A note on supermodular sublattices in
finite relatively complemented lattices
Andrei Krokhin and
Benoit Larose
Abstract. A sublattice in a lattice is called supermodular if, for every two elements, one
of which belongs to the sublattice, at least one of their meet and join also belongs to the
sublattice. In this note, we describe supermodular sublattices in products of relatively
complemented lattices.
1. Introduction and Preliminaries
Let L be a lattice. A function f : L → R is called supermodular if it satisfies
f (x)+f (y) ≤ f (x∧y)+f (x∨y) for all x, y ∈ L. Supermodular functions on lattices
play an important role in mathematical economics and operations research (see,
e.g., [6]). This concept has recently found a new application in computer science,
namely in the study of complexity of certain important optimisation problems called
maximum constraint satisfaction problems (or Max CSP), see [1, 2, 4, 5]. An
instance of Max CSP consists of a set of variables and a collection of constraints
which are applied to certain specified subsets of these variables; the goal is to
find values for the variables which maximize the number of simultaneously satisfied
constraints. In this setting, each constraint is given by a 0-1 valued function on some
direct power of a fixed finite set D. Max CSP is computationally hard in general,
but there is good evidence (see the above mentioned papers for details) that all easy
subproblems based on restricting the form of constraints can be described as those
where all constraints are supermodular functions on suitable powers of some lattice
on D. Hence, 0-1 valued functions which are supermodular on direct powers of
finite lattices are of particular importance in this context. An explicit description of
such functions can help design efficient algorithms for certain subproblems of Max
CSP, as it happened, e.g., in [5] when the finite lattice in question is a diamond. It
turns out that these functions give rise to a very natural lattice-theoretic concept
which, to the best of our knowledge, has not been studied before, and which, when
thoroughly studied, may facilitate progress in classifying the complexity of Max
CSP.
Definition 1.1. A sublattice L0 of a lattice L is called supermodular (in L) if, for
any x ∈ L0 and y ∈ L, at least one of the elements x ∧ y, x ∨ y belongs to L0 .
The name for such sublattices is justified by the following lemma.
1
2
ANDREI KROKHIN AND BENOIT LAROSE
Lemma 1.2. Let L be a lattice and let f be a 0-1 valued function on L. Then f is
supermodular on L if and only if the set Sf = {x ∈ L | f (x) = 1} is a supermodular
sublattice of L.
Proof. Assume that f is supermodular on L. Then, for any x, y ∈ Sf , we have
f (x) = f (y) = 1, so f (x) + f (y) = 2 ≤ f (x ∧ y) + f (x ∨ y), and f (x ∧ y) =
f (x ∨ y) = 1 implying that Sf is a sublattice. Similarly, if x ∈ Sf and y ∈ L then
f (x ∧ y) + f (x ∨ y) ≥ 1, and so at least one of x ∧ y, x ∨ y belongs to Sf . The other
direction is similar.
¤
We now present the basic terminology and notation we shall use in the sequel.
Let L be a finite lattice. As usual we denote its smallest and largest elements by 0
and 1 respectively. An atom a of L is an element that covers 0, i.e. such that 0 < a
and 0 < a0 ≤ a implies a0 = a. A coatom is an element covered by 1. An interval
is a subset of L of the form {x : a ≤ x ≤ b} for some a, b ∈ L. A subset I of L is
an order ideal of L if a ∈ I whenever a ≤ b for some b ∈ I; an order filter is defined
dually. An ideal (filter) of L is an order ideal (filter) which is also a sublattice of
L. The principal ideal generated by b is the ideal ↓ (b) = {x : x ≤ b}; the principal
filter ↑ (a) is defined dually. It is well-known that every ideal and every filter in a
finite lattice is principal. Let ↓ (b1 , . . . , bl ) denote the order ideal consisting of all
x ∈ L such that x ≤ bi for some 1 ≤ i ≤ l; define the order filter ↑ (a1 , . . . , ak )
similarly. A subset C of L is convex if b ∈ C whenever a ≤ b ≤ c for some a, c ∈ C.
Prototypical examples of supermodular sublattices in any lattice are its ideals,
filters, and (set-theoretic) unions of an ideal and a filter. Note that it is possible
that supermodular sublattices in some finite lattice L have much less structure than
those in its direct powers, as the next example shows.
