Extrinsic horizontal path lifting As a follow up question to my previous question about the orthonormal frame bundle, I would like to understand a simple example explicitly. 
Let $\mathbb{S}^2$ be written extrinsically as $$\mathbb{S}^2 = \{x\in\mathbb{R}^3|\|x\|=1\}$$ and let an arbitrary smooth path $w:[0,1]\to\mathbb{R}^2$ be given. 
The ultimate goal is to lift $w$ to a path $\gamma:[0,1]\to\mathbb{S}^2$ which has the same "energy", i.e. $$ \langle\dot{w},\dot{w}\rangle_{\mathbb{R}^2} \stackrel{!}{=} g_\gamma(\dot{\gamma},\dot{\gamma}) $$ on $[0,1]$, where $g$ is the Riemannian metric on $\mathbb{S}^2$ (which, as written here, is induced by the Euclidean metric on $\mathbb{R}^3$. 
I presume eventually there would be some choice of (arbitrary) initial conditions and a (1st oder?) ODE to solve in order to obtain a path $\gamma:[0,1]\to\mathbb{S}^2\subseteq\mathbb{R}^3$. 
I tried to follow this in a systematic way according to the prescription of: 1) building an orthonormal frame bundle $O\mathbb{S}^2$ on top of $\mathbb{S}^2$, 2) lifting $w$ to a horizontal path $\tilde{\gamma}:[0,1]\to O\mathbb{S}^2$, and 3) projecting down from $O\mathbb{S}^2$ to $\mathbb{S}^2$. I tried to do all of this extrinsically without using charts, and that's where I got stuck (perhaps this is a pointless endeavor, but I thought one point of using the frame bundle is to work with global objects rather than within charts).
Question 1: Is there a better procedure to achieve this goal rather than follow the horizontal path lifting? Perhaps something more explicit in this particular setting.
Question 2: How to follow the horizontal path lifting procedure extrinsically in this case? Here's how I got stuck:


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*Define the orthonormal frame bundle extrinsically as $$ O\mathbb{S}^2 = \{ (x,A) \in \mathbb{R}^3\times\mathbb{R}^{9} | x\in\mathbb{S}^2 \land A \in O(3) \text{ s.t. }Ax=x\}\,. $$ In the case of the sphere it's easy to picture that the fiber is one dimensional ($\dim(O(2))=1$) and amounts to the angle by which to rotate a basis of a 2D tangent space to each point on the sphere.

*Now we need to define the tangent bundle of this, $$TO\mathbb{S}^2 = \{ (x,A,v_x,v_a) \in \mathbb{R}^3\times\mathbb{R}^{9}\times\mathbb{R}^3\times\mathbb{R}^{9} | (x,A)\in O\mathbb{S}^2\land \langle x,v_x\rangle+\langle A,v_a\rangle=0\}\,.$$ and its horizontal sub-bundle $HO\mathbb{S}^2 = ???$, find the two vector fields $H_1,H_2$ that build a global frame for $HO\mathbb{S}^2$, I guess they are called the canonical horizontal vector fields. This is the step where I get stuck because as far as I know, to check that a curve $u:[0,1]\to O\mathbb{S}^2$ is horizontal, I need to verify the equation $$ \nabla_{\dot{x}} v = 0 $$ for all columns $v$ in $A$ which are not equal to $x$, where $(x,A)=u$. Here $\nabla$ is the covariant derivative, which I understand in this extrinsic description is just the gradient along a vector projected to the tangent space of the manifold. So if $P_x = I - x\otimes x^\ast$, then the covariant derivative of two vector fields $a,b$ equals $$(\nabla_a b)(x) = P_x (a_j \partial_j b)(x)\,.$$ Using this interpretation I find the equation for a horizontal curve to equal $$ P_x \dot{v}(x) = 0 $$ for any column $v$ in $A$ not equal to $x$. This stopped making sense to me. 
How to find $H_1,H_2$ in this description? Is there any point to write them as elements of $TO\mathbb{S}^2 \subseteq \mathbb{R}^3\times\mathbb{R}^{9}\times\mathbb{R}^3\times\mathbb{R}^{9}$?


*Solve the ODE $$\dot{\tilde{\gamma}} = \sum_{i=1}^2 H_i(\tilde{\gamma}) \dot{w_i}$$ for $\tilde{\gamma}$ and project $\tilde{\gamma}\mapsto\gamma$. Here the notation $H_i(\tilde{\gamma})$ means evaluate the vector field $H_i$ at the base point $\tilde{\gamma}$.

 A: So I think I have an answer, but instead of using the ODE in Step 3, it uses a simpler equation that implies it: $$ \dot{w} = \tilde{\gamma}^{-1}\dot{\gamma}\,. $$
Here $w:[0,1]\to\mathbb{R}^2$ is a given curve, $\gamma:[0,1]\to\mathbb{S}^2$ is the unknown curve, and $\tilde{\gamma}$ is the horizontal curve in $O\mathbb{S}^2$ lifted from $\gamma$.
It turns out that it is after all quite easy to write the horizontal curve in $\tilde{\gamma}$ induced by a given $\gamma$ if one uses spherical coordinates (and later on one may switch back to Cartesian coordinates if need be). Then if $\theta,\varphi:[0,1]\to\mathbb{R}$ parametrizes the curve $\gamma$ in spherical coordinates, find $\psi:[0,1]\to\mathbb{R}$ out of the equation $$ \dot{\psi} = -\dot{\varphi}\cos(\theta)\,. \tag{H}$$
Then $\psi$ gives the angle of rotation compared with the standard orthonormal frame on $T_\gamma\mathbb{S}^2$ given by the (co-moving) orthonormal frame $\hat{\theta},\hat{\varphi}$.
Then for each $t\in[0,1]$, $\tilde{\gamma}(t)$ may be viewed as a map $$ \tilde{\gamma}(t):\mathbb{R}^2\to T_{\gamma(t)}\mathbb{S}^2 $$ which is in fact an isometric isomorphism by construction. In our case, parametrized by $\psi$, it is given by $$ \mathbb{R}^2\ni v\mapsto (R_\psi v)_1\hat{\theta}+(R_\psi v)_2\hat{\varphi} $$
where $$R_\psi=\begin{bmatrix}\cos(\psi) && -\sin(\psi) \\ \sin(\psi)&&\cos(\psi)\end{bmatrix}$$ is the $2\times 2$ rotation matrix associated with $\psi$. Hence there is an easy way to write the inverse map $$ \tilde{\gamma}(t)^{-1}:T_{\gamma(t)}\mathbb{S}^2\to\mathbb{R}^2 $$ which is given by $$ y_\theta \hat{\theta} + y_\varphi\hat{\varphi} \mapsto R_\psi^{-1}\begin{bmatrix}y_\theta \\ y_\varphi\end{bmatrix}\in\mathbb{R}^2\,. $$
Now $$\dot{\gamma} = \dot{\theta}\hat{\theta}+\sin(\theta)\dot{\varphi}\hat{\varphi}$$ and so this finally yields the following ODE to be solved for the unknowns $\theta,\varphi$: $$ \dot{w} = R_\psi^{-1}\begin{bmatrix}\dot{\theta} \\ \sin(\theta)\dot{\varphi}\end{bmatrix} $$ where $\psi$ is also a function of $\theta,\varphi$ via (H).
