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Arrow Relations in Lattices of Integer Partitions

Asma’a Almazaydeh aalmazaydeh@ttu.edu.jo Mike Behrisch behrisch@logic.at Edith Vargas-García111This author gratefully acknowledges support by the Asociación Mexicana de Cultura A.C. edith.vargas@itam.mx Andreas Wachtel222This author gratefully acknowledges support by the Asociación Mexicana de Cultura A.C. andreas.wachtel@itam.mx
Abstract

We give a complete characterisation of the single and double arrow relations of the standard context 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) of the lattice n\mathcal{L}_{n} of partitions of any positive integer nn under the dominance order, thereby addressing an open question of Ganter, 2022.

keywords:
Integer partition, Lattice, Dominance order, Arrow relation , Subdirect representation
MSC:
[2020] 05A17, 11P81, 06B23, 06A07, 06B05, 06B15
\affiliation

[TTU]organization=Department of Mathematics, Tafila Technical University, addressline=PO Box 179, postcode=66110, city=Tafila, country=Jordan \affiliation[TUWien]organization=Institute of Discrete Mathematics and Geometry, TU Wien, addressline=Wiedner Hauptstr. 8–10, postcode=1040, city=Vienna, country=Austria \affiliation[ITAM]organization=Department of Mathematics, ITAM, addressline=Río Hondo 1, city=Ciudad de México, postcode=CP 01080, country=Mexico

{graphicalabstract}
[Uncaptioned image]
{highlights}

Type 4 partitions allow up-arrows, which fail to be down-arrows, to all types 14.

Type 4 partitions allow down-arrows, which fail to be up-arrows, to all types 14.

Type 4 partitions have double arrows only to those of type 4 and vice versa.

Partitions of types 13 are double arrow related only to types 13 and vice versa.

For n3n\geq 3 there are exactly 2n42n-4 one-generated arrow-closed (1×1)(1\times 1)-subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

1 Introduction

Integer partitions have captivated mathematicians for centuries, starting as early as 1674 with Leibniz investigating the number p(n)p(n) of ways in which a natural number nn can be partitioned, that is, expressed as a sum of a non-increasing sequence of positive integer summands, see [14, p. 37]. Recursive presentations of p(n)p(n), for example, following from Euler’s pentagonal number theorem, are well known, and the search for more explicit formulæ or approximations for p(n)p(n) culminated with the celebrated asymptotic expressions given by Hardy and Ramanujan [11] and with Rademacher’s representations by convergent series [16, 17].

The partitions of a given integer nn\in\mathbb{N} can be ordered by dominance, that is, by pointwise comparing their sequences of partial sums; the resulting ordered set carries the structure of a finite (hence complete) lattice n\mathcal{L}_{n}, see [5]. The cardinalities of these lattices n\mathcal{L}_{n}, that is, the numbers p(n)=|n|p(n)=\left\lvert\mathcal{L}_{n}\right\rvert of (unrestricted) partitions of nn, grow fast as nn is increasing, the asymptotics shown in [11] to be p(n)exp(π2n/3)4n3p(n)\sim\frac{\exp\bigl{(}\pi\sqrt{2n/3}\bigr{)}}{4n\sqrt{3}} for nn\to\infty. Their size alone suggests an increasing complexity of the lattices n\mathcal{L}_{n} for larger values of nn, wherefore splitting them up into smaller parts would be an important step towards enhancing our understanding of their structure. Fortunately, formal concept analysis and lattice theory offer techniques for this task in the form of subdirect representations of complete lattices, see, e.g., [10, Chapter 4]. These are embeddings of the given lattice n\mathcal{L}_{n} into a direct product of smaller lattices, the subdirect factors, such that for each coordinate the corresponding projection is surjective. Of course, the subdirect factors may themselves be again subdirectly representable by even smaller factors, leading eventually to the concept of subdirect irreducibility: a lattice is subdirectly irreducible if in each subdirect representation at least one of the coordinate projections is not only surjective, but bijective, that is, an isomorphism onto the corresponding subdirect factor. The most efficient decompositions are hence given by representations using subdirectly irreducible factors. Such decompositions exist for any doubly founded (in particular any finite) complete lattice, such as n\mathcal{L}_{n}, see [10, Theorem 18].

Formal concept analysis [10] offers a powerful framework to study complete lattices 𝕃\mathbb{L} (up to isomorphism) as lattices of Galois closed sets of a suitable Galois connection between completely join-dense and completely meet-dense subsets of the lattice. This is part of the basic (or fundamental) theorem of formal concept analysis, see [10, Theorem 3]. The Galois connection is, up to isomorphism, induced by the order relation of the lattice between the elements of the two dense subsets, and this inducing binary relation is usually represented in a tabular form, called formal context. A canonically derived complete lattice, the concept lattice, is isomorphic (anti-isomorphic) to the lattice of Galois closed sets and serves to represent the given complete lattice 𝕃\mathbb{L}. For finite lattices 𝕃\mathbb{L} there is, up to isomorphism, a unique way of representation, namely through the so-called standard context, which is given by the order relation between all completely join-irreducible and complete meet-irreducible elements of 𝕃\mathbb{L}, see [10, Proposition 12]. Formal concept analysis further defines binary relations ,\mathrel{\swarrow},\mathrel{\nearrow} and ={\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}}={\mathrel{\swarrow}}\cap{\mathrel{\nearrow}} as certain subsets of the complement of the relation represented by a formal context, cf. [10, Definition 25]. These arrow relations appear in the ‘empty cells’ of the context table, and from them one may determine so-called one-generated arrow-closed subcontexts. This is done by adding attributes (resp. objects) pointed to by arrows ensuing from objects (resp. attributes) already appearing in the subcontext until the configuration stabilises. According to [10, Proposition 62] the one-generated arrow-closed subcontexts of the standard context of a finite lattice 𝕃\mathbb{L} give subdirectly irreducible concept lattices, and taking sufficiently many of them, one may construct a subdirect decomposition of 𝕃\mathbb{L}, see [10, Proposition 61]. A thorough understanding of the arrow relations in the standard context is hence a crucial step towards systematically obtaining subdirect decompositions of n\mathcal{L}_{n}.

For the lattice n\mathcal{L}_{n} of partitions of an integer nn\in\mathbb{N}, the sets of completely join-irreducible and completely meet-irreducible elements were described by Brylawski [5]; moreover, a very intuitive understanding of the covering relation in n\mathcal{L}_{n} (and thus of the irreducibles) was later given in [13]. Recursive and non-recursive constructions of the standard contexts 𝕂(n)\mathbb{K}\left(\mathcal{L}_{n}\right) for increasing values of nn were studied in [3] and [9]. In [3], supported by [4], also the non-embeddability of 𝕂(9)\mathbb{K}\left(\mathcal{L}_{9}\right) into 𝕂(10)\mathbb{K}\left(\mathcal{L}_{10}\right) was argued, the proof of which was later refined in [8] for the symmetric case, extending previous work in [9, 2].

We have computationally determined the standard contexts of n\mathcal{L}_{n} for parameters n60n\leq 60 and our results show some curious patterns regarding the appearing one-generated arrow-closed subcontexts and the corresponding subdirectly irreducible factors, cf. [15] for a limited prospect. In order to be able, at a later stage, to substantiate these experimental results with rigid proofs, we aim in this paper at a complete characterisation of the arrow relations in 𝕂(n)\mathbb{K}\left(\mathcal{L}_{n}\right) for every n+n\in\mathbb{N}_{+}, a question that was raised in [9, p. 40]. Our work has partially evolved in parallel with [7], which also mentions some of the more basic results of this article and has, for example, inspired the graphical presentation of 11\mathcal{L}_{11} including all arrows in Fig. 6. We first describe all double arrows of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) in Theorems 4.10, 4.17 and 4.23. After that we provide characterisations of all down-arrows that fail to be up-arrows in Theorems 5.2, 5.4 and 5.6, with a summary in Corollary 5.7; then we use partition conjugation to obtain the dual results, i.e., up-arrows without down-arrows, in Corollaries 5.5 and 5.8. Our knowledge regarding arrows is schematically summarised in Tables 1 and 2, and illustrated within the lattice 11\mathcal{L}_{11} in Fig. 6. As a proof of concept we finally show in Section 6 how our characterisations can be used to determine and count, for each n+n\in\mathbb{N}_{+}, all one-generated arrow-closed one-by-one subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}), which correspond to two-element subdirectly irreducible lattice factors of n\mathcal{L}_{n}.

2 Preliminaries

2.1 Lattices and ordered sets

Throughout the text we write :={0,1,2,}\mathbb{N}\mathrel{\mathop{:}}=\left\{0,1,2,\ldots\right\} for the set of natural numbers, and we set +:={0}\mathbb{N}_{+}\mathrel{\mathop{:}}=\mathbb{N}\setminus\left\{0\right\}. An ordered set is a pair =(P,)\mathbb{P}=\left(P,\leq\right) where PP is a set and P×P{\leq}\subseteq P\times P is a reflexive, antisymmetric and transitive binary relation on it. For a,bPa,b\in P we have aba\geq b if and only if bab\leq a, and we write a<ba<b for aba\leq b and aba\neq b as usual. Moreover, we say that aa is covered by bb in \mathbb{P} (or bb covers aa in \mathbb{P}), symbolically aba\prec b, if a<ba<b and ax<ba\leq x<b implies x=ax=a for each xPx\in P. The dual of \mathbb{P} is the ordered set (P,)\left(P,\geq\right). An order-isomorphism φ:\varphi\colon\mathbb{P}\to\mathbb{Q} is given by a map φ:PQ\varphi\colon P\to Q such that for all a,bPa,b\in P the inequality aba\leq b in \mathbb{P} is equivalent to φ(a)φ(b)\varphi(a)\leq\varphi(b) in \mathbb{Q}. The ordered sets \mathbb{P} and \mathbb{Q} are then said to be (order-)isomorphic. An order-antiisomorphism or dual isomorphism φ:\varphi\colon\mathbb{P}\to\mathbb{Q} is an order isomorphism between \mathbb{P} and the dual of \mathbb{Q}, and \mathbb{P} and \mathbb{Q} are then called antiisomorphic or dually isomorphic; \mathbb{P} is self-dual (or autodual) if it is dually isomorphic to itself. Furthermore, we define x:={yyx}\mathop{\downarrow}x\mathrel{\mathop{:}}=\left\{y\in\mathbb{P}\mid y\leq x\right\} as the principal down-set of xPx\in P, and x:={yxy}\mathop{\uparrow}x\mathrel{\mathop{:}}=\left\{y\in\mathbb{P}\mid x\leq y\right\} as the principal up-set of xx.

A complete lattice is an ordered set 𝕃=(L,)\mathbb{L}=(L,\leq) where any subset SLS\subseteq L has a greatest common lower bound SL\bigwedge S\in L (called infimum of SS) and a least common upper bound SL\bigvee S\in L (called supremum of SS). It is customary to write a1aka_{1}\wedge\dotsm\wedge a_{k} for {a1,,ak}\bigwedge\left\{a_{1},\dotsc,a_{k}\right\} and a1aka_{1}\vee\dotsm\vee a_{k} for {a1,,ak}\bigvee\left\{a_{1},\dotsc,a_{k}\right\} for finitely many elements a1,,akLa_{1},\dotsc,a_{k}\in L. A subset DLD\subseteq L is called (completely) join-dense in 𝕃\mathbb{L} if for every xLx\in L there is a subset SDS\subseteq D with x=Sx=\bigvee S; DD is (completely) meet-dense in 𝕃\mathbb{L} if for each xLx\in L there is some SDS\subseteq D with x=Sx=\bigwedge S. We say that aLa\in L is completely join-irreducible, denoted by \bigvee-irreducible, if for all SLS\subseteq L such that a=Sa=\bigvee S we necessarily have that aSa\in S. Dually, aLa\in L is completely meet-irreducible, or \bigwedge-irreducible, if for all SLS\subseteq L such that a=Sa=\bigwedge S we must have aSa\in S. For a finite non-empty lattice, it is sufficient to check this condition for the empty and all two-element subsets SS of LL. That is, for finite LL\neq\emptyset, an element aLa\in L is completely join-irreducible, if aa is not the minimum element of LL and for all b,cLb,c\in L the condition a=bca=b\vee c implies that a=ba=b or a=ca=c. This happens exactly if aa covers exactly one element below it, cf. [10, Proposition 2]. Dually, for a finite lattice, aLa\in L is completely meet-irreducible if it is not the top element of LL and for all b,cLb,c\in L with a=bca=b\wedge c it follows that b=ab=a or c=ac=a. This condition is met precisely if aa has exactly one upper cover. For a complete lattice 𝕃\mathbb{L}, we denote by 𝒥(𝕃)\mathcal{J}\left(\mathbb{L}\right) and by (𝕃)\mathcal{M}\left(\mathbb{L}\right) the sets of all completely join-irreducible elements, and of all completely meet-irreducible elements of 𝕃\mathbb{L}, respectively. Elements of 𝒥(𝕃)(𝕃)\mathcal{J}\left(\mathbb{L}\right)\cap\mathcal{M}\left(\mathbb{L}\right) are called doubly completely irreducible. In the lattice diagrams shown in Fig. 3 and 6, the completely irreducible elements have been highlighted using half or completely filled nodes.

A (complete) lattice 𝕃=(L,)\mathbb{L}=(L,{\leq}) is supremum-founded if, for any two x<yx<y from LL, the set {pL|py&px}\left\{\left.p\in L\ \vphantom{p\leq y\mathbin{\&}p\nleq x}\right|\ p\leq y\mathbin{\&}p\nleq x\right\} contains a \leq-minimal element; the dual property that for any x<yx<y in LL the set {pL|xp&yp}\left\{\left.p\in L\ \vphantom{x\leq p\mathbin{\&}y\nleq p}\right|\ x\leq p\mathbin{\&}y\nleq p\right\} includes a \leq-maximal element is called infimum-founded. The lattice 𝕃\mathbb{L} is doubly founded if it is both supremum-founded and infimum-founded, see [10, p. 33]. Every chain-finite and hence every finite lattice is doubly founded, cf. [10, p. 33 et seqq., Fig. 1.11, p. 35].

2.2 Notions of formal concept analysis

Formal concept analysis (FCA) is a theoretical framework that harnesses the powers of general abstract Galois theory and the structure theory of complete lattices for data analysis and many other applications. At its core lies the notion of a Galois connection between (the power sets of) sets GG and MM, induced by a binary relation IG×MI\subseteq G\times M. This data is collected in a formal context 𝕂=(G,M,I)\mathbb{K}=(G,M,I), and the elements of GG and MM are given the interpretative names objects and attributes, respectively. The set II is called the incidence relation, and (g,m)I(g,m)\in I is usually written as g𝐼mg\mathrel{I}m and read as ‘object gg has attribute mm’. When GG and MM are finite, the context 𝕂\mathbb{K} is often given as a cross table, where the objects form the rows, the attributes label the columns, and the crosses represent the characteristic function of II on G×MG\times M. Formal concept analysis extends the ‘prime notation’ for the Galois derivatives, which is common in classical Galois theory [12, Chapter V, Theorem 2.3 et seqq.], to general formal contexts K=(G,M,I)K=(G,M,I). For every set AGA\subseteq G of objects, A:={mMgA:g𝐼m}A^{\prime}\mathrel{\mathop{:}}=\left\{m\in M\mid\forall g\in A\colon g\mathrel{I}m\right\} assigns to it the set of attributes commonly shared by all objects of AA. Dually, for BMB\subseteq M, the set B:={gGmB:g𝐼m}B^{\prime}\mathrel{\mathop{:}}=\left\{g\in G\mid\forall m\in B\colon g\mathrel{I}m\right\} contains exactly those objects possessing all the attributes in BB. A formal concept is a pair (A,B)(A,B) where the extent AGA\subseteq G and the intent BMB\subseteq M are sets that are mutually Galois closed: A=BA=B^{\prime} and B=AB=A^{\prime}. Intents of the form B={g}B=\left\{g\right\}^{\prime} with gGg\in G are called object intents and are written as gg^{\prime}, for short; dually extents A={m}=:mA=\left\{m\right\}^{\prime}\mathrel{\mathopen{=}{\mathclose{:}}}m^{\prime} with mMm\in M are referred to as attribute extents. The equivalent conditions A1A2A_{1}\subseteq A_{2} and B1B2B_{1}\supseteq B_{2} define an order (A1,B1)(A2,B2)(A_{1},B_{1})\leq(A_{2},B_{2}) on the set 𝔅(𝕂)\mathfrak{B}(\mathbb{K}) of all formal concepts. 𝔅¯(𝕂)=(𝔅(𝕂),)\underline{\mathfrak{B}}(\mathbb{K})=(\mathfrak{B}(\mathbb{K}),\leq) becomes a complete lattice under this order, the concept lattice of 𝕂\mathbb{K}. The fundamental theorem of formal concept analysis [10, Theorem 3] states that, in fact, every complete lattice is a concept lattice, up to isomorphism. Namely, if 𝕃\mathbb{L} is a complete lattice, then 𝕃𝔅¯(L,L,)\mathbb{L}\cong\underline{\mathfrak{B}}(L,L,{\leq}). For finite lattices, which are always complete, this construction can be improved: 𝕃𝔅¯(𝕂)\mathbb{L}\cong\underline{\mathfrak{B}}(\mathbb{K}), where 𝕂=(𝒥(𝕃),(𝕃),)\mathbb{K}=(\mathcal{J}\left(\mathbb{L}\right),\mathcal{M}\left(\mathbb{L}\right),{\leq}) is the standard context of 𝕃\mathbb{L} [10, Proposition 12]. This applies to partition lattices n\mathcal{L}_{n} in particular.

A central notion for this paper will be the arrow relations of a formal context, which fill up some of the empty cells in the cross table.

Definition 2.1 ([10, Definition 25]).

If (G,M,I)(G,M,I) is a context, gGg\in G an object, and mMm\in M an attribute, we write

gm\displaystyle g\mathrel{\swarrow}m :g ̵I ̵m, and for all hG with gh we have h𝐼m;\displaystyle\,\mathrel{\mathop{:}}\Longleftrightarrow\,g\mathrel{\mbox{\hbox to0.0pt{\hbox{{\char 32\relax}}\hss}$I$\raisebox{1.1625pt}{\hbox{{\char 32\relax}}}}}m,\text{ and for all }h\in G\text{ with }g^{\prime}\subsetneqq h^{\prime}\text{ we have }h\mathrel{I}m;
gm\displaystyle g\mathrel{\nearrow}m :g ̵I ̵m, and for all nH with mn we have g𝐼n;\displaystyle\,\mathrel{\mathop{:}}\Longleftrightarrow\,g\mathrel{\mbox{\hbox to0.0pt{\hbox{{\char 32\relax}}\hss}$I$\raisebox{1.1625pt}{\hbox{{\char 32\relax}}}}}m,\text{ and for all }n\in H\text{ with }m^{\prime}\subsetneqq n^{\prime}\text{ we have }g\mathrel{I}n;
gm\displaystyle g\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m :gm and gm.\displaystyle\,\mathrel{\mathop{:}}\Longleftrightarrow\,g\mathrel{\nearrow}m\text{ and }g\mathrel{\swarrow}m.

Thus, gmg\mathrel{\swarrow}m if and only if gg^{\prime} is maximal among all object intents which do not contain mm; dually we have gmg\mathrel{\nearrow}m if and only if mm^{\prime} is maximal among all attribute extents which do not contain gg.

We will now derive a useful characterisation of the arrow relations in standard contexts of doubly founded lattices.

Remark 2.2.

Consider the (standard) context 𝕂(𝕃)=(𝒥(𝕃),(𝕃),)\mathbb{K}(\mathbb{L})=\left(\mathcal{J}\left(\mathbb{L}\right),\mathcal{M}\left(\mathbb{L}\right),\mathord{\leq}\right) of any complete lattice 𝕃=(L,)\mathbb{L}=(L,\mathord{\leq}). Note that for g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right) and m(𝕃)m\in\mathcal{M}\left(\mathbb{L}\right) we have

g\displaystyle g^{\prime} ={m(𝕃)|gm}\displaystyle=\left\{\left.m\in\mathcal{M}\left(\mathbb{L}\right)\ \vphantom{g\leq m}\right|\ g\leq m\right\} =g(𝕃)\displaystyle=\mathop{\uparrow}g\cap\mathcal{M}\left(\mathbb{L}\right) =:g(𝕃),\displaystyle\mathrel{\mathopen{=}{\mathclose{:}}}g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}},
m\displaystyle m^{\prime} ={g𝒥(𝕃)|gm}\displaystyle=\left\{\left.g\in\mathcal{J}\left(\mathbb{L}\right)\ \vphantom{g\leq m}\right|\ g\leq m\right\} =m𝒥(𝕃)\displaystyle=\mathop{\downarrow}m\cap\mathcal{J}\left(\mathbb{L}\right) =:m𝒥(𝕃).\displaystyle\mathrel{\mathopen{=}{\mathclose{:}}}m{\downarrow_{\mathcal{J}\left(\mathbb{L}\right)}}.

