2 Fixed typos

Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the exterior covariant derivative of the connection $1$-form' to get the curvature.

Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$.

Now, given a representation $\rho:G\to V$, \text{GL}(V)$, where$V$is a vector space, one can define an associated vector bundle$E = P\times_\rho V$. Using$\omega$, it is possible to define an 'exterior covariant derivative operator'$D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$(where$A^p\to M$is the bundle of alternating (i.e., 'exterior')$p$-forms on$M$). The operator${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$then turns out to be linear over the$C^\infty$alternating forms, so it it is determined by its value when$p=0$, i.e., by${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of$\text{End}(E)\otimes A^2$, i.e., a$2$-form with values in$\text{End}(E)$. The formula for this section, when pulled back to$P$, can now be expressed in terms of$\Omega$in the usual way. In particular,${D_\omega}^2$vanishes identically if$\Omega$does. Note that one does not take the 'exterior covariant derivative' of the$1$-form$\omega$anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of$E$. I suspect that what you may be trying to do is interpret$\omega$as a$1$-form on$P$with values in the trivial bundle$P\times\frak{g}$and then say that the curvature is the 'exterior covariant derivative' of$\omega$. However, to make this work, you have to specify a connection on the trivial bundle$P\times\frak{g}$, which amounts to choosing a$1$-form$\eta$on$P$that takes values in$\text{End}(\frak{g})$, so that \text{End}(\frak{g})$ and setting $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the right' answer?

1

Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the exterior covariant derivative of the connection $1$-form' to get the curvature.

Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$.

Now, given a representation $\rho:G\to V$, where $V$ is a vector space, one can define an associated vector bundle $E = P\times_\rho V$. Using $\omega$, it is possible to define an 'exterior covariant derivative operator' $D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$ (where $A^p\to M$ is the bundle of alternating (i.e., 'exterior') $p$-forms on $M$). The operator ${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$ then turns out to be linear over the $C^\infty$ alternating forms, so it it determined by its value when $p=0$, i.e., by ${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of $\text{End}(E)\otimes A^2$, i.e., a $2$-form with values in $\text{End}(E)$. The formula for this section, when pulled back to $P$, can now be expressed in terms of $\Omega$ in the usual way. In particular, ${D_\omega}^2$ vanishes identically if $\Omega$ does.

Note that one does not take the 'exterior covariant derivative' of the $1$-form $\omega$ anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of $E$.

I suspect that what you may be trying to do is interpret $\omega$ as a $1$-form on $P$ with values in the trivial bundle $P\times\frak{g}$ and then say that the curvature is the 'exterior covariant derivative' of $\omega$. However, to make this work, you have to specify a connection on the trivial bundle $P\times\frak{g}$, which amounts to choosing a $1$-form $\eta$ on $P$ that takes values in $\text{End}(\frak{g})$, so that $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the right' answer?