SURFACES IN HYPERSPACE. 329 



Cones I and II. We may now state that: The locus of intersection 

 of consecutive normal spaces N^ generate a cone {thus all the consecutive 

 normal spaces pass through a point). This cone is a quadric cone. For 

 if r*$^^T = 0, where $"^ is a self conjugate dyadic, be the equation 

 of the cone described by a, which we shall call Cone I, the normal to 

 the tangent plane is determinable from the equation r'4>~'«(/r = 

 and r«$~' = n or r = $*n. Hence the locus of n is n'$*n = 0, 

 the reciprocal cone to Cone I. This reciprocal cone we shall call 

 Cone II; its vertex is at 0', not at 0. As the Cones I and II are recip- 

 rocal, we can infer not only that the normals to the tangent plaiies to I 

 generate II hut that reciprocalli/ the normals to the tangent planes to II 

 generate I. The special case previously treated where lies in the 

 plane of the indicatrix falls under the general case because 0' has 

 retreated to infinity and consequently Cone II becomes a cylinder. 



In case n > 5 we may, if we desire, restrict ourselves to the normal 

 space iVs in which the indicatrix lies. We shall then have precisely 

 the relations just proved for the case n = 5. But when n > 5 the 

 equations (92') have additional solutions in the rest of the total normal 

 space Nn-2 external to the particular N^ . The adjacent normal spaces 

 Nn-2 intersect in a space iV„_4 which is perpendicular to N3 and contains 

 in N^ an element in Cone II. 



38. The fundamental dyadic * . The forms of $ and $-^ which 

 determine Cones II and I may be found in the general case as follows. 

 Let h', Hl', 8' be the reciprocal set to h, H-, and 6, and consider 



<^ = c(hh -\i-\i-- 55), $-1 = c-i(h'h' - ix'|i' - 5'5'), 



where c is any constant. The vectors h + 8 lie on r'$"^T = 0. But 



(h + 5) . (h'h' - |i>' - 5'5') . (h + 5) = 0. 



Moreover the expression \i-\i- + 58 is in^'ariant of the system X as may 

 be seen from (89) where the accents denote new values of |^ and 5 not 

 the reciprocal set as here. As h is independent of X, the diadics $ 

 and $-^ are independent of X and any value of h + 5 will satisfy 

 r . $-1 . r = 0. The expression written down for $ is therefore correct. 



The dyadic $ may be expressed directly and simply in terms of the 

 vector coefficients of the form '^. Consider the dyadic 12 which is 

 the discriminant matrix of ^, namely. 



Q 



yii yi2 

 y2i y22 p 



yiiy-22 - yi2y2i . (93) 



