TITLE: Compute a parallel transport
QUESTION [5 upvotes]: 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 ?

REPLY [2 votes]: You can always set up the coordinate as
 $(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, and choose the geodesic to be $\phi=0$ and $p=(1, 0, 0)$ be a point at the equatorial. Note that $X=\frac{\partial}{\partial\theta}=(\cos\theta\cos\phi, \cos\theta\sin\phi, -sin\theta)$ and $Y=\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=(-\sin\phi, \cos\phi, 0)$ are both parallel along the geodesic $\phi=0$ near p.  
Now let $\xi\in T_pS^2$, decompose $\xi=aX_p+bY_p$, then the parallel transport of $\xi$ along $\phi=0$ is $aX+bY$.  
Since the sphere is homogeneous you can always expect p and the geodesic to be arranged like this.