( 255 ) 
In the first place it is evident that the ten “spaces through six 
lines” with the spacial Coordinates 
(aj a) ....(—1, —1, i ay el) 
(artan erf "0, OR) 10 OF 10) 
OO Eeen GO, LIE 0, 0, 0) 
fand ete (1, i 1, —1, 1) 
(aig BBP: an «She G2) Ly 0, 0, 040) 
Cr EE NP je On 0) 
(aprag)he We € 1, =1,. —1, Bones) 
fea. wea ky 1, —1, —1, 1) 
(agaz;).... (—l, 1, —1, eh) 
te ee on sf SL) SSL, 1, —l, 1) 
are double spaces of the enveloped space, so that this must be of 
order four and cannot admit of an eleventh double space, because 
a space of order two cannot be of class three. 
Moreover it is evident, that the fifteen “points in three lines” 
are points for which the tangent planes of the curved space, passing 
through it, envelope a conic space degenerated into three nets of 
planes (all planes through a line). So the three lines as, cs, ej cut: 
each other in the point 
ys By SS Py Se dd 
with the equation 
Pi + po + pz + ps = 0 
and the combination of this with the equation of the cubic envelope 
causes the latter to be transformed by elimination of p, into 
(pa + ps) (Ps + Pi) (pi + Po) = 0, 
which in connection with the first furnishes the planes 
Po + ps3 = 9 Ps. Biss pi + po = 9 
Pyt Pa 0) pe + p= 04 P3 + pa =0 
with the axes ej, ¢3, az. 