Example 1.3. Let L be a finite chain. Then it is easy to check that every sublattice
of L is supermodular. However, it can be easily derived from Lemma 4.4 of [2] that
a sublattice L0 of L2 is supermodular if and only if L0 = T1 ∪ T2 is the disjoint union
of two sublattices T1 and T2 (one of which may be empty) such that T1 is an ideal,
or a filter, or a union of an ideal and a filter in L2 , and T2 is S × L or L × S for
some sublattice S of L.
2
Recall that a lattice L is relatively complemented if every interval of L is complemented, i.e. for all a ≤ x ≤ b there exists a ≤ y ≤ b such that a = x ∧ y and
x ∨ y = b. Our main result is an explicit description of supermodular sublattices in
products of finite relatively complemented lattices. Note that every finite relatively
complemented lattice is a direct product of simple relatively complemented lattices
([3]).
2. Results
Lemma 2.1. A sublattice X of a finite relatively complemented lattice L is supermodular if and only if L \ X is convex.
A NOTE ON SUPERMODULAR SUBLATTICES
3
Proof. Assume that X is supermodular and let a ≤ x ≤ b with a, b 6∈ X . Let
x0 be the complement of x in [a, b]. Then x 6∈ X , since otherwise at least one of
a = x∧x0 , b = x∨x0 would belong to X . Hence L\X is convex. The other direction
is straightforward.
¤
Lemma 2.2. A non-empty subset X of a finite relatively complemented lattice L is
a supermodular sublattice if and only if one of the following conditions is satisfied:
(1) X is an ideal of L, or
(2) X is a filter of L, or
(3) there exist atoms d1 , . . . , dk and coatoms e1 , . . . , el of L such that, for all
x ∈ L, x 6∈ X if and only if di ≤ x ≤ ej for some i, j, and furthermore
(a) for every 1 ≤ i ≤ k, ↑ (di )\ ↓ (e1 , . . . , el ) is a sublattice of L, and
(b) for every 1 ≤ j ≤ l, ↓ (ej )\ ↑ (d1 , . . . , dk ) is a sublattice of L.
Proof. (⇒) Suppose that X is neither an ideal nor a filter; in particular it contains
both 0 and 1, since L \ X is convex by Lemma 2.1. Let u be a minimal element in
L \ X . We show that u is an atom. Indeed, if 0 < x < u, let x0 be a complement
of x in [0, u]; then by choice of u we have that x, x0 ∈ X , so u = x ∨ x0 ∈ X , a
contradiction. A similar argument shows that every maximal element of L \ X is
a coatom of L. The first claim in (3) then follows easily. Now consider x, y ≥ di
such that x, y 6≤ ej for all j. Clearly x ∨ y satisfies both conditions, and clearly
x ∧ y ≥ di . Since X is a sublattice, we have that x ∧ y ∈ X so by the first claim
x ∧ y 6≤ ej for all j, which proves (3a). The property (3b) is dual.
(⇐) Trivially, every ideal and every filter of L is a supermodular sublattice. So
now suppose that X satisfies (3): we must show that X is a supermodular sublattice.
In fact, L \ X is obviously convex, so we only need to show that it is a sublattice.
Note that an element x ∈ L belongs to X if and only if x 6≥ di for all i or x 6≤ ej
for all j, or both. Let x, y ∈ X . Suppose without loss of generality that x 6≤ ej for
all j (the other case is dual). Then obviously x ∨ y 6≤ ej for all j, so x ∨ y ∈ X .
Assume for contradiction that x ∧ y 6∈ X . Then we have that di ≤ x ∧ y ≤ ej for
some i, j. Then obviously di ≤ x, y. Since y ∈ X , we have that y 6≤ ej for all j.
Then both x and y are in a sublattice of X described in (3a) and hence x ∧ y ∈ X ,
a contradiction.
¤
Theorem 2.3. Let n ≥ 2 and let Li , 1 ≤ i ≤ n, be finite relatively complemented
lattices, with at least two elements each. A subset X of L = L1 × . . . × Ln is a
supermodular sublattice if and only X is of one of the following types:
(1) an ideal, a filter, or a union of an ideal and a filter in L;
(2) there exists an index 1 ≤ i ≤ n and a supermodular sublattice Xi of Li such
that
X = L1 × . . . Li−1 × Xi × Li+1 × . . . Ln .
