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:

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 $\tilde{\gamma}$.