This allows us to reformulate the definition of the arrow relations of 𝕂(𝕃)\mathbb{K}(\mathbb{L}) in terms of the up-sets and down-sets of 𝕃\mathbb{L}. Consider again g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right) and m(𝕃)m\in\mathcal{M}\left(\mathbb{L}\right). Then we have

gm\displaystyle g\mathrel{\swarrow}m I:=gm&h𝒥(𝕃){g}:\displaystyle\stackrel{{\scriptstyle I\mathrel{\mathop{:}}={\leq}}}{{\iff}}g\nleq m\mathbin{\&}\forall h\in\mathcal{J}\left(\mathbb{L}\right)\setminus\left\{g\right\}\colon g\displaystyle g^{\prime} h\displaystyle\subsetneqq h^{\prime} \displaystyle\implies h\displaystyle h m,\displaystyle\leq m,
gm&h𝒥(𝕃){g}:\displaystyle\iff g\nleq m\mathbin{\&}\forall h\in\mathcal{J}\left(\mathbb{L}\right)\setminus\left\{g\right\}\colon g(𝕃)\displaystyle g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}} h(𝕃)\displaystyle\subsetneqq h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}} \displaystyle\implies m\displaystyle m h,\displaystyle\in\mathop{\uparrow}h,
gm\displaystyle g\mathrel{\nearrow}m I:=gm&a(𝕃){m}:\displaystyle\stackrel{{\scriptstyle I\mathrel{\mathop{:}}={\leq}}}{{\iff}}g\nleq m\mathbin{\&}\forall a\in\mathcal{M}\left(\mathbb{L}\right)\setminus\left\{m\right\}\colon m\displaystyle m^{\prime} a\displaystyle\subsetneqq a^{\prime} \displaystyle\implies g\displaystyle g a,\displaystyle\leq a,
gm&a(𝕃){m}:\displaystyle\iff g\nleq m\mathbin{\&}\forall a\in\mathcal{M}\left(\mathbb{L}\right)\setminus\left\{m\right\}\colon m𝒥(𝕃)\displaystyle m{\downarrow_{\mathcal{J}\left(\mathbb{L}\right)}} a𝒥(𝕃)\displaystyle\subsetneqq a{\downarrow_{\mathcal{J}\left(\mathbb{L}\right)}} \displaystyle\implies g\displaystyle g a.\displaystyle\in\mathop{\downarrow}a.

The following sufficient condition will lead to our main tool to establish arrow relations in standard contexts, in particular in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

Lemma 2.3.

Let 𝕃=(L,)\mathbb{L}=(L,{\leq}) be any complete lattice. Consider any g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right) with unique lower cover g~L\tilde{g}\in L and m(𝕃)m\in\mathcal{M}\left(\mathbb{L}\right) with unique upper cover m~L\tilde{m}\in L such that gmg\nleq m. Then in the context 𝕂(𝕃)=(𝒥(𝕃),(𝕃),)\mathbb{K}(\mathbb{L})=(\mathcal{J}\left(\mathbb{L}\right),\mathcal{M}\left(\mathbb{L}\right),{\leq}) the following implications regarding arrow relations hold.

  1. 1.

    If g~m\tilde{g}\leq m and there is a set S(𝕃)S\subseteq\mathcal{M}\left(\mathbb{L}\right) such that g=Sg=\bigwedge S, then gmg\mathrel{\swarrow}m.

  2. 2.

    If m~g\tilde{m}\geq g and there is a set S𝒥(𝕃)S\subseteq\mathcal{J}\left(\mathbb{L}\right) such that m=Sm=\bigvee S, then gmg\mathrel{\nearrow}m.

  3. 3.

    If g~m\tilde{g}\leq m and g(𝕃)g\in\mathcal{M}\left(\mathbb{L}\right) (thus doubly completely irreducible), then gmg\mathrel{\swarrow}m.

  4. 4.

    If m~g\tilde{m}\geq g and m𝒥(𝕃)m\in\mathcal{J}\left(\mathbb{L}\right) (thus doubly completely irreducible), then gmg\mathrel{\nearrow}m.

Proof.

We only prove statement (1), for (2) is completely dual; (3) follows by setting S:={g}S\mathrel{\mathop{:}}=\left\{g\right\} in (1), and (4) by setting S:={m}S\mathrel{\mathop{:}}=\left\{m\right\} in (2). By assumption we have gmg\nleq m. To show gmg\mathrel{\swarrow}m, according to Remark 2.2, we take any h𝒥(𝕃){g}h\in\mathcal{J}\left(\mathbb{L}\right)\setminus\left\{g\right\} and assume g(𝕃)h(𝕃)g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}\subsetneqq h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}. From the hypothesis of (1) we have g=Sg=\bigwedge S with S(𝕃)S\subseteq\mathcal{M}\left(\mathbb{L}\right), thus we infer Sg(𝕃)h(𝕃)S\subseteq g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}\subseteq h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}. Therefore, hh is a common lower bound of the elements of SS, hence hS=gh\leq\bigwedge S=g. Since hgh\neq g, we have h<gh<g, and, as g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right), we infer hg~mh\leq\tilde{g}\leq m, as it is required for gmg\mathrel{\swarrow}m. ∎

We shall need the following statement for infimum/supremum founded complete lattices, which can be read from [10, Fig. 1.11, p. 35]. For completeness, we provide a proof of this fundamental fact.

Lemma 2.4.

Let 𝕃=(L,)\mathbb{L}=(L,{\leq}) be a complete lattice with completely join-irreducibles 𝒥(𝕃)\mathcal{J}\left(\mathbb{L}\right) and meet-irreducibles (𝕃)\mathcal{M}\left(\mathbb{L}\right).

  1. 1.

    If 𝕃\mathbb{L} is infimum-founded, then (𝕃)\mathcal{M}\left(\mathbb{L}\right) is completely meet-dense.

  2. 2.

    If 𝕃\mathbb{L} is supremum-founded, then 𝒥(𝕃)\mathcal{J}\left(\mathbb{L}\right) is completely join-dense.

Proof.

Part (2) follows from (1) by duality, thus we only show the latter. Consider any g𝕃g\in\mathbb{L} and define the subset S:={y(𝕃)|gy}(𝕃)S\mathrel{\mathop{:}}=\left\{\left.y\in\mathcal{M}\left(\mathbb{L}\right)\ \vphantom{g\leq y}\right|\ g\leq y\right\}\subseteq\mathcal{M}\left(\mathbb{L}\right). By its construction, SS satisfies gSg\leq\bigwedge S. Let us assume for a contradiction that g<S=:hg<\bigwedge S\mathrel{\mathopen{=}{\mathclose{:}}}h. By infimum-foundedness, there is hence a maximal element xLx\in L with the properties gxg\leq x but hxh\nleq x. If x(𝕃)x\in\mathcal{M}\left(\mathbb{L}\right), then xSx\in S and thus h=Sxh=\bigwedge S\leq x, being a contradiction. Therefore, we consider now x(𝕃)x\notin\mathcal{M}\left(\mathbb{L}\right). This means there must exist a subset WL{x}W\subseteq L\setminus\left\{x\right\} with x=Wx=\bigwedge W. It follows for each wWw\in W that xwx\leq w, and, in fact, gx<wg\leq x<w since wxw\neq x. From the maximality of xx we infer now that hwh\nleq w fails, i.e., hwh\leq w. Since this holds for all wWw\in W, we conclude hW=xh\leq\bigwedge W=x, which is again a contradiction. Both contradictions show g=h=Sg=h=\bigwedge S. ∎

For our purposes the following characterisation of the arrow relations is the most appropriate one since the lattice n\mathcal{L}_{n} is finite, thus all its chains have only a finite number of elements, and it hence is doubly founded.

Proposition 2.5.

Let 𝕃=(L,)\mathbb{L}=(L,{\leq}) be a doubly founded complete lattice, e.g., a finite lattice. In the formal context 𝕂(𝕃)=(𝒥(𝕃),(𝕃),)\mathbb{K}(\mathbb{L})=(\mathcal{J}\left(\mathbb{L}\right),\mathcal{M}\left(\mathbb{L}\right),{\leq}) the arrow relations can be characterised as follows (cf. Fig. 1). Consider any g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right) with unique lower cover g~L\tilde{g}\in L and m(𝕃)m\in\mathcal{M}\left(\mathbb{L}\right) with unique upper cover m~L{\tilde{m}\in L}. Then we have

  1. 1.

    gmgm&g~mg\mathrel{\swarrow}m\iff g\nleq m\mathbin{\&}\tilde{g}\leq m;

  2. 2.

    gmgm&m~gg\mathrel{\nearrow}m\iff g\nleq m\mathbin{\&}\tilde{m}\geq g.

Proof.

We shall only prove (1) since (2) is completely dual. To show ‘\Longleftarrow’ we use infimum-foundedness, which implies that the set (𝕃)\mathcal{M}\left(\mathbb{L}\right) is infimum-dense, see Lemma 2.4(1). This means that every gLg\in L can be written as S\bigwedge S for some set S(𝕃)S\subseteq\mathcal{M}\left(\mathbb{L}\right), namely, we may take S={y(𝕃)|gy}S=\left\{\left.y\in\mathcal{M}\left(\mathbb{L}\right)\ \vphantom{g\leq y}\right|\ g\leq y\right\}. This is true, in particular, for all g𝒥(𝕃)g\in\mathcal{J}\left(\mathbb{L}\right), hence gmg\nleq m, g~m\tilde{g}\leq m and Lemma 2.3(1) imply gmg\mathrel{\swarrow}m.

For the converse implication, let us assume that gmg\mathrel{\swarrow}m holds. This implies gmg\nleq m by Remark 2.2. By infimum-foundedness, there is pLp\in L that is maximal with respect to the property g~p\tilde{g}\leq p but gpg\nleq p. Let U:={zL|z>p}U\mathrel{\mathop{:}}=\left\{\left.z\in L\ \vphantom{z>p}\right|\ z>p\right\}. For every zUz\in U we have z>pg~z>p\geq\tilde{g}, thus, in order to not violate the maximality of pp, the element zz must fail the property gzg\nleq z, that is, gzg\leq z must hold. Therefore, gg is a common lower bound for the elements of UU, and hence we have gUg\leq\bigwedge U. As gpg\nleq p, we know that Up\bigwedge U\neq p, thus in fact, U>p\bigwedge U>p since all zUz\in U are above pp. Consequently, if p=Tp=\bigwedge T for some subset TLT\subseteq L, then pTp\in T, for otherwise TUT\subseteq U and thus p<UTp<\bigwedge U\leq\bigwedge T. This shows that p(𝕃)p\in\mathcal{M}\left(\mathbb{L}\right). Since 𝕃\mathbb{L} is supremum-founded, the set 𝒥(𝕃)\mathcal{J}\left(\mathbb{L}\right) is supremum-dense, see Lemma 2.4(2). Thus we can write g~=S\tilde{g}=\bigvee S for some set S𝒥(𝕃)S\subseteq\mathcal{J}\left(\mathbb{L}\right). For every hSh\in S we have hS=g~<gh\leq\bigvee S=\tilde{g}<g, wherefore g(𝕃)h(𝕃)g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}\subseteq h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}. As p(𝕃)p\in\mathcal{M}\left(\mathbb{L}\right) and pg~p\geq\tilde{g}, we have ph(𝕃)p\in h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}, but certainly pg(𝕃)p\notin g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}} since pgp\ngeq g. Thus, ph(𝕃)g(𝕃)p\in h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}\setminus g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}, i.e., g(𝕃)h(𝕃)g{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}\subsetneqq h{\uparrow_{\mathcal{M}\left(\mathbb{L}\right)}}. Now gmg\mathrel{\swarrow}m and Remark 2.2 imply hmh\leq m. As hSh\in S was arbitrary, we conclude that g~=Sm\tilde{g}=\bigvee S\leq m. ∎

Refer to caption
Fig. 1: Graphical representation of Proposition 2.5; solid edges represent the covering relation.

In Section 1 we explained that so-called arrow-closed subcontexts (of the standard context) are a key ingredient in order to obtain subdirect decompositions of finite lattices. We now provide concrete definitions as far as they are needed in this paper. A subcontext of a context 𝕂=(G,M,I)\mathbb{K}=(G,M,I) is a context 𝕂~=(H,N,J)\tilde{\mathbb{K}}=(H,N,J) where HGH\subseteq G, NMN\subseteq M and J=I(H×N)J=I\cap(H\times N). For a clarified context 𝕂\mathbb{K}, that is, g=hg^{\prime}=h^{\prime} implies g=hg=h, and m=nm^{\prime}=n^{\prime} implies m=nm=n for all g,hGg,h\in G and m,nMm,n\in M, such a subcontext 𝕂~\tilde{\mathbb{K}} is arrow-closed if for all hHh\in H, mMm\in M, nNn\in N and gGg\in G the condition gmg\mathrel{\nearrow}m implies mNm\in N, and gng\mathrel{\swarrow}n implies gHg\in H, see [10, Definition 46]. Note that the standard context of a finite lattice 𝕃\mathbb{L} is always clarified and reduced [10, Proposition 12]. For a finite clarified context 𝕂=(G,M,I)\mathbb{K}=(G,M,I) and G1GG_{1}\subseteq G and M1MM_{1}\subseteq M there is always a smallest arrow-closed subcontext 𝕂~=(H,N,I(H×N))\tilde{\mathbb{K}}=(H,N,I\cap(H\times N)) of 𝕂\mathbb{K} with G1HG_{1}\subseteq H and M1NM_{1}\subseteq N. It can be obtained by constructing the directed graph (G˙M,()1)\left(G\mathbin{\dot{\cup}}M,{\mathrel{\nearrow}}\cup{(\mathrel{\swarrow})^{-1}}\right) and considering the (not necessarily strongly) connected directed components [x][x] of each xG1˙M1x\in G_{1}\mathbin{\dot{\cup}}M_{1}. One then forms xG1˙M1[x]\bigcup_{x\in G_{1}\mathbin{\dot{\cup}}M_{1}}[x], which can be written in a unique way as H˙NH\mathbin{\dot{\cup}}N with HGH\subseteq G and NMN\subseteq M. In particular, starting from G1={g}G_{1}=\left\{g\right\} and M1=M_{1}=\emptyset with gGg\in G (or dually from G1=G_{1}=\emptyset and M1={m}M_{1}=\left\{m\right\} with mMm\in M), we get the one-generated arrow-closed subcontexts of 𝕂\mathbb{K}, cf. [10, Section 4.1]. Note that if the context 𝕂\mathbb{K} is reduced, we may always concentrate on using either only objects or attributes for constructing all its one-generated arrow-closed subcontexts.

2.3 Integer partitions

Our aim is to study the arrow relations of the standard context of the lattice n\mathcal{L}_{n} of positive integer partitions, which is formed by the sets 𝒥(n)\mathcal{J}\left(\mathcal{L}_{n}\right) and (n)\mathcal{M}\left(\mathcal{L}_{n}\right). First, we define formally, what a partition and the dominance order is.

Definition 2.6.

An (ordered) partition of a number nn\in\mathbb{N} is an nn-tuple a:=(a1,,an)a\mathrel{\mathop{:}}=\left(a_{1},\dotsc,a_{n}\right) of natural numbers such that

a1a2an0 and n=a1+a2++an.a_{1}\geq a_{2}\geq\dotsm\geq a_{n}\geq 0\qquad\text{ and }\qquad n=a_{1}+a_{2}+\dotsm+a_{n}.

If there is k{1,,n}k\in\left\{1,\dotsc,n\right\} such that ak1a_{k}\geq 1 and ai=0a_{i}=0 for all i>ki>k, we also allow for the partition aa to be written in the form (a1,a2,,ak)\left(a_{1},a_{2},\dotsc,a_{k}\right), where we have deleted the zeros at the end.

For example, (5,4,1,1,0,0,0,0,0,0,0)(5,4,1,1,0,0,0,0,0,0,0) is a partition of 1111 because 5410{5\geq 4\geq 1\geq 0} and 5+4+1+1=115+4+1+1=11. By removing the trailing zeros we can represent it more compactly as (5,4,1,1)(5,4,1,1). Graphically, we can illustrate a partition using a diagram drawn with small squares or ‘bricks’ arranged in a downward ladder shape (cf. Fig. 2), which is known as Ferrers diagram (usually Ferrers diagrams are drawn rotated clockwise by a 9090 degrees angle, but the chosen presentation is more useful to us, cf. Definition 2.9). Given a partition aa of nn, one obtains the conjugated or dual partition aa^{*} in the sense of [5] and [9] as a=(a1,,an)a^{*}=(a_{1}^{*},\dotsc,a_{n}^{*}) where ai:=|{1jnaji}|a_{i}^{*}\mathrel{\mathop{:}}=\left\lvert\{1\leq j\leq n\mid a_{j}\geq i\}\right\rvert for all i{1,,n}i\in\left\{1,\dotsc,n\right\}. The Ferrers diagram of aa^{*} can be seen from the diagram of aa by reading it by rows, from bottom to top. For instance, the partition (5,4,1,1)(5,4,1,1) of 11=5+4+1+111=5+4+1+1 has the Ferrers diagram shown in Fig. 2, and its conjugate consists of 44 bricks from the first row, 22 bricks from the second to fourth, and 11 brick from the fifth row. We thereby obtain the partition (4,2,2,2,1)(4,2,2,2,1); its Ferrers diagram is also depicted in Fig. 2.

Refer to caption
Fig. 2: Ferrers diagrams. Left: of g=(5,4,1,1)g=(5,4,1,1); right: of the conjugated (dual) partition g=(4,2,2,2,1)g^{*}=(4,2,2,2,1).

We denote the set of all partitions of an integer nn\in\mathbb{N} by Part(n)\operatorname{Part}(n). From the construction of the conjugate via the Ferrers diagram, it is easy to see that (a)=a\left(a^{*}\right)^{*}=a holds for every aPart(n)a\in\operatorname{Part}(n). Therefore, partition conjugation is an involutive permutation of Part(n)\operatorname{Part}(n). One may order the set Part(n)\operatorname{Part}(n) in different ways, for example, lexicographically, or pointwise. In this paper we are interested in the dominance order established by Brylawski [5].

Definition 2.7 ([5]).

Let a=(a1,,an)a=\left(a_{1},\dotsc,a_{n}\right) and b=(b1,,bn)b=\left(b_{1},\dotsc,b_{n}\right) be two partitions of nn\in\mathbb{N}. We define the dominance order between aa and bb by setting aba\geq b if and only if i=1jaii=1jbi\sum_{i=1}^{j}a_{i}\geq\sum_{i=1}^{j}b_{i} holds for all j{1,,n}j\in\left\{1,\dotsc,n\right\}.

Brylawski showed in [5, Proposition 2.2] that the set Part(n)\operatorname{Part}(n) forms a lattice under the dominance order; further arguments for this were given in [6, Chapter 3] and [9, Section 2]. We denote by n\mathcal{L}_{n} the lattice of all partitions of nn\in\mathbb{N} with the dominance order. It follows from [5, Proposition 2.8] that for partitions a,bna,b\in\mathcal{L}_{n} we have aba\leq b if and only if aba^{*}\geq b^{*}, that is, partition conjugation is an order-antiautomorphism of n\mathcal{L}_{n}, making n\mathcal{L}_{n} is self-dual.

Brylawski in [5, Proposition 2.3] characterised precisely two possibilities for downward movement along the covering relation of n\mathcal{L}_{n}. These were later given a more intuitive geometric interpretation as transition rules by Latapy and Phan [13, Fig. 2, p. 1358], which we are going to follow in this article. The subsequent definitions are required for this.

Definition 2.8 (cf. [13, p. 1358]).

Let n,jn,j\in\mathbb{N} and 1j<n1\leq j<n. The partition a=(a1,,an)na=\left(a_{1},\dotsc,a_{n}\right)\in\mathcal{L}_{n} has

  1. 1.

    a cliff at jj if ajaj+12a_{j}-a_{j+1}\geq 2;

  2. 2.

    a slippery step at jj if there is \ell\in\mathbb{N} such that 2nj2\leq\ell\leq n-j and aj1=aj+1==aj+1=aj++1a_{j}-1=a_{j+1}=\dots=a_{j+\ell-1}=a_{j+\ell}+1.

For example, the left Ferrers diagram in Fig. 2 has a cliff in the second position (j=2j=2) and its dual has a cliff at j=1j=1 and a slippery step (with =2\ell=2) at j=4j=4.

Definition 2.9 (Transition rules, cf. [5, 13]).

Let nn\in\mathbb{N} and ana\in\mathcal{L}_{n}.

  1. 1.

    If a=(,k+1,k,,k,k1,)a=(\dots,k+1,k,\dotsc,k,k-1,\dots) for some kk\in\mathbb{N} with 1k<n1\leq k<n has a slippery step, then the brick at the slippery step may slip across the step to give a~=(,k,k,,k,k,)\tilde{a}=(\dotsc,k,k,\dotsc,k,k,\dotsc). The subsequent illustration shows the application of such a transition to a slippery step at position jj (with ij+2i\geq j+2).

    [Uncaptioned image]
  2. 2.

    If a=(,k,kh,)a=(\dotsc,k,k-h,\dotsc) for some k,hk,h\in\mathbb{N} with 2hkn2\leq h\leq k\leq n has a cliff, then the brick may fall from the cliff to give a~=(,k1,kh+1,)\tilde{a}=(\dotsc,k-1,k-h+1,\dotsc). Again, this is illustrated with a cliff of height 33 at position jj.

    [Uncaptioned image]
Lemma 2.10 ([5, Proposition 2.3] and [13]).

The set of lower covers of a partition gng\in\mathcal{L}_{n} consists exactly of all partitions g~\tilde{g} that can be obtained from gg by applying one of the two transition rules (1) or (2) described in Definition 2.9.

Since the completely join-irreducible elements of n\mathcal{L}_{n} have exactly one lower cover, according to Lemma 2.10, the partitions in 𝒥(n)\mathcal{J}\left(\mathcal{L}_{n}\right) are precisely those to which exactly one of the transition rules (1) or (2) applies. Therefore, the \bigvee-irreducible partitions of nn can be characterised as those that have exactly one cliff (and no slippery step) or exactly one slippery step (and no cliff). Based on this Brylawski [5] split the set 𝒥(n)\mathcal{J}\left(\mathcal{L}_{n}\right) into four categories, extending this by conjugation to (n)\mathcal{M}\left(\mathcal{L}_{n}\right). To be able to express our results more compactly, in the following statement we have slightly modified the borders between the different types of irreducibles compared to [5], making them in particular disjoint.

