partial x^{\alpha}}{\partial y^{\sigma}}\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right)}_{=\delta^{\alpha}_{\rho}}-\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial x^{\alpha}}{\partial y^{\sigma}}\right)\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right]\\ &=V^{\rho}\omega_{\alpha}\left[\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial^2x^{\alpha}}{\partial y^{\sigma}\partial y^{\mu}}+0-\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial^2x^{\alpha}}{\partial y^{\mu}\partial y^{\sigma}}\right]\\ &=0, \end{align} where in the final step, we used the fact that $x^{\alpha}$ is a smooth function, so its mixed partials relative to the $y$ coordinates (and this is a very important point; it is important that both partials are taken with respect to the same coordinate system… that is the only case in which the Schwarz theorem on symmetry of mixed partials from multivariable analysis can be adapted to the manifold setting!) are equal.
Finally, I should remark that this has nothing to do with Lorentz transformations at all. Lie derivatives are defined on arbitrary smooth manifolds, independently of any Riemannian/Lorentzian metric, so you can talk about Lie derivatives even if you can’t talk about SR/GR/Lorentzian geometry/covariant derivatives etc. The only thing I’ve used in my above manipulations are index juggling, and that the matrices $\frac{\partial x}{\partial y}$ and $\frac{\partial y}{\partial x}$ are inverses (a purely analysis fact), and finally, equality of mixed partials in a given coordinate system (a direct consequence of an analysis fact).