Luigi Santocanale
email: luigi.santocanale@lif.univ-mrs.fr

Friedrich Wehrung
email: wehrung@math.unicaen.fr

VARIETIES OF LATTICES WITH GEOMETRIC DESCRIPTIONS

Keywords: July 1, 2011. 2010 Mathematics Subject Classification. Primary 06B20. Secondary 06B23, 06B05, 06B15, 06B35, 06C05 Lattice, complete, algebraic, dually algebraic, ideal, filter, upper continuous, lower continuous, modular, join-semidistributive, point, seed, spatial, strongly spatial, cover, irredundant cover, tight cover, minimal cover

published or not. The documents may come

Introduction

An element p in a lattice L is completely join-irreducible, or a point, if there is a largest element smaller than p. We say that L is spatial if every element of L is a (possibly infinite) join of points. In such a case, elements of L can be identified with certain sets of points of L. If, in addition, L is algebraic, then we say that we have a geometric description of L. When dealing with equational properties of lattices, the geometric description enables to prove representation results that may have been very hard to obtain otherwise.

A prominent illustration of such methods is given in Herrmann, Pickering, and Roddy [START_REF] Herrmann | A geometric description of modular lattices[END_REF], where it is proved that every modular lattice L embeds into some algebraic and spatial lattice L that satisfies the same identities as L-we say that L embeds into L within its variety. In particular, as L is modular, so is L. This is used in [START_REF] Herrmann | A geometric description of modular lattices[END_REF] to prove that a lattice L embeds into the subspace lattice of a vector space over an arbitrary field iff L is modular and 2-distributive.

Nevertheless it was not known whether every lattice embeds, within its variety, into some algebraic and spatial lattice. (This question is stated in the comments following Semenova and Wehrung [START_REF] Semenova | Sublattices of lattices of order-convex sets, III. The case of totally ordered sets[END_REF]Problem 4].) We find a counterexample to that question in Theorem 9.5. This counterexample is join-semidistributive-in fact, it generates a variety of join-semidistributive lattices.

Yet even for join-semidistributive lattices, there are many situations where lattices enjoy geometric descriptions. Such geometric descriptions are massively used in Semenova and Wehrung [START_REF] Semenova | Sublattices of lattices of order-convex sets, I. The main representation theorem[END_REF][START_REF] Semenova | Sublattices of lattices of order-convex sets. II. Posets of finite length[END_REF][START_REF] Semenova | Sublattices of lattices of order-convex sets, III. The case of totally ordered sets[END_REF] or Semenova and Zamojska-Dzienio [START_REF] Semenova | On lattices embeddable into lattices of orderconvex sets. Case of trees[END_REF] for descriptions of lattices of order-convex subsets of various kinds of posets. Denote by Co(P ) the lattice of all order-convex subsets of a poset P , and by SUB the class of all lattices that can be embedded into some Co(P ). It is proved in Semenova and Wehrung [START_REF] Semenova | Sublattices of lattices of order-convex sets, I. The main representation theorem[END_REF] that SUB is a finitely based variety of lattices. It is asked in Semenova and Wehrung [START_REF] Semenova | Sublattices of lattices of order-convex sets, III. The case of totally ordered sets[END_REF]Problem 4] whether every lattice in SUB can be embedded, within its variety, into some algebraic and spatial lattice. We prove in the present paper that this is the case-we actually get a stronger result (Theorem 5.3): For every positive integer n, every n-distributive lattice can be embedded, within its variety, into some algebraic and spatial lattice. As all lattices of the form Co(P ) are 2-distributive, this solves the question above. The lattices obtained in Theorem 5.3 are actually strongly spatial (cf. Definition 3.5), which means spatial plus the fact that minimal join-covers between points are well-behaved. For modular lattices, spatial implies strongly spatial (cf. Theorem 7.4) and so the distinction is immaterial.

A further consequence of Theorem 5.3 is that the variety of all n-distributive lattices is generated by its finite members, thus it has a decidable word problem for free lattices (Theorem 6.3). This solves two problems contained in Huhn [START_REF] Huhn | On nonmodular n-distributive lattices: the decision problem for identities in finite n-distributive lattices[END_REF][START_REF] Huhn | On nonmodular n-distributive lattices. I. Lattices of convex sets[END_REF].

We also prove that every lattice which is either well-founded or bi-algebraic is strongly spatial (Corollary 3.13). Hence our main counterexample (Theorem 9.5) extends the result, proved by the second author in [START_REF] Wehrung | Sublattices of complete lattices with continuity conditions[END_REF], that not every lattice can be embedded into some bi-algebraic lattice.

Basic concepts

We set

Q ↓ X := {q ∈ Q | (∃x ∈ X)(q ≤ x)} , Q ↑ X := {q ∈ Q | (∃x ∈ X)(q ≥ x)} , Q X := {q ∈ Q | (∃x ∈ X)(q < x)} ,
for all subsets X and Q in a poset P . We also set Q ↓ a := Q ↓ {a}, Q ↑ a := Q ↑ {a}, and Q a := Q {a}, for each a ∈ P . A subset X of P is a lower subset of P if X = P ↓ X.

A subset X of P refines a subset Y of P , in notation X ≤ ref Y , if X ⊆ P ↓ Y . (As the present work touches upon algebraic and continuous lattices, where x ≪ y denotes the "way-below" relation, we shall stray away from the usual notation X ≪ Y for the refinement relation on subsets.) We shall also write X < ref Y for the conjunction of X ≤ ref Y and X = Y .

For elements x and y in a poset P , let x ≺ y hold (in words, "x is a lower cover of y", or "y is an upper cover of x") if x < y and there is no z ∈ P such that x < z < y. An element p in a join-semilattice L is join-irreducible if p = X implies that p ∈ X, for any finite subset X of L (in particular, taking X := ∅, this rules out p being the zero element of L); completely join-irreducible-from now on a point, if the set of all elements smaller than p has a largest element, then denoted by p * . Of course, every point is join-irreducible;

-an atom of L if L has a zero element and 0 ≺ p. We denote by J(L) (J c (L), At(L), respectively) the set of all join-irreducible elements (points, atoms, respectively) of L. Trivially, At(L) ⊆ J c (L) ⊆ J(L).

We say that a subset Σ of L is join-dense in L if every element of L is a join of elements of Σ. Equivalently, for all a, b ∈ L with a b, there exists x ∈ Σ such that x ≤ a and x b. An element a in L is compact (resp., countably compact ) if for every nonempty directed (resp., countable nonempty directed) subset D of L with a join, a ≤ D implies that a ∈ L ↓ D. We say that L is spatial if the set of all points of L is join-dense in L; atomistic if the set of all atoms of L is join-dense in L; compactly generated if the set of all compact elements of L is join-dense in L; algebraic if it is complete and compactly generated; bi-algebraic if it is both algebraic and dually algebraic. It is well known that every dually algebraic lattice is spatial-see Gierz et al. [START_REF] Gierz | Continuous Lattices and Domains[END_REF]Theorem I.4.25] or Gorbunov [START_REF] Gorbunov | Algebraic theory of quasivarieties[END_REF]Lemma 1.3.2].

A lattice L is upper continuous if the equality a ∧ D = (a ∧ D) (where a ∧ D := {a ∧ x | x ∈ D}) holds for every a ∈ L and every nonempty directed subset D of L with a join. Lower continuity is defined dually. Let L be a (∨, 0)-semilattice. We denote by Σ ∨ the set of all finite joins of elements of Σ (0 included), for each subset Σ ⊆ L. Furthermore, we denote by Id L the ideal lattice of L. Dually, for a (∧, 1)-semilattice L, we denote by Fil L the lattice of all filters (i.e., dual ideals) of L, partially ordered under reverse inclusion.

A lattice is • n-distributive (where n is a positive integer) if it satisfies the identity

x ∧ 0≤i≤n y i = 0≤i≤n x i ∧ 0≤j≤n, j =i y j , (2.1)
• join-semidistributive if it satisfies the quasi-identity

x ∨ y = x ∨ z =⇒ x ∨ y = x ∨ (y ∧ z) . We define a sequence (p n | n < ω) of ternary lattice terms by p 0 (x, y, z) := y , p n+1 (x, y, z) := y ∧ (x ∨ p n (x, z, y)) ,
for all n < ω .

Observe that the lattice inclusions y ∧ z ≤ p n (x, y, z) ≤ y, for n < ω, are valid in all lattices. Denote by (SD n ∨ ) the lattice identity p n (x, y, z) ≤ x ∨ (y ∧ z). In particular, distributivity is equivalent to (SD 1 ∨ ), and (SD n ∨ ) implies (SD n+1 ∨ ). If a lattice satisfies (SD n ∨ ) for some n, then it is join-semidistributive; the converse holds for finite lattices, see Jónsson and Rival [START_REF] Jónsson | Lattice varieties covering the smallest nonmodular variety[END_REF] or Jipsen and Rose [START_REF] Jipsen | Varieties of Lattices[END_REF]Section 4.2].

We shall denote by con L (x, y) the least congruence of a lattice L that identifies elements x, y ∈ L. For a congruence θ of L, we shall write x ≡ θ y instead of (x, y) ∈ θ and x ≤ θ y instead of (x ∨ y, y) ∈ θ.

We shall denote by 2 = {0, 1} the two-element lattice and by ω the chain of all non-negative integers. For any poset P , we shall set P -:= P \ {0} if P has a least element, and P -:= P otherwise.

E ⊆ L -. We set iCov(p) := E ⊆ L -finite | E covers p irredundantly , iCov = (p) := E ⊆ L -finite | E joins to p irredundantly = iCov(p) ∩ Cov = (p) .
The elements of iCov(p) are called the irredundant join-covers of p, while the elements of iCov = (p) are called the irredundant join-representations of p.

• E covers (resp., joins to) p tightly if E ⊆ L -, E covers p (resp., joins to p), and p x ∨ (E \ {u}) for each u ∈ E and each x < u. We set

tCov(p) := E ⊆ L -finite | E covers p tightly , tCov = (p) := E ⊆ L -finite | E joins to p tightly = tCov(p) ∩ Cov = (p) .
The elements of tCov(p) are called the tight join-covers of p, while the elements of tCov = (p) are called the tight join-representations of p. • E covers (resp., joins to) p minimally if E ⊆ L, E covers p (resp., joins to p), and p ≤ X and X ≤ ref E implies that E ⊆ X for each finite subset X of L. Observe that both conditions imply that E ⊆ L -. We set

mCov(p) := E ⊆ L -finite | E covers p minimally , mCov = (p) := E ⊆ L -finite | E joins to p minimally = mCov(p) ∩ Cov = (p) .
The elements of mCov(p) are called the minimal join-covers of p, while the elements of mCov = (p) are called the minimal join-representations of p.

Minimality comes in more than one way in defining mCov(p):

Lemma 3.2 (folklore). The minimal join-covers of p are exactly the ≤ ref -minimal elements of iCov(p).

We define similarly irredundant covers for vectors, by saying, for example, that a family (a i | i ∈ I) (where I is finite) covers p irredundantly if the set {a i | i ∈ I} covers p irredundantly and the map i → a i is one-to-one. A similar definition applies to tight, resp. minimal, covers, resp. join-representations.

Observe that we allow the possibility of trivial join-covers p ≤ X, that is, those X such that p ∈ L ↓ X.

