An open problem on congruences of finite lattices

TL;DR

This paper surveys the ongoing research on the properties and characterization of congruences in SPS lattices, a special class of slim, planar, semimodular lattices, highlighting significant contributions and open problems in the field.

Contribution

It provides a comprehensive overview of the development and current state of the problem of characterizing congruences in SPS lattices, including major contributions by Gábor Czédli.

Findings

The Two-cover Property for SPS lattices was established.

Over 50 papers have advanced understanding of congruences in SPS lattices.

Open problems remain in fully characterizing congruences of SPS lattices.

Abstract

Let $L$ be a planar semimodular lattice. We call $L$ \emph{slim}, if it has no $\mthree$ sublattice. Let us define an \emph{SPS lattice} as a slim, planar, semimodular lattice $L$. In 2016, I proved a property of congruences of SPS lattices (Two-cover Property) and raised the problem of characterizing them. Since then, more than 50 papers have been published contributing to this problem. In this survey, I provide an overview of this field with major contributions by G\'abor Cz\'edli.

Equations

$$\mathfrak{r}={\mathfrak{r}}_0,{\mathfrak{r}}_1,\dots ,{\mathfrak{r}}_n=\mathfrak{q}$$

$$1_{{\mathfrak{r}}_0}\ge 1_{{\mathfrak{r}}_1}\ge \cdots \ge 1_{{\mathfrak{r}}_n}.$$

$$\text{lc(D)}\vee \text{rc(D)}$$

$$\text{lc(D)}\wedge \text{rc(D)}$$

$$\psi :x\mapsto (x\wedge \text{lc(L)},x\wedge \text{rc(L)})$$

$$\mathrm{col}Z\prec \mathrm{col}X,\mathrm{col}Y.$$

$$X\stackrel{\text{dn}}{\sim }{S}_X\stackrel{\text{ ex}}{\text{rotatebox}160.0\curvearrowleft }{T}_X,Y\stackrel{\text{dn}}{\sim }{S}_Y\stackrel{\text{ ex}}{\text{rotatebox}160.0\curvearrowleft }{T}_Y,Z\in {\mathcal{T}}_X\cap {\mathcal{T}}_Y.$$

Full text

In 2016, I proved a property of congruences of SPS lattices (Two-cover Property) and raised the problem of characterizing them.

Since then, more than 50 papers have been published contributing to this problem. In this survey, I provide an overview of this field with major contributions by Gábor Czédli.

1. Introduction

Paul Erdős defined a mathematician as a machine, where the input is (espresso) coffee and the output is theorems and conjectures. He thought that raising problems was as important as proving theorems.

Here is an illustration. My book [35], General Lattice Theory, was published in 1978. A year after its publication, I wrote an article (see [36]), 1979 problem update. The article has 21 references affecting 40 problems proposed in the book. This book and its second and third editions (see [37] and [38]) have 850 references in the Mathematical Reviews, and many of them are connected to open problems these books propose.

This survey focuses on a single open problem on congruences of finite lattices; I proposed it in a 2016 paper (see [43]), as related in my book [51].

1.1. Notation

We use the notation as in my book [51]. Part I of this book can be freely downloaded: arXiv:2104.06539

1.2. Outline

Birkhoff duality helps in computing in large distributive lattices; this topic is introduced in Section 2. Section 3 recalls how this field started. A crucial result of G. Grätzer, H. Lakser, and E. T. Schmidt [58] is stated in Section 4. Section 5 formulates the Problem. The first known property of congruences of finite lattices, the Two-cover Property, is given in Section 6.

There are three major tools required to tackle the Problem, they are described in Section 7.1 Sections 7.2, and 7.3. Section 8 has the four major properties of G. Czédli. The 3P3C property is discussed in Section 9.

To illustrate how the tools are used, Section 10 sketches some proofs. Finally, Section 11 formulates a related problem.

2. Finite distributive lattices

Small distributive lattices are easy to compute in. Larger ones, not so much. Birkhoff duality (see G. Birkhoff [2]) provides a solution.

