I am an analyst struggling through some geometry used in physics.


**Some background:** For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$ (a "principal connection"). This is a Lie algebra-valued 1-form. 


As for the curvature two-form, either you see definitions with no explanation at all, e.g. "The curvature is given by $\Omega = d\omega + \frac{1}{2}[\omega,\omega]$". This is obviously less than ideal for improving one's intuitive appreciation. Or one defines something called the exterior covariant derivative $D$ (see [wiki](http://en.wikipedia.org/wiki/Exterior_covariant_derivativecurvature?)) and then the curvature is simply the exterior covariant derivative of the connection one-form. 


**The issue:** I can't get round the following observation though: From the point of view of the manifold $P$, $\omega$ is just a one-form with values in some vector space which happens to be $\mathfrak{g}$. Usually when you need to covariantly differentiate such an object, you would need a connection in a bundle $E \to P$ with fibre $\mathfrak{g}$, no? Then $\omega$ would be an $E$-valued one-form on $P$, *i.e.* in $\Gamma(E) \otimes\Omega^1(P)$, and you can differentiate covariantly in the normal way using the connection. Why is this scenario different?


The exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some sort of covariant differentiation $D$ which differentiates forms $\eta$ taking values in $\mathfrak{g}$ and for which $D^2$ is some sort of curvature... but of $P \to M$ and not of the bundle in which $\eta$ is taking values. Isn't this strange? Or is this indeed just how things are? This prompts my more precise question: 



> 
> Is the Lie algebra-valued curvature
>  two-form on a principal bundle P the
>  curvature of some vector bundle over P with fibre $\mathfrak{g}$?
> 
> 
> 



Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P?
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 $\rho:G\to \text{GL}(V)$, 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 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 $\text{End}(\frak{g})$ 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?


