$\parbox{0.45\textwidth}{\begin{flushleft}\vspace{-\baselineskip} {\L... ...tet}\\ \textbf{Matematiska Institutionen}\\ Thomas Erlandsson \end{flushleft}}$

DIFFERENTIAL GEOMETRY MN1 FALL 1999

PROBLEM 12 PAGE 1

The Poincaré Upper Half Plane Model of $\,H^2$

The surface $\,H^2\,$ is in the Poincaré upper half plane model the set

$$\,U=\{(u,v) \in R^2\vert\,v >0\}\,$$

with the Riemann metric

$$\,ds^2= \frac{du^2+dv^2}{v^2}.$$

Using Gauss' equation we find immediately that this surface has constant Gauss curvature
$\,K=-1.$ The line element $\,ds^2\,$ in $\,H^2\,$ is equal to the euclidean line element $\,du^2+dv^2\,$ multiplied by a strictly positive function. Therefore an angle measured with respect to the Riemann metric coincides with the euclidean angle.

The geodesics in the upper half plane model of $\,H^2\,$ are the euclidean circles and straight lines which meet the boundary $\,v=0\,$ orthogonally. This can be shown in the following way:

In $\,H^2\,$ we get $\,g_{11}=1/v^2,\quad g_{12}=0,\quad g_{22}=1/v^2$ and it follows that

$$\,\Gamma^{1}_{11}=\Gamma^{1}_{22}= \Gamma^{2}_{12}= 0,\quad\Gamma^{2}_{11}= -\Gamma^{2}_{22}= -\Gamma^{1}_{21}=1/v.$$

The differential equations of the geodesics can therefore be written

$$\ddot u-\frac{2\dot u \dot v }{v}=0,\quad \ddot v+\frac{{\dot u}^2 -{\dot v }^2}{v}=0.$$

If $\,\dot u =0\,$ then $\,u=$ constant. In this case it is clear that the geodesic is a straight euclidean line orthogonal to $\,v=0.$

If $\,\dot u \ne 0\,$ we get from the first equation that $\ln(\dot u/v^2)\,=$ constant so $\,\dot u = cv^2\ne 0\,$ for some constant $\,c.$ In the same way we get from the second equation that $\,{\dot u}^2+{\dot v}^2=bv^2 >0\,$ for some constant $\,b.$ By combining these equations we get $\,(dv/du)^2={\dot v}^2/{\dot u}^2 =b/c^2v^2-1.$ Therefore $\,(u-a)^2+v^2=b/c^2\,$ for some constant $\,a.$ This is a circle with centre on $\,v=0$ and so meets $\,v=0\,$ orthogonally.

To Problem 12 page 2

To Problem 12 page 3

Tillbaka till Differentialgeometri MN1