Startseite Classification lattices are geometric for complete atomistic lattices
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Classification lattices are geometric for complete atomistic lattices

  • Hua Mao EMAIL logo
Veröffentlicht/Copyright: 13. Juli 2017
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Open Mathematics
Aus der Zeitschrift Open Mathematics Band 15 Heft 1

Abstract

We characterize complete atomistic lattices whose classification lattices are geometric. This implies an proper solution to a problem raised by S. Radeleczki in 2002.

Keywords: Geometric; Classification system; Classification lattice; Complete atomistic lattice
MSC 2010: 06B05; 06B15; 06C10

1 Introduction and Preliminaries

Given a complete atomistic lattice L (see [1, p.251,Definition 285]), a subset C of L \ {0} is called a classification system of L if

(C1) c ∧ d = 0 for any two distinct elements c, d ∈ C, and

(C2) x = ∨{x ∧ c | c ∈ C} holds for each x ∈ L.

This concept, not only for the atomistic case, is due to Radeleczki [2], and it has motivations in the theory of concept lattices (see [2, 3]). Let Cls(L) stand for the collection of all classification systems of L. For C, D ∈ Cls(L), we let C ≤ D if and only if for each c ∈ C there is a d ∈ D such that c ≤ d. By [2, Theorem 4.2], this relation turns Cls(L) into a complete lattice.

Radeleczki [2] provides two interesting and attractive open problems for classification systems. Mao [4] solves one of them completely. Another open problem asks to characterize those complete atomistic lattices whose classification lattices are geometric. Though this problem is partly solved by Mao [5], it is unsolved completely to date. This problem is related to the application of concept lattices to one of the main problems in group technology (see [2, 6-8]). Actually, the problem is also related with the study of CD-bases of a lattice (see [9,10]), or to the investigation of the decomposition systems of a closure system (cf. [11,12]). In other words, the solution of this problem may have useful applications in several related fields. Based on these discussions, we will reveal the answer to this problem in this paper.

1.1 Definitions and properties

To find the answer of the open problem, we first review some definitions and properties what we need later on. In what follows, for more basic concepts of lattice theory and classification systems, the reader is kindly referred to [1,13] and [2] respectively.

Some definitions

  1. [1, p.1] A nonempty set equipped with a relation (reflexivity, antisymmetry, and transitivity) is called a poset.

  2. [1, p.9] A poset (K, ≤) is a lattice if sup{a, b} and inf{a, b} exist for all a, b ∈ K.

  3. [1, p.28] The lattices (K, ≤) and (K′, ≤) are isomorphic (in symbols, (K, ≤) ≅ (K′, ≤)) if the map ϕ : K → K′ is a bijection such that a ≤ b in (K, ≤) if and only if ϕ(a) ≤ ϕ(b) in (K′, ≤).

  4. [13, p.11] N5 = {a, b, c, d, e} is a lattice which satisfies the following conditions:

    (p.1) e < x < d for any x ∈ {a, b, c}; (p.2) c < a; (p.3) a||b and c||b.

  5. [1, p.35] An interval in a lattice K is a subset of K of the form [a, b] = {x ∈ K | a ≤ x ≤ b} for some a, b, ∈ K with a ≤ b.

  6. [13, p.5 & 1,p.4] The length of a finite chain n is defined to be n — 1; the length l(P) of a poset P is defined as the least upper bound of the lengths of the chains in P.

    [1, p.4] The height of an element a ∈ K, denoted by h(a), is the length of the order {x ∈ K | x ≤ a}.

    That a covers b in a poset P is defined in [1, p.6].

    The diagram of a finite poset is defined in [1, p.6].

  7. [1, p.50] A lattice K is called complete if ∨H and ∧H exist for any subset H ⊆ K, in which ∨H and ∧H are seen [1, p.5].

  8. [1, p.101] An element a of a lattice K is an atom if a covers 0 (i.e the least element in K if it exists). A lattice K is called atomistic if every element is a join of atoms.

  9. [1, p.329] A lattice K is called semimodular if it satisfies for all a, b ∈ K, a ≺ b ⇒ a∨b ≺ b∨c or a∨c = b∨c.

  10. [1, p.52] Let K be a complete lattice and let a be an element of K. Then a is called compact if a ≤ ∨X, for any X ⊆ K, implies that a ≤ ∨X1 for some finite X1 ⊆ X.

  11. [1, p.342] A lattice K is called geometric if the lattice K is complete, K is atomistic, all atoms are compact, and K is semimodular.

Some notations

  1. Let K be a lattice.

    A(K) denotes the set of atoms of K; A(x) = {a | a ∈ A(K), a ≤ x};
    x||y if x ≰ y and y ≰ x; x ∦ y if x||y is not true; x ≺ y implies that y covers x;
    h(x) denotes the height of x ∈ K; [a, b] = {y ∈ K | a ≤ y ≤ b}.
    ℱ2 = {x ∈ K | h(x) = 2 in K}; ℱ2(z) = {d ∈ ℱ2 | d ≤ z for any z ∈ K.
    Let K have a greatest element 1K, then we naturally define h(K) = h(1K).

  2. L denotes a complete atomistic lattice; 0 and 1 denote its least and greatest element, respectively.

    Sx {x} ∪ {a ∈ A(L) | a ∧ x = 0} for any x ∈ L \ {0}; S0 = ∧{S |S ∈ Cls(L)}.

Some properties

  1. [1, p.342,Corollary 390] An interval of a geometric lattice is again a geometric lattice.

  2. [13, p.13,Theorem 12] Up to isomorphism, any non-modular lattice contains the lattice N5 as a sublattice.

    [1, p.109,Theorem 102] A lattice K is modular if and only if it does not contain a N5.

  3. [2] S0 ∈ Cls(L) and S1 = {1} ∈ Cls(L) hold, and Sx ∈ Cls(L) holds for any x ∈ L \ {0}.

  4. [5] Cls(L) possesses the following properties.

    (4.1) Let x, y ∈ L \ {0}. Then x ≤ y ⇔ Sx ≤ Sy. Further, x < y ⇔ Sx < Sy.

    (4.2) A(Cls(L)) = {Sd | d ∈ ℱ2}.

1.2 Answer to the cases of |𝓕2| ≤ 1

This subsection will reveal the answers to the open problem if a complete atomistic lattice L satisfies |ℱ2| ≤ 1. In fact, |ℱ2| ≤ 1 means |ℱ2| = 0 or |ℱ2| = 1. We will search out the answers to the open problem correspondent to the above two cases respectively in this subsection.

Case I

|ℱ2| = 0.

Answer: Cls(L) is geometric if and only if h(L) = 1.

The reason is the following.

|ℱ2| = 0 means h(L) = 1 or h(L) = 0.

If h(L) = 1. Then we obtain L = {0, 1} such that 0 ≺ 1 and A(L) = {1}. This implies Cls(L) = {S0} = {S1}.

So, Cls(L) is a geometric lattice.

If h(L) = 0. Then we obtain L = {0}. This implies Cls(L) to be empty. So, Cls(L) is not a lattice according to the definition of lattice in Subsection 1.1.

Conversely, if Cls(L) is geometric. According to Cls(L) ≠ ∅ and the above analysis, we may be assured h(L) = 1.

Case II

|ℱ2| = 1.

Answer: Cls(L) is geometric if and only if h(L) = 2.

The reason is the following.

From |ℱ2| = 1, we obtain h(L) ≥ 2.

If h(L) = 2. According to the atomic of L, we obtain L = {0, 1} ∪ A(L) such that 0 < 1 and A(L) ≠ ∅. From these results, we confirm Cls(L) = {S0, S1} with S0 ≺ S1. So, Cls(L) is a geometric lattice.