For a finite ordered set $P$, call $A\subseteq P$ a down set iff $x\in A$ and $y\leq x$ in $P$, imply that $y\in A$. For $H\subseteq P$, there is a smallest down set containing $H$, namely, \{\,x\mid x\leq h,\ \text{for some h\in H}\,\}; we use the notation $\downarrow\!H$ for this set. If $H=\{a\}$, we write $\downarrow\!a$ for $\downarrow\!\{a\}$. Let $\operatorname{Dn}P$ denote the set of all down sets ordered by set inclusion. Then $\operatorname{Dn}P$ is a finite distributive lattice.

Conversely, for a finite distributive lattice $D$, let $P$ denote the ordered set of join-irreducible elements of $D$.

Birkhoff duality**.**

1. (i)

The assignment $P\to\operatorname{Dn}P$ is a bijection between finite ordered sets and finite distributive lattices. It assigns $a\in P$ to $\downarrow\!a\in\operatorname{Dn}P$. 2. (ii)

The inverse map is $D\to\textup{J}(D)$. It assigns $a\in D$ to $\textup{id}(a)\cap P$.

Figure 1 provides an example. The finite distributive lattice $D$ on the left has 20 elements and a fairly complicated structure. The corresponding finite ordered set $P$ on the right has only 6 elements.

For a finite lattice $L$, the congruences form a finite distributive lattice, $\operatorname{Con}L$. The corresponding finite ordered set is the ordered set of join-irreducible congruences, $\textup{J}(\operatorname{Con}L)$. For $a\prec b$ in $L$, we can form $\textup{con}(a,b)$, the smallest congruence of $L$ collapsing $a$ and $b$. Then $\textup{con}(a,b)\in\textup{J}(\operatorname{Con}L)$ and conversely.

3. How it started

The first result in this field is in a 1942 paper, N. Funayama and T. Nakayama [34]: the congruence lattice of a lattice is distributive. So the congruence lattice of a finite lattice is a finite distributive lattice.

The converse was discovered by R. P. Dilworth, unpublished, circa 1944 (see the book, K. P. Bogart, R. Freese, and J. P. S. Kung, eds, [3] for a discussion).

Theorem 1**.**

Every finite distributive lattice $D$ can be represented as the congruence lattice of a finite lattice $L$.

The first proof of the this result was published in 1962 (see G. Grätzer and E. T. Schmidt [59]); it represented $D$ with a sectionally complemented lattice $L$.

4. Representing with semimodular latices

Can we represent every finite distributive lattice $D$ as $\operatorname{Con}L$, the congruence lattice of a finite semimodular lattice $L$?

Or, equivalently, can we represent every finite ordered set $P$ as $\textup{J}(\operatorname{Con})L$, the ordered set of join-irreducible congruences of a finite semimodular lattice $L$?

Indeed, we can, as verified in G. Grätzer, H. Lakser, and E. T. Schmidt [58].

Theorem 2**.**

Every finite distributive lattice $D$ can be represented as the congruence lattice of a finite planar semimodular lattice $L$.

It was a great surprise to us that the proof yielded a planar lattice.

We now provide a Proof-by-Picture, utilizing the semimodular lattice $\mathsf{S}_{8}$, see Figure 2, to represent the ordered set $P$ of Figure 3 as $\textup{J}(\operatorname{Con})L$ of a finite semimodular lattice $L$

A colored lattice is a finite lattice (some) of whose edges (prime intervals) are labeled so that if the edges $\mathfrak{p}$ and $\mathfrak{q}$ are of the same color, then $\textup{con}(\mathfrak{p})=\textup{con}(\mathfrak{q})$. These labels represent equivalence classes of edges defining the same join-irreducible congruence.

To represent $P$, we start out by constructing the colored lattices $A_{a}$ and $A_{b}$ (see Figure 4). We obtain $A_{a}$ by gluing two copies of $\mathsf{S}_{8}$ together, coloring it with $\{a,b,d\}$ so that $\textup{con}(a)>\textup{con}(b)$ is accomplished in the top $\mathsf{S}_{8}$ of $A_{a}$ and $\textup{con}(a)>\textup{con}(d)$ is accomplished in the bottom $\mathsf{S}_{8}$ of $A_{a}$. The lattice $A_{b}$ is $\mathsf{S}_{8}$ colored by $\{b,c\}$, so that $\textup{con}(b)>\textup{con}(c)$ in $A_{b}$.

Observe that the lattice $A_{a}$ takes care of all $a\succ x$ orderings; in the example, there are only two. We could do three coverings by gluing three copies of $\mathsf{S}_{8}$ together, and so on.

