How to determine four points with cross-ratio 3 on four given lines in $\mathbb{C}P^2$

Question

I'm following an introductory course in geometry and this exercise came up: Suppose you have four lines in $\mathbb{C}P^2$. $a \leftrightarrow x_0=0$, $b \leftrightarrow x_1=0$, $c \leftrightarrow x_0 + x_1 =0$, $d \leftrightarrow x_0+x_1+x_2 =0$. Determine all lines l so that the cross-ratio (A,B,C,D)=3 if A,B,C and D are the intersections of l and a,b,c and d respectively. Can someone please help me with this? Thank you!

Answer 1

Your special case

In your case, $a,b,c$ are concurrent, since they all pass through the point $Q=(0,0,1)$. Therefore you can compute the intersections between these and $d$. Let's call those intersections $A'$ through $C'$. You can find a unique point $D'$ on $d$ such that $(A',B';C',D')=3$. Every line passing through $D'$ will satisfy your condition. This is because $A,B,C,D$ and $A',B',C',D'$ will be related via a perspectivity from $Q$.

General concept

I found the above answer after thinking a lot more complicated thoughts. Although you don't need this for your answer, below is the approach to take for cases where the three lines are not concurrent.

Let $A,B,C,D,P$ be points in the plane, and $a,b,c,d$ be the lines joining $P$ to $A,B,C,D$ resp. Then if you fix $A$ through $D$, the set of all possible locations such that $(a,b;c,d)=\lambda$ for a fixed constant $\lambda$ will be a conic passing through $A,B,C,D$. The general form of your problem is dual to this, so the answer will be some dual conic, i.e. the set of lines tangent to a certain conic.

1. Start with computing the six points of intersection between your lines.
2. Form three pairs from these points of intersection, in such a way that if one point was the intersection of two of your lines, then the other point in the pair will be the intersection of the other two.
3. Create $3\times3$ matrices of degenerate conics, one for each pair, by multiplying the column vector of one point by the row vector of the other. If such a matrix is called $M$, then a line $g$ will belong to that degenerate conic algebraically if it satisfies $g^TMg=0$ and geometrically if it passes trough either of the two points from which the matrix was obtained.
4. Symmetrize the matrices by adding them to their own transpose.
5. Label the resulting matrices as $M_0$, $M_1$ and $M_\infty$ depending on what cross ratio you get from a line belonging to each of these conics. Since each such line will pass through one of the points of intersection, some of the points $A$ through $D$ from your definition will coincide, thus resulting in one of the three special cross ratios.
6. Normalize your matrices (multiplying them with a scalar) such that $M_0+M_\infty=M_1$ holds.
7. $M_0 + 3M_\infty$ is the matrix you want. You can use that to characterize the set of all lines satisfying your requirement.
8. For a more geometric interpretation, take the adjoint (or inverse) of this matrix. It describes the corresponding primal conic, i.e. a set of points instead of a set of lines. You might be more familiar with such a representation. Your solution is the set of lines tangent to that conic.

The pencil of conics through four given points forms a real projective line. By duality, so does a pencil of conics tangent to four lines. What the above procedure does is formulate a projective basis for this family, in such a way that you can obtain an element for a given cross ratio using a simple linear combination.

Relation between general and special case

The general case outlined above can be applied to your special case as well, with a bit of care. Each pair of intersection points would contain $Q$, therefore every linear combination of these degenerate conics would again be degenerate, with $Q$ as one of its factors. For $\lambda=3$ your conic would be equivalent to $D'Q^T$, describing the set of lines which pass either through $D'$ or through $Q$. A line passing through $Q$ satisfies the condition $[AC][BD]=3[AD][BC]$, which is mostly equivalent to $(A,B;C,D)=3$ but in this case it is satisfied by both sides being zero, which would lead to the undefined division $\frac00$ in the common formulation. So in a certain sense, lines through $Q$ satisfy the requirement, although I wouldn't include this in an official answer to your question.