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Class Note for MATH 796 at KU 6

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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 17 views.

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Date Created: 02/06/15
Wednesday 13008 Modular Lattices De nition A lattice L is modular if for every x y z E L with x S 2 l xVyzxVyzl Note For all lattices if x S 2 then x y 2 S x y Some basic facts and examples 1 Every sublattice of a modular lattice is modular Also if L is distributive and x S 2 E L then xVyz aczyz xVyz so L is modular 2 L is modular if and only if Lquot is modular Unlike the corresponding statement for distributivity this is completely trivial because the de nition of modularity is invariant under dualization 3 N5 is not modular With the labeling below we have a S b but aVcbaO a aVcblb b C a 4 M5 E H3 is modular However H4 is not modular exercise Modular lattices tend to come up in algebraic settings 0 Subspaces of a vector space 0 Subgroups of a group 0 Rsubmodules of an Rmodule Elgl if X Y Z are subspaces of a vector space V with X Q Z then the modularity condition says that XYnZXYnZ Proposition 1 Let L be a lattice TFAE l L is modular 2 For allxyz E L ifxE y22 then x xVy2 2 For allxyzE L ifxE yyz then x aczy 3 For all y 2 E L there is an isomorphism of lattices MadelynV2 given byai HlVy bAZEb Proof 1 gt 2 is easy if we take the de nition of modularity and assume in addition that ac 2 y 2 then the equation becomes ac ac y 2 For 2 gt 1 suppose that 2 holds Let X Y Z E L with X 3 Z Note that YAZgXVYZ ZVZZ so applying 2 with y Y 2 Z ac X Y Z gives XVYZ XVYZY Z XvYMZ as desired 2 ltgt 2 because modularity is a selfdual conditionl Finally 3 is equivalent to 2 and 2 togetherl D Theorem 2 Let L be a lattice 1 L is modular and only it contains no sublattice isomorphic to N5 2 L is distributive and only if it contains no sublattice isomorphic to N5 or M5 Proof Both gt directions are easy because N5 is not modular and M5 is not distributivel Suppose that ac y 2 is a triple for which modularity failsl One can check that xVy xVy z xly is a sublattice details left to the reader Suppose that L is not distributivel If it isn t modular then it contains an N5 so there s nothing to prove If it is modular then choose ac y 2 such that acyzgt acyac2l You can then show that 1 this inequality is invariant under permuting ac y z 2 95A y Vz 3 A2 and the two other lattice elements obtained by permuting ac y 2 form a cochain and 3 the join respl meet of any of two of those three guys is equal Hence we have constructed a sublattice of L isomorphic to M5 D Semimodular Lattices De nition A lattice L is upper semimodular if for all z y E L 2 xAylty 9 sz Here s the idea Consider the interval as y x y C L xVy xy H L is semimodular then the interval has the property that if the southeast relation is a cover then so is the northwest relation L is lower semimodular if the converse of 2 holds for all z y E L Lemma 3 UL is modular then it is upper and lower semimodular Proof If xy lt y then the sublattice acy y has only two elements H L is modular then by condition 3 of Proposition 1 we have x y y E 9595 y so 95 lt x y Hence L is upper semimodular A similar argument proves that L is lower smimodular D In fact upper and lower semimodularity together imply modularity To make this more explicit we will show that each of these three conditions on a lattice L implies that it is ranked and moreover for all x y E L the rank function r satis es rx y rx y S rx ry if L is upper semimodular rx y rx y 2 rx ry if L is lower semimodular rx y rx y rx ry if L is modular

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