Timeline for How to recognize if a lattice is distributive? [closed]
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| Jul 30, 2016 at 14:29 | comment | added | Tri | Professor Stanley writes above, "I don't know whether the following is true: let $L$ be a finite lattice with maximal chain of length $n$. If $L$ has exactly $n$ join-irreducibles and $n$ meet-irreducibles, then $L$ is distributive." If we read this as "some maximal chain has length $n$", consider the non-modular 5-element lattice $N_5$. We are in the territory of Professor Markowsky's "extremal lattices." If we read this as "every maximal chain has length $n$", the aforementioned Theorem 3.1 of Professor Markowsky says the lattice is distributive. | |
| Jul 30, 2016 at 13:14 | comment | added | Tri | Also, there is the nice result of Professor George Markowsky, that a finite lattice is distributive with n join-irreducibles if and only if if has length n, satisfies the Jordan-Dedekind chain condition, and has n join-irreducibles and n meet-irreducibles (Theorem 3.1 on page 180 of Algebra Universalis, volume 11). I have it on good authority Farley regrets not referring to this paper, since I hear he did initially use the theorem when proving the results in the Farley-Schmidt paper (although Dr.Schmidt---and, in the paper, Farley and Schmidt---used a different approach). | |
| Jul 30, 2016 at 12:50 | comment | added | Tri | I believe the Farley-Schmidt result Professor Stanley (or someone claiming to be Professor Stanley) mentions has as a hypothesis that the poset (it need not be a lattice) be bounded and graded of rank at least 3 (but I could be wrong). | |
| Jan 26, 2015 at 2:02 | comment | added | Richard Stanley | The condition that the posets of join- and meet-irreducibles are isomorphic is certainly not sufficient for distributivity, e.g., a $k$-element antichain, $k>2$, with a top and bottom adjoined. I don't know whether the following is true: let $L$ be a finite lattice with maximal chain of length $n$. If $L$ has exactly $n$ join-irreducibles and $n$ meet-irreducibles, then $L$ is distributive. | |
| Jan 18, 2015 at 5:44 | comment | added | მამუკა ჯიბლაძე | Ooops, not antiisomorphic but isomorphic in fact... | |
| Jan 17, 2015 at 7:34 | comment | added | მამუკა ჯიბლაძე | Reconsidering after these comments - yes I also agree there might be interesting approaches. One thought as an example: it is relatively easy to read out posets of join- and meet-irreducibles from the Hasse diagram. Now for distributivity it is certainly necessary that these posets be antiisomorphic. Presently I cannot find out whether this is also sufficient. What is certainly sufficient (and also necessary) is that this antiisomorphism be established through splitting (mutually complementary principal ideals/filters). This might be more tedious to verify from the picture but still possible. | |
| Jan 16, 2015 at 0:26 | comment | added | Joseph Van Name | I would be glad to vote to reopen this question if it were reworded a bit since there could be some interesting answers. | |
| Jan 15, 2015 at 16:27 | comment | added | Richard Stanley | 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. | |
| Jan 15, 2015 at 3:35 | history | closed |
Emil Jeřábek Benjamin Steinberg Chris Godsil Dima Pasechnik Karl Schwede |
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| Jan 14, 2015 at 16:36 | comment | added | JohnDoe | Okay, everything is clear now. Thanks everyone. | |
| Jan 14, 2015 at 16:32 | comment | added | Pace Nielsen | Isn't (ii) distributive? Note that a sublattice must have the sames meets and joins as the original lattice. | |
| Jan 14, 2015 at 16:17 | vote | accept | JohnDoe | ||
| Jan 14, 2015 at 16:17 | comment | added | JohnDoe | So, none of these three are distributive. Am I right? | |
| Jan 14, 2015 at 16:11 | comment | added | მამუკა ჯიბლაძე | not all of these are lattices in fact | |
| S Jan 14, 2015 at 15:59 | history | suggested | Aaron Maroja | CC BY-SA 3.0 |
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| Jan 14, 2015 at 15:50 | answer | added | arsmath | timeline score: 6 | |
| Jan 14, 2015 at 15:48 | comment | added | Tony Huynh | 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 | |
| Jan 14, 2015 at 15:47 | review | Close votes | |||
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| Jan 14, 2015 at 15:43 | review | Suggested edits | |||
| S Jan 14, 2015 at 15:59 | |||||
| Jan 14, 2015 at 15:23 | history | edited | JohnDoe | CC BY-SA 3.0 |
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| Jan 14, 2015 at 15:20 | review | First posts | |||
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| Jan 14, 2015 at 15:16 | history | asked | JohnDoe | CC BY-SA 3.0 |