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There is a correspendence between Lawvere theories $L$ and finitary monads $\mathbb{T}_L$ (associated to $L$), due to Lawvere: the category $Mod(L)$ of models of $L$ (in $\mathbf{Set}$) is equivalent to the Eilenberg-Moore category $\mathbb{T}_L$-$\mathbf{Alg}$ of algebras over the monad $\mathbb{T}_L$.

Now, this category $\mathbb{C}=Mod(L)$ is a locally finitely presentable category (by definition). Write $\mathbb{C}_f$ for the (full) subcategory of finitely presentable objects of $\mathbb{C}$. Can $\mathbb{C}_f$ be constructed somehow from the Kleisli category $\mathbf{Kl}(\mathbb{T}_L)$ of the monad $\mathbb{T}_L$? It seems to me that there must be some relationship between the two (at first I thought $\mathbb{C}_f=\mathbf{Kl}(\mathbb{T}_L)^{\text{op}}$, but this is definitely not the case).

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Kevin Carlson is right, of course. But you can construct the category of finitely presentable algebras as the free reflexive coequalizer completion of the Lawvere theory itself, though. See Theorem 7.3 in Adámek, Rosický, and Vitale's Algebraic Theories.

You can also distinguish the free algebras $A$ which are finitely presentable in $\mathbb{T}_L-\mathbf{Alg}$ from those which are not by the fact that $\