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Let $V$ and $W$ be classes of algebraic structures, and suppose we have some canonical way of constructing objects of $W$ from objects of $V$. Let's call this construction $C$, so that for all $A\in V$, $C(A)\in W$.

Then by a representation theorem, I mean a way of associating to each $B\in W$ an object $U(B)\in V$ such that $B$ embeds into $C(U(B))$.


  • $V = \operatorname{Sets}$, $W = \operatorname{Groups}$. For $X$ a set, let $C(X) = S_X$, the symmetric group on $X$. Then for any group $G$, we can take $U(G)$ to be the underlying set of $G$, and $G$ embeds in $S_{U(G)}$.

  • $V = \operatorname{Abelian Groups}$, $W = \operatorname{Rings}$. For $V$ a set, let $C(V) = \text{End}(V)$, the endomorphism ring of $V$. Then for any ring $R$, we can take $U(R)$ to be the additive group of $R$, and $R$ embeds in $\text{End}(U(R))$.

  • $V = \operatorname{Sets}$, $W = \operatorname{Boolean Algebras}$. For $X$ a set, let $C(X) = \mathcal{P}(X)$, the power set of $X$. Then for any Boolean algebra $B$, we can take $U(B)$ to be the set of ultrafilters on $B$, and $B$ embeds in $\mathcal{P}(U(B))$.

This setup does not seem to fit nicely into a categorical framework: In the first two examples above, the construction $C$ is not a functor, at least from the usual categories of sets and abelian groups. $C$ is a functor if we only force it to respect isomorphisms (by considering $V$ as a category with invertible morphisms as arrows), but we care about a noninvertible morphism in $W$ (the embedding $B\rightarrow C(U(B))$, so we won't be able to make $U$ into a functor to $V$. Finally, even if we could make $C$ and $U$ functors, it doesn't seem that the embeddings $B\rightarrow C(U(B))$ would cohere into a natural transformation $\operatorname{id}_W\rightarrow C\circ U$.

Of course, the third example is much better behaved, categorically speaking. I am aware of the family of examples related to this one, as described, for example, in Johnstone's book Stone Spaces.

But do we know anything about the general algebraic setup? In particular are there general conditions under which representation theorems must exist? If we know that a representation theorem exists, is there any sense in which we can compute what it must be?

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You might find "lattice of interpretability types of varieties" to be related. The literature on it is small enough (as well as the active group of researchers on the subject) that you can likely find connections from this area to your more general topic, or at the very least one of the researchers may supply you with a connection. However, the representation or intepretation is not necessarily a functor. If you need a categorical perspective, someone here may suggest Lawvere theories as an alternate direction of investigation. – The Masked Avenger Jul 16 '13 at 6:31
I took the liberty of textifying some things you had in mathmode - I hope this is OK with you. – Yemon Choi Jul 16 '13 at 6:38
@TheMaskedAvenger Thanks for the pointer - I hadn't thought of representation theorems as interpretations, because we usually embed the object of $W$ into some higher order (as opposed to first-order or equationally definable) object obtained from the object of $V$. But maybe in the right logic with the right signature... I don't need a categorical interpretation, I was just surprised that the obvious one didn't work! – Alex Kruckman Jul 16 '13 at 23:45

I'm not aware of a general theory, but the first two examples are corollaries of the Yoneda Lemma. For the first one, consider groups as categories with one object (and observe that you get a similar representation for monoids); for the second, consider rings as linear categories with one object (and use the enriched Yoneda Lemma). You could also view partial orders as categories in the usual way and thereby embed them into power sets.

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Yes, and it would be great if the Yoneda Lemma were somehow responsible for all representation theorems in algebra... but I don't see how to extend its reach far enough. Pleasant observation: The Yoneda embedding is itself an example of the setup I described. Take $V=W=\text{Categories}$, $C(X)=$ the category of contravriant functors $X\rightarrow \text{Set}$, and D the identity. – Alex Kruckman Jul 16 '13 at 16:05

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