Let $\nabla$ be a linear connection in a vector bundle $\pi:E\to M$. Fix $X\in \mathfrak{X}(M)$ and a local frame $s_1,\ldots,s_k\in \Gamma(E_U)$. It seems to me that the plan is to use linear algebra ideas to study the operator $\nabla_X$ in $\Gamma(E)$, great. So one writes $$\nabla_Xs_j = \omega^i_{\hspace{.5ex}j}(X)s_i$$and proves that the $\omega^i_{\hspace{.5ex}j}$ are one-forms. Wonderful.
Now comes my question. Take another local frame $\overline{s_1},\ldots,\overline{s_k}\in\Gamma(E_U)$. Write $\overline{s_j} = h^i_{\hspace{.5ex}j}s_i $ for some convenient smooth functions $h^i_{\hspace{.5ex}j}$. Repeating the above discussion for this new frame, and fiddling around, I find out that $$\overline{\omega^i_{\hspace{.5ex}j}} h^k_{\hspace{.5ex}i}= \omega^k_{\hspace{.5ex}i}h^i_{\hspace{.5ex}j} + {\rm d}h^k_{\hspace{.5ex}j}.$$ I expected the first term but not the ${\rm d}h^k_{\hspace{.5ex}j}$ (apart from the Leibniz rule, I mean). What is the interpretation for the extra term?
Ivo, in heuristic terms, the second term is there because the frames are twisting relative to one another. (In the most rudimentary setting, if I watch you walking in a bus, I see a different speed if the bus is moving relative to me.) But you can see it even in the simplest case of surfaces. Suppose the new (orthonormal) frame $\overline e_1,\overline e_2$ is at a (variable) angle $\theta$ from the original frame $e_1,e_2$. Then the rate at which $\overline e_1$ twists toward $\overline e_2$ is given by the sum of the rate at which $e_1$ twists toward $e_2$ and the rate at which $\theta$ is changing---all of this along some path or direction, instantaneously.
By the way, if you solve for the matrix $\overline\omega$, multiplying by $h^{-1}$ on the right, you should recognize $dh\cdot h^{-1}$ as the (right-invariant) Maurer-Cartan form of the structure group (the orthogonal group in the case I was alluding to).