(3) there exist an index 1 ≤ i ≤ n, an element ai and an order ideal Ii in Li
such that Ii ∪ ↑ (ai ) is a sublattice in Li , and, for all 1 ≤ j ≤ n with j 6= i,
elements cj ∈ Lj , such that X = I ∪ F where
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ANDREI KROKHIN AND BENOIT LAROSE
I
F
=
=
↓ (c1 ) × . . . × ↓ (ci−1 ) × Ii × ↓ (ci+1 ) × . . . × ↓ (cn ),
L1 × . . . × Li−1 × ↑ (ai ) × Li+1 × . . . × Ln .
(4) there exist an index 1 ≤ i ≤ n, an element ai and an order filter Fi in Li
such that ↓ (ai ) ∪ Fi is a sublattice in Li , and, for all 1 ≤ j ≤ n with j 6= i,
elements cj ∈ Lj , such that X = I ∪ F where
I
F
= L1 × . . . × Li−1 × ↓ (ai ) × Li+1 × . . . × Ln ,
= ↑ (c1 ) × . . . × ↑ (ci−1 ) × Fi × ↑ (ci+1 ) × . . . × ↑ (cn ).
Proof. It is easy to verify that all subsets of L described in (1)-(4) are supermodular
sublattices. Now assume that X is a supermodular sublattice of L: we show that
it has of the forms described above. For all i, let 0i and 1i denote the least and the
greatest element of Li , respectively.
We may assume X is neither an ideal nor a filter in L. We show that, for all
1 ≤ i ≤ n, there exist an order ideal Ii and
Fi of Li such that X can
Qnan order filter Q
n
be represented as X = I ∪ F where I = i=1 Ii and F = i=1 Fi . By Lemma 2.2,
there exist atoms d1 , . . . , dk and coatoms e1 , . . . , el of L such that, for all x ∈ L,
x ∈ X if and only if x 6≥ di for all 1 ≤ i ≤ k or x 6≤ ej for all 1 ≤ j ≤ l. Every atom
of L has the form (01 , , . . . , , 0i−1 , ai , 0i+1 , , . . . , 0n ) for some index i and some atom
ai of Li , and a dual statement holds for coatoms of L. Let Ui denote the set of all
atoms a in Li such that, for some 1 ≤ j ≤ k, dj = (01 , , . . . , , 0i−1 , a, 0i+1 , , . . . , 0n ),
and let Ii = {x ∈ Li | x 6≥ a for all a ∈ Ui } (note that Ii = Li when Ui = ∅).
Dually, let Vi denote the set of all coatoms b in Li such that, for some 1 ≤ j ≤ l,
ej = (11 , , . . . , , 1i−1 , b, 1i+1 , , . . . , 1n ), and let Fi = {x ∈ Li | x 6≤ b for all b ∈ Vi }.
It is Q
clear that, for allQ
i, Ii is an order ideal and Fi is order filter in Li . Moreover,
n
n
I = i=1 Ii and F = i=1 Fi form an order ideal and an order filter, respectively,
of L. Furthermore, an element x ∈ L belongs to I if and only if x 6≥ di for all
1 ≤ i ≤ k, and again the dual statement holds for F . Thus, we have proved that
X = I ∪ F.
If all order ideals Ii and all order filters Fi are principal then, clearly, X is a
union of an ideal and a filter in L. Assume now that there is an order ideal Ii which
is not principal. Let u and v be distinct maximal elements in Ii and consider the
tuples (01 , , . . . , , 0i−1 , u, 0i+1 , , . . . , 0n ) and (01 , , . . . , , 0i−1 , v, 0i+1 , , . . . , 0n ). These
tuples belong to I, but their join does not. Hence their join must belong to F
which implies there every order filter Fj , except possibly Fi , contains 0j and so is
equal to Lj . This implies that either all order ideals Ij , j 6= i, are principal or else
X = L. Now if Fi is not principal then the dual argument shows that every order
ideal Ij , except possibly Ii is equal to Lj , in which case X is of type 2. On the
other hand, if Fi is principal then X is of type 3. Dually, one can show that if, for
some i, the order filter Fi is not principal then X is of type 2 or 4. This completes
the proof.
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A NOTE ON SUPERMODULAR SUBLATTICES
5
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Department of Computer Science, University of Durham, Durham, DH1 3LE, UK,
[email protected]
Department of Mathematics and Statistics, Concordia University, Montréal, Qc,
Canada, H3G 1M8, [email protected]