The $n$-cells of a modular lattice are antichains

Question

The terminology "$n$-cell" is made up by me, and I'd love to hear if this has an official name.

Given a poset $(X,\le)$, define the set $A_n$ of $n$-cells of $X$ recursively as follows:

• An element is a $0$-cell iff it is the bottom element: $x\in A_0\iff\forall y,x\le y$.
• An element is an $n+1$-cell iff it covers some $n$-cell: $x\in A_{n+1}\iff\exists y\in A_n,y<:x$.

(Recall that $x<:y$, read "$x$ covers $y$", if $x<y$ and $\lnot\exists z:x<z<y$.) Note that a $1$-cell is the same thing as an atom, so this is a generalization to "higher-dimensional atoms".

One curious property of the Hilbert lattice, or any other vector subspace lattice for that matter, is that $A_n$ is an antichain for each $n$ (because $A_n$ is the set of subspaces of dimension $n$). On the other hand, this is not always the case for an arbitrary lattice. For example, in the pentagon lattice $N_5$ with atoms $a,x$ and coatoms $a,b$, we have $A_2=\{b,1\}$ which is not an antichain since $b\le 1$.

Does this property (which I will call "cell-graded" for lack of a better term) follow from the assumption that the lattice is modular?

Edit: I've now beat myself over the head with several failed classification theorems for modular lattices in attempt to prove that a modular lattice is cell-graded, with no luck. However, these attempts and the counterexamples that are generated halfway during the "proof" have only made me more sure that I was correct with my original claim:

Claim: The $n$-cells of a modular lattice are antichains.

Answer 1

(This answer has been cleaned up from the earlier stream-of-consciousness style now that the proof is complete, see the revision log.)

Edit: Amazingly, it turns out that these theorems, up to and including the main theorem in the OP, can be generalized still further, from modular lattices to weakly semimodular lattices.

A lattice is said to be a Birkhoff lattice or weakly semimodular lattice if $a:>a\land b<:b$ implies $a<:a\lor b:>b$. (In other words, both sides of the diamond must be covering before it can be "pushed up".) By contrast, an upper semimodular lattice satisfies $a\land b<:b\to a<:a\lor b$ (only one side is needed to push the relation up), and a modular lattice has isomorphisms between $[a\land b,b]$ and $[a,a\lor b]$, so the arrow goes in both directions.

Let $a<:^nb$ denote that there is a covering sequence of length $n$ from $a$ to $b$. Obvious properties of this relation include:

• $a<:^mb<:^nc\implies a<:^{m+n}c$
• $a<:^0b\iff a=b$
• $a<:^1b\iff a<:b$
• $a\in A_n\leftrightarrow 0<:^na$ (assuming $0$ exists; if not, the main theorem is trivial).

The Birkhoff law also generalizes to weak covering $a\le:b$ (which means $a<:b$ or $a=b$): if $a:\ge c\le:b$, then $a\le:a\lor b:\ge b$. Furthermore, by an induction in first $m$ then $n$, we have that if $a:>^mc<:^nb$, then $a<:^ia\lor b:>^jb$ for some $i\le m$, $j\le n$.

Theorem 1: If $M$ is a weakly semimodular lattice, then any two finite covering sequences from $a$ to $b$ have the same length, or in our notation, if $a<:^mb$ and $a<:^nb$, then $m=n$. (The statement from Roman is much weaker, assuming that $M$ is upper semimodular and has no infinite chains, and replaces "finite covering sequence" with "maximal chain", which is equivalent when there are no infinite chains.)

Proof: By induction. If $n=0$, then $a=b$ is the only possible chain. Suppose it is true for chains of length $<n$, and $a<:^nb$ and $a<:^mb$. We must have $m>0$ because $n>0$ iff $a<b$ iff $m>0$, so separate off the first elements as $a<:x<:^{n-1}b$ and $a<:y<:^{m-1}b$. If $x=y$, then the induction hypothesis implies that $n-1=m-1$, so we are done. Otherwise, $x\land y=a$, so the Birkhoff law implies that $x<:x\lor y:>y$. The Birkhoff law also implies that $x\lor y<:^kb\lor y=b$ for some $k\le n-1$. Then $x<:^{n-1}b$ and $x<:^{k+1}b$, and $y<:^{m-1}b$ and $y<:^{k+1}b$, so $n-1=k+1=m-1$ by the induction hypothesis. $$\tag*{$\blacksquare$}$$

This generalizes a bit:

Corollary: If $a<:^mb\le c$ and $a<:^nc$, then $m\le n$ and $b<:^{n-m}c$.

The Birkhoff law gives $b=a\lor b<:^kc\lor b=c$ for some $k\le n$, so $a<:^{m+k}c$. Then Theorem 1 gives $m+k=n$, so $m\le n$ and $b<:^{n-m}c$.

With this result in hand, the original question is simple:

Theorem 2: If $M$ is a weakly semimodular lattice, then the set $\{x\mid a<:^nx\}$ is an antichain (and in particular, $A_n$ is an antichain).