Form the glued sum $S$ of $A_{a}$ and $A_{b}$; all the covers of $P$ are taken care of in $S$. There is only one problem: $S$ is not a colored lattice; in this example, if $\mathfrak{p}$ is a prime interval of color $b$ in $S_{b}$ (as in Figure 4) and $\mathfrak{q}$ is a prime interval of color $b$ in $S_{b}$ (as in Figure 4), then in $S$ we have $\textup{con}(\mathfrak{p})\wedge\textup{con}(\mathfrak{q})=\mathbf{0}$. Of course, we should have $\textup{con}(\mathfrak{p})=\textup{con}(\mathfrak{q})=\textup{con}(b)$. We accomplish this by extending $S$ to the lattice $L$ of Figure 5. In $L$, the black-filled elements form the sublattice $S$.

As you see, we extend $S$ by adding to it a distributive “grid.” The right corner is $\mathsf{C}_{5}^{2}$ colored by $\{a,b,c,d\}$; each of the four covering squares colored by the same color twice are made into a cover-preserving $\mathsf{M}_{3}$. This makes the coloring behave properly in the right corner. In the rest of the lattice we do the same: we look for a covering square colored by the same color twice, and make it into a cover-preserving $\mathsf{M}_{3}$. This makes $L$ into a colored lattice: any two prime intervals of the same color generate the same congruence.

Finally, we remember the element $e\in P$. We add a “tail” to the lattice and color it $e$.

The resulting lattice $L$ is planar and semimodular and $\textup{J}(\operatorname{Con}L)$ is isomorphic to $P$ with the isomorphism $x\mapsto\textup{con}(x)$ for $x\in\{a,b,c,d,e\}$.

5. The Problem

According to Theorem 2, every finite distributive lattice $D$ can be represented as the congruence lattice of a finite planar semimodular lattice $L$. In the construction, we strongly utilize $\mathsf{M}_{3}$ sublattices. Could we do the construction without them?

Let us call a finite planar semimodular lattice $L$ slim, if it has no $\mathsf{M}_{3}$ sublattice. Let us define an SPS lattice as a slim, finite, planar semimodular lattice $L$ without an $\mathsf{M}_{3}$ sublattice. (SPS stands for Slim, Planar, Semimodular.)

In the paper [43], I proposed the following.

Problem**.**

Characterize the congruence lattices of SPS lattices.

6. The Two-cover Property

A finite ordered set $P$ satisfies the Two-Cover Condition (see my paper [43]), if any element of $P$ has at most two covers. Figure 7 shows the diagram of an ordered set $F_{3}$ failing the Two-Cover Condition. A finite ordered set $P$ satisfies the Two-Cover Condition iff $F_{3}$ has no cover-preserving embedding into $P$.

The following is the main result of my paper [43].

Theorem 3** (Two-Cover Theorem).**

The ordered set of join-irreducible congruences of an SPS lattice $L$ has the Two-Cover Condition.

The Two-Cover Condition is a necessary condition. G. Czédli [7] proves that it is not sufficient. My paper [41] provides a short proof.

Let us say that a finite ordered set $P$ satisfies the Two-max Condition, if $P$ has at least two maximal elements.

In my note [45], I prove the following very easy result.

Theorem 4** (Two-max Theorem).**

The ordered set of join-irreducible congruences of a nontrivial SPS lattice $L$ has the Two-max Condition.

It follows that the three-element chain $\mathsf{C}_{3}$, which satisfies the Two-Cover Condition, cannot be represented as the congruence lattice of an SPS lattice; of course, an example of minimal size.

G. Czédli [13] discovered four major properties of the congruence lattices of SPS lattices. We’ll discuss them in Section 8. We present the sixth major property in Section 9, see G. Czédli and G. Grätzer [23].

G. Czédli maintains a list of papers making contributions to this problem, 52 as of this writing, see

http://www.math.u-szeged.hu/~czedli/m/listak/publ-psml.pdf

They are all listed in the References.

