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For definitions, see Section 1 of Chapter 3 of Richard Stanley, Enumerative Combinatorics, Volume I (second edition). Also see Section 8 of Chapter II of Garrett Birkhoff, Lattice Theory (third edition).

A poset is semimodular if whenever an element x is covered by distinct elements y and z, there exists an element w covering y and z. It is known that finite semimodular posets with a least element are ranked.

Can you find a finite (ranked) semimodular poset with ranks 0 through 2 looking like the figure below, with the additional property that every rank 2 interval has either 3 or 4 elements (i.e., is a 3-element chain or a diamond) except for the interval [0,d]?

enter image description here

Robert A. Proctor and Lindsey M. Scoppetta, "d-Complete Posets: Local Structural Axioms, Properties, and Equivalent Definitions," Order 36 (2019), 399-422.

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  • $\begingroup$ Not a critique of your question or anything, but I've never heard of "semimodular" used outside the context of lattices. $\endgroup$ Commented Apr 2, 2024 at 13:22
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    $\begingroup$ @SamHopkins: see for instance ams.org/journals/proc/1971-028-02/S0002-9939-1971-0276144-6/… $\endgroup$ Commented Apr 2, 2024 at 14:30

1 Answer 1

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In GraphViz notation,

digraph {
    rankdir="BT";
    edge [arrowhead=none];
    0 -> a; 0 -> b; 0 -> c;
    a -> ab; a -> ac; a -> d;
    b -> ab; b -> bc; b -> d;
    c -> ac; c -> bc; c -> d;
    ab -> D; ab -> AB;
    ac -> D; ac -> AC;
    d -> AB; d -> AC; d -> BC;
    bc -> D; bc -> BC;
    D -> A; D -> B;
    AB -> A; AB -> B;
    AC -> A; AC -> C;
    BC -> B; BC -> C;
    A -> 1;
    B -> 1;
    C -> 1;
}

Rendered by dot

It has a pleasing symmetry (with one tiny exception, where $C$ doesn't cover $D$) which I've tried to bring out with the labels.

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    $\begingroup$ This is a beautiful poset, but the interval $[D, 1]$ has five elements. I believe it can be fixed by adding one more level. $\endgroup$ Commented Apr 3, 2024 at 12:32
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    $\begingroup$ @JukkaKohonen, I missed that when trying to increase the symmetry. The fix is simply to make $C$ not cover $D$. $\endgroup$ Commented Apr 3, 2024 at 13:33
  • $\begingroup$ Thank you. This appears to resolve Conjecture 4.3 (negatively) in the above paper of Proctor and Scoppetta. $\endgroup$ Commented Apr 4, 2024 at 4:03

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