338 
MATHEMATICS: H. S. WHITE 
ing planes of a second. This second curve K^, however, is completely 
determined by the first, Cz, and the relation 
F (X, m) = S U (X) . gi (m) = 0; 
for there is only one way of expressing a given bicubical function 
F (X, ix) Hnearly in terms of four independent cubics /i (X),/2 (X),/3 (X), 
A(x). 
Such a bicubical function F (X, ju) contains 15 arbitrary coefficients, 
besides one multiplicative constant. If it relates cubic curves whose 
relative situation is like that of the C3 and Kz mentioned above, we may 
say that the relation F (X, m) admits a solution of period 7, or briefly, 
that it admits a A 7. A special kind of (3, 3) relation is that which factors 
into three (1, 1) relations or projectivities: 
F (X, ix) (X, m) . (X, m) . X (X, m), 
the triply bilinear relation. Of this special kind there is a sub-species 
which admits a A 7. 
To prove the theorem stated above, viz., that every (3, 3) relation 
which admits one A 7 must necessarily admit a simple infinity of A 7*5, I ^ 
proceed by counting the number of free constants in each of the four 
classes of (3.3) relations which have just now been noticed. 
The first class, the general (3, 3) relation, contains 15 constants, in- 
cluding 3 that might have been deducted for Hnear transformation 
of either X or /z. The second class, that admitting one A 7 (or more 
than one), contains the three constants of a linear transformation and 
apparently 7 others, since according to the former theorem cited above 
the 7 points on the C3 can be chosen at random. If however the pres- 
ence of one A 7 should imply 00 1 others, the number of free constants 
would reduce to 9:- call the number 10— R, where R = 0 or 1. The 
third class contains 9 constants, 3 for each of the colHneations involved. 
The use of the fourth class is probably novel, at least in this connection ; 
it contains 3 constants. The proof of this is the essential part of the 
demonstration. 
The argument is now most easily stated geometrically. In a linear 
(flat) space of 15 dimensions, two contained algebraic varieties or spreads 
of 5 and of k dimensions respectively must have in common a spread of 
at least s -\- k— 15 dimensions. Here s = 10 — R, k = 9, and the com- 
mon part or intersection is of 3 dimensions. Hence 10 — i?+9— 15< 3, 
01 1 < R. But we had J? ^ 1, therefore = 1, as asserted in the 
theorem. 