Lemma 2.11 (cf. [5, Corollary 2.5]).

For nn\in\mathbb{N} the completely join-irreducible partitions of 𝒥(n)\mathcal{J}\left(\mathcal{L}_{n}\right) can be categorised into four types where bb\in\mathbb{N}, d,+d,\ell\in\mathbb{N}_{+}:

  1. 1.

    (k,k,,k)(k,k,\dotsc,\overset{\ell}{k}) for k2k\geq 2.

  2. 2.

    (k,k,,k𝑏,k1,k1,,k1b+)(k,k,\dotsc,\overset{b}{k},k-1,k-1,\dotsc,\overset{b+\ell}{k-1}) for k2k\geq 2, b1b\geq 1.

  3. 3.

    (k,1,1,,1d+1)(k,1,1,\dotsc,\overset{d+1}{1}) for k3k\geq 3.

  4. 4.

    (k+1,k+1,,k+1𝑏,k,,kb+,1,1,,1b+l+d)(k+1,k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,1,\dotsc,\overset{b+l+d}{1}) for k3k\geq 3, b+2b+\ell\geq 2.

Also the \bigwedge-irreducible elements can be split into four groups where c+c\in\mathbb{N}_{+}:

  1. 1.

    (t,t,,t𝑐)(t,t,\dotsc,\overset{c}{t}) for t1t\geq 1, c2c\geq 2, i.e., tt appears at least twice.

  2. 2.

    (t,t,,t𝑐,r)(t,t,\dotsc,\overset{c\vphantom{1}}{t},r) for t>r1t>r\geq 1.

  3. 3.

    (a,1,1,,1c+1)(a,1,1,\dotsc,\overset{c+1}{1}) for a2a\geq 2, c2c\geq 2, i.e., there are at least two 11s.

  4. 4.

    (a,t,t,,tc+1,r)(a,t,t,\dotsc,\overset{c+1}{t},r) for a>t>r0a>t>r\geq 0, t,c2t,c\geq 2, i.e., t2t\geq 2 appears at least twice.

Observe that for each (L,J){(1,1),(2,2),(3,3),(4,4)}(L,J)\in\left\{(\ref{typeA},\ref{typeI}),(\ref{typeB},\ref{typeII}),(\ref{typeC},\ref{typeIII}),(\ref{typeD},\ref{typeIV})\right\} we have that if g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) has type LL, then gg^{*} has type JJ, and if m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) has type JJ, then mm^{*} has type LL. Therefore, the pairs (L,J)(L,J) of categories of completely irreducible elements of n\mathcal{L}_{n} are completely dual to each other.

We mentioned that n\mathcal{L}_{n} is a self-dual lattice under partition conjugation as involutive antiautomorphism, see [5, Proposition 2.8]. Using this fact we obtain the following simple but useful results to switch down and up-arrows.

Lemma 2.12.

Let g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) with duals gg^{*} and mm^{*}, respectively. Then it holds that

gmmg.g\mathrel{\swarrow}m\iff m^{*}\mathrel{\nearrow}g^{*}.
Proof.

The proof is a routine calculation exploiting the involutive antiautomorphism :nn{}^{*}\colon\mathcal{L}_{n}\to\mathcal{L}_{n} and the duality of the involved concepts. ∎

Similarly, one can prove the following lemma.

Lemma 2.13.

Let g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) with duals gg^{*} and mm^{*}, respectively. Then it holds that

gmmg.g\mathrel{\nearrow}m\iff m^{*}\mathrel{\swarrow}g^{*}.

Combining Lemma 2.12 and Lemma 2.13, we obtain the following corollary.

Corollary 2.14.

Let g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) with duals g(n)g^{*}\in\mathcal{M}\left(\mathcal{L}_{n}\right) and m𝒥(n)m^{*}\in\mathcal{J}\left(\mathcal{L}_{n}\right). Then it holds that

gmmg.g\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m\iff m^{*}\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*}.

To obtain some familiarity with the completely irreducible elements of n\mathcal{L}_{n} and the characterisations concerning the arrow relations given in Subsection 2.2, we consider the lattice 7\mathcal{L}_{7} shown in Fig. 3.

Refer to caption
Fig. 3: Lattice 7\mathcal{L}_{7} with all arrows. Up-arrows are shown with open arrow tips ending in \bigwedge-irreducibles, and down-arrows with filled arrow tips ending in \bigvee-irreducibles; single up-arrows are coloured green, single down-arrows red.
Example 2.15.

For n=7n=7 we have from Lemma 2.11, cf. also [9, Proposition 2], that

𝒥(7)\displaystyle\mathcal{J}\left(\mathcal{L}_{7}\right) ={(2,1,,1),(2,2,1,1,1),(2,2,2,1),(3,1,,1),(3,2,2),(3,3,1),(4,1,1,1),(4,3),(5,1,1),(6,1),(7)},\displaystyle=\left\{(2,1,\dotsc,1),(2,2,1,1,1),(2,2,2,1),(3,1,\dotsc,1),(3,2,2),(3,3,1),(4,1,1,1),(4,3),(5,1,1),(6,1),(7)\right\},
(7)\displaystyle\mathcal{M}\left(\mathcal{L}_{7}\right) ={(1,,1),(2,1,,1),(2,2,2,1),(3,1,,1),(3,2,2),(3,3,1),(4,1,1,1),(4,3),(5,1,1),(5,2),(6,1)}.\displaystyle=\left\{(1,\dotsc,1),(2,1,\dotsc,1),(2,2,2,1),(3,1,\dotsc,1),(3,2,2),(3,3,1),(4,1,1,1),(4,3),(5,1,1),(5,2),(6,1)\right\}.

The standard context 𝕂(7)=(𝒥(7),(7),)\mathbb{K}(\mathcal{L}_{7})=\left(\mathcal{J}\left(\mathcal{L}_{7}\right),\mathcal{M}\left(\mathcal{L}_{7}\right),\leq\right) is presented in Fig. 4. In both Fig. 4 and Fig. 3 we have indicated the arrow relations. Note that n=7n=7 is the first case where single arrows appear. In fact there is exactly one single up-arrow and one single down-arrow in 𝕂(7)\mathbb{K}(\mathcal{L}_{7}) (where the respective opposite arrow is not present). We shall investigate this in the case of the down-arrow (4,1,1,1)(3,2,2)(4,1,1,1)\mathrel{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathrel{\swarrow}}(3,2,2), shown in red in Fig. 3. The green up-arrow can then be explained by self-duality of 7\mathcal{L}_{7} using Lemma 2.13.

The partition g:=(4,1,1,1)g\mathrel{\mathop{:}}=(4,1,1,1) is completely join-irreducible of type 3, while m=(3,2,2)m=(3,2,2) is completely meet-irreducible of type 4. To explain the down-arrow we shall use Proposition 2.5(1). Clearly, gmg\nleq m since the condition on the partial sums already fails with 434\nleq 3 in the first position. The partition gg has a single cliff in the first position, from which a brick may fall to obtain the unique lower cover g~=(3,2,1,1)\tilde{g}=(3,2,1,1), cf. transition rule (2) and Fig. 3. Indeed, the lattice diagram confirms that g~m\tilde{g}\leq m, which can also be checked with Definition 2.7. Thus, by Proposition 2.5(1) we conclude gmg\mathrel{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathrel{\swarrow}}m. Moreover, that this arrow is not a double arrow, can also be explained with the help of Proposition 2.5. Namely, the unique upper cover of mm is m~=(3,3,1)\tilde{m}=(3,3,1) since we may let the brick fall from the cliff in the second position of m~\tilde{m} to get mm, cf. transition rule (2). Since gm~g\nleq\tilde{m}, we have gmg\mathrel{\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\mathrel{\ooalign{$\mathrel{\nearrow}$\cr${\scriptstyle-\mkern 8.5mu}$}}}m by Proposition 2.5(2).

The fact that there is a unique up-arrow (and a unique down-arrow) in 𝕂(7)\mathbb{K}(\mathcal{L}_{7}) that fails to be a double arrow, and why non-double arrows appear for n=7n=7 for the first time, cannot be explained yet. This requires the characterisations in Sections 4 and 5.

𝕂(7)\mathbb{K}\left(\mathcal{L}_{7}\right) 1111111 211111 31111 2221 4111 322 331 511 43 52 61
7 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}
61 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times
511 ×\times \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times
43 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times
4111 ×\times \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathrel{\swarrow} \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times
331 \color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\mathrel{\nearrow} \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times
322 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times ×\times
31111 ×\times \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times ×\times ×\times
2221 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times
22111 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times
211111 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times ×\times
Fig. 4: Standard context 𝕂(7)=(𝒥(7),(7),)\mathbb{K}\left(\mathcal{L}_{7}\right)=(\mathcal{J}\left(\mathcal{L}_{7}\right),\mathcal{M}\left(\mathcal{L}_{7}\right),{\leq}) with arrow relations shown.

3 General facts concerning the dominance order of integer partitions

We start with a lemma giving a simple sufficient condition for the dominance order among partitions.

Lemma 3.1.

Let n,j,k,mn,j,k,m\in\mathbb{N} be such that j,k,mnj,k,m\leq n and let a=(a1,,ak,0,),b=(b1,,bm,0,)na=(a_{1},\dotsc,a_{k},0,\dots),b=(b_{1},\dotsc,b_{m},0,\dots)\in\mathcal{L}_{n}. If s:=i=1jai=i=1jbis\mathrel{\mathop{:}}=\sum_{i=1}^{j}a_{i}=\sum_{i=1}^{j}b_{i} and aibia_{i}\geq b_{i} for each ii\in\mathbb{N} with 2ij2\leq i\leq j, while aibia_{i}\leq b_{i} for each ii\in\mathbb{N} with j<i<mj<i<m, then aba\leq b.

Proof.

Let l{1,,n}l\in\left\{1,\dotsc,n\right\}. If ljl\leq j, then we have i=1lai=si=l+1jaisi=l+1jbi=i=1lbi\sum_{i=1}^{l}a_{i}=s-\sum_{i=l+1}^{j}a_{i}\leq s-\sum_{i=l+1}^{j}b_{i}=\sum_{i=1}^{l}b_{i} since biaib_{i}\leq a_{i} for 2l+1ij2\leq l+1\leq i\leq j. Let now l>jl>j. If l<ml<m, we have i=1lai=s+i=j+1lais+i=j+1lbi=i=1lbi\sum_{i=1}^{l}a_{i}=s+\sum_{i=j+1}^{l}a_{i}\leq s+\sum_{i=j+1}^{l}b_{i}=\sum_{i=1}^{l}b_{i} since aibia_{i}\leq b_{i} for j<il<mj<i\leq l<m. Otherwise, mlnm\leq l\leq n, and then we have i=1lain=i=1mbi=i=1lbi\sum_{i=1}^{l}a_{i}\leq n=\sum_{i=1}^{m}b_{i}=\sum_{i=1}^{l}b_{i} as a,bna,b\in\mathcal{L}_{n}. ∎

As an easy corollary we have the situation where bb is not longer than aa and lies pointwise above aa.

Corollary 3.2.

Let n,k,mn,k,m\in\mathbb{N} and let a=(a1,,ak),b=(b1,,bm)na=(a_{1},\dotsc,a_{k}),b=(b_{1},\dotsc,b_{m})\in\mathcal{L}_{n}. If aibia_{i}\leq b_{i} for all 1i<m1\leq i<m, then aba\leq b.

Proof.

This follows from Lemma 3.1 for j=0j=0 and s=iai=0=ibis=\sum_{i\in\emptyset}a_{i}=0=\sum_{i\in\emptyset}b_{i}. ∎

We may also have the somewhat opposite situation.

Corollary 3.3.

Let n,k,mn,k,m\in\mathbb{N} and let a=(a1,,ak),b=(b1,,bm)na=(a_{1},\dotsc,a_{k}),b=(b_{1},\dotsc,b_{m})\in\mathcal{L}_{n}. If aibia_{i}\geq b_{i} for all 2im2\leq i\leq m, then aba\leq b.

Proof.

For 2im2\leq i\leq m we have aibia_{i}\geq b_{i}; for m<inm<i\leq n we have ai0=bia_{i}\geq 0=b_{i}. Hence aibia_{i}\geq b_{i} holds for all 2in2\leq i\leq n. Thus we can apply Lemma 3.1 with j=nj=n and s=i=1nai=n=i=1bis=\sum_{i=1}^{n}a_{i}=n=\sum_{i=1}b_{i} to get aba\leq b. ∎

We now introduce two natural geometric parameters of partitions.

Definition 3.4.

Let nn\in\mathbb{N} and p=(p1,p2,)np=(p_{1},p_{2},\dots)\in\mathcal{L}_{n}.

  1. 1.

    The height of pp is the value of its first entry, that is, ht(p):=p1\operatorname{ht}(p)\mathrel{\mathop{:}}=p_{1}.

  2. 2.

    For hh\in\mathbb{N} we define n(h):={pn|ht(p)h}\mathcal{H}_{n}(h)\mathrel{\mathop{:}}=\left\{\left.p\in\mathcal{L}_{n}\ \vphantom{\operatorname{ht}(p)\leq h}\right|\ \operatorname{ht}(p)\leq h\right\} to be the set of partitions of height at most hh.

  3. 3.

    The length, synonymously width, of pp is the value of the first index jj\in\mathbb{N} such that pj+1=0p_{j+1}=0, that is to say, len(p):=min{j|pj+1=0}\operatorname{len}(p)\mathrel{\mathop{:}}=\min\left\{\left.j\in\mathbb{N}\ \vphantom{p_{j+1}=0}\right|\ p_{j+1}=0\right\}.

  4. 4.

    For ww\in\mathbb{N} we define 𝒲n(w):={pn|len(p)w}\mathcal{W}_{n}(w)\mathrel{\mathop{:}}=\left\{\left.p\in\mathcal{L}_{n}\ \vphantom{\operatorname{len}(p)\leq w}\right|\ \operatorname{len}(p)\leq w\right\} to be the set of partitions of length (width) at most ww.

Lemma 3.5.

For n+n\in\mathbb{N}_{+} the length of pnp\in\mathcal{L}_{n} is the value of the largest index j1j\geq 1 where pj1p_{j}\geq 1, that is, len(p)=max{j+|pj>0}\operatorname{len}(p)=\max\left\{\left.j\in\mathbb{N}_{+}\ \vphantom{p_{j}>0}\right|\ p_{j}>0\right\}.

Proof.

For n+n\in\mathbb{N}_{+} we have p1=ht(p)1p_{1}=\operatorname{ht}(p)\geq 1, thus the maximum w:=max{j+|pj>0}w\mathrel{\mathop{:}}=\max\left\{\left.j\in\mathbb{N}_{+}\ \vphantom{p_{j}>0}\right|\ p_{j}>0\right\} exists. We have pw>0p_{w}>0, but, by maximality, pw+1=0p_{w+1}=0. By antitonicity of pp we have pjpw>0p_{j}\geq p_{w}>0 for 1jw1\leq j\leq w, thus ww is the first index where pw+1=0p_{w+1}=0. Hence len(p)=w\operatorname{len}(p)=w. ∎

Lemma 3.6.

For n,wn,w\in\mathbb{N} we have 𝒲n(w)={pn|pw+1=0}\mathcal{W}_{n}(w)=\left\{\left.p\in\mathcal{L}_{n}\ \vphantom{p_{w+1}=0}\right|\ p_{w+1}=0\right\}.

Proof.

Let pnp\in\mathcal{L}_{n} and ww\in\mathbb{N}. If pw+1=0p_{w+1}=0, then the minimality in the definition of length implies len(p)w\operatorname{len}(p)\leq w, thus p𝒲n(w)p\in\mathcal{W}_{n}(w). Conversely, if j:=len(p)wj\mathrel{\mathop{:}}=\operatorname{len}(p)\leq w, then 0=pj+1pw+100=p_{j+1}\geq p_{w+1}\geq 0 by the antitonicity of pp. ∎

For parameter values at least nn the concept of bounded height or width trivialises.

Lemma 3.7.

For n,hn,h\in\mathbb{N} such that hnh\geq n, we have n(h)=𝒲n(h)=n\mathcal{H}_{n}(h)=\mathcal{W}_{n}(h)=\mathcal{L}_{n}.

Proof.

For every pnp\in\mathcal{L}_{n} we have ht(p)=p1nh\operatorname{ht}(p)=p_{1}\leq n\leq h, thus pn(h)p\in\mathcal{H}_{n}(h). If ph+1>0p_{h+1}>0, then for every 1jh+11\leq j\leq h+1 we would have pjph+11p_{j}\geq p_{h+1}\geq 1, thus nj=1h+1pjj=1h+11=h+1n+1n\geq\sum_{j=1}^{h+1}p_{j}\geq\sum_{j=1}^{h+1}1=h+1\geq n+1, which is a contradiction. Therefore, ph+1=0p_{h+1}=0 and p𝒲n(h)p\in\mathcal{W}_{n}(h) by Lemma 3.6. ∎

In a similar fashion, for n+n\in\mathbb{N}_{+}, we have n(0)=𝒲n(0)=\mathcal{H}_{n}(0)=\mathcal{W}_{n}(0)=\emptyset since p1=0p_{1}=0 implies p=(0,0)p=(0,0\dots) and thus n=0n=0. Therefore, for n+n\in\mathbb{N}_{+} we may safely restrict our attention to parameter values 1hn1\leq h\leq n to bound the height or width of a partition, which is, of course, clear from the geometric intuition via the Ferrers diagrams.

It is useful to observe that height and width are dual concepts with respect to partition conjugation.

Lemma 3.8.

For nn\in\mathbb{N} and pnp\in\mathcal{L}_{n} we have ht(p)=len(p)\operatorname{ht}(p^{*})=\operatorname{len}(p) and len(p)=ht(p)\operatorname{len}(p^{*})=\operatorname{ht}(p). Moreover, for ww\in\mathbb{N} we have p𝒲n(w)p\in\mathcal{W}_{n}(w) exactly if pn(w)p^{*}\in\mathcal{H}_{n}(w).

Proof.

Let w:=len(p)w\mathrel{\mathop{:}}=\operatorname{len}(p) such that p=(p1,,pw)p=(p_{1},\dotsc,p_{w}) with pj1p_{j}\geq 1 for 1jw1\leq j\leq w. From the definition of the conjugate we infer that the first entry of pp^{*} is ww, hence ht(p)=w=len(p)\operatorname{ht}(p^{*})=w=\operatorname{len}(p). The second equality follows from ht(p)=len(p)\operatorname{ht}(p^{*})=\operatorname{len}(p) since p=p{p^{*}}^{*}=p. Finally, for ww\in\mathbb{N} we have p𝒲n(w)p\in\mathcal{W}_{n}(w) if and only if ht(p)=len(p)w\operatorname{ht}(p^{*})=\operatorname{len}(p)\leq w, i.e., if and only if pn(w)p^{*}\in\mathcal{H}_{n}(w). ∎

We may also observe a basic necessary condition following from the dominance order: the dominated partition is always at least as long as the dominating one.

Lemma 3.9.

Let nn\in\mathbb{N} and a,bna,b\in\mathcal{L}_{n}. If len(a)<len(b)\operatorname{len}(a)<\operatorname{len}(b), then aba\nleq b.

Proof.

Assume l:=len(a)<w:=len(b)l\mathrel{\mathop{:}}=\operatorname{len}(a)<w\mathrel{\mathop{:}}=\operatorname{len}(b). Then al+1=0a_{l+1}=0 and n=i=1lain=\sum_{i=1}^{l}a_{i}. Since l<wl<w we have l+1wl+1\leq w, whence bl+1bw1b_{l+1}\geq b_{w}\geq 1. Thus i=1lbi<i=1l+1bin=i=1lai\sum_{i=1}^{l}b_{i}<\sum_{i=1}^{l+1}b_{i}\leq n=\sum_{i=1}^{l}a_{i}, demonstrating aba\nleq b. ∎

The following lemma describes the largest partition of a given least length \ell.

Lemma 3.10.

Let n,n,\ell\in\mathbb{N} with n1n\geq\ell\geq 1, let pnp\in\mathcal{L}_{n} and set m:=(n(1),1,,1)m\mathrel{\mathop{:}}=(n-(\ell-1),1,\dotsc,\overset{\ell}{1}). If len(p)\operatorname{len}(p)\geq\ell, then pmp\leq m; if len(p)<\operatorname{len}(p)<\ell, then pmp\nleq m.

Proof.

Take pnp\in\mathcal{L}_{n} with len(p)\operatorname{len}(p)\geq\ell, that is, p=(p1,,p,)p=(p_{1},\dotsc,p_{\ell},\ldots) with pip1p_{i}\geq p_{\ell}\geq 1 for 1i1\leq i\leq\ell. Thus, for 2i2\leq i\leq\ell we have mi=1pim_{i}=1\leq p_{i}, hence pmp\leq m by Corollary 3.3. Furthermore, if pp is shorter than mm, then pmp\nleq m by Lemma 3.9. ∎

There is also a second largest partition of a given least length \ell.

Corollary 3.11.

