Is every connection locally flat for an other connection?

Question

Consider a $C^{\infty}$ connection $d_A = d+A$ on the unit ball $B^n\subset \mathbb{R}^n$. Does there exists another connection $d_{\tilde{A}} = d+\tilde{A}$ such that $d_{\tilde{A}} A = 0$? That is to say, can we find a second connection which makes $A$ flat?

I am interested in the local version, that is to say, I can reduce the radius of the ball if necessary. Hence, I imagine that there should not be any topological issue.

We can enlarge the setting a little bit: given a section $s$ of a vector bundle over $B^n$, can we find a connection $\nabla$ on this vector bundle such that $\nabla s=0$?

Answer 1

This was already explained by Ben McKay, so I am only elaborating on the point made. Let $\mathscr{M}$ be a connected smooth manifold and $E = \mathbf{R}^k \times \mathscr{M}$ be a trivial vector bundle. If $s_0 \in \Gamma(E)$ is nowhere vanishing, then there exists a smooth bundle isomorphism $\tau \in \Gamma(\operatorname{End} E)$ such that $\tau s_0$ is constant. This suggests we define the connection $\mathrm{d}^\tau = \tau^{-1} \circ \mathrm{d} \circ \tau$. Indeed, as a composition of $\mathbb{R}$-linear maps it is also $\mathbb{R}$-linear, and for any $g \in C^\infty(\mathscr{M})$ and $s \in \Gamma(E)$, $$ \mathrm{d}^\tau (gs) = \tau^{-1} \mathrm{d}(\tau g s) = \tau^{-1} \big(\mathrm{d}g \otimes \tau s + g\hspace{1pt} \mathrm{d}(\tau s) \big) = \mathrm{d}g \otimes s + g \hspace{1pt}\mathrm{d}^\tau(s), $$ so the Leibniz rule holds. By construction, $s_0$ is parallel with respect to $\mathrm{d}^\tau$, and the same holds true for $\mathrm{d}^{\alpha \tau}$, where $\alpha \in \Gamma(\mathrm{GL}(\mathbf{R}^k) \times \mathscr{M})$ is any constant section.

On the other hand, if $s_0$ vanishes at a point, then a connection for which $s_0$ is parallel exists if and only if $s_0$ is identically zero (which implies $s_0$ is parallel with respect to any connection).