Past exam of the short options course of the University of Oxford / 2015 / 3

Use the cross-ratio convention compatible with the permutation identities in this paper. Represent a projective point $P_i\in {\mathbb{CP}}^1$ by a nonzero vector $p_i\in {\mathbb{C}}^2$, and write $[p_i,p_j]=\det (p_i,p_j)$. DefineThe denominator is nonzero because the four projective points are distinct. Rescaling any one representative multiplies numerator and denominator by the same nonzero scalar, so the definition is independent of the chosen homogeneous coordinates. In finite affine coordinates it readswith the point at infinity of the complex projective line handled by the homogeneous formula.

A projective transformation is represented by an invertible matrix $A\in {\mathrm{GL}}_2(\mathbb{C})$. Since $[A{p}_i,A{p}_j]=\det (A)[p_i,p_j]$, the two factors of $\det (A)$ in the numerator cancel the two in the denominator. Therefore every projective transformation preserves this cross-ratio. This also shows that the formula descends from matrices to the projective linear group, whose representatives differ by a nonzero common scalar.

The needed general-position theorem is sharp triple transitivity of the projective linear group on the projective line: for any two ordered triples of distinct projective points, exactly one projective transformation takes the first triple to the second.

In particular, let $T_P$ take $(P_0,P_1,P_2)$ to $(\infty ,0,1)$. In these coordinates,so cross-ratio invariance gives $T_P(P_3)=\lambda$, where $\lambda =(P_0{P}_1:P_2{P}_3)$. Distinctness implies $\lambda \ne 0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized toFor a second quadruple $(Q_0,Q_1,Q_2,Q_3)$ with the same cross-ratio, its normalizing map $T_Q$ yields the same quadruple. ConsequentlyThis proves the equivalence. Uniqueness on the first three points also shows that the projectivity carrying one ordered quadruple to the other is unique. The cross-ratio normalization of four projective points therefore gives a complete invariant for this ordered classification.

Choose three distinct projective points $A,B,C$. Their images under the injective map $\tau$ are distinct. Sharp triple transitivity gives a projective transformation $S$ agreeing with $\tau$ on these three points. The map $\sigma =S^{-1}\circ \tau$ fixes $A,B,C$ and preserves their cross-ratios.

For any other projective point $P$, invariance givesLet $T$ normalize $(A,B,C)$ to $(\infty ,0,1)$. By part (b), the left side is $T(\sigma (P))$ and the right side is $T(P)$. Since $T$ is a bijection, $\sigma (P)=P$. It fixes the original three points too, so $\sigma$ is the identity map everywhere. ThereforeNo continuity or prior surjectivity assumption was needed. This is the rigidity assertion that cross-ratio preserving maps are projective.

By invariance under projective transformations and part (b), it is enough to calculate on the normalized quadruple $(\infty ,0,1,\lambda )$. Direct evaluation of the cross-ratio givesFor instance, $(0\infty :1\lambda )=(0-1)/(0-\lambda )=1/\lambda$, and $(\infty 1:0\lambda )=(1-\lambda )/(1-0)$. ThusAll denominators are legitimate because $\lambda \notin \{0,1\}$. The first identity explains why some distinct permutations always give the same cross-ratio.

The cross-ratio permutation orbit is generated by the two operations $\lambda \mapsto 1/\lambda$ and $\lambda \mapsto 1-\lambda$ from part (d). The full list isThese are all the possibilities: the four permutations generated by interchanging both pairs and by exchanging the two pairs leave the cross-ratio unchanged. They form the Klein four-group inside the symmetric group on four symbols, leaving at most $24/4=6$ different values; the displayed operations attain all six rational expressions.

There are fewer than six values exactly when a nonidentity operation in this six-element action fixes $\lambda$: if two operations give the same value, composing one with the inverse of the other gives such a fixed point. For real $\lambda \ne 0,1$, the fixed-point equations areThe other two equations reduce to ${\lambda }^2-\lambda +1=0$, which has no real root. HenceEach exceptional value has the same three-element orbit $\{-1,\frac{1}{2},2\}$. Every other permitted real value gives six different values.

For the geometric interpretation, put $P_0=[1:0]=\infty$ and $P_i=[x_i:1]$ for $i>0$. ThenThe three exceptional cases becomeThus the three finite points are collinear and equally spaced, with one their midpoint, even when the $x_i$ are viewed as points of the complex plane. Equivalently they form an arithmetic progression in some order. These are the harmonic conjugate points configurations with the fourth point equal to the point at infinity of the complex projective line.