Let $M$ be a smooth manifold and $\pi:P\to M$ a principal $G$ bundle over $M$. Suppose that $P$ is equipped with a connection one form $\omega$. We can define an exterior covariant derivative on $P$ via $D\eta = d\eta \circ \operatorname{hor}$, where $\operatorname{hor}$ is the horizontal part of any vector in $TP$. For $G$-equivariant functions $\phi :P\to F$ (that is $\phi(p\cdot g) =\rho(g^{-1})\phi(p)$, where $\rho:G\to F$ is a representation of $G$), and indeed for $G$-equivariant forms as well (but we don't need them), the exterior covariant derivative takes the simple form $$D\phi = d \phi + \rho_{*e}(\omega)\wedge \phi,$$where the wedge product can be defined in a natural way. These $G$-equivariant functions are in one-to-one correspondence with local sections of the associated bundle $Q :=P \times _G F$ over an open set $U$ of $M$, and they contain the same information. The question is then, how does one go from the exterior covariant derivative on $G$-equivariant functions to the covariant derivative of local sections of the associated bundle $Q$? One could naturally take the covariant derivative to simply be the local section $\sigma$ associated to $D\phi(X)$, where $X \in TP$, but this is not a covariant derivative on $Q$, since $X$ is not in $TM$. I have also seen the covariant derivative defined by pulling back $D\phi$ by a local section $s$ of $P$ over an open set $U$, but this approach doesn't make much sense to me. This approach would make more sense if the correspondence mentioned above was between global sections and $G$-equivariant functions, in which case we just take the local representation of the section (i.e. $s^*(D\phi)$). If anyone could clarify, it would be very helpful.
How do we go from a covariant derivative on a principal bundle to a covariant derivative on an associated bundle
connectionsdifferential-geometryprincipal-bundles
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So from my point of view what a covariant derivative should be or should do I'd say that indeed you need a linear structure.
So in principle, when passing from simply vector space-valued functions $M\to W$ (or sections in the trivial bundle $M\times W$, resp.) to sections in a not-necessarily trivial bundle $E\to M$ with fiber-type $W$ you run into troubles defining a derivative of such functions. Usually this is paraphrased as that it is intrinsically "not possible to compare points in different fibers". However, it is possible to chose a covariant derivative, and thereby (at least locally) a frame which is "constant" or parallel to $M$ (cf. my comment above). Once you everywhere have distinguished such a notion of parallelity you can now indeed take derivatives and view it as "how a function changes compared to what we call constant", which is of course just the action of the covariant derivative. For this notion you need a vector space structure.
On the other hand, on a principal bundle you also have a notion of parallel in the sense of everywhere horizontal but as you don't have a vector space structure in the fibers, so "comparison with a constant section" works a bit different, particularly not by something having similar properties as a derivative in the above sense. Rather would you do something like @Deane in the comments above, but I'd say this is not a "derivative" in the sense of what a derivative should do.
With the help of @ahersh23's answer I think I got it now:
It is $$\mathrm{d}\left(\rho(g(x))^{-1}\phi'\right)(X(x)) =\rho(g(x))^{-1}\mathrm{d}\phi'(X(x)) -\rho_\ast\left((g^\ast\mu_G)(X(x))\right)\rho(g(x))^{-1}\phi'(x),$$ so that we find $$\left(\nabla^{A_{s'}}_X\phi'\right)(x)=\rho(g(x))^{-1}\mathrm{d}\phi'(X(x))+\rho_\ast\left(\mathrm{Ad}\left(g(x)^{-1}\right)A_{s'}(X(x))\right)\rho(g(x))^{-1}\phi'(x).$$ Now the adjoint representation of $G$ is defined by $$\mathrm{Ad}:G\rightarrow\mathrm{GL}(\mathfrak{g}),\,g\mapsto\left(L_g\circ R_{g^{-1}}\right)_\ast.$$ With this we find for $Y\in\mathfrak{g}$ and $h\in G$ $$\rho_\ast\left(\mathrm{Ad}(h^{-1})Y\right) =\left(\rho\circ L_{h^{-1}}\circ R_h\right)_\ast(Y) =\left(\rho(h^{-1})\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\left(\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\rho_\ast(Y)\rho(h),$$ where the second equal sign follows from the fact that $(\rho\circ L_{h^{-1}})(g)=\rho(h^{-1}g)=\rho(h^{-1})\rho(g)$ and the last equal sign follows analogously.
Best Answer
In the description of a principal connection via a connection form, this is not so easy to see. The point is that the combination you use to define $D\phi$ is chosen in such a way that it vanishes on any fundamental vector field (if $\phi$ is a form rather than a function, you have to assume that $\phi$ itself is horizontal). Hence $D\phi(X)$ depends only on the projection of $X$ to $TM$. Otherwise put, for a vector field on $M$, you can choose any lift to a $G$-invariant vector fields on $P$ and insert this into to $D\phi$ to get the function corresponding to the covariant derivative of the section determined by $\phi$. This also leads to a description in terms of local sections.
Alternatively, you can move towards the description of principal connections as distributions. Any tangent vector $X$ on $M$ can be uniquely lifted to tangent vectors along the appropriate fiber of $P$ which are horizontal in the sense that they are annihilated by $\omega$. Applying this point-wise to a smooth vector field on $M$, one obtains a horizontal vector field on $P$ which in addition is automatically $G$-invariant. Using this to differentiate $\phi$ directly produces the equivariant function corresponding to the covariant derivative of $s$ in the direction of the initial vector field.