139] AN INTRODUCTORY MEMOIR UPON QUANTICS. 229 
15. Suppose now that $ is a covariant of ©, then the operation performed 
upon any covariant of U gives rise to a covariant of the system 
m m' 
(*3fc y. y',-- •)•••)> 
(f> •••$«, y, • ••)> (%', y, •••$#', y', •••)> &c - 
To prove this it is to be in the first instance noticed, that as regards (£, 77, y,...), &c. 
we have = ?/3|, &c. Hence considering {xd y }, &c. as referring to the quantic JJ, 
the operation % {xd y \ — xd y will be equivalent to [xdy\ + — xd y , and therefore every 
covariant of the system must be reduced to zero by each of the operations 
59 = {xdy} + ydg — xd y . 
This being the case, we have 
59. = 59<b + 59 ($), 
. 59 = 3>59 + <3? (59), 
equations which it is obvious may be replaced by 
59. = iDd? + yd$ (<E>), 
$>. 59 = 3>59 + <I> {{xdy}), 
and consequently (in virtue of the theorem) by 
59. <f> = 59$ + yds (3>), 
d?. 59 = $59 + \yd^ (d>); 
and we have therefore 
59.3> — 4>. 59 = — ({?70^} — yd$) (<I>) ; 
or, since 4> is a covariant of ©, we have 59 • d> = <E>. 59. And since every covariant 
of the system is reduced to zero by the operation 59, and therefore by the operation 
d>. 59, such covariant will also be reduced to zero by the operation 59. d>, or what is 
the same thing, the covariant operated on by d>, is reduced to zero by the operation 
59 and is therefore a co variant, i.e. d> operating upon a covariant gives a covariant. 
16. In the case of a quantic such as U — 
m m 
(*&&, y\x, y')...\ 
we may instead of the new sets (£, y), (£', y)... employ the sets (y, — x), (y',—x'), &c. 
The operative quantic © is in this case defined by the equation © U = 0, and if 
be, as before, any covariant of ©, then d> operating upon a covariant of U will give 
a covariant of JJ. The proof is nearly the same as in the preceding case; we have 
instead of the equation d> ({xdy}) = (d>) the analogous equation 
{{xdy}) = - {xdy} (d>), 
where on the left-hand side [xd y ] refers to JJ, but on the right-hand side [xd y } refers 
to ©, and instead of 5® = {xdy\ + yd% — xd y we have simply 59 = {xd y } — xd y .