378 HITCHCOCK. 



the number of terms in the double polyadic, frequently also the order 

 of the polyadics. Thus ((AM, BN)) S is symbolic of ((<pi, <p2))s- 

 W hen the order of the polyadic is given we may also treat with polyads 

 in many operations as symbolic of polyadics. Thus if we have double 

 dyadics in any space we may often write aa' for A, mm' for M, etc. 

 in which case a bar may be used to separate antecedent from conse- 

 quent in the double polyads, or ((aa' | mm', bb' | nn')) s may be taken 

 symbolically for the much more general ((<p u ^))s- Hence if <pi and 

 <pi are double dyadics we may write 



((fpi, <P2))s = 2(ama'm'bnb'n'- ana'n'b mb'm') (72) 



and may even omit the summation sign in many operations. It is of 

 interest to compare this vector expression with (64) which was based 

 on the notion of the dyadic as a matrix. 



11. Invariants regarded as Products. 



The examples of the last article are sufficient to indicate that the 

 scalars (36) are invariants of polyadic systems and that they are 

 subject to a considerable variety of transformations. There is one 

 more method by which they may be calculated, based on a concept of 

 these scalars which may serve to widen considerably our view of the 

 formal laws to which double polyadics are amenable. Returning to 

 the earlier conception of a double polyadic as a sum of unit double 

 polyads E,-Ea- each multiplied by a scalar, as in (25), consider scalars 

 of the form ((£,£*, E r E,)) s , that is E,: E A E r : E s - E,: E s E r : E k . 

 The unit polyads E, and E r are antecedents while E^ and E s are conse- 

 quents. Wesee that the scalar is equal to +1 when each antecedent 

 is equal to its own consequent, but unlike the other antecedent and 

 consequent; is equal to — 1 when each antecedent differs from its own 

 consequent but is equal to the other consequent; and is otherwise 

 zero. That is 



((EiE i} E k E k )) s = + 1, ((E,E,, E,E ! )) 5 = - 1, ((E,E„ E r E,)) s = 

 ((E t -E,- t E,E,n s = 0, ((E,E„, E,.E r )) s = 0, ((E.E,, E,E,)) 5 = 

 ((E^,, E,E,)) S = 0, ((E f E fcj E r E s )) 5 = 



(73) 



on the understanding that subscripts are unequal unless denoted by 

 the same letter. 



