544 Proceedings of the Poyal Society of Edinburgh. [Sess. 
This proves that a skew determinant of even order is a perfect square, 
a 2 b 2 . . . . , and one of odd order is 0 a 2 b 2 = 0 . (The determinant in question 
is [%]•) 
The vertical line | =K is not wanted with fictorlinities, though it is 
with multilinities. We will cease to use it. 
Rotational Unities . — If 0 is a rotational fictorlinity, = and 0 is 
really the rigid replacement (where fictors replace factors) under a new 
name, and we might help ourselves somewhat by previous results, but 
refrain. 
Since 0 / 0 = l, <fip is never zero. Operating then by 0( )0 _1 we get also 
00' = 1. Put \js for the colinity part J(0 + 0 / ) of 0 and x for the skew part 
J (0 — 00 - Since 0'0 = 1 = 00', 
(0' - 0)(0' + 0) = 0' 2 - 0 2 = (0' + 0)(0' ~ 0) 
and (0' + 0) 2 — 2 = 0' 2 + 0 2 = (0' — 0) 2 + 2, 
or ^x=X^»^-X 2 = 1 ( 64 > 
Let x be given by ( 60 ). If a = 0 , x l i = X 1 2 = ®- Hence from ( 64 ) 
0 2 l i = q, 0 2 t 2 = t 2 . 
Hence 0q = 0q — ± q, 0t 2 = 0 t 2 = — L 2- 
[There is actually an ambiguity of sign, but it need only affect a single 
i, as I leave the reader to verify at the end of the treatment of the rota- 
tional linity. Strictly, we ought to consider the necessity in certain cases 
of perversion with respect to a single i, but for simplicity I will ignore 
this.] 
When a is not zero, it is easy to verify from ( 60 ) and ( 64 ) that without 
exception 
0q= J (1 -a 2 ).t v 0t 2 = x/(! ~ a 2 )' L 2’ 
or putting a — sin 0 , b = sin O', , 
0q = q cos 6 + i 2 sin 0 0'q = q cos 6 - q sin 0 | 
0q = - q sin 6 + 1 2 cos 0 0't 2 = q sin # + i 2 cos 0 ( 
(opposite circular variations). Hence we have 
0P —ppp~ l .... 
where p is the even fictor product 
p — (cos + qq _1 sin J#)(cos + qt 3 _1 sin J 0 ') .... 
Since (qq _1 ) 2 = — 1 , we have, putting q _1 = aq, iffqt 3 _1 = a) 2 , 
p = e^i e«2 .... 
Also, though in general r 1 *" 1 would not be equal to e“ 1+ “ 2 , yet in the 
( 65 ) 
( 66 ) 
(67) 
