Möbius transformations

Now we turn to studying the connection between Möbius transformations and conic sections. The transformations we consider are those that preserve the unit disk $\{\vert\zeta\vert<1\}$. It is straightforward to prove that all such Möbius transformations correspond to matrices of the form

$$m=\begin{bmatrix}\alpha & \beta \\ \overline\beta & \overline\alpha\end{bmatrix}$$

for complex numbers $\alpha$, $\beta$ satisfying $\vert\alpha\vert^2-\vert\beta\vert^2=1$. These conditions define the special unitary group $SU(1,1)$, which is conjugate to the group of $2\times2$ real matrices of determinant 1, denoted $SL(2,\mathbb{R})$.

Given such a matrix $m$, the Möbius transform $m(z)=\frac{\alpha\zeta+\beta}{\overline\beta\zeta+\overline\alpha}$ maps the unit disk into itself in such a way that it maps arcs orthogonal to the unit circle to arcs orthogonal to the unit circle. These transforms are in fact the isometries of the Poincaré disk. We can interpret them as mappings of the hyperboloid $\mathcal{H}$ by using the mapping $F$ defined in the previous section. This comes down to the calculation of

$$F^{-1}\circ m \circ F(x,y,z)$$

We leave this just as a calculation to be carried out by the reader (a computer algebra system would help!). The result is a matrix multiplication

$$\left[ \begin{array}{ccc} 1+2b_1^2-2a_2^2& 2b_1b_2-2a_1a_2& 2b_1... ..._2& 1+2b_1^2+2b_2^2 \end {array} \right] \begin{bmatrix}x\\ y\\ z\end{bmatrix}$$

where we have written $\alpha=a_1+ia_2$ and $\beta=b_1+ib_2$ in terms of their real and imaginary parts. Remarkably, this is a linear map, defined by a matrix we shall denote by $h(m)=h(\alpha,\beta)$. Given the relation

$$1=\vert\alpha\vert^2-\vert\beta\vert^2=a_1^2+a_2^2-b_1^2-b_2^2$$

it is possible to calculate that $h(m)^T Jh(m)=J$, in other words that $h(m)\in SO(2,1)$.

Thus, we have ended up with a homomorphism $h:SU(1,1)\to SO(2,1)$. This mapping is not injective; it is easy to see that $h(-\alpha,-\beta)=h(\alpha,\beta)$. In fact, it is a 2-to-1 mapping.