六次曲面不能有66个节点
A sextic surface cannot have 66 nodes
摘要：
设S是复射影3-空间中的曲面，只有结点作为奇点。假设S的度数为6。我们证明了S可以拥有的最大节点数是65。这方面的简要历史如下。Basset证明了S最多可以有66个节点。Catanese和Ceresa和Stagnaro构建了具有64个节点的六次曲面。Barth最近展示了一个65节点的六次曲面。我们通过证明S不能有66个节点来完成这个故事。设f：S~ --> S是奇点的极小分解。S上的节点集合N是偶数，如果存在S~上的因子Q使得2 Q ~ f^{-1}（N）。我们证明了S上一个非空的偶数节点集的大小必须为24、32、40、56或64。这个结果是证明66节点六次曲线不存在的关键。我们不知道一个六次曲面是否可以有一个大小为56或64的偶数节点集。大偶节点集的存在或不存在与下面的消失问题有关。设S是CP ^3中的一个s次法曲面。设D是S上的Weil因子，使得对某些r \in \Q，D Q-有理等价于rH.在什么情况下我们有H^1（O_S（D））= 0？例如，当r < 0时成立。对于s=4和r=0，H^1可以不为零。对于s=6和r=0，如果存在一个56或64节点的偶集，则H^1可以是非零的。H^1的消失还与P^3中集论完全相交的线性正规性、二次正规性等有关。
英文摘要：
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.
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