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This is a follow-up to a recent question of mine:

For $n\in\mathbb{N}$ let $C(n)$ be the smallest $k$ such that every bounded lattice with cardinality $\le n$ which is isomorphic to the congruence lattice of some finite algebra, is isomorphic to the congruence lattice of some algebra with cardinality $\le k$.

What do we know about the growth rate of $C(n)$?

I'm especially interested in lower bounds, since right now my main interest is pedagogical (explaining why the finite congruence representation problem is difficult). Keith Kearnes' answer to my above-linked question uses a theorem of Bunn/Grow/Insall/Thiem which implies that for infinitely many $n$ we have $C(n)\ge 2(n-3)$. However, I don't know whether $C$ is known to grow faster than linear.

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What do we know about the growth rate of $C(n)$?

We know the exact value of $C(n)$ if, in its definition, we restrict to the class of distributive lattices. Otherwise we only have partial results. Let me say a few words.

Claim.
If $L$ is a finite distributive that has a maximal chain $\ell_1<\ell_2<\cdots < \ell_n$, then

  1. $L$ is representable as the congruence lattice of an algebra of size $n$, and
  2. $L$ is not representable as the congruence lattice of any algebra of size $< n$.

Reasons:
For for the first item, observe that if $L$ is a finite distributive lattice with a specified maximal chain $\gamma: \ell_1<\ell_2<\cdots < \ell_n$, then it has a $0,1$-embedding into a Boolean lattice $B$ in such a way that the chain $\gamma$ remains a maximal chain. Next, any Boolean lattice $B$ with an $n$-element maximal chain is isomorphic to a power-set lattice $\mathscr{P}(S)$ where $S=\{s_1,\ldots,s_{n-1}\}$ is a set of size $n-1$. Assume that $0\notin S$. Next, $\mathscr{P}(S)$ can be embedded into the lattice $\Pi(\{0\}\cup S)$ of partitions of $\{0\}\cup S$ by mapping $U\subseteq S$ to the partition where $\{0\}\cup U$ is one cell and the other cells are singletons. This is a $0,1$-embedding. Altogether we have $0,1$-embeddings
$L\leq B\cong \mathscr{P}(S)\leq \Pi(\{0\}\cup S)$,
which compose to a $0,1$-embedding of $L$ into a partition lattice on a set of size $|\{0\}\cup S|=n$.

From here we cite

Quackenbush, R.; Wolk, B.
Strong representation of congruence lattices.
Algebra Universalis 1 (1971/72), 165-166.

which shows that any $0,1$-distributive sublattice $L$ of the lattice of the lattice of equivalence relations on an $n$-element set is a representation of $L$ as a congruence lattice.

For the second item of the claim, it is impossible to represent any lattice $L$ with a maximal chain of $n$-elements as a congruence lattice on a set of size $k<n$, since the full lattice of partitions on a $k$-element set does not have a chain of $n$ elements.

In particular, if we relativize the definition of $C(n)$ to the class of distributive lattices, we get the following. Let $C_{\textrm{Dist}}(n)$ be the least $k$ such that every bounded distributive lattice with cardinality $\leq n$ which is isomorphic to the congruence lattice of some finite algebra, is isomorphic to the congruence lattice of some algebra with cardinality $k$. With this definition and the preceding remarks, we get $C_{\textrm{Dist}}(n)=n$. (If you look at all bounded distributive lattices with cardinality $\leq n$, they are all representable, and the hardest to represent is the $n$-element chain. This can be represented as the congruence lattice of an $n$-element algebra, but of no smaller algebra.)

What I said above about distributive lattices can be extended a little bit. A finite distributive lattice is a sublattice of a power of the $2$-element lattice, $\mathbf{2}$. The same kinds of results mentioned above can be proved for the smallest nondistributive lattices $M_3$ and $N_5$. Namely, $M_3$ can be represented as a congruence lattice on a $3$-element set, $N_5$ can be represented as a congruence lattice on a $4$-element set, and any sublattice of a power of either representation is a congruence representation. For this see

Snow, John W.
Every finite lattice in $\mathscr{V}(M_3)$ is representable.
Algebra Universalis 50 (2003), no. 1, 75-81.

and

Snow, John W.
Subdirect products of hereditary congruence lattices.
Algebra Universalis 54 (2005), no. 1, 65-71.

