SURFACES IN HYPERSPACE. 327 



rxN = 0. If (It = i,ds is an infinitesimal displacement along the 

 surface and N + f/N the normal at its extremity, the equation of the 

 adjacent normal space becomes 



(r - dr) X (N + f/N) = or rxrfN - rfrxN = 0. 



The intersection of the two normal spaces is determined by the 

 simultaneous equations 



rxN = 0, r X ,- - |xN = 0. 

 ds 



If we take complements we may write these equations 



r-M = 0, r.'^-|-M = 0. (92) 



ds 



The first equation merely states that r is perpendicular to M and we 

 shall therefore consider only such values of r in the second equation. 

 From (73), (80), (82), we have, 



r-(a + 7Tl)x-n + |x(|i _ ^1) - l-dx-n) = 0, 



or r- (ax-q + |x|jl) -)- -q = 0, 



— T'ar\ + r-Hi| = - t\. 



Hence r-a=l, r-jx = 0. (92') 



Special Cases. Consider first the case n = 4. Here the indicatrix 

 is a conic in a plane through the surface-point 0. The vector a runs 

 from to a point of this conic. If we lay off from the radius of 

 curvature instead of the curvature itself, we get a point Q which is 

 the inverse of P with respect to 0. The locus of Q is therefore a 

 bicircular quartic. If we draw through Q a line perpendicular to a, 

 we have a line for which r • a = 1 ; and the point where this line cuts 

 the perpendicular from upon \i; or upon the tangent to the indicatrix 

 at P, is a point P' which is the common solution of (92') and which 

 therefore is the point of intersection of the normal plane N with the 

 adjacent normal plane in the direction §. 



If we consider the triangles 0PM and OQP' we see that OMOP' = 

 OQOP = 1. Hence P' and M are inverse points. But the locus 



