Higher Order Cross-Ratios
Created | Updated Dec 29, 2008
The cross-ratio R of the points A, B, C, D is R(A,B,C,D)=(A-C)(B-D)/(A-D)/(B-C)=r. Let f(r)=1-r and g(r)=1/r and consider the functions r, f(r), gf(r), fgf(r), gfgf(r), .... Now 1-(A-C)(B-D)/(A-D)/(B-C)=(A-B)(C-D)/(A-D)/(C-B), so f(r) effects the transformation R(A,B,C,D)=>R(A,C,B,D). Here the middle two points have been switched. Now 1/[(A-B)(C-D)/(A-D)/(C-B)]=(A-D)(C-B)/(A-B)/(C-D), so gf(r) effects the transformation R(A,C,B,D)=>R(A,C,D,B). Here the points to be switched have shifted to the right by one. Similarly, fgf(r) effects the transformation R(A,C,D,B)=>R(B,C,D,A). Again the points to be switched have shifted to the right by one (in a cyclic sense). Continuing this process gives R(B,C,D,A)=>R(C,B,D,A)=>R(C,D,B,A)=>R(C,D,A,B)=>...=>R(A,B,C,D). The original configuration is arrived at after twelve transformations, however the last six transformations equal the first six. All the formal values of R(A,B,C,D) are obtained by the repeated composition of two functions. Consider the equation x2-x+1=0. If x1, x2 are roots, then x1+x2=1 and x1x2=1 (the symmetric equations) and hence x2=1-x1 and x1=1/x2. (By a suitable change of sign, the cross-ratio could be made to correspond to the cube roots of unity.) This is (or at least can be taken to be) the origin of the functions f and g. (If f(r)=a-r and g(r)=b/r, that is, the functions corresponding to x2-ax+b=0, it is not difficult to show that repetition in r, f(r), gf(r), fgf(r), ... does not generally depend on r.) Apparently, this can be extended to cubic and higher power equations. Consider the analogous functions for the equation x3-x2+x-1=0, i.e., f(x1,x2)=1-x1-x2, g(x1,x2)=(1-x1x2)/(x1+x2), h(x1,x2)=1/(x1x2). The analogous sequence is r1, r2, f(r1,r2), g(r2,f(r1,r2)), h(f(r1,r2),g(r2,f(r1,r2))), .... (Note that two initial values must be specified.) For some arbitrarily chosen r1, r2 values, the sequence appears to repeat after the 18th function (the computation is especially prone to error due to the inexact divisions). (Note that 6=3*2! and 18=3*3!.) The analogous definition of the cross ratio of 6 points is R(A,B,C,D,E,F)=(A-C)(B-E)(D-F)/(A-F)/(B-C)/(D-E). This ratio is preserved under a projective transformation, but this gives only one initial value for the sequence and what the other value should be isn't clear. (My requirements for higher order cross-ratios are that they be preserved under projective transformations and that some systematic relationship exists between the formal values and the values in the sequence.) Note that the points do not appear symmetrically in the denominator. This is the motivation for redefining the cross ratio. Define R1(A,B,C,D) by (A-C)(B-D)/(A-B)/(C-D) and R2(A,B,C,D,E,F,G,H) by (A-D)(B-G)(C-F)(E-H)/(A-B)/(C-D)/(E-F)/(G-H). Similar results for R1(A,B,C,D) hold, that is, r=>f(r) effects the transformation R1(A,B,C,D)=>R1(A,B,D,C), f(r)=>gf(r) effects the transformation R1(A,B,D,C)=>R1(A,D,B,C), gf(r)=>fgf(r) effects the transformation R1(A,D,B,C)=>R1(D,A,B,C), etc. That is, rather than start in the middle with points to be switched and move to the right by one, start on the right with points to be switched and move to the left by one. Forming the sequence for second order cross-ratios, i.e., expressions containing the term R2(A,B,C,D,E,F,G,H), requires three initial values. (The functions to be employed are f(x1,x2,x3)=1-x1-x2-x3, g(x1,x2,x3)=(1-x1x2-x2x3-x1x3)/(x1+x2+x3), h(x1,x2,x3)=(1-x1x2x3)/(x1x2+x1x3+x2x3), i(x1,x2,x3)=1/(x1x2x3).) We can now let R1(A,B,C,D)=r1, R1(E,F,G,H)=r2, and R2(A,B,C,D,E,F,G,H)=r3. Much computation gives;
(1) f(r1,r2,r3)=R2(A,B,C,D,H,G,F,E)-R1(A,B,C,D)R1(H,G,F,E)
This is a result analogous to that for the redefined first order cross-ratio. There the first f function flipped the last half of (A,B,C,D). Here the first f function flips the last half of (A,B,C,D,E,F,G,H) and there is a second term consisting of the product of the corresponding first order cross-ratios. The next step is to compute g(R1(E,F,G,H), R2(A,B,C,D,E,F,G,H), R2(A,B,C,D,H,G,F,E)-R1(A,B,C,D)R1(H,G,F,E)). This laborious task has yet to be done. So far, this appears to be an appropriate definition of a second order cross-ratio.
(1) f(r1,r2,r3)=R2(A,B,C,D,H,G,F,E)-R1(A,B,C,D)R1(H,G,F,E)
This is a result analogous to that for the redefined first order cross-ratio. There the first f function flipped the last half of (A,B,C,D). Here the first f function flips the last half of (A,B,C,D,E,F,G,H) and there is a second term consisting of the product of the corresponding first order cross-ratios. The next step is to compute g(R1(E,F,G,H), R2(A,B,C,D,E,F,G,H), R2(A,B,C,D,H,G,F,E)-R1(A,B,C,D)R1(H,G,F,E)). This laborious task has yet to be done. So far, this appears to be an appropriate definition of a second order cross-ratio.
