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I would like to compute the cross-ratio of the points $A,B,C,D \in \mathbb{RP}^2$, in the projective plane, given by: $$ A=(0:1:2) \quad B=(1:2:3) \quad C=(2:3:4) \quad D=(3:4:5) $$ First I want to demonstrate that they are all collinear. I believe this is the case as they all satisfy $x-2y+z=0$. So the corresponding lines in $\mathbb{R}^3$ are coplanar, which I believe to be the defintion of collinearity in the real projective plane.

Now in order to compute the cross-ratio, I consider the line $l\subset\mathbb{R}^3$ given by $l=\{(x,y,z)\in\mathbb{R}^3 : x-2y+z=0\ \text{and}\ y=12\ \}$ and determine the $x$ coordinates of the points at which the lines in $\mathbb{R}^3$ corresponding to the four given points intersect the line $l$. One sees by inspection that this is at $x=0,6,8,9$ for the points $A,B,C,D$ respectively. And hence the cross ratio is computed as: $$ [A,B,C,D] = \frac{8-0}{8-6} : \frac{9-6}{9-0} = \frac{4}{3} $$
I am not entirely certain that this is the correct approach, as I have been unable to find resources where the cross-ratio of points on the real projective plane is computed in this manner. I would therefore highly appreciate any comment on my result and any reference to relevant resources where similar problems are encountered.

Bowei Tang
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Heinrich Wagner
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1 Answers1

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What you’ve done is fine. You’ve basically constructed a colineation between the lines $1:-2:1$ and $0:1:-12$ and implicitly defined projective bases for these two lines. All of this preserves cross-ratios.

You could instead work directly with the homogeneous coordinates of the given points. If you choose some point $O$ not on the line, then $$[A,B;C,D] = {[O,A,C][O,B,D]\over[O,A,D][O,B,C]}.$$ (In case you haven’t come across this notation before, the brackets represent determinants.) Taking $O=1:0:0$, this expands into $${\begin{vmatrix}1&0&0\\0&1&2\\2&3&4\end{vmatrix} \begin{vmatrix}1&0&0\\1&2&3\\3&4&5\end{vmatrix} \over \begin{vmatrix}1&0&0\\0&1&2\\3&4&5\end{vmatrix} \begin{vmatrix}1&0&0\\1&2&3\\2&3&4\end{vmatrix}} = {(-2)\cdot(-2) \over (-3)\cdot(-1)} = \frac43,$$ which agrees with your computation.

Jürgen Richter-Gebert’s Perspectives on Projective Geometry has proofs of the above and many other properties of cross-ratios. It’s an excellent resource if you’re delving into projective geometry.

amd
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