Let n,+n,\ell\in\mathbb{N}_{+} with 2,n2\leq\ell,n-\ell. Then m=(n(1),1,,1)𝒥(n)m=(n-(\ell-1),1,\dotsc,\overset{\ell}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 3 and has m~=(n,2,1,,1)\tilde{m}=(n-\ell,2,1,\dotsc,\overset{\ell}{1}) as its unique lower cover (the part 1,,11,\dotsc,1 appearing as of 3\ell\geq 3); moreover any pn{m}p\in\mathcal{L}_{n}\setminus\left\{m\right\} with len(p)\operatorname{len}(p)\geq\ell satisfies pm~p\leq\tilde{m}.

Proof.

Since n(1)3n-(\ell-1)\geq 3 and 2\ell\geq 2, we see that mm is of type 3, and we can obtain the unique lower cover m~\tilde{m} by applying transition rule (2) to the cliff in the first position of mm. If pnp\in\mathcal{L}_{n} has length len(p)\operatorname{len}(p)\geq\ell, then pmp\leq m by Lemma 3.10. If we additionally assume pmp\neq m, then p<mp<m and thus pm~p\leq\tilde{m}, for m~\tilde{m} is the unique lower cover of mm. ∎

We can also identify a largest partition with a given height bound.

Lemma 3.12.

Let n,h+n,h\in\mathbb{N}_{+} with hnh\leq n and factorise n=wh+rn=wh+r where r,wr,w\in\mathbb{N}, 0r<h0\leq r<h. Then w1w\geq 1 and every pn(h)p\in\mathcal{H}_{n}(h) satisfies pm:=(h,,h𝑤,r)n(h)p\leq m\mathrel{\mathop{:}}=(h,\dotsc,\overset{w}{h},r)\in\mathcal{H}_{n}(h).

Proof.

Since n=wh+rn=wh+r, we have mnm\in\mathcal{L}_{n}. If w=0w=0, then n=r<hnn=r<h\leq n; thus w1w\geq 1 and therefore ht(m)=h\operatorname{ht}(m)=h, i.e., mn(h)m\in\mathcal{H}_{n}(h). By antitonicity, for each 1iw1\leq i\leq w we have pip1=ht(p)h=mip_{i}\leq p_{1}=\operatorname{ht}(p)\leq h=m_{i}, hence pmp\leq m by Corollary 3.2. ∎

By duality we have a least partition of a given bounded width.

Lemma 3.13.

Let n,w+n,w\in\mathbb{N}_{+}, wnw\leq n and factorise n=κw+bn=\kappa w+b where κ,b\kappa,b\in\mathbb{N}, 0b<w0\leq b<w. Then κ1\kappa\geq 1 and every p𝒲n(w)p\in\mathcal{W}_{n}(w) satisfies pg=(κ+1,,κ+1𝑏,κ,,κ𝑤)p\geq g=(\kappa+1,\dotsc,\overset{b}{\kappa+1},\kappa,\dotsc,\overset{w}{\kappa}).

Proof.

By Lemma 3.8, for pnp\in\mathcal{L}_{n} we have p𝒲n(w)p\in\mathcal{W}_{n}(w) if and only if pn(w)p^{*}\in\mathcal{H}_{n}(w). Therefore, by Lemma 3.12, we obtain pm=(w,,w𝜅,b)p^{*}\leq m=(w,\dotsc,\overset{\kappa}{w},b), and since conjugation is an involutive antiautomorphism of n\mathcal{L}_{n}, this inequality implies p=pm=(κ+1,,κ+1𝑏,κ,,κ𝑤)=gp={p^{*}}^{*}\geq m^{*}=(\kappa+1,\dotsc,\overset{b}{\kappa+1},\kappa,\dotsc,\overset{w}{\kappa})=g. ∎

Corollary 3.14.

Let bb\in\mathbb{N}, k,+k,\ell\in\mathbb{N}_{+} and set n:=b(k+1)+kn\mathrel{\mathop{:}}=b(k+1)+\ell k. The least partition in 𝒲n(b+)\mathcal{W}_{n}(b+\ell) has the form g=(k+1,,k+1𝑏,k,,kb+)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). If b+<nb+\ell<n, then g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1 and k2k\geq 2, or gg is of type 2.

Proof.

We set w:=b+1w\mathrel{\mathop{:}}=b+\ell\geq 1. Since k1k\geq 1 we have n=b(k+1)+kb+=wn=b(k+1)+\ell k\geq b+\ell=w, moreover n=b+(b+)k=b+wkn=b+(b+\ell)k=b+wk. Since 0b<b+=w0\leq b<b+\ell=w, this provides the factorisation of nn modulo ww, whence Lemma 3.13 shows that the least element of 𝒲n(w)\mathcal{W}_{n}(w) is g=(k+1,,k+1𝑏,k,,kb+)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). Furthermore, if b1b\geq 1, then g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 2. If otherwise b=0b=0, then =b+<n=k\ell=b+\ell<n=\ell k implies k>1k>1, i.e., k2k\geq 2. Therefore, g=(k,,k)g=(k,\dotsc,\overset{\ell}{k}) is of type 1. ∎

4 Description of the double arrow relation in the standard context (𝒥(n),(n),)\left(\mathcal{J}\left(\mathcal{L}_{n}\right),\mathcal{M}\left(\mathcal{L}_{n}\right),\leq\right).

It is our aim to describe the arrow relations of the standard context 𝕂(n)\mathbb{K}\left(\mathcal{L}_{n}\right). To do this, we start by giving a complete characterisation of all arrow relations of 𝕂(7)\mathbb{K}\left(\mathcal{L}_{7}\right). To this end we present the following lemmata, which are true for n\mathcal{L}_{n} in general. The first lemma is quite elementary, but useful, as it describes the top part of n\mathcal{L}_{n}.

Lemma 4.1.

Let nn\in\mathbb{N} such that n2n\geq 2.

  1. 1.

    Then (n)𝒥(n)(n)\in\mathcal{J}\left(\mathcal{L}_{n}\right) and (n1,1)(n)(n-1,1)\in\mathcal{M}\left(\mathcal{L}_{n}\right) is its unique lower cover.

  2. 2.

    If n3n\geq 3, then (n1,1)(n)𝒥(n)(n-1,1)\in\mathcal{M}\left(\mathcal{L}_{n}\right)\cap\mathcal{J}\left(\mathcal{L}_{n}\right).

  3. 3.

    If n4n\geq 4, then (n2,2)(n)(n-2,2)\in\mathcal{M}\left(\mathcal{L}_{n}\right) is the unique lower cover of (n1,1)(n-1,1). Hence the top part of n\mathcal{L}_{n} has the shape depicted in Fig. 5.

Refer to caption
Fig. 5: Chain at the top of n\mathcal{L}_{n} for n4n\geq 4; edges denote the covering relation.
Proof.

The top partition (n)(n) is completely join-irreducible, since n2n\geq 2 ensures that it is of type 1. Letting a brick fall from the cliff, we obtain via transition rule (2) its unique lower cover (n1,1)(n-1,1). The latter partition is completely meet-irreducible, for it is of type 1 when n=2n=2 and of type 2 if n3n\geq 3. This shows part (1).

Let us now assume that n3n\geq 3. If n=3n=3, then (n1,1)(n-1,1) is completely join-irreducible of type 2; moreover, it is of type 3 when n4n\geq 4. Part (2) has been demonstrated.

Finally let us impose n4n\geq 4. By applying transition rule (2), that is, by letting a brick fall from the cliff in the first position of (n1,1)(n-1,1), we obtain (n2,2)(n-2,2) as its lower cover, which is unique by (2). Moreover, (n2,2)(n-2,2) is completely meet-irreducible of type 1 if n=4n=4 and of type 2 if n5n\geq 5. ∎

Lemma 4.2.

For n2n\geq 2, we have g=(n)𝒥(n)g=(n)\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(n1,1)(n)m=\left(n-1,1\right)\in\mathcal{M}\left(\mathcal{L}_{n}\right), and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

Proof.

From Lemma 4.1(1), we have g=(n)𝒥(n)g=(n)\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(n1,1)(n)m=\left(n-1,1\right)\in\mathcal{M}\left(\mathcal{L}_{n}\right), and clearly g~=m\tilde{g}=m is the unique lower cover of gg and m~=g\tilde{m}=g is the unique upper cover of mm. Since gmg\nleq m, Proposition 2.5 directly implies gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. ∎

Lemma 4.3.

For n4n\geq 4 we have g=(n1,1)𝒥(n)g=\left(n-1,1\right)\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(n2,2)(n)m=\left(n-2,2\right)\in\mathcal{M}\left(\mathcal{L}_{n}\right), and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

Proof.

As n4n\geq 4, we know from Lemma 4.1 that g𝒥(n)(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right)\cap\mathcal{M}\left(\mathcal{L}_{n}\right), having g~:=m(n)\tilde{g}\mathrel{\mathop{:}}=m\in\mathcal{M}\left(\mathcal{L}_{n}\right) as its unique lower cover. At the same time m~=g\tilde{m}=g is thus the (unique) upper cover of mm. Since gmg\nleq m, Proposition 2.5 directly yields gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. ∎

Applying Corollary 2.14 to Lemmata 4.2 and 4.3, respectively, we obtain two additional double arrows in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

Lemma 4.4.

In the standard context 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) of n\mathcal{L}_{n} the following facts hold:

  1. 1.

    For n2n\geq 2 we have g=(2,1,,1n1)𝒥(n)g=(2,1,\dotsc,\overset{n-1}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(1,1,,1𝑛)(n)m=(1,1,\dotsc,\overset{n}{1})\in\mathcal{M}\left(\mathcal{L}_{n}\right), and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m.

  2. 2.

    For n4n\geq 4 we have g=(2,2,1,,1n2)𝒥(n)g=(2,2,1,\dotsc,\overset{n-2}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(2,1,,1n1)(n)m=(2,1,\dotsc,\overset{n-1}{1})\in\mathcal{M}\left(\mathcal{L}_{n}\right), and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m.

Proof.

To prove the first statement, let g=(2,1,,1n1)g=(2,1,\dotsc,\overset{n-1}{1}) and m=(1,1,,1𝑛)m=(1,1,\dotsc,\overset{n}{1}). Then m=(n)m^{*}=\left(n\right) and g=(n1,1)g^{*}=\left(n-1,1\right). From Lemma 4.2, we infer that m𝒥(n)m^{*}\in\mathcal{J}\left(\mathcal{L}_{n}\right), g(n)g^{*}\in\mathcal{M}\left(\mathcal{L}_{n}\right) and mgm^{*}\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*}. Applying Corollary 2.14 to the latter expression, we obtain gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, as required; moreover, using partition conjugation, it follows that g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right). Similarly, combining Corollary 2.14 with Lemma 4.3, one can prove the second statement. ∎

Lemma 4.5.

Let b+b\in\mathbb{N}_{+}, dd\in\mathbb{N}; set n=2b+dn=2b+d. Then we have g=(2,,2𝑏,1,,1b+d)𝒥(n)g=(2,\dotsc,\overset{b}{2},1,\dotsc,\overset{b+d}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 if d=0d=0 and of type 2 if d1d\geq 1. Its unique lower cover is g~=(2,,2b1,1,,1b+d,1)\tilde{g}=(2,\dotsc,\overset{b-1}{2},1,\dotsc,\overset{b+d}{1},1).

Proof.

If d=0d=0, then gg is of type 1 and has got a unique cliff at position bb, from which a brick may fall and produce g~=(2,,2b1,1,1)\tilde{g}=(2,\dotsc,\overset{b-1}{2},1,1), see transition rule (2). If d1d\geq 1, then gg has got no cliffs but a unique slippery step at position bb. Therefore its unique lower cover is obtained by applying transition rule (1) to gg, that is, by letting a brick slip across the step at position bb to obtain g~=(2,,2,1𝑏,,1b+d,1)\tilde{g}=(2,\dotsc,2,\overset{b}{1},\dotsc,\overset{b+d}{1},1). ∎

Lemma 4.6.

Let b,db,d\in\mathbb{N} with b2b\geq 2; set n:=2b+dn\mathrel{\mathop{:}}=2b+d. Then we have m=(b,1,,1b+d,1)(n)m=(b,1,\dotsc,\overset{b+d}{1},1)\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 3, having m~=(b,2,1,,1b+d)\tilde{m}=(b,2,1,\dotsc,\overset{b+d}{1}) as its unique upper cover, where the part 1,,11,\dotsc,1 only appears for b+d3b+d\geq 3, that is, d1d\geq 1, or d=0d=0 and b3b\geq 3.

Proof.

As b2b\geq 2, we have m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 3. Since partition conjugation is an order-antiautomorphism of n\mathcal{L}_{n}, the unique upper cover of mm can be obtained by conjugating the unique lower cover of m=(b+d+1,1,1,,1𝑏)𝒥(n)m^{*}=(b+d+1,1,1,\dotsc,\overset{b}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right). Since b2b\geq 2, the join-irreducible mm^{*} has got a unique cliff in its first position and no slippery step. Applying transition rule (2), we obtain its lower cover (b+d,2,1,,1𝑏)(b+d,2,1,\dotsc,\overset{b}{1}), which we can conjugate to obtain m~=(b,2,1,,1b+d)\tilde{m}=(b,2,1,\dotsc,\overset{b+d}{1}), the unique upper cover of mm. Knowing that there is a unique upper cover m~\tilde{m} of mm from m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right), the fact that m~\tilde{m} looks as given before can also be verified by checking that mm is a lower cover of m~\tilde{m}. Indeed, by simply applying transition rule (1) to the slippery step in the second position of m~\tilde{m} when b+d3b+d\geq 3, or by applying transition rule (2) to the cliff in the second position of m~=(b,2)=(2,2)\tilde{m}=(b,2)=(2,2) when d=0d=0 and b=2b=2, we see that we get mm. ∎

Lemma 4.7.

Let b,db,d\in\mathbb{N}, b2b\geq 2, and set n:=2b+dn\mathrel{\mathop{:}}=2b+d. Then, for g:=(2,,2𝑏,1,,1b+d)g\mathrel{\mathop{:}}=(2,\dotsc,\overset{b}{2},1,\dotsc,\overset{b+d}{1}) and m:=(b,1,,1b+d,1)m\mathrel{\mathop{:}}=(b,1,\dotsc,\overset{b+d}{1},1) we have gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m; moreover, m=(d+b+1,1,,1𝑏)g=(d+b,b)m^{*}=(d+b+1,1,\dotsc,\overset{b}{1})\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*}=(d+b,b).

Proof.

Clearly, the statement about mm^{*} and gg^{*} will follow from the one about gg and mm by Corollary 2.14. We therefore let n=2b+dn=2b+d with b2b\geq 2, d0d\geq 0 and consider gg and mm. We have that gmg\nleq m because the partial sums up to index bb exhibit the relation 2b>2b1=b+b12b>2b-1=b+b-1. From Lemma 4.5 we know that g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) has the unique lower cover g~=(2,,2b1,1,,1b+d,1)\tilde{g}=(2,\dotsc,\overset{b-1}{2},1,\dotsc,\overset{b+d}{1},1); from Lemma 4.6 we infer that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) has the unique upper cover m~=(b,2,1,,1b+d)\tilde{m}=(b,2,1,\dotsc,\overset{b+d}{1}).

First we prove that gmg\mathrel{\nearrow}m. We shall verify that gm~g\leq\tilde{m} by comparing the sequences of partial sums. For the sum up to the index j=1j=1 we have 2b2\leq b by hypothesis. The sequence of partial sums of m~\tilde{m} continues after bb with the terms (min(b+j,n))j2\left(\min(b+j,n)\right)_{j\geq 2}. If 2jb2\leq j\leq b, then the partial sum of gg up to index jj yields 2j=j+jj+b2j=j+j\leq j+b since jbj\leq b. For indices jbj\geq b both partial sums coincide with the value j+bj+b, until they reach the common maximum nn. Hence m~g\tilde{m}\geq g, and thus we may rely on Proposition 2.5(2) to show that gmg\mathrel{\nearrow}m.

Our next goal is to demonstrate that gmg\mathrel{\swarrow}m. For this we first show that g~m\tilde{g}\leq m by comparing the sequences of partial sums. For mm we obtain the sequence (min(n,b1+j))j1(\min(n,b-1+j))_{j\geq 1}, while for g~\tilde{g} it begins with (2j)1j<b(2j)_{1\leq j<b} and then continues with (min(n,b1+j))jb(\min(n,b-1+j))_{j\geq b}. For 1jb11\leq j\leq b-1 we have 2j=j+jb1+j2j=j+j\leq b-1+j, and for indices jbj\geq b both sequences coincide. Thus we have established g~m\tilde{g}\leq m, and hence Proposition 2.5(1) yields gmg\mathrel{\swarrow}m. ∎

Corollary 4.8.

Let dd\in\mathbb{N}, b+b\in\mathbb{N}_{+} and set n:=2b+dn\mathrel{\mathop{:}}=2b+d. Under these assumptions we have the two double arrow relations g=(2,,2𝑏,1,,1b+d)m=(b,1,,1b+d+1)g=(2,\dotsc,\overset{b}{2},1,\dotsc,\overset{b+d}{1})\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m=(b,1,\dotsc,\overset{b+d+1}{1}) and m=(b+d+1,1,,1𝑏)g=(b+d,b)m^{*}=(b+d+1,1,\dotsc,\overset{b}{1})\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*}=(b+d,b).

Proof.

If b=1b=1, the first statement follows from Lemma 4.4(1), while for b2b\geq 2 it follows from Lemma 4.7. The second statement follows by dualisation via Corollary 2.14. ∎

Fig. 4 shows the arrow relations of 𝕂(7)\mathbb{K}\left(\mathcal{L}_{7}\right). Using Lemmata 4.2, 4.3 and 4.4, one can verify some of those:

(7)\displaystyle\left(7\right) (6,1)\displaystyle\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\left(6,1\right) by Lemma 4.2 (6,1)\displaystyle\left(6,1\right) (5,2)\displaystyle\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\left(5,2\right) by Lemma 4.3
(2,1,1,1,1,1)\displaystyle\left(2,1,1,1,1,1\right) (1,1,1,1,1,1,1)\displaystyle\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\left(1,1,1,1,1,1,1\right) by Lemma 4.4 (2,2,1,1,1)\displaystyle\left(2,2,1,1,1\right) (2,1,1,1,1,1)\displaystyle\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\left(2,1,1,1,1,1\right) by Lemma 4.4

All remaining double arrows of 𝕂(7)\mathbb{K}\left(\mathcal{L}_{7}\right) will be collected in the following remark, whose proof will illustrate several similar cases. In subsequent sections we introduce parameters to prove general statements and avoid such repetitions.

Remark 4.9.

We consider Fig. 3 and the context given in Fig. 4. The first double arrow (2,2,2,1)(3,1,1,1,1)(2,2,2,1)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(3,1,1,1,1) and its dual (5,1,1)(4,3)(5,1,1)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(4,3) are confirmed by Lemma 4.7 for b=3b=3 and d=1d=1. It remains to verify

a)(3,1,1,1,1)(2,2,2,1)(4,3)(5,1,1)\displaystyle\text{a)}\quad\begin{aligned} (3,1,1,1,1)&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(2,2,2,1)\\ (4,3)&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(5,1,1)\end{aligned} b)(4,1,1,1)(3,3,1)(3,2,2)(4,1,1,1)\displaystyle\text{b)}\quad\begin{aligned} (4,1,1,1)&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(3,3,1)\\ (3,2,2)&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(4,1,1,1)\end{aligned} c)(3,3,1)(3,2,2)\displaystyle\text{c)}\quad(3,3,1)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(3,2,2)
Proof.

First, we verify a). Using the definition or inspecting Fig. 4, we get g:=(3,1,1,1,1)m:=(2,2,2,1)g\mathrel{\mathop{:}}=\left(3,1,1,1,1\right)\nleq m\mathrel{\mathop{:}}=\left(2,2,2,1\right). Clearly, we have g𝒥(7)g\in\mathcal{J}\left(\mathcal{L}_{7}\right) of type 3 and m(7)m\in\mathcal{M}\left(\mathcal{L}_{7}\right) of type 2. Applying transition rule (2), the unique lower cover of gg is g~=(2,2,1,1,1)\tilde{g}=(2,2,1,1,1), which lies below mm. Hence Proposition 2.5(1) entails gmg\mathrel{\swarrow}m. To prove the relation gmg\mathrel{\nearrow}m, we note that the unique upper cover of mm is m~=(3,2,1,1)\tilde{m}=(3,2,1,1), which is the conjugate of the unique lower cover of m=(4,3)m^{*}=(4,3), as explained in the proof of Lemma 4.6. Using the definition or inspecting Fig. 3, we observe gm~g\leq\tilde{m}. Then, as gmg\nleq m, Proposition 2.5(2) entails gmg\mathrel{\nearrow}m. We conclude that gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. Furthermore, since m=(4,3)m^{*}=(4,3) and g=(5,1,1)g^{*}=(5,1,1), Corollary 2.14 gives the second double arrow in a), namely (4,3)(5,1,1)(4,3)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(5,1,1).

Next, we verify b). We repeat the arguments of a) and get g:=(4,1,1,1)m:=(3,3,1)g\mathrel{\mathop{:}}=\left(4,1,1,1\right)\nleq m\mathrel{\mathop{:}}=\left(3,3,1\right) with g𝒥(7)g\in\mathcal{J}\left(\mathcal{L}_{7}\right) of type 3 and m(7)m\in\mathcal{M}\left(\mathcal{L}_{7}\right) of type 2. Applying the same transition rules yields the unique lower cover g~=(3,2,1,1)\tilde{g}=(3,2,1,1) and the unique upper cover m~=(4,2,1)\tilde{m}=(4,2,1). Also, the definition or inspecting Fig. 3 gives gm~g\leq\tilde{m} and g~m\tilde{g}\leq m. Therefore, since gmg\nleq m, Proposition 2.5 confirms gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. Furthermore, since m=(3,2,2)m^{*}=(3,2,2) and g=(4,1,1,1)g^{*}=(4,1,1,1), Corollary 2.14 gives the second double arrow in b), namely (3,2,2)(4,1,1,1)(3,2,2)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(4,1,1,1).