It is a straightforward exercise to verify that the containments

mCov(p) ⊆ tCov(p) ⊆ iCov(p) ⊆ Cov(p)
hold, with none of the converse containments holding as a rule, with easy counterexamples for finite lattices. A similar comment applies to the containments mCov

= (p) ⊆ tCov = (p) ⊆ iCov = (p) ⊆ Cov = (p).
Lemma 3.3 (folklore). Let p be an element in a join-semilattice L. Then every element

E of mCov(p) is contained in J(L). Furthermore, if p is compact, then E is contained in J c (L).
Proof. If one of the elements of E is the zero of L, the set X := E \ {0} belongs to Cov(a) and refines E, thus contains E, a contradiction; so all elements of E are nonzero. If q ∈ E is not join-irreducible, then q = x ∨ y for some x, y < q, so the set X := {x, y} ∪ (E \ {q}) refines E and belongs to Cov(a), and so E ⊆ X, and thus q ∈ X, a contradiction. Therefore, E ⊆ J(L). Now assume that p is compact and that an element q of E is not a point. As q is join-irreducible, the set D := L q is nonempty directed with join q. Setting b := (E \ {q}), it follows that p ≤ E = b ∨ D, thus, as p is compact, there exists x ∈ D such that p ≤ b ∨x. It follows that the set X := {x} ∪(E \ {q}) belongs to Cov(p), thus, as X refines E, we get E ⊆ X, and thus q ∈ X, a contradiction.

The following result originates in the proof of Nation [START_REF] Nation | An approach to lattice varieties of finite height[END_REF]Theorem 3.2]. Lemma 3.4. Let p be an element in a complete and lower continuous lattice L. Then every element of Cov(p) can be refined to some element of tCov(p).

Proof. Let E = {p 1 , . . . , p m } ∈ Cov(p), with m < ω, and set

X := (x 1 , . . . , x m ) ∈ L m | p ≤ m i=1
x i and x i ≤ p i for each i , ordered componentwise. It follows from the completeness and lower continuity of L that every nonempty chain of X has a meet, which also belongs to X. By Zorn's Lemma, X has a minimal element, say (q 1 , . . . , q m ). Then F := {q 1 , . . . , q m } \ {0} refines E, and it belongs to tCov(p).

In case p is a nonzero element in an infinite complete atomless Boolean lattice L, then mCov(p) is empty. In particular, tCov(p) cannot be replaced by mCov(p) in the statement of Lemma 3.4. However, searching in which particular cases this can be done leads to fascinating problems. Partial (and nontrivial) positive answers are introduced in Theorem 3.12 and Lemma 5.1.

Join-seeds were introduced in Semenova and Wehrung [START_REF] Semenova | Sublattices of lattices of order-convex sets, III. The case of totally ordered sets[END_REF], in a context where all lattices were 2-distributive. The seeds that we define here are related. Definition 3.5. A subset Σ in a join-semilattice L is -a pre-seed if for each p ∈ Σ and each X ∈ Cov(p), there exists I ∈ Cov(p) contained in Σ such that I ≤ ref X; -a quasi-seed if it is a join-dense pre-seed contained in J(L); -a seed if it is join-dense, contained in J(L), and for each p ∈ Σ and each X ∈ Cov(p), there exists

I ∈ mCov(p) contained in Σ such that I ≤ ref X. We say that L is strongly spatial if J c (L) is a seed in L.
As a seed is join-dense, every strongly spatial lattice is spatial. In the distributive case, the two statements are equivalent: indeed, every algebraic, distributive, and spatial lattice is obviously strongly spatial (join-irreducible elements have no nontrivial join-covers). This is also the case for modular lattices, but the proof is harder, see Theorem 7.4. Another easy case of strong spatiality is provided by the following result. Proposition 3.6. Every algebraic atomistic lattice is strongly spatial.

Proof. The points of an algebraic atomistic lattice L are exactly its atoms. Now if A is a finite cover of an atom p of L, it follows from the compactness of p together with the fact that each element of A is a join of atoms that there exists a finite cover X of p, consisting only of atoms, refining A. Now every irredundant cover Y of p contained in X refines A and belongs to mCov(p).

It is easy to see that in the non-modular case, algebraic and spatial does not imply strongly spatial. For example, let

ω ∂ := {n * | n < ω} with 0 * > 1 * > 2 * > • • • . Then the lattice L := ω ∂ ∪ {0, c} ,
with the only new relations 0 < c < 0 * and 0 < n * for each n < ω, is algebraic and spatial although not strongly spatial. This example is 2-distributive, and not dually algebraic. The latter observation is also a consequence of the forthcoming Corollary 3.13. In order to prepare for the proof of that result, we shall establish a few lemmas with independent interest. From Lemma 3.7 to Proposition 3.11 we shall fix an element p in a join-semilattice L.

Lemma 3.7. Let A 0 , A 1 ∈ tCov(p) such that A 1 ≤ ref A 0 .
Then for each a ∈ A 0 , the set A 1 ↓ a joins to a tightly. Moreover, each element of A 0 contains an element of A 1 and

A 0 = A 1 . Proof. Set a := (A 1 ↓ a), for each a ∈ A 0 . The assumption that A 1 refines A 0 means that A 1 = (A 1 ↓ a | a ∈ A 0 ), thus p ≤ A 1 = (a | a ∈ A 0 ). As a ≤ a
for each a and as A 0 ∈ tCov(p), it follows that a = a for each a ∈ A 0 . In particular, a contains an element of

A 1 . Now let a ∈ A 0 and b ∈ A 1 ↓ a, and let y ≤ b such that a = y ∨ ((A 1 ↓ a) \ {b}). Set B := A 1 \ (A 1 ↓ a). As p ≤ A 1 = (A 1 ↓ a) ∨ B = a ∨ B = y ∨ (A 1 ↓ a) \ {b}) ∪ B while ((A 1 ↓ a) \ {b}) ∪ B is contained in A 1 \ {b}, we obtain the inequality p ≤ y ∨ (A 1 \ {b}), which, as A 1 ∈ tCov(p)
, implies that y = b. This proves that A 1 ↓ a joins to a tightly. Furthermore,

A 0 = (A 1 ↓ a) | a ∈ A 0 (by the paragraph above) = (A 1 ↓ a | a ∈ A 0 ) = A 1 (because A 1 ≤ ref A 0 ) .
We shall use later the following immediate consequence of Lemma 3.7.

Corollary 3.8. Let A ∈ tCov(p). If A ⊆ J(L), then A ∈ mCov(p). Lemma 3.9. Let A 0 , A 1 ∈ tCov(p) such that A 1 ≤ ref A 0 . Then the sets A 1 ↓ x, for x ∈ A 0 , are pairwise disjoint.
Proof. Suppose that there are x ∈ A 0 and z ∈ (A 1 ↓ x) ∩ B where

B := (A 1 ↓ y | y ∈ A 0 \ {x}) . From A 1 ≤ ref A 0 it follows that A 1 = (A 1 ↓ x) ∪ B.
Furthermore, it follows from Lemma 3.7 that the element x ′ := ((A 1 ↓ x) \ {z}) (defined as a new zero element in case L has no zero and

A 1 ↓ x = {z}) is (strictly) smaller than x since, by Lemma 3.7, x = A 1 ↓ x tightly, while B = (A 0 \ {x}). Now we compute p ≤ A 1 = (A 1 ↓ x) ∪ B = ((A 1 ↓ x) \ {z}) ∪ B (because z ∈ B) = x ′ ∨ (A 0 \ {x}) ,
which contradicts the assumption that A 0 covers p tightly.

Lemma 3.10.

Let A 0 , A 1 ∈ tCov(p) such that A 1 ≤ ref A 0 . Then |A 0 ∩ H| ≤ |A 1 ∩ H| for each lower subset H of L. In particular, |A 0 ↓ a| ≤ |A 1 ↓ a| for each a ∈ L.
Proof. For each u ∈ A 0 ∩ H, it follows from Lemma 3.7 that A 1 ↓ u is nonempty; pick an element f (u) there. It follows from Lemma 3.9 that f is one-to-one. As the range of f is contained in A 1 ∩ H, the first conclusion follows. The second conclusion is a particular case of the first one, with H := L ↓ a.

Although we shall not use the following result in the proof of Theorem 3.12, we record it for its independent interest.

Proposition 3.11 (Interpolation property for tight covers). Let

A 0 , A 1 , A 2 ∈ tCov(p) such that A 2 ≤ ref A 1 ≤ ref A 0 and let (a 0 , a 2 ) ∈ A 0 × A 2 . If a 2 ≤ a 0 , then there exists a 1 ∈ A 1 such that a 2 ≤ a 1 ≤ a 0 . Proof. As a 2 ∈ A 2 and A 2 ≤ ref A 1 , there exists a 1 ∈ A 1 ↑ a 2 . Likewise, as a 1 ∈ A 1 and A 1 ≤ ref A 0 ,
there exists a ∈ A 0 ↑ a 1 . Now a 2 belongs to both sets A 2 ↓ a 0 and A 2 ↓ a, thus, by Lemma 3.9, a = a 0 ; and thus a 1 ≤ a 0 . Theorem 3.12. Let p be an element in a complete, lower continuous lattice L. If either p is countably compact or L is either well-founded or dually well-founded, then every join-cover of p can be refined to some minimal join-cover of p.

Proof. Let C ∈ Cov(p), we wish to refine C to an element of mCov(p). By Lemmas 3.2 and 3.4, it suffices to prove that tCov(p), endowed with the refinement order, is well-founded. Suppose otherwise. There exists a sequence A = (A n | n < ω) from tCov(p) such that the inequality A n+1 < ref A n holds for each n < ω.

Say that an element x ∈ L is A-reducible if there exists a natural number k such that |A k ↓ x| ≥ 2. By Lemma 3.10, the sequence

(|A k ↓ x| | k < ω) is nondecreasing, thus the A-reducibility of x is equivalent to saying that |A k ↓ x| ≥ 2 for all large enough k < ω. Now we set B n := x ∈ A n | x is A-reducible , for each n < ω .
Observe that by Lemma 3.7,

A n ↓ x = {x} for all m ≤ n < ω and all x ∈ A m \ B m . (3.1)
In particular, if

B n = ∅, then A n ⊆ A n+1 , thus A n = A n+1
as each of these sets covers p tightly, which contradicts the assumption that A n+1 < ref A n . Therefore, B n is nonempty. Furthermore, for each n < ω, there exists k > n such that |A k ↓ x| ≥ 2 for each x ∈ B n . An easy inductive argument yields a strictly increasing sequence (n i | i < ω) of natural numbers, with n 0 = 0, such that for all i < ω and all x ∈ B ni , the set A ni+1 ↓ x has at least two elements. Set A ′ := (A ni | i < ω). As, by Lemma 3.10, the notions of A-reducibility and A ′ -reducibility are equivalent, we may replace A by A ′ and thus assume that n i = i for each i < ω. Hence For all m < n < ω and for all

x ∈ B m , |A n ↓ x| ≥ 2 . (3.2) Claim. B n ∩ A n+1 = ∅ and B n+1 refines B n , for each n < ω. Proof of Claim. Let u ∈ B n ∩ A n+1 . It follows from Lemma 3.7 that A n+1 ↓ u cov- ers u tightly, but u ∈ A n+1 , thus A n+1 ↓ u = {u}, which contradicts the assumption that u ∈ B n together with (3.2). Hence B n ∩ A n+1 = ∅. Now let v ∈ B n+1 . There exists u ∈ A n such that v ≤ u. If u / ∈ B n , then, by (3.1), A n+1 ↓ u = {u}, thus v = u, a contradiction as u is A-irreducible while v is A-reducible; so u ∈ B n .
Claim. Now we consider the set T of all finite sequences x = (x 0 , x 1 , . . . , x n ), where n < ω (the length of x), x i ∈ B i for each i ≤ n, and x i+1 ≤ x i for each i < n. By our Claim, it follows that x i+1 < x i for each i < n. Furthermore, for each positive integer n, we may pick x n ∈ B n (because B n = ∅), then, using our Claim, x n-1 ∈ B n-1 such that x n ≤ x n-1 , and so on by induction; we get a finite sequence (x 0 , x 1 , . . . , x n ) ∈ T . As this can be done for every n, the set T is infinite.

As each A n is finite, every element of T has only finitely many upper covers for the initial segment ordering. By König's Theorem, T has an infinite branch, say (x n | n < ω). As this branch is a (strictly) decreasing sequence, L is not wellfounded; thus, by assumption, either L is dually well-founded or p is countably compact.

