3
$\begingroup$

It is easy to think up interesting, natural models of the algebraic theory presented as follows, such that in these models, $x^\dagger$ is always the multiplicative inverse of $x$.

Question. What are some interesting, natural models in which $x^\dagger$ is not necessarily the multiplicative inverse?

Sorts. $U$

Functions. $$\wedge : U \times U \rightarrow U, \qquad \vee : U \times U \rightarrow U$$

$$1 : U, \qquad \times : U \times U \rightarrow U, \qquad (x \mapsto x^{\dagger}) : U \rightarrow U$$

Axioms. (I write $\times$ concatenatively).

  1. $(U,\wedge,\vee)$ is a distributive lattice
  2. $(U,1,\times)$ is a commutative monoid
  3. Multiplication distributes over both $\wedge$ and $\vee$
  4. $1^\dagger = 1$
  5. $(xy)^\dagger = x^\dagger y^\dagger$
  6. $x^{\dagger\dagger} = x$
  7. $(x \wedge y)^\dagger = x^\dagger \vee y^\dagger$
  8. $(x \vee y)^\dagger = x^\dagger \wedge y^\dagger$

Examples of such things in which $x^\dagger$ is always the multiplicative inverse.

  • $\mathbb{Q}_{>0}$ or $\mathbb{R}_{>0}$ with $\wedge$ and $\vee$ interpreted as $\mathrm{min}$ and $\mathrm{max}$ respectively.

  • $\mathbb{Z}$ with $(1,\times,x \mapsto x^\dagger)$ interpreted as $(0,+,x \mapsto -x)$, and $\wedge$ and $\vee$ interpreted as $\mathrm{min}$ and $\mathrm{max}$ respectively.

  • Cartesian products of these

$\endgroup$
5
  • $\begingroup$ 3. means ... $a(x \vee y) = ax \vee ay$ and similarly for $\wedge$ ?? $\endgroup$ Dec 26, 2014 at 20:04
  • $\begingroup$ @GeraldEdgar, yes, that is what I mean. $\endgroup$ Dec 27, 2014 at 0:28
  • $\begingroup$ consider relational algebras. $\endgroup$ Dec 28, 2014 at 6:25
  • 1
    $\begingroup$ @TheMaskedAvenger I'm not picking up on your hint at all. You mean where elements of $U$ are suitable binary relations and the monoid multiplication is given by relational composition? Unless I'm missing something, it's rare that relational composition distributes over meets, and commutativity of multiplication imposes a pretty tight restriction as well. And what would $\dagger$ be that interchanges meet and join? $\endgroup$
    – Todd Trimble
    Dec 29, 2014 at 14:21
  • $\begingroup$ There are a lot of changes one can ring on such algebras. I misread the problem, not noticing the join meet interchange. I thought dagger might serve as reverse (transpose?) and then one could pick a commutative and distributive subalgebra. But for the interchange, I don't know now. I still think something can be done with relational algebras, because Tarski et al did it with relational and cylindric algebras. $\endgroup$ Dec 29, 2014 at 17:32

1 Answer 1

2
$\begingroup$

not a solution
Anything that satisfies 1,2,3; then define $x^\dagger = 1$ for all $x$.

added December 27
Try this example...
$U := \mathbb Z \times \mathbb Z$.

  • multiplication is performed by componentwise addition, $(a,b)(c,d) = (a+c,b+d)$

  • the lattice operations are also performed componentwise, $(a,b) \vee (c,d) = (a\vee c,b\vee d)$, $(a,b) \wedge (c,d) = (a\wedge c,b\wedge d)$

  • but reflect the symmetry, $(a,b)^\dagger = (-b,-a)$

Now there are lots of things to check, to see if it really works.

$\endgroup$
4
  • 1
    $\begingroup$ This is not involutive. $\endgroup$ Dec 26, 2014 at 19:51
  • $\begingroup$ i.e. 6 fails. OK. $\endgroup$ Dec 26, 2014 at 19:52
  • 1
    $\begingroup$ Another observation is that $x^\dagger = x$ doesn't give anything interesting, because this forces meets to coincide with joins, by 7 or 8. $\endgroup$ Dec 26, 2014 at 19:53
  • 2
    $\begingroup$ The December 27 answer apparently works. $\endgroup$
    – Todd Trimble
    Dec 28, 2014 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.