Finally, to prove g:=(3,3,1)m:=(3,2,2)g\mathrel{\mathop{:}}=(3,3,1)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m\mathrel{\mathop{:}}=(3,2,2), observe that g𝒥(7)g\in\mathcal{J}\left(\mathcal{L}_{7}\right) of type 4 and m(7)m\in\mathcal{M}\left(\mathcal{L}_{7}\right) of type 4. Since g1+g2=6>m1+m2g_{1}+g_{2}=6>m_{1}+m_{2}, we get gmg\nleq m. Furthermore, the unique lower cover g~=(3,2,2)\tilde{g}=(3,2,2) of gg is obtained by applying transition rule (2). This gives both g~=m\tilde{g}=m and g=m~g=\tilde{m} where m~\tilde{m} is the unique upper cover of mm. We conclude that gmg\nleq m, g~m\tilde{g}\leq m and gm~g\leq\tilde{m}, wherefore Proposition 2.5 entails gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. ∎

4.1 Double arrows involving completely join-irreducibles of type 1 or 2

The subsequent theorem exhibits the following one-to-one relation: For every g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 or 2 there is exactly one m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) such that gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. This unique mm has the form m=(a,1,,1)m=(a,1,\dotsc,1) and for each mm of this shape there is exactly one p𝒥(n)p\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 or 2 such that pmp\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, namely p=gp=g.

Theorem 4.10.

Let bb\in\mathbb{N}, k,,n+k,\ell,n\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k and set a=nba=n-b-\ell. Then g=(k+1,,k+1𝑏,k,,kb+)𝒥(n)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k})\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1 for b=0b=0 and of type 2 for b1b\geq 1, m=(a,1,,1b+,1)(n)m=(a,1,\dotsc,\overset{b+\ell}{1},1)\in\mathcal{M}\left(\mathcal{L}_{n}\right), and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m.

Moreover, the previously described mm are the only m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfying gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m for given g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 or 2.

Proof.

First, we consider the cases b+=1b+\ell=1 and k=1k=1 individually. If b+=1b+\ell=1, then we get =1\ell=1, b=0b=0, hence 1=b+<n=k1=b+\ell<n=k (since b=0b=0), and thus g=(k)=(n)g=(k)=(n) is of type 1 and m=(n1,1)m=(n-1,1). Hence the case b+=1b+\ell=1 was proved in Lemma 4.2. Moreover, if k=1k=1, then 0<nb=b0<n-b-\ell=b, i.e., b1b\geq 1, and gg is of type 2, wherefore this case was proved in Corollary 4.8.

Accordingly, it remains to show the statement for k2k\geq 2 and b+2b+\ell\geq 2. Then, as k2k\geq 2, we have g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) because it is completely join-irreducible of type 1 (for b=0b=0) and of type 2 (for b1b\geq 1). Moreover, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 3 as a=nb=b(k+1)+kb3b+2b=b+b+2a=n-b-\ell=b(k+1)+\ell k-b-\ell\geq 3b+2\ell-b-\ell=b+b+\ell\geq 2 due to k2k\geq 2 and b+2b+\ell\geq 2. Since gg is ‘shorter’ than mm, Lemma 3.9 asserts gmg\nleq m.

The unique lower cover of gg is g~=(k+1,,k+1𝑏,k,,k,k1b+,1)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},1), where the middle part k,,kk,\dotsc,k only appears for 2\ell\geq 2. The partition g~\tilde{g} is obtained by letting a brick fall from the cliff in the last position of gg (transition rule (2)). The unique upper cover of mm is m~=(a,2,1,,1b+)\tilde{m}=(a,2,1,\dotsc,\overset{b+\ell}{1}), where, to obtain mm, we can let a brick fall from the cliff in the second position if b+=2b+\ell=2, or let a brick slip across the slippery step if b+3b+\ell\geq 3 and the part 1,,11,\dotsc,1 is present. We shall prove that g~m\tilde{g}\leq m and gm~g\leq\tilde{m}; then Proposition 2.5 will entail gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. By Lemma 3.10, mm is the largest partition of length b++1=len(g~)b+\ell+1=\operatorname{len}(\tilde{g}); hence g~m\tilde{g}\leq m. By Corollary 3.11, m~\tilde{m} is the second largest partition of length b+=len(g)b+\ell=\operatorname{len}(g), while, according to Lemma 3.10, p:=(n(b+1),1,,1b+)p\mathrel{\mathop{:}}=(n-(b+\ell-1),1,\dotsc,\overset{b+\ell}{1}) is the largest. Since k2>1k\geq 2>1 we have gpg\neq p; hence gm~g\leq\tilde{m}.

That there are no other m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) in relation gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m with gg of type 1 or 2 will be proved in the subsequent lemmata, the proof being concluded with Corollary 4.16. ∎

Lemma 4.11.

Let bb\in\mathbb{N}, k,,n+k,\ell,n\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k and set g:=(k+1,,k+1𝑏,k,,kb+)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfies gmg\mathrel{\swarrow}m, then len(m)=b++1\operatorname{len}(m)=b+\ell+1 and the final entry of mm is mb++1=1m_{b+\ell+1}=1.

Proof.

Under the given conditions, gg is completely join-irreducible of type 1 or 2, see Corollary 3.14. Let us assume that gmg\mathrel{\swarrow}m with m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right). If len(m)b+\operatorname{len}(m)\leq b+\ell, then m𝒲n(b+)m\in\mathcal{W}_{n}(b+\ell), and thus Corollary 3.14 shows that gmg\leq m, in contradiction to gmg\mathrel{\swarrow}m. Therefore, len(m)b++1\operatorname{len}(m)\geq b+\ell+1. The unique lower cover g~\tilde{g} of gg satisfies len(g~)=b++1\operatorname{len}(\tilde{g})=b+\ell+1. By Proposition 2.5(1), our assumption gmg\mathrel{\swarrow}m entails g~m\tilde{g}\leq m, wherefore len(g~)<len(m)\operatorname{len}(\tilde{g})<\operatorname{len}(m) is impossible by Lemma 3.9. Consequently, we conclude len(m)len(g~)=b++1\operatorname{len}(m)\leq\operatorname{len}(\tilde{g})=b+\ell+1 and hence len(m)=len(g~)=b++1\operatorname{len}(m)=\operatorname{len}(\tilde{g})=b+\ell+1, which also yields mb++11m_{b+\ell+1}\geq 1. Moreover, we have n1=i=1b+g~ii=1b+mi=nmb++1n-1=\sum_{i=1}^{b+\ell}\tilde{g}_{i}\leq\sum_{i=1}^{b+\ell}m_{i}=n-m_{b+\ell+1} due to g~m\tilde{g}\leq m and g~b++1=1\tilde{g}_{b+\ell+1}=1; thus mb++11m_{b+\ell+1}\leq 1, and therefore mb++1=1m_{b+\ell+1}=1. ∎

Lemma 4.12.

Let bb\in\mathbb{N}, k,,n+k,\ell,n\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k and set g:=(k+1,,k+1𝑏,k,,kb+)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1, 2 or 3 and satisfies gmg\mathrel{\swarrow}m, then mm is of the form as described in Theorem 4.10.

Proof.

We assume that the given gg and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfy gmg\mathrel{\swarrow}m. From Lemma 4.11 we know that len(m)=b++1\operatorname{len}(m)=b+\ell+1 and that mb++1=1m_{b+\ell+1}=1. If mm is of type 1, 3, or 2 with len(m)=2\operatorname{len}(m)=2, then mm is of the form m=(a,1,,1b++1)m=(a,1,\dotsc,\overset{b+\ell+1}{1}) with a1a\geq 1. Since its total sum amounts to n=a+(b+)n=a+(b+\ell), we have a=nba=n-b-\ell and mm is exactly of the form described in Theorem 4.10.

Finally, let us consider the case that mm is of type 2 with len(m)3\operatorname{len}(m)\geq 3 and mm has the form m=(κ,,κb+,1)m=(\kappa,\dotsc,\overset{b+\ell}{\kappa},1) where b+2b+\ell\geq 2 and κ2\kappa\geq 2. Thus, if k=1k=1, then gmg\leq m by Corollary 3.2, which contradicts gmg\mathrel{\swarrow}m. Therefore, we know that k2k\geq 2. Hence the unique lower cover g~\tilde{g} of gg is obtained by letting a brick fall from the cliff in position b+2b+\ell\geq 2, and thus g~=(k+1,,k+1𝑏,k,,k,k1b+,1)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dots,k,\overset{b+\ell}{k-1},1) with its first entry satisfying g~1k\tilde{g}_{1}\geq k. Now gmg\mathrel{\swarrow}m and Proposition 2.5(1) entail g~m\tilde{g}\leq m, hence kg~1m1=κk\leq\tilde{g}_{1}\leq m_{1}=\kappa. Using this, we can finally estimate i=1b+1mi=nκ1nk1<nk=i=1b+1g~i\sum_{i=1}^{b+\ell-1}m_{i}=n-\kappa-1\leq n-k-1<n-k=\sum_{i=1}^{b+\ell-1}\tilde{g}_{i}, which implies the contradiction g~m\tilde{g}\nleq m. Therefore, the third case, where mm is of type 2 with len(m)3\operatorname{len}(m)\geq 3, is impossible and the lemma has been shown. ∎

Corollary 4.13.

Partitions g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 2 are not arrow-related to any m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 2.

Proof.

First, we prove that gg of type 2, mm of type 2 and gmg\mathrel{\swarrow}m is impossible. Note that any arrow relation for g=(k+1,,k+1𝑏,k,,kb+)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}) of type 2 requires len(m)>len(g)=b+\operatorname{len}(m)>\operatorname{len}(g)=b+\ell, because otherwise Corollary 3.14 yields m𝒲n(b+)m\in\mathcal{W}_{n}(b+\ell) and gmg\leq m, which by Proposition 2.5 prevents any arrow. Therefore, we have len(g)2\operatorname{len}(g)\geq 2 and thus len(m)3\operatorname{len}(m)\geq 3. The proof of Lemma 4.12 confirms that a relation gmg\mathrel{\swarrow}m with mm of type 2 with len(m)3\operatorname{len}(m)\geq 3 is impossible.

Finally, if gg of type 2 and mm of type 2 were such that gmg\mathrel{\nearrow}m, then the conjugates mm^{*} of type 2 and gg^{*} of type 2 would satisfy mgm^{*}\mathrel{\swarrow}g^{*} by Lemma 2.13, but this is impossible as shown before. ∎

The following simple observation will be used more than once.

Lemma 4.14.

Let a,t,r,ca,t,r,c\in\mathbb{N} be such that a>t>r0a>t>r\geq 0 and c2c\geq 2 and set n:=a+ct+rn\mathrel{\mathop{:}}=a+ct+r. In n\mathcal{L}_{n} we have the covering relation m:=(a,t,,tc+1,r)m~:=(a,t+1,t,,t,t1c+1,r)m\mathrel{\mathop{:}}=(a,t,\dotsc,\overset{c+1}{t},r)\prec\tilde{m}\mathrel{\mathop{:}}=(a,t+1,t,\dotsc,t,\overset{c+1}{t-1},r). Furthermore, m~\tilde{m} is the unique upper cover of mm.

Proof.

If c=2c=2, we apply transition rule (2) to the cliff in the second position of m~=(a,t+1,t1,r)\tilde{m}=(a,t+1,t-1,r), yielding m=(a,t,t,r){m=(a,t,t,r)}. If c3c\geq 3, we apply rule (1) to the slippery step in the second position of m~=(a,t+1,t,,t,t1,r){\tilde{m}=(a,t+1,t,\dotsc,t,t-1,r)}. In both cases the result is mm, and thus Lemma 2.10 yields mm~m\prec\tilde{m}. Moreover, mm is completely meet-irreducible of type 3 (if t=1t=1) and 4 (if t2t\geq 2), wherefore m~\tilde{m} is the unique upper cover. ∎

Lemma 4.15.

Let bb\in\mathbb{N}, k,,n+k,\ell,n\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k and set g:=(k+1,,k+1𝑏,k,,kb+)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4 and satisfies gmg\mathrel{\swarrow}m, then gmg\mathrel{\nearrow}m fails.

Proof.

We assume that m=(a,t,,t,r)m=(a,t,\dotsc,t,r) of type 4 satisfies gmg\mathrel{\swarrow}m. Then Lemma 4.11 implies that len(m)=b++1\operatorname{len}(m)=b+\ell+1 and that the last non-zero entry of mm equals 11. Hence if r=0r=0, then t=1t=1, which is forbidden for mm of type 4; we therefore know r1r\geq 1 and actually r=1r=1 by Lemma 4.11. We thus have m=(a,t,,tb+,1)m=(a,t,\dotsc,\overset{b+\ell}{t},1) with b+3b+\ell\geq 3 and a>t2a>t\geq 2. By Lemma 4.14, the unique upper cover of mm is m~=(a,t+1,t,,t,t1b+,1)\tilde{m}=(a,t+1,t,\dotsc,t,\overset{b+\ell}{t-1},1), where the part t,,tt,\dotsc,t appears as of b+4b+\ell\geq 4. Since len(g)=b+<b++1=len(m~)\operatorname{len}(g)=b+\ell<b+\ell+1=\operatorname{len}(\tilde{m}), we have gm~g\nleq\tilde{m} by Lemma 3.9. According to Proposition 2.5(2), this renders gmg\mathrel{\nearrow}m impossible. ∎

The following result shows that Theorem 4.10 indeed describes all double arrows gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m involving partitions gg of type 1 or 2.

Corollary 4.16.

Let bb\in\mathbb{N}, k,,n+k,\ell,n\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k and set g:=(k+1,,k+1𝑏,k,,kb+)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}). Every m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) in relation gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m is of the form described in Theorem 4.10.

Proof.

Under the assumption gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, Lemma 4.15 implies that mm is not of type 4. Therefore, it is of type 13, and Lemma 4.12 establishes the desired claim. ∎

4.2 Double arrows involving completely join-irreducibles of type 3

The following result exhibits further one-to-one double arrows gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, originating from any g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3, and covering two exceptional cases. These are (n)(n) and (2,1,,1)(2,1,\ldots,1), which are also discussed in Theorem 4.10.

Theorem 4.17.

Let n,k,t,rn,k,t,r\in\mathbb{N} be such that 2kn2\leq k\leq n and n=t(k1)+rn=t(k-1)+r where 0rk20\leq r\leq k-2. We have t1t\geq 1, g=(k,1,,1nk+1)𝒥(n)g=(k,1,\dotsc,\overset{n-k+1}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(k1,,k1𝑡,r)(n)m=(k-1,\dotsc,\overset{t}{k-1},r)\in\mathcal{M}\left(\mathcal{L}_{n}\right) and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m.

Moreover, if g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 3, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m holds, then mm must be of the shape as described before.

Proof.

The proof of gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m relies on showing mgm^{*}\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*} and then applying Corollary 2.14. The conjugate of mm is m=(t+1,,t+1𝑟,t,,tk1)m^{*}=(t+1,\dotsc,\overset{r}{t+1},t,\dotsc,\overset{k-1}{t}), the one of gg is g=(n(k1),1,,1𝑘)g^{*}=(n-(k-1),1,\dotsc,\overset{k}{1}). We infer that t1t\geq 1, for otherwise n=t(k1)+r=rk2n2n=t(k-1)+r=r\leq k-2\leq n-2. As rr\in\mathbb{N}, =k1r1\ell=k-1-r\geq 1 and k1n1<nk-1\leq n-1<n, we can apply Theorem 4.10 to infer mgm^{*}\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}g^{*}.

To show, for g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3 that there are no other double arrows will require a series of little results, being complete with Corollary 4.22. ∎

Lemma 4.18.

Let n,kn,k\in\mathbb{N} be such that 2kn2\leq k\leq n and set g:=(k,1,,1nk+1)g\mathrel{\mathop{:}}=(k,1,\dotsc,\overset{n-k+1}{1}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfies gmg\mathrel{\swarrow}m or gmg\mathrel{\nearrow}m, then m1=k1m_{1}=k-1.

Proof.

Let us assume that m1km_{1}\geq k. In this case Corollary 3.2 directly implies gmg\leq m, which, by Proposition 2.5, makes gmg\mathrel{\swarrow}m and gmg\mathrel{\nearrow}m impossible.

Let us now assume that m1k2m_{1}\leq k-2. We work with the first entries of the unique lower cover g~\tilde{g} of gg and the unique upper cover m~\tilde{m} of mm. We have ht(m)=m1k2<k1=g~1\operatorname{ht}(m)=m_{1}\leq k-2<k-1=\tilde{g}_{1}, hence g~m\tilde{g}\nleq m. Thus, by Proposition 2.5(1), gmg\mathrel{\swarrow}m is excluded. Likewise, we have ht(m~)m1+1k1<k=g1\operatorname{ht}(\tilde{m})\leq m_{1}+1\leq k-1<k=g_{1}, hence gm~g\nleq\tilde{m}. Now Proposition 2.5(2) excludes gmg\mathrel{\nearrow}m. ∎

Corollary 4.19.

Let n,kn,k\in\mathbb{N} be such that 2kn2\leq k\leq n and set g:=(k,1,,1nk+1)g\mathrel{\mathop{:}}=(k,1,\dotsc,\overset{n-k+1}{1}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1 or 2 and satisfies gmg\mathrel{\swarrow}m or gmg\mathrel{\nearrow}m, then mm is of the form as described in Theorem 4.17.

Lemma 4.20.

Partitions g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3 are not arrow-related to any m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 3.

Proof.

Let g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) be of type 3, that is g=(k,1,,1d+1)g=(k,1,\dotsc,\overset{d+1}{1}) with k3k\geq 3. Suppose that mm is of type 3, with gmg\mathrel{\swarrow}m or gmg\mathrel{\nearrow}m. Then m=(a,1,,1c+1)m=(a,1,\ldots,\overset{c+1}{1}) with a,c2a,c\geq 2, and Lemma 4.18 implies a=k1a=k-1 and c+1=d+2c+1=d+2. Then the unique upper cover of mm is m~=(k1,2,1,,1𝑐)\tilde{m}=(k-1,2,1,\dotsc,\overset{c}{1}) by Lemma 4.14. Hence m~g\tilde{m}\ngeq g, and Proposition 2.5(2) proves gmg\mathrel{\ooalign{$\mathrel{\nearrow}$\cr${\scriptstyle-\mkern 8.5mu}$}}m. Therefore gmg\mathrel{\swarrow}m. Now, the unique lower cover of gg is g~=(k1,2,1,,1d+1)\tilde{g}=(k-1,2,1,\dotsc,\overset{d+1}{1}) since k3k\geq 3. Then, we get len(g~)=d+1<c+1=len(m)\operatorname{len}(\tilde{g})=d+1<c+1=\operatorname{len}(m), and thus Lemma 3.9 yields g~m\tilde{g}\nleq m, contradicting gmg\mathrel{\swarrow}m by Proposition 2.5(1). ∎

Lemma 4.21.

Let n,kn,k\in\mathbb{N} be such that 2kn2\leq k\leq n and set g:=(k,1,,1nk+1)g\mathrel{\mathop{:}}=(k,1,\dotsc,\overset{n-k+1}{1}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 3 or 4, then gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m is impossible.

Proof.

By Lemma 4.20, gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m with mm of type 3 is impossible. Therefore, aiming for a contradiction, we assume that gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m with mm of type 4. Then len(m)3\operatorname{len}(m)\geq 3, and, by Lemma 4.18, we must have m1=k1m_{1}=k-1, i.e., m=(k1,t,,t,r)m=(k-1,t,\dotsc,\overset{\ell}{t},r) where 0r<t<k10\leq r<t<k-1, t2t\geq 2 and 3\ell\geq 3. Now, by Lemma 4.14, the upper cover of mm is m~=(k1,t+1,t,,t,t1,r)\tilde{m}=(k-1,t+1,t,\dotsc,t,t-1,r), implying that g1=k>k1=m~1g_{1}=k>k-1=\tilde{m}_{1}. Therefore, gm~g\nleq\tilde{m}, and consequently Proposition 2.5(2) excludes gmg\mathrel{\nearrow}m, in contradiction to gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. ∎

We can now finally observe that for gg of type 3 all double arrows gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m were characterised in the first part of Theorem 4.17.

Corollary 4.22.

Let n,kn,k\in\mathbb{N} be such that 2kn2\leq k\leq n and set g:=(k,1,,1nk+1)g\mathrel{\mathop{:}}=(k,1,\dotsc,\overset{n-k+1}{1}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfies gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, then mm is of the form as described in Theorem 4.17.

Proof.

Assuming gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, Lemma 4.21 yields that mm cannot be of type 4 nor of type 3. Therefore, mm is of type 1 or 2, whence Corollary 4.19 implies that mm is of the form as given in Theorem 4.17. ∎

4.3 Double arrows involving completely join-irreducibles of type 4

Theorem 4.23.

Let bb\in\mathbb{N}, k,d,+k,d,\ell\in\mathbb{N}_{+} with k3,b+2k\geq 3,b+\ell\geq 2 and set n:=b(k+1)+k+dn\mathrel{\mathop{:}}=b(k+1)+\ell k+d. Now, choose 2tmin(k1,d+1)2\leq t\leq\min(k-1,d+1), set a:=b+(b+)(kt)+t1a\mathrel{\mathop{:}}=b+(b+\ell)(k-t)+t-1 and decompose na=ct+rn-a=ct+r with 0r<t0\leq r<t and cc\in\mathbb{N}. Then we have g:=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)𝒥(n)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m:=(a,t,,tc+1,r)(n)m\mathrel{\mathop{:}}=(a,t,\dotsc,\overset{c+1}{t},r)\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4 and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m.

