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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 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.