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:

  1. 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.

  2. 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 $TO\mathbb{S}^2 \subseteq \mathbb{R}^3\times\mathbb{R}^{9}\times\mathbb{R}^3\times\mathbb{R}^{9}$?

  1. 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 $\tilde{\gamma}$.
  • $\begingroup$ I haven't read the whole question but to talk about lifting you need a map $\mathbb S^2\to \mathbb R^2$. Which map are you using? Without a map the word "lift" has no meaning. In general if a map is a Riemannian submersion, then you can lift any path in the base to a horizontal path in the total space. Any Riemannian submersion of manifolds of the same dimension is a local isometry. Round sphere is not locally isometric to the Euclidean plane. What you call "energy" is actually the square of the speed. It depends on how you parametrize the curve. Any curve can be parametrize by unit speed. $\endgroup$ – Igor Belegradek May 6 '20 at 14:46
  • $\begingroup$ @IgorBelegradek, thanks for the comment. I understand this depends on the parametrization, but I exactly want to capture that (i.e. not allow for any parametrization), but stick to a given one that whomever gave me $w$ chose. When I say lift (technically) I mean a lift of curves from the sphere to the orthonormal frame bundle. I think my usage of the word 'lift' to denote taking a curve in Euclidean space to the sphere was technically incorrect. $\endgroup$ – PPR May 6 '20 at 14:56
  • $\begingroup$ You need to edit the question so that it makes sense mathematically. Otherwise, it may get closed as "unclear what you are asking". It seems this is what you want: fix a unit speed parametrization of a curve in $\mathbb S^2$ and $\mathbb R^2$. Then any other parametrization differs from the fixed one by a self-map $\phi$, and then the new speed will be $|\phi^\prime|$. This is true both in $\mathbb S^2$ and $\mathbb R^2$. Am I missing something? $\endgroup$ – Igor Belegradek May 6 '20 at 15:21
  • $\begingroup$ @IgorBelegradek, as far as I understood from the construction of the orthonormal frame bundle, there is a 1-1 correspondence between paths in $\mathbb{R}^2$ and paths in $\mathbb{S}^2$ in an isometric way, which doesn't require charts. The question is how to realize this 1-1 map explicitly, thinking extrinsically of the sphere embedded in $\mathbb{R}^3$, starting from a curve in $\mathbb{R}^2$. In your comment you fix two curves on the two spaces, but it's not clear (to me) how to relate the two $\endgroup$ – PPR May 6 '20 at 15:32
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    $\begingroup$ Maybe the chapter 'rolling without twisting or slipping' in Appendix B of Sharpe's Differential Geometry book might be an interesting read, it sounds similar to what you are trying to do, and can be applied extrinsically. $\endgroup$ – S.Surace May 8 '20 at 10:21

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).


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