= Solution
By invariance under <projective transformations> and part (b), it is enough to calculate on the normalized quadruple $(\infty,0,1,\lambda)$. Direct evaluation of the <cross-ratio> gives
$$
\begin{aligned}
(0\,\infty:\lambda\,1)&=\lambda,&
(1\,\lambda:\infty\,0)&=\lambda,\\
(0\,\infty:1\,\lambda)&=\lambda^{-1},&
(\infty\,1:0\,\lambda)&=1-\lambda.
\end{aligned}
$$
For instance, $(0\,\infty:1\,\lambda)=(0-1)/(0-\lambda)=1/\lambda$, and $(\infty\,1:0\,\lambda)=(1-\lambda)/(1-0)$. Thus
$$
\boxed{(P_0P_1:P_2P_3)=(P_1P_0:P_3P_2)=(P_2P_3:P_0P_1),}
$$
$$
\boxed{(P_0P_1:P_2P_3)=(P_1P_0:P_2P_3)^{-1},}
$$
$$
\boxed{(P_0P_2:P_1P_3)=1-(P_0P_1:P_2P_3).}
$$
All denominators are legitimate because $\lambda\notin\{0,1\}$. The first identity explains why some distinct permutations always give the same <cross-ratio>.
Back to article page
