( 627 ) 
Mathematics. — “A group of algebraic complexes of rays”. By 
Prof. JAN DE VRIES. 
§ 1. Supposing the rays a of a pencil (A, @) to be projective to 
the curves 6" of order n, passing through 7? fixed points, By, of the 
plane 8, we shall regard the complex of the rays resting on homolo- 
gous lines. For n =1 we evidently find the tetrahedral complex. 
Out of any point P we project (A,a@) on 8 in a pencil (4’, 8), 
generating with the pencil (47) a curve c’+!. So we have a complex 
of order (n +1). 
Evidently the curve emt! does not change when the point P is 
moved along the right line AA’; so the intersections of the op” cones 
of the complex (7?) with the plane 8 belong to a system o*. It is 
easy to see that they form a net. 
For, if such a curve ct! is to contain the point X and if Die 
is the eurve through Bj, and X, and ax the ray conjugate to it 
through A, the point A’ must be situated on the right line connecting 
X with the trace of ax on the plane 8. In like manner a 
second point through which c”tt must pass, gives a second right 
line containing A’. The curve et! being determined as soon as A! 
is found, one curve ct! can be brought through two arbitrary 
points of 9. 
On the right line @ the given pencils determine a (1, 7)-corre- 
spondence; its (n +1) coincidences C, are situated on each c”t!. 
So the net has (n° +7 +1) fixed base-points *). 
§ 2. When A’ moves along a right line a’ situated in 8 and 
cutting the plane a in S, the curve ct! will always have to pass 
through the 7 points Dj, which @ has in common with the curve 
bn conjugate to the ray AS. It then passes through (n + 1)* fixed 
points, so it describes a pencil comprised in the net. 
To the 387° nodes of curves belonging to that pencil must be 
counted the # points of intersection of ¢8 with that c” passing through 
the points Bj, and Dy. Hence a’ contains, besides |S, (3 2? — 7) points 
A’ for which the corresponding curve c+! possesses a node. 
If A’ coincides with one of the base-points 4; then the projective 
pencils (A) and (6") generate a c+! possessing in that point B a 
node. According to a well known property B is equivalent to 
two of the nodes appearing in the pencil (c*+!) which is formed 
1) To determine this particular net one can choose arbitrarily but 4 2(z-++-3)— 1 
points B and three points C. 
43 
Proceedings Royal Acad. Amsterdam. Vol. VII. 
