# What is the purely synthetic construction for producing a projective collineation, given two lines and three points on each line?

It is a well-known fact (sometimes called the fundamental theorem of projective geometry) that given two lines and three points on each of the two lines, there is a unique projectivity between the two lines. It also appears to be a known fact that this projectivity can be extended to a collineation of the entire real projective plane.

I'm looking for a specific recipe (draw these lines here, construct the intersection, draw these other lines, etc.) for how to actually implement the collineation, without needing to pass to coordinates. It feels like there's almost such a recipe in various sources, (e.g., Coxeter, partly Richter-Gebert) but I haven't been able to quite make it work.

(Specifically, I want to be able to select a line, three points on the line, another line, three points on that line, and then any other 7th point, and be able to construct the image point via a tool in GeoGebra.)

Coxeter, in his book Projective Geometry, writes about one and two dimensional projective mappings.

In the second edition, Theorem 4.12 is the fundamental theorem of one dimensional projectivities, and is the one to which you refer (projectivities determined by three points on each of two lines).

Theorem 6.13 is the fundamental theorem of (two-dimensional) projective collineations - basically that a projective transformation is determined by two complete quadrilaterals $$DEFPQR$$ and $$D'E'F'P'Q'R'$$. The correspondence between $$DEF$$ and $$D'E'F'$$ is like the correspondence that determines a 1D projectivity. But the extra correspondence between $$DQR$$ and $$D'Q'R'$$ adds more information. Altogether, it adds up to the usual determination of a projective transformation by specifying the mappings of four points.

The diagram below, from Coxeter's book, summarizes the synthetic construction for a projective collineation that maps a line $$a=XY$$ to $$a'=X'Y'$$. Here the construction for the 1D projectivity is used twice, once for each of $$X \rightarrow X'$$ (using $$DEF\rightarrow D'E'F'$$) and $$Y \rightarrow Y'$$ (using $$DQR\rightarrow D'Q'R'$$). There are infinitely many collineations between 3 pairs of points. So you can pick any 8th point as the image of the 7th point and get a collineation that maps the 4 points to their images.

You can determine a unique collineation when given 4 pairs points and their images in general position. I'm not sure if there's a purely synthetic way to determine this collineation though. The analytic way to do this is given in the answer to another question.

Note: A collineation induces a unique projectivity on a line and its image under the said collineation but the reverse is not true. A projectivity does not induce a unique collineation on the plane.

• Good to see that point mentioned here (strangely not upvoted for years). However, due to preservation of cross-ratio, there is uniqueness in how the points of the first line are mapped to the second line when prescribing three point mappings. Perhaps that's what the question meant. Apr 8 at 3:11

If you are looking for a straightedge-only construction of a cross-ratio-preserving map between the points of two lines, here is one. I call it Pappus-like because it uses the Pappus line. Given different lines $$L,L'$$ with pairwise distinct points $$A,B,C$$ on $$L\setminus L'$$ and pairwise distinct points $$A',B',C'$$ on $$L'\setminus L$$, first construct the Pappus line $$\overline{RS}$$ as shown above.

Then, given $$D$$ on $$L$$, construct $$T$$ as the point of intersection of $$\overline{C'D}$$ with $$\overline{RS}$$, then $$D'$$ as the point of intersection of $$\overline{CT}$$ with $$L'$$.

In the latter two steps you can use $$(A,A')$$ or $$(B,B')$$ in place of $$(C,C')$$; the result for $$D'$$ is the same as long as $$D$$ remains on $$L$$.

In particular, the resulting map $$D\mapsto D'$$ maps $$A\mapsto A'$$, $$B\mapsto B'$$ (using Pappus's theorem) and $$C\mapsto C'$$.

The proof that the cross-ratio of $$A,B,C,D$$ equals that of $$A',B',C',D'$$ is left as an exercise to you. Note that the transformation according to this recipe is singular because its image is constrained to $$L'$$. Therefore you cannot simply argue algebraically with the multiplicativity of determinants to deduce that cross-ratio is preserved. There are ways around that obstacle, but I prefer direct calculation using a homogenous coordinate system where $$A,B,A',B'$$ have coordinate tuples $$(\pm 1:\pm 1:\pm 1)$$.

• Turns out that the figure in @brainjam's answer uses Pappus lines too. Apr 8 at 3:23