Subject: categories: Re: Laws

From: [email-masked] (Peter Selinger)
Date: Tue, 8 Aug 2006 02:08:40 -0300 (ADT)

Hi Tom,

a lot is known about this. I will leave it to more qualified others to
give the category-theoretic account.  In set-like language, the answer
to your question is provided by universal algebra.

Denote by Th(G) the theory associated to a particular algebra G (over
a given signature). More generally, to a class of algebras S (all over
the same, from now on fixed, signature), associate Th(S), the theory
of all those equations satisfied by all the algebras in S.  Also, to a
given theory T, let V(T) be the class of all algebras satisfying the
equations in T (also called a variety of algebras).

Birkhoff's HSP theorem states that a class C of algebras is of the
form V(T), for some T, if and only if C is closed under isomorphism,
and under the operations of taking quotient algebras, subalgebras, and
cartesian products. (HSP stands for "homomorphic image, subalgebra,
product").

As a direct consequence, let C=V(Th(G)), the class of all groups
satisfying those equations that a particular group G satisfies. Then C
is precisely the class of groups that can be obtained, up to
isomorphism, from G by repeatedly taking quotients, subalgebras, and
cartesian products. [Proof: certainly, the right-hand side is
contained in C.  Conversely, by the HSP theorem, the right-hand side
class is of the form V(T), for some T.  Since G is in the class, T can
only contain equations that hold in G, thus T is a subset of Th(G). By
contravariance of the "V" operation, it follows that C=V(Th(G)) is a
subset of V(T)].

Moreover, since a subalgebra of a quotient is a quotient of a
subalgebra, and a cartesian product of quotients [subalgebras] is a
quotient [subalgebra] of a cartesian product, the three HSP operations
can be taken in this particular order: Thus, a group satisfies all the
equations that G satisfies, if and only if it is isomorphic to a
quotient of a subalgebra of some (possibly infinite) product G x ... x G.

There are generalizations to properties other than equational ones,
but I don't remember them as well. A "Horn clause" is an implication
between equations, or more precisely a property of the form (forall
x1...xn)(P1 and ... and Pn => Q), where P1,...,Pn,Q are equations.  Of
course, every equation is trivially a Horn clause (for n=0), but not
the other way around. A typical example of a Horn clause is cancellability,
the property xz=yz => x=y. (This holds in groups, but not in monoids,
and cannot be expressed equationally in monoids, because it is not
preserved under quotients).

If you want to consider the class of algebras (in general smaller than
V(Th(G))) that satisfy all the Horn clauses that G satisfies, then you
have to drop the homomorphic images. I believe that the algebras in
question will be precisely the subalgebras of products of G, but
someone might correct me if I remember this wrongly.

-- Peter

Tom Leinster wrote:
>
> Dear category theorists,
>
> Here's something that I don't understand.  People sometimes talk about
> algebraic structures "satisfying laws".  E.g. let's take groups.  Being
> abelian is a law; it says that the equation xy = yx holds.  A group G
> "satisfies no laws" if
>
>     whenever X is a set and w, w' are distinct elements of the free
>     group F(X) on X, there exists a homomorphism f: F(X) ---> G
>     such that f(w) and f(w') are distinct.
>
> For example, an abelian group cannot satisfy no laws, since you could take
> X = {x, y}, w = xy, and w' = yx.  There are various interesting examples
> of groups that satisfy no laws.
>
> To be rather concrete about it, you could define a "law satisfied by G" to
> be a triple (X, w, w') consisting of a set X and elements w, w' of F(X),
> such that every homomorphism F(X) ---> G sends w and w' to the same thing.
>  A law is "trivial" if w = w'.  Then "satisfies no laws" means "satisfies
> only trivial laws".
>
> You could then say: given a group G, consider the groups that satisfy all
> the laws satisfied by G.  (E.g. if G is abelian then all such groups will
> be abelian.)  This is going to be a new algebraic theory.
>
> What bothers me is that I feel there must be some categorical story I'm
> missing here.  Everything above is very concrete; for instance, it's
> heavily set-based.  What's known about all this?  In particular, what's
> known about the process described in the previous paragraph, whereby any
> theory T and  T-algebra G give rise to a new theory?
>
> Thanks,
> Tom
>
>
>
>