8 KANSAS UNIVERSITY SCIENCE BULLETIN. 



xs ; the three planes y, z and ui each intersect the monoid in the line 

 yz twice, a line of kind III, and a transversal. A plane through the 

 double line and either of the two ordinary lines also meets the monoid 

 in a transversal ; the monoid thus has six transversals. 



h). If the monoid has a line of kind 1(1,2), the superior cone 

 has three double edges which are simple edges on the inferior cone 

 and a simple edge which is a double edge on the inferior cone. The 

 monoid has four ordinary lines on it and has in general no trans- 

 versals. 



c). If the monoid has two lines of kind 1(1,2), the inferior cone 

 must break up into a quadric cone and a plane. The equation of the 

 monoid will then be of the form 



U4 + U2U1S = ; 

 where U4, U2 and ui are homogeneous functions of x, y and z of de- 

 grees equal to the subscripts. The single plane of the inferior cone 

 meets the monoid in four lines, of which two are the lines of kind I 

 and two are the two ordinary lines necessary to make up the twelve. 

 This monoid has no transversals. 



d). If the monoid has three lines of kind 1(1,2), the inferior 

 cone breaks up into three planes not having a line in common. The 

 equation of the monoid will then be of the form 



U4 -f uiviwis = ; 

 where U4, ui, vi and wi are homogeneous functions of x, y and z of 

 degrees equal to the subscripts. Each plane of the inferior cone in- 

 tersects the monoid in a line of kind III and two lines of kind I. 

 There are no transversals. 



e). If the monoid has a line of kind 1(1, 3) the equation is of the 



form 



U4 + U3Z + U2z"^ + uiz^ + z^ + viWiti s = ; 



where vi, wi and ti are homogeneous functions of x, y and z of degrees 

 equal to the subscripts. Each of the three planes of the inferior cone 

 meets the monoid in the line of kind I, the line of kind III counting 

 twice and an ordinary line on the monoid. There are no transversals. 



/). If the monoid has a double line of kind 1(2, 3), the equation 

 of the monoid will be of the form 



U1V3 + viwitis = ; 

 where ui, vi, wi, vs and ti are homogeneous functions of x, y and z of 

 degrees equal to the subscripts. There are of course no other lines 

 on the monoid passing through the vertex; there is however one 

 transversal, and it passes through all three double points. 



