# projective geometry and relationship of cross-ratios

Define for pairwise different points $P_i=[v_i]$ the cross-ratio $\operatorname{CR}(P_1,P_2,P_3,P_4) = \frac{\det(v_1,v_2)}{\det(v_2,v_3)}\cdot\frac{\det(v_3,v_4)}{\det(v_4,v_1)}$ on $\mathbb{KP^1}$ where $\det$ is a determinant function on the underlying vector space. $\operatorname{CR}$ is independent of the choice of $\det$ and scaling of the $v_i$.

Show

$$\operatorname{CR}(P_1,P_2,P_3,P_4) = \operatorname{CR}(P_3,P_2,P_1,P_4) \implies \operatorname{CR}(P_1,P_2,P_3,P_4) = -1$$

Using that $\det$ is alternating, I showed that $\operatorname{CR}(P_3,P_2,P_1,P_4) = (\operatorname{CR}(P_1,P_2,P_3,P_4))^{-1}$ and hence from $\operatorname{CR}(P_1,P_2,P_3,P_4) = \operatorname{CR}(P_3,P_2,P_1,P_4) = (\operatorname{CR}(P_1,P_2,P_3,P_4))^{-1}$ concluded that $\operatorname{CR}(P_1,P_2,P_3,P_4) \in \{-1,1\}$.

But how to eliminate now the option $1$?

Based on that I am asked tho show for pairwise points $P_1,P_2,P_3,P_4 \in \mathbb{R}$ (how should I interpret this here?), that the following are equivalent:

a. $\operatorname{CR}(P_1,P_2,P_3,P_4) = -1$ (what is CR here? we actually defined it for $\mathbb{KP^1}$)

b. either $P_3$ or $P_4$ lies between $P_1$ and $P_2$ and $\frac{|P_1-P_3|}{|P_3-P_2|} = \frac{|P_1-P_4|}{|P_4-P_2|}$

I do not even get the formulation of this second part..

Funny, I did the proof which excludes $1$ in a lecture today. The basic idea is that you can show that if the cross ratio is $1$, then two points must coincide. Without loss of generality you may assume

$$v_1=\begin{pmatrix}a\\1\end{pmatrix} \qquad v_2=\begin{pmatrix}b\\1\end{pmatrix} \qquad v_3=\begin{pmatrix}c\\1\end{pmatrix} \qquad v_4=\begin{pmatrix}d\\1\end{pmatrix}$$

You can do this because the cross ratio is independent from your choice of basis, so you will always find a basis such that none of your four points is $(1,0)^T$, and then you can choose representants as above.

With this choice you get

\begin{align*} \operatorname{CR}(P_1,P_2,P_3,P_4) &= 1\\ \frac{(a-b)(c-d)}{(b-c)(d-a)} &= 1\\ (a-b)(c-d)&=(b-c)(d-a) \\ ac-ad-bc+bd&=bd-ba-cd+ca \\ -ad-bc&=-ba-cd \\ ab-ad&=bc-cd \\ a(b-d)&=c(b-d) \end{align*}

From this you can conclude that the cross ratio will be $1$ if and only if $P_1=P_3$ or $P_2=P_4$. So as your statement guarantees four distinct points, you know that the cross ratio cannot be $1$.

Note that these $a,b,c,d\in\mathbb R$ above are the inhomogenous desciption your second part talks about. Adding a one in the last coordinate is one possible way of homogenizing them, thus turning them into coordinates in $\mathbb{RP}^1$. So this difference computation is almost like mine, except that it uses absolute values so it has to check for order separately.