Is a poset an algebra?

Question

I encountered some nice material about so-called universal algebras.

E.g. a lattice is an algebra (in the sense of universal algebra), and can be presented as a tuple $\langle L,\wedge,\vee\rangle$ where $\wedge$ and $\vee$ are binary operations on $L$.

I am not (yet) familiar with that stuff, and one of the first questions that arose was:

Can a partial order be presented as an algebra?

I think not because I cannot really find operations on it with some (finite) arity.

Am I correct in my thinking?

More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?

Answer 1

A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.

In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,\leq)$, one defines an algebra $(A,\cdot)$, where $$ a \cdot b = \begin{cases} a &\text{if } a \leq b,\\ b &\text{otherwise.} \end{cases} $$ The poset can be recovered from the algebra by making $a \leq b$ iff $a\cdot b = a$.

In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,\leq, 1)$ with a top element they define two algebraic structures: $$ a \to b = \begin{cases} 1 &\text{if } a \leq b,\\ b &\text{otherwise.} \end{cases} $$ and $$ a \cdot b = \begin{cases} b &\text{if } a \leq b,\\ 1 &\text{otherwise.} \end{cases} $$ These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.

Answer 2

No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=\{a\leq b,c\leq d\}$ with the equivalence relation $\sim$ generated by $b\sim c$. The quotient by this equivalence relation is the poset $Q=\{a\leq [b]\leq d\}$, where by transitivity $a\leq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:P\to Q$ to $p^{-1}(a\leq d)\to a\leq d$. Now, $p^{-1}(a\leq d)$ is just the discrete poset $\{a,d\}$. But the morphism $\{a,d\}\to\{a\leq d\}$ is not the quotient map for the restriction of $\sim$ to $\{a,d\}$. Indeed, it's not the quotient by any equivalence relation at all.

This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $\sim$ and we restrict the quotient map $A\to A/{\sim}$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $G\to G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $g\sim h\iff gh^{-1}\in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $\sim$, so the restriction of the quotient map is indeed a quotient map.

To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $X\times X$ such that the diagonal map $X\to X\times X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $X\times X\to X\times X$ that interchanges the factors (symmetry), and the pullback of $R\times_X R\subset X\times X\times X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.

So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.

Answer 3

From the question:

More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?

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There is a theorem of algebra that says

Theorem. A bijective morphism is an isomorphism.

That is, if $\varphi: A\to B$ is a homomorphism that is 1-1 and onto, then $\varphi^{-1}$ is also a homomorphism. More categorically, the theorem says that, if $\mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.

This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.

You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.