Moreover, if g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 4, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m holds, then mm must be of the shape as described before.

Note that if m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4, then its dual partition m𝒥(n)m^{*}\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 4. Since Theorem 4.23 exhibits double arrow relationships between \bigvee-irreducibles of type 4 and \bigwedge-irreducibles of type 4, the dual result of Theorem 4.23 under partition conjugation is contained within the original statement.

Moreover, Theorem 4.23 represents a one-to-many double arrow relationship, as can be observed from the example g(4,3,3)g\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(4,3,3) and g(5,2,2,1)g\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(5,2,2,1) where g=(4,4,1,1)g=(4,4,1,1) in both cases.

Proof.

First, we check that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4 by verifying the inequalities cb+2c\geq b+\ell\geq 2 and ak+bt+13a\geq k+b\geq t+1\geq 3. We have ak=b+(b+1)(kt)1=(b1+)(kt)+b1b+(b+2)ba-k=b+(b+\ell-1)(k-t)-1=(b-1+\ell)(k-t)+b-1\geq b+(b+\ell-2)\geq b since kt1k-t\geq 1 and b+2b+\ell\geq 2, hence ak+bkt+13a\geq k+b\geq k\geq t+1\geq 3. For the other inequality we see that

a+(b+1)t=b+(b+)k1=b(k+1)+k1=nd1nt=a+ct+rt<a+ct,a+(b+\ell-1)t=b+(b+\ell)k-1=b(k+1)+\ell k-1=n-d-1\leq n-t=a+ct+r-t<a+ct,

where we have used that d+1td+1\geq t and rt<0r-t<0. Thus, due to t>0t>0, we have b+1<cb+\ell-1<c, i.e., b+<c+1b+\ell<c+1 or 2b+c2\leq b+\ell\leq c. Hence mm is completely meet-irreducible of type 4. We also observe that gg is completely join-irreducible of type 4 since b+2b+\ell\geq 2, k3k\geq 3 and d,1d,\ell\geq 1.

We have that gmg\nleq m because summing up the first b+cb+\ell\leq c entries of gg gives b(k+1)+k=ndb(k+1)+\ell k=n-d, which is larger than the corresponding value of mm, namely a+(b+1)t=nd1a+(b+\ell-1)t=n-d-1.

The unique lower cover of gg is g~=(k+1,,k+1𝑏,k,,k,k1b+,2,1,,1b++d)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},2,1,\dotsc,\overset{b+\ell+d}{1}), in which the part k,,kk,\dotsc,k only appears for 2\ell\geq 2 and 1,,11,\dotsc,1 only appears for d2d\geq 2. We obtain g~\tilde{g} from gg by letting the brick in position b+b+\ell fall from the cliff (transition rule (2)). Since k1t2k-1\geq t\geq 2, Lemma 3.1 with position j:=b+j\mathrel{\mathop{:}}=b+\ell and common sum s=b(k+1)+k1=nd1=a+(b+1)ts=b(k+1)+\ell k-1=n-d-1=a+(b+\ell-1)t directly implies g~m\tilde{g}\leq m. Therefore, Proposition 2.5(1) yields gmg\mathrel{\swarrow}m.

By Lemma 4.14, the unique upper cover of mm is m~=(a,t+1,t,,t,t1c+1,r)\tilde{m}=(a,t+1,t,\dotsc,t,\overset{c+1}{t-1},r). Since b+cb+\ell\leq c, kt+1k\geq t+1 and 1t11\leq t-1, applying Lemma 3.1 with position j:=b+j\mathrel{\mathop{:}}=b+\ell and the common sum s=b(k+1)+k=a+(b+1)t+1s=b(k+1)+\ell k=a+(b+\ell-1)t+1 shows gm~g\leq\tilde{m}. Thus Proposition 2.5(2) yields gmg\mathrel{\nearrow}m.

That, for given g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4, there are no other m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) than the previous with gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m will follow from the following results, the proof being complete with Corollary 4.27. ∎

The following lemma in particular applies to completely join-irreducible elements gg of any type.

Lemma 4.24.

Let b,db,d\in\mathbb{N}, k,+k,\ell\in\mathbb{N}_{+} and set n:=b(k+1)+k+dn\mathrel{\mathop{:}}=b(k+1)+\ell k+d. For g:=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) we have that m^b+:=i=1b+mind\hat{m}_{b+\ell}\mathrel{\mathop{:}}=\sum_{i=1}^{b+\ell}m_{i}\geq n-d implies gmg\leq m.

Proof.

As k1k\geq 1, we have 1j:=b+b(k+1)+k=ndm^j1\leq j\mathrel{\mathop{:}}=b+\ell\leq b(k+1)+\ell k=n-d\leq\hat{m}_{j}. There are κ,r\kappa,r\in\mathbb{N} with r<jr<j such that m^j\hat{m}_{j} decomposes modulo jj as m^j=jκ+r\hat{m}_{j}=j\kappa+r. As r<jr<j, we have jk=bk+kb(k+1)+k=ndm^j=jκ+r<j(κ+1)jk=bk+\ell k\leq b(k+1)+\ell k=n-d\leq\hat{m}_{j}=j\kappa+r<j(\kappa+1), hence k<κ+1k<\kappa+1, i.e., kκk\leq\kappa. If k=κk=\kappa, then b+jk=b(k+1)+k=ndm^j=r+jκ=r+jkb+jk=b(k+1)+\ell k=n-d\leq\hat{m}_{j}=r+j\kappa=r+jk and so brb\leq r; otherwise κk+1\kappa\geq k+1. In both cases g˘:=(k+1,,k+1𝑏,k,,k𝑗)\breve{g}\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{j}{k}) lies pointwise below p:=(κ+1,,κ+1𝑟,κ,,κ𝑗)p\mathrel{\mathop{:}}=(\kappa+1,\dotsc,\overset{r}{\kappa+1},\kappa,\dotsc,\overset{j}{\kappa}). Now we have p,m˘:=(m1,,mj)𝒲m^j(j)p,\breve{m}\mathrel{\mathop{:}}=(m_{1},\dotsc,m_{j})\in\mathcal{W}_{\hat{m}_{j}}(j), and since m^j=κj+r\hat{m}_{j}=\kappa j+r with 0r<j0\leq r<j and 1jm^j1\leq j\leq\hat{m}_{j}, Lemma 3.13 implies pm˘p\leq\breve{m}. Hence we obtain i=1sgi=i=1sg˘ii=1spii=1sm˘i=i=1smi\sum_{i=1}^{s}g_{i}=\sum_{i=1}^{s}\breve{g}_{i}\leq\sum_{i=1}^{s}p_{i}\leq\sum_{i=1}^{s}\breve{m}_{i}=\sum_{i=1}^{s}m_{i} for each ss with 1sj1\leq s\leq j. If len(m)>j\operatorname{len}(m)>j, then for each j<ilen(m)j<i\leq\operatorname{len}(m) we have gi1mig_{i}\leq 1\leq m_{i}, and thus i=1sgi=i=1jgi+i=j+1sgii=1jmi+i=j+1smi=i=1smi\sum_{i=1}^{s}g_{i}=\sum_{i=1}^{j}g_{i}+\sum_{i=j+1}^{s}g_{i}\leq\sum_{i=1}^{j}m_{i}+\sum_{i=j+1}^{s}m_{i}=\sum_{i=1}^{s}m_{i} for any s>js>j. Therefore, we conclude gmg\leq m. ∎

Also the following lemma applies to any possible completely join-irreducible partition gg.

Lemma 4.25.

Let b,db,d\in\mathbb{N}, k,+k,\ell\in\mathbb{N}_{+} with k2k\geq 2 and set n:=b(k+1)+k+dn\mathrel{\mathop{:}}=b(k+1)+\ell k+d. If k=2k=2 and d1d\geq 1, we additionally assume b=0b=0. Define g:=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}) and let m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) be such that gmg\mathrel{\swarrow}m. Then m^b+:=i=1b+mi=n(d+1)<n\hat{m}_{b+\ell}\mathrel{\mathop{:}}=\sum_{i=1}^{b+\ell}m_{i}=n-(d+1)<n.

Proof.

Let j:=b+j\mathrel{\mathop{:}}=b+\ell and assume gmg\mathrel{\swarrow}m. By Lemma 4.24, we must have m^jnd1\hat{m}_{j}\leq n-d-1 because m^jnd\hat{m}_{j}\geq n-d would entail gmg\leq m, in contradiction to gmg\mathrel{\swarrow}m, see Proposition 2.5(1). On the other hand, we know from Proposition 2.5(1), that g~m\tilde{g}\leq m holds for the unique lower cover g~\tilde{g} of gg, wherefore, we have m^ji=1jg~i=:N\hat{m}_{j}\geq\sum_{i=1}^{j}\tilde{g}_{i}\mathrel{\mathopen{=}{\mathclose{:}}}N. It only remains to observe that the latter sum indeed amounts to n(d+1)n-(d+1). We do this by considering three different cases: if d=0d=0, then g=(k+1,,k+1𝑏,k,,k𝑗)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{j}{k}), g~=(k+1,,k+1𝑏,k,,k,k1𝑗,1)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{j}{k-1},1) and N=b(k+1)+k1=n1=nd1N=b(k+1)+\ell k-1=n-1=n-d-1. Next, if d1d\geq 1 and k=2k=2, then b=0b=0, j=j=\ell, n=k+d=2+dn=\ell k+d=2\ell+d and g=(2,,2,1,,1+d)g=(2,\dotsc,\overset{\ell}{2},1,\dotsc,\overset{\ell+d}{1}), g~=(2,,2j1,1,,1+d,1)\tilde{g}=(2,\dotsc,\overset{j-1}{2},1,\dotsc,\overset{\ell+d}{1},1) and N=21=b(k+1)+k1=nd1N=2\ell-1=b(k+1)+\ell k-1=n-d-1. The last case is d1d\geq 1 and k3k\geq 3. Now gg is of the general form given in the statement and its unique lover cover is g~=(k+1,,k+1𝑏,k,,k,k1b+,2,1,,1b++d)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},2,1,\dotsc,\overset{b+\ell+d}{1}) where the part k,,kk,\dotsc,k only appears for 2\ell\geq 2 and 1,,11,\dotsc,1 only for d2d\geq 2. Again, we get N=b(k+1)+k1=nd1N=b(k+1)+\ell k-1=n-d-1 as n=b(k+1)+k+dn=b(k+1)+\ell k+d. Therefore, we conclude that in all cases mentioned in the lemma the inequalities n>nd1m^jN=nd1n>n-d-1\geq\hat{m}_{j}\geq N=n-d-1, i.e., n>nd1=m^jn>n-d-1=\hat{m}_{j}, hold. ∎

The subsequent lemma shows in particular that if g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 4, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1, 2 or 3, then gmg\mathrel{\swarrow}m fails.

Lemma 4.26.

Let bb\in\mathbb{N}, k,,d+k,\ell,d\in\mathbb{N}_{+} be such that k3k\geq 3, b+2b+\ell\geq 2 and set n:=b(k+1)+k+dn\mathrel{\mathop{:}}=b(k+1)+\ell k+d. Consider g:=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}) and any m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right). If gmg\mathrel{\swarrow}m, then there are a,t,c,ra,t,c,r\in\mathbb{N} such that n=a+ct+rn=a+ct+r, m=(a,t,,tc+1,r)m=(a,t,\dotsc,\overset{c+1}{t},r), a>t>r0a>t>r\geq 0, c2c\geq 2, 2tk12\leq t\leq k-1 and a=b+(b+)(kt)+t1a=b+(b+\ell)(k-t)+t-1. In particular, mm is of type 4. Moreover, if mm is such that t>d+1t>d+1, then r=d+12r=d+1\geq 2, b+=c+1b+\ell=c+1 and gmg\mathrel{\nearrow}m fails.

Proof.

Suppose that gmg\mathrel{\swarrow}m; then Proposition 2.5(1) entails g~m\tilde{g}\leq m for the unique lower cover g~\tilde{g} of gg. Since b+2b+\ell\geq 2, we thus have g1=g~1m1=:hg_{1}=\tilde{g}_{1}\leq m_{1}\mathrel{\mathopen{=}{\mathclose{:}}}h. If mm were of type 1 or 2, then m=(h,,h,r)m=(h,\dotsc,h,r) for some 0r<h0\leq r<h, and thus mm would be the largest member of n(h)\mathcal{H}_{n}(h) (see Lemma 3.12). Since ht(g)=g1h\operatorname{ht}(g)=g_{1}\leq h, we would have gn(h)g\in\mathcal{H}_{n}(h), hence gmg\leq m, contradicting gmg\mathrel{\swarrow}m by Proposition 2.5(1).

Therefore, mm is neither of type 1 nor 2, thus is must be of type 3 or 4. Hence it has the form m=(a,t,,tc+1,r)m=(a,t,\dotsc,\overset{c+1}{t},r) where a>t>r0a>t>r\geq 0 and c2c\geq 2. The total sum of mm being nn, we note that n=a+ct+rn=a+ct+r. If tk+1t\geq k+1, then a>tk+1>k3>1a>t\geq k+1>k\geq 3>1 and hence gmg\leq m by Corollary 3.2. By Proposition 2.5(1), this is in contradiction to gmg\mathrel{\swarrow}m, wherefore tkt\leq k. As k3k\geq 3, Lemma 4.25 shows that the partial sum of mm up to position b+b+\ell is m^b+=n(d+1)<n\hat{m}_{b+\ell}=n-(d+1)<n. Since m^b+<n\hat{m}_{b+\ell}<n, we must have b+c+1b+\ell\leq c+1, hence m^b+=a+(b+1)t=n(d+1)=a+ct+r(d+1)\hat{m}_{b+\ell}=a+(b+\ell-1)t=n-(d+1)=a+ct+r-(d+1) and m^b+1=n(d+1)t\hat{m}_{b+\ell-1}=n-(d+1)-t. Since gmg\mathrel{\swarrow}m, Proposition 2.5(1) yields g~m\tilde{g}\leq m for the unique lower cover g~\tilde{g} of gg. Thus, n(d+k)=ni=b+b++dgi=ni=b+b++dg~i=i=1b+1g~im^b+1=n(d+1)tn-(d+k)=n-\sum_{i=b+\ell}^{b+\ell+d}g_{i}=n-\sum_{i=b+\ell}^{b+\ell+d}\tilde{g}_{i}=\sum_{i=1}^{b+\ell-1}\tilde{g}_{i}\leq\hat{m}_{b+\ell-1}=n-(d+1)-t, i.e., tk1t\leq k-1. Since t>r0t>r\geq 0, we have t1t\geq 1. With the intent to derive a contradiction, we assume that t=1t=1, hence m=(a,1,,1)m=(a,1,\dotsc,1). We know that g~=(k+1,,k+1𝑏,k,,k,k1b+,2,1,,1b++d)m\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},2,1,\dotsc,\overset{b+\ell+d}{1})\leq m, thus d1d\geq 1 and Lemma 3.9 imply len(g)=len(g~)len(m)\operatorname{len}(g)=\operatorname{len}(\tilde{g})\geq\operatorname{len}(m); hence our assumption t=1t=1 and Lemma 3.10 yield gmg\leq m, contradicting gmg\mathrel{\swarrow}m by Proposition 2.5(1). Therefore, 2tk12\leq t\leq k-1, and mm has type 4. Moreover, from m^b+=a+(b+1)t=n(d+1)\hat{m}_{b+\ell}=a+(b+\ell-1)t=n-(d+1) we infer

a\displaystyle a =n(d+1)(b+1)t=(b(k+1)+k+d)d1(b+)t+t\displaystyle=n-(d+1)-(b+\ell-1)t=(b(k+1)+\ell k+d)-d-1-(b+\ell)t+t
=b+b(kt)+(kt)+t1=b+(b+)(kt)+t1.\displaystyle=b+b(k-t)+\ell(k-t)+t-1=b+(b+\ell)(k-t)+t-1.

From m^b+=a+(b+1)t=a+ct+r(d+1)\hat{m}_{b+\ell}=a+(b+\ell-1)t=a+ct+r-(d+1) we infer that d+1+(b+1)t=ct+rd+1+(b+\ell-1)t=ct+r and hence d+1r(modt)d+1\equiv r\pmod{t}. If we now additionally assume that d+1<td+1<t, then d+1=rd+1=r, and we may cancel this common summand from both sides of the equation, leading to (b+1)t=ct(b+\ell-1)t=ct, or b+1=cb+\ell-1=c, i.e., b+=c+1b+\ell=c+1. Consequently, we have m=(a,t,,tb+,d+1)m=(a,t,\dotsc,\overset{b+\ell}{t},d+1), being of type 4 with d+12d+1\geq 2, and thus the unique upper cover of mm is m~=(a,t+1,t,,t,t1b+,d+1)\tilde{m}=(a,t+1,t,\dotsc,t,\overset{b+\ell}{t-1},d+1), see Lemma 4.14. Therefore, we obtain i=1b+m~i=n(d+1)<nd=b(k+1)+k=i=1b+gi\sum_{i=1}^{b+\ell}\tilde{m}_{i}=n-(d+1)<n-d=b(k+1)+\ell k=\sum_{i=1}^{b+\ell}g_{i} and hence gm~g\nleq\tilde{m}. By Proposition 2.5(2) gmg\mathrel{\nearrow}m fails. ∎

The following corollary shows that all double arrows gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m involving g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4 were described in Theorem 4.23. In particular, dualising the statements of Theorem 4.23 yields double arrow relations that are already described as a part of that theorem.

Corollary 4.27.

Let bb\in\mathbb{N}, k,d,+k,d,\ell\in\mathbb{N}_{+} be such that k3k\geq 3, b+2b+\ell\geq 2 and define n:=b(k+1)+k+dn\mathrel{\mathop{:}}=b(k+1)+\ell k+d, as well as g:=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)g\mathrel{\mathop{:}}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}). If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfies gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, then mm is of type 4 of the shape as shown in Theorem 4.23.

Proof.

Let us assume that gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. According to Lemma 4.26, the partition mnm\in\mathcal{L}_{n} is of the form m=(a,t,,tc+1,r)m=(a,t,\dotsc,\overset{c+1}{t},r) with a>t>r0a>t>r\geq 0, c2c\geq 2, na=ct+rn-a=ct+r, 2tk12\leq t\leq k-1, and a=b+(b+)(kt)+t1a=b+(b+\ell)(k-t)+t-1. Moreover, if t>d+1t>d+1, then gmg\mathrel{\nearrow}m would fail by Lemma 4.26, contradicting gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m. Therefore, we have 2tmin(k1,d+1)2\leq t\leq\min(k-1,d+1), whence mm is of type 4 with exactly the shape as given in Theorem 4.23. ∎

We summarise our understanding of the double arrow relation in Table 1.

Result g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) Arrow m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right)
Theorem 4.10: (k,,k)(k,\dotsc,\overset{\ell}{k}),  k2k\geq 2 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (n,1,,1,1)(n-\ell,1,\dotsc,\overset{\ell}{1},1)
Theorem 4.10: (k+1,,k+1𝑏,k,,kb+)(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}),  k1k\geq 1 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (nb,1,,1b+,1)(n-b-\ell,1,\dotsc,\overset{b+\ell}{1},1)
Theorem 4.17: (k,1,,1nk+1)(k,1,\dotsc,\overset{n-k+1}{1}),  k2k\geq 2 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (k1,,k1𝑐,r)(k-1,\dotsc,\overset{c}{k-1},r)
Theorem 4.23: (k,,k,1,,1+d)(k,\dotsc,\overset{\ell}{k},1,\dotsc,\overset{\ell+d}{1}),  k3k\geq 3, 2\ell\geq 2 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (a,t,,tc+1,r)(a,t,\dotsc,\overset{c+1}{t},r),  c2c\geq 2
Theorem 4.23: (k+1,,k+1𝑏,k,,kb+,1,,1b++d)(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}),  k3k\geq 3 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (a,t,,tc+1,r)(a,t,\dotsc,\overset{c+1}{t},r),  c2c\geq 2
Table 1: A summary of the \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} characterisation results where b,,d,c,a,t+b,\ell,d,c,a,t\in\mathbb{N}_{+} with ak>t>r0a\geq k>t>r\geq 0 and t2t\geq 2. The precise value of aa is stated in Theorem 4.23.

5 Results regarding single arrows

Since results for up-arrows can be deduced from down-arrows by partition conjugation and Lemma 2.13, we focus here on the complete description of the relation gmg\mathrel{\swarrow}m where g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right). As, by definition, gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m implies gmg\mathrel{\swarrow}m, the results from Section 4 yield part of the description of the down-arrows. Subsequently, we need to concentrate on finding the remaining ‘single’ down-arrows, where gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m. Our strategy will be to derive necessary conditions that follow from gmg\mathrel{\swarrow}m, to isolate the cases that were already covered in Section 4, and then to prove in the remaining cases from a subset of the necessary conditions that gmg\mathrel{\swarrow}m actually holds. We organise our investigations according to the different types g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) may have.

For g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 or 2, Lemma 4.12 confirms that all down-arrows involving mm of types 1, 2 and 3 are contained in Theorem 4.10 and hence are double arrows. Therefore, we here only discuss down-arrows between gg of type 1 or 2 and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4. In this respect, Lemma 5.1 gives necessary conditions for the down-arrows; these necessary conditions will be shown to be sufficient in Theorem 5.2.