If h(L) > 2. In view of |ℱ2| = 1, we may assume ℱ2 = {d}. Considering the atomic of L with |ℱ2| = 1 and h(1) = h(L), we may affirm that for any x ∈ L \ ({0, 1, d} ∪ A(d)), we receive x ∈ A(L) or d < x < 1. Thereby, we may decide Cls(L) = {S0, S1 Sd} ∪ {Sx | x ∈ L \ ({0, 1, d} ∪ A(d)) and d < x < 1} satisfying A(Cls(L)) = {Sd}. Furthermore, S1 is not a join of some atoms in Cls(L). Hence, Cls(L) is not geometric.

Conversely, if Cls(L) is geometric. According to the above analysis for h(L) > 2, we may be assured h(L) = 2.

Combining the above answers for |ℱ2| ≤ 1, we determine that in what follows, we will consider |ℱ2| > 1 to discover the answer for the open problem in [2].

1.3 Organization

The rest of the paper is organized as follows. Section 2 will give the answer to the open problem for the case of |ℱ2| > 1. The final part of this paper, Section 3, concludes this paper and leaves room for our future work.

2 Main results

This section answers the open problem for a complete atomistic lattice L which satisfies |ℱ2| > 1.

Before we provide the answer, we present some properties with regards to L which we will use to find the answer to the open problem.

Lemma 2.1

(1) Sa = S0 if and only if a ∈ A(L).

(2) Let x, y ∈ L \ (A(L) ∪ 0) and x ≠ y. Then x < y ⇔ Sx < Sy.

Furthermore, for any x, y ∈ L\ (A(L) ∪ 0), we obtain x ≤ y ⇔ Sx ≤ Sy.

Proof

Firstly, we prove item (1).

Since L is atomistic, we may easily obtain that A(L) ⊆ L\ {0} holds and A(L) satisfies both of (C1) and (C2). That is to say, A(L) ∈ Cls(L) holds.

Let S ∈ Cls(L) \ A(L). Then, we affirm that S contains an element x satisfying x ∉ A(L) according to S ⊆ L \ {0} and S ≠ A(L). Thus, by the definition of classification system and the atomic of L, we obtain A(L) < S in Cls(L).

Therefore, S0 = A(L) holds.

Since Sa = {a} ∪ {x ∈ A(L) | x ≠ a} for any a ∈ A(L), we may easily obtain Sa = A(L). Hence, we obtain Sa = S0.

According to Sx = {x} ∪ {a ∈ A(L) | a ≰ A(x)} for any x ∈ L \ {0} with the above A(L) = S0, we may easily attain that if Sx = S0 holds, then x ∈ A(L) holds.

Secondly, we prove item (2).

It will be finished by routine verification of the property (4.1) in Subsection 1.1.

Let x, y ∈ A(L) and x ≠ y. Then we may easily see that Sx = Sy = S0 = A(L) since Lemma 2.1 (1). Even though, in the following, we still write S0 not A(L) when we consider Cls(L). The reason is that we hope to keep the coherence since Sz ∈ Cls(L) holds for any z ∈ L\ (A(L) ∪ 0) in view of property (3) in Subsection 1.1.□

Taking Lemma 2.1 (2) and Lemma 3.1(3.1) in [10], we will obtain the following corollary.

Corollary 2.2

(1) Let x, y ∈ L\ (A(L) ∪ 0).

Then Sx = Sy ⇔ x = y. In addition, x||y ⇔ Sx||Sy.

(2) Let x, y ∈ L\ {0}. Then

x ≤ y ⇒ Sx ≤ Sy;   x < y ⇒ Sx < Sy.

Proof

At first, we prove Sx = Sy ⇔ x = y in item (1).

If x = y, then it is easy to see Sx = Sy by the definitions of Sx and Sy.

Conversely, let Sx = Sy. According to Sx = {x}∪{a ∈ A(L) | a∧x = 0}, Sy = {y}∪{b ∈ A(L) | b∧y = 0} and x ∉ A(L) ∪ 0, we confirm x ∉ {b ∈ A(L) | b ∧ y = 0}. Considering Sx = Sy, we decide x ∈ {y}. Hence, we obtain x = y.

Secondly, we prove x||y ⇔ Sx||Sy in item (1).

Considering Lemma 2.1 (1), we may be assured Sx, Sy ∈ Cls(L)\{S0}.

x||y implies the true of x ≠ y, x ≮ y and y ≮ x.

Using the above and x ≠ y, we find Sx ≠ Sy. Using x ≮ y (y ≮ x) and Lemma 2.1 (2), we find Sx ≮ Sy (Sy ≮ Sx). Therefore, we know Sx||Sy.

Conversely, if Sx||Sy. This means Sx ≠ Sy, Sx ≮ Sy and Sy ≮ Sx. Applying the ‘first part’ above and Sx ≠ Sy, we find x ≠ y. Applying Sx ≮ Sy (Sy ≮ Sx) and Lemma 2.1 (2), we find x ≮ y (y ≮ x). Therefore, we know x||y.

Next, we will prove item (2).

If x = y. Then we easily know Sx = Sy.

If x, y ∈ L \ (A(L) ∪ 0). Then by Lemma 2.1 (2) and item (1) above, we obtain x ≤ y ⇔ Sx ≤ Sy, and x < y ⇔ Sx < Sy.

If x, y ∈ A(L) and x ≠ y hold, or y ∈ A(L) and x ∉ A(L) ∪ 0 hold. Then it is easy to see that both of x < y and x ≤ y will not happen. Thus, we do not consider the needed results.

If x ∈ A(L) and y ∉ A(L) ∪ 0. Then x ≤ y will mean x ∈ A(y). Let b ∈ A(L) \ A(y). We may easily see that b ∧ y = 0 since L is atomistic. This implies b ∈ Sy according to Sy = {y} ∪ {a ∈ A(L)| a ∧ y = 0}. In addition, Sx = A(L) = S0 holds in light of Lemma 2.1 (1). Hence, Sx < Sy holds. That is to say, Sx ≤ Sy is correct.

Actually, the three of x ∈ A(L), y ∉ A(L) ∪ 0 and x ≤ y imply x < y. So, we easily obtain S0 < Sy. That is to say, Sx < Sy holds since Sx = S0 holds by Lemma 2.1 (1). ☐

Let L be a complete atomistic lattice with at least two elements of height 2. Based on the Subsection 1.2 and the above discussions, in answering the problem for L we will consider two statuses:

First status: there is x ∧ y ≠ 0 for some x, y ∈ L \ (A(L) ∪ 0 ∪ 1) and x||y.

Second status: there are x ∧ y = 0 for any x, y ∈ L \ (A(L) ∪ 0 ∪ 1) and x||y.

In fact, Since |ℱ2| > 1, we obtain |L \ (A(L) ∪ 0 ∪ 1)| > 1. Thus, we may easily find that only one of the above statuses will happen for L.

2.1 Answer for ‘First status’

In this subsection, we will give an answer to the open problem for the complete atomistic lattices that satisfy the conditions in First status.

Through this subsection, LCls(L) denotes L \ A(L). We may easily see that LCls(L) with the same order in L is a sublattice of L. We still denote this sublattice as LCls(L).

The main result in this subsection is the following Theorem 2.3, that is, an answer to the open problem for ‘First status’.

Theorem 2.3

Let L satisfy the conditions in ‘First status’. Then Cls(L) is geometric if and only if LCls(L) satisfies the following conditions.

  1. For any m, n ∈ L \ (A(L) ∪ 0 ∪ 1) with m||n, m ∧ n ≠ 0 holds.

  2. LCls(L) is geometric.

Before giving the proof of Theorem 2.3, we present two lemmas as preparations.

Lemma 2.4

Let L satisfy the conditions in ‘First status’. If Cls(L) is geometric. Then the following statements hold.

(1) Cls(L) = {Sx | x ∈ L \ {0}}.

