I know that a Boolean lattice must be distributive.
But what with these lattices? Are these distributive?
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How to recognize which lattices are distributive or not only by looking on their diagrams? Is it even possible?
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8$\begingroup$ Yes, it's possible. A lattice is distributive if and only if it does not contain the diamond lattice or the pentagon lattice as a sublattice. See en.wikipedia.org/wiki/Distributive_lattice $\endgroup$– Tony HuynhJan 14, 2015 at 15:48
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2$\begingroup$ not all of these are lattices in fact $\endgroup$– მამუკა ჯიბლაძეJan 14, 2015 at 16:11
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$\begingroup$ So, none of these three are distributive. Am I right? $\endgroup$– JohnDoeJan 14, 2015 at 16:17
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$\begingroup$ Isn't (ii) distributive? Note that a sublattice must have the sames meets and joins as the original lattice. $\endgroup$– Pace NielsenJan 14, 2015 at 16:32
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2$\begingroup$ I am not sure this question deserves to be on hold. The question of how to recognize which lattices are distributive from their diagram is rather interesting. In addition to the classical result about $M_3$ and $N_5$, there is the result of Farley and Schmidt that a finite lattice is distributive if and only if every open interval is either an antichain or is connected, and every interval of rank three is distributive. (Up to isomorphism, there are five distributive lattices of rank three.) See Enumerative Combinatorics, vol. 1, 2nd ed., Exercise 3.31. $\endgroup$– Richard StanleyJan 15, 2015 at 16:27
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1 Answer
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There is of course the classic criterion of excluded lattices. A lattice is distributive if does not contain either $M_3$ or $N_5$ (see here for definitions). An easier criterion to check for large lattices is Birkhoff's two chain theorem: if a lattice is generated by two chains, then it is distributive. (The converse is not true.) You can find this in Birkhoff's book Lattice Theory.
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1$\begingroup$ Birkhoff's theorem states that a modular lattice generated by two chains is distributive. It is not true that any lattice generated by two chains is distributive. $\endgroup$ Jan 15, 2015 at 2:51
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$\begingroup$ You're completely right. My subconscious bias in favor of lattices of modules: revealed. $\endgroup$– arsmathJan 15, 2015 at 9:07
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$\begingroup$ (Professor George Markowsky has asked me to make the following comment.) Theorem 2 from Some Combinatorial Aspects of Lattice Theory, Proc. Univ. of Houston Lattice Theory Conf., 1973, 36-68 states: THEOREM 2. Let L be a finite lattice. The following are equivalent. (a) L has length n, satisfies the Jordan-Dedekind chain condition, has n join-irreducible elements and n meet-irreducible elements. (b) L has n join-irreducible elements, and every connected (maximal) chain between I and 0 has length n. (c) L is distributive and has n join-irreducible elements. $\endgroup$– TriAug 2, 2016 at 6:50
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$\begingroup$ (continuing Professor Markowsky's comment) In a Hasse diagram of a finite lattice, the join-irreducible elements are the nodes that have only one edge going downward, while the meet-irreducible elements are the nodes that have only one edge going upward from them. For the examples given: (i) has 4 join-irreducibles (a, b, c, d), 3 meet-irreducibles (c, d, e) and all maximal chains have length 3 so it is not distributive. (ii) has 3 join-irreducibles (a, b, d), 3 meet-irreducibles (a, c, d), and all maximal chains have length 3 so it is distributive. $\endgroup$– TriAug 2, 2016 at 6:51
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$\begingroup$ (concluding Professor Markowsky's comment) It is the Cartesian product of the two-element chain and the three-element chain. (iii) is not a lattice since a and b do not have a join. $\endgroup$– TriAug 2, 2016 at 6:51