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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

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  • $\begingroup$ 3. means ... $a(x \vee y) = ax \vee ay$ and similarly for $\wedge$ ?? $\endgroup$ – Gerald Edgar Dec 26 '14 at 20:04
  • $\begingroup$ @GeraldEdgar, yes, that is what I mean. $\endgroup$ – goblin Dec 27 '14 at 0:28
  • $\begingroup$ consider relational algebras. $\endgroup$ – The Masked Avenger Dec 28 '14 at 6:25
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    $\begingroup$ @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 '14 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. $\endgroup$ – The Masked Avenger Dec 29 '14 at 17:32
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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.

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    $\begingroup$ This is not involutive. $\endgroup$ – goblin Dec 26 '14 at 19:51
  • $\begingroup$ i.e. 6 fails. OK. $\endgroup$ – Gerald Edgar Dec 26 '14 at 19:52
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    $\begingroup$ Another observation is that $x^\dagger = x$ doesn't give anything interesting, because this forces meets to coincide with joins, by 7 or 8. $\endgroup$ – goblin Dec 26 '14 at 19:53
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    $\begingroup$ The December 27 answer apparently works. $\endgroup$ – Todd Trimble Dec 28 '14 at 11:22

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