Set

y n := (A n \ {x n }) for each n < ω. As A n ∈ tCov(p), we get p ≤ x n ∨ y n and (∀z < x n )(p z ∨ y n ) . (3.3) Furthermore, let u ∈ A n \ {x n } and let v ∈ A n+1 ↓ u. If v = x n+1
, then, by Lemma 3.9, x n = u, a contradiction; hence v ∈ A n+1 \ {x n+1 }. By joining over all the possible u-s and v-s, we obtain, using Lemma 3.7, that

y n = (A n \ {x n }) = (A n+1 ↓ u | u ∈ A n \ {x n }) ≤ (A n+1 \ {x n+1 }) = y n+1 . Now set x := (x n | n < ω) (directed meet) and y := (y n | n < ω) (directed join). From (3.3) it follows that p ≤ x n ∨ y for each n < ω, thus, as L is lower continuous, p ≤ x ∨ y.
If L is dually well-founded, then there exists m < ω such that y m = y, so p ≤ x ∨ y m . If L is not dually well-founded, then, by assumption, p is countably compact, thus, as p ≤ x ∨ y = n<ω (x ∨ y n ) (directed join), there exists m < ω such that p ≤ x ∨ y m . Hence the latter conclusion holds in every case, which, as x ≤ x m+1 < x m , contradicts (3.3). Corollary 3.13. Let L be a lattice. If L is either well-founded or bi-algebraic, then it is strongly spatial.

Proof. A direct inductive argument (within L) easily shows that if L is well-founded, then L is spatial. This conclusion also holds in case L is bi-algebraic, because then L is dually algebraic. Furthermore, if L is bi-algebraic, then, by Proposition 2.1, every point of L is compact. The conclusion then follows from Theorem 3.12.

Not every algebraic strongly spatial lattice is bi-algebraic. For example, the lattice Co(ω) of all order-convex subsets of the chain ω of all natural numbers is algebraic and atomistic, thus strongly spatial (cf. Proposition 3.6). On the other hand, it is not dually algebraic-in fact, by Wehrung [30, Corollary 12.5], Co(ω) cannot be embedded into any bi-algebraic lattice.

From quasi-seeds to algebraic spatial lattices

The following easy result relates quasi-seeds to spatial lattices. Proposition 4.1. Let L be a compactly generated lattice. Then L is spatial iff J c (L) is a quasi-seed in L.

Proof. If J c (L) is a quasi-seed in L, then it is join-dense, thus L is spatial. Conversely, assume that L is spatial. Let p ∈ J c (L) and let X ∈ Cov(p), we must find

I ∈ Cov(p) contained in J c (L) such that I ≤ ref X.
As p is a point, it is compact (cf. Proposition 2.1); as p ≤ X and each element of X is a join of points, we may assume that each element x ∈ X is a finite join of points, so x = I x where I x is a finite subset of J c (L). It follows that the set I := x∈X I x belongs to Cov(p), is contained in J c (L), and refines X.

In the statement of the following lemma we make use of (covariant) Galois connections, see for example Gierz et al. [START_REF] Gierz | Continuous Lattices and Domains[END_REF]. Lemma 4.2. Let Σ be a subset in a (∨, 0)-semilattice L. We define mappings ε : Id Σ ∨ → Id L and π : Id L → Id Σ ∨ by the rules

ε(A) := L ↓ A , for each A ∈ Id Σ ∨ , (4.1) π(B) := B ∩ Σ ∨ , for each B ∈ Id L . (4.2)
Then the following statements hold: (i) The pair (ε, π) is a Galois connection between Id Σ ∨ and Id L.

(ii) π • ε is the identity on Id Σ ∨ . In particular, π is surjective.

(iii) The subset Σ is a pre-seed in L iff π is a lattice homomorphism.

Proof. (i). We must prove that ε(A) ⊆ B iff A ⊆ π(B), for each A ∈ Id Σ ∨ and each B ∈ Id L. This is trivial.

(ii). We must prove that A = Σ ∨ ↓ A, for each A ∈ Id Σ ∨ . This is trivial.

(iii). Assume first that Σ is a pre-seed. It follows from (i) and (ii) that π is a surjective meet-homomorphism. (The map ε is also a one-to-one join-homomorphism, but we will not use this fact.) Hence it suffices to prove that π Conversely, assume that π is a lattice homomorphism. The assignment a → π(L ↓ a) = Σ ∨ ↓ a then defines a (∨, 0)-homomorphism ψ : L → Id Σ ∨ . Let p ∈ Σ, let n be a positive integer, and let a 0 , . . . , a n-1 ∈ L such that p ≤ i<n a i . Hence p belongs to ψ i<n a i = i<n ψ(a i ), that is, there are x i ∈ ψ(a i ), for i < n, such that p ≤ i<n x i . For each i < n, there exists a finite subset X i of Σ ↓ a i such that x i = X i . Therefore, the set X := i<n X i is contained in Σ, refines {a i | i < n}, and p ≤ i<n x i = X. Therefore, Σ is a pre-seed. Lemma 4.3. Let L be a 0-lattice with a quasi-seed Σ and set S := Σ ∨ . Then Id S is an algebraic and spatial lattice, which contains a copy of L as a 0-sublattice, and which belongs to the same lattice variety as L. Furthermore, if Σ is a seed in L, then Id S is strongly spatial.

(A ∨ B) is contained in π(A) ∨ π(B), for all A, B ∈ Id L. As π(A ∨ B) is an ideal of Σ ∨ ,
Proof. Denote by η : L ֒→ Id L, a → L ↓ a the canonical lattice embedding. As, by Lemma 4.2, the natural map π : Id L ։ Id S is a lattice homomorphism, the composite ψ := π •η is also a lattice homomorphism; it is obviously zero-preserving. Note that ψ(a) = S ↓ a, for each a ∈ L. As Σ is join-dense in L, it follows immediately that ψ is a 0-lattice embedding from L into Id S. As Id S is a homomorphic image of Id L and the latter belongs to the lattice variety generated by L (cf. Grätzer [START_REF] Grätzer | General Lattice Theory[END_REF]Lemma I.4.8]), so does Id S.

The lattice Id S, being the ideal lattice of a (∨, 0)-semilattice, is an algebraic lattice. Now we claim that S ↓ p is a point of Id S , with lower cover S p , for each p ∈ Σ .

(4.3) Indeed, as p is join-irreducible, the subset S p is an ideal of S; furthermore, every ideal of S properly contained in S ↓ p is contained in S p, which completes the proof of our claim. Now let A, B ∈ Id S such that A ⊆ B and let a ∈ A \ B. As a is a finite join of elements of Σ, one of those elements belongs to A \ B. This proves that the set P := {S ↓ p | p ∈ Σ} is join-dense in Id S. In particular, every point of Id S, being a join of elements of P , belongs to P . Therefore, P is the set of all points of Id S. As P is join-dense in Id S, it follows that Id S is spatial.

Finally assume that Σ is a seed in L. We must prove that P is a seed in Id S. Let p ∈ J c (L), let n be a positive integer, and let A 0 , . . . , A n-1 be ideals of S such that S ↓ p ⊆ i<n A i (i.e., p ∈ i<n A i ) in Id S. It follows that there are a 0 ∈ A 0 , . . . , a n-1 ∈ A n-1 such that p ≤ i<n a i . As Σ is a seed in L, there exists I ∈ mCov(p) contained in Σ such that I ≤ ref {a 0 , . . . , a n-1 }; the latter relation implies that {S ↓ q | q ∈ I} refines {A i | i < n}. From p ≤ I it follows that S ↓ p ⊆ q∈I (S ↓ q). If this relation is not a minimal join-covering, then, by (4.3) applied to S ↓ q for all q ∈ I, there exist q ∈ I and q ′ < q such that p ≤ q ′ ∨ (I \ {q}), which contradicts I ∈ mCov(p). We have thus proved that {S ↓ q | q ∈ I} is a minimal join-cover of S ↓ p in Id S which refines {A i | i < n}.

Seeds in n-distributive lattices

The following lemma is a slight improvement of Nation [25, Theorem 3.2], and its proof is virtually the same.

Lemma 5.1. Let n be a positive integer and let p be a join-irreducible element in a complete lower continuous n-distributive lattice L. Then every element of Cov(p) can be refined to some element of mCov(p).

Proof. As L is n-distributive and p is join-irreducible, every element of iCov(p) must have at most n elements. Now let E ∈ Cov(p). By the above observation, there exists a maximal-sized irredundant join-cover P of p refining E, and further, denoting by m the cardinality of P , m ≤ n. Furthermore, every X ∈ iCov(p) refining P must have exactly m elements.

It follows from Lemma 3.4 that there exists

Q ∈ tCov(p) such that Q ≤ ref P . Every R ∈ iCov(p) refining Q refines P , so |Q| = |R| = m.
Furthermore, for each r ∈ R, there exists f (r) ∈ Q such that r ≤ f (r), and p ≤ R ≤ r∈R f (r). As the range of f is contained in Q and Q ∈ iCov(p), f is surjective, and thus it is a bijection from R onto Q. Now p ≤ R = q∈Q f -1 (q) with f -1 (q) ≤ q for each q. As Q ∈ tCov(p), f -1 (q) = q for each q, and therefore Q = R. By Lemma 3.2, it follows that Q ∈ mCov(p).

Corollary 5.2. Let n be a positive integer and let L be a dually algebraic ndistributive lattice. Then J(L) and J c (L) are both seeds of L.

Proof. As L is dually algebraic, it is spatial, thus J c (L) (and thus also J(L)) is join-dense in L. Now let p ∈ J(L) and let X ∈ Cov(p). As L is dually algebraic, it is complete and lower continuous, thus it follows from Lemma 5.1 that there exists I ∈ mCov(p) such that I ≤ ref X. By Lemma 3.3, I is contained in J(L). Furthermore, if p is a point, then it is compact (cf. Proposition 2.1), thus, by Lemma 3.3, I is contained in J c (L).

Theorem 5.3. Let n be a positive integer. Then every n-distributive lattice L can be embedded, within its variety, into an algebraic and strongly spatial lattice L such that if L has a least element (a largest element, both a least and a largest element, respectively), then so does L.

Proof. The lattice L can be embedded, zero-preservingly in case L has a zero, into its filter lattice Fil L, via the assignment a → L ↑ a; furthermore, Fil L generates the same variety as L (cf. Section I.3 and Lemma I.4.8 in Grätzer [START_REF] Grätzer | General Lattice Theory[END_REF]). As Fil L is dually algebraic, we may assume that L is dually algebraic. By Corollary 5.2, J(L) is a seed in L. By Lemma 4.3, the ideal lattice L of J(L) ∨ is algebraic and strongly spatial while it generates the same variety as L. This takes care of everything except the preservation of the largest element of L if it exists. However, as every ideal of an algebraic (resp., strongly spatial) lattice is algebraic (resp., strongly spatial), the latter point is easily taken care of by replacing L by the principal ideal generated by the largest element of L.

Generation of the variety of all n-distributive lattices

In Lemmas 6.1 and 6.2, we shall fix a positive integer n and a n-distributive, algebraic, and spatial lattice L. The set P of all finite subsets of J c (L), partially ordered by inclusion, is directed. For each P ∈ P, the (∨, 0)-subsemilattice P ∨ of L generated by P is a finite lattice, although not necessarily a sublattice of L. We shall denote by ∧ P the meet operation in P ∨ . Likewise, we shall denote by t P the interpretation of a lattice term t in the lattice P ∨ . Lemma 6.1. The lattice P ∨ is n-distributive, for each P ∈ P.

Proof. By Nation [25, Theorem 3.1], it suffices to prove that for every p ∈ J(P ∨ ), every irredundant join-cover X of p in P ∨ contains at most n elements. Now observe that J(P ∨ ) = P . As P ∨ is a (∨, 0)-subsemilattice of L, X is also an irredundant join-cover of p in L. As p ∈ J(L) and L is n-distributive, |X| ≤ n.

