$E\to B$ is a vector bundle. $\nabla$ is a connection of this bundle. Choose a local trivialization of $E$, then we can write a formula of $\nabla$ with respect to this local trivialization: $\nabla(s)=\text{d}s+\omega\otimes s$. I know that if we change the local trivialization by some section $g\in\Gamma(\text{Aut}(E))$, then $\omega\to\text{d}g\cdot g^{-1}+g\omega g^{-1}$.
My question is that for any connection, can we choose a local trivialization such that $\nabla(s)=\text{d}s$? I think to solve this we need to find a section $g\in\Gamma(\text{Aut}(E))$ such that $\text{d}g+g\omega=0$. But I can't solve the equation.
I also try to go this way. I know that a path $\gamma$ in the base space $B$ can be lifted to $E$ and it defines a isomorphism $E_0\to E_1$ ($E_x$ means the fiber over $\gamma(x)$). Suppose our connection is flat, locally two paths with same end points give same isomorphism. So we can use this fact to choose section $g\in\Gamma(\text{Aut}(E))$. But I'm not sure if it is right.
$\nabla(s)=ds$ implies that the curvature and the torsion form vanish on the trivialization. This is not always positive. Take $B=S^2$ and $E$ the tangent bundle of $S^2$. The sectional curvature of the Levi-Civita connection associated to the canonical differentiable metric on $S^2$ is 1.