Timeline for Sperner property of a distributive lattice associated to a divisor poset and the free distributive lattice
Current License: CC BY-SA 4.0
14 events
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Oct 9, 2023 at 22:52 | comment | added | Tri | Dr. Larry Harper's student worked on this. Is the Free Distributive Lattice (FD(n)) Sperner? (talk by L. Harper) Tuesday, March 23, 2021 Abstract: For many years one of the most challenging problems of order theory has been to prove that FD(n) is Sperner – that the largest (middle) rank is the largest antichain. That conjecture has resisted so many good ideas that we had to consider the possibility that it is false for some large n. In this lecture we present a heuristic argument showing that it is at least asymptotically true. escholarship.org/uc/item/8wh6f7rc | |
Sep 30, 2023 at 6:32 | comment | added | Fabius Wiesner | I know nothing about the subject, but just in case you overlooked it, I have found the sequence OEIS A341633. If it is not related to the question, I will cancel the comment. | |
Sep 29, 2023 at 15:31 | comment | added | Richard Stanley | I can't think of any poset which is known to be Sperner but for which we don't know a "good" description of its elements. To my knowledge, no good description is known for the order ideals of $P_n$ or for a product of more than two chains (of length at least one). | |
Sep 29, 2023 at 14:20 | comment | added | Sam Hopkins | FWIW, I just tested in Sage and $J(3 \times 3 \times 3)$ is also Sperner, so maybe my guesses are all wrong. | |
Sep 29, 2023 at 12:37 | comment | added | Mare | @SamHopkins I added the question about the free distributive lattice as question 2. | |
Sep 29, 2023 at 12:37 | history | edited | Mare | CC BY-SA 4.0 |
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Sep 29, 2023 at 12:36 | comment | added | Sam Hopkins | Interesting. Okay, so maybe there is hope then, I really don't know. | |
Sep 29, 2023 at 12:36 | comment | added | Mare | @SamHopkins J(2x2x2x2) is Sperner but J(2x2x2x2x2) is too big for the computer... | |
Sep 29, 2023 at 12:35 | comment | added | Sam Hopkins | My guess would be that already $J(2\times 2 \times 2 \times 2)$ is not Sperner, though I have never thought deeply about this. | |
Sep 29, 2023 at 12:34 | comment | added | Sam Hopkins | I don't know the exact classification. But even in the case when $P=m\times n$ is the product of two chains, that the distributive lattice $J(P)$ (in this case often denoted $L(m,n)$) is Sperner is a famous result of Stanley from his paper "Weyl Groups, the Hard Lefschetz Theorem, and the Sperner Property" (math.mit.edu/~rstan/pubs/pubfiles/42.pdf), though I think nowadays there are also proofs of this that avoid the Hard Lefschetz theorem. | |
Sep 29, 2023 at 12:33 | comment | added | Mare | @SamHopkins For example is it known when the free distributive lattice (J(P) for P Boolean) is Sperner? | |
Sep 29, 2023 at 12:29 | comment | added | Mare | @SamHopkins Is it known precisely when your examples are Sperner? Sadly the Sperner property seems hard to check with a computer when the poset has a high number of elements. | |
Sep 29, 2023 at 12:25 | comment | added | Sam Hopkins | If we instead took $P$ to be the divisors of $n$ ordered by divisibility, this would give us a product of chains corresponding to the prime factorization of $n$. Now, when dealing with $J(P)$ (the distributive lattice of order ideals), usually it behaves nicely for $P$ a product of two chains, and maybe some other small cases like $P = 2 \times a \times b$. But by the time we get to $P = 3 \times 3 \times 3$ the good behavior usually breaks. So you may want to check an example that contains $3 \times 3 \times 3$. I guess $n=4^3=64$ would be the smallest example like that. | |
Sep 29, 2023 at 12:19 | history | asked | Mare | CC BY-SA 4.0 |