An ideal triangle in the Poincaré disk model is a triangle with vertices in $\mathbb{R} \cup \{\infty\}$. I have to prove that there exists an isometry between every two ideal triangles. I know that isometries in this space are moebius transforms $\phi_A$ with $A \in SL(2, \mathbb{R})$.
The transformation $\phi_A = \frac{(z – z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ maps our vertices $z_1, z_2, z_3$ to $0, 1$ and $\infty$.
So with this two transformations like the above, $\phi_A$ and $\phi_B$ map two ideal triangles to the ideal triangle with vertices $0,1$ and $\infty$ and $\phi_A^{-1}\circ\phi_B$ maps ideal triangle B to ideal triangle A. All nice and well, but $det(A)=(z_2-z_1)(z_2-z_3)(z_1-z_3)$ is not necessarily equal to $1$, as is required for $A \in SL(2, \mathbb{R})$. Where did I go wrong?
Best Answer
Note that, given a Möbius transform, $\phi(z)=\frac{az+b}{cz+d},$ then it's also true that $\phi(z)=\frac{\lambda(az+b)}{\lambda (cz+d)}$ for any $\lambda\in \mathbb{R}$. Now, the determinant the two corresponding matrices are $ad-bc$ and $\lambda^2(ad-bc)$. Accordingly, a Möbius transform is defined by a matrix up to a scaling of its determinant.
Accordingly, you could write out your transformation and then scale numerator and denominator by the inverse squareroot of the determinant and get the same Möbius transform (as a map), but in the form you want, where the coefficients come from an $SL(2,\mathbb{R})$-matrix.