Understanding cross ratio and harmonic conjugates

One of the most basic aspects of projective geometry is that the projective transformations act transitively on triples of distinct points in the projective line. That is, if $P,Q,R$ are three distinct points on a projective line, and $P',Q',R'$ are three other distinct points, then there exists a projective transformation of the line mapping $P$ to $P'$, $Q$ to $Q'$, and $R$ to $R'$.

Therefore, when considering the relationship between four points $P,Q,R,S$, it makes sense to use a projective transformation to map the first three points $P,Q,R$ to some "standard position". Thus usual choice is to map $P$ to $0$, $Q$ to $1$, and $R$ to $\infty$, in which case the image of $S$ is the cross ratio $$ (P,Q;R,S) \;=\; \frac{(S-P)(Q-R)}{(Q-P)(S-R)}. $$ Another way of saying this is that the map $$ x \;\mapsto\; (P,Q;R,x) $$ is the unique projective transformation of $\mathbb{R}P^1$ that maps $P$ to $0$, $Q$ to $1$, and $R$ to $\infty$.

In terms of perspective drawing, if $P$, $Q$, and $R$ are points on a line $L$ in a painting, where $R$ lies on the "horizon", then the cross ratio defines a linear scale on $L$, using the distance from $P$ to $Q$ as a unit of length.

Originally I'd have written a three-part answer, but now that Jim Belk wrote one of them, I'll concentrate on the others.

Cross ratio

Start with the fact that a projective transformation can be expressed as a non-singular matrix, which is multiplied with the homogeneous coordinates of the points. In order to obtain an invariant, you need something where that matrix multiplication has no effect but instead cancels out. But as I'll show below, an even stronger motivation might be that your possible invariants only depend on the points involved, not on their homogeneous representatives used in the computation.

What operations do make sense? There are not too many reasonable operations one can perform on two-element vectors. You can sum them, subtract them and so on, but in those cases you obtain another vector. We'd like to make things simpler, so we look for something which takes some vectors and produces a scalar. The determinant is a good candidate here. I'll use brackets $[\ldots]$ to denote determinants. Since your vectors have two elements, and determinants are only defined for square matrices, you know that each such bracket will contain (the representatives of) two points.

Start with any two points, and you can form the determinant $[AB]$. This value depends on the representatives chosen for your points. You have two options: either write an equation like $[AB]=0$ or write a fraction in such a way that any scalar coefficients in front of your vectors will cancel. The equation actually checks whether $A$ and $B$ denote the same point, which certainly is an invariant property, but not what we want just now. So either of these fractions is an option:

$$\frac{[AB]}{[AB]}=1 \qquad \frac{[AB]}{[BA]}=-1$$

It turns out that with two points, the invariants are too invariant: they produce fixed values and therefore make no statement at all about the point configuration. So let's take a third point, and you can form things like

$$\frac{[AB][AC]}{[AB][BC]}\qquad\frac{[AB][AC]}{[AC][BA]}=-1$$

But these are again problematic. The first depends on representatives, it scales proportional with $A$ and reciprocal with $B$. The second is again fixed. Phrased as requirements, we need every point to occur the same number of times in numerator and denominator, and we need numerator and denominator to be substantially different, not just terms exchanged and columns swapped within a determinant. No matter how you try, you can't achieve both of this with three points.

So take a fourth point. Make sure you have all four points in numerator and denominator, so factors cancel. Also make sure that the pairs of points in a common determinant are different for numerator and denominator. The result will be a cross ratio of your four points, although you might still hit any of the six permutations. The canonical one would be

$$(A,B;C,D)=\frac{[AC][BD]}{[AD][BC]}$$

And there you are: this is the easiest number you can attach to the minimal number of points which is not constant and does not depend on representatives. And since you have the same number of determinants in numerator and denominator (since you have the same number of points occurring the same number of times), you can factor out the determinant of a projective transformation and cancel that as well, so you really have an invariant.

I'm not claiming that the above is the historic development, I'm just motivating why the cross ratio is not as obscure as it might seem.

Harmonic conjugates

So now that we know that cross ratios make sense algebraically, you might want to construct them as well. And you'll notice that you have a hard time doing so, depending on the actual value you aim for. You might start looking at the simplest values.

One approach is by considering the cross ratio of one points if the others are fixed at $\infty$, $0$ and $1$. The designated point at $\infty$ allows you to express affine concepts, the point $0$ fixes an origin and $1$ fixes a scale, so that choice is kind of understandable as well.

So you could start looking for special cases of cross ratios, and after some twiddling around you'll also try three equidistant points. So your free point is either at $\frac12$, between $0$ and $1$, or it is at $2$ or at $-1$, with one of the other points exactly in between. Nice easy cross ratios, but can you construct them?

In the Euclidean case, you surely can construct equidistant points. You can even do it using only tools which have a counterpart in projective planar geometry: join, meet and parallel. One possible construction using congruent triangles would be this:

Now you can turn that construction into its projective counterpart, by moving the point at infinity to a finite position, and also the line at infinity which is needed for the parallels. You end up with this:

And there you have it: a well-defined cross ratio which you can not only compute but also construct.

Now one can show that a finite sequence of such harmonic point constructions can construct any point with rational cross ratio to three given points. So any operation which preserves harmonic points will most likely also preserve cross ratios. At least over $\mathbb Q$ and $\mathbb R$ this is the case, although the latter requires some more work. On the other hand, everything which preserves lines will preserve harmonic point constructions. So this gives us a kind of geometric counterpart to why cross ratios are projective invariants.