(2) (2.1) Let x, y ∈ A(L) and x ≠ y. Then

Sx ∨ Sy = S0 < Sx∨y;

Sx ∧ Sy = S0 ≠ Sx∧y.

(2.2) Let x ∈ A(L) and y ∈ L \ (A(L) ∪ 0). Then

Sx ∨ Sy = Sx∨y and Sx ∧ Sy = Sx if x < y;

Sx ∨ Sy < Sx∨y and Sx ∧ Sy = S0 ≠ Sx∧y if x||y.

(2.3) Let x, y ∈ L \ (A(L) ∪ 0) and x ≠ y. Then

Sx∨y = Sx ∨ Sy;

S0 = Sx∧Sy if x ∧ y ≠ 0;

Sx∧y = Sx ∧ Sy if x ∧ y ≠ 0.

Proof

We prove item (1).

Otherwise, there is S ∈ Cls(L) \ {Sx | x ∈ L \ {0}}. Using Lemma 2.1 (1), Sx = S0 holds for any x ∈ A(L). Thus, there must exist y, z ∈ S \ (A(L) ∪ 0) satisfying y ≠ z and y||z according to the definition of Sx (∀x ∈ L \ (A(L) ∪ 0)) in Subsection 1.1. Hence, we may determine y ∧ z = 0 since S satisfies (C1).

From y, z ∈ S \ (A(L) ∪ 0), we obtain h(y, h(z) ≥ 2 where h is the height function of L. So, we believe ℱ2(y) ≠ ∅ and ℱ2(z) ≠ ∅. According to y ∧ z = 0, we decide ℱ2(y) ∩ ℱ2(z) = ∅.

We will continue the discussion in the following two steps. Let d ∈ ℱ2(y) and e ∈ ℱ2(z).

Step 1

Let Sde = {d, e} ∪ {p ∈ A(L) | p ≰ A(d) and p ≰ A(e)}. We will prove Sde ∈ Cls(L).

We may easily gain d ∧ e = 0 and d||e since 0 ≤ d ∧ e ≤ y ∧ z = 0 and y||z. We also obtain d ∧ p = e ∧ p = 0 since p ∈ Sde satisfies p ∉ A(d) and p ∉ A(e) with d = ⋁ a ∈ A ( d ) a a n d e = ⋁ q ∈ A ( e ) q. This implies that Sde satisfies (C1).

We distinguish three cases to demonstrate Sde to satisfy (C2). Let g ∈ L \ {0}.

Case 1

g ≤ d.

This implies g ∧ e = g ∧ p = 0 for any p ∈ A(L) \ (A(d) ∪ A(e)) since d ∧ e = 0 and p ∉ A(d). Thus, we may obtain g = (g ∧ d) ∨ (g ∧ e) ∨ ( ⋁ p ∈ S d e ∖ { d , e } ( g ∧ p ) ).

Case 2

g ≤ e.

Similarly to Case 1, we can obtain g = ( g ∧ e ) ∨ ( ⋁ q ∈ S d e ∖ { e } ( g ∧ q ) ).

Case 3

g ≰ d and g ≰ e.

We may easily find A(g) = (A(g) ∩ A(d)) ∪ (A(g) ∩ A(e)) ∪ (A(g) ∩ {q ∈ A(L) | q ∉ A(d) and q ∉ A(e)} and g = ∨A(g) = ( ⋁ u ∈ A ( g ) ∩ A ( d ) u ) ∨ ( ⋁ w ∈ A ( g ) ∩ A ( e ) w ) ∨ ( ⋁ q ∈ A ( L ) ∖ ( A ( d ) ∪ A ( e ) ) q ).

If g ≰ d. Then it is easy to see g||d or d < g.

In fact, g||d means ⋁ u ∈ A ( g ) ∩ A ( d ) u = g ∧ d . d < g means A(d) ⊂ A(g). Furthermore, we obtain ⋁ u ∈ A ( g ) ∩ A ( d ) u = ⋁ u ∈ A ( d ) u = d = g ∧ d .

Similarly, we can affirm ⋁ w ∈ A ( g ) ∩ A ( e ) w = g ∧ e.

Moreover, g = (g ∧ d) ∨ (g ∧ e) ∨ ( ⋁ q ∈ A ( L ) ∖ ( A ( d ) ∪ A ( e ) ) ( g ∧ q ) ) is accepted. Summing up Cases 1, 2 and 3, we can decide that Sde satisfies (C2). Therefore, Sde ∈ Cls (L) holds.

Step 2

From the definition of Sde in Step 1, we may easily determine Sde ∉ {Sx | x ∈ L \ {0}} since d, e ∉ A(L) ∪ 0.

In addition, we also obtain S0 < Sd < Sde and S0 < Se < Sde ≤ S.

Step 3

According to the conditions in ‘First status’, we may confirm that there exist m, n ∈ L \ (A(L) ∪ 0 ∪ 1) such that m||n and m ∧ n ≠ 0.

We will analyze the relationships among Sm, Sn and [S0, Sde] in the following Parts (I) and (II).

Part (I)

If both of Sm and Sn belong to [S0, Sde].

Then we may easily find Sm ≤ Sde and Sn ≤ Sde. By the definition of “≤” in Cls(L), we receive m ≤ d or m ≤ e by Lemma 2.1 (2). At the same time, we also receive n ≤ d or n ≤ e. Thus, for clarity, we will distinguish four cases from Case I1 to Case I4 to continue our discussion. In fact, there are only the following four cases for m and n according to the above discussions.

Case I1

m ≤ d and n ≤ e.

Case I2

m ≤ e and n ≤ d.

Case I3

m ≤ d and n ≤ d.

Case I4

m ≤ e and n ≤ e.

Suppose that Case I1 happens. Then, we obtain 0 ≤ m ∧ n ≤ d ∧ e = 0. So, we receive m ∧ n = 0. This is a contradiction to m ∧ n ≠ 0. Thus, we determine that Case I1 will not happen.

Similarly to the discussion for Case I1, we obtain that Case I2 will not happen. Suppose that Case I3 happens. If m = d. Then n ≤ d follows n ≤ m. This is a contradiction to m||n. Thus, we determine m < d. Analogously, n < d holds.

According to d ∈ ℱ2(y) and m < d, we obtain h(m) < 2. This implies m ∈ A(L)∪0. But this is a contradiction to m ∈ L \ (A(L) ∪ 0 ∪ 1). Thus, we can say m ≮ d.

Analogously, we can also find n ≮ d. Shortly, Case I3 can not happen.

Similarly to the discussion for Case I3, we obtain that Case I4 can not happen.

These discussions imply that if Sm ∈ [S0, Sde] (Sn ∈ [S0, Sde]), then Sn ∉ [S0, Sde] (Sm ∉ [S0, Sde]).

Part (II)

If Sm ∉ [S0, Sde] but Sn ∈ [S0, Sde].

Sm ∉ [S0, Sde] means Sm ≰ Sde. Thus, we obtain Sde < Sm or Sde||Sm. For clarity, we will use the following Cases II1 and II2 to discuss.

Case II1

Sde < Sm.

We find Sd, Se < Sm since we may easily see that Sd, Se < Sde. This implies d < m and e < m according to Lemma 2.1(2).

In view of the property (4.2) in Subsection 1.1 and d, e ∈ ℱ2, we find Sd, Se ∈ A(Cls(L)). However, we easily find out Sd ∨ Se = Sde. Thus, from the geometric of Cls(L) and the property (4.2) in Subsection 1.1, we confirm that there exists a c ∈ ℱ2 satisfying Sc < Sm and Sc||Sde since Sde < Sm. This follows d||c and e||c. Hence, we obtain 𝒞 ≠ ∅ in which 𝒞 = ℱ2(m) \ {d,e}. We also easily find Sx ≰ Sde for any x ∈ 𝒞.

