Colloquium publications.

88

THE MADISON COLLOQUIUM.

22. For n = 0 and m odd, we may delete the terms a31 from
f3 by use of I1Km. First, let m = 1 and apply transformation
(51); we get
82l = 1' + = 3', 23 = ~a',
(82)
R' = fill' + f2~2' + (fi + f3) 3' + (XI'X2'X3' +- x2X22')q.
Thus 4 = 0. Since f3 = I + 12q, condition fi + f3 = f3' gives
I = I2(a1bl + c2b2 + a3b3 + a2ao3 + ala2).
Add to this the relation obtained by permuting the subscripts
1, 2. Thus
0 = I2(bi + bz + a2o3 + alaa).
The increment under (22) is I2(bl + a3 + a2a3) = 0. Now I1
is of the form x + yA + zA, where x, y, z are constants. From
the terms in blb2, we get y = 0. Then x = z = 0. The only
covariants are therefore I1K.
Second, let m > 1. Then KW(-')1/2 is of the form (76) with
f3 = a3q + a3, by (811). Hence we may set
f3= I + cq + dqA (c, d constants).
In R given by (76), let g denote the coefficient of
(83) X1X2x3 ' X2X32m-3.
In the function derived from R by the transformation (51), the
term corresponding to (83) has the coefficient g + fi, since by
(82) the hi parts contribute only one such term, that from
l N3. Now
f = I + cq' + dq'A, q' = b2b3 + (b2 + b3)al.
When g is given the notation (56), g' - g = fi is the function
(57). But a3b1 occurs in fl only in J and AJ and in them with
the linearly independent multipliers (62). Hence
I = n1(A + 1) + n2A.
The coefficient of a3 in fi is now
nlalao2 + n2(ala2 + b3) + dq'ala2 = p(b3 + a2).