A counterexample to Orlik’s conjecture
A counterexample to Orlik’s conjecture
复制标题
奥尔利克猜想的反例
DOI：
10.1090/s0002-9939-1993-1134624-x
复制
发表时间：
1993
期刊：
影响因子：
--
通讯作者：
V. Reiner
中科院分区：
文献类型：
--
作者：
Paul H. Edelman;V. Reiner
我们提出了一个反例的猜想Orlik的限制一个自由的超平面安排，它的超平面之一是自由的。I.定义和反例本说明提出了一个反例的猜想Orlik超平面安排。在下文中，我们只给出陈述猜想和反例所必需的那些定义。有关超平面排列理论的更多背景，请参见[或]。设X是通过Rd中原点的超平面（余维为1的子空间）的有限集合，对于A中的每个超平面H，设'H是多项式环S = R[x1，.，Xd]，其在H上消失（因此'H被唯一定义为纯量倍数）。i-导子模Der（A '）被定义为所有导子6：S S的集合，其性质是：对于A中的所有H，6（iH）可被'H整除。Der（A）是多项式环S上的模，我们说A是自由排列，如果它是S上的自由模。给定A中的任何超平面H，我们定义限制排列AIH为子空间H（将H视为Rd-i）内的排列，其超平面是X与H的超平面的所有交点。猜想（Orlik [T，问题2;或，第86页]）。对于自由排列A中的任何超平面H，限制排列“IH”是自由的。这个猜想的一个反例是由R5中的21个超平面组成的排列A，它由方程xi = 0（i= 1，2，3，4，5）和X1+ 82 X2 + 83 X3 + 84 X4 + e85 X5 =0（其中ei=+l（i））定义。利用计算机代数软件包MAGNETLAY，证明了Der（Z）是一个自由S-模，其S-基分别由5个齐次导子（0，4，4，4，4）组成。然后使用MATHEMATICA对这些推导进行双重检查，以满足Saito的游离度标准[或，定理9.7]。选择H为定义为“H =”的超平面，编辑于1991年9月24日收到，修订版于1991年11月6日收到。1991年数学学科分类。第52 B30集? 1993美国数学学会0002-9939/93 $1.00 + $.25每页
We present a counterexample to the conjecture by Orlik that the restriction of a free hyperplane arrangement to one of its hyperplanes is free. I. DEFINITIONS AND THE COUNTEREXAMPLE This note presents a counterexample to a conjecture by Orlik on hyperplane arrangements. In what follows, we give only those definitions necessary to state the conjecture and the counterexample. For more background on the theory of hyperplane arrangements, see [Or]. Let X be a finite set of hyperplanes (subspaces of codimension one) passing through the origin in Rd, and for each hyperplane H in A, let 'H be the linear form in the polynomial ring S = R[x1, ... , Xd] that vanishes on H (so that 'H is uniquely defined up to a scalar multiple). The module of i-derivations Der(A') is defined to be the set of all derivations 6: S S with the property that 6(iH) is divisible by 'H for all H in A. Der(A) is a module over the polynomial ring S, and and we say A is a free arrangement if it is a free module over S. Given any hyperplane H in A, we define the restriction arrangement AIH to be the arrangement within the subspace H (thinking of H as Rd-i ) whose hyperplanes are all of the intersections of hyperplanes of X with H. Conjecture (Orlik [T, Problem 2; Or, p. 86]). For any hyperplane H in a free arrangement A, the restriction arrangement 'IH is free. A counterexample to this conjecture is given by the arrangement A consisting of the 21 hyperplanes in R5 defined by the equations xi = 0 for i=1,2,3,4,5 and X1+82X2+83X3+84X4+e85X5=0 where ei=+l foreach i. Using the computer algebra package MACAULAY, it was checked that Der(Z) is a free S-module with an S-basis consisting of five homogeneous derivations of degrees (0, 4, 4, 4, 4), respectively. These derivations were then double-checked to satisfy Saito's criterion for freeness [Or, Theorem 9.7] using MATHEMATICA. Choosing H to be the hyperplane defined by 'H = Received by the editors September 24, 1991 and, in revised form, November 6, 1991. 1991 Mathematics Subject Classification. Primary 52B30. ? 1993 American Mathematical Society 0002-9939/93 $1.00 + $.25 per page