417] THE THEORY OF SCREWS IN NON-EUCLIDIAN SPACE.
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For these equations must be linear and if X2, X.3, Xt are all zero then
«1, «;2, x3, xt become xi, xi, xi, xi as they ought to do, and similarly for the
others, whence we get
X-J x" x(n xi'" Xi = Xi Xi" xi" xi"'
x<2 x" X2" x"" x2 ^2 xi" xi'"
x3' x3 x3 xi'" X3 x3" xi" xi'"
xi xi' xi" xi'" Xi x4" x^" xi'"
We may write this result thus
HXi = Xi X^2 Xi
Xi" x" xi' xi'
xi" •X2 xi" x"'
xi'" 0C2"' x"" xi'"
Let us now suppose that the vertices of this new tetrahedron are the
double points of a homography defined by the equations
yi = (11) «1 + (12) «2 + (13) «3 + (14) xt,
y2 = (21) «j + (22) + (23) x3 + (24)a>4,
2/s = (31) + (32) x2 4- (33) xs + (34) x4,
yt = (41) «j + (42) x2 + (43) x3 + (44) x4.
We have to solve the biquadratic
(11) -P (12) (13)
(21) (22) - p (23)
(31) (32) (33) -p
(41) (42) (43)
(14)
(24)
(34)
(44) - p
= 0.
Let the roots be p,, p2, p3, pt. Then we have
Pixi = (11)«/ + (12) x2' + (13) x3' + (14) xi,
pixi = (21) xi + (22) xi + (23) xi + (24) xi,
pixi = (31) x^ + (32) xi + (33) xi + (34) xi,
piXi = (41) Xi + (42) xi + (43) xi + (44)
with similar equations for xi', x"', xi'", x,", &c.