ar X iv : 0 70 6 . 43 58 v 1 [ m at h . A G ] 2 9 Ju n 20 07 ON WAHL ’ S PROOF OF μ ( 6 ) = 65
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چکیده
In this note we present a short proof of the following theorem of D. Jaffe and D. Ruberman: Theorem [Ja-Ru]. A sextic hypersurface in P3 has at most 65 nodes. The bound is sharp by Barth’s construction [Ba] of a sextic with 65 nodes. Following Beauville [Be], to a set of n nodes on a surface is associated a linear subspace of F (where F is the field with two elements) whose elements corresponds to the so-called even subsets of the set of the nodes. Studying this code Beauville proved that the maximal number of nodes of a quintic surface is 31. The same idea was used by Jaffe and Ruberman, but their proof is not so short as the one of Beauville, partly because at that time a complete understanding of the possible cardinalities of an even set of nodes was missing. Almost at the same time, J. Wahl [Wa] proposed a much shorter proof of the same result. He proved indeed the following (see the beginning of the next section for the missing definitions) Theorem [Wa]. Let V ⊂ F66 be a code, with weights in {24, 32, 40}. Then dim(V ) ≤ 12. He claimed that Jaffe-Ruberman’s theorem follows as a corollary since the code associated to a nodal sextic has dimension at least n− 53 (see section 1 of [Ca-To] for this computation). In fact, he used an incorrect result stated by Casnati and Catanese in [Ca-Ca], asserting that the possible cardinalities of an even set of nodes on a sextic are only 24, 32 and 40. Recently Catanese and Tonoli showed indeed Theorem [Ca-To]. On a sextic nodal surface in P3, an even set of nodes has cardinality in {24, 32, 40, 56}. Note however that [Ca-To] used a result by Jaffe and Ruberman, namely that there is no even set of nodes of cardinality 48.
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