Background: a Lawvere theory $T$ is a category with finite products such that each object is a power of a fixed object $x$. Given a Lawvere theory $T$, the category $\text{Mod}_T$ of models of $T$ is the category of product-preserving functors $T \to \text{Set}$. Any such category is locally finitely presentable (if I'm not mistaken). Gabriel-Ulmer duality says, among other things, that locally finitely presentable categories are categories of models of generalizations of Lawvere theories. Do we have something like this result for categories of models of Lawvere theories (in the sense that we can identify some categorical properties that are satisfied precisely by such categories)?

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Adámek and Rosický [On sifted colimits and generalized varieties] have shown that a category $\mathcal{C}$ is equivalent to the category of models for a (finitary) Lawvere theory in $\mathbf{Set}$ if and only if it satisfies these conditions:

  1. $\mathcal{C}$ is locally small and cocomplete.
  2. There exists a small family $\mathcal{G}$ of objects in $\mathcal{C}$ such that, for every object $G$ in $\mathcal{G}$, the hom-functor $\mathcal{C}(G, -) : \mathcal{C} \to \mathbf{Set}$ preserves sifted colimits, and for every object $A$ in $\mathcal{C}$, there exists a small sifted diagram of objects in $\mathcal{G}$ whose colimit in $\mathcal{C}$ is $A$.

These conditions are easily seen to be analogues of the conditions for $\mathcal{C}$ to be locally finitely presentable, except for replacing "filtered" (or "directed") everywhere by "sifted".

Incidentally, any such $\mathcal{C}$ is locally finitely presentable, and we may also replace "cocomplete" in condition (1) with "complete", just as for locally finitely presentable categories.

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    $\begingroup$ Great! Am I correct in guessing that by "Lawvere theory" here you mean a multisorted Lawvere theory, and that if I was only interested in Lawvere theories with one sort the second condition should be "there exists an object in $C$ such that..."? $\endgroup$ – Qiaochu Yuan Jun 10 '13 at 23:20
  • $\begingroup$ Yes I believe Zhen is talking about the multisorted case and you will get the single-sorted case by taking a single object (a small projective generator). Note the algebras over a multi/singly-sorted theory correspond to cocomplete algebraic categories with a set of projective generators/a single projective generator in the sense of Quillen. $\endgroup$ – Justin Noel Jun 11 '13 at 7:05
  • $\begingroup$ @Qiaochu: Yes, this is multi-sorted. It will not do to take $\mathcal{G}$ to be a single object: this already fails for concrete examples like $\mathbf{Ab}$. I think instead you need to ask that $\mathcal{G}$ be generated by a single object under finite coproducts – but I have not checked. Certainly it is enough to ask that the full subcategory of strongly finitely presentable objects have this property, because this subcategory becomes the Lawvere theory. $\endgroup$ – Zhen Lin Jun 11 '13 at 7:39
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    $\begingroup$ A small subtlety: if small categories with finite products are your "theories" and algebraic categories are your "models", then you can only recover the theory from the models up to Morita equivalence. If your Lawvere theory was the finitely-generated free algebras, then the strongly finitely presentable algebras are the retracts of these. So if we want a precise duality, we need our theories to be idempotent-complete as well as having finite produts. This complication doesn't arise for Gabriel-Ulmer duality because finitely-complete categories are automatically idempotent-complete. $\endgroup$ – Tim Campion May 9 '14 at 2:13
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    $\begingroup$ A duality that actually involves classical 1-sorted Lawvere theories rather than their idempotent completions is given in Adamek, Rosicky, and Vitale's book Algebraic Theories, Thm 11.39, but they consider concrete categories, i.e. the forgetful functor to $\mathbf{Set}$ is part of the data. Barring such a move, I suspect it might be impossible to get a functorial duality because there autoequivalences of the category of algebras which don't fix the free algebras. $\endgroup$ – Tim Campion May 9 '14 at 2:45

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