Geodesics in Poincare Disk

Question

I would like to find the geodesics in the Poincare disk. I know that the metric is $$\frac{dx^2+dy^2}{(1-x^2-y^2)^2}$$ so $$s=\int \frac{\sqrt{1+y'^2}}{1-x^2-y^2}\, dx$$ Then I try to find y(x) using the Euler-Lagrange equations. The ODE in which I end up is: $$yy'^2+xy'+y=0$$ I can't solve the above ODE. I also tried using polar coordinates: $$s= \int \frac{\sqrt{r^2+r'^2}}{1-r^2}\,d\theta$$ but the expression I get is also not easy to handle with.

Answer 1

One method is to use symmetry:

1. First prove (using your ODE, if you like) that the only geodesic with a vertical tangent vector is a vertical geodesic $x(t)=$constant, $y(t) = e^{\pm t + C}$.

2. Next prove that the metric is invariant under the action of the group $SL(2,\mathbb{R})$ acting by fractional linear transformations. For the most pedestrian was to do this, consider a fractional linear transformation on the upper half plane of the form $$w = \frac{az+b}{cz+d} \quad\text{(with $a,b,c,d \in \mathbb R$, $ad-bc=1$)} $$ Assign variables for the real and complex parts $z=x+iy$ and $w=x'+iy'$. Rewrite the above equation as a system of two equations in two unknowns $$x' = f(x,y), \qquad y' = g(x,y) $$ and prove that $$\frac{(dx')^2 + (dy')^2}{(y')^2} = \frac{dx^2 + dy^2}{y^2} $$ There are probably slicker ways to do this.

3. One should also prove that the fractional linear action of $SL(2,\mathbb R)$ acts transitively on points and on unit tangent vectors in the upper half plane. Transitivity on unit tangent vectors at $z=i$ is proved using the matrix $$\begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ which represents a rotation of the unit tangent circle at $z=i$ by the angle $\theta/2$. Transitivity on points is proved by setting $z=i$, choosing $w$ arbitrarily, and finding $a,b,c,d$.

4. Finally, transform the equations for vertical geodesics in part 1 using the factional linear transformation in part 2. This will give you the general equations for geodesics.

Answer 2

$$y = \pm i \sqrt{2} x $$

Is a solution. It can be motivated (without the use of any assumptions) through a lot of work, but I'm trying to first see if using it I can find a general form, Namely I'm trying to find a general $f$ such that if

$$ r(r')^2 + xr' + r = 0$$

then

$$ f(r)(f(r'))^2 + xf(r)r' + f(r) = 0 $$

Your equation is also equivalent to

$$ (w')^2 + 2xw' + w = 0$$

where $w = y^2$