# Showing that the geodesics in hyperbolic upper half-plane model are half circles

I am currently working on some differential geometry and I came across the following question.

Let $$H = \{(x,y)\in \mathbb{R}^2|y>0\}$$, this is the hyperbolic plane (the upper half-plane model). The first fundamental form is $$\frac{dx^2+dy^2}{y^2}.$$ Write down the geodesic equation of the hyperbolic plane and show that geodesics in $$H$$ are either straight lines with constant $$x$$ or a half circle with center on the $$x$$-axis.

I have found that the geodesic equations are $$\frac{d}{dt}\bigg[ \frac{\overset{.}{x}}{y^2} \bigg] =0$$ $$\frac{d}{dt}\bigg[ \frac{\overset{.}{y}}{y^2} \bigg] = \frac{-1}{y^3}(\overset{.}{x}^2+\overset{.}{y}^2).$$

From this its easy to see that $$\overset{.}{x} = Cy^2$$ for some constant $$C$$, and so if $$C=0$$ we find the straight-line solutions. However I am struggling to show that there are also circular solutions. Using the second geodesic equation I am also able to show that

$$\overset{..}{y} = \frac{\overset{.}{y}^2}{y}-\frac{\overset{.}{x}^2}{y}.$$

However I have not been able to make any more progress after this.

• Check out do Carmo's Riemannian Geometry. It is left as an exercise that the Mobius transformations $M(z)=\frac{az+b}{cz+d}$ with $ad-bc=1$ are isometries of $\mathbb{H}^2$ (actually they are all the isometries). Then use the fact that isometries take geodesics to geodesics. Since you already know that vertical lines are geodesics, prove that their image through any $T$ is either a vertical line, or a half circle orthogonal to the $x-$axis. – Laz Oct 1 '18 at 3:47
• In the course that I am doing we haven't really been worked with the mobius transformation, So i am thinking there is another approach – Jandré Snyman Oct 1 '18 at 3:49
• Another is to compute what is exactly the covariant derivative in $\mathbb{H}^2$, and rather that solving $\frac{D}{dt} \gamma^{'}=0$, parametrize these circles by arc lenght and prove their acceleration is $0$. – Laz Oct 1 '18 at 3:53
• Anyway, what tools do you have at your disposal? So that I can try to take a suitable approach. – Laz Oct 1 '18 at 3:55
• Basically we have only the geodesic equations. We have covered covariant derivatives but very briefly, I have covered the mobius transformation in another course but not in this one – Jandré Snyman Oct 1 '18 at 3:57

## 1 Answer

Let the geodesic has unit speed, so $$\frac{\dot{x}^2+\dot{y}^2}{y^2}=1$$ i.e., $$\dot{y}^2=y^2-\dot{x}^2.$$ You know $$\dot{x}=Cy^2$$, so this gives $$\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2=\frac{\dot y^2}{\dot x^2}=\frac{y^2}{\dot x^2}-1=\frac{R^2}{y^2}-1=\frac{R^2-y^2}{y^2}$$ where $$R^2=C^{-2}$$ (remember we are assuming $$C\neq 0$$). Solving the last equation gives the circle centered at $$x$$-axis of radius $$R$$, and of course only half of it lies on the upper half space.