7. The Tools

7.1. The Swing lemma

For the edges $\mathfrak{p},\mathfrak{q}$ of an SPS lattice $L$, we define a binary relation: $\mathfrak{p}$ swings to $\mathfrak{q}$, written as \mathfrak{p}\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}\mathfrak{q}, if $1_{\mathfrak{p}}=1_{\mathfrak{q}}$, this element covers at least three elements, and $0_{\mathfrak{q}}$ is neither the left-most nor the right-most element covered by $1_{\mathfrak{p}}=1_{\mathfrak{q}}$. We call the element $1_{\mathfrak{p}}=1_{\mathfrak{q}}$ the hinge of the swing. If $0_{\mathfrak{p}}$ is either the left-most or the right-most element covered by the hinge, then we call the swing external, in formula, \mathfrak{p}\stackrel{{\scriptstyle\textup{ ex}}}{{\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}}}\mathfrak{q}. Otherwise, the swing is internal, in formula, \mathfrak{p}\stackrel{{\scriptstyle\textup{\,in}}}{{\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}}}\mathfrak{q}. See Figure 8 for two examples; in the first, the hinge covers three elements and \mathfrak{p}\stackrel{{\scriptstyle\textup{ ex}}}{{\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}}}\mathfrak{q}; in the second, the hinge covers five elements and \mathfrak{p}\stackrel{{\scriptstyle\textup{\,in}}}{{\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}}}\mathfrak{q}.

Lemma 5** (Swing Lemma).**

Let $L$ be an SPS lattice and let $\mathfrak{p}$ and $\mathfrak{q}$ be distinct edges in $L$. Then $\mathfrak{q}$ is collapsed by $\textup{con}(\mathfrak{p})$ iff there exists a edge $\mathfrak{r}$ and sequence of pairwise distinct edges

[TABLE]

such that $\mathfrak{p}$ is up perspective to $\mathfrak{r}$, and $\mathfrak{r}_{i}$ is down perspective to or swings to $\mathfrak{r}_{i+1}$ for $i=0,\dots,n-1$. In addition, the sequence (1) also satisfies

[TABLE]

The Swing Lemma is easy to visualize using skiing analogies.

1. (i)

Perspectivity up is “climbing” (see Figure 12). 2. (ii)

Perspectivity down is “sliding” (see Figure 12). 3. (iii)

Swinging is “turning” (see Figure 12).

So we get from $\mathfrak{p}$ to $\mathfrak{q}$ by climbing once and then alternating sliding and swinging. In the example of Figure 12, we climb up from $\mathfrak{p}$ to $\mathfrak{r}=\mathfrak{r}_{0}$, swing from $\mathfrak{r}_{0}$ to $\mathfrak{r}_{1}$, slide down from $\mathfrak{r}_{1}$ to $\mathfrak{r}_{2}$, swing from $\mathfrak{r}_{2}$ to $\mathfrak{r}_{3}$, and finally slide down from $\mathfrak{r}_{3}$ to $\mathfrak{r}_{4}$.

7.2. Natural diagrams and $\mathcal{C}_{1}$-diagrams

As in my joint paper with E. Knapp [52], we call the planar semimodular lattice $L$ rectangular, if its left boundary chain has exactly one doubly-irreducible element, lc(L), and its right boundary chain has exactly one doubly-irreducible element, rc(D), and these elements are complementary, that is,

[TABLE]

We call a slim rectangular lattice an SR lattice.

In this section, we consider nice diagrams for SR lattices.

Natural diagrams for SR lattice

We follow my joint paper with E. Knapp [56] published in 2010. For an SR lattice $L$, let $\textup{C}_{\textup{l}}(L)$ be the lower left and $\textup{C}_{\textup{r}}(L)$ the lower right boundary chain of $L$, respectively.

We regard the grid $G=\textup{C}_{\textup{l}}(L)\times\textup{C}_{\textup{r}}(L)$ as a planar lattice, with $\textup{C}_{\textup{l}}(L)=\textup{C}_{\textup{l}}(G)$ and $\textup{C}_{\textup{r}}(L)=\textup{C}_{\textup{r}}(G)$. Then the map

[TABLE]

is a meet-embedding of $L$ into $G$; the map $\psi$ also preserves the bounds. Therefore, the image of $L$ under $\psi$ in $G$ is a diagram of $L$; we call it the natural diagram representing $L$.

$\mathcal{C}_{1}$-diagrams

This research tool, introduced by G. Czédli in 2017, has been playing an important role in recent papers (see G. Czédli [12]–[17], my joint paper with G. Czédli [23], and also G. Grätzer [47]; for the definition, see G. Czédli [12] and G. Grätzer [47]).

