Colloquium publications.

68

THE MADISON COLLOQUIUM.

The pfaffians C2, **, Cm are unaltered modulo 2, while
(10) C1'C1-+C2, u2'-u2+-U, uit'ui (i+2) (mod 2).
Hence K is unaltered modulo 2. Note that
0 C12 C13 * Celm UC1
C12 0 C23 '** C2m U2
(11) K2 -......... (mod 2).
Clm C2m C3m *~ 0 Um
U1 U2 Us. Um. m 0
We saw that C1, * * *, Cm are cogredient with x1, * *, xm. This
is evident from the fact that the apex is covariantively related
to q(x). Hence if we substitute C1 for xi, *., Cm for Xm in (1),
we obtain the formal invariant
(12) qm(C) = ZcijCiCj + ZbiC2 (i, i = 1, * *, m; i < j).
If this invariant vanishes, the apex is on the locus, which is
then a cone. Indeed, by (2), every point on the line joining (C)
to a point on q(x)= 0 lies on the latter. Hence q(x) can be
transformed into a form in m - 1 variables and hence has the
discriminant zero. To argue algebraically, let new variables be
chosen so that the apex becomes (0,.., 0, 1). The polar of any
point (y) passes through the apex. Taking zl = 0,.., Zm-l = 0,
Zm = 1 in (4), we see that the polar (3') becomes cimYl + *
+ Cm-lmym-l, which must vanish for arbitrary y's. Hence
bmxm2 is the only term of (1) involving Xm. But the apex is on
the locus. Hence bm = 0 and q(x) is free of Xm. The converse
is obvious from (5).
Whether m is odd or even, q(x) has the invariant
(13) Am = n(ci+ 1) (i,j 1,, m; i <j).
This is evidently true by (9) or as follows. If Am E 1 (mod 2),
every cij 0 and q (2bix)2; while if Am = 0, at least one
cj is not congruent to zero, and q is not a double line.
Hence the product Amq(x) is a covariant; in fact, the square