I am following the text here which explains that we are transporting a vector $\mathbf{v}$ around a closed circular loop (C) on a sphere under the constraints $\mathbf{e}_3\cdot\mathbf{v}=0$ and $\mathbf{e}_3 \times d\mathbf{v}=0$ where $\mathbf{e}_3$ is the local normal vector to the surface of the sphere. At each point we can write $\mathbf{v}$ in the local coordinate frame $[\mathbf{e}_1,\mathbf{e}_2]$ as
$$\mathbf{v}(t)=\cos\alpha(t)\mathbf{e}_1(t)+\sin\alpha(t)\mathbf{e}_2(t)$$
where $t$ parameterises the curve. Then, apparently, by using the constraints and this expanded form of $\mathbf{v}$ we realise that
$$d\alpha=\mathbf{e}_1\cdot d\mathbf{e}_2 \equiv \omega_{21}$$
but it seems like there's a bunch of omitted steps to get to this result which I don't understand. Subsequently the total angle $\alpha(C)$ is the integral of $\omega_{21}$ and is then linked to geometrical phases etc etc but I need to understand the omitted steps first.
Thanks very much for any help
Looking at the paper, the actual condition is $e_3 \times dv = 0$. We compute: $$ dv = d\alpha ( -e_1\sin{\alpha}+e_2\cos{\alpha} )+de_1 \cos{\alpha}+de_2 \sin{\alpha} $$ Now dot with $e_1$. Since $dv \parallel e_3$, we have $e_1 \cdot dv=0$, and $$ 0 = d\alpha ( -e_1 \cdot e_1 \sin{\alpha}+e_1 \cdot e_2\cos{\alpha} )+ e_1 \cdot de_1 \cos{\alpha}+ e_1 \cdot de_2 \sin{\alpha}. $$ We know $e_1 \cdot e_2= 0$ (definition), and $$ 1 = e_i \cdot e_i \implies 0 = 2e_i \cdot de_i, $$ so the equation cancels down to $$ 0 = (-d\alpha +e_1 \cdot de_2) \sin{\alpha}, $$ which implies the given equation. Similarly, dotting with $e_2$ gives the same thing multiplied by $\cos{\alpha}$, so $d\alpha = e_1 \cdot de_2$ for any value of $\alpha$.