%0 Journal Article
%T A sextic surface cannot have 66 nodes
%A David B. Jaffe
%A Daniel Ruberman
%J Mathematics
%D 1995
%I arXiv
%X Let S be a surface in complex projective 3-space, having only nodes as singularities. Suppose that S has degree 6. We show that the maximum number of nodes which S can have is 65. An abbreviated history of this is as follows. Basset showed that S can have at most 66 nodes. Catanese and Ceresa and Stagnaro constructed sextic surfaces having 64 nodes. Barth has recently exhibited a 65 node sextic surface. We complete the story by showing that S cannot have 66 nodes. Let f: S~ --> S be a minimal resolution of singularities. A set N of nodes on S is even if there exists a divisor Q on S~ such that 2Q ~ f^{-1}(N). We show that a nonempty even set of nodes on S must have size 24, 32, 40, 56, or 64. This result is key to showing the nonexistence of the 66 node sextic. We do not know if a sextic surface can have an even node set of size 56 or 64. The existence or nonexistence of large even node sets is related to the following vanishing problem. Let S be a normal surface of degree s in CP^3. Let D be a Weil divisor on S such that D is Q-rationally equivalent to rH, for some r \in \Q. Under what circumstances do we have H^1(O_S(D)) = 0? For instance, this holds when r < 0. For s=4 and r=0, H^1 can be nonzero. For s=6 and r=0, if a 56 or 64 node even set exists, then H^1 can be nonzero. The vanishing of H^1 is also related to linear normality, quadric normality, etc. of set-theoretic complete intersections in P^3.
%U http://arxiv.org/abs/alg-geom/9502001v1