4:22 Dr. L. Silberstein on the Relation between 



The particular form of the function l(n) will depend on the 

 kind of the contemplated straight line, or of the space to 

 which it belongs : Riemannian, Euclidean, or Lobatchev- 

 skyan *. In the Euclidean case the question would be 

 answered by simply equating the cross-ratio of the indices 

 [as in (1), with ?ii = 0, n 2 = n, n s =l, ?i 4 = co] to the double 

 ratio of the corresponding lengths and by solving for L 

 'But the following way, giving a formula for all three cases, 

 seems preferable. 



As is well-known, the Riemannian plane f of curvature 



^ has all the properties of a Euclidean sphere of radius R, 



the Riemannian straight lines being replaced by geodesies 

 (great circles) ; and all Lobatchevskyan formulae follow 

 from the corresponding Riemannian ones on replacing R 

 by iR. 



Consider therefore an ordinary sphere of radius R, and 

 let the previous OUT be a segment of a great circle drawn 

 on it. Since, as can easily be proved, the theorem of 

 Desargues holds for spherical triangles, v. Staudt's con- 

 struction can be repeated on the sphere, replacing straight 

 lines by great circles, the only caution being that T must 

 not be the antipode of 0. Thus we can imagine the scale 

 of Staudtians set up on the geodesic arc OUT. To prevent 

 two points of this segment receiving the same index n, take 

 OUT smaller than a half great-circle. Then giving to U 

 the index 1, the points of this segment will be covered in a 

 one-to-one correspondence by the indices w = to oo . (Each 

 of the negative n-values will then belong to two distinct, 

 antipodal points of the supplementary segment, but this 

 will not invalidate our result.) 



Now, from the centre of the sphere let four rays be 

 drawn to the points 0, 1, n, co of the segment. Then, 6 being 

 the ordinary geographical longitude reckoned from 0, we 

 shall have l = R0, l l = R0 1 , l n =110^, and since the cross- ratio 

 of the four rays is equal to that of the four points, we have, 



by (1), 



0-1 n-co _ J^ _ sin Q l sin(fl-0J 

 — go * n— 1 ~" 1— n~sm0 x ' sin ((9 — 6{) 

 Thus, after easy reductions, and replacing 6 by l/R, 



cot k = l{ ootl i + {n - 1)oot R}- ■ ■ (2 > 



* Notice that, according to a famous theorem due to Schur, every 

 projective space is a space of constant curvature, and vice versa. 

 t Or better, a limited portion of it. 



