newson: projective transformations. 49 



line xA^B a one-dimensional hyperbolic transformation whose 

 characteristic cross-ratio is k, and in the pencil through A a one- 

 dimensional transformation whose cross-ratio is k,. T' also in- 

 duces along the invariant line 1, and through the invariant point B 

 parabolic transformations whose characteristic constants are re- 

 spectively a' and a. These four one-dimensional transformations 

 are not independent and our next problem is to determine what 

 relations exist among them. 



8. The Cross-Ratios k and k, are Equal. It can be shown 

 that the two hyperbolic transformations, one along AB and the 

 other through A, are both characterized by the same cross-ratio k 

 when the elements are taken in the right order. The transforma- 

 tion T' and its invariant figure constitute the limiting form of a 

 transformation of the first type with invariant triangle (ABC). In 

 the case of the invariant triangle (ABC) it was shown in Art. 2 

 that the product of the cross-ratios k^ ky kj, is unity when they 

 are reckoned in the same order around the triangle. In the triangle 

 let C be made to coincide with A, then the ratio ky along CA 



becomes unity; this gives the relation k^ ki,= i or k^^^j — . Now 



k is the cross-ratio of the pencil A(BCPP,) and ki5=(ABXXj); 

 interchanging the rays A and B in the range we have have 



. ^^(BAXXj). Hence the characteristic cross-ratio of the range 



along AB is the same as that of the pencil through A, the order of 

 the elements being as follows k=^(BAXX j)=A(ACPP, ) where 

 X and X, are a pair of corresponding points on the line AB. 



9. The Two Characteristic Constants Have the Relation 



a^^=pu. We now go on to establish the relation between a, and a, 

 the characteristic constants of the two parabolic transformations 

 along C and through B respectively. 



Let the angle lAB be <^ and let the distance AB be d. From 

 the theory* of one-dimensional projective transformation we have 



I 1 



AC^ AC 



and 



a=cotP^BA— cotPBA=:cot^, — cot< 

 In the triangle ABC, Fig. 4, we have 



sin^^ sin2<^ sin-<^ 



(8) 



AC" BC" AC'-f AB2— 2AC.AB cos<^ 



*K. U. Quarterly, VdI. vii, ISils. p. l:i(), 



