The cross-ratio

Here is a nice invariant from classical geometry that I’d never heard of before today.

The action of \text{GL}_2(\mathbb{C}) on \mathbb{C}^2 restricts to an action on \mathbb{CP}^1; this is all that a Möbius transformation really is.  Now \text{GL}_2(\mathbb{C}) is four-dimensional, but there’s a one-dimensional subspace corresponding to scaling which stabilizes each point of the projective line, so we may as well quotient this out and get a \text{PGL}_2(\mathbb{C})-action.

In fact, this is the projective automorphism group of \mathbb{CP}^1, hence the name PGL; its elements are called projectivities.

\text{PGL}_2(\mathbb{C}) group is three-dimensional, so we would expect that with our free degrees of freedom we could send any three points to any three other points, and indeed we can: the action is 3-transitive.  On the other hand, the dimension of \text{PGL}_2(\mathbb{C}) implies that it can’t possibly be 4-transitive.

What this says is that, up to projectivities, any three or fewer points on the complex projective line look like any other set of the same cardinality, but there are sets of four or more points which are essentially “different.”  In particular, given a set of n > 3 points in \mathbb{CP}^1, we ought to be able to find an invariant which determines whether a projectivity takes one set to the other.

Let’s consider the case n = 4.  Obviously we’ve got some latitude in determining this invariant up to a constant, so let’s just decree that R(0, 1, \infty, x) = x.  Consequently, for any element M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in PGL_2(\mathbb{C}), we have

R(M \cdot 0, M \cdot 1, M \cdot \infty, M \cdot x) = x.

Expanding the expression,

\displaystyle R\left(\frac{b}{d}, \; \frac{a+b}{c+d}, \; \frac{a}{c}, \;\frac{ax+b}{cx+d}\right) = x.

So all we need to do is, given a system of equations

\displaystyle p = \frac{b}{d}, \; q = \frac{a+c}{b+d}, \; r = \frac{a}{c}, s = \frac{ax+b}{cx+d},

figure out how to solve for x as a rational function of p, q, r, and s.  This isn’t too bad — note first that  ax + b = s(cx + d), or (a - cs) x = ds - b, so

x = \displaystyle \frac{ds - b}{a-cs} = \frac{ds - dp}{cr - cs} = \frac{d}{c} \frac{s-p}{r-s}

Now we just need to express d/c in terms of p, q, and r. Inspired by the previous expression, it’s not hard to determine that we can write d/c = (q-s)/(p-q), so altogether we get

R(p, q, r, s) = \displaystyle \frac{(q-r)(s-p)}{(p-q)(r-s)}.

Actually, as the negative reciprocal would provide just as good an invariant, let’s redefine R slightly to get all the variables in a nice, alphabetical order:

R(p, q, r, s) = \displaystyle \frac{(p-q)(r-s)}{(p-s)(q-r)}.

This function is the classical “cross-ratio” of the four points p, q, r, and s.  As we can see from the formula, it’s a ratio of ratios of distances between points.

Of course the point of the preceding is to provide one justification for why we should expect such an invariant to exist, and how we could determine it.  In fact, the cross-ratio and its significance to projective geometry was known already to Pappus of Alexandria around AD 300.

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One Response to The cross-ratio

  1. Mike McLaury says:

    you left out some (): d(s-p)/c(r-s)

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