Colloquium publications.

90

THE MADISON COLLOQUIUM.

As in ~ 18, X = 0. Thus 12 and I are the products of A + A + 1
by k2, kl, so that
S = (k + k2b3) ( + A + 1) + a21 + ki(b1b3 + bloa2 + aia2)
(87)
+ k2b3(a2bl + ala2) + k5A(blb3 + bl).
For n odd, S is the increment to S1 under (50) and hence has
no term containing a3bl. If t is the coefficient of J in I1, a3bl
occurs in (87) only in ta2J and in the final part, being multiplied
by ta2alb3 and k5saia2(b3 + 1), respectively. Hence t 3 k5 - 0.
Since S is of the form (57), the coefficient of b1 must vanish if
al 0. Thus
kl(b3 + a2) + k2b3a2 = 0, k1 k-2 0.
Now S = a2I1 = a2(u + vA) must vanish for al - 0, b3 - a2
by (57); then = a2(b2 + a3), so that u = v = 0, S = 0. Any
covariant with m = 1 and n odd differs from one of rank > 1 by
a linear function of KLn, AKL.
24. For m = 1, n = 4v, we may delete a2I1 from (85) by use
of IiKQ2V. Set fz = B2 + * * * + Bnxan. Then (51) replaces
(76) by
R =?2[SX3n + S1X3n —1l + (S + S2)x3n-2Xl2 + * *] + J3f3
+ ('l + a3)[Bn(x3n + X3n-4xl +..)
+ Bn-lX2( 1 + 2 3n-2X1 + ( + + *)+ (X + X12X2).
Since S1 is the increment of S2, it is a linear combination of the
functions (74). By use of L-3Ql, Ln-3K2 and their products by
A and A, we may, without disturbing S, delete from Si
b3+ala2+1, Ab3, A+b3A+b3ala2, a2(b3+1), a2A(b3+1).
Hence we may set
SI = tl(b3 + a2) + t2b3ala2 + t3Aa2 + t4(b3 + a2)J.
Applying (ala2)(blb2) to S and SX, we obtain Bn and Bn1_.
Let 1 be the coefficient of X2X3-"1 in /. By the coefficient of