Suppose $a<:^n b$, $a<:^n c$, and $b\le c$. Then the corollary gives $b<:^{n-n}c$, so $b=c$. $$\tag*{$\blacksquare$}$$

All of the results that hold for $A_n$ also hold for the more general $\{x\mid a<:^nx\}$, which is simply a matter of looking at $A_n$ within the sublattice $[a,\infty)$. In particular, $\bigcup_n A_n$ is downward closed, and there is at least one covering sequence of the appropriate length between comparable elements. It is also closed under joins, because if $a\in A_m$ and $b\in A_n$, then $a<:^ka\lor b$ for some $k\le n$, hence $a\lor b\in A_i$ for some $i=m+k\le m+n$. Thus $\bigcup_n A_n$ is an ideal and a sublattice.

We can also prove that $\bigcup_n A_n$ has no bounded infinite chains, using theorem 2. Suppose that there is an infinite chain $C$ in $\bigcup_n A_n$ with a top element $a\in A_n$. Then $C\cap A_i$ must be infinite for some $i$ (because the union up to $i=n$ gives $C$), so in particular there is an $A_i$ that is not an antichain.

Answer 2

A couple of relevant results can be found in chapter 4 of Steven Roman – Lattices and Ordered Sets. Using this, we can prove your claim under the additional assumption that there are no infinite chains. This also shows how your idea fits in with more conventional lattice theory vocabulary.

Claim. Let $M$ be an upper semimodular lattice with no infinite chains. Then the $n$-cells $A_n \subseteq M$ are antichains.

Without loss of generality, we may assume that $M$ has a least element $0$, for otherwise the sets $A_n$ are empty. For $x\in M$ we define $\text{height}(x) \in \mathbb Z_{\geq 0} \cup \{\infty\}$ as $$ \text{height}(x) = \sup\{n\in\mathbb Z_{\geq 0} \ : \ \text{there exists a chain}\ 0 = a_0 < a_1 < \cdots < a_n = x\ \text{of length $n$}\}. $$ Since we assumed that $M$ has no infinite chains, we have the following result.

Proposition. Every $x\in M$ is an $n$-cell for some $n\in\mathbb Z_{\geq 0}$.

Proof. The statement clearly holds for $0$. Suppose, for the sake of contradiction, that we have $x > 0$ and $y \not<: x$ for all $y \in M$. Then in particular we have $0 \not<: x$, so we may choose $a_1 \in M$ with $0 < a_1 < x$. Now we also have $a_1 \not<: x$, so we may choose $a_2 \in M$ with $a_1 < a_2 < x$. Proceeding this way we find an infinite chain $0 < a_1 < a_2 < \cdots$, contrary to our assumption that $M$ has no infinite chains. Thus, for every $x > 0$ there is some $y \in M$ such that $y <: x$ holds.

Now we can construct a chain $\cdots <: x_2 <: x_1 <: x$ where each step is covering. Since $M$ has no infinite chains, the process must terminate at some point, so we have $$ 0 = x_n <: x_{n-1} <: \cdots <: x_2 <: x_1 <: x. \tag*{$\Box$} $$

Roman furthermore proves (in theorem 4.14) that the following holds for $a < b$:

1. A chain from $a$ to $b$ is maximal if and only if every step is covering. Here we really use that there are no infinite chains. The proof is analogous to proof of the above proposition: if a step $x_i < x_{i+1}$ in the chain is not covering, then either there is an element that can be added to the chain, or there are infinitely many elements between $x_i$ and $x_{i+1}$ already contained in the chain.
2. If there is a maximal chain from $a$ to $b$ of length $n$, then all maximal chains from $a$ to $b$ have length $n$. (Here we really use that $M$ is upper semimodular.) The chain $\{a,b\}$ is contained in a maximal chain, which must be finite since there are no infinite chains, so we see that every maximal chain from $a$ to $b$ has the same finite length. (This is called the Jordan–Dedekind chain condition.)

In particular all maximal chains from $0$ to $x$ have the same finite lenght, so it follows that $\text{height}(x)$ is finite for all $x \in M$. Furthermore, we have $x \in A_n$ if and only if $\text{height}(x) = n$, hence $$ A_n = \{x \in M \ : \ \text{height}(x) = n\}. $$ Now the result follows easily. Suppose we have $a < b$, then we may extend the chain $\{a,b\}$ to a maximal chain $a = c_0 <: c_1 <: \cdots <: c_{k-1} <: c_k = b$, so we have $\text{height}(b) = \text{height}(a) + k$ with $k \geq 1$. Thus, in general, if $x$ and $y$ are comparable and $x \neq y$, then we have $\text{height}(x) \neq \text{height}(y)$. Since all elements of $A_n$ have height $n$, they cannot be comparable.

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For the general case, where $M$ is allowed to have infinite chains, I still have no idea what the answer might be. The assumption is used at various points throughout the proof, so I wouldn't be surprised if the result fails to hold in general modular lattices. On the other hand, we may also be able to restrict to a sublattice that has no infinite chains. The union of all $A_n$ may already have an infinite chain, but maybe we can look at the sublattice generated by $\bigcup_{n=0}^k A_n$? Not sure if this is going to work. Anyway, just wanted to show my progress. :-)