About the antisymmetry of the curvature matrix in vector bundle Suppose we have a vector bundle $\pi:  E \rightarrow M $ with fiber $F_p = \pi^{-1}(p) = \mathbb{R}^d$. $$D: Γ(E) \rightarrow Γ(TM^* \otimes E) $$ is connection in the bundle. $K = D \circ D : Γ(E) \rightarrow Γ(\Lambda^2TM^* \otimes E \otimes E^*) $ is curvature two-form. As we know $$ D(e) = e \, \omega,$$ $$ K(e) = e \, \Omega,$$ where $e = (e_1, \ldots, e_n)$ is local frame. $$\Omega = d\omega + \omega \wedge \omega.$$
Does the matrix of curvature of connectivity in the local frame always be antisymmetric $\Omega_{ij}= - \Omega_{ji}$? If so, how to prove it?

No, this is true if you have a Riemannian metric on $E$ and the frame $e$ is an orthonormal frame.
