50 ANALYTICAL TKEATMENT OF THE POLYTOPES EEGULARLY 



tains a zero, whilst by the second process the symbols of the second 

 column are deduced from those of the first column placed on the 

 same line. In the third column we find the partition cycle 1312 

 proceeding one step to the right at every application of the second 

 process. 



So, if the case considered is that of an existiug net — which 

 question does not yet interest us here — , this net must admit seven 

 different constituents, i. e. three pairs of oppositely orientated ones 



(3222100) , (3321110) 

 (3221000) , (3332110) 

 (4322100) , (4432210) 



and the central symmetric (4322210). 



This example leads us to a general rule about the number of 

 groups of constituents any net is to have; we state it in the form of: 



Theorem XIV. "In general the number of kinds of constituents 

 of a net of JSf n is n -}- 1; in the case r =. 1 it is n." 



The proof of the general case runs as follows. The zero symbol 

 0Zi> <?2> • • •> Sn-if Çn> 0) of the central polytope we start with passes 

 either into (r, q x , q 2 , . • . , q n -±> 0) or into (r — 1 , q ± — 1 5 ^ 2 — 1 > 

 . . ., q n -i — 1,0) according to q n being either zero or one. So, in 

 continuing the application of the two processes of art. 27, at each 

 step the digit q x moves one place to the right and comes back to 

 its original place after n -j- 1 moves. Moreover it reappears there 

 with its original value q ±m For the increase by r at the jump from 

 the rear to the front is exactly counterbalanced by the loss of a 

 unit every time when of two unequal adjacent digits the right hand 

 one jumps to the fore, this loss occurring exactly r times; indeed, in 

 the circular permutation of the digits from the left to the right — 

 executed for simplicity for a moment without increasing or decreasing — 

 the zero at the end has to be replaced successively by 1 , by 2 , ... 

 by r — 1 and finally in the case q ± = r — 1 by zero, in the case 

 q i = r by r. So after n -j- 1 moves the original zero symbol recurs 

 and the total process has come to a close. 1 ) 



In the exceptional case r = 1 we find only q ± = 1 , i. e. the zero 

 symbol of any constituent can only contain units and zeros. So we can 



n u — 1 



start with the simplex (1 00 ... 0) and find successively (11 00 . . . 0), 



M -2 



n 



(111 00. . .0), etc. But when we have to pass from (11. . . .1 0) 



') The process may come to a close sooner. Compare for this exception page 57. 



