WILLIAMS. — GEOMETRY ON RULED QUARTIC SURFACES. 41 



in a point where the plane meets A and once where the plane is tangent 

 to the scroll ; therefore the section lias five double points and the plane 

 is a double tangent plane. The conic and either generator lie on differ- 

 ent sheets at the point where they meet K, and, regarding them as lying 

 on the scroll, they do not intersect there. An arbitrary generator meets 

 the plane once, and, since it does not meet either generator in the plane, it 

 meets the conic once, and therefore every conic is a 2 X . Any plane 

 through one and only one generator cuts out a plane cubic, a 3 U having 

 a double point on A" and passing once through each of the points where 

 the generator meets A' and D. Any plane that does not contain D, K, 

 or any generator, cuts out a plane quartic, a 4 1} having three double 

 points, two on A' and one on D. 



The double conic A is the complete intersection of its plane with the 

 scroll, aud every plane quartic is the complete intersection of its plane 

 with the scroll, aud therefore formula (1) holds for A" and for every plane 

 quartic (Theorem II). A plane cubic is cut out by a plane through a 

 single generator, D is cut out by a plane through two generators, and 

 every conic is cut out by a plane through two generators, and therefore 

 formula (1) holds for every plane cubic, for D, and for every conic 

 (Theorem III). Formula (1) holds, therefore, for all plane curves on 

 the scroll. 



4. Twisted Cubic, 3 X . — Since a < - , every twisted cubic is a 3^ We 



have seen that there are a points of a a on A, i. e. there are three points 

 of the twisted cubic on K, and, since we have two points at our disposal 

 in the determination of the quadric that cuts out the twisted cubic, we 

 can take these two points on A" and thus make the quadric contain K. 

 The residual, which is of order 5, will then consist of K, which counts 

 for 4, and one generator, and therefore formula (1) holds for every 

 twisted cubic on the scroll (Theorem III). 



5. Twisted Quartics, 4 2 and 4 X . — Since a < --, every twisted quartic 



JU 



is either a 4 2 or a 4 1# A 4 2 may be cut out by a quadric ; every generator 

 will then meet the quadric twice on the quartic curve 4 2 and cannot meet 

 it again without lying on it ; consequently, every generator that has on 

 it a point of the residual must lie on the quadric and form part of the 

 residual, and the residual therefore consists of four generators ; therefore 

 formula (1) holds for every 4 2 (Theorem III). For the 4 n we take 

 v = 3 ; then r = 8, and we have 19 — 13 = 6 points at our disposal in 

 the determination of the cubic surface £ l3) that cuts out the 4 X . There 