Let S′ = {d, e} ∪ {g | g ∈ 𝒞} ∪ {a ∈ A(L) | a ∉ A(d) and a ∉ A(e), and a ∉ A(g) for any g ∈ 𝒞}. We may determine S′ ∈ Cls(L) according to the definition of Cls(L). In addition, we also determine S′ < Sm since d, e, g ∈ ℱ2(m) with d, e < m and g ≤ m for any g ∈ 𝒞.

Additionally, let M ∈ A(Cls(L)) and M ≤ Sm. According to the property (4.2) in Subsection 1.1, we may decide M = Sx0 for some x0 ∈ ℱ2. Owing to the item (2) in Lemma 2.1, the property (4.2) in Subsection 1.1 and Sd < Sm, we may decide m ∉ ℱ2. Thus, we obtain Sx0 ≠ Sm. So, Sx0 < Sm holds. In view of Lemma 2.1 (2), we can indicate x0 < m. This implies x0 ∈ ℱ2(m). Hence, we find x0 ∈ {d, e} ∪ {g | g ∈ 𝒞}. Therefore, we can infer to Sx0 < S′.

Considering the property (4.2) in Subsection 1.1, we may confirm that Sm is not a join of atoms in Cls(L). This is a contradiction to the geometric of Cls(L).

Case II2

Sde||Sm.

If m ≤ d (or m ≤ e; or d, e ≤ m). Then, we decide Sm ≤ Sd ≤ Sde (or Sm ≤ Se; or Sde ≤ Sm) according to Lemma 2.1 (2). No matter which results happens, it is a contradiction to Sde||Sm.

If d||m and e||m. We may easily decide Sd, Se, Sx < Su in which x ∈ ℱ2(m) and u = d ∨ e ∨ m. Analogously to the discussion for Case II1, we will find that Su is not a join of atoms in Clx(L) since S″ < Su and the property (4.2) in Subsection 1.1 in which S″ = {d, e} ∪ {x | x ∈ ℱ2(m)} ∪ {y | y ∈ ℱ2(u) and y ∉ {d, e} ∪ {x | x ∈ ℱ2(m)} ∪ {a ∈ A(L) | a ≰ A(d), and a ≰ A(e), and in addition, a ≰ A(m) ∈ Cls(L). This is a contradiction to the geometric of Cls(L).

If d||m and e ∦m. Actually, we find m ≰ e. This implies e < m since e ∦m. Therefore, we confirm e ∈ ℱ2(m) and d ∉ ℱ2(m). Moreover, ℱ2(m) ≠ ∅ holds. We attain Se < Sm since Lemma 2.1 (2) and the property (4.2) in Subsection 1.1. We may easily find S d ∨ ( ⋁ x ∈ F 2 ( m ) x ) ∈ C l s ( L ). Similarly to the discussion for Case II1, we can state that S d ∨ ( ⋁ x ∈ F 2 ( m ) x ) is not a join of atoms in Cls(L) since S‴ < S d ∨ ( ⋁ x ∈ F 2 ( m ) x ) and the property (4.2) in Subsection 1.1 in which S ‴ = d ∪ x | x ∈ F 2 ( m ) ∪ y | y ∈ F 2 ( d ∨ ( ⋁ x ∈ F 2 ( m ) x ) ) a n d y ∉ { d } ∪ { x | x ∈ F 2 ( m ) } ∪ a ∈ A ( L ) | a ≰ A ( d ∨ ( ⋁ x ∈ F 2 ( m ) x ) ) ∈ C l s ( L ). This is a contradiction to the geometric of Cls(L).

The above Case II1 and Case II2 show that if Sm ∉[S0, Sde] and Sn ∈ [S0, Sde] hold, then Cls(L) is not geometric.

Analogously, if Sn ∉ [S0, Sde] and Sm ∈ [S0, Sde] hold, then Cls(L) is not geometric.

Combining the discussions in Parts (I) and (II), we may state that neither m nor n exist. This is a contradiction to the conditions in ‘First status’.

Therefore, we may be assured that the supposition of S ∈ Cls(L) \ {Sx | x ∈ L \ {0}} does not hold.

Furthermore, we demonstrate Cls(L) = {Sx | x ∈ L \ {0}}.

We prove item (2). We prove (2.1).

Let x, y ∈ A(L). This implies x ∧ y = 0 and 2 ≤ h(x ∨ y). By Lemma 2.1 (1), we indicate Sx = Sy = S0. 2 ≤ h(x ∨ y) implies S0 < Sx∨y.

In addition, Sx∧y does not exist since Sx∧y = {0} does not satisfy (C2). Therefore, we can decide the correctness of (2.1).

We prove (2.2).

x ∈ A(L) implies Sx = S0 by Lemma 2.1 (1). y ∈ L \ (A(L) ∪ 0) implies 2 ≤ h(y) and Sy ∈ Cls(L) \ {S0} by Lemma 2.1 (1). Thus, we may easily find Sx < Sy. So, we obtain Sx ∨ Sy = Sy and Sx ∧ Sy = Sx.

If x < y holds. Then this implies x ∨ y = y and x ∧ y = x. Thus, we can receive Sx ∨ Sy = Sy = Sx∨y and Sx ∧ Sy = Sx = Sx∧y.

If x||y holds. Then this implies x ∧ y = 0 and x, y < x∨y with h(y) < h(x∨y). Considering Lemma 2.1 (2), we find S0 = Sx < Sx∨y and Sy < Sx∨y. However, Sx ∨ Sy = S0 ∨ Sy = Sy and Sx ∧ Sy = S0 ∧ Sy = S0 = Sx. Considering S0 ∈ Cls(L), we may obtain Sx ∨ Sy < Sx∨y and Sx ∧ Sy = S0 ≠ Sx∧y.

We prove (2.3).

We may easily find x, y ≤ x ∨ y ∈ L\ (A(L) ∪ 0). This implies Sx, Sy, Sx∨y ∈ Cls(L) \ {S0}.

Since Cls(L) is a lattice, we may easily obtain Sx, Sy ≤ Sx ∨ Sy ∈ Cls(L) \ {S0}. In addition, considering the above item (1), we obtain Sx ∨ Sy = Sb for some b ∈ L \ (A(L) ∪ 0). Thus, we may decide x ≤ b and y ≤ b since Sx, Sy ≤ Sb and Lemma 2.1 (2). Furthermore, we obtain x ∨ y ≤ b. So, we get Sx∨y ≤ Sb.

Additionally, we may find x, y ≤ x∨y since L is a lattice. Using Lemma 2.1 (2), we obtain Sx, Sy ≤ Sx∨y. So, we may be assured Sx ∨ Sy ≤ Sx∨y since Cls(L) is a lattice. Therefore, we have demonstrated Sx ∨ Sy = Sx∨y.

Next, we will consider the other correspondent statements in item (2.3).

Sx ∧ Sy ∈ Cls(L) holds since Cls(L) is a lattice. According to item (1), we confirm Sx ∧ Sy = Sc for some c ∈ L \ {0}. In addition, we easily find S0 ≤ Sx ∧ Sy ≤ Sx, Sy.

When x ∧ y = 0. If c ∉ A(L), then according to Lemma 2.1 (2), we determine c ≤ x, y. So, c ≤ x ∧ y holds. This means 2 ≤ h(c) ≤ h(x∧y). Thus, we obtain x∧y ≠ 0. This is a contradiction to the supposition of x∧y = 0. Moreover, we determine c ∈ A(L). Considering Lemma 2.1 (1), we indicate Sx ∧ Sy = S0.

When x ∧ y ≠ 0. This means Sx∧y ∈ Cls(L). From x ∧ y ≤ x, y and Corollary 2.2, we may find Sx∧y ≤ Sx, Sy. So, we decide Sx∧y ≤ Sx ∧ Sy.

