Class Note for MATH 796 at KU 2
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Date Created: 02/06/15
Monday 33 Crosscuts Recall that the Mobius algebra AL of a lattice L is the vector space of formal Clinear combinations of elements of L with multiplication given by the meet operation We showed last time that AL E ClLl as rings because the elements Ex Eyltx uy 95 satisfy EEEy 61151 where 61y is the Kronecker delta 7 De nition 1 Let L be a lattice A A 7 An upper crosscut of L is a set X C L 1 such that if y E L X 1 then y lt x for some as E X i A lower crosscut of L is a set X C L such that if y E L X 0 then y gt x for some as E X Proposition 1 Rota s Crosscut Theorem Let L be a nite lattice and let X be an upper crosscut Then 1 L Z Ulyl YgX Y6 Dually ifX is a lower crosscut then 1b ML Z U yl YgX VYi Proof Let x E L We have the following equation in the Mobius algebra of L my tel Ham Mm 25y cEX acEX ygac yEY where Y y E L l y i x for all x E X But Y because X is an upper crosscuti That is 2 Hdix Ei 2min acEX yEL Therefore On the other hand 3 Haw ZHWAA xeX AgX Now extracting the coefficient of O on the righthand sides of 2 and 3 yields la The proof of lb is similar D We already know that un 71 What if L is distributive but not Boolean Proposition 2 Let L be a dist ributiue lattice that is not Boolean Then uL 0 Proof The set X of atoms of L is a lower crosscuti Since L is not Boolean it has a joinirreducible element b E Xi But b i V X because b i a for all a E X this is Lemma 2 of the lecture notes from 12807 In particular V X y i so the sum on the righthand side of lb is empty U More generally ML 0 if L is a lattice in which i is not a join of atoms or dually if 0 is not a meet of coatomsi The crosscut theorem will also be useful in studying hyperplane arrangementsi A Topological Application Proposition 3 Let P be a chain nite bounded poset and let cl 0x0ltx1ltltxll the number of chains of lengthi between O and Then 4 MP ZPIYCV 1 Proof The incidence algebra makes the proof almost triViali Recall that cl 7110 7 ll0 Since sufficiently high powers of n vanish 17 n 1 10 is a perfectly good equation in P Therefore 2c ZWM 17n 10 1 C1611 Mali D 10 10 The expression 4 looks like an Euler characteristici lndeed let P be a nite poseti The order complex of P is de ned as the simplicial complex A AP Whose vertices are elements of P and Whose faces are chains of Pi Then Proposition 3 implies that uP is the reduced Euler characteristic of Al Another nice fact is the following result due to l Folkman 1966 Whose proof used the crosscut theoremi Theorem 4 Let L be a geometric lattice of rank r and let P L 0 Then ZlML ifi r 7 2 0 otherwise H1APZ 2 where H denotes reduced simplicidl homology Hyperplane Arrangements De nition 2 Let K be a eld and n 2 1 A linear hyperplane in Kquot is a vector subspace ofcodimension 1 An a ine hyperplane is a translate of a linear hyperplane A hyperplane arrangement A is a nite collection of distinct hyperplanes The number n is called the dimension of A Example 1 The lefthand arrangement A1 is linear it consists of the lines 95 0 y 0 and x y The righthand arrangement A2 is af ne it consists of the four lines 95 y x 7y y l and y 71 Each hyperplane is the zero set of some linear form so their union is the zero set of the product of those 3 linear forms We can specify an arrangement concisely by that product called the de ning polynomial of A as an algebraic variety in fact For example the de ning polynomials of A1 and A2 are avgx 7 y and avgx 7 y 7 1 respectively Example 2 Here are some 3D arrangements pictures produced using Maple The Boolean arrangement Q7 consists of the coordinate hyperplanes in nspace so its de ning polynomial is acle 957 Here s 3 The braid arrangement Q7 consists of the hyperplanes x1 7 xj in nspace so its de ning polynomial is H x 7 xj 1 lltjS7l Here s Brg Every hyperplane in 7 contains the line spanned by the allones vector If we project R4 to the quotient by that line then A4 ends up looking like this The Intersection Poset De nition 3 Let A C K be an arrangement lts intersection poset LA is the poset of all intersections of subsets of A ordered by reverse inclusion This poset always has a 0 element namely K It has a 1 element if and only if HeA H y ll such an arrangement is called central Proposition 5 Let A C K be an arrangement The following are equivalent 0 A is central 0 A is a translation of a linear arrangement 0 LA is a geometric lattice Proof Linear arrangements are central because every hyperplane contains 6 E K Conversely if A is central and p E HeA H then translating everything by 7p produces a linear arrangement If A is central then LA is bounded It is a joinsemilattice with join given by intersection hence it is a lattice Indeed it is a geometric lattice it is clearly atomic and it is submodular because it is a sublattice of the chain nite modular lattice LK 7 that is the dual of the lattice of all subspaces of B When A is central we may as well assume linear the matroid associated with LA is naturally represented by the normal vectors to the hyperplanes in A Therefore all of the tools we have developed for looking at lattices and matroids can be applied to study hyperplane arrangements The dimension of an arrangement cannot be inferred from the intersection poset For example if A1 is as above then LA1 E LBr3 but dimA1 2 and dim Brg 3 A more useful invariant of A is its rank rank A de ned as the rank of LAi Equivalently de ne W C K to be the subspace spanned by the normal vectors up Then rankA dim W De nition 4 An arrangement A is essential if rankA dim A In general the essentialization essA is the arrangement H n W l H e A C W Equivalently if V WL HeA H then rankA n 7 dim V and we could de ne the essentialization of A as a quotient HV l HEA C K V Observe that essA is essential and that LA is naturally isomorphic to LessAi
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