On sextic surfaces having only nodes (preliminary report)
  Let S be a surface in CP^3, having only nodes as singularities. Let pi: 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 ~ pi^{-1}(N).   Suppose that S has degree 6. It is known (Basset) that S cannot have 67 or more nodes. It is also known (Barth) that S can have 65 nodes. It is not known if S can have 66 nodes. Likewise, it is not known exactly what sizes can occur for an even set of nodes on S.   We show that an nonempty even set of nodes on S must have size 24, 32, 40, 56, or 64. We do not know if the sizes 56 and 64 can occur. We show that if S has 66 nodes, then it must have an even set of 64 nodes, and it cannot have an even set of 56 nodes. THUS IF ONE COULD RULE OUT THE CASE OF A 64 NODE EVEN SET, IT WOULD FOLLOW THAT S CANNOT HAVE 66 NODES.   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. Hard copy is available from the author. E-mail to jaffe@cpthree.unl.edu. 
David B. Jaffe