If Sx∧y = Sx ∧ Sy holds. Then we obtain the needed result.

If Sx∧y < Sx ∧ Sy happens. We will distinguish two cases 2.3.1 and 2.3.2 to discuss.

Case 2.3.1

x ∧ y ∉ A(L).

The hypothesis implies S0 < Sx∧y since Lemma 2.1 (1). Furthermore, we obtain c ∈ L \ (A(L) ∪ 0) since S0 < Sx∧y < Sc and Lemma 2.1 (1). Taking Sc = Sx ∧ Sy ≤ Sx, Sy with Lemma 2.1 (2) together, we find c ≤ x and c ≤ y. So, we obtain c ≤ x ∧ y. Furthermore, we obtain Sc ≤ Sx∧y since Lemma 2.1 (2). This is a contradiction to Sx∧y < Sx ∧ Sy.

Case 2.3.2

x∧y ∈ A(L).

The hypothesis implies S0 = Sx∧y since Lemma 2.1 (1). Furthermore, we obtain c ∈ L \ (A(L) ∪ 0) since S0 < Sc and Lemma 2.1 (1). Combining x, y, c ∈ L \ (A(L) ∪ 0) with Corollary 2.2 and Sc = Sx ∧ Sy ≤ Sx, Sy, we may find c ≤ x, y. So, we believe c ≤ x ∧ y. Therefore, c ∈ A(L) ∪ 0 holds since x ∧ y ∈ A(L). This means Sc = S0 according to Lemma 2.1 (1). This is a contradiction to S0 < Sc.

The above Case 2.3.1 and Case 2.3.2 show that Sx∧y < Sx ∧ Sy will not happen.

Therefore, we confirm Sx∧y = Sx ∧ Sy since we have proved Sx∧y ≤ Sx ∧ Sy. □

Corollary 2.5

Let L be defined in Lemma 2.4. If Cls(L) is geometric, then Cls(L) is isomorphic to LCls(L).

Proof

Define a map π : LCls(L) → Cls(L) as

π ( x ) = S x , f o r x ∈ L C l s ( L ) ∖ { 0 } S 0 , f o r x = 0 .

We will demonstrate that π is a lattice isomorphism between two lattices LCls(L) and Cls(L).

By item (1) in Lemma 2.4, we find that π is a surjection. In light of both of items (1) and (2) in Lemma 2.1 with Corollary 2.2, we find that π is an injection.

Let x, y ∈ LCls(L) and x ≠ y.

If x = y = 0. Then we indicate π(x) = π(y) = π(x ∧ y) = π(x ∨ y) = S0. So, both of π(x) ∧ π(y) = π(x ∧ y) and π(x) ∨ π(y) = π(x ∨ y) hold.

If x = 0 and y ≠ 0. Then we indicate x ∧ y = 0, x ∨ y = y, π(x) = S0 and π(y) = Sy. Furthermore, we obtain π(x ∨ y) = π(y) = Sy = S0 ∨ Sy = π(x) ∨ π(y) and π(x ∧ y) = π(0) = S0 = S0 ∧ Sy = π(x) ∧ π(y).

If x ≠ 0 and y ≠ 0. Then we indicate x, y ∈ L \ (A(L) ∪ 0). By item (2.3) in Lemma 2.4, we easily obtain π(x ∧ y) = π(x) ∨ π(y).

If x ∧ y = 0, then we obtain S0 = Sx∧Sy by item (2.3) in Lemma 2.4. Thus, π(x ∧ y) = Sx ∧ Sy = π(x)∧π(y) is accepted.

If x ∧ y ≠ 0, then we obtain Sx∧y = Sx ∧ Sy by item (2.3) in Lemma 2.4. Thus π(x ∧ y) = Sx∧y = Sx ∧ Sy = π(x) ∧ π(y) is accepted.

Therefore, we have demonstrated π(x ∧ y) = π(x)∧π(y).

Summarizing the above, we affirm that π is a lattice isomorphism between LCls(L) and Cls(L) □

Lemma 2.6

Let L satisfy the conditions in ‘First status’. If L satisfies both of the following conditions, then Cls(L) is geometric.

  1. Let m, n ∈ L \ (A(L) ∪ 0 ∪ 1) and m||n. Then m ∧ n ≠ 0 holds.

  2. LCls(L) is geometric.

Proof

We will prove the needed results by the following three steps.

Step 1

We prove Cls(L) = {Sx | x ∈ L \ (A(L) ∪ 0)}.

Using Lemma 2.1 (1), we find {Sx | x ∈ L \ (A(L) ∪ 0)} ∪ {S0} = {Sx | x ∈ L \ {0}}. We may easily obtain {Sx | x ∈ L \ {0}} ⊆ Cls(L). Let S ∈ Cls(L) \ {Sx | x ∈ L \ {0}}. Then, considering Lemma 2.1 (1) and the definition of Cls(L), we find that there are at least two elements d, e ∈ S ⊆ L \ (A(L) ∪ 0 ∪ 1) with d||e. This implies d ∧ e = 0 by (C1). But this is a contradiction to the given condition (1). Thus, we decide Cls(L) = {Sx | x ∈ L \ {0}}.

Step 2

We prove Cls(L) ≅ LCls(L) as

Define a map π : LCls(L) → Cls(L) as

π ( x ) = S x , f o r x ∈ L C l s ( L ) ∖ { 0 } S 0 , f o r x = 0 .

We may easily find that π is well defined by Lemma 2.1 and Corollary 2.2.

We will prove that π is a lattice isomorphism. According to Step 1, we may confirm that π is a surjection.

We may easily decide that 0 < x in Cls(L) means S0 < Sx in Cls(L) by Lemma 2.1 (1) and Lemma 2.1 (2). Combining Step 1, we obtain that in Cls(L), S0 < S follows S = Sz for some z ∈ LCls(L) \ {0}.

Let x, y ∈ LCls(L) \ {0} with x ≠ y. According to Corollary 2.2 (2) and Lemma 2.1 (2), we obtain

x ≤ y i n L C l s ( L ) ⇔ π ( x ) ≤ π ( y ) i n C l s ( L ) .(*)

and also obtain x < y in LCls(L) ⇔ π(x) < π(y) in Cls(L)\{S0}. Therefore, we may state that π is an injection.

Let m, n ∈ L. We will prove that π keeps the two operators-meet ∧ and join ∨ in lattices by the following Parts (I1) and (I2).

Part (I1)

Let m = 0 and n ∈ LCls(L). This implies m ∨ n = n since LCls(L) is a lattice with 0 as the least element.

If n = 0. Then it is ease to see π(m ∨ n) = π(m ∧ n) = π(0) = S0 = π(m) ∨ π(n) = π(m) ∧ π(n).

If n ≠ 0. By Step 1 and Lemma 2.1, the formula π(n) = Sn > S0 is accepted. Thus, we obtain π(m ∨ n) = π(n) = Sn = S0 ∨ Sn = π(m) ∨ π(n) and π(m ∧ n) = π(0) = S0 = S0 ∧ Sn = π(m) ∧ π(n).

Part (I2)

Let m, n ∈ LCls(L) \ {0}.

Since m, n ≤ m ∨ n in LCls(L), we receive Sm, Sn ≤ Sm∨n. This implies Sm ∨ Sn ≤ Sm∨n.

Additionally, from Step 1 and Sm ∨ Sn ∈ Cls(L), we affirm that there is c ∈ LCls(L) satisfying Sm ∨ Sn = Sc. According to Sm ≤ Sc and the definition of LCls(L), we confirm m ≤ c and n ≤ c. So, c ∈ LCls(L) \ {0} holds. Furthermore, we gain m ∨ n ≤ c. Hence, we decide Sm∨n ≤ Sc since the expression (٭). This means Sm∨n ≤ Sm ∨ Sn. Therefore, we determine Sm∨n = Sm ∨ Sn. Thus, we can express π(m ∨ n) = Sm∨n = Sm ∨ Sn = π(m) ∨ π(n).