In the diagram of an SR lattice $K$, a normal edge (line) has a slope of $45\degree$ or $135\degree$. Any edge (line) of slope strictly between $45\degree$ and $135\degree$ is steep. A normal edge (line) has a slope of $45\degree$ or $135\degree$. A peek sublattice in $K$ is an $\mathsf{S}_{7}$ sublattice whose top three does are covering.

Definition 6**.**

A diagram of an SR lattice $L$ is a ${\mathcal{C}}_{1}$-diagram, if the middle edge of a peek sublattice is steep and all other edges are normal.

Theorem 7**.**

Every SR lattice $L$ has a ${\mathcal{C}}_{1}$-diagram.

This was proved in G. Czédli [12]. My note [47] presents a short and direct proof.

It turns out that the two approaches to nice diagrams of SR latices are the same (see my paper [50]).

Theorem 8**.**

Let $L$ be a SR lattice. Then a natural diagram of $L$ is a ${\mathcal{C}}_{1}$-diagram. Conversely, every ${\mathcal{C}}_{1}$-diagram is natural.

7.3. Lamps

Let $L$ be a SR lattice. On the edges of $L$ with meet-irreducible bottoms, the internal swing, \mathfrak{p}\stackrel{{\scriptstyle\textup{\,in}}}{{\mathbin{\raisebox{2.0pt}{\rotatebox{160.0}{\curvearrowleft}}}}}\mathfrak{q}, is a reflexive, symmetric, and transitive binary relation. So these edges are partitioned into equivalence classes; G. Czédli [13] calls them lamps.

The concept of a lamp is a crucial tool for the discovery of most deep properties of congruence lattices of SPS lattices. It is a very deep and technical concept, and it will await Czédli to write an introduction to the field in the style of this paper.

G. Czédli [18] also utilizes lamps to find an infinite family of independent properties of congruence lattices of SPS lattices.

There are a number of ways to view how congruences spread in SP lattices. The first approach uses trajectories, see G. Czédli [6] and my paper [44]. The second is the Swing Lemma. The third is lamps. They are very similar in some ways, but quite different technically.

8. Four major properties

This section introduces four major properties of congruences of SR lattices, as introduced in G. Czédli [13].

Let $K$ be a slim, planar, semimodular lattice with at least three elements and let $\mathcal{P}$ be the ordered set of join-irreducible congruences of $K$.

8.1. The No Child Property

See G. Czédli [13], illustrated in Figure 15.

Theorem 9** (No Child Property).**

Let $x\neq y\in\mathcal{P}$ and let $z$ be a maximal element of $\mathcal{P}$. Let us assume that both $x$ and $y$ are covered by $z$ in $\mathcal{P}$. Then there is no element $u\in\mathcal{P}$ such that $u$ is covered by $x$ and $y$.

8.2. The Four-crown Two-pendant Property

See G. Czédli [13] and illustrated in Figure 16.

Theorem 10** (Four-crown Two-pendant Property).**

There is no cover-preserving embedding of the ordered set $\mathcal{R}$ of Figure 16 into $\mathcal{P}$ satisfying the property that any maximal element of $\mathcal{R}$ maps into a maximal element of $\mathcal{P}$.

8.3. The Partition Property

See G. Czédli [13].

Theorem 11** (Partition Property).**

The set of maximal elements of $\mathcal{P}$ can be partitioned into two nonempty subsets such that no two distinct elements in the same subset have a common lower cover.

8.4. The Maximal Cover Property

See G. Czédli [13].

Theorem 12** (Maximal Cover Propery).**

If $x\in\mathcal{P}$ is covered by a maximal element $y\in\mathcal{P}$, then $y$ is not the only cover of $x$.

8.5. Translation

“You pays your money and you takes your choice.” Aldous Huxley, Brave New World

Czédli and I use different names for some of these properties. Translation:

 

 

Czédli                  **Grätzer

** 

Forbidden Marriage Property        No Child Property

Dioecious Maximal Elements Property Maximal Cover Property

Bipartite Maximal Elements Property   Partition Property

9. The 3P3C Property

The last major property we discuss is in G. Czédli and G. Grätzer [49].

Theorem 13** (Three-pendant Three-crown Property).**

The ordered set of Figure 17 has no cover presering embedding into $\mathcal{P}$.

