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Throughout all posets will be finite.

Let $P$ be a poset, and let $\mathcal{A}(P)$ denote the set of antichains of $P$. We give $\mathcal{A}(P)$ a partial order whereby $A \leq A'$ iff for all $x \in A$ there exists $y\in A'$ such that $x\leq y$. It's easy to see that $\mathcal{A}(P)$ is isomorphic to the distributive lattice of order ideals of $P$ via the map that associates an antichain to the order ideal it generates. Hence by Birkhoff's Fundamental Theorem of (Finite) Distributive Lattices, all (finite) distributive lattices arise as $\mathcal{A}(P)$ in this way.

Now for $k\in\mathbb{N}$, let $\mathcal{A}_k(P) := \{A \in\mathcal{A}(P)\colon \#A=k\}\subseteq \mathcal{A}(P)$, with its induced poset structure.

Note that $\mathcal{A}_1(P)\simeq P$; hence $\{\mathcal{A}_1(P)\colon \textrm{$P$ a poset}\}=\{\textrm{all posets $P$}\}$. Even more trivially we have $\{\mathcal{A}_0(P)\colon \textrm{$P$ a poset}\}=\{\textrm{the $1$ element poset}\}$.

On the other hand, let $m_P:=\mathrm{max}\{m\colon m=\#A \textrm{ for some $A\in\mathcal{A}(P)$}\}$. Then $\mathcal{A}_{m_P}(P)$ is always a distributive lattice, and in fact $\{\mathcal{A}_{m_P}(P)\colon \textrm{$P$ a poset}\}=\{\textrm{all (finite) distributive lattices $L$}\}$ (see the references in my previous MO question Map on class of all finite posets coming from maximal sized antichains).

Question: What is known about which classes of posets arise as $\mathcal{A}_k(P)$ for other values of $k$? For instance, $\mathcal{A}_2(P)$ or $\mathcal{A}_{m_P-1}(P)$?

One could consider even a more general setup.

Namely, for $S\subseteq \mathbb{N}$, we could let $\mathcal{A}_S(P) := \{A \in\mathcal{A}(P)\colon \#A\in S\} \subseteq \mathcal{A}(P)$, and then ask about which posets arise as $\mathcal{A}_S(P)$ for various subsets $S$. For example, posets of the form $\mathcal{A}_{\{0,1,\ldots,m_P\}}(P)=\mathcal{A}(P)$ are exactly the (finite) distributive lattices by Birkhoff's theorem.

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  • $\begingroup$ Good question. Is there a command to get $A_{{m_P} -1}(P)$ or more generally $A_k(P)$ for a given poset $P$ using Sage (Im not too experienced yet with Sage)? Looking at some examples might reveal some properties. $\endgroup$
    – Mare
    Commented Sep 15, 2020 at 19:02
  • $\begingroup$ @Mare: it should be completely straightforward to code, but I don't think there's an existing command like that. But also, it's not so obvious to me that looking at small examples would tell you much: e.g., to get a given distributive lattice as the maximum sized antichains of a poset requires a poset of a very particular form. $\endgroup$
    – Sam Hopkins
    Commented Sep 15, 2020 at 19:04
  • $\begingroup$ I find this interesting. We go from all posets to all distributive lattices... I have not thought about it much, but I wonder if the classes of posets you get are weakly increasing (in some sense I'm not pinning down, since $m_P$ of course depends on $P$). I wonder if, when you have joins and meets existing, you have the distributive law. P.S. Trotter's Axiom: "All posets are finite." $\endgroup$
    – Tri
    Commented Sep 17, 2020 at 23:33
  • $\begingroup$ @Tri: Consider the graded poset $P$ with ranks of sizes 2, 3, and 2, and all relations between adjacent ranks. The size 2 antichains of $P$ form a $5$-element non-distributive lattice consisting of a minimum, maximum, and $3$ incomparable elements. $\endgroup$
    – Sam Hopkins
    Commented Sep 18, 2020 at 3:30
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    $\begingroup$ I threw in the vague phrase "in some sense," but maybe one way to make it concrete is this: I was wondering if, as you go from $A_{m_P}$ (for all $P$) to $A_{m_P-1}$ (for all $P$) to $A_{m_P-2}$ (for all $P$)..., you get less and less "distributive lattice-like"---maybe one interpretation of this is that you don't get all distributive lattices, but you don't get the class of all finite posets and the class of posets you get grows larger and its members less and less like distributive lattices; and maybe something similar happens in the opposite direction although it won't be exactly analogous $\endgroup$
    – Tri
    Commented Sep 19, 2020 at 5:40

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