We will prove π(m ∧ n) = π(m) ∧ π(n).

When m ∧ n ≠ 0. We obtain Sm∧n, Sm, Sn ∈ Cls(L). From m ∧ n ≤ m, n, we may easily find Sm∧n ≤ Sm, Sn. Moreover, Sm∧n ≤ Sm ∧ Sn holds. So, there is b ∈ LCls(L) \ {0} satisfying Sb = Sm ∧ Sn. Thus, we obtain b ≤ m, n since Sb ≤ Sm, Sn. Meanwhile, we receive m ∧ n ≤ b since Sm∧n ≤ Sb. Furthermore, we find b = m ∧ n. Hence, Sm∧n = Sm ∧ Sn holds. That is to say, π(m ∧ n) = π(m) ∧ π(n) is correct.

When m ∧ n = 0.

If m ≤ n. Then according to m, n ∈ LCls(L) \ {0}, we determine m ∧ n = m ≠ 0. This is a contradiction to m ∧ n = 0.

If m||n. Then according to m, n ∈ LCls(L) \ {0} and m||n with the given condition (1), we determine m ∧ n ≠ 0. This is a contradiction to m ∧ n = 0. Therefore, up to now, we have demonstrated π(m ∧ n) = Sm∧n = Sm ∧ Sn = π(m) ∧ π(n).

Summing up, we determine that π is a lattice isomorphism. That is to say, Cls(L) ≅ LCls(L) holds.

Step 3

We prove that Cls(L) is geometric.

Considering the given condition (2) with Step 2, we may state that Cls(L) is geometric. Combining Lemmas 2.4 and 2.6, we may demonstrate the true of Theorem 2.3 as follows. ☐

Proof of Theorem 2.3

(⇒) If the condition (1) is not true then there are two elements x, y ∈ L \ (A(L) ∪ 0 ∪ 1) with x||y satisfying x ∧ y = 0. Hence, we obtain S = {x, y} ∪ {z ∈ A(L) | z ∉ A(x) and z ∉ A(y)} ∈ Cls(L).

However, considering Lemma 2.4(1), we confirm S ∉ Cls(L). This is a contradiction to the above S ∈ Cls(L). Therefore, the condition (1) is accepted. Using Corollary 2.5 and the geometric of Cls(L), we accept the condition (2).

(⇐) Applying Lemma 2.6, the needed result is followed. ☐

The following examples will express the correctness of Theorem 2.3.

Example 2.7

Let L1 be shown in Figure 1. It is easy to see that L1 is complete and atomistic. In addition, we also easily explore that L1 satisfies the conditions for ‘First status’. LCls(L) is shown as Figure 2. According to the definition of geometric lattice in Subsection 1.1, LCls(L1) is geometric.

Figure 1 diagram of L1
Figure 1

diagram of L1

Figure 2 diagram of LCls(L1)
Figure 2

diagram of LCls(L1)

We may easily see that the diagram of Cls(L) is in Figure 3, in which S0 = {a, b, c, d}, S1 = {1}, S(bc) = {(bc), a, d}, S(ab) = {(ab), c, d}, S(cd) = {(cd, a, b} and S(ab)(cd) = {(ab, (cd)}. From the definition of geometric lattice in Subsection 1.1, we decide that Cls(L1) is not geometric. In fact, (ab), (cd) ∈ L1 \ (A(L1) ∪ 0) satisfies (ab) ∧ (cd) = 0. In other words, L1 does not satisfy the given condition (1) in Theorem 2.3.

Figure 3 diagram of Cls(L1)
Figure 3

diagram of Cls(L1)

Example 2.8

Let L2 be shown as Figure 4. We may easily demonstrate that L2 satisfies the conditions in ‘First status’ and the condition (1) in Theorem 2.3.

Figure 4 diagram of L2
Figure 4

diagram of L2

The diagram of LCls(L2) is in Figure 5. We may confirm the geometry of LCls(L2). Hence, using Theorem 2.3, Cls(L2) is geometric.

Figure 5 diagram of LCls(L2)
Figure 5

diagram of LCls(L2)

The diagram of Cls(L2) is in Figure 6, in which S(ab) = {(ab), c}, S(ac) = {(ac), b}, S(bc) = {(bc), a}, S0 = {a, b, c} and S1 = {1}. We may easily prove the geometry of Cls(L2). This is the same to the result from Theorem 2.3.

Figure 6 diagram of Cls(L2)
Figure 6

diagram of Cls(L2)

2.2 Answer for ‘Second status’

This subsection will discover the answer to the open problem for the complete atomistic lattices satisfying the conditions in ‘Second status’.

The main result in this subsection is the following theorem.

Theorem 2.9

Let L satisfy the conditions in ‘Second status’. Then, Cls(L) is geometric if and only if L satisfies the following conditions.

  1. L is modular.

  2. L is compact.

  3. x = ⋁ d ∈ F 2 ( x ) d holds for any x ∈ L \ (A(L) ∪ 0).

To get Theorem 2.9, we need the following series of lemmas.

Lemma 2.10

Let L satisfy the conditions in ‘Second status’. If Cls(L) is geometric, then the following statements hold.

  1. L is modular.

  2. L is compact.

  3. x = ⋁ d ∈ F 2 ( x ) d holds for any x ∈ L \ (A(L) ∪ 0).

Proof

To prove item (1). Suppose that L is not modular. Then in view of the property (2) in Subsection 1.1, L contains a sublattice L′ which is isomorphic to N5. Let L′ = {a ∧ b, a, b, c, a ∨ b} with a ∧ b < c < a < a ∨ b, a ∧ b < b < a ∨ b, a||b and c||b. Thus, we obtain h′(a) ≥ 2 and h′(a ∨ b) ≥ 3 in which h′ is the height function of L′. Certainly, we may affirm h(a) ≥ 2 and h(a ∨ b) ≥ 3 since h′(a) ≥ 2 and h′(a ∨ b) ≥ 3 in which h is the height function in L. We will distinguish the following Cases 1 and 2 to continue the discussions.

Case 1

a ∧ b = 0.

Suppose that in L, x ∈ A(L) holds for any x ∈ [0, a ∨ b] with x||a. Then we obtain [S0, Sa∨b] ⊆ Cls(L) to be {S0, Sa} ∪ {Sy | a < y < a ∨ b} ∪ {Sz | 0 < z < a and z ∉ A(L)} ∪ {Sa∨b}. From the definition of Cls(L) and Lemma 2.1, we may easily find S0 < St < Ss for any t, s ∈ [0, a ∨ b] \ (A(L) ∪ 0) and t < s. Therefore, we obtain |A([S0, Sa∨b])| = 1 since the property (4.2) in Subsection 1.1 and h′(a) ≥ 2. We may assume A([S0, Sa∨b]) = {SA}. Then, Sa∨b is not a join of atoms in [S0, Sa∨b] since SA < Sa∨b. This implies that [S0, Sa∨b] is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since Cls(L) is geometric.

Suppose that there is d ∈ [0, a ∨ b] satisfying d ||a and d ∉ A(L). This implies h′(d) ≥ 2. Moreover, we decide S0 < Sa and S0 < Sd, and besides, we find Sa, Sd < Sa∨Sd = Sad = {a, d}∪{x ∈ A(L) | x ≰ A(a) and x ≰ A(d)} < Sa∨d ≤ Sa∨b.

