A sextic surface cannot have 66 nodes

TL;DR

This paper proves that a degree 6 surface in complex projective 3-space cannot have 66 nodes, completing the classification of maximum nodes on sextic surfaces and exploring related divisor and cohomology properties.

Contribution

It establishes the nonexistence of a sextic surface with 66 nodes and characterizes even node sets, advancing the understanding of singularities on sextic surfaces.

Findings

Maximum nodes on a sextic surface is 65.

Even node sets have sizes 24, 32, 40, 56, or 64.

Nonexistence of 66-node sextic surface confirmed.

Abstract

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.