Parallel transport $w$ along $\gamma$: Explicit computation

connectionsdifferential-geometryriemannian-geometry

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.

Best Answer

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:

  1. 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).
  2. Calculate the components of the metric with respect to this coordinate system. For my choice, you have $$ x_r(r,t) = \left( \cos t, \sin t, -\frac{r}{\sqrt{1-r^2}} \right), \qquad x_t(r,t) = \left( -r \sin t, r \cos t, 0 \right), \\ \left< x_r, x_r \right> = \frac{1}{1 - r^2}, \qquad \left< x_t, x_t \right> = r^2, \qquad \left< x_r, x_t \right> = 0. $$
  3. 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 $a(t),b(t)$.
  4. 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 $$ \frac{Dv}{dt} = \left( a'(t) + \mathbf{r}(\mathbf{r}^2 - 1) b(t) \right) x_r|_{\gamma(t)} + \left( b'(t) + \frac{a(t)}{\mathbf{r}} \right) x_t|_{\gamma(t)}. $$
  5. 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 $$ a(t) = \mathbf{r} \sqrt{1 - \mathbf{r}^2} \sin \left( \sqrt{1 - \mathbf{r}^2} t \right) + a(0) \cos \left( \sqrt{1 - \mathbf{r}^2} t \right), \\ b(t) = -\frac{a(0)}{\mathbf{r} \sqrt{1 - \mathbf{r}^2}} \sin \left( \sqrt{1 - \mathbf{r}^2} t \right) + b(0) \cos \left( \sqrt{1 - \mathbf{r}^2} t \right).$$
  6. 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 $v'(t)$.
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