Let {xt, t ∈ 𝕋} = {y | y ∈ [0, a ∨ b], y ≺ a ∨ b, h(y) ≥ 2}. Then we may be assured |𝕋| ≥ 1 since h′(a) ≥ 2 and h′(d) ≥ 2 follows h(a) ≥ 2 and h(d) ≥ 2 respectively. Additionally, for any x, z ∈ {xt, t ∈ 𝕋} and x ≠ z, x||z holds since the definitions of {xt, t ∈ 𝕋}. In addition, we confirm x ∧ z = 0 since the conditions in ‘Second status’.

Let Sxt, t∈𝕋 = {xt, t ∈ 𝕋} ∪ {x ∈ A(L) | x ∉ A(a) and x ∉ A(xt) for any t ∈ 𝕋}. For any S ∈ [S0, Sa∨b] \ {Sa∨b}, we may easily find S ≤ Sxt,t∈𝕋 < Sa∨b. This implies Sa ≤ Sxt,t∈𝕋 < Sa∨b for any Sa ∈ A([S0, Sa∨b]). Thus, we decide ⋁ S a ∈ A ( [ S 0 , S a ∨ b ] ) S a ≤ S x t , t ∈ T . Hence, we may state that Sa∨b is not a join of atoms in [S0, Sa∨b].

Furthermore, we confirm that the interval [S0, Sa∨b] ⊆ Cls(L) is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since Cls(L) is geometric.

Case 2

a ∧ b ≠ 0.

The supposition of a ∧ b ≠ 0 and a ∧ b < b taken together implies h(b) ≥ 2. Considering the conditions in ‘Second status’ with a||b and a, b < a ∨ b ≤ 1, we find a ∧ b = 0. This is a contradiction to the supposition of a ∧ b ≠ 0.

To summarize the whole discussion in Case 1 and Case 2, we can express that L′ does not exist. Furthermore, L is modular according to the property (2) in Subsection 1.1.

To prove item (2).

Otherwise, we can suppose that there is a ∈ L \ {0} and a ≤ ∨i∈𝕀xi for some xi ∈ L \ {0}, (i ∈ 𝕀) such that for any finite subset J ⊆ 𝕀, a ≰ ∨j∈Jxj holds. Thus, we may find ∨j∈Jxj < ∨i∈𝕀xi. Thereby, we confirm |𝕀| ≮ ∞.

In fact, if xs ≤ xt holds for some s, t ∈ 𝕀, then ∨i∈𝕀xi = ∨i∈𝕀\{s}xi is accepted. Hence, we can suppose that for any s, t ∈ 𝕀, s||t holds.

Let x1, x2 ∈ {xi, i ∈ 𝕀} satisfy x1 ≠ x2. Since a ≰ ∨j∈Jxj holds for any finite subset J ⊆ 𝕀, we may be assured that there is x3 ∈ {xi, i ∈ 𝕀} \ {x1, x2} satisfying x3||x1 ∨ x2. Thus, we obtain that the sublattice {0, x1, x3, x1 ∨ x2, x1 ∨ x2 ∨ x3} in L is isomorphic to N5. Therefore, L is not a modular. This is a contradiction to item (1).

Moreover, L must be compact.

To prove item (3).

We fulfill the proof with the following Cases 3.1 and 3.2.

Case 3.1

x ∈ ℱ2. The result is trivial.

Case 3.2

x ∉ ℱ2. Then we confirm ⋁ d ∈ F 2 ( x ) d ≤ x and ℱ2(x) ≠ ∅ since x ∈ L \ (A(L) ∪ 0).

If ⋁ d ∈ F 2 ( x ) d = x holds. This is the needed result.

If ⋁ d ∈ F 2 ( x ) d < x holds.

Since L is atomistic, we decide x = ( ⋁ d ∈ F 2 ( x ) d ) ∨ ( ⋁ a i ∈ A ( x ) a n d a i ≰ d f o r a n y d ∈ F 2 , i ∈ I a i ) satisfying {ai | ai ∈ A(x) and ai ≰ d for any d ∈ ℱ2, i ∈ I} ≠ ∅. According to item (2), we obtain that there are some dj ∈ ℱ2(x)(j = 1,...,n < ∞) and at ∈ {ai | ai ∈ A(x) and ai ≰ d for any d ∈ ℱ2; i ∈ I}(t = 1,...,m < ∞) satisfying ⋁ d ∈ F 2 ( x ) d = d 1 ∨ . . . ∨ d n a n d ⋁ a i ∈ A ( x ) a n d a i ≰ d f o r a n y d ∈ F 2 , i ∈ I a i = a 1 ∨ . . . ∨ a m .

In addition, owing to d1 ∈ ℱ2, we obtain that there is b ∈ A(x) satisfying b ≺ d1. Thus, we infer that the sublattice {x, d1, b, a1, 0} in L is isomorphic to N5. Hence, L is not modular. This is a contradiction to item (1).

Therefore, we demonstrate ⋁ d ∈ F 2 ( x ) d = x . ☐

We will reveal the construction of Cls(L).

Lemma 2.11

Let L satisfy the conditions in ‘Second status’. If L satisfies the following statements (1) and (2), then Cls(L) = {Sx | x ∈ L \ {0}} holds.

  1. L is modular.

  2. x = ⋁ d ∈ F 2 ( x ) d holds for any x ∈ L \ (A(L) ∪ 0).

Proof

According to Lemma 2.1 (1), we can obtain {Sx | x ∈ L \ (A(L) ∪ 0)} ∪ {S0} = {Sx | x ∈ L \ {0}}. It is easy to see {Sx | x ∈ L \ {0}} ⊆ Cls(L).

Let S ∈ Cls(L) \ {Sx | x ∈ L \ {0}}. Then, by S ≠ S0 and Lemma 2.1 (1), we affirm that there exists y ∈ S such that y ∉ A(L) ∪ 0.

Let {yj, j ∈ J} = {y | y ∈ S \ (A(L) ∪ 0)}. Owing to (C1), we obtain yα||yβ for any α, β ∈ J and α ≠ β. According to S ∉ {Sx | x ∈ L \ (A(L) ∪ 0)} ∪ {S0} and Lemma 2.1 (1), we obtain |J| ≥ 2. In addition, we may easily receive Syj < S for any j ∈ J in view of the definition of Cls(L).

Since |J| ≥ 2, we can select y1, y2 ∈ S\(A(L)∪0) with y1 ≠ y2. Thus, we confirm y1 || y2 and h(y1), h(y2) ≥ 2 since yj ∉ A(L) ∪ 0 follows h(yj) ≥ 2 (j = 1, 2) in which h is the height function in L.

Combining S ∈ Cls(L) and (C1), we obtain y1 ∧ y2 = 0. Considering the atomic of L with h(y1), h(y2) ≥ 2 and y1 || y2 and y1 ∧ y2 = 0, we obtain A(y1) \ A(y2) ≠ ∅ and A(y2) \ A(y1) ≠ ∅. Let a1 ∈ A(y1) \ A(y2) and a2 ∈ A(y2) \ A(y1). Then we decide that the sublattice {y1 ∨ y2, y1, a1, a2, 0} of L is isomorphic to N5. This is a contradiction to the given condition (1) according to property (2) in Subsection 1.1.

Therefore, we determine Cls(L) = {Sx | x ∈ L \ {0}}. ☐

Lemma 2.12

Let L satisfy the conditions in ‘Second status’ and the statements

  1. L is modular.

  2. L is compact.

  3. x = ⋁ d ∈ F 2 ( x ) d holds for any x ∈ L \ (A(L) ∪ 0).

Then ⋁ y ∈ Y S y = S ∨ y ∈ Y y holds for any {y | y ∈ 𝔜} ⊆ L\ (A(L) ∪ 0).

Proof

Since L is complete, we obtain ∨y∈𝔜y ∈ L. Let b = ∨y∈𝔜y. Then, we confirm S∨y∈𝔜y ∈ Cls(L).

