TITLE: How to use the Lie derivative to "perform" a parallel transport along a curve
QUESTION [9 upvotes]: Setup
Consider a metric, for example that of a sphere with fixed radius $R$, i.e.
$$ds^2 = R^2 d\theta^2 + R^2\sin^2\theta^2d\varphi^2,$$
and a curve on that sphere $\gamma = (\theta_0, \varphi)$, where $\theta_0 = const.$ and $\varphi\in[0,2\pi)$, together with a vector $X_0=(X^\theta_0,X^\varphi_0)$. I can parallel transport this vector and in this specific example even determine its components at every specific point of the curve (for a derivation see Section 2 of these lecture notes), with the result
$$
\begin{aligned} 
X^{\theta}(\varphi) &=X_{0}^{\theta} \cos \left(\varphi \cos \theta_{0}\right)+X_{0}^{\varphi} \sin \theta_{0} \sin \left(\varphi \cos \theta_{0}\right), \\ 
X^{\varphi}(\varphi) &=X_{0}^{\varphi} \cos \left(\varphi \cos \theta_{0}\right)-\frac{X_{0}^{\theta}}{\sin \theta_{0}} \sin \left(\varphi \cos \theta_{0}\right). 
\end{aligned}
$$
My Question:
In an exam I was asked to reproduce the above result using the Lie derivative. I think the principle behind this is to use the fact that the Lie derivative can be seen as the infinitesimal generator of the push-forward, since by definition
$$(L_YT)_p := \left.\frac{d}{dt}(\varphi_{-t}^*T_{\varphi_t(p)})\right|_{t=0},$$
if $Y$ is the vector field which generates the flow $\varphi_t$. Intuitively I would say that one can now take the above vector $X$ and just "push it forward" using the Lie derivative in the direction of the curve, i.e. in the direction of the vector field $\frac{d\gamma}{dt} = (0, d\varphi/dt)$.
The issue is that I don't really know how one is supposed to do that explicitly. How exactly do I act with the Lie derivative on $X$ and what am I supposed to do with this expression?
Side note:
It's been a while since the exam, so I had time to think about the problem. Unfortunately I don't really see how this is supposed to work. I always thought of parallel transport and Lie derivative as two distinct things that each on their own have applications for certain problems.

REPLY [5 votes]: NOTE: I am taking $R=1$ for convenience.
Although it's not clear to me what was intended by the exam question, what I said at the end of my comment does lead to a solution. Let $X=\dfrac{\partial}{\partial\theta}$ and $Y=\dfrac1{\sin\theta}\dfrac{\partial}{\partial\phi}$ be the unit tangent vector fields along the longitudes and lines of latitudes, respectively. Since the longitudes are geodesics, $\nabla_X X = 0$; since $Y$ makes a constant angle with $X$, $\nabla_X Y = 0$ as well (i.e., $Y$ is parallel along the longitudes). 
Now, since the Levi-Civita connection $\nabla$ is torsion-free, we have the "well-known formula"
$$\nabla_X Y - \nabla_Y X = [X,Y] = L_X Y.$$
Thus, we have
$$\nabla_Y X = L_X Y = -L_Y X.$$
This will tell you the rate at which $X$ is turning as you parallel translate along the $Y$-curve, i.e., the line of latitude, and from this you can compute parallel translation along the line of latitude.
(For reference, $L_Y X = -\cot\theta\, Y$. Look familiar?)