When you move to $M_4$, it starts to get complicated, as is explained in an answer to an earlier question. In fact, it is not hard to produce a $0,1$-embedding of $M_q$ into the lattice of partitions of a $q$-element set whenever $q$ is an odd prime, but the least size congruence representation of $M_{p+1}$ has size $2p$ when $p$ is an odd prime. Thus, if $p$ and $q$ are odd primes satisfying $p<q<2p$, then there is a $0,1$-embedding of $M_{p+1}$ into $M_q$ and then into the lattice of partitions of a $q$-element set, but there is no congruence representation of $M_{p+1}$ on a $q$-element set.

Let me jump ahead to identify what seems to be the hardest part of this problem. In

Pálfy, Péter Pál; Pudlák, Pavel
Congruence lattices of finite algebras and intervals in subgroup lattices of finite groups.
Algebra Universalis 11 (1980), no. 1, 22-27.

Palfy and Pudlak isolate three properties of a finite bounded lattice $L$ of size strictly greater than $2$:

(A) $L$ is simple.
(B) For any nonzero $x\in L$ there exist $y_1, y_2\in L$ such that $x\vee y_1=x\vee y_2=1$ and $y_1\wedge y_2=0$.
(C) Any $x\in L$ that is not zero and not an atom dominates at least $4$ atoms.

Palfy and Pudlak show two things about these properties. They show that if $L$ is any finite lattice, then it can embedded as an upper interval $[u,1]$ in a lattice $L'$ which satisfies (A), (B), and (C) and has size $|L'|=5|L|+1$. If one can represent $L'$ as a congruence lattice of a finite algebra $\mathbf{A}$, then one can represent $L$ as the congruence lattice of $\mathbf{A}/u$. Thus, $L'$ (satisfying (A), (B), (C)) is only slightly larger than $L$ (not necessarily satisfying (A), (B), (C)), yet any upper bound on the size of a minimal congruence lattice representation for $L'$ is also an upper bound on the size of a minimal congruence lattice representation for $L$.

Second, they show that if $L$ satisfies (A), (B), and (C) and $L$ is representable as the congruence lattice of a finite algebra $\mathbf{A}$, then there is representation of $L$ as the congruence lattice of a transitive $G$-set $\mathbf{B}$ satisfying $|B|\leq |A|$. This says that, if $L$ is sufficiently complicated and representable as a congruence lattice, then it is representable as the congruence lattice of a transitive $G$-set for some finite group $G$. Such $G$-sets are isomorphic to $G/H$ under the action of left multiplication by elements of $G$ for some subgroup $H\leq G$. Hence, if $L$ satisfies the conditions (A), (B), and (C), and $L$ is representable as the congruence lattice of a finite algebra, then a minimal congruence representation may be assumed to be of the form of a transitive $G$-set $G/H$ of size $|B|=[G:H]$, and the lattice $L$ will be isomorphic to interval in $\textrm{Sub}(G)$ of subgroups containing $H$. If we try to estimate $C(n)$ for some $n$ equal to the size of some Palfy-Pudlack lattice (defined as satisfying their conditions (A), (B), (C)), we find that we are asking for the least index $[G:H]$ in a finite group if the interval in $\textrm{Sub}(G)$ of subgroups containing $H$ has size $n$.

The work of Palfy and Pudlak suggests that the hard lattices to represent are those satisfying (A), (B), and (C). If $L$ is such a lattice, which is representable as a congruence lattice and it has size $n$, then to compute $C(n)$ we must be able to determine $[G:H]$ where $G/H$ is the smallest algebra affording a congruence representation of $L$. This has been attempted for the sequence of lattices $L=M_n$, for which see

Baddeley, Robert; Lucchini, Andrea
On representing finite lattices as intervals in subgroup lattices of finite groups.
J. Algebra 196 (1997), no. 1, 1-100.

They give $G$-set congruence lattice representations for $M_n$ for $n$ of the form $1, 2, q+1, q+2, ((q^t+1)/(q+1))+1$, $q$ a prime power. I believe that it is still unknown whether $M_n$ is representable for other values of $n$.

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  • $\begingroup$ This is amazing, thank you! I will point my students to this (and your earlier answer as well). $\endgroup$ Commented Mar 14, 2023 at 16:38

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