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By a Lawvere theory, I mean a finite-product category $\mathsf{T}$ equipped with a distinguished object, such that every object of $\mathsf{T}$ can be expressed as a finite product of the distinguished object.

Now let $X$ denote an object of a finite-product category $\mathbf{C}$. Then $X$ gives rise to a Lawvere theory $\mathrm{Lawv}(X)$ with distinguished object $X$, defined as the full subcategory of $\mathbf{C}$ induced by the objects $\{X^n \mid n:\mathbb{N}\}.$

Examples.

  • Viewing $2$ as an object of $\mathbf{Set}$, we see that the arrows of $\mathbf{Lawv}(2)$ are "truth tables." Hence $\mathbf{Lawv}(2)$ is the Lawvere theory of Boolean algebras.

  • Let $\mathbb{K}$ denote a field. Write $U_{\mathrm{Mod}}(\mathbb{K})$ for the underlying $\mathbb{K}$-module. Then $\mathbf{Lawv}(U_{\mathrm{Mod}}(\mathbb{K}))$ is the Lawvere theory of $\mathbb{K}$-modules.

Question. Is it true that for all Lawvere theories $\mathsf{T}$, there exists a Lawvere theory $\mathsf{S}$ and an $\mathsf{S}$-algebra $X$ such that $\mathrm{Lawv}(X) \cong \mathsf{T}$?

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    $\begingroup$ What the two examples have in common is a good theory of dualisation. After all, the opposite of any Lawvere theory embeds in the category of algebras in a canonical way. $\endgroup$ – Zhen Lin Jul 28 '15 at 2:32
  • $\begingroup$ In that second example, did you mean "Lawvere theory of $\mathbb{K}$-modules"? $\endgroup$ – David Roberts Jul 28 '15 at 2:32
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    $\begingroup$ Since the Lawvere theory for a variety of algebras is given by the opposite of the category of finitely presented free algebras, shouldn't $X$ be the "free algebra on one generator" (whatever that is)? $\endgroup$ – David Roberts Jul 28 '15 at 2:35
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    $\begingroup$ @David: no, that already fails in the case of groups. The way you recover a Lawvere theory from the free algebra over it on one generator $X$ is by taking the opposite of the full subcategory on the finite coproducts of $X$, not by taking the full subcategory on the finite products of $X$. $\endgroup$ – Qiaochu Yuan Jul 28 '15 at 7:27
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    $\begingroup$ @GiorgioMossa, is that clearer? We require it to be a full subcategory, if that helps. $\endgroup$ – goblin Jul 28 '15 at 12:38
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I can elaborate on the second example a bit more:

Let $T$ be a Lawvere theory. Then $T$ is commutative iff $T \cong Lawv(\text{hom}_T(x,-))$, where $x$ is the generic object in $T$.

A Lawvere theory is commutative if all its $n$-ary operations commute with each each other. A morphism of $T$-algebras is a natural transformation between functors; the naturality condition is precisely the requirement to commute with all $n$-ary operations. Your example of modules over a field (more generally a commutative semi-ring) is the prime example of a commutative theory.

Now commutative Lawvere theories are extremely useful, but also very restrictive. I will think about the general case some more.

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