Denote by a (P ) the largest element of P ∨ below a, for each a ∈ L and each P ∈ P. For a vector a = (a 1 , . . . , a k ), we shall use the notation a (P ) := ((a 1 ) (P ) , . . . , (a k ) (P ) ). Lemma 6.2. The following statements hold, for every lattice term t with k variables and every vector a = (a 1 , . . . , a k ) ∈ L k :

(i) t P ( a (P ) ) ≤ t Q ( a (Q) ) holds for all P ⊆ Q in P;

(ii) t( a) = t P ( a (P ) ) | P ∈ P in L.

We shall abbreviate the conjunction of (i) and (ii) above by the notation t( a) = ↑ t P ( a (P ) ) | P ∈ P . (6.1)

Proof. (i). We argue by induction on the complexity of t. The result is obvious in case t is a projection. Now, supposing we have proved the statement for terms t 0 and t 1 , we must prove it for both terms t 0 ∧ t 1 and

t 0 ∨ t 1 . Let a ∈ L k , set b R i := t R i ( a (R) ) for i ∈ {0, 1} and R ∈ {P, Q}. It follows from the induction hypothesis that b P i ≤ b Q i for all i ∈ {0, 1}. Now b P 0 ∧ P b P 1 is an element of P ∨ , thus of Q ∨ , contained in both elements b Q 0 and b Q 1 , thus also in b Q 0 ∧ Q b Q 1 ; this proves that (t 0 ∧ t 1 ) P ( a (P ) ) = b P 0 ∧ P b P 1 ≤ b Q 0 ∧ Q b Q 1 = (t 0 ∧ t 1 ) Q ( a (Q) ) . Next, b P i ≤ b Q i ≤ b Q 0 ∨ b Q 1 for each i ∈ {0, 1}, thus b P 0 ∨ b P 1 ≤ b Q 0 ∨ b Q 1 ; this proves that (t 0 ∨ t 1 ) P ( a (P ) ) = b P 0 ∨ b P 1 ≤ b Q 0 ∨ b Q 1 = (t 0 ∨ t 1 ) Q ( a (Q) ) . (ii)
. Again, we argue on the complexity of t. In case t is a projection, we must prove the relation a = a (P ) | P ∈ P , for each a ∈ L . (6.2)

Trivially a ≥ a (P ) for each P . Conversely, let b ∈ L such that a (P ) ≤ b for each P ∈ P, we must prove that a ≤ b. As J c (L) is join-dense in L, it suffices to prove that p ≤ b for each point p such that p ≤ a. Setting P := {p}, we obtain that p = a (P ) ≤ b. This completes the proof of (6.2).

For the induction step, let t 0 and t 1 be lattice terms for which the induction hypothesis holds at a vector a ∈ L k , that is,

t i ( a) = ↑ t P i ( a (P ) ) | P ∈ P , for each i ∈ {0, 1} . (6.3)
We must prove the inequality

t( a) ≤ ↑ t P ( a (P ) ) | P ∈ P , (6.4)
for each t ∈ {t 0 ∧ t 1 , t 0 ∨ t 1 }. We first deal with the meet. Let p ∈ J c (L) lying below t 0 ( a) ∧ t 1 ( a). For each i ∈ {0, 1}, as p is a compact element lying below the right hand side of (6.3), which is a directed join by (i), there exists P i ∈ P such that p ≤ t Pi i ( a (Pi) ). Set P := P 0 ∪ P 1 ∪ {p}. It follows from (i) that p ≤ t Pi i ( a (Pi) ) ≤ t P i ( a (P ) ), for each i ∈ {0, 1}; and thus, as p ∈ P , we get p ≤ (t 0 ∧ t 1 ) P ( a (P ) ). Now suppose that t = t 0 ∨ t 1 . For each i ∈ {0, 1},

t i ( a) = ↑ t P i ( a (P ) ) | P ∈ P ≤ ↑ t P ( a (P ) ) | P ∈ P ,
thus, by forming the join of those inequalities for i ∈ {0, 1}, we obtain (6.4).

We do not claim that the lattice P ∨ satisfies every lattice identity satisfied by L (in fact, easy examples show that this is not the case as a rule). However, we can still prove the following result. Theorem 6.3. Let n be a positive integer. Then the variety D n of all n-distributive lattices is generated by its finite members. Consequently, the word problem for free lattices in D n is decidable. Proof. Let s and t be lattice terms, say of arity k, such that every finite lattice in D n satisfies the identity s = t. We must prove that every lattice in D n satisfies that identity. By Theorem 5.3, it suffices to prove that every n-distributive, algebraic, and spatial lattice L satisfies the identity. Define the directed poset P as above. It follows from Lemma 6.1 that P ∨ is n-distributive, for each P ∈ P. Thus, by assumption, P ∨ satisfies the identity s = t. Therefore, for each vector a ∈ L k , s( a) =

Seeds in modular lattices

Lemma 7.1. Let p, p, and a be elements in a modular lattice L with p and p both join-irreducible, and let Q ∈ mCov(a) satisfying the following conditions:

(i) p ≤ p ∨ a is a tight join-cover; (ii) p / ∈ Q; (iii) p ≤ ({p} ∪ Q) irredundantly.
Then {p} ∪ Q belongs to mCov(p).

Proof. Set b := Q. We first prove the following statement:

p x ∨ b for each x < p . (7.1)
Suppose that p ≤ x ∨ b. As x < p and a ≤ b, it follows from Zassenhaus' Butterfly Lemma (cf. Grätzer [START_REF] Grätzer | General Lattice Theory[END_REF]Theorem IV.1.13]) that the sublattice of L generated by {x, p, a, b} is distributive. Hence, setting u

:= p ∧ (x ∨ b) = x ∨ (p ∧ b), we get p ≤ (p ∨ a) ∧ (x ∨ b) = u ∨ a, with u ≤ p.
By the minimality assumption on p in (i), it follows that u = p, thus, as p is join-irreducible and x < p, we get p ≤ b. As a ≤ b, it follows from (i) that p ≤ p ∨ a ≤ b = Q, thus, by (iii), p ∈ Q, which contradicts (ii). This completes the proof of (7.1). Now let q ∈ Q, set Q ′ := Q \ {q} and a ′ := Q ′ (we may set it equal to a new zero element of L in case Q ′ = ∅ and L has no zero), we must prove the statement p p ∨ y ∨ a ′ for each y < q .

(7.2)

Suppose that p ≤ p ∨ y ∨ a ′ . As p ≤ p ∨ a, it follows from the modularity of L that p ≤ (p ∨ a) ∧ (p ∨ y ∨ a ′ ) = p ∨ (a ∧ (p ∨ y ∨ a ′ )). As a ∧ (p ∨ y ∨ a ′ ) ≤ a, it follows from the minimality assumption on a in (i) that a ≤ p ∨ y ∨ a ′ . As a ≤ Q = q ∨ a ′ , it follows from the modularity of L that a ≤ (q ∨ a ′ ) ∧ (p ∨ y ∨ a ′ ) = q ∧ (p ∨ y ∨ a ′ ) ∨ a ′ = q ∧ (p ∨ y ∨ a ′ ) ∨ Q ′ ,
with q∧(p∨y ∨a ′ ) ≤ q. Hence, as Q ∈ mCov(a), we get that q ≤ p∨y ∨a ′ . As y < q, it follows from the modularity of L that q = q ∧(p∨y ∨a ′ ) = y ∨(q ∧(p ∨a ′ )). As q is join-irreducible and y < q, it follows that q ≤ p ∨ a ′ , thus a ≤ Q = q ∨ a ′ ≤ p ∨ a ′ , and thus p ≤ p ∨ a ≤ p ∨ a ′ = p ∨ Q ′ , which contradicts the combination of (ii) and (iii). This completes the proof of (7.2). The combination of (7.1) and (7.2) (the latter being stated for each q ∈ Q) means that {p} ∪ Q ∈ tCov(p). As moreover {p} ∪ Q ⊆ J(L), it follows from Corollary 3.8 that {p} ∪ Q ∈ mCov(p).

As in Herrmann, Pickering, and Roddy [START_REF] Herrmann | A geometric description of modular lattices[END_REF], the collinearity relation in a modular lattice L is defined on the join-irreducible elements of L by letting col(p, q, r) hold if p, q, r are pairwise incomparable and p ∨ q = p ∨ r = q ∨ r. Lemma 7.2. Let p, q, r be join-irreducible elements in a modular lattice L. Then col(p, q, r) implies that p ≤ q ∨ r is a tight cover.

Proof. Let, say, p ≤ x ∨ r with x < q. As q ≤ p ∨ r, it follows that q ≤ x ∨ r. Using x < q and the modularity of L, we obtain q = x ∨ (q ∧ r), and thus, by the join-irreducibility of q, we get q ≤ r, a contradiction. Lemma 7.3. Let n be a positive integer and let p, a 1 , . . . , a n be elements in a modular spatial lattice L with p a point and p ≤ a 1 ∨ • • • ∨ a n irredundantly. Then there are points q 1 ≤ a 1 , . . . , q n ≤ a n such that {q 1 , . . . , q n } is a minimal cover of p.

Proof. We argue by induction on n. For n = 1 the result is trivial. For n = 2, it follows from Herrmann, Pickering, and Roddy [15, Lemma 2.2] that there are points q 1 ≤ a 1 and q 2 ≤ a 2 such that col(p, q 1 , q 2 ) holds. By Lemma 7.2, it follows that {q 1 , q 2 } ∈ mCov(p). Now suppose that the result holds for n and suppose that

p ≤ a 0 ∨ a 1 ∨ • • • ∨ a n irredundantly , (7.3)
with p a point. By the case n = 2, there are points q 0 ≤ a 0 and q ≤ a 1 ∨ • • • ∨ a n such that {q 0 , q} ∈ mCov(p). If q ≤ i∈[1,n]\{j} a i for some j ∈ [1, n], then, as p ≤ q 0 ∨ q ≤ a 0 ∨ q, we obtain that p ≤ i∈[0,n]\{j} a i , a contradiction; hence we obtain that q ≤ a 1 ∨ • • • ∨ a n irredundantly . (7.4) By the induction hypothesis, there are points q 1 ≤ a 1 , . . . , q n ≤ a n such that the set Q := {q 1 , . . . , q n } belongs to mCov(q). If p ≤ Q, then p ≤ 1≤i≤n a i , which contradicts (7.3); hence p Q, but p ≤ q 0 ∨ q ≤ q 0 ∨ Q, and thus

q 0 / ∈ Q. If p ≤ q 0 ∨ (Q \ {q j }) for some j ∈ [1, n], then p ≤ i∈[0,
n]\{j} a i , a contradiction; hence p ≤ ({q 0 }∪Q) irredundantly. Now it follows from Lemma 7.1 that {q 0 } ∪ Q = {q 0 , q 1 , . . . , q n } belongs to mCov(p).

Theorem 7.4. Every modular spatial lattice is strongly spatial.

Proof. Let p be a point in a modular spatial lattice L and let X ∈ Cov(p). There exists a subset Y of X such that p ≤ Y irredundantly. By Lemma 7.3, Y can be refined to an element of mCov(p).

As Herrmann, Pickering, and Roddy prove in [START_REF] Herrmann | A geometric description of modular lattices[END_REF] that every modular lattice can be embedded, within its variety and zero-preservingly in case there is a zero, into some algebraic and spatial lattice, we thus obtain the following.

Corollary 7.5. Every modular lattice L can be embedded, within its variety, into some algebraic strongly spatial lattice. If L has a zero, then the embedding can be taken zero-preserving.

