# proof of converse of Möbius transformation cross-ratio preservation theorem

Suppose that $a,b,c,d$ are distinct. Consider the transform $\mu$ defined as

 $\mu(z)={(b-d)(c-d)\over(c-b)(z-d)}-{b-d\over c-b}.$

Simple calculation reveals that $\mu(b)=1$, $\mu(c)=0$, and $\mu(d)=\infty$. Furthermore, $\mu(a)$ equals the cross-ratio of $a,b,c,d$.

Suppose we have two tetrads with a common cross-ratio $\lambda$. Then, as above, we may construct a transform $\mu_{1}$ which maps the first tetrad to $(\lambda,1,0,\infty)$ and a transform $\mu_{2}$ which maps the first tetrad to $(\lambda,1,0,\infty)$. Then $\mu_{2}^{-1}\circ\mu_{1}$ maps the former tetrad to the latter and, by the group property, it is also a Möbius transformation.

Title proof of converse of Möbius transformation cross-ratio preservation theorem ProofOfConverseOfMobiusTransformationCrossratioPreservationTheorem 2013-03-22 17:01:51 2013-03-22 17:01:51 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 30E20