﻿Combination with Homogeneous Functions. 707 



vanish identically, an orthogyral Fp may be said to be mixed. 

 These two types are clearly not restricted to homogeneous 

 vectors. I shall speak o£ the scalar SpFp as the associated 

 scalar of the vector Fp. 



From (13) we may now deduce a variety of simple con- 

 sequences. First, any homogeneous vector whose associated 

 scalar vanishes identically is orthogyral. This is evident from 

 the mere form of (13) except when n=— 1. In this case it 

 can be shown by actual expansion of WV/ot, as in (15). 



Second, if a homogeneous orthogyral vector be divided 

 by its associated scalar, the resulting vector is lamellar, for we 

 have 



vv (VpVVVp + VSpFlX =o, 

 (. SpFp j 



by actual expansion, if SF\7F = ? i. e. if F is orthogyral. 

 Exception must be made of conical vectors. 



It further appears from (13) that the theory of orthogyral 

 vectors may be connected with that of algebraic plane curves. 

 Suppose n, the degree of the homogeneous vector Fp, to be a 

 positive integer, and let Fp be orthogyral of mixed type, 

 and let its components be polynomials. Let x, y, and z, the 

 usual coordinates of a point p in space, define a point in a 

 plane in homogeneous coordinates. Then the associated 

 scalar, SpFp, will define, by its vanishing, a plane curve, of 

 degree n + 1. I shall now show that if Fp is orthogyral the 

 curve defined by its associated scalar must have n double 

 points. The condition that Fp shall be orthogyral is that 

 the scalar product of the two vectors Y\?Fp and \/$>pFp 

 shall vanish identically. Call the components of S/SpFp 

 X, Y, and Z, and those of WFo X l5 Y 1? and Z x . Expanding 

 the scalar product we have the condition 



XX X + Y Yi + ZZ X = 0, identically. 



These six components define six curves. Wherever X and Y 

 both vanish, either Z or Z 1 must vanish. But Zj meets X 

 or Y at most in n(n — 1) points, the product of their degrees*. 

 Hence at the remaining n 2 — n(n — 1) intersections of X and 

 Y we must also have Z vanish. That is, \/$pFp vanishes at 

 n points, and the curve $pFp = Q has n double points. 



For example, let n=l. The associated scalar, being of 

 the second degree, defines a conic. It must have one double 



* If Z 1} X, and Y happen to have a common factor, we may make a 

 new choice of our coordinate system. It is easy to complete the formal 

 proof. 



2 Z 2 



