Parallel Transport along a curve We had this homework assignment for our geometry course, and we couldn't figure it out, any ideas on how to do this:

Consider the Poincare model of Lobachevsky plane, 
$H^2=\left\lbrace{  (x,y):\quad x \in \mathbb{R},  \quad y > 0,\quad  dl^2 = (dx^2 + dy^2)/y^2  }\right\rbrace$
  
  
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*Show that in the course of parallel transport along the curve $\gamma = \left\lbrace x(t) = t,  y = y_0 = \text{constant} > 0 \right\rbrace $, vectors rotate uniformly with angular velocity $1/y_0$.
  

 A: I realize this is an old question, but I was searching for a related problem and I found a simple solution. First, some theory:
Let $E_1 = (y,0)$ and $E_2 = (0,y)$ be an orthonormal basis for $T_{(x,y)} \mathbb{H}^2$ the connection form $\omega_{12}$ can be calculated like any connection form in a conformal manifold with ruler $g$, by: $$\omega_{12} = g_y \theta_1 + g_x \theta_2$$
Where $\theta_1$ and $\theta_2$ are the dual 1-forms of $E_1,E_2$, $\theta_1 = \frac{dx}{g}$ and $\theta_2 = \frac{dy}{g}$, in terms of the dual forms of $\mathbb{R}^2$.
But since $g_x = 0$ and $g_y = 1$, we have $\omega_{12} = \frac{dx}{g}$, but the coordinate expression for $\gamma'(t) $ is $(1,0)$ and then, $\omega_{12} ( \gamma'(t) ) = \frac{1}{g(\gamma(t))} = \frac{1}{y_0}$.
Now, $Y$ can be written as 
$$Y = c \cos \phi E_1 + c\sin \phi E_2$$
Where $\phi = \angle Y,E_1$. Using the condition $Y'(t) =0$ we arrive at 
$$\phi'(t) = - \omega_{12} (\gamma'(t))$$ And therefore
$\phi(t) = \phi(0) - \int_0^t \frac{1}{k} d\eta$, $\phi(t) = \phi(0) - t \frac{1}{y_0}$
So $Y = c \cos \left(\phi(0) - t \frac{1}{y_0} \right )E_1 + c \sin \left(\phi(0) - t \frac{1}{y_0} \right )E_2$
That is, vectors rotate uniformly with angular velocity $\frac{1}{y_0}$
You can see in the image below several of those vector fields with initial vector of $(1,0)$ in the points of the form $(0,y_0)$ with $0<y_0<5$. 

A: Let me preface this answer by admitting I don't know enough about the definition of parallel transport invoving an affine connection, which would likely be the more rigorous way to show this. I use an intuitive approach of parallel transport being effected by keeping the same angle with a geodesic. I came up with an explanation which makes sense to me, and would love feedback from anyone knowing more.
At a point $P=\gamma(t)$, the geodesic through $P$ going along the horizontal direction of travel is the part of the circle of radius $1/y_0$ centered at $(t,0)$ lying in the upper half-plane. So the unit vector along the geodesic has angular velocity $-1/y_0$ radians per second, negative since it is moving clockwise as we move to the right along $\gamma$. Since our unit vector in the $\gamma'$ direction is a constant vector pointing to the right, this means that, with respect to the geodesics used in parallel transport, $\gamma'$ is rotating at angualr velocity of $+1/y_0$.
