178 



PHILLIPS AND MOORE. 



Consequently Mi and il/g are linear functions of the same pair of 

 completely perpendicular planes ki2 and ^-34. We could obtain a 

 similar conclusion if il/2 was not self-complementary or anti-self- 

 complementary. If both belong to these special types, let 



Ml = a {kri + A-34). 

 From (34) we then have 



Ml * 3/2 = a [(Xi3 - X.24) A-23 + (Xi4 + X03) A-24 + (X23 + X14) ksi + 



(X24 ~ X13) /i4i. 



If this is zero 



Xl3= X24, Xi4= — X23. 



Hence ilf 2 is anti-self-complementary and so 



il/2 = X12 (ki2 — ksi) -f Xi3 (/»-i3 + koi) + Xi4 (A'i4 — /i-23). 



By direct multiplication we get 



Ml ■ il/2 = il/i X il/2 = Ml* Mo = 0, 

 Ml X Ml = Ml- Ml = 2 a2 > 0, 



M2 X il/2 = - il/2 •il/2 = - 2 (X12" + Xu' + X14-) < 0. 

 Hence 



(Ml + X il/2) X {Ml -f X il/2) = Ml X Ml -h X2 il/2 X il/2 = 



•lias two real roots X. If then 



TTi = il/i + X Mo, To = il/i - X iV/2, 



TTi and TTo are perpendicular planes and Mi and il/2 linear functions of 

 them. Hence in any case if the star product of two two-vectors in four 

 dimensions is zero, they are expressible as lir^ear functions of the same 

 pair of completely perpendicular planes. Incidentally we have shown 

 that the star product, cross product and dot product of a self-comple- 

 mentary and an anti-self-complementary two-vector are all zero. 



Let -Ki , TTo , TTs' be the complements of tti, tto, tts. We shall now 

 prove the identity 



TTi * (tto * 7r3) = (7ri-7r3)7r2— (7ri-7r2)7r3+(7ri-7r3')7r2' — (7ri-7r2')7r3'. (38) 