Lemma 5.1.

Let bb\in\mathbb{N}, n,k,+n,k,\ell\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k. The partition g=(k+1,,k+1𝑏,k,,kb+)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}) is of type 1 or 2. If m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4 and satisfies gmg\mathrel{\swarrow}m, then k,b+3k,b+\ell\geq 3 and m=(a,t,,tb+,1)m=(a,t,\dotsc,\overset{b+\ell}{t},1) with a>ta>t and 2tk12\leq t\leq k-1.

Proof.

Lemma 4.11 yields len(m)=b++1\operatorname{len}(m)=b+\ell+1 and that m=(a,t,,t,r)m=(a,t,\dotsc,t,r) with a>t>r0a>t>r\geq 0 and t2t\geq 2 must end with 11. If r=0r=0, then this implies the contradiction t=1t=1, hence r1r\geq 1 and thus r=1r=1 by Lemma 4.11. Therefore, m=(a,t,,tb+,1)m=(a,t,\dotsc,\overset{b+\ell}{t},1) with a>t2a>t\geq 2, and since two values tt must appear, we obtain b+3b+\ell\geq 3. Finally, we prove tk1t\leq k-1 in two steps. First, we cannot have tk+1t\geq k+1 because under this assumption the partial sum of mm exceeds (k+1)(b+)+1>n(k+1)(b+\ell)+1>n as a>tk+1a>t\geq k+1. Thus, we have 2tk2\leq t\leq k, i.e., k2k\geq 2, wherefore gg has a cliff in position b+b+\ell, and by transition rule (2) the unique lower cover of gg is g~=(k+1,,k+1𝑏,k,,k,k1b+,1)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},1). Our assumption gmg\mathrel{\swarrow}m and Proposition 2.5(1) imply g~m\tilde{g}\leq m, hence we get nk=i=1b+1g~ii=1b+1mi=nt1n-k=\sum_{i=1}^{b+\ell-1}\tilde{g}_{i}\leq\sum_{i=1}^{b+\ell-1}m_{i}=n-t-1, i.e., tk1t\leq k-1. Therefore, kt+13k\geq t+1\geq 3, finishing the proof. ∎

We now show that the conditions in Lemma 5.1 are sufficient for a down-arrow between gg of type 1 or 2 and mm of type 4, and that this arrow is not a double arrow.

Theorem 5.2.

Let bb\in\mathbb{N}, k,t,+k,t,\ell\in\mathbb{N}_{+} with k,b+3k,b+\ell\geq 3 and 2tk12\leq t\leq k-1; set n=b(k+1)+kn=b(k+1)+\ell k and a=n1(b+1)ta=n-1-(b+\ell-1)t. Then the partitions g=(k+1,,k+1𝑏,k,,kb+)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k}) and m=(a,t,,tb+,1)m=(a,t,\dotsc,\overset{b+\ell}{t},1) satisfy g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) being of type 1 or 2, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) being of type 4 with at+2a\geq t+2, and gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m.

Proof.

As k3k\geq 3 we have that g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1 (if b=0b=0) or 2 (if b1b\geq 1). In Lemma 5.1 we saw that the unique lower cover of gg is g~=(k+1,,k+1𝑏,k,,k,k1b+,1)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},1). Since gg is shorter than mm, Lemma 3.9 yields gmg\nleq m. Next, we verify that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4. First, given b+3b+\ell\geq 3, we get that tt appears (b+1)2(b+\ell-1)\geq 2 times. Next, we have mnm\in\mathcal{L}_{n} since a=n1(b+1)ta=n-1-(b+\ell-1)t. Furthermore, from n=(b+)(k1)+2b+n=(b+\ell)(k-1)+2b+\ell, tk1t\leq k-1 and b+3b+\ell\geq 3, we get at=n(b+)t1n(b+)(k1)1=2b+1=(b+)+b12a-t=n-(b+\ell)t-1\geq n-(b+\ell)(k-1)-1=2b+\ell-1=(b+\ell)+b-1\geq 2, hence at+2>ta\geq t+2>t and mm is of type 4.

We now prove g~m\tilde{g}\leq m. Note that the partial sums of both g~\tilde{g} and mm reach n1n-1 at position b+b+\ell. Therefore, since g~ik1t=mi\tilde{g}_{i}\geq k-1\geq t=m_{i} for 2ib+2\leq i\leq b+\ell, Lemma 3.1 with j=b+j=b+\ell gives g~m\tilde{g}\leq m. Hence Proposition 2.5(1) yields gmg\mathrel{\swarrow}m. Finally, as nk(b+)3(b+)>b+n\geq k(b+\ell)\geq 3(b+\ell)>b+\ell, Lemma 4.15 shows that gmg\mathrel{\nearrow}m in Theorem 5.2 is impossible. ∎

Lemmata 4.12, 5.1 and Theorem 5.2 characterise all m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) in relation gmg\mathrel{\swarrow}m with gg of type 1 or 2.

We are now starting to deal with gmg\mathrel{\swarrow}m for partitions g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3. Theorem 4.17 and Corollary 4.19 cover the case when m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1 or 2. Here the down-arrow automatically is a double arrow. We therefore focus in the subsequent two results on mm of types 3 or 4. As done earlier we first derive necessary conditions from gmg\mathrel{\swarrow}m, which we then prove to be sufficient for a down-arrow that is not a double arrow.

Lemma 5.3.

Let k,d+k,d\in\mathbb{N}_{+} with k3k\geq 3, set n=k+dn=k+d and let g=(k,1,,11+d)𝒥(n)g=(k,1,\dotsc,\overset{1+d}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right). Suppose gmg\mathrel{\swarrow}m with m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 3 or 4, then k4k\geq 4, m=(k1,t,,tc+1,r)m=(k-1,t,\dotsc,\overset{c+1}{t},r) for r,t,cr,t,c\in\mathbb{N} with 0r<tk20\leq r<t\leq k-2, t2t\geq 2, c2c\geq 2, i.e., mm is of type 4, and d+12td+1\geq 2t.

Proof.

By Lemma 4.18, mm starts with m1=k1=n(d+1)m_{1}=k-1=n-(d+1). Since mm is of type 3 or 4, we have m=(m1,t,,tc+1,r)m=(m_{1},t,\dotsc,\overset{c+1}{t},r) where k1=m1>t>r0k-1=m_{1}>t>r\geq 0 and c2c\geq 2. From gmg\mathrel{\swarrow}m we infer by Proposition 2.5(1) that g~m\tilde{g}\leq m, where g~\tilde{g} denotes the unique lower cover of gg. Hence we obtain k+1=(k1)+2=g~1+g~2m1+m2=k1+tk+1=(k-1)+2=\tilde{g}_{1}+\tilde{g}_{2}\leq m_{1}+m_{2}=k-1+t, i.e., t2t\geq 2 and mm is of type 4. As m1=n(d+1)m_{1}=n-(d+1), the total sum of mm equals n=n(d+1)+ct+rn=n-(d+1)+ct+r, thus d+1=ct+rct2td+1=ct+r\geq ct\geq 2t since r0r\geq 0 and c2c\geq 2. Finally, we get k4k\geq 4 from 2tk22\leq t\leq k-2. ∎

Theorem 5.4.

Let k,d,t+k,d,t\in\mathbb{N}_{+} with k4k\geq 4, 2tk22\leq t\leq k-2, 2td+12t\leq d+1 and set n=k+dn=k+d. Let us decompose n(k1)=ct+rn-(k-1)=ct+r with c,rc,r\in\mathbb{N} such that 0r<t0\leq r<t. Then c2c\geq 2, and for g=(k,1,,1d+1)g=(k,1,\dotsc,\overset{d+1}{1}) and m=(k1,t,,tc+1,r)m=(k-1,t,\dotsc,\overset{c+1}{t},r) we have g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) being of type 3, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) being of type 4, and gmg\mathrel{\swarrow}m, but not gmg\mathrel{\nearrow}m.

Observe that g=(5,1,1,1,1,1)g=(5,1,1,1,1,1) satisfies g(4,3,3)g\mathrel{\swarrow}(4,3,3) and g(4,2,2,2)g\mathrel{\swarrow}(4,2,2,2). Hence Theorem 5.4 represents a one-to-many relationship. The number of distinct m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) such that gmg\mathrel{\swarrow}m with a fixed gg grows with nn, e.g., if n=16n=16 and g=(7,1,,1)g=(7,1,\dotsc,1), then gmg\mathrel{\swarrow}m for m{(6,5,5),(6,4,4,2),(6,3,3,3,1),(6,2,2,2,2,2)}m\in\left\{(6,5,5),(6,4,4,2),(6,3,3,3,1),(6,2,2,2,2,2)\right\}.

Proof.

First, we check that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right). By the definition of cc and rr, we have mnm\in\mathcal{L}_{n}. Moreover, since t>rt>r, n=k+dn=k+d and d+12td+1\geq 2t, we obtain (c1)t>ct+r2t=n(k1)2t=d+12t0(c-1)t>ct+r-2t=n-(k-1)-2t=d+1-2t\geq 0, which yields c1>0c-1>0, i.e., c2c\geq 2. Thus, mm is completely meet-irreducible of type 4. We have gmg\nleq m as the first entries are g1=k>k1=m1g_{1}=k>k-1=m_{1}. Since k4k\geq 4 and d1d\geq 1, we observe that gg is completely join-irreducible of type 3 and that its unique lower cover is g~=(k1,2,1,,1d+1)\tilde{g}=(k-1,2,1,\dotsc,\overset{d+1}{1}), which is obtained from gg by letting the brick in position 11 fall from the cliff (transition rule (2)). Since m1=k1=g~1m_{1}=k-1=\tilde{g}_{1} and mi=t2>1>0m_{i}=t\geq 2>1>0 for 2ic+12\leq i\leq c+1, we obtain g~m\tilde{g}\leq m from Corollary 3.2. Thus, Proposition 2.5(1) confirms gmg\mathrel{\swarrow}m.

Finally, by Lemma 4.14, we get m~=(k1,t+1,t,,t,t1c+1,r)\tilde{m}=(k-1,t+1,t,\dotsc,t,\overset{c+1}{t-1},r) for the upper cover of mm, which means that m~1=k1<k=g1\tilde{m}_{1}=k-1<k=g_{1} and thus m~g\tilde{m}\ngeq g. Hence gmg\mathrel{\nearrow}m becomes impossible by Proposition 2.5(2). ∎

Corollary 4.19, Lemma 5.3 and Theorem 5.4 characterise all m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) in relation gmg\mathrel{\swarrow}m with gg of type 3. Dualising Theorem 5.4 with the help of Lemmata 2.12 and 2.13 yields the following result.

Corollary 5.5.

Let k,d,t+k,d,t\in\mathbb{N}_{+} with k4k\geq 4, 2tk22\leq t\leq k-2 and 2td+12t\leq d+1 and set n=k+dn=k+d. Moreover, decompose n(k1)=ct+rn-(k-1)=ct+r with c,rc,r\in\mathbb{N} such that 0r<t0\leq r<t. Then c2c\geq 2, and abbreviating κ:=c+1\kappa\mathrel{\mathop{:}}=c+1 we have m=(κ+1,,κ+1𝑟,κ,,κ𝑡,1,,1k1)g=(d+1,1,,1𝑘)m^{*}=\big{(}\kappa+1,\dotsc,\overset{r}{\kappa+1},\kappa,\dotsc,\overset{t}{\kappa},1,\dotsc,\overset{k-1}{1}\big{)}\mathrel{\nearrow}g^{*}=\big{(}d+1,1,\dotsc,\overset{k}{1}\big{)}, but not mgm^{*}\mathrel{\swarrow}g^{*}, where mm^{*} is of type 4 and gg^{*} is of type 3.

It remains to discuss g=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)𝒥(n)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dots,\overset{b+\ell+d}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4 where bb\in\mathbb{N}, k,,d+k,\ell,d\in\mathbb{N}_{+}, k3k\geq 3, b+2b+\ell\geq 2 and n=b(k+1)+k+dn=b(k+1)+\ell k+d. Then we may deduce via Lemma 4.26 from gmg\mathrel{\swarrow}m for some m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) that m=(a,t,,tc+1,r)nm=(a,t,\dotsc,\overset{c+1}{t},r)\in\mathcal{L}_{n} is of type 4 with n=a+ct+rn=a+ct+r, a>t>r0a>t>r\geq 0, c2c\geq 2, a=b+(b+)(kt)+t1a=b+(b+\ell)(k-t)+t-1 and 2tk12\leq t\leq k-1. In this situation there are two cases: the first is that td+1t\leq d+1, in which Theorem 4.23 entails that gmg\mathrel{\swarrow}m is not only present, but that it is actually a double arrow. The second case is that d+1<td+1<t, in which Lemma 4.26 implies t>r=d+12t>r=d+1\geq 2, b+=c+13b+\ell=c+1\geq 3 and that gmg\mathrel{\nearrow}m is impossible. Under these necessary conditions, the following result shows that gmg\mathrel{\swarrow}m is actually present (and clearly is not a double arrow).

Theorem 5.6.

Let bb\in\mathbb{N}, k,,d,t+k,\ell,d,t\in\mathbb{N}_{+} with b+3b+\ell\geq 3 and d+1<tk1d+1<t\leq k-1; set n=b(k+1)+k+dn=b(k+1)+\ell k+d and a=b+(b+)(kt)+t1a=b+(b+\ell)(k-t)+t-1, and consider g=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}) and m=(a,t,,tb+,d+1)m=(a,t,\dotsc,\overset{b+\ell}{t},d+1). Then we have g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4 with k4k\geq 4, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4 with at+2a\geq t+2, and gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m.

Proof.

Since d+2tk1d+2\leq t\leq k-1 and d1d\geq 1, we have t3t\geq 3 and kd+34k\geq d+3\geq 4. Moreover, we observe

a+(b+1)t+d+1=b+(b+)(kt)+t1+(b+)tt+d+1=b+(b+)k+d=b+bk+k+d=b(k+1)+k+d=n,a+(b+\ell-1)t+d+1=b+(b+\ell)(k-t)+t-1+(b+\ell)t-t+d+1=b+(b+\ell)k+d\\ =b+bk+\ell k+d=b(k+1)+\ell k+d=n,

whence mnm\in\mathcal{L}_{n}. We check that a=b+(b+)(kt)+t1b+b++t10+3+t1=t+2>ta=b+(b+\ell)(k-t)+t-1\geq b+b+\ell+t-1\geq 0+3+t-1=t+2>t as kt1k-t\geq 1 and b+3b+\ell\geq 3. Hence, as a>t>d+12a>t>d+1\geq 2 and b+3b+\ell\geq 3, we conclude that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 4. Since k3k\geq 3, ,d1\ell,d\geq 1 and b+2b+\ell\geq 2, it is also clear that g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4. Furthermore, we have gmg\nleq m because the partial sums up to position b+b+\ell satisfy g^b+=nd>n(d+1)=m^b+\hat{g}_{b+\ell}=n-d>n-(d+1)=\hat{m}_{b+\ell}. Applying transition rule (2) to the cliff in position b+b+\ell of gg, we obtain its unique lower cover g~=(k+1,,k+1𝑏,k,,k,k1b+,2,1,,1b++d)\tilde{g}=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,k,\overset{b+\ell}{k-1},2,1,\dotsc,\overset{b+\ell+d}{1}) as d1d\geq 1, with 1,,11,\dotsc,1 appearing for d2d\geq 2. Note that the partial sums of both g~\tilde{g} and mm reach n(d+1)n-(d+1) at position b+b+\ell. Therefore, as g~ik1t=mi\tilde{g}_{i}\geq k-1\geq t=m_{i} for 2ib+2\leq i\leq b+\ell, Lemma 3.1 with j=b+j=b+\ell yields g~m\tilde{g}\leq m, and hence Proposition 2.5(1) implies gmg\mathrel{\swarrow}m.

Finally, since k3k\geq 3, b+2b+\ell\geq 2, m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right), gmg\mathrel{\swarrow}m and t>d+1t>d+1, Lemma 4.26 implies that gmg\mathrel{\nearrow}m is impossible. ∎

Lemma 4.26, Corollary 4.27 and Theorem 5.6 characterise all m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) in relation gmg\mathrel{\swarrow}m with gg of type 4.

The following summary combines Theorems 5.2 and 5.6, covering gg of types 1, 2 or 4.

Corollary 5.7.

Let b,db,d\in\mathbb{N}, k,+k,\ell\in\mathbb{N}_{+} with b+3b+\ell\geq 3 and kd+3k\geq d+3; set n=b(k+1)+k+dn=b(k+1)+\ell k+d. Furthermore, choose tt\in\mathbb{N} with d+1<tk1d+1<t\leq k-1 and set a:=b+(b+)(kt)+t1a\mathrel{\mathop{:}}=b+(b+\ell)(k-t)+t-1. Then we have g=(k+1,,k+1𝑏,k,,kb+,1,,1b++d)𝒥(n)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{\hbox to0.0pt{$\scriptstyle b+\ell+d$\hss}}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right), m=(a,t,,tb+,d+1)(n){m=(a,t,\dotsc,\overset{b+\ell}{t},d+1)\in\mathcal{M}\left(\mathcal{L}_{n}\right)} of type 4 with at+2a\geq t+2, and gg and mm satisfy gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m.

Observe that g=(4,4,4)g=(4,4,4) satisfies g(5,3,3,1)g\mathrel{\swarrow}(5,3,3,1) and g(7,2,2,1)g\mathrel{\swarrow}(7,2,2,1). Hence Corollary 5.7 represents a one-to-many relationship. Similarly to Theorem 5.4 the number of distinct m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) such that gmg\mathrel{\swarrow}m with a fixed gg grows with nn, e.g., if n=15n=15 and g=(5,5,5)g=(5,5,5), then gmg\mathrel{\swarrow}m for m{(10,2,2,1),(8,3,3,1),(6,4,4,1)}m\in\left\{(10,2,2,1),(8,3,3,1),(6,4,4,1)\right\}.

Proof.

Note that m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is always of type 4 while g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1 (if b=d=0b=d=0), of type 2 (if b1b\geq 1, d=0d=0), of type 4 (if d1d\geq 1).

For d=0d=0 the hypotheses and partitions gg and mm simplify to those of Theorem 5.2, since d+1<td+1<t is equivalent to 2t2\leq t. Therefore, Theorem 5.2 proves gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m.

For d1d\geq 1 the hypotheses restate those of Theorem 5.6. Hence gmg\mathrel{\swarrow}m but not gmg\mathrel{\nearrow}m. ∎

Dualising Corollary 5.7 via Lemmata 2.12 and 2.13 produces the following result.

Corollary 5.8.

Let b,db,d\in\mathbb{N}, k,,t+k,\ell,t\in\mathbb{N}_{+} with b+3b+\ell\geq 3, kd+3k\geq d+3 and d+1<tk1d+1<t\leq k-1; further set n=b(k+1)+k+dn=b(k+1)+\ell k+d and a=b+(b+)(kt)+t1a=b+(b+\ell)(k-t)+t-1. Then at+2a\geq t+2, and abbreviating κ:=b+\kappa\mathrel{\mathop{:}}=b+\ell we have

m=(κ+1,,κ+1d+1,κ,,κ𝑡,1,,1𝑎)g=(κ+d,κ,,κ𝑘,b),but notmg,m^{*}=\big{(}\kappa+1,\dotsc,\overset{d+1}{\kappa+1},\kappa,\dotsc,\overset{t}{\kappa},1,\dotsc,\overset{a}{1}\big{)}\mathrel{\nearrow}g^{*}=\big{(}\kappa+d,\,\kappa,\dotsc,\overset{k}{\kappa},\,b\big{)},\quad\text{but not}\quad m^{*}\mathrel{\swarrow}g^{*},

where m𝒥(n)m^{*}\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 4 and g(n)g^{*}\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1 (if b=d=0b=d=0), 2 (if d=0,b1d=0,b\geq 1) or 4 (if d1d\geq 1).

Refer to caption
Fig. 6: Lattice n\mathcal{L}_{n} for n=11n=11 with all arrows.

We finally collect our knowledge about single arrows in Table 2. Moreover, with Fig. 6 depicting 11\mathcal{L}_{11} we aim to illustrate for the concrete case n=11n=11 the various up, down and double arrow relations appearing in Tables 1 and 2.

