parallel transport $w$ along $\gamma$: Explicit computation Let $S^2 = \{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 +z^2 =1\}$ . For fixed $t \in [0,1)$ consider the curve $\gamma(t) = (r\cos(t), r\sin(t), \sqrt{1-r^2})$, $t\in [0,2\pi]$. Take $w\in T_{\gamma(0)} S^2$.
I want to compute the parallel transport $w$ along $\gamma$.

Naively I know parallel transport $w$ along $\gamma$ as follow:
Let $x(u,v) = \gamma(t)$, then $w = ax_u + b x_v$, then
\begin{align}
  \frac{Dw}{dt} &= \left(a' + \Gamma^1{}_{11} au' + \Gamma^1{}_{12} av' + \Gamma^{1}{}_{12} bu' + \Gamma^1{}_{22} bv' \right) x_u  \\
  & \quad + \left( b'+ \Gamma^2{}_{11} au' + \Gamma^2{}_{12} a v' + \Gamma^2{}_{12} bu' + \Gamma^2{}_{22} bv' \right) x_v 
\end{align}
where $\Gamma^{i}{}_{jk}$ are Chritoffel symobl.
So assuming my Riemannian metric as $ds^2=dx^2+dy^2+dz^2 = r^2dr^2+ r^2\sin^2(\theta) d\theta^2$, I can compute $\Gamma$.
But I am having trouble computing $v$ and corresponding $a,b$. Since $w \in T_{\gamma(0)} S^2$, I guess $w$ should pass through the point $\gamma(0) = (r,0,\sqrt{1-r^2})$

From $x(u,v) = \gamma(t) = (r\cos(t),r\sin(t),\sqrt{1-r^2})$, Can I idenitfy $(u,v) = (r,t)$? But in this case $u$ is independent of $t$ so my $u'$ all vanishes...

Is my approach correct? (How to formulate $a,b$ out of $\gamma(t)$?) I am familiar with the covariant derivatives acting on tensors, $\nabla_{\mu} x^{\nu} = \partial_{\mu} x^{\nu} + \Gamma^{\nu}{}_{\mu \rho} x^{\rho}$ like in General relativity, but not familiar with these differential geometry notations so having trouble expliict computations.
 A: This is an exercise one really should do by hand in order to understand all the details involved but let me give you some pointers:

*

*Choose some coordinate system $x$ whose image covers (almost all of) the image of the curve $\gamma(t)$. In your case, I would choose
$$ x(r,t) = \left( r \cos t, r \sin t, \sqrt{1 - r^2} \right) $$
because then the curve $\gamma$ takes the form $\gamma(t) = x(\mathbf{r},t)$ for a fixed $0 < \mathbf{r} < 1$. I will use the bold notation in order to emphasize that $\mathbf{r}$ is fixed (instead of using $r_0$ or something like that which leads to uglier equations).

*Calculate the components of the metric with respect to this coordinate system. For my choice, you have
$$ x_r(r,t) = \left( \cos t, \sin t, -\frac{r}{\sqrt{1-r^2}} \right), \qquad
x_t(r,t) = \left( -r \sin t, r \cos t, 0 \right), \\
\left< x_r, x_r \right> = \frac{1}{1 - r^2}, \qquad \left< x_t, x_t \right> = r^2, \qquad \left< x_r, x_t \right> = 0. $$

*Denote by $v = v(t)$ a parallel vector field along $\gamma$ with $v(0) = w$ and write it in terms of the basis for the tangent space induced by your coordinate system as
$$v(t) = a(t) x_r|_{\gamma(t)} + b(t) x_t|_{\gamma(t)} = a(t) x_r(\mathbf{r}, t)+ b(t) x_t (\mathbf{r}, t). $$
You need to solve the equation $\frac{Dv}{dt}$ which will give you a system of first order differential equations in terms of the components $a(t),b(t)$.

*In order to compute $\frac{Dv}{dt}$, you can compute the Christoffel symbols with respect to the coordinate system $x$, express $\gamma'(t)$ in terms of $x_r,x_t$ (we have $\gamma'(t) = x_t(\mathbf{r},t)$) and use the formula for the covariant derivative. Alternatively, a faster method in this case is to calculate the regular derivative $v'(t)$ of
$$ v(t) = a(t) x_r(\mathbf{r},t) + b(t) x_t(\mathbf{r},t) $$
and project it orthogonally onto the tangent space $T_{\gamma(t)}S^2$. The final result is
$$ \frac{Dv}{dt} = \left( a'(t) + \mathbf{r}(\mathbf{r}^2 - 1) b(t) \right) x_r|_{\gamma(t)} + \left( b'(t) + \frac{a(t)}{\mathbf{r}} \right) x_t|_{\gamma(t)}. $$

*The equation $\frac{Dv}{dt} = 0$ translates into the system of equations
$$ a'(t) = \mathbf{r} \left( 1 - \mathbf{r}^2 \right) b(t), \qquad b'(t) = -\frac{a(t)}{\mathbf{r}}. $$
Plug one of the equations into the other to get a second order equation for $a(t)$ which you can easily solve. Do the same for $b(t)$. Writing everything in terms of the initial conditions $a(0),b(0)$, you will get
$$ a(t) = \mathbf{r} \sqrt{1 - \mathbf{r}^2} \sin \left( \sqrt{1 - \mathbf{r}^2} t \right) + a(0) \cos \left( \sqrt{1 - \mathbf{r}^2} t \right), \\
b(t) = -\frac{a(0)}{\mathbf{r} \sqrt{1 - \mathbf{r}^2}} \sin \left( \sqrt{1 - \mathbf{r}^2} t \right) + b(0) \cos \left( \sqrt{1 - \mathbf{r}^2} t \right).$$

*Finally, do some sanity checks. For example, the initial condition $w = \gamma'(0)$ corresponds to $a(0) = 0, b(0) = 1$. In this case,
$$ v(t) = \mathbf{r} \sqrt{1 - \mathbf{r}^2} \sin \left( \sqrt{1 - \mathbf{r}^2} t \right) x_r \left( \mathbf{r}, t \right) + \cos \left( \sqrt{1 - \mathbf{r}^2} t \right) x_t \left( \mathbf{r}, t \right). $$
You can see that if $\mathbf{r} \in \{ 0, 1\}$ then $v(t)$ is constant which is expected because when $\mathbf{r} = 1$ the curve $\gamma(t)$ is a geodesic while if $\mathbf{r} = 0$ the curve $\gamma(t)$ is constant. When $0 < \mathbf{r} < 1$, you can see that the cosine of the angle between $v(2\pi)$ and $v(0)$ is $\cos \left( \sqrt{ 1 - \mathbf{r}^2} 2\pi \right)$ so as $\mathbf{r}$ gets closer to $0$ (and the curve $\gamma$ gets closer and closer to the north pole), $v(0)$ rotates almost completely around the tangent $v'(t)$.

