405] 
AN EIGHTH MEMOIR ON QUANTICS. 
189 
But the theory may be presented under a better form ; take for instance the cubic, 
co lc 
viz. writing - and j for x and k respectively, we have (a, 6, c, d^Jcy + lx, ly — kx) 3 
k 3 
k 2 l 
kl 2 
¿ 3 
X 3 
- d, 
3c, 
-36, 
a 
3c, 
— 66 + 3d, 
3a — 6c, 
36 
a-i/ 2 
-36, 
3a — 6c, 
66 - 3d, 
3c 
2/ 3 
a, 
36, 
3c, 
d 
a bipartite cubic function (*]£&, l) 3 (x, y) 3 \ and the determinant formed out of the 
matrix is at once seen to be an invariant of this bipartite cubic function. 
Assume now that we have identically 
(a, b, c, djx, y) 3 =(a!, b', c', d'\\(x + iy), %(x-iy)) 3 , 
viz. this equation written under the equivalent form 
(a', b', o', d'\X, F) 3 = (a, b, c, d#X + Y, i (X - F)) 3 , 
determines (a, b', c', d') as linear functions of (a, b, c, d), it in fact gives 
a = (a, b, c, d}£ 1, — i) 3 = a — 3bi + 3a 2 — di 3 , 
b' —(a, b, c, d$ 1, — if (1, i) —a — bi— ci 2 + di 3 , 
c' = (a, b, c, (¿$1, — i) (1, i) 2 = a + bi — ci 2 — di 3 , 
d! = (a, b, c, d$l, i) 3 = a + 3bi + 3ci 2 + di 3 , 
then observing that ky + lx ±i (ly — kx) = (x ± iy) (+ ik + l), we have 
(a, b, c, d'fyky + lx, ly - kx) 3 = (a', b', c', d'\\ (x + iy) (- ik +1), \(x - iy) (ik + l)) 3 , 
and if in the expression on the right-hand side we make the linear transformations 
x + iy = x' C2, — ik + l = k' V2, 
x — iy = — iy' V2, ik + 1 = — iV V2, 
which are respectively of the determinant +1, the transformed function is 
= (a, V, c', d'\k'x, -I'yJ, 
that is, we have 
(a, b, c, d\ky-(-lx, ly — kx) 3 = (a, 6', c', d'\k'x, — Vy'f.