A class of lattices with a bounded distributive lattice parameter

In this section we shall prepare the ground for the construction of a lattice without any variety-preserving extension to an algebraic and spatial one, cf. Theorem 9.5. Notation 8.1. Let D be a nontrivial bounded distributive lattice. We set

x -:= 0 , if x < 1 1 , if x = 1 , for each x ∈ D .
Furthermore, we set

L(D) := 2 × D × 2 × 2 , K(D) := (x 0 , x 1 , x 2 , x 3 ) ∈ L(D) | (∀i < j < k in {0, 1, 2, 3})(x - i ∧ x - k ≤ x j )
, and we define a map γ : L(D) → L(D) by setting

γ(x) j := x j ∨ x - i ∧ x - k | i < j < k , for each j ∈ {0, 1, 2, 3}
. We shall denote by ∨ c the join operation in L(D). Hence (x ∨ c y) i = x i ∨ y i , for all x, y ∈ L(D) and all i < 4 .

(8.1)

We shall denote by q i the element of L(D) with 1 at the i-th place and 0 elsewhere, for each i ∈ {0, 1, 2, 3}. Furthermore, we set xq 1 := (0, x, 0, 0), for each x ∈ D.

Obviously 0q 1 = 0, 1q 1 = q 1 , and every element either of the form q i or of the form xq 1 belongs to K(D).

In Lemma 8.2 and Corollary 8.3 we shall fix a nontrivial bounded distributive lattice D.

Lemma 8.2. The element γ(x) is the least element of K(D) above x, for each x ∈ L(D).

Proof. It is obvious that γ is isotone, that γ(x) ≥ x for each x ∈ L(D), and that γ(x) = x iff x ∈ K(D). Hence it suffices to prove that y := γ(x) belongs to K(D), for each x ∈ L(D). Let i < j < k in {0, 1, 2, 3}, we must prove that if y i = y k = 1, then y j = 1. From y i = 1 it follows that there exists i ′ ≤ i such that x i ′ = 1. Similarly, from y k = 1 it follows that there exists

k ′ ≥ k such that x k ′ = 1. From i ′ < j < k ′ it follows that y j ≥ x - i ′ ∧ x - k ′ = 1.
As L(D), endowed with the componentwise ordering, is a bounded lattice, we obtain the following.

Corollary 8.3. The set K(D), endowed with the componentwise ordering, is a closure system in L(D). In particular, it is a bounded lattice, and also a meet-subsemilattice of L(D). (i) The equality con K(B) (0, q 2 ) = con K(B) (0, xq 1 ) holds for each x ∈ B \ {0}.

(ii) The lattice K(B) is subdirectly irreducible, with minimal nonzero congruence con K(B) (0, q 1 ).

Proof. (i). Let θ be a congruence of K(B). Suppose first that q 2 ≡ θ 0. From q 1 ≤ q 0 ∨ q 2 it follows that q 1 ≤ θ q 0 , but q 1 ∧ q 0 = 0, thus q 1 ≡ θ 0. This implies in turn that xq 1 ≡ θ 0 for each x ∈ B. Finally suppose that xq 1 ≡ θ 0 for some x ∈ B\{0}. Denote by y the complement of x in B. From q 2 ≤ q 1 ∨q 3 = xq 1 ∨yq 1 ∨q 3 it follows that q 2 ≤ θ yq 1 ∨ q 3 . From y < 1 it follows that yq 1 ∨ q 3 = (0, y, 0, 1), thus q 2 ∧ (yq 1 ∨ q 3 ) = 0, and thus q 2 ≡ θ 0. (ii). It suffices to prove that q 1 ≡ θ 0 for each nonzero congruence θ of K(B). There are x < y in K(B) such that x ≡ θ y. As z = i<4 (z ∧ q i ) for each z ∈ K(B) we may assume that y ≤ q i for some i < 4. If i = 2 then x = 0 and y = q 2 , thus q 2 ≡ θ 0, and thus, by (i), q 1 ≡ θ 0 and we are done. If i = 1, then, as B is Boolean, there exists z ∈ K(B) ↓ q 1 such that x ∨ z = y while x ∧ z = 0; observe that 0 < z ≤ q 1 and z ≡ θ 0. It follows from (i) above that q 1 ≡ θ 0. If i = 0 then q 0 ≡ θ 0, but q 1 ≤ q 0 ∨ q 2 , thus q 1 ≤ θ q 2 , but q 1 ∧ q 2 = 0, thus q 1 ≡ θ 0. If i = 3 then q 3 ≡ θ 0, thus, as q 2 ≤ q 1 ∨ q 3 , we get q 2 ≤ θ q 1 , but q 1 ∧ q 2 = 0, thus q 2 ≡ θ 0, and thus, by (i) above, q 1 ≡ θ 0. Proof. Let n be a positive integer and let x 1 , . . . , x n ∈ K(D). Denote by C the 0, 1-sublattice of D generated by {(x 1 ) 1 , . . . , (x n ) 1 }. Then the sublattice of K(D) generated by {x 1 , . . . , x n } is contained in K(C), which has at most 2 2 n +3 elements. As this bound depends of n only, the local finiteness statement about the variety follows easily by using a standard argument of universal algebra.

Local distributivity of lattices of the form K(D)

Throughout this section we shall denote by K the class of all lattices of the form K(B), for nontrivial Boolean lattices B, and by V the lattice variety generated by K. Furthermore, we denote by B 0 the (Boolean) lattice of all subsets of ω that are either finite or cofinite. It is in fact easy to verify that V is generated by K(B 0 ). Lemma 9.1. Let D be a nontrivial bounded distributive lattice and let a ∈ K(D). If K(D) ↓ a is not distributive, then a is either equal to (1, 1, 1, 0), or to (0, 1, 1, 1), or to (1, 1, 1, 1).

Proof. There are x, y, z ∈ K(D) ↓ a such that (x ∨ y) ∧ z > (x ∧ z) ∨ (y ∧ z). Set t := x ∨ c y; observe that t ≤ a. It follows from the distributivity of D that x ∨ y > t (cf. (8.1)), that is, (x∨y) j > t j for some j < 4. By Lemma 8.2, x∨y = γ(t). Hence, by the definition of γ, there are i < j and k > j such that t i = t k = 1. From t ≤ a it follows that a i = a k = 1. As a ∈ K(D), it follows that a takes the value 1 on at least three consecutive elements of {0, 1, 2, 3}. The conclusion follows.

Corollary 9.2. Every subdirectly irreducible member of V satisfies the sentence

(∀x, y, z, t 1 , t 2 ) x ∨ y ∨ z < t 1 < t 2 =⇒ (x ∨ y) ∧ z = (x ∧ z) ∨ (y ∧ z) . (9.1)
Proof. As both (1, 1, 1, 0) and (0, 1, 1, 1) are lower covers of (1, 1, 1, 1) in K(D), it follows from Lemma 9.1 that every member of K satisfies (9.1). By Los' Ultraproduct Theorem, every ultraproduct of members of K also satisfies (9.1), and thus every sublattice of such an ultraproduct satisfies (9.1). By Jónsson's Lemma, for every subdirectly irreducible member L of V, there are a sublattice L of an ultraproduct of members of K and a surjective lattice homomorphism f : L ։ L. Now let x, y, z, t 1 , t 2 ∈ L such that x ∨ y ∨ z < t 1 < t 2 , with respective preimages x, y, z, t 1 , t 2 ∈ L under f . Necessarily,

x ∨ y ∨ z < x ∨ y ∨ z ∨ t 1 < x ∨ y ∨ z ∨ t 1 ∨ t 2 ,
thus, as L satisfies (9.1), (x ∨ y) ∧ z = (x ∧ z) ∨ (y ∧ z). By applying the homomorphism f , we obtain that (x ∨ y) ∧ z = (x ∧ y) ∨ (x ∧ z).

Corollary 9.3. Let D be a nontrivial bounded distributive lattice and let L be a subdirectly irreducible member of V containing K(D) as a sublattice. Then L ↓ q 1 is distributive.

Proof. This follows immediately from the inequalities q 1 < q 1 ∨ q 2 < q 1 ∨ q 2 ∨ q 3 (in K(D), thus in L) together with Corollary 9.2.

Lemma 9.4. Let B be a nontrivial Boolean lattice and let L be a lattice in V containing K(B) as a sublattice. Then (x ∨ y) ∧ z = (x ∧ z) ∨ (y ∧ z) holds for all x, y ∈ K(B) ↓ q 1 and all z ∈ L.

Proof. By replacing z by z ∧ q 1 , we may assume that z ≤ q 1 . It follows from Birkhoff's Subdirect Representation Theorem that L is a subdirect product of subdirectly irreducible lattices. Hence it suffices to prove that the relation

(x ∨ y) ∧ z ≡ ψ (x ∧ z) ∨ (y ∧ z) (9.2)
holds for each completely meet-irreducible congruence ψ of L. Set L ψ := L/ψ and denote by p ψ : L ։ L ψ the canonical projection. We separate cases.

Case 1. p ψ ↾ K(B) is one-to-one. It follows from Corollary 9.3 that L ψ ↓ (q 1 /ψ) is distributive. The relation (9.2) follows. Case 2. p ψ ↾ K(B) is not one-to-one. It follows from Lemma 8.4 that q 1 ≡ ψ 0 K(B) , thus, as x, y ∈ K(B) ↓ q 1 , we get x ≡ ψ y ≡ ψ 0 K(B) . The relation (9.2) follows trivially. This concludes the proof. Theorem 9.5. There is no algebraic spatial lattice L in V that contains K(B 0 ) as a sublattice.

Proof. Set K := K(B 0 ), p n := (0, {n} , 0, 0), and u n := (0, ω \ {n} , 0, 0) for each n < ω. Observe that p n and u n both belong to K and p n ∨ u n = q 1 in K. As L is spatial, there exists a point x of L such that x ≤ q 2 and x 0 K . As x ≤ q 2 ≤ q 1 ∨ q 3 and x is compact while L is spatial, there are a positive integer m and points y 1 , . . . , y m ≤ q 1 of L such that

x ≤ y 1 ∨ • • • ∨ y m ∨ q 3 . (9.3)
Set I := {i ∈ {1, . . . , m} | (∃n < ω)(y i ≤ p n )}. We may assume that I = {1, . . , r} for some r ∈ {0, . . . , m}. There exists ℓ < ω such that

y 1 ∨ • • • ∨ y r ≤ p 0 ∨ • • • ∨ p ℓ-1 ; it follows that y 1 ∨ • • • ∨ y r ≤ u ℓ .
(9.4) As y i ≤ q 1 = p ℓ ∨ u ℓ , for each i ∈ {1, . . . , m}, it follows from Lemma 9.4 that y i = (y i ∧ p ℓ ) ∨ (y i ∧ u ℓ ), and thus, by the join-irreducibility of y i , either y i ≤ p ℓ or y i ≤ u ℓ . It follows that y i ≤ u ℓ for each i > r, and so

y r+1 ∨ • • • ∨ y m ≤ u ℓ . (9.5)
By putting (9.4) and (9.5) together, we obtain the inequality

y 1 ∨ • • • ∨ y m ≤ u ℓ . (9.6)
By putting (9.3) and (9.6) together, we obtain

x ≤ (0, ω \ {ℓ} , 0, 1) .

As x ≤ q 2 = (0, 0, 1, 0), it follows that x ≤ 0 K , a contradiction.

10. Join-semidistributivity of the lattices K(D)

In this section we shall prove that all lattices K(D) are join-semidistributive. In fact we shall prove a stronger statement, namely that all lattices K(D) satisfy the lattice identity (inclusion) (SD 3 ∨ ). The optimality of the "3" superscript is a consequence of the following easy statement. It is easy to compute that p 2 (x, y, z) = (0, b, 0, 0) while x ∨ (y ∧ z) = (1, a, 0, 0), so p 2 (x, y, z) x ∨ (y ∧ z).

Proposition 10.2. The lattice K(D) satisfies (SD 3 ∨ ), for any nontrivial bounded distributive lattice D.

