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Question 1:Is there a classification of finite lattices which admit a multiplication making them into a finite multiplicative lattices? (see https://encyclopediaofmath.org/wiki/Multiplicative_lattice for a definition)? Is there a good way to find all multiplications on a finite lattices that make it into a multiplicative lattice via a computer?

If it helps, we can assume that the lattices are distributive in the classical sense.

Question 2: Is there such a classification if we drop the commutativity assumption in the definition?

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    $\begingroup$ For question 2, after you drop the commutativity assumption, left distributivity and right distributivity are distinct notions. Did you want the finite lattices to just be left distributive, or did you want two-way distributivity? $\endgroup$
    – Joseph Van Name
    Commented Feb 15, 2022 at 17:12
  • $\begingroup$ @JosephVanName I want to find all lattices that have a non-commuattive multiplicative lattice structure. If it makes the problem easier, we can assume that those lattices that we start with are distributive (in the classical sense). $\endgroup$
    – Mare
    Commented Feb 15, 2022 at 17:14
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    $\begingroup$ In order to have another term to search for, it should be helpful to know that multiplicative lattices are also known as commutative quantales. $\endgroup$
    – Tobias Fritz
    Commented Jun 9, 2022 at 11:47

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Assuming distributivity, the answer to the first part of Question 1 is simple:

Every finite distributive lattice $L$ admits a multiplication, namely the meet operation.

  1. Meet is commutative and associative.
  2. The top acts as the multiplicative identity: $1 \wedge a = a$ for all $a \in L$.
  3. For all $a \in L$ and $B \subseteq L$, the condition $a \wedge \left(\bigvee_{b \in B} b\right) = \bigvee_{b \in B} (a \wedge b)$ follows from distributivity and finiteness.

But there may be also other multiplications. For example, the three-element chain $0 < 1 < 2$ admits two multiplications, whose multiplication tables are:

[0 0 0]  [0 0 0]
[0 0 1]  [0 1 1]
[0 1 2], [0 1 2]

(the latter is the meet).

The four-element chain $0 < 1 < 2 < 3$ admits six lattice multiplications:

[0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]  [0 0 0 0]
[0 0 0 1]  [0 0 0 1]  [0 0 0 1]  [0 0 1 1]  [0 1 1 1]  [0 1 1 1]
[0 0 0 2]  [0 0 1 2]  [0 0 2 2]  [0 1 2 2]  [0 1 1 2]  [0 1 2 2]
[0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3], [0 1 2 3]

The four-element diamond $0 < \{1,2\} < 3$ admits only one lattice multiplication, namely the meet.

These were found with a relatively brute-force search. How to find all lattice multiplications efficiently, I do not know.

For chains of at most 14 elements, the number of different multiplications is in OEIS A030453 "Number of linearly ordered Abelian monoids of size n (semi-groups with greatest element of the corresponding chain as neutral element); triangular norms on an n-chain."

There are some nondistributive, even nonmodular lattices that admit a multiplication. The smallest examples have $6$ elements:

Two six-element nondistributive lattices that admit multiplication

Of course we recognize them as the smallest nondistributive lattices (diamond $M_3$ and pentagon) with an augmented top. The multiplication is a bit moot: $a \cdot b = 0$ when $a,b<1$, and $a \cdot 1 = 1 \cdot a = a$ (where $0$ denotes bottom and $1$ denotes top). In fact, any lattice that has only one coatom admits this kind of multiplication.

So we have at least two categories of finite lattices that admit a multiplication:

  1. Distributive lattices
  2. One-coatom lattices
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