1
$\begingroup$

The hyperbolic plane is the set { $(x,y)∈R^2: y>0$ } on which the first fundamental form is $\frac{dx^2 + dy^2}{y^2}$. Show that a geodesic of the hyperbolic plane is either a line of fixed $x$, or (part of) a half circle with centre on the $x$-axis.

Consider geodesics starting on the line $y=1$. Two such geodesics are called parallel if they start in the same direction. Show that there are parallel geodesics which intersect.

==

Things I know:

  1. The Christoffel symbols $\Gamma^1_{11} = \Gamma^1_{22} = \Gamma^2_{12} = 0$, $\Gamma^1_{12} = \Gamma^2_{22} = -\frac{1}{y}$, $\Gamma^2_{11} = \frac{1}{y}$, and hence the geodesic equation.

  2. The parametrisation of a geodesic starting at $(x_0,y_0)$ in the direction $(y_0a, y_0b)$ is $\gamma(t) = (x_0 + \frac{ay_0 sinh(t)}{cosh(t)-bsinh(t)}, \frac{y_0}{cosh(t)-bsinh(t)})$ with $a^2 + b^2 =1$.

What can I do next to figure out this problem? Thanks!!

$\endgroup$
  • 1
    $\begingroup$ "Two such geodesics are called parallel if they start in the same direction. Show that there are parallel geodesics which intersect." I suppose they want you to find two geodesics starting from different points on the line $y=1$. The problem is, you cannot talk about "the same direction" at different points... Unless they want you to interpret "same direction" as viewing each tangent space as $\Bbb R^2$ and "same direction" means same vector from this tangent space, which is really bad, because there is already a notion is "parallel" in the context of Riemannian geometry. $\endgroup$ – edm Oct 26 '18 at 15:03
0
$\begingroup$

Remark: I interpreted the formulation "in the same direction" very literal, i.e. the starting vectors $(ay_0,by_0)$ are the same. However, as edm points out in his comment under the question, this might be very confusing.

The direct way is to choose two geodesics that start on $y=1$ with the same initial tangent vector and hope that they will intersect. By taking the two starting points not to far apart, this will work.

But there is an easier way. The real thing we need to know is how these geodesics look like. The geodesics look like (parts of) semicircles that intersect the $x$-axis orthogonally. This can be deduced from your parametrisation by noting that $$ \left( x-\Bigl(x_0 + y_0 \frac{b}{a}\Bigr)\right)^2 + y^2 = \frac{y_0^2}{a^2}. $$ Using this fact you now asily see that there are parallel geodesics that intersect: take a geodesic that passes through the line $y=1$ and translate it a bit to the right (or left).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.