Proof. We must prove that p 3 (x, z, y) ≤ x ∨ (y ∧ z) for all x, y, z ∈ K(D). By replacing x by x ∨ (y ∧ z), then z by z ∧ (x ∨ y), we may assume without loss of generality that y ∧ z ≤ x and z ≤ x ∨ y . ( Suppose first that x 0 = 1. If x 3 = 1, then, as x ∈ K(D), x = 1, which contradicts (10.2); so x 3 = 0. As z ≤ x ∨ y we obtain that z 3 ≤ y 3 , but y ∧ z ≤ x and thus z 3 = z 3 ∧ y 3 ≤ x 3 , and so z 3 = 0. If x 2 = 1 then, as x 0 = 1 and x < 1, we get that x = (1, 1, 1, 0) ≥ z, which contradicts (10.2); hence x 2 = 0. So far we have proved that x = (1, x 1 , 0, 0). Now from (10.3) it follows that y (1, 1, 0, 0) and z

(1, 1, 0, 0). If either y 2 = z 2 = 1 or y 3 = z 3 = 1, then either y 2 ∧ z 2 = 1 or y 3 ∧ z 3 = 1, a contradiction as y ∧ z ≤ x and x = (1, x 1 , 0, 0). As z 3 ≤ y 3 , the only remaining possibility is y 3 = z 2 = 1 and y 2 = z 3 = 0. As y 3 = 1 and y ∈ K(D) it follows that y 0 = 0. Similarly, as z 2 = 1, z ∈ K(D), and x z, we obtain that z 0 = 0. We have thus proved that x = (1, x 1 , 0, 0) , y = (0, y 1 , 0, 1) , z = (0, z 1 , 1, 0) .

It follows that

x ∨ z = (1, 1, 1, 0), then that y ∧ (x ∨ z) = (0, y 1 , 0, 0), then that x ∨ (y ∧ (x ∨ z)) = (1, x 1 ∨ y 1 , 0, 0). Therefore, p 3 (x, z, y) = (0, z 1 ∧ (x 1 ∨ y 1 ), 0, 0). As D is distributive, z 1 ∧ x 1 ≤ x 1 , and y 1 ∧ z 1 ≤ x 1 , we obtain that p 3 (x, z, y) ≤ x.
The remaining case is x 0 = 0. As z ≤ x ∨ y we get z 0 ≤ y 0 , thus z 0 = y 0 ∧ z 0 ≤ x 0 = 0, and thus z 0 = 0, and therefore (x ∨ z) 0 = 0. By (10.3), it follows that x ∨ z = (0, 1, 1, 1). In particular, (10.4) and as y ∧ z ≤ x, we get

x 1 ∨ z 1 = x 3 ∨ z 3 = 1 . (10.4) If x ∨ y = (1, 1, 1, 0), then x 3 = y 3 = 0, thus (as z ≤ x ∨ y) z 3 = 0, thus, by (10.4), x 3 = 1, contradicting x ∨ y = (1, 1, 1, 0). If x ∨ y = (0, 1, 1, 1), then x 1 ∨ y 1 = x 3 ∨ y 3 = 1, thus, by
x 1 = x 1 ∨ (y 1 ∧ z 1 ) = (x 1 ∨ y 1 ) ∧ (x 1 ∨ z 1 ) = 1 ,
and similarly, x 3 = 1, so x ≥ (0, 1, 1, 1) ≥ z, contradicting (10.2). The only remaining possibility is x 0 = 0 and x ∨ y = (1, 1, 1, 1). From x 0 = 0 it follows that y 0 = 1. If y 3 = 1 then y = 1, which contradicts (10.2); so y 3 = 0. As x ∨ y = (1, 1, 1, 1), we get x 3 = 1. If y 2 = 1, then, as y 0 = 1 and y 3 = 0, we get One possible illustration of the hopelessness of the desired task is the following result (cf. Wehrung [START_REF] Wehrung | Sublattices of complete lattices with continuity conditions[END_REF]Corollary 12.4]): The subspace lattice of an infinite-dimensional vector space cannot be embedded into any bi-algebraic lattice.

Another illustration of the interplay between rings and lattices about those questions is given by Faith's example (cf. Cozzens and Faith [1, Example 5.14]) of a von Neumann regular ring R which is not a "left V-ring", the latter meaning that not every left ideal is an intersection of maximal left ideals. It is then not hard to prove that if L denotes the lattice of all principal right ideals of R, which is complemented modular (cf. Maeda [START_REF] Maeda | Kontinuierliche Geometrien[END_REF]Section VI.4] or Goodearl [8, Theorem 2.3]), then Fil L is not atomistic. It follows that the set of all atoms of Fil L is not join-dense in Fil L (thus it is not a quasi-seed) and that not all points of Fil L are atoms (for Fil L is spatial).

Frink's example and Faith's example are both coordinatizable (i.e., each of them is isomorphic to the lattice of all principal right ideals of a regular ring), but they are not identical. While Frink's example is a continuous geometry, Faith starts with an infinite-dimensional right vector space V over a field F , then considers the ideal S in the endomorphism ring End V F consisting of all endomorphisms with finite rank, then defines R as the subalgebra of End V F generated by S and the identity.

11.2. Word problems in various classes of lattices. The geometric description provided for n-distributive lattices in Theorem 5.3 is the key tool for establishing the result, stated in Theorem 6.3, that the variety of all n-distributive lattices is generated by its finite members, and thus has a decidable word problem for free lattices. On the other hand, the corresponding results for modular lattices do not hold (cf. Freese [START_REF] Freese | The variety of modular lattices is not generated by its finite members[END_REF][START_REF] Freese | Free modular lattices[END_REF] and Herrmann [START_REF] Herrmann | On the word problem for the modular lattice with four free generators[END_REF]). It is even observed on Freese [4, page 90] that the free lattice on five generators in the variety of all n-distributive modular lattices, for n ≥ 4, has an undecidable word problem. As every modular lattice embeds into some algebraic, modular, and spatial lattice (Herrmann, Pickering, and Roddy [START_REF] Herrmann | A geometric description of modular lattices[END_REF]), the result of Theorem 6.3 for n-distributive lattices appears a bit as a fluke. A look at the axiomatization of the abstract projective geometries associated with algebraic spatial lattices described in Herrmann, Pickering, and Roddy [START_REF] Herrmann | A geometric description of modular lattices[END_REF]Section 3], in particular the so-called Triangle Axiom, shows that the existential quantifier involved in that axiom prevents us from expressing an infinite projective space as a "limit" of finite projective spaces in any satisfactory way. Due to Lemma 6.1, this obstacle does not appear in the case of n-distributive lattices.

However, a further look at positive decidability results obtained for other classes of modular lattices shows that this existential quantifier alone is not sufficient to prevent decidability to occur. Here is a sample of such results, the third one being of more hybrid nature due to the extra operation symbol for complementation:

-Hutchinson and Czédli [START_REF] Hutchinson | A test for identities satisfied in lattices of submodules[END_REF] characterize those rings R for which the word problem for free lattices in the variety generated by all subspace lattices of left R-modules is decidable. This class of rings includes all fields, and also the ring Z of all integers as well as its quotient rings Z/mZ for positive integers m (cf. Herrmann [START_REF] Herrmann | On the equational theory of submodule lattices[END_REF], Herrmann and Huhn [START_REF] Herrmann | Zum Wortproblem für freie Untermodulverbände[END_REF]). -Herrmann and Huhn [START_REF] Herrmann | Zum Wortproblem für freie Untermodulverbände[END_REF] also prove that the word problem for free lattices in the variety generated by all complemented modular lattices is solvable.

-By using results about von Neumann regular rings proved by Goodearl, Menal, and Moncasi [START_REF] Goodearl | Free and residually artinian regular rings[END_REF], Herrmann and Semenova [START_REF] Herrmann | Existence varieties of regular rings and complemented modular lattices[END_REF] prove that the variety generated by complemented Arguesian lattices with an extra unary operation symbol for complementation is generated by its finite members, and thus that the word problem for free lattices with complementation in the variety generated by those structures is decidable. The latter decidability result extends to the variety generated by all complemented modular lattices with a unary operation symbol for complementation, although residual finiteness is replaced by residual finite length. This suggests that in the modular world, the spatial theory alone is probably far from sufficient for settling residual finiteness and word problem matters. 11.3. Open problems. Problem 1. Can every algebraic and spatial lattice be embedded, within its variety, into some algebraic and strongly spatial lattice?

For our next problem, we shall consider the identities β ′ m , given in Nation [START_REF] Nation | An approach to lattice varieties of finite height[END_REF], that characterize, among finite lattices, those lattices without D-sequences of length m + 1 (where D denotes join-dependency). Nation proves in [START_REF] Nation | An approach to lattice varieties of finite height[END_REF]Section 5] that for fixed m, the variety of lattices defined by β ′ m is locally finite. Problem 2. Prove that every lattice satisfying β ′ m for some m can be embedded, within its variety, into some algebraic and strongly spatial lattice.

Our next problem asks for a semidistributive analogue of Theorem 9.5. Problem 3. Construct a semidistributive lattice that cannot be embedded, within its variety, into any algebraic spatial lattice. Can such a lattice be locally finite, or generate a variety of semidistributive lattices, or both? Problem 4. Let p be a compact element in an algebraic, modular, spatial lattice L. Can every join-cover of p be refined to a minimal join-cover of p? By Theorem 7.4, the conclusion of Problem 4 holds in case p is a point of L.

Problem 5. Can every modular lattice be embedded, within its quasivariety, into some algebraic (spatial, algebraic spatial, respectively) lattice?

Proposition 2 . 1

 21 (folklore). Every compactly generated lattice L is upper continuous and every point of L is compact. Proof. Let a ∈ L and let D be a nonempty directed subset of L with a join, we must prove that a ∧ D ≤ (a ∧ D). Let c ∈ L compact with c ≤ a ∧ D. As c ≤ D and c is compact, there exists d ∈ D such that c ≤ d. It follows that c ≤ a ∧ d ≤ (a ∧ D). As L is compactly generated, the upper continuity of L follows. Now let p be a point of L and let D be a directed subset of L with a join such that p ≤ D. If p / ∈ L ↓ D, then p ∧ x ≤ p * for each x ∈ D, thus, by using the upper continuity of L, we get p = p ∧ D ≤ p * , a contradiction.

  it is generated by its intersection with Σ; thus it suffices to prove that each element p ∈ Σ ∩ π(A ∨ B) belongs to π(A) ∨ π(B). By the definition of the map π, the element p belongs to A ∨ B, that is, there exists (a, b)∈ A × B such that p ≤ a ∨ b.As {a, b} ∈ Cov(p) and Σ is a pre-seed, there exists X ∈ Cov(p) contained in Σ such that X ≤ ref {a, b}. The latter relation means that X = (X ↓ a) ∪ (X ↓ b). Setting a ′ := (X ↓ a) and b ′ := (X ↓ b), it follows that p ≤ X = a ′ ∨ b ′ . As a ′ ∈ π(A) and b ′ ∈ π(B), the desired conclusion follows.

↑

  s P ( a (P ) ) | P ∈ P (by Lemma 6.2) = ↑ t P ( a (P ) ) | P ∈ P (because P ∨ satisfies s = t) = t( a) (by Lemma 6.2) , which concludes the proof of the first statement. The decidability statement then follows from McKinsey [24, Theorem 3].

Lemma 8 . 4 .

 84 The following statements hold, for any nontrivial Boolean lattice B.

Lemma 8 . 5 .

 85 The lattice K(D) is locally finite, for each nontrivial bounded distributive lattice D. In fact, K(D) generates a locally finite lattice variety.

Proposition 10 . 1 .

 101 The lattice K(B) does not satisfy (SD 2 ∨ ), for any Boolean lattice B with more than two elements.Proof. There are nonzero complementary elements a, b ∈ B. We set x := (1, a, 0, 0) , y := (0, b, 0, 1) , z := (0, a, 1, 0) .

3 .

 3 Seeds, algebraic lattices, (strongly) spatial lattices Definition 3.1. For an element p and a finite subset E in a join-semilattice L, we say that

• E covers (resp., joins to) p if p ≤ E (resp., p = E). We set

Cov(p) := {E ⊆ L finite | E covers p} , Cov = (p) := {E ⊆ L finite | E joins to p} .

