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

DIFFERENTIAL GEOMETRY MN1 FALL 1999

PROBLEM 12 PAGE 2

The isometries of H² are well-known maps in the upper half plane model. Let $\,SL(2, R)\,$ be the special linear group in dimension 2, i.e. the group of all real $\,(2\times 2)$-matrices with determinant $\,=1.$ $\,SL(2, R)\,$ acts on $\,H^2\,$ in the following way. Let $\,z=u+iv.$ The points $\,(u,v)\,$ in the upper half plane correspond to $\,z=u+iv,\,\,\,v>0.$ If

$$g = \left(\begin{array}{rrrr} a&b\\ c&d\end{array}\right)\in SL(2, R),$$

let

$$gz=\frac{az+b}{cz+d}.$$

Proposition

The group $\,SL(2, R)\,$ acts as a group of isometries on $\,H^2.$

Proof: Let $\,u+iv=z\,$ och

$$\,\frac{az+b}{cz+d}=\tilde z.$$

If we write $\,dz\,d\bar z\,$ for $\,du^2+dv^2\,$ the line element of $\,H^2\,$ can be written

$$ds^2(z)=\frac{-4dz\,d\bar z}{(z-\bar z)^2}, \qquad \bar z=u-iv.$$

As $\,d\tilde z=d((az+b)(cz+d))=dz/((cz+d)^2\,$ it follows that $\,ds^2(z)=ds^2(\tilde z),$ which means that $\,z\mapsto\tilde z\,$ is an isometry.

a)

Calculate the arc length of the geodesic $\,c(t)=(r\cos t, r\sin t),\,\,0 < t < \pi\,$ starting from the top of the half circle , $\,t=\pi/2.$ $\qquad$ (Result: $\vert\ln \tan \frac{t}{2}\vert$)

Calculate also the arc length of the geodesic $\,u=u_0\,$ from $\,v=a\,$ till $\,v=b.$
(Result: $(\vert\ln \frac{a}{b}\vert$)

b)

Calculate the geodesic curvature $\,k_{g}\,$ of the curve $\,v=1.$ $\quad$ (Result: $\,k_{g}=1$)

c)

A vector is parallel translated the hyperbolic distance $\,d\,$ along the curve $\,v=1.$ Calculate the angle the vector has turned during this translation.

To Problem 12 page 1

To Problem 12 page 3

Tillbaka till Differentialgeometri MN1