I am wondering about generalisations of the concept of equivalence relations to suplattices.

Here is my motivation: Given a set $X$. The powerset $\mathcal{P}(X)$ is a suplattice. For suplattices there is a tensor product, giving $\mathcal{P}(X)\otimes\mathcal{P}(X)=\mathcal{P}(X\times X)$. Now we can define the diagonal $\Delta:=\left\{(x,x)\mid x\in X\right\}\in\mathcal{P}(X\times X)$, which is an equivalence relation. Take a general suplattice $M$. There is a tensor product $M\otimes M$, which can be embedded into the suplattice $\mathcal{P}(M\times M)$ (reference). The elements of $\mathcal{P}(M\times M)$ can be projected to the first (or to the second) component and you can take the supremum of the resulting subset of $M$. Thus you get projections $\pi_0,\pi_1\colon M\otimes M\to M$, preserving suprema.

In this context you could call a $\Delta\in M\otimes M$ an “equivalence relation in $M$ if it satisfies these conditions:

  • Symmetry: For all $a,b\in M$ such that $a\otimes b\le\Delta$ we have $b\otimes a\le\Delta$.
  • Transitivity: For all $a,b,c\in M$ such that $a\otimes b\le\Delta$ and $b\otimes c\le\Delta$ we have $a\otimes c\le\Delta$.
  • Reflexivity: The projections of the relation are maximal: $\pi_0(\Delta)=\pi_1(\Delta)=\top$ where $\top\in M$ is the top of the lattice.

Clearly, the diagonal is an equivalence relation in $\mathcal{P}(X)$. However,

I am interested in ways to construct such relations in non-trivial cases (i. e. where the lattice is not the powerset of a set and $\Delta$ should not be the top itself). Is there such a concept in the literature? Has it been studied? Or do you have any comment? Non-trivial examples (maybe there are even canonical examples?) or objections regarding the soundness of the definition are welcome.

My original motivation for regarding these concepts is the definition of the semantics of equality in certain logics interpreted in $\mathcal{P}(X\times\ldots\times X)$ using the diagonal—I am not an expert in the theory of suplattices etc. Best regards

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    $\begingroup$ Be careful: the embedding of $M \otimes M$ into $P(M \times M)$ is a right adjoint (in the locally ordered 2-category of posets and order-preserving maps), not a left adjoint. So I'm not seeing how you're getting sup-preserving maps $M \otimes M \to M$ "for free". $\endgroup$ – Todd Trimble Jun 2 '13 at 2:55
  • $\begingroup$ You are right, it does not work automatically, thanks for the warning. An example: Using the explicit representation of the tensor product given in the paper referenced above, $\pi_0(a)$ would be $\top$ for all $a$. You have to “disregard representations of the form $a\otimes\bot$”. Using a similar embedding of $M\otimes M$ into $\mathcal{P}((M\setminus\left\{\bot\right\})\times (M\setminus\left\{\bot\right\})$ it should work. Or define $\pi_0(x):=\bigvee_{a\otimes b\le x,b\ne\bot} a$. $\endgroup$ – The User Jun 2 '13 at 11:06

It seems like a more natural way to generalize equivalence relations from $P(X)$ to lattices is to look at partitions. One defines a partition of a lattice L(or Boolean algebra, frame, or whatever) to be a subset $p\subseteq L\setminus\{0\}$ such that $\bigvee p=1$ and where $a\wedge b=0$ whenever $a,b\in p,a\neq b$. The obvious advantage of partitions in this case would be that the definition of a partition is much simpler and it does not involve any kinds of tensor products. I personally have researched partitions on Boolean algebras and frames and I have found them quite useful and interesting since they relate Boolean algebras to diverse topics such as uniform spaces, the ultrapower construction, pro-sets, and point-free topology. A good question to ask is if there is a one-to-one correspondence between partitions on lattices and the equivalence relations as you defined them.

  • $\begingroup$ Thanks. A natural choice to translate a partition $p$ into the tensor product is $\bigvee_{a\in p} a\otimes a$. Symmetry and reflexivity (using the more precise definition of the projection given in the comment above) are obvious. However, I currently cannot prove transitivity. $\endgroup$ – The User Jun 2 '13 at 15:14

A suplattice automatically has all meets, and a suplattice where meets distribute over all joins is a locale. The category of partial (i.e. nonreflexive) equivalence relations and functional relations valued in a suplattice is equivalent to the topos of sheaves over that locale. These are Grothendieck toposes which have been studied intensively.

Let me spell out the equivalence. Starting with a sheaf $F$ over a locale $L$, let a local section of $F$ is just a member of $F(p)$ for some $p\in L$. For each pair of local sections $f,g$, there is a greatest $p\in L$ such that $f_p = g_p$ ($f_p$ is the restriction of $f$ to $p$). This determines an $L$-valued partial equivalence relation $e$ on local sections. The reason that $e(f,f)$ isn't always $\top$, is because $f$ may not be a global section. For a morphism of sheaves $\eta:F\to G$, and local sections $f$ of $F$ and $g$ of $G$, there is a greatest $p$ such that $\eta_p(f_p) = g_p$, and this determined an $L$-valued functional relation.

For each $p\in L$ there is a trivial partial equivalence relation $Y(p) = (1,p)$, where $p:1\times 1\to L$ is constantly $p$. One weak inverse of the equivalence above sends each set with partial equivalence $(X,e)$ to the sheaf $F$ where $F(p)$ is the set of functional relations $Y(p) \to (X,e)$.

For examples of suplattice valued equivalence relations, have a look at Booleans valued models, fuzzy sets and complete Heyting algebra valued sets.

  • $\begingroup$ Thanks, that seems to be an interesting relationship. But do you see any relationship between (partial) suplattice valued equivalence relations and (maybe nonreflexive) equivalence relations defined in the tensor product of suplattices? $\endgroup$ – The User Jun 2 '13 at 15:22
  • $\begingroup$ It is the same connection as with the power sets: $L^X\otimes L^X = L^{X\times X}$ for some tensorproduct of suplattices. Hence a suplattice valued equivalence relation is an equivalence relation in a suplattice. $\endgroup$ – Wouter Stekelenburg Jun 3 '13 at 9:36

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