Suppose that $G$ is a Lie group and $\pi:P\rightarrow X$ is a principal $G$-bundle, where $X$ is taken to be a closed manifold. Denote the affine space of connections for this principal bundle as $\mathcal{A}_P$, to which we endow a differential structure thinking of it as a Fréchet manifold.
Consider now a smooth path of connections, $$ \omega:I\longrightarrow \mathcal{A}_p, $$ $I = [0,1]\subset \mathbb{R}$. Denote by $f:X\times I \rightarrow X$ the projection into the first factor and set $Q =f^*P$ the pullback principal bundle over $X\times I$.
This may be a silly question, but can we 'paste' the different $\omega(t),t\in I$, to obtain a connection on $Q$? Intuitively, of course, one would want $\omega(t)$ to act as a connection on the slice $P\times\{t\}$, but what should be the correct way to act on tangent vectors of $TQ$ that have a 'timeward' component?
I am convinced the formula must involve the derivative with respect to $t$ of $\omega$, $$ \dot{\omega}(t) = \left.\frac{d}{ds}\right|_{s=t} \omega(s) \in \Omega^1_{\text{Ad}}(P,\mathfrak{g}) \cong \Omega^1(X,\text{ad}(P)), $$ where $\Omega^1_{\text{Ad}}(P,\mathfrak{g})$ is the vector space of tensorial forms of type $\text{Ad}$ and $\text{ad}(P) = P \times_{\text{Ad}} \mathfrak{g}$ is the adjoint bundle, but how specifically?
Furthermore, how does the curvature of the 'pasted' connection form on $Q$ look like?