The elements of Cov(p) are called the join-covers of p, while the elements of Cov = (p) are called the join-representations of p.

• E covers (resp., joins to) p irredundantly if E covers p (resp., joins to p) and p (E \ {u}) for each u ∈ E. Observe that both conditions imply that

  .1)Elementary computations show that if two of the elements x, y, z are comparable, then p 2 (x, y, z) ≤ x, thus p 3 (x, z, y) ≤ x. Hence we shall assume from now on that x , y , z are pairwise incomparable . (10.2) If K(D)↓(x∨y) is distributive, then, as z ≤ x∨y, we get that z = (z∧x)∨(z∨y) ≤ x, contradicting (10.2). If K(D)↓(x∨z) is distributive, then, setting y ′ := p 2 (x, y, z) = y ∧ (x ∨ z), we get that y ′ = (y ′ ∧ x) ∨ (y ′ ∧ z) ≤ x, and thus p 3 (x, z, y) ≤ x. Hence

we may assume that neither K(D) ↓ (x ∨ y) nor K(D) ↓ (x ∨ z) is distributive. By Lemma 9.1, this implies that {x ∨ y, x ∨ z} ⊆ {(1, 1, 1, 0), (0, 1, 1, 1), (1, 1, 1, 1)} .

(10.3)

y = (1, 1, 1, 0); as y ∧ z ≤ x and x 3 = 1, we get z ≤ x, which contradicts (10.2); so y 2 = 0. So far we have obtained that x = (0, x 1 , x 2 , 1) , y = (1, y 1 , 0, 0) , z = (0, z 1 , z 2 , z 3 ) .

It follows that x ∨ z has the form (0,

and so p 3 (x, z, y) ≤ x.

As (SD 3

∨ ) is a lattice-theoretical identity implying join-semidistributivity, we obtain the following corollary. 11. Discussion 11.1. Frink's Embedding Theorem. This theorem (cf. Frink [START_REF] Frink | Complemented modular lattices and projective spaces of infinite dimension[END_REF], see also Grätzer [11, Section IV.5]) provides a 0, 1-lattice embedding construction of any complemented modular lattice into some geomodular (i.e., algebraic, atomistic, and modular) lattice. Jónsson observed in [START_REF] Jónsson | Modular lattices and Desargues' Theorem[END_REF] that the lattice constructed by Frink generates the same variety as L.

Let us outline Frink's construction. We start with a complemented modular lattice L. Denote by L := Fil L the filter lattice of L and by P the set of all atoms of L (i.e., the elements of P are the maximal proper filters of L). By the general properties of modular lattices (cf. Crawley and Dilworth [2, Section 4.1]), P ∨ is an ideal of L; in particular, it satisfies all the identities satisfied by L, which are the same as all the identities satisfied by L. Now Id P ∨ is an algebraic, atomistic, modular (that is, geomodular) lattice, belonging to the variety generated by L, with set of atoms P := {L ↓ p | p ∈ P }. Then P , being the set of all atoms in an algebraic atomistic lattice, is a seed (cf. Proposition 3.6). In order to conclude the proof, Frink proves that the canonical map ε : L ֒→ Id P ∨ , x → p ∈ P ∨ | x ∈ p is a lattice homomorphism. The hard core of that task amounts to proving the following:

Let x, y ∈ L \ {0} and let r ∈ P such that x ∨ y ∈ r. Then there are p, q ∈ P such that x ∈ p, y ∈ q, and p ∩ q ⊆ r. Conceivably, the proof of Frink's Theorem could have been dealt with in a more expeditious manner if it had been possible to prove that Fil L is complemented. However, as Frink observes on Frink [6, page 462], this is not the case as a rule, for example for L a continuous geometry. Later on the same page, Frink expresses the hope that ideals and filters could be combined in the same construction. We quote the corresponding paragraph on Frink [6, page 462]:

To be sure, one could consider the dual concept of the lattice D(L), namely the lattice of all ideals (multiplicative ideals) instead of dual ideals. Among these would be found maximal ideals, which would serve as complements of maximal dual ideals, if both kinds of ideal could be combined in a single lattice.


Roberto Sarmiento
email: sarmientoroberto@yahoo.com.mx

Keywords: MANUFACTURING STRATEGY, OPERATIONS MANAGEMENT Keywords (user): deterministic analyses, probabilistic analyses trade-off model, cumulative capabilities model, compatibilities, deterministic analyses, probabilistic analyses

A note on "trade A note on "trade----off and compatibility between off and compatibility between off and compatibility between off

A note on "trade A note on "trade A note on "trade A note on "trade----off and compatibility between off and compatibility between off and compatibility between off and compatibility between performance: definitions and empirical evidence" performance: definitions and empirical evidence" performance: definitions and empirical evidence" performance: definitions and empirical evidence" 1 Introduction

The question of whether a firm can achieve market-leading performance across a number of competitive criteria (e.g.; delivery, quality, flexibility) has been a topic of considerable interest in the field of strategic/operations management.

The seminal works by [START_REF] Skinner | Manufacturing-Missing Link in Corporate Strategy[END_REF][START_REF] Skinner | The Focused Factory[END_REF] have inspired a growing number of investigations that have looked at different aspects of what has been termed as the "trade-off model". Over the years, research on this and related topics (e.g.;

Ferdows and de [START_REF] Ferdows | Lasting improvements in manufacturing performance: In search of a new theory[END_REF][START_REF] Schmenner | On theory in operations management[END_REF][START_REF] Safizadeh | Revisiting alternative theoretical paradigms in manufacturing strategy[END_REF][START_REF] Da Silveira | Exploring the trade-off concept[END_REF]Corbbett and Claridge, 2002;[START_REF] Da Silveira | Improving trade-offs in manufacturing: Methods and illustration[END_REF][START_REF] Narasimhan | An exploratory study of manufacturing practice and performance interrelationships[END_REF][START_REF] Miltenburg | Setting manufacturing strategy for a factory-within-a-factory[END_REF] have resulted in new advances and in a better understanding of the different aspects involved in the attainment of high levels of performance by manufacturing/service firms.

A key issue in this theme is the methodologies and rationales by which investigators come to the conclusion that two or more competitive criteria are in a trade-off and/or a compatibility (e.g.; cumulative capability) relationship. Recent literature reviews [START_REF] Rosenzweig | Tradeoffs in manufacturing? A metaanalysis and critique of the literature[END_REF][START_REF] Sarmiento | Manufacturing capabilities and performance: a critical analysis and review[END_REF] show that quantitative studies using probabilistic methodologies, rationales and estimates based on statistics are a preferred approach in this line of research.

Analysts have usually relied on linear regression/correlation analysis and related methodologies (e.g.; path analysis) to determine trade-offs/cumulative capabilities between different pairs of competitive criteria. For example, "a positive significant coefficient indicates a cumulative capability, while a negative significant coefficient indicates a trade-off" (Flynn and Flynn, 2004, p. 446).

General, overall and probabilistic approaches to the study of these themes are important. Nevertheless, the literature is almost void of approaches that can be used to assert the presence of trade-offs/compatibilities between multiple Skinner, 1992;[START_REF] Schmenner | On theory in operations management[END_REF] argue that trade-offs and compatibilities are not necessarily mutually exclusive concepts that belong to different frameworks. Also, [START_REF] Sarmiento | Natural laws and the trade-off model: implications for research and policy-advising[END_REF] shows that the main principle of the trade-off model rests on the analysis of whether individual and identifiable firms can achieve market-leading level of performance across multiple competitive criteria. Therefore, based on these recent investigations, we try to build on [START_REF] Filippini | Trade-off and compatibility between performance: Definitions and empirical evidence[END_REF] work by proposing an improved rationale and methodology by which the trade-offs and compatibilities concepts are unified under a single framework when individual firms are the units of analysis.

The rest of the paper is organized as follows: section 2 presents various definitions, clarifications and background necessary for the investigation. Section 3 details our new methodology and also includes several illustrative examples.

Finally, section 4 offers some conclusions and venues for future research. Nonetheless, their methodology still considers trade-offs and compatibilities as belonging to two different frameworks and relationships. As such, it does not take into account newer developments that contradict those long-held views. To understand the limitations of their approach and hence the way it can be improved, we quote various definitions that Filippini et al utilize, and a brief reexamination of such statements follows:

"…the underpinning concept is that in a trade-off situation, high-levels of performance over several performance types are not reached.

Compatibility should be seen as the possibility of achieving high values over several types of performance". (Filippini et al, 1998, p. 3383).

Clarification 1: While it is true that the underlying fundament of the cumulative capabilities (e.g.; compatibilities) model is that of the achievement of high levels of performance across several competitive criteria, the core principle behind the trade-off model states that no manufacturing system can attain an industry-leading type of achievement along all performance types. As Skinner (1996, p. 6) writes: "… one system cannot be outstanding enough at meeting all criteria to create competitive advantage". Also, "(T)o ever assume that a plant or service operation can function without trade-offs is absurd" (Skinner, 1992, p. 20). The preceding statements imply that, to truly be a competing concept to Skinner's, the cumulative capabilities model should aim at explaining how firms can achieve industry-leading performance across all capabilities, since the tradeoff model denies that possibility. Put differently, firms could very well achieve excellent, high levels of performance in three, four or more capabilities. But as long as they are not able to attain that type of achievement along all criteria, the trade-off concept will not be refuted (Note: Skinner initially proposed seven performance criteria by which a manufacturing system should be assessed.

Nevertheless, this should not be seen as a static and unchanging set. For example, new criteria could be added due to market forces and/or new technologies. See Skinner, 1992Skinner, , 1996, for a more detailed discussion).

"It can therefore be said that two types of performance are in trade-off if the achievement of high values in one type of performance means low values in another. A trade-off situation is characterized, therefore, by high levels of difference between the two types of performance". (Filippini et al, 1998, p. 3383).

Clarification 2: Skinner (1992) defines a trade-off situation as one in which a high, industry-leading level of performance in one capability is attained at the expense of lesser (and not necessarily "low") levels of achievement in one or more performance types. The difference between Skinner's and Filippini et al's definitions of a compromised relationship is subtle, but important. By Skinner's definition, the gap in performance between two capabilities that are compromised need not be high, as Filippini et al propose. The difference only needs to be noticeable enough so that a gap in competitive power between the two capabilities can be distinguished. This will be fully explained in section 3.

"On the other hand, it can be said that two types of performance are compatible if both have high values. A situation of compatibility is characterized, therefore, by low value difference between performance types and by the presence of high values in general". (Filippini et al, 1998, p. 3383).

Clarification 3: As commented before, a gap in performance between two capabilities can also be explained as a trade-off situation, even if the difference is low. As long as the gap is distinguishable enough so that a differential in competitive power can be observed between two capabilities, it can be validly deemed as a compromised relationship. In light of this, a compatibility situation can only be considered as such when two or more capabilities observe common 6 levels of performance. Such a compatibility can be seen at the lower, in-between and high/industry-leading levels of achievement. This will be further clarified in section 3.

Another key and underlying point concerns the main focus of the trade-off concept. While an assessment and analysis of general trends and commonalities (e.g.; significant and positive correlation coefficients) are important, the validity of the trade-off model rests on an analysis of whether individual and identifiable firms can attain market-leading performance across multiple, and eventually all, competitive criteria. This can be derived from statements such as "… one system cannot be outstanding enough at meeting all criteria to create competitive advantage" (Skinner, 1996, p. 6).

We now discuss briefly key concepts and research that support our investigation. Although Skinner (1992) (1998) attempted to offer such a framework. Nonetheless, their methodology has some deficiencies that stem mainly from the fact that they do not consider tradeoffs/compatibilities as belonging to the same framework and relationship.

In the next section, we attempt to improve on [START_REF] Filippini | Trade-off and compatibility between performance: Definitions and empirical evidence[END_REF] methodology. Specifically, we will modify it using the concepts by Skinner (1992) and [START_REF] Schmenner | On theory in operations management[END_REF], and also the work by Sarmiento (2009) (e.g.; footnote2).

