98 Proceedings of the Roycd Irish Academy. 



is imposed besides qKq = scalar, and qlq = scalar. Here it may be 

 noted that, 



Xg-^ = {Kq)-' as qq'^ = 1 = K^^Kq. 



32. Another and simpler way of investigating the structure of the 

 functions P as depending on the nature of q is to change the signs of 

 all the units in the equation Vq = qp. 



Manifestly, if (^ is odd in the units, qp is even, and therefore Vq is 

 even, and P must be odd. If q is even in the units, P must still be 

 odd in the units. 



Grenerally, let q = q' -\- q", and P = P' + P", where q' and P' are 

 odd, and q" and P" even in the units ; then 



{q' + q")p = {V' + V"){q' ^q"), 



and on change of sign of all the units, 



(^q'-f)p = i^-p'-V"){q'-q"). 



Hence, adding and subtracting, 



q'p = Vq' + V"q", and q"p = Y'q" + F"q'. 



If, for all values of p, P is odd in the units, either q' or q" must 

 vanish ; for otherwise 



q'pq'-^ = P' = q"pq"-\ and hence q"~'^q'p = pq"~^q' ; 



that is, q"~'^q' must be commutative with every unit, and therefore 



q"-'^q' - X = sccdar, or q' = xq" ; 



but this is an impossible equation, since q' is odd in the units, and q" 

 even. 



It seems to be impossible for P to be even in the units for all 

 values of p. 



33. In the particular case in which q is a linear vector (ai), p is 

 changed into its reflection with respect to ai by the operator ai ( ) ai~^ 

 For splitting p into two parts, p' parallel, and p" perpendicular to ai, 



o-ipaf^ = ai (p' + p") af 1 = (p' - p") aiai"^ = p' - p"- 



Thus the part parallel to ai is unchanged, and the part perpendicular" 

 to it is reversed in direction. 



Reflecting p in succession to two lines aj and a, in order, and sup- 

 posing p' to be the component of p in the plane of these two vectors^ 

 while p" is the perpendicular component, 



a^aipaf^a^'^ = OaOi (p' + p") ai~^ao~^ = p" + a2aip'ax~^a2~^ 