10. Some proofs

In this section, I sketch the proofs of the Partition and the Two-pendant Four-crown Properties of congruences of SR lattices, based on my paper [48].

The Partition Property

We start with a lemma.

Lemma 14**.**

Let $X$ and $Y$ be distinct edges on the upper-left boundary of $K$. Then there is no edge $Z$ of $K$ such that

[TABLE]

Proof.

By way of contradiction, let $Z$ be an edge such that (3) holds. Since $X$ and $Y$ are on the upper-left boundary, there exist normal-up edges $S_{X},S_{Y}$ and steep edges $T_{X},T_{Y}$ such that

[TABLE]

The third formula implies that $T_{X}=T_{Y}$. Since $S_{X}$ and $S_{Y}$ are normal-up, it follows that $X=Y$, contradicting the assumption that $X$ and $Y$ are distinct. ∎

The set of maximal elements of $\mathcal{P}$ is the same as the set of colors of edges in the upper boundaries. The latter we can partition into the set of edges $\mathcal{L}$ in the upper-left boundary and the set of edges $\mathcal{R}$ in the upper-right boundary. Let $X$ and $Y$ be distinct edges in $\mathcal{L}$. By Lemma 14, there is no edge $Z$ of $K$ such that $\operatorname{col}Z\prec\operatorname{col}X,\operatorname{col}Y$. By symmetry, this verifies the Partition Property.

The Two-pendant Four-crown Property

By way of contradiction, assume that the ordered set $\mathcal{R}$ of Figure 16 is a cover-preserving ordered subset of $\mathcal{P}$, where $a,b,c,d$ are maximal elements of $\mathcal{P}$. So there are edges $A,B,C,D$ on the upper boundary of $K$, so that $\operatorname{col}A=a$, $\operatorname{col}B=b$, $\operatorname{col}C=c$, $\operatorname{col}D=d$. By left-right symmetry, we can assume that the edge $A$ is on the upper-left boundary of $K$. Since $p\prec a,b$ in $\mathcal{P}$, it follows from Lemma 14 that the edge $B$ is on the upper-right boundary of $K$, and so is $D$. Similarly, $C$ is on the upper-left boundary of $K$.

Because of the automorphisms of the ordered set $\mathcal{R}$, it is sufficient to deal with one case only: $C$ is below $A$ and $B$ is below $D$.

So there is a peak sublattice $\mathsf{S}_{7}$ with middle edge $P$ (as in Figure 18) so that $A$ and $B$ are down-perspective to the upper-left edge and the upper-right edge of this peak sublattice, respectively. We define, similarly, the edge $Q$ for $C$ and $B$, the edge $S$ for $A$ and $D$, the edge $R$ for $C$ and $D$, and the edge $U$ for $R$ and $P$. Finally, $v\prec q,s$ in $\mathcal{R}$, therefore, there is a peak sublattice $\mathsf{S}_{7}$ with middle edge $V$ with upper-left edge $V_{l}$ and the upper-right edge $V_{r}$ so that $S\stackrel{{\scriptstyle\textup{dn}}}{{\thicksim}}V_{l}$ and $Q\stackrel{{\scriptstyle\textup{dn}}}{{\thicksim}}V_{r}$, or symmetrically. Then $V$ is in both the peak sublattices containing $\mathsf{S}_{7}$, $S$, and $Q$, respectively, a contradiction. This concludes the proof of the Two-pendant Four-crown Property.

Of course, the diagram in Figure 18 is only an illustration. The grid could be much larger, the edges $A,C$ and $B,D$ may not be adjacent, and there maybe lots of other elements in $K$.

11. One more problem

Let $D$ be a finite distributive lattice of the form $D^{h}\dotplus\mathsf{B}_{2}$, the glued sum of a finite distributive lattice $D^{h}$ with $\mathsf{B}_{2}$. One can think of $D^{h}$ as $D$ with the hat removed.

Recall that a rectangular lattice $L$ is a patch lattice, if the corners are dual atoms. It is easy to verify that a rectangular lattice $L$ is a patch lattice iff $\operatorname{Con}L$ is a glued sum of a finite distributive lattice $\operatorname{Con}^{h}L$ with $\mathsf{B}_{2}$.

Patch Problem**.**

Characterize the congruence lattice of a patch lattice $L$ in terms of the properties of the finite distributive lattice $\operatorname{Con}^{h}L$.