From Sy ∈ Cls(L) (∀y ∈ 𝔜) and the completion of Cls(L), we receive ⋁ y ∈ Y S y ∈ C l s ( L ) . Combining Lemma 2.11, we decide ⋁ y ∈ Y S y = S z for some z ∈ L \ (A(L) ∪ 0) or ⋁ y ∈ Y S y = S 0 .

We distinguish two cases to continue the proof.

Case 1

⋁ y ∈ Y S y = S z for some z ∈ L \ (A(L) ∪ 0).

Considering y ≤ S∨y∈𝔜y for any y ∈ 𝔜 with Lemma 2.1 (2) and Corollary 2.2 (2), we decide Sy ≤ S∨y∈𝔜y. Hence, we affirm ⋁ y ∈ Y S y ≤ S ∨ y ∈ Y y . That is to say, Sz ≤ Sb holds. Using Lemma 2.1 (2) and Corollary 2.2, we obtain z ≤ b.

Since L is compact, we may be assured b = ∨y∈𝔜y = y1 ∨ ... ∨ yn for some yj ∈ 𝔜 with n < ∞, (j = 1,..., n).

In view of the selection of z, we obtain Sy ≤ Sz for any y ∈ 𝔜. Combining Lemma 2.1 (2) and Corollary 2.2 (2), we decide y ≤ z for any y ∈ 𝔜. Hence, we get yj ≤ z, (j = 1,..., n). So, we also get y1 ∨ ... ∨ yn ≤ z. That is, b ≤ z holds. Therefore, we produce Sb ≤ Sz according to Lemma 2.1 (2).

Thereby, we determine Sb = Sz.

Case 2

⋁ y ∈ Y S y = S 0 .

Since Sy ≤ ⋁ y ∈ Y S y = S 0 and Lemma 2.1 (1), we obtain y ∈ A(L). So, we believe {y | y ∈ 𝔜} ⊆ A(L). This is a contradiction to {y | y ∈ 𝔜} ⊆ L \ (A(L) ∪ 0). ☐

By extension of Lemmas 2.11 and 2.12, we will obtain the following result.

Lemma 2.13

Let L satisfy the conditions in ‘Second status’ and the following conditions.

  1. L is modular.

  2. L is compact.

  3. x = ⋁ d ∈ F 2 ( x ) d holds for any x ∈ L \ (A(L) ∪ 0).

Then, Cls(L) is geometric.

Proof

From the given three conditions, we confirm that L is geometric in view of the definition of geometric lattice in Subsection 1.1.

We prove that the complete lattice Cls(L) is geometric with the following steps.

Step 1

We will prove that Cls(L) is atomistic.

In virtue of Lemma 2.11, we find Cls(L) = {Sx | x ∈ L \ {0}}. With the given condition (3), we obtain Sy = S∨d∈ℱ2(y)d for any y ∈ L \ (A(L) ∪ 0). Combining Lemma 2.12, we receive S ∨ d ∈ F 2 ( y ) d = ⋁ d ∈ F 2 ( y ) S d

By the property (4.2) in Subsection 1.1, we decide {Sd | d ∈ ℱ2(y)} ⊆ A(Cls(L)) for any y ∈ L\(A(L)∪0).

Hence, we may state that Sy is the join of some atoms in Cls(L) for any y ∈ L \ (A(L) ∪ 0). Therefore, we indicate that Cls(L) is atomistic.

Step 2

We will prove that all atoms in Cls(L) are compact.

Let S ∈ A(Cls(L)). Then by the property (4.2) in Subsection 1.1, we obtain S = Sd for some d ∈ ℱ2.

Suppose S d ≤ ⋁ j ∈ J S x j for some xj ∈ L(j ∈ 𝕁). According to Lemma 2.12, we determine ⋁ j ∈ J S x j = S ∨ j ∈ J x j . These imply Sd ≤ S∨j∈𝕁xj. By Lemma 2.1 (2), we obtain d ≤ ∨j∈𝕁xj. Since L is compact, we obtain d ≤ x1 ∨ ... ∨ xn for some {x1,...., xn} ⊆ {xj, j ∈ 𝕁} and n < ∞. Thereby, combining Lemma 2.1 (2) and Lemma 2.12, we confirm Sd ≤ Sx1∨...∨xn = Sx1 ∨ ... ∨ Sxn. Thus, we can express that Sd is compact.

Therefore, every atom in Cls(L) is compact.

Step 3

We will prove that Cls(L) is modular.

Suppose that Cls(L) is not modular. Then in Cls(L), there is a sublattice {A, B, C, A ∧ C, A ∨ C} which is isomorphic to N5, where A||C, B||C, A ∧ C < A < B < A ∨ C and A ∧ C < C < A ∨ C.

According to Lemma 2.1, Lemma 2.11 and A ∧ C < A, B with A ∧ C ∈ Cls(L), we own Sa = A, Sb = B and Sc = C for some a, b, c ∈ L \ (A(L) ∪ 0). In light of Lemma 2.12, we may be assured A ∨ C = Sa ∨ Sc = Sa∨c.

Additionally, combining A||C, B||C, A < B and Lemma 2.1 with Corollary 2.2, we affirm a||c and b||c; a < b; b, c < a ∨ c.

From a||c and a, c ∈ L \(A(L)∪0), we confirm a ∧ c = 0 since L satisfies the conditions in ‘Second status’.

Taking Lemma 2.1 (2) and Corollary 2.2, we may be assured that the sublattice {0, a ∨ c, a, b, c} in L satisfies a||c and b||c, and 0 < a < b < a ∨ c and 0 < c < a ∨ c. This implies that the sublattice {0, a ∨ c, a, b, c} in L is isomorphic to N5. Thus, we find that L is not modular. This is a contradiction to the given condition (1). ☐

The proof of Theorem 2.9 follows.

Proof of Theorem 2.9

Combining all the results in Lemmas 2.10 and 2.13, we can express that Theorem 2.9 holds. ☐

The correctness of Theorem 2.9 has been proved. Hence, the following example will show that if a complete atomistic lattice L3 satisfies the conditions in ‘Second status’, but does not satisfy the conditions in Theorem 2.9, then Cls(L3) will not be geometric. This will show that the importance of Theorem 2.9 to decide the geometric of Cls(L) for a complete atomistic lattice L.

Example 2.14

Let L3 be defined as Figure 7. We may easily reveal that L3 satisfies all of conditions in ‘Second status’ and the conditions (2) and (3) in Theorem 2.9.

Figure 7 diagram of L3
Figure 7

diagram of L3

Since {1, (ab), b, 0, c} is N5 up to isomorphism. This means that L3 is not modular in view of property (2) in Subsection 1.1. By Theorem 2.9, we demonstrate that Cls(L3) is not geometric.

Actually, the diagram of Cls(L3) is shown as Figure 8, in which S(ab) = {(ab), c, d}, S(cd) = {(cd), a, b}, S(ab)(cd) = {(ab), (cd)}, S0 = {a, b, c, d} and S1 = {1}. We may easily find the non-geometry of Cls(L3) since S1 is not the join of atoms in Cls(L3). This is the same consequence of that by Theorem 2.9.

Figure 8 diagram of Cls(L3)
Figure 8

diagram of Cls(L3)

3 Conclusion

Utilizing the results in Subsection 1.2 with Theorems 2.3 and 2.9, we completely respond to Radeleczki’s open problem which hope to characterize complete atomistic lattice whose classification lattice is geometric in [2]. For the answer of the open problem, we will discover the applied fields, such as concept lattice theory, in the future.

Acknowledgement

This research is supported by National Nature Science Foundation of China (61572011).

References

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Received: 2016-9-10
Accepted: 2017-5-25
Published Online: 2017-7-13

© 2017 Mao

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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Geometric; Classification system; Classification lattice; Complete atomistic lattice
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