Let $\pi: P \rightarrow M$ be a principal $G$-bundle with connection $A$ and $E = P \times_\rho V$ an associated vector bundle to $P$ where $V$ is a vector space and $\rho: G \rightarrow GL(V)$ is a representation of $V$. For a section $\Phi$ of $E$ and a vector field $X \in \mathfrak{X}(M)$ on $M$ let $\nabla_X^A \Phi$ denote the covariant derivative of $\Phi$.
Now let $U \subset M$ and $s: U \rightarrow P$ a local section, and $A_s = s^* A$ a local Lie-algebra valued connection 1-form on $U$ defined via the pullback under $s$. Let $\phi: U \rightarrow V$ be the map such that the section $\Phi$ is given by $[s, \phi]$, which is known to be unique. Then we have that
$$\nabla_X^A\Phi(x) = [s(x), d\phi(X) + \rho_*\big(A_s(X)\big)\phi(x)] \in E_x.$$
I have a minor question about the above formula, which is the meaning of $d\phi$. This notation seems to be standard but what is this supposed to represent? If it means the differential why do we not use $\phi_*$? I assume this is more than just notation as the formula uses both the $d(\cdot)$ and $(\cdot)_*$ notation.
This notation arises when taking the following time derivative:
$$\nabla_X^A\Phi = \frac{d}{dt}\Big|_{t=0} [s(x), \rho(g(t))\phi(\gamma(t))] = [s(x), \rho(\dot{g}(0))\phi(x) + d\phi(X)]$$
where $g(t)$ is some smooth curve in $G$ and $\gamma$ is a curve such that $\gamma(0) = x$ and $\dot{\gamma}(0) = X$.
In the proof I am following they also use this notation again as follows:
$$\frac{d}{dt}\Big|_{t=0} s(\gamma(t)) = ds(X).$$
What is the difference between $d(\cdot)$ and $(\cdot)_*$ here?
Best Answer
I took these ideas from Sharpe's Differential geometry book (see e.g. chapter 1, section 5: differential forms. i.e. page 52). $\def\R{\mathbb R} \def\id{\mathrm{id}}$
I will start by recalling some conventions. In the following everything will be smooth. Remember that a curve $\gamma:(-\epsilon,\epsilon)\to M$ is a curve with $p=\gamma(0)$ induces a tangent vector $\gamma'(0)\in T_pM$, i.e. a differential operator that acts on functions $f:M\to\R$ by the equation $$\gamma'(0).f = (f\circ\gamma)'(0)\in\R.$$ In fact, $T_pM$ is the vector space of all such operators, and these vector spaces assemble to form the tangent bundle $TM$ of $M$.
A function $\phi:M\to N$ induces, at every point $p\in M$, a linear function $\phi_{*,p}:T_pM\to T_{f(p)}N$ called the pushforward along $\phi$ at $p$ defined by $$\phi_*(\gamma'(0))=(\phi\circ\gamma)'(0)$$ i.e. the vector induced by the new curve $\phi\circ\gamma:(-\epsilon,\epsilon)\to N$. These functions assemble into a new smooth function $\phi_*:TM\to TN$, called the pushforward along $\phi$.
Now, in particular, if $N=\R$, then a function $f:M\to\R$ produces the pushforward $f_*:TM\to T\R$. But the tangent bundle $T\R$ has an identification with $\R\times\R$ that sends a tangent vector $\gamma'(0)\in T_t\R$ to the pair $(t,\gamma'(0).\id_\R)$. In other words, every tangent vector $\gamma'(0)$ at $t\in\R$ is of the form $\gamma'(0)=\alpha \partial_t$ for some unique $\alpha\in\R$, (where $\partial_t\in T_t\R$ is the tangent vector induced by the curve $\beta:\R\to\R$ given by $\beta(s)=s+t$; note that the $+t$ part is there only to ensure $\beta(0)=t$) and then the identification $T\R\simeq\R\times\R$ sends $$\gamma'(0)=\alpha\partial_t \mapsto (t,\alpha).$$ In yet another words, the first component records the base point $t$ and the second component records the speed $\alpha$ at which the curve $\gamma$ passes through $t$. One could say that $t$ is the "point part" of $\gamma(0)$, since $t=\gamma(0)$, while $\alpha$ is the "vector part", since $\gamma'(0)=\alpha\partial_t$.
Now take $f:M\to\R$ and fix a point $p\in M$. The elements $v=\gamma'(0)\in T_{f(p)}\R$ get identified with the ordered pairs of the form $(f(p),\alpha)$. Since $f_{*,p}:T_pM\to T_{f(p)}\R$ is linear, when you compose with this identification, $f_{*,p}$ takes the form $$v\mapsto (f(p),f'(p)v)$$ for a unique linear map $f'(p):T_pM\to\R$. Then the exterior derivative $df$ of $f$ is a function $df:TM\to\R$ defined by the formula $df(v)=f'(p)v$.
In short, the value of the exterior derivative $df$ at a vector $v\in TM$ definitely depends on the base point of $v$, but this information about the base point is not kept after the evaluation: you only keep the number $df(v)\in\R$.
Similarly, if instead of choosing $N=\R$ you choose $N=V$ for some vector space $V$, you get an identification $TV\simeq V\times V$ and then the exterior derivative $df:$ of a function $f:M\to V$ is given by the vector part of $f_*:TM\to TV$ (i.e. the second component under the identification above).
Hope it helps.