Demonstration of Cross Ratio in

Let $\ensuremath{{\bf p}_{i}} = (X_i,1), i = 1,\ldots, 4$ be four points on the projective line. (In this demonstration, we will only consider finite points, although the cross ratio holds for infinite points as well.) Define the distance $\Delta_{ij}$ between two points i and j as $\Delta_{ij} = \vert X_i - X_j\vert$. What we want to show is that the cross ratio

$$Cr(\ensuremath{{\bf p}_{1}} ,\ensuremath{{\bf p}_{2}} ;\ensur... ...4}} ) = \frac{\Delta_{13}\Delta_{24}}{\Delta_{14}\Delta_{23}}$$

is preserved under projective projection of the points.

A point $\ensuremath{{\bf p}_{i}}$ is projected through a $2 \times 2$transformation matrix T to a new point $\ensuremath{{\bf p}_{i}} '= (t_{11}X_i+t_{12}, t_{21}X_i+t_{22})$. Therefore, the new coordinate $\ensuremath{{\bf p}_{i}} ' = (X'_i, 1)$ is defined by:

$$X'_i = \frac{t_{11}X_i+t_{12}}{t_{21}X_i+t_{22}}.$$

Then, the distance between two points X'_i and X'_j is

$$\begin{eqnarray*}\Delta'_{ij} & = & \vert X'_i - X'_j\vert \\ & = & \left\vert\... ...X_i - X_j)} {(t_{21}X_i+t_{22})( t_{21}X_j+t_{22})}\right\vert, \end{eqnarray*}$$

where $\det(T) = t_{11}t_{22}-t_{12}t_{21}$. The ratio between two distances, one from a point X'_i to another point X'_j, and another from the same point X'_i to a third point X'_k, is

$$\begin{eqnarray*}\frac{\Delta'_{ij}}{\Delta'_{ik}} & = & \frac{\vert X'_i - X'_... ...cdot \frac{ t_{21}X_k+t_{22}} { t_{21}X_j+t_{22}} \right\vert, \end{eqnarray*}$$

which is the original ratio $\Delta_{ij}/\Delta_{ik}$, multiplied by a constant that is dependent only upon the coordinates X_j and X_k. A similar ratio $\Delta_{lj}/\Delta_{lk}$, taken with respect to another point X_l, has this same constant, and therefore dividing the two ratios causes the constants to cancel:

$$\begin{eqnarray*}Cr(\ensuremath{{\bf p}_{1}} ,\ensuremath{{\bf p}_{2}} ;\ensurem... ...\\ & = & \frac{\Delta_{13}\Delta_{24}}{\Delta_{14}\Delta_{23}}, \end{eqnarray*}$$

showing that the cross ratio is unaffected by projection.

Suppose that one of the points, say $\ensuremath{{\bf p}_{1}}$, is at infinity (i.e., its second coordinate is zero). Then, dividing by the second coordinate (which is what we normally do to transform the point into the required form) yields $X_1 = \infty$. Substituting into the above formula yields:

$$Cr(\ensuremath{{\bf p}_{1}} ,\ensuremath{{\bf p}_{2}} ;\ensur... ...\ensuremath{{\bf p}_{4}} ) = \frac{\Delta_{24}}{\Delta_{23}},$$

since the terms with $\infty$ cancel each other. (Technically speaking, we must take the limit of the equation as X₁ tends to $\infty$, but the result is the same.) Similarly, if any of the other points are at infinity, we simply cancel the terms containing the point, and the result is the cross ratio. Remember that at most one point may be at infinity, because the points must be distinct, and there is only one point at infinity on the projective line.