292 ON SYMMETRIC FUNCTIONS AND SEMIN VARIANTS. [932 
(ce oo c 3 ) beginning with the non-unitary term ce. We may, in fact, write 6 = 0; we 
thus have 
[eZ 2 ] = — 2g — ‘SOce + 20 d 2 , 
[c 3 ] = 3 g — 45 ce + 60 d 2 + .90c 3 , 
giving 3 [d 2 ] + 2 [c 3 ] = — 180 (ce — d 2 — c 3 ), and then ce — d 2 — c 3 , putting therein for c, d, e 
the values c — b 2 , d — 3be + 2b 3 , e — 46iZ + 66 2 c — 36 4 , gives the complete value ut supra, 
ce — d 2 — b 2 e + 2bed — c 3 , and we thus see d priori that this contains no term bf, but 
in fact begins with ce. And in carrying out this process for any higher given 
weight, it is proper also to arrange the non-unitary terms not in AO but in CO, 
and then in each case beginning with the terms highest in CO and eliminating as 
many as possible of these terms we obtain the sharp seminvariant. Consider for 
instance the weight 12: taking the finals in AO, we have here 
(to oo g 2 ), (to oo cf 2 ), (to go e 3 ), (to co b 2 / 2 ), ... 
the initials in CO are to, ck, dj, ei, ... and it might at first sight appear that the 
foregoing process of elimination would lead to the forms (to co g 2 ), (ck oo cf 2 ), (dj oo e 3 ), 
(ei oo b 2 f 2 ), ...; we in fact have the form (moog 2 )\ and if from (mcog 2 ) and (meoef 2 ) 
we eliminate to, we obtain the form (ck oo cf 2 ); but we cannot have a form (dj oo e 3 ) 
(for a form beginning with dj is of necessity of the degree 4 at least); what 
happens is that when from (to oo g 2 ), (m oo cf 2 ) and (to oo e 3 ) we eliminate to and ck, 
the next term dj disappears of itself, and (the following term ei not disappearing) 
the resulting form is (ei oo e 3 ): to obtain a form beginning with dj we must use 
the fourth form (to oo b 2 f 2 ), and we thence obtain (dj oo b 2 f 2 ). Arranging the initials 
in CO and the finals in AO, we thus have 
to g 2 
ck cf 2 
d J\/ e3 
ei b 2 f 2 
that is, we have the sharp semin variants to oo g 2 , ck oo cf 2 , ei oo e 3 , dj oo b 2 f 2 , ...; these 
are the results given by the MacMahon linkage as will be explained further on, but 
I will first approach the question from a different side. 
42. It has been seen that we have A, = + 2bd c + 3c0d + ..., as the annihilator 
of a seminvariant. Considering in the first place the entire set of terms, say for 
the weight 6, g (ao) b 6 , we assume for a seminvariant the sum of these each 
multiplied by an arbitrary coefficient; the number of coefficients is equal to the 
number of terms of g (ao) b 6 . Operating with A, we obtain a function of the next 
inferior weight 5, containing all the terms of Dg(ao)b 6 , that is, of f(ao)b 5 , each