290 G. H. KNIBBS. 



Fig. 16 is therefore the required representation. The only type 

 of surface upon which hyperbolic space of 2-dimensions can be 

 delineated, is that indicated in Fig. 17. It is consequently evident 

 from considerations of symmetry, that the completed or closed 

 symmetrical surface, if such exist, must be a tore of some form, 1 

 Let RCS, Figs. 18, 19, be the axis of the ring: then the principal 

 meridians of the surface will be as shewn by the heavy lines in 

 Fig. 18, these are sections of the surface by planes containing 

 the axis RCS. The equator of the surface would be the section 

 perpendicular to that axis through MOM', see both figures, and is 

 shewn by the continuous heavy line inside the tore. 



The short or incomplete equidistant lines are not geodesies, but 

 parallels of latitude. The radii of curvature in the meridian and 

 at right-angles thereto, at any point P, are respectively PQ = p, 

 and PR = ytt; at a point like Q, they are respectively QQ' = cr, and 

 QS = v; hence for a succession of points on these surfaces, we 

 must have 



H'iPi = / x 2i°2 = e ^ c * = • • • = v i°"i = v 2°"2 = e ^ c (25) 



if the curvature is to be constant. 



In passing along the curves OK and KQ, [x and v necessarily 

 increase as K and Q are approached, each becoming finally infinite, 

 hence ultimately p and o- will be infinitesimal. Consequently it is 

 not possible to continue the curve starting at O beyond the points 

 K and L as it must be continually convex towards the axis of the 

 tore, and therefore concave outwards beyond the points K and L. 



We can now see that it is not possible for the KM surface to 

 be identical with OK unless CO is infinite; in other words a finite 

 and completely symmetrical hyperbolic surface does not exist. 2 



1 See Figs. 1, 5, 18. 



2 So far as I am aware this has not been previously demonstrated. I 

 give therefore the proof, which depends on the fact that KOL is a cycloid, 

 if OC be infinite; KML being its evolute. The equation of the cycloid, 

 O being the origin at its vertex, is 



y = a cos~ l a ~ x + V(2ax-x 2 ) 



where a is the radius of the generating circle Op = x, pV — y. From this 

 we deduce 



