Let $\mathbb{S}^{2} \subset \mathbb{R}^{3}$ be the $2$-sphere ($\mathbb{S}^{2} = \left\{ (x,y,z) \in \mathbb{R}^3, \; x^2+y^2+z^2 = 1 \right\}$). Let $p \in \mathbb{S}^{2}$ and $\xi \in T_{p}S^{2} = \left\{ p \right\}^{\perp} \simeq \mathbb{R}^2$. When I wanted to compute explicitly the parallel transport of $\xi$ along a geodesic $\gamma$ (such that $\gamma(0)=p$), I used the following "trick" : if $T_{\gamma,0,t}(\xi)$ denotes the parallel transport of $\xi$ from the point $p$ to $\gamma(t)$, along $\gamma$, then :
$$\forall t, \; T_{\gamma,0,t}(\xi) \in T_{\gamma(t)}\mathbb{S}^{2}$$
and an orthogonal basis of $T_{\gamma(t)}\mathbb{S}^{2}$ is given by : $(e_{1}(t),e_{2}(t))=(\gamma'(t), \gamma(t) \wedge \gamma'(t))$. As a consequence, $T_{\gamma,0,t}(\xi)$ writes :
$$ \forall t, \; T_{\gamma,0,t}(\xi) = \alpha(t) e_{1}(t) + \beta(t) e_{2}(t) $$
And, since the parallel transport is an isometry,
$$ \left\langle T_{\gamma,0,t}(\xi),\gamma'(t) \right\rangle = \left\langle \xi,\gamma'(0) \right\rangle = \alpha^{2}(t) \Vert e_{1}(t) \Vert^{2} \tag{1}$$
and
$$ \Vert T_{\gamma,0,t}(\xi) \Vert^{2} = \Vert \xi \Vert^{2} = \alpha^{2}(t) + \beta^{2}(t) \tag{2}$$
$(1)$ and $(2)$ allow to determine $\alpha$ and $\beta$.
But I think the method does not generalize to higher dimension.. Here is my idea : If I want to compute the parallel transport in $\mathbb{S}^{3} \subset \mathbb{R}^{4}$ using the same method, I could determine an orthogonal basis of $T_{\gamma(t)} \mathbb{S}^{3}$, say $(e_{1}(t),e_{2}(t),e_{3}(t))$ and write :
$$ T_{\gamma,0,t}(\xi) = \alpha_{1}(t) e_{1}(t) + \alpha_{2}e_{2}(t) + \alpha_{3}(t) e_{3}(t) $$
but I don't know how I would determine $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$ since the relations $\left\langle T_{\gamma,0,t}(\xi),\gamma'(t)\right\rangle = \left\langle \xi,\gamma'(0) \right\rangle$ and $\Vert T_{\gamma,0,t}(\xi) \Vert^{2} = \Vert \xi \Vert^{2}$ do not give enough information (mainly because two equations are not enough to determine the three $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$)… Is there a way out with this method or shall I compute the parallel transport by solving (when possible) the differential equation ?
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$Here's a slightly different viewpoint (even for the $2$-sphere). The point $p$ in $S^{n}$ and the normalized initial velocity $v$ of the geodesic $\gamma$ may be viewed as orthogonal unit vectors in $\Reals^{n+1}$, and they span a real $2$-plane $N$ (for "normal space") through the origin.
The tangent space $T_{p} S^{n}$ is identified with the orthogonal complement of $p$, and has an orthogonal decomposition $\operatorname{span}(v) \oplus B$, ($B$ for "binormal space"). An arbitrary vector $\xi$ in $T_{p} S^{n}$ decomposes: let $av = a\gamma'(0)$ denote the tangential component, and $b$ the component of $\xi$ in $B$. (In your notation, $av = \alpha(0) e_{1}(0)$ and $b = \beta(0) e_{2}(0)$. Incidentally, your functions $\alpha$ and $\beta$ are constant.)
The geodesic $\gamma$ has explicit parametrization $$ \gamma(t) = (\cos t) p + (\sin t) v. $$ The tangent space of $S^{n}$ at $q = \gamma(t)$ is spanned by $\gamma'(t)$ and $B$, and under parallel transport the $B$ component of $\xi$ is constant. (This is the crucial point.)
Consequently, the parallel transport of $\xi$ along $\gamma$ from $p$ to $q$ is $a\gamma'(t) + b$. (This expresses the geometric fact that parallel transport along a great circle "depends" only on the subspace of $\Reals^{n+1}$ spanned by $p$, $v$, and $\xi$.)
The preceding isn't all that different from your idea; it merely "agglomerates" the $B$ component of $\xi$ instead of expressing it in terms of an orthonormal basis of $T_{p} S^{n}$.