## A NEW LATTICE CONSTRUCTION: THE BOX PRODUCT (2005)

Citations: | 5 - 1 self |

### BibTeX

@MISC{Grätzer05anew,

author = {G. Grätzer and F. Wehrung},

title = {A NEW LATTICE CONSTRUCTION: THE BOX PRODUCT},

year = {2005}

}

### OpenURL

### Abstract

In a recent paper, the authors have proved that for lattices A and B with zero, the isomorphism Conc(A ⊗ B) ∼ = Conc A ⊗ Conc B, holds, provided that the tensor product satisfies a very natural condition (of being capped) implying that A ⊗ B is a lattice. In general, A ⊗ B is not a lattice; for instance, we proved that M3 ⊗ F(3) is not a lattice. In this paper, we introduce a new lattice construction, the box product for arbitrary lattices. The tensor product construction for complete lattices introduced by G. N. Raney in 1960 and by R. Wille in 1985 and the tensor product construction of A. Fraser in 1978 for semilattices bear some formal resemblance to the new construction. For lattices A and B, while their tensor product A ⊗ B (as semilattices) is not always a lattice, the box product, A □B, is always a lattice. Furthermore, the box product and some of its ideals behave like an improved tensor product. For example, if A and B are lattices with unit, then the isomorphism Conc(A □B) ∼ = Conc A ⊗ Conc B holds. There are analogous results for lattices A and B with zero and for a bounded lattice A and an arbitrary lattice B. A join-semilattice S with zero is called {0}-representable, if there exists a lattice L with zero such that Conc L ∼ = S. The above isomorphism results yield the following consequence: The tensor product of two {0}-representable semilattices is {0}-representable.

### Citations

34 | Applied lattice theory: formal concept analysis. Preprints http://wwwbib.mathematik.tudarmstadt.de/MathNet /Preprints/Listen/pp97.html
- Ganter, Wille
- 1997
(Show Context)
Citation Context ...emilattices); (ii) can be characterized with maps from one lattice to the other; (iii) have an “Isomorphism Theorem” for their (compact) congruence (semi) lattices. We refer to B. Ganter and R. Wille =-=[2]-=-, G. Grätzer, H Lakser, and R. W. Quackenbush [6], R. W. Quackenbush [13], G. N. Raney [14], Z. Shmuely [15], R. Wille [17], and the authors’ papers [9]–[12], for more information. More interestingly,... |

19 |
The structure of Galois connections
- Shmuely
- 1974
(Show Context)
Citation Context ... Theorem” for their (compact) congruence (semi) lattices. We refer to B. Ganter and R. Wille [2], G. Grätzer, H Lakser, and R. W. Quackenbush [6], R. W. Quackenbush [13], G. N. Raney [14], Z. Shmuely =-=[15]-=-, R. Wille [17], and the authors’ papers [9]–[12], for more information. More interestingly, it seems that formally similar results for two different types of tensor products do not seem to imply each... |

17 | Simultaneous representations of semilattices by lattices with permutable congruences
- Tůma, Wehrung
(Show Context)
Citation Context ...f there exists a lattice (resp., a lattice with zero) L with permutable congruences such that Conc L ∼ = D. There are non p-representable distributive {∨, 0}-semilattices, see J. T˚uma and F. Wehrung =-=[16]-=-. Furthermore, the second author of the present paper proved the following result: Let A and B be lattices with permutable congruences. If A⊠B is defined, then A ⊠ B has permutable congruences. In par... |

15 |
Tensorial decompositions of concept lattices, Order 2
- Wille
- 1985
(Show Context)
Citation Context ...tices (Definition 2.1). For lattices A and B, the box product, A □ B, is always a lattice. If A and B are finite, then A □ B is isomorphic to the complete tensor product A ̂⊗ B considered in R. Wille =-=[17]-=-, see also Section 11. We also introduce an ideal A⊠B of A□B; we shall call A⊠B the lattice tensor product of A and B. The ideal A ⊠ B can be defined if A and B have a zero, or if either A or B is bou... |

13 | Proper congruence-preserving extensions of lattices
- Grätzer, Wehrung
- 1999
(Show Context)
Citation Context ...s, to compute that (5.4) (5.5) M3〈L〉 = { 〈x, y, z〉 ∈ L 3 | x = ˆy ∧ ˆz, y = ˆx ∧ ˆz, z = ˆx ∧ ˆy }, N5〈L〉 = { 〈x, y, z〉 ∈ L 3 | x = z ∧ (x ∨ y) }. In particular, M3〈L〉 has the same meaning here as in =-=[9]-=-. Thus it suffices to prove that M3〈M3〉 ̸= M3[M3] and that N5〈N5〉 ̸= N5[N5]. But it is easy to verify that 〈p, q, r〉 ∈ M3[M3] − M3〈M3〉, 〈c, b, a〉 ∈ N5[N5] − N5〈N5〉. By using (5.2) and (5.4), it is als... |