A methodology and illustrative examples

To explain our arguments, we utilize the figures by Filippini et al with their original interpretation. We then offer alternative understandings and conclusions using an improved methodology based on the more recent theoretical developments and clarifications discussed in section 2. The two columns that are divided into quartiles represent any two manufacturing capabilities (e.g.; quality, cost), while the black dots represent the various positions a firm can achieve along these performance types. Filippini et al characterize as "strong" those situations in which the two performance metrics are placed at the "top quartile" levels in the case of compatibilities, and at the "top quartile" and "bottom quartile" levels in terms of trade-offs. The rest of the situations are explained as "general". It is also noted that, consistent with previous investigations, Filippini et al also understand trade-offs and compatibilities (e.g.; cumulative capabilities) as belonging to different frameworks and situations. This is reflected in the scenarios depicted in figure 1, which are excluded from figure 2, and vice versa.

____________

We now move on to improve on Filippini et al's approach. First of all, we delineate a methodology by which the presence of compatibilities/trade-offs between pairs of competitive criteria can be asserted: criteria (e.g.; "quality" and "cost") of an individual and identifiable firm (e.g.; firm "X"). If there is evidence of an industry-leading type of performance in at least one competitive criterion, go to stage 2. If there is no evidence of a marketleading level of achievement in any of the criteria, no statement regarding the existence of trade-offs between them can be made2

Stage 2: Compare the performance of firm "X" along a pair of competitive criteria (e.g.; cost and quality). If both capabilities observe "industry-leading" type of performance, it can be asserted that no trade-offs exist, because no gap with respect to market-leading type of performance between the criteria is observed.

This means that firm "X" was capable of attaining best-in-industry type of performance in one competitive criterion (e.g.; quality) without causing a decrease in achievement in the other one (e.g.; cost). If one of the criteria observes "less-than-industry-leading" type of performance, go to stage 3.

Stage 3: When one competitive criterion observes market-leading performance and the other one does not, it can be asserted plausibly that there is a trade-off between the two criteria. This is because a trade-off signifies a difference in competitive power between multiple capabilities. Thus, it could be argued that the observed gap in achievement could be the result of a compromised relationship between the two criteria. This is consistent with Skinner's definition of a trade-off situation (1992). Moreover, common levels of performance between the criteria could also be assessed.

It can be seen that all the situations included in Filippini et al's two frameworks can be also described using our new methodology. We now offer Using our improved methodology, the "general compatibility" situation can be described as one in which firm "X" achieved compatibilities between the two performance types up to the "upper quartile" level, and that a trade-off with respect to the "top quartile" level of achievement (i.e.; a gap in performance representing a differential in competitive power) is also observed. As explained before, for a trade-off to be deemed as such, the difference in performance between two capabilities does not need to be "high". Thus, in our improved methodology and rationale, levels of compatibilities and differences with respect to the achievement of an industry-leading type of performance are acknowledged. Figure 3 illustrates our proposal.

____________

Take in figure 3 ____________ Let us imagine another hypothetical case (e.g.; firm "Y") that observes a "strong compatibility" scenario between the two competitive criteria (e.g.; any and all of the 15 firms identified in Filippini et al's figure 3, p. 3393). This is represented in figure 1 by the continuous and horizontal line connecting the two top quartiles. This relationship would also fit perfectly within our new approach.

Nevertheless, as opposed to the "general compatibility" case explained before, in the "strong compatibility" situation, no trade-offs between the two capabilities can we propose to reinterpret it as a scenario in which firm "Z" observes a compatibility at the "lower quartile" level between the two performance types, and also a trade-off with respect to the "top quartile" level of achievement, as there is a distinguishable gap in performance between the two capabilities. Figure 5 shows this: ____________ Take in figure 5 ____________ From our proposed methodology and the above examples, it is evident that the only scenario in which it is possible to state that no trade-off is present is when the two performance types are situated at the "top quartile" level at the same time. In all other cases, varying combinations of both trade-offs and/or compatibilities can be asserted. This is also consistent with the results by [START_REF] Sarmiento | Natural laws and the trade-off model: implications for research and policy-advising[END_REF]. Our improved methodology can also utilize likert-type scales.

These instruments are widely-used by researchers to collect information in survey-based studies. In those investigations, respondents are usually asked to perceptually assess their firm's performance across various capabilities (e.g.;

delivery, flexibility, cost) compared to their competitors using a 5 or a 7 pointscale. The upper end of the scale normally describes the highest level of performance (e.g.; 5= "our delivery performance is the best in industry"), while the rest of the points in the scale (e.g.; 4, 3, 2, 1) represent lesser levels of achievement.

Like Filippini et al's methodology, our new approach can also be applied to cases in which more than two performance types are included. Using a 5 point likert-type scale, let us suppose that a firm observes a "5" and "5" levels of achievement along "cost" and "quality", a "4" level of performance in "flexibility", and a "3" level of achievement in "delivery". We would interpret the observed scenario as a firm that exhibits no trade-offs between "cost" and "quality". We would also say that compatibilities are observed up to the "3" level of achievement across the analyzed capabilities, a level after which there are distinguishable and varying differences in performance between some capabilities (e.g.; cost and flexibility), something which signals the presence of trade-offs amongst various competitive criteria.

4

Conclusions and venues for future research We think that there are opportunities for future research using our proposed methodology. For example, as recent literature reviews (Sarmiento et al, 2010, Rosenzweig and[START_REF] Rosenzweig | Tradeoffs in manufacturing? A metaanalysis and critique of the literature[END_REF]

  on "trade-off and compatibility between performance: definitions and empirical evidence"

2

  Trade-offs and compatibilities: definitions, clarifications and background[START_REF] Filippini | Trade-off and compatibility between performance: Definitions and empirical evidence[END_REF] investigation offers an important departure from previous research. Their study, contrary to widely-used quantitative and probabilistic methods, is the first attempt to establish the existence of trade-offs and compatibilities by means of a deterministic approach using single firms as the units of analysis. They base their methodology on an assessment of the actual levels of performance attained by individual firms across various levels of performance, as well as differentials, are analyzed, and the existence of compatibilities and trade-offs are thus asserted.

  information regarding the performance along competitive

  showing how the presence of trade-offs and/or compatibilities can be assessed using a single framework. For example, let us imagine a manufacturing system (e.g.; firm "X") that observes an "upper quartile" level of achievement in one capability and a "top quartile" type of performance in the other capability (e.g.; any and all of the 35 firms identified in Filippini et al's figure3, p. 3393). This would represent a "general compatibility" relationship (figure1) as represented by the diagonal and dotted line connecting a top quartile with the upper quartile in the opposite column.

  scenario (e.g.; firm "Z") can be used to show that the cases included in Filippini et al's figure 2 can also be explained in light of our new proposal. For example, when one of the performance types is situated at the "top quartile" level, and the other one is at the "lower quartile" level (i.e.; a "general trade-off" situation shown in figure2represented by a dotted and diagonal line),

Filippini

  et al's (1998) novel methodology aids a researcher to assert in a deterministic (i.e.; precise) manner the presence of compatibilities and trade-offs between two or more capabilities in individual firms. Their work offers a different perspective from the widely-used quantitative and probabilistic approaches utilized in research that investigates whether a firm is capable of achieving an industry-leading type of performance across several capabilities. By taking into consideration more recent theoretical developments, we have tried to improve on their original approach.

Figure

  Figure 2. Trade-off situations ( strong definition, general definition) (adapted from Filippini et al, 1998)

  sum up, our investigation borrows from recent developments in this line of research. For example, from works such asSkinner (1992) and[START_REF] Schmenner | On theory in operations management[END_REF] we know that, as opposed to widely-held views, compatibilities (e.g.; cumulative capabilities) and trade-offs are two concepts that can be integrated under a single framework and relationship. From[START_REF] Sarmiento | Natural laws and the trade-off model: implications for research and policy-advising[END_REF], we understand that the trade-off model is a plausibly falsifiable (i.e.; deterministic/precise) concept. Its main premise rests on the analysis of whether identifiable and observable firms can achieve excellent levels of performance across a number of capabilities. As such, to estimate the validity of the trade-off model, a researcher necessitates a framework that can guide her/him in the analysis of individual firms in order to make assessments regarding the existence of trade-offs and/or compatibilities between multiple competitive criteria. Filippini et al

	it can also envelop probabilistic and deterministic theories, hypotheses,
	estimates, statements, cumulative capabilities, compatibilities, etc.
	F o
	r
	P
	e e r
	R e v i e w
	O n
	l y
	7

had already considered the possibility that compatibilities and trade-offs could explain more fully the relationships between two or more competitive criteria, it is

[START_REF] Schmenner | On theory in operations management[END_REF]

who formally attempt to integrate the two apparently competing concepts into a single framework. Their theory of performance frontiers gives a new light into the nature of compatibility and trade-off relationships between several criteria. In short, they suggest that while it is possible for a firm to achieve compatibilities (e.g.; cumulative capabilities) up to a certain level across several performance metrics, structural factors will impede the achievement of a high, industry-leading type of performance in all of them, something which is consistent with the underlying principle of the trade-off model explained before. More recently, Sarmiento (2009), utilizing Karl Popper's falsificationism

(Popper, 2002 a & b)

, demonstrates that the trade-off model is a plausibly falsifiable concept 1 . As such,

1

The trade-off model is presented in a similar (if not identical) way to that of the natural laws, as it negates the existence of a single firm that is capable of achieving industry-leading performance across all competitive criteria. This means that the trade-off model is a plausibly falsifiable concept. See

[START_REF] Sarmiento | Natural laws and the trade-off model: implications for research and policy-advising[END_REF] Popper (2002a, p. 48)

for more details.

To

uk International Journal of Production Research

  Skinner, W. 1992. Missing the links in manufacturing strategy. In Voss, C.A. (Ed), Manufacturing Strategy: Process and Content, Chapman and Hall, London, 13-

	concentrating on general trends in samples of firms (e.g.; the estimation of trade-Figures
	offs/cumulative capabilities using linear regression/correlation analysis).
	Relatively very few research (e.g.; case studies) focusing on individual firms have 25.
	been performed thus far. In our view, this means there is a particular need for Top quartile Top quartile
	studies that take individual firms as the unit of analysis in order to investigate the Skinner, W. 1996. Manufacturing strategy on the "S" curve. Production and
	potential existence of trade-offs and compatibilities. By integrating newer Operations Management, 5 (1), 3-14. Upper quartile Upper quartile
	F o theoretical developments into Filippini et al's (1998) pioneering work, we have provided researchers with a tool that can help them to determine the presence of trade-offs/compatibilities when individual firms are the units of analysis. We hope F o F o F o Lower quartile Lower quartile F o Bottom quartile Bottom quartile
	r future investigators will integrate our framework into their studies of the trade-off r r r r
	P model and related concepts. P P P P Figure 1. Compatibility situations (	strong definition,	general definition)
	e e r e e r e e r e e r r e e (adapted from Filippini et al, 1998)
		R e v i e w R e v i e w R e v i e w R e v i e w R e v i e w
		O n O n O n O n O n
			l y l y l y l y l y
			13 16

demonstrate, the study of trade-offs and compatibilities has been dominated by quantitative investigations Page 17 of 20 http://mc.manuscriptcentral.com/tprs Email: ijpr@lboro.ac.

http://mc.manuscriptcentral.com/tprs Email: ijpr@lboro.ac.ukInternational Journal of Production Research

A trade-off relationship between two capabilities implies the achievement of a market-leading level of performance in one of them (e.g.; quality) at the expense of lower levels in the other one (e.g.; cost). Thus, when neither of the analysed capabilities has achieved an excellent type of performance, it is impossible/illogical to assert the presence of a trade-off relationship between the two competitive criteria. Nevertheless, we think that assessments regarding potential compatibilities (i.e.; common levels of performance) between the two capabilities could still be made.