Finding Connection Forms on a surface I'm trying to understand connection forms on Riemannian Manifolds and I am kind of confused by this example that I found in a paper I am reading


I am pretty sure I understand how $w_1,w_2$ were found, but $w_{12}$ is the connection form and I've been searching through quite a few sources and I am just not understanding how they find it. From my readings, I found the connection form is found by calculating $w_{12}(v) = \nabla_v e_1 \cdot e_2$ where $v$ is an element in the tangent space, so a vector field. And $\nabla_v e_1= \frac{d}{dt}e_1(p+tv)(0)$ for a point $p$ on our manifold, but then I am confused are we evaluating a vector field $e_1$ in terms of another vector field, similar to the Lie bracket for example? Then with respect to the dot product, we use the Riemannian metric defined by $w_1$ and $w_2$ to find it I think. But I am just kind of confused about this whole thing, I am fairly new to Riemannian Geometry and this idea of a connection form really confuses me. If somebody could help me with this example or any other example that would be great, I am having a lot of difficulties understanding this but it seems like such a simple concept...
 A: Using some rather compact notation, we have that if $\mathfrak{e}=\mathfrak{v}R_\phi$, then $$\omega = R_{-\phi}\theta R_\phi + R_{-\phi} {\rm d}(R_\phi),$$because this is the general transformation law for local connection $1$-forms relative to two different frames, in terms of the change-of-basis matrix. Now, this becomes $$\begin{pmatrix} 0 & \omega_{12} \\ -\omega_{12} & 0\end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix}\begin{pmatrix} 0 & \theta_{12} \\ -\theta_{12} & 0\end{pmatrix}\begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi\end{pmatrix}+ \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix}\begin{pmatrix} -\sin\phi & -\cos\phi \\ \cos\phi & -\sin\phi\end{pmatrix}\,{\rm d}\phi. $$Looking only at the $(1,2)$-entry, the desired formula follows after applying $\cos^2\phi+\sin^2\phi=1$.

Proof of transformation law. If $(M,\nabla)$ is any affine manifold (i.e., $M$ is equipped with an arbitrary affine connection $\nabla$; it may have torsion, $M$ doesn't need to have a metric, we're doing it freestyle), and $\mathfrak{e}$ and $\mathfrak{v}$ are local frames with connection $1$-form matrices $\theta$ and $\omega$, and related via $\mathfrak{e} = \mathfrak{v}A$ for some non-singular matrix of functions $A$, we'll show that $$\theta = A^{-1}\omega A + A^{-1}{\rm d}A.$$
Namely, our assumptions fully written are that $$e_j = \sum_i a^i_{\,j}v_i,\quad \nabla e_j = \sum_i\theta^i_{\,j}\otimes e_i,\quad \nabla v_j = \sum_i \omega^i_{\,j}\otimes v_i.$$We'll write the latter two simply as $$\nabla\mathfrak{e} = \mathfrak{e}\theta\quad\mbox{and}\quad \nabla\mathfrak{v} = \mathfrak{v}\omega.$$Thus, we compute $$\mathfrak{v}A\theta = \mathfrak{e}\theta = \nabla\mathfrak{e} = \nabla(\mathfrak{v}A) = (\nabla\mathfrak{v})A+\mathfrak{v}\,{\rm d}A = \mathfrak{v}\omega A +\mathfrak{v}\,{\rm d}A = \mathfrak{v}(\omega A + {\rm d}A), $$so linear independence of $\mathfrak{v}$ gives $$A\theta = \omega A + {\rm d}A \implies \theta = A^{-1}\omega A + A^{-1}\,{\rm d}A.$$Same proof gives same transformation law for connection $1$-forms in an arbitrary vector bundle with connection, and two local trivializations.

You can try to read my short notes on Cartan computations, pages 11 through 14 of my notes on principal bundles, and the book by Loring Tu (Amazon link).