12 |
On complete congruence lattices of complete lattices
- GRÄTZER, LAKSER
- 1991
(Show Context)
Citation Context ...eem to have nothing in common: (11.1) equates tensor products of two distributive {∨, 0}-semilattices, while (11.2) equates tensor products of arbitrary complete lattices. It was proved in G. Grätzer =-=[3]-=- (see G. Grätzer and H. Lakser [5] for the shortest proof and G. Grätzer and E. T. Schmidt [8] for the strongest result) that Con∞ A can be any complete lattice. In general, the constructions of compl... |

11 | The structure of tensor products of semilattices with zero
- Grätzer, Lakser, et al.
- 1981
(Show Context)
Citation Context ...result of this field is the isomorphism (1.1) Conc(A ⊗ B) ∼ = Conc A ⊗ Conc B we proved in [10] for capped tensor products; this generalizes the result of G. Grätzer, H. Lakser, and R. W. Quackenbush =-=[6]-=- for finite lattices. This isomorphism does not always make sense because A ⊗ B is not a lattice, in general; in [11] and [12], we provided examples, for instance, M3 ⊗ F(3) is not a lattice (this sol... |

10 |
Nonmodular varieties of semimodular lattices with a spanning M3. Special volume on ordered sets and their applications (L’Arbresle
- Quackenbush
- 1982
(Show Context)
Citation Context ...not always make sense because A ⊗ B is not a lattice, in general; in [11] and [12], we provided examples, for instance, M3 ⊗ F(3) is not a lattice (this solved a problem proposed in R. W. Quackenbush =-=[13]-=-). In [12], we solved a problem of E. T. Schmidt and the first author: does every lattice have a proper congruence-preserving extension. In earlier papers, such an extension for a distributive lattice... |

9 |
Tight Galois connections and complete distributivity
- Raney
- 1960
(Show Context)
Citation Context ...e an “Isomorphism Theorem” for their (compact) congruence (semi) lattices. We refer to B. Ganter and R. Wille [2], G. Grätzer, H Lakser, and R. W. Quackenbush [6], R. W. Quackenbush [13], G. N. Raney =-=[14]-=-, Z. Shmuely [15], R. Wille [17], and the authors’ papers [9]–[12], for more information. More interestingly, it seems that formally similar results for two different types of tensor products do not s... |

5 | Lattice Theory. Second Edition, Birkhäuser - General - 1998 |

5 |
products and transferability of semilattices, Canad
- Tensor
- 1999
(Show Context)
Citation Context ...lar, A ⊗ B is an algebraic lattice. The {∨, 0}-semilattice tensor product A ⊗ B is defined as the {∨, 0}-semilattice of all compact elements of A ⊗ B. The relationship between A ⊗ B (as in [6], [10], =-=[11]-=- but not as in [1]) and the lattice tensor product A ⊠ B is quite mysterious. Note that while A ⊗ B may not be a lattice (see [11] and [12]), A ⊠ B is always a lattice. Both A ⊗ B and A ⊠ B are {∨, 0}... |

4 |
products of lattices with zero, revisited
- Tensor
(Show Context)
Citation Context ...zero such that Conc L ∼ = S. The above isomorphism results yield the following consequence: The tensor product of two {0}-representable semilattices is {0}-representable. 1. Introduction In our paper =-=[10]-=-, we recalled in detail the introduction of tensor products of lattices in the seventies. The main result of this field is the isomorphism (1.1) Conc(A ⊗ B) ∼ = Conc A ⊗ Conc B we proved in [10] for c... |

3 |
The tensor product of semilattices, Algebra Universalis 8
- Fraser
- 1978
(Show Context)
Citation Context ...s. This holds for all X ∈ c ∗ , thus y ≤ Pc ∗(⃗ b). Hence, which concludes the proof. 〈x, y〉 ∈ Pc(⃗a) ⊠ Pc ∗(⃗ b) ⊆ K, □ Lemma 4.4 implies two important purely arithmetical formulas (see G. A. Fraser =-=[1]-=- and [10]). Lemma 4.5. Let A and B be lattices with zero. Let a0, a1 ∈ A and b0, b1 ∈ B. Then (a0 ⊠ b0) ∩ (a1 ⊠ b1) = (a0 ∧ a1) ⊠ (b0 ∧ b1), (a0 ⊠ b0) ∨ (a1 ⊠ b1) = (a0 ⊠ b0) ∪ (a1 ⊠ b1) ∪ ((a0 ∨ a1) ... |