THE DYADICS IN THREE DIMENSIONS. 427 



L into a line or complex r such that r is a linear function of p and q. 

 Hence p and q are polar with respect to r (see §5). This is true for 

 every pair of planes ^i and ^2 passing through L. To find the complex 

 or line into tvhich [AC] [BD] transforms L, ice theief ore find the complex 

 or line with respect to xohich p and q are polar whatever pair of planes 

 are taken through L. 



As a second case consider a correlation AB and coUineation Cy,Ai, 

 Bi, Ci being points and 7 a plane. The double product 



[AC] (By) 



is a complex or line. In this case (By) is a number and so X must be a 

 number. K Pi, P2, P3, P4 are the four vertices of a tetrahedron we 

 may take 



X = (P1P2P3P4). 



Transform Pi by the coUineation C7 and the opposite plane [P2P3P4] 

 by AB. Let the join of the resulting points be pi. Proceed in the 

 same way with the other vertices of the tetrahedron and the planes 

 opposite them. The above general discussion shows that [AC](By)X, 

 and so, [AC]{By) is a complex or line belonging io the congruence deter- 

 mined by the four lines thus obtained. This is true for every tetrahedron 



Pi. P2, Psj P4. 



As a final illustration consider the case of a one-one dyadic AB and a 

 two-two dyadic pq, The double product 



[Ap] [Bq] 



is a three-three dyadic. Let X be any point and ^1, ^2, ^s three planes 

 through it. Let 



Pi = A{B^,), ri = ^(91^2^3). 



The join of Pi and r^ is a plane [Prri]. Permuting ^1, ^2, ^3 we get three 

 such planes. The three planes intersect in a point. By the above 

 general discussion that point is on the plane into which [Ap] [Bq\ trans- 

 forms X. This is true for every set of three planes through X. 



16. Double products with idemfactors. In particular the 

 double product of a dyadic $ and an idemfactor is an invariant or 

 covariant of $. This is a special case of the preceding general dis- 

 cussion. For example let Aa be the dyadic determining the identical 

 point coUineation and let 



^ = Bfi 