Result g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) Arrow m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right)
Theorem 5.4: (k,1,,1d+1)(k,1,\dotsc,\overset{d+1}{1}),  k4k\geq 4, d2t1d\geq 2t-1 ,\mathrel{\swarrow},\mathrel{\ooalign{$\mathrel{\nearrow}$\cr${\scriptstyle-\mkern 8.5mu}$}} (k1,t,,tc+1,r)(k-1,t,\dotsc,\overset{c+1}{t},r)
Corollary 5.5: (κ+1,,κ+1𝑟,κ,,κ𝑡,1,,1k1)\big{(}\kappa+1,\dotsc,\overset{r}{\kappa+1},\kappa,\dotsc,\overset{t}{\kappa},1,\dotsc,\overset{k-1}{1}\big{)},  κ:=c+1\kappa\mathrel{\mathop{:}}=c+1 ,\mathrel{\nearrow},\mathrel{\ooalign{$\mathrel{\swarrow}$\cr\raise 1.29167pt\hbox{${\scriptstyle-\mkern 5.0mu}$}}} (d+1,1,,1𝑘)\big{(}d+1,1,\dotsc,\overset{k}{1}\big{)}
Corollary 5.7: (k+1,,k+1𝑏,k,,kb+,1,,1b++d)(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k},1,\dotsc,\overset{b+\ell+d}{1}),  kd+3k\geq d+3, dt2d\leq t-2 ,\mathrel{\swarrow},\mathrel{\ooalign{$\mathrel{\nearrow}$\cr${\scriptstyle-\mkern 8.5mu}$}} (a,t,,tb+,d+1)(a,t,\dotsc,\overset{b+\ell}{t},d+1)
Corollary 5.8: (κ+1,,κ+1d+1,κ,,κ𝑡,1,,1𝑎),\big{(}\kappa+1,\dotsc,\overset{d+1}{\kappa+1},\kappa,\dotsc,\overset{t}{\kappa},1,\dotsc,\overset{a}{1}\big{)},κ:=b+\kappa\mathrel{\mathop{:}}=b+\ell ,\mathrel{\nearrow},\mathrel{\ooalign{$\mathrel{\swarrow}$\cr\raise 1.29167pt\hbox{${\scriptstyle-\mkern 5.0mu}$}}} (κ+d,κ,,κ𝑘,b)\big{(}\kappa+d,\,\kappa,\dotsc,\overset{k}{\kappa},\,b\big{)}
Table 2: A summary of the arrow characterisation results for \mathrel{\nearrow} and \mathrel{\swarrow} where b,db,d\in\mathbb{N}, c,,t+c,\ell,t\in\mathbb{N}_{+}, b+3b+\ell\geq 3 and c,t2c,t\geq 2. There are different additional restrictions on tt that have to be taken from the respective results. Rows 1 and 2, and 3 and 4 are duals of each other and are subject to the same conditions.

We are now ready to discuss once more the single arrows in Example 2.15. The first nn\in\mathbb{N} where a single arrow appears in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) is n=7n=7. This is justified by our Theorem 5.4 since its assumptions require k4k\geq 4 and 42td+14\leq 2t\leq d+1, which gives n=k+d4+3=7n=k+d\geq 4+3=7. Moreover, Corollary 5.7 can only be applied for n9n\geq 9 as it requires b+3b+\ell\geq 3 and kd+3k\geq d+3, wherefore n=b+(b+)k+d3k3(d+3)9n=b+(b+\ell)k+d\geq 3k\geq 3(d+3)\geq 9 as b,d0b,d\geq 0. The dual results Corollary 5.5 and Corollary 5.8 are also valid for n7n\geq 7 and n9n\geq 9, respectively.

Finally, for n=7n=7, if d=3d=3, then the restriction 42td+1=44\leq 2t\leq d+1=4 of Theorem 5.4 shows that t=2t=2, wherefore the partition g=(4,1,1,1)g=(4,1,1,1) satisfies gmg\mathrel{\swarrow}m with a unique m(7)m\in\mathcal{M}\left(\mathcal{L}_{7}\right). Hence, considering the dual arrow (justified by Corollary 5.5) there exist exactly two single-directional arrows in 𝕂(7)\mathbb{K}(\mathcal{L}_{7}). These two arrows are the ones exhibited in Example 2.15 and illustrated in Fig. 3.

6 One-generated arrow-closed one-by-one subcontexts

We have implemented a brute-force algorithm (based on Definition 2.1) and an algorithm that uses the information given in Tables 1 and 2 to discover one-generated arrow-closed subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}). Both implementations suggest that as of n3n\geq 3 there exist exactly 2n42n-4 arrow-closed (1×1)(1\times 1)-subcontexts. To demonstrate the applicability of our general characterisations of the arrow relations in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) to the original problem of describing one-generated arrow-closed subcontexts (with the ultimate intent to obtain subdirect decompositions), we prove in this section that 2n42n-4 is the correct number of one-generated arrow-closed subcontexts of format (1×1)(1\times 1).

Our first lemma deals precisely with all arrow-closed subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) generated by g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of types 1 or 2.

Lemma 6.1.

Let b,,k+b\in\mathbb{N},\ell,k\in\mathbb{N}_{+} be such that b+<n=b(k+1)+kb+\ell<n=b(k+1)+\ell k. Then g=(k+1,,k+1𝑏,k,,kb+)𝒥(n)g=(k+1,\dotsc,\overset{b}{k+1},k,\dotsc,\overset{b+\ell}{k})\in\mathcal{J}\left(\mathcal{L}_{n}\right) generates the following arrow-closed (1×1)(1\times 1)-subcontext of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) with m=(nb,1,,1b+,1)(n)m=(n-b-\ell,1,\dotsc,\overset{b+\ell}{1},1)\in\mathcal{M}\left(\mathcal{L}_{n}\right).

m=(nb,1,,1b+,1)g=(k+1,,k+1𝑏,k,,kb+)\begin{array}[]{|c||c|}\hline\cr&m=(n-b-\ell,1,\ldots,\overset{b+\ell}{1},1)\\ \hline\cr\hline\cr g=(k+1,\ldots,\overset{b}{k+1},k,\ldots,\overset{b+\ell}{k})&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\\ \hline\cr\end{array}
Proof.

Under the given conditions, gg is of type 1 or 2, see Corollary 3.14. We apply the procedure to arrow-close the context. The summarising Tables 1 and 2 show that gm=(nb,1,,1b+,1)g\mathrel{\nearrow}m=(n-b-\ell,1,\dotsc,\overset{b+\ell}{1},1) by Theorem 4.10 (in fact gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m) and that this mm is the only one in relation gmg\mathrel{\nearrow}m. It remains to justify that no other partition pgp\neq g satisfies pmp\mathrel{\swarrow}m for the given mm. Considering the summarising Tables 1 and 2 and the shape of m=(a,1,,1)m=(a,1,\dotsc,1), we realise that Theorem 4.10 describes all p𝒥(n)p\in\mathcal{J}\left(\mathcal{L}_{n}\right) such that pmp\mathrel{\swarrow}m (the cases m=(1,,1)m=(1,\dotsc,1) and m=(n1,1)m=(n-1,1) can be handled by Theorem 4.17, as well, and yield the same p=gp=g as Theorem 4.10). Therefore, the partitions satisfying pmp\mathrel{\swarrow}m are exclusively of type 1 or 2 and have length b+=len(m)1b+\ell=\operatorname{len}(m)-1. Then the uniqueness of the decomposition n=(b+)k+bn=(b+\ell)k+b modulo b+b+\ell where 0b<b+0\leq b<b+\ell entails that pp coincides with the given gg. ∎

We now dualise Lemma 6.1 and thereby cover in particular all arrow-closed subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) generated by any g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3.

Lemma 6.2.

Let dd\in\mathbb{N}, k+k\in\mathbb{N}_{+} with k2k\geq 2, set n=k+dn=k+d and decompose n=c(k1)+rn=c(k-1)+r with c,rc,r\in\mathbb{N} such that 0rk20\leq r\leq k-2. Then the partition g=(k,1,,1d+1)𝒥(n)g=(k,1,\dotsc,\overset{d+1}{1})\in\mathcal{J}\left(\mathcal{L}_{n}\right) generates the following arrow-closed (1×1)(1\times 1)-subcontext of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) with m=(k1,,k1𝑐,r)(n)m=(k-1,\dotsc,\overset{c}{k-1},r)\in\mathcal{M}\left(\mathcal{L}_{n}\right).

m=(k1,,k1𝑐,r)g=(k,1,,1d+1)\begin{array}[]{|c||c|}\hline\cr&m=(k-1,\ldots,\overset{c}{k-1},r)\\ \hline\cr\hline\cr g=(k,1,\ldots,\overset{d+1}{1})&\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}\\ \hline\cr\end{array}
Proof.

This result is the exact dual of Lemma 6.1. An explicit proof can be obtained by dualising the argument of Lemma 6.1 where applications of Theorem 4.10 need to be replaced by Theorem 4.17 and types 1 and 2 become types 1 and 2. The lemma can also be proved directly using the arrow-closing procedure and the tables. In fact, if m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of type 1 or 2, then Tables 1 and 2 confirm that Theorem 4.17 describes all p𝒥(n)p\in\mathcal{J}\left(\mathcal{L}_{n}\right) such that pmp\mathrel{\swarrow}m (there are (only) two cases, p=(n)p=(n) and p=(2,1,,1)p=(2,1,\dotsc,1), which are handled by Theorem 4.10 and 4.17 simultaneously). ∎

Proposition 6.3.

For nn\in\mathbb{N}, n3n\geq 3 there are exactly 2n42n-4 one-generated arrow-closed (1×1)(1\times 1)-subcontexts in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

Proof.

Considering Lemmata 6.1 and 6.2 we may count the arrow-closed (1×1)(1\times 1)-subcontexts. First, Lemma 6.2 allows to choose the parameter k{2,,n}k\in\left\{2,\ldots,n\right\}. Hence, n1n-1 choices are available, and each leads to a distinct gg and thus a distinct (1×1)(1\times 1)-subcontext. Moreover, Lemma 6.1 is exactly the dual of Lemma 6.2, wherefore we have once more n1n-1 choices, and thus n1n-1 distinct subcontexts, available. The only g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) that are captured by Lemma 6.1 and 6.2 at the same time are either g=(n)g=(n), or have length at least two and thus, according to Lemma 6.2, end with 11, have height at least 22 and satisfy g1>g2g_{1}>g_{2}. Now, according to the shapes of gg considered in Lemma 6.1, the height must be equal to 22 since gg ends in 11, and so g=(2,1,1,,1)g=(2,1,1,\dotsc,1). We therefore have to reduce the sum 2(n1)2(n-1) by 22 and thus have at least 2n42n-4 distinct arrow-closed (1×1)(1\times 1)-subcontexts generated by g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of types 1, 2 or 3.

Finally, if gg is of type 4 with k3k\geq 3 and b+2b+\ell\geq 2, then Theorem 4.23 exhibits some m4(n)m_{\text{\ref{typeIV}}}\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4 with gm4g\mathrel{\nearrow}m_{\text{\ref{typeIV}}}, and Corollary 5.5 yields some m3(n)m_{\text{\ref{typeIII}}}\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 3 with gm3g\mathrel{\nearrow}m_{\text{\ref{typeIII}}}. Since the type classes are disjoint, we therefore have two distinct m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) such that gmg\mathrel{\nearrow}m, and hence arrow-closed subcontexts generated by gg of type 4 cannot be of format 1×11\times 1. Consequently, the subcontexts shown in Lemmata 6.1 and 6.2 are the only ones of this shape. ∎

7 Conclusion

Arrow vs. type For First example Result
1 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 1 n=2n=2 (2)(2) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (1,1)(1,1) Theorem 4.10
2 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 1 n3n\geq 3 (2,1)(2,1) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (1,1,1)(1,1,1) Theorem 4.10
1 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 2 n3n\geq 3 (3)(3) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (2,1)(2,1) Theorem 4.10
1 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 3 n4n\geq 4 (2,2)(2,2) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (2,1,1)(2,1,1) Theorem 4.10
2 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 3 n5n\geq 5 (2,2,1)(2,2,1) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (2,1,1,1)(2,1,1,1) Theorem 4.10
3 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 1 n4n\geq 4 (3,1)(3,1) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (2,2)(2,2) Theorem 4.17
3 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 2 n5n\geq 5 (3,1,1)(3,1,1) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (2,2,1)(2,2,1) Theorem 4.17
4 \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} 4 n7n\geq 7 (3,3,1)(3,3,1) \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}} (3,2,2)(3,2,2) Theorem 4.23
3 \mathrel{\swarrow} 4 n7n\geq 7 (4,1,1,1)(4,1,1,1) \mathrel{\swarrow} (3,2,2)(3,2,2) Theorem 5.4
1 \mathrel{\swarrow} 4 n9n\geq 9 (3,3,3)(3,3,3) \mathrel{\swarrow} (4,2,2,1)(4,2,2,1) Corollary 5.7
2 \mathrel{\swarrow} 4 n10n\geq 10 (4,3,3)(4,3,3) \mathrel{\swarrow} (5,2,2,1)(5,2,2,1) Corollary 5.7
4 \mathrel{\swarrow} 4 n13n\geq 13 (4,4,4,1)(4,4,4,1) \mathrel{\swarrow} (5,3,3,2)(5,3,3,2) Corollary 5.7
4 \mathrel{\nearrow} 3 n7n\geq 7 (3,3,1)(3,3,1) \mathrel{\nearrow} (4,1,1,1)(4,1,1,1) Corollary 5.5
4 \mathrel{\nearrow} 1 n9n\geq 9 (4,3,1,1)(4,3,1,1) \mathrel{\nearrow} (3,3,3)(3,3,3) Corollary 5.8
4 \mathrel{\nearrow} 2 n10n\geq 10 (4,3,1,1,1)(4,3,1,1,1) \mathrel{\nearrow} (3,3,3,1)(3,3,3,1) Corollary 5.8
4 \mathrel{\nearrow} 4 n13n\geq 13 (4,4,3,1,1)(4,4,3,1,1) \mathrel{\nearrow} (4,3,3,3)(4,3,3,3) Corollary 5.8
Table 3: Arrow relation patterns between different types of partitions with least nn\in\mathbb{N} such that these occur in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}).

In this paper, for every nn\in\mathbb{N}, we characterised all arrow relations appearing in the standard context of the lattice n\mathcal{L}_{n} of partitions of nn under the dominance order. When looking at Tables 1 and 2, collecting our results, some curious patterns with respect to the three arrow shapes and the partition types connected by arrows of a specific form arise. These will be summarised subsequently; more detailed information for which values of nn these patterns appear and which result justifies them, can seen from Table 3. Inspecting Table 3, we note that all of these connections can be observed in 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) as soon as n13n\geq 13. The following patterns caught our eye:

  1. 1.

    If g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1, 2 or 3 and m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) satisfies gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, then mm is not of type 4 and mm is unique. Conversely, if m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) is of types 1, 2 or 3 and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m with g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right), then gg is not of type 4 and it is also unique. Therefore, the double arrows establish a one-to-one relationship between elements of the union of the type classes 13 and the union of the classes 13.

    More specifically, if g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) is of type 1 or 2 such that g(n)g\neq(n), g(2,1,,1n1)g\neq(2,1,\dotsc,\overset{n-1}{1}) and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, then Theorem 4.10 implies that mm is of type 3, and the whole class of type 3 partitions is exhausted by those gg. The two exceptional double arrow relations are (n)(n1,1)(n)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(n-1,1) with gg of type 1 and mm of type 2 (or 1 for n=2n=2), and (2,1,,1)(1,,1)(2,1,\dotsc,1)\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}(1,\dotsc,1) with gg of type 2 and mm of type 1. In particular, g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1 is never \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}-connected to m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 1 (or 4) when n3n\geq 3, and gg of type 2 is never \mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}-connected to mm of type 2 (or 4). In fact, gg of type 2 is never connected via any kind of arrow to mm of type 2, as shown in Corollary 4.13. Having observed the bijective correspondence given by the double arrow relation, this leaves that the g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 3 correspond in a one-to-one way to the m(n){(n1,1),(1,,1)}m\in\mathcal{M}\left(\mathcal{L}_{n}\right)\setminus\left\{(n-1,1),(1,\dotsc,1)\right\} of types 1 or 2. This can also be verified directly using Theorem 4.17 for parameters 2an22\leq a\leq n-2. Note also that gg of type 3 is not even \mathrel{\nearrow} or \mathrel{\swarrow}-related to mm of type 3, as shown in Lemma 4.20.

  2. 2.

    If g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right), m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) and gmg\mathrel{\ooalign{$\nearrow$\cr$\swarrow$}}m, then gg is of type 4 if and only if mm is of type 4, see Table 1. However, this correspondence is not unique because for sufficiently large values of nn, kk and dd, Theorem 4.23 allows for several choices of the parameter tt determining distinct shapes of m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4. Dually, a given mm of type 4 may only be double arrow related to g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4, and these gg may not be unique.

  3. 3.

    Every up-arrow originating in g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 1, 2 or 3 is actually a double arrow described in Table 1. This is so because Table 2 does not contain any up-arrows of such type. Conversely, every down-arrow connecting g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) with some m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 1, 2 or 3 is necessarily a double arrow. This implies that g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of types 1, 2 or 3 receive down-arrows from m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) that are not double-arrows only if mm is of type 4. Dually, g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) are in relation gmg\mathrel{\nearrow}m and gmg\mathrel{\ooalign{$\mathrel{\swarrow}$\cr\raise 1.29167pt\hbox{${\scriptstyle-\mkern 5.0mu}$}}}m with m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of types 1, 2 or 3 only if gg is of type 4.

    On the other hand, g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of type 4 exhibit up-arrows to m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of all four types that fail to be double arrows; likewise m(n)m\in\mathcal{M}\left(\mathcal{L}_{n}\right) of type 4 satisfy gmg\mathrel{\swarrow}m but gg\mathrel{\ooalign{$\mathrel{\nearrow}$\cr${\scriptstyle-\mkern 8.5mu}$}} with g𝒥(n)g\in\mathcal{J}\left(\mathcal{L}_{n}\right) of all four types. All of these single directional arrow relationships are in general not one-to-one as Theorem 5.4 and Corollary 5.7 exhibit a flexible parameter range for tt with fixed and sufficiently large k,dk,d and nn. This happens as of n10n\geq 10 for Theorem 5.4 and as of n12n\geq 12 for Corollary 5.7.

As a side-product of Proposition 6.3, we discovered that the number of ways a natural number nn can be decomposed yielding partitions of type 1 or 2 is exactly n1n-1, for this is the same as the number of partitions of nn of the shape (a,1,,1)(a,1,\dotsc,1) with 2an2\leq a\leq n. This is an example of a bijective proof as they are typical for counting the number of restricted partitions in the combinatorial theory of integer partitions, see, e.g., [1, Section 2.2].

Finally, in order to demonstrate the applicability of our general descriptions, we started by characterising all one-generated arrow-closed subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) of format 1×11\times 1. We determined that their number is exactly 2n42n-4, see Proposition 6.3. Working on this we observed that also the characterisation of other arrow-closed subcontexts of small shape seems to be within reach of our results. We leave the complete description of the one-generated arrow-closed subcontexts of 𝕂(n)\mathbb{K}(\mathcal{L}_{n}) as a task for future investigation.

References

  • [1] G. E. Andrews, K. Eriksson, Integer partitions, Cambridge University Press, Cambridge, 2004. doi:10.1017/CBO9781139167239.
  • [2] M. Behrisch, Symmetric embeddings between standard contexts of lattices of integer partitions [dataset], Zenodo (Dec. 2021). doi:10.5281/zenodo.5810903.
  • [3] M. Behrisch, A. Chavarri Villarello, E. Vargas-García, Representing partition lattices through FCA, in: A. Braud, A. Buzmakov, T. Hanika, F. Le Ber (Eds.), Formal Concept Analysis – 16th International Conference, ICFCA 2021, Strasbourg, France, June 29 – July 2, 2021, Proceedings, Vol. 12733 of Lecture Notes in Artificial Intelligence, Springer, Cham, 2021, pp. 3–19. doi:10.1007/978-3-030-77867-5_1.
  • [4] M. Behrisch, E. Vargas-García, Non-embeddability of standard contexts of lattices of integer partitions [dataset], Zenodo (Dec. 2021). doi:10.5281/zenodo.5805981.
  • [5] T. Brylawski, The lattice of integer partitions, Discrete Math. 6 (3) (1973) 201–219. doi:10.1016/0012-365X(73)90094-0.
  • [6] A. Chavarri Villarello, El retículo de particiones de enteros positivos y su contexto estándar, Undergraduate honors thesis, ITAM, Río Hondo 1, Ciudad de México (Jul. 2020).
  • [7] S. A. Díaz Miranda, Retículos de números naturales y sus relaciones de flechas, Undergraduate honors thesis, ITAM, Río Hondo 1, Ciudad de México (Mar. 2024).
  • [8] L. A. Franco Solorio, Encajes entre contextos estándar de los retículos de particiones de números enteros, Undergraduate honors thesis, ITAM, Río Hondo 1, Ciudad de México (Mar. 2022).
  • [9] B. Ganter, Notes on integer partitions, International Journal of Approximate Reasoning 142 (2022) 31–40. doi:10.1016/j.ijar.2021.11.004.
  • [10] B. Ganter, R. Wille, Formal concept analysis. Mathematical foundations, Springer, Berlin, 1999, translated from the 1996 German original by Cornelia Franzke. doi:10.1007/978-3-642-59830-2.
  • [11] G. H. Hardy, S. Ramanujan, Asymptotic formulæ in combinatory analysis, Proc. London Math. Soc. (2) 17 (1) (1918) 75–115. doi:10.1112/plms/s2-17.1.75.
  • [12] T. W. Hungerford, Algebra, Vol. 73 of Graduate Texts in Mathematics, Springer, New York, NY, 2011, reprint of the 1974 original. doi:10.1007/978-1-4612-6101-8.
  • [13] M. Latapy, T. H. D. Phan, The lattice of integer partitions and its infinite extension, Discrete Math. 309 (6) (2009) 1357–1367. doi:10.1016/j.disc.2008.02.002.
  • [14] D. Mahnke, Leibniz auf der Suche nach einer allgemeinen Primzahlgleichung, Bibl. Math. (3) XIII (1912–1913) 29–61.
    URL https://www.ophen.org/pub-102519
  • [15] A. Páramo Pitol, Factores subdirectamente irreducibles de pequeños retículos de particiones, Undergraduate honors thesis, ITAM, Río Hondo 1, Ciudad de México (Jul. 2022).
  • [16] H. Rademacher, On the partition function p(n)p(n), Proc. London Math. Soc. (2) 43 (4) (1937) 241–254. doi:10.1112/plms/s2-43.4.241.
  • [17] H. Rademacher, On the expansion of the partition function in a series, Ann. of Math. (2) 44 (3) (1943) 416–422. doi:10.2307/1968973.