My problem relates to space form and geodesics in Riemman Geometry.Let $D=${$ x^ {2} + y^ {2} \leqslant 1$} with Poincare metric $g=\frac{4}{(1-x^ {2}-y^{2})^{2}}(dxdx+dydy)$
(1)Show that D is space form of curvature -1:
In my opinion, this is easy to calculate: ${K}=-\frac{1}{2 \mathrm{~h}} \Delta \log h
$ , here $h=\frac {4}{(1-x^ {2}-y^ {2})^ {2}}$ thus K=-1.
The second one concerns me most:
(2)Let $\gamma$ be a geodesic in $D$, $p \notin \gamma$ , then there exists exactly two geodesics $\zeta_{1}$ and $\zeta_{2}$ which satisfies:$\zeta_{1}(0)$=$\zeta_{2}(0)$=p, with no intersection to $\gamma$:
(a)for any geodesic $\zeta(0)$=p , and $\dot{\zeta}(0)$ between $\dot{\zeta_{1}}(0)$ and $\dot{\zeta_{2}}(0)$. It has no intersection with $\gamma$.
(b) )for any geodesic $\zeta(0)$=p, and $\dot{\zeta}(0)$ between $-\dot{\zeta_{1}}(0)$ and $\dot{\zeta_{2}}(0)$ . It must intersect with $\gamma$.
Rk: As i know, the geodesic in D is the arc orthogonal to D, thus i just transform this problem to a Euclidean geometry problem, can you give me some advice?
I think you should draw a picture of this. I prefer the Poincaré upper half-plane model so I will work there. There you can assume, that $\gamma$ is a vertical line through the origin. Now you have two candidate to $\zeta_1$ and $\zeta_2$, one of them is the vertical line through $p$, the other is the half circle through $p$ and the origin. After you have the two candidates, I hope it will be easier to finish the problem.