In order to better understand connections on principal bundles, I would like to compute the Ereshmann distribution (horizontal distribution) and the connection 1-form associated on the frame bundle of $S^2$, explicitly in coordinates. But I am stuck!
I did the following:
I took the coordinates for $S^2$: given $$U=\Big\{ (\phi,\theta ): \theta \in [0,2\pi) , \phi \in \Big]0 ,\pi[ \Big[ \Big\}$$ I took $\varphi: U \to S^2$ defined by $\varphi(\phi, \theta)= \big(r \cos(\theta ) \sin(\phi) , r \sin(\theta)\sin (\phi) , r \cos(\phi)\big)$ The metric is given by $$g_{\phi,\theta}(\phi,\theta)= \begin{bmatrix} r^2 & 0\\ 0 & r^2\sin (\phi)^2 \\ \end{bmatrix} $$ and an orthonormal basis by $$e_{\phi}:= \frac{1}{r}\partial_{\phi} \ \ \ \ \ \ \ e_{\theta}:=\frac{1}{r \sin(\phi)}\partial_{\theta} $$ The dual basis is then $e_{\phi}^*:= r d \phi ,e_{\theta}^*:= r \sin(\phi)d\theta $. I used the first Cartan equation to compute the connection matrix (on $S^2$) in local coordinates. The first Cartan equation gives$$ \begin{cases} de_{\phi}^*=-\omega_{1}^{1} \wedge e_{\phi}^* -\omega_{2}^1 \wedge e_{\theta}^* \\ de_{\theta}^* =-\omega_{1}^{2} \wedge e_{\phi}^*- \omega_{2}^{2} \wedge e_{\theta}^* \end{cases} $$ For a Levi Civita connection we have $\omega_{1}^{1}=\omega_{2}^{2}=0$ e $\omega_{1}^{2}=-\omega_{2}^{1}$ then if $\omega^{1}_{2}=a_{\phi }d\phi +a_{\theta}d\theta $ we obtain
$$ \begin{cases} d \big(r d \phi \big)= - \big( a_{\phi }d\phi +a_{\theta}d\theta \big) \wedge \Big( r\sin(\phi)d\theta \Big) \\ d\big( r\sin(\phi)d\theta \big) = \big(a_{\phi}d\phi +a_{\theta}d\theta \big)\wedge \Big( r d \phi \Big) \end{cases} $$ that is $$ \begin{cases} 0 = - a_{\phi } r\sin(\phi)d\phi \wedge d\theta \\ r\cos(\phi)d\phi \wedge d\theta = a_{\theta} r d\theta \wedge d \phi \end{cases} $$ then $a_{\phi }=0 ,a_{\theta} = -\cos(\phi)$ thus$$\omega^{1}_{2}= -\cos(\phi) d\theta $$
I'm quite sure the computation is correct (I've computed the Gaussian curvature!). Let's now consider the frame bundle $GL(2, \mathbb{R} ) \to \text{Fr}(TS^2)$. From the connection matrix I've computed, it should be possible to compute the connection 1-form $\omega$ on $\text{Fr}(TS^2)$ with values in $\mathfrak{gl}(2,\mathbb{R})$ and from that one the horizontal distribution by imposing the equations for $X$ given by $\omega(X)=0$. I read as much books and notes as possible on that thing, but I still don't understand how to get $\omega$!
The best reference I've found is "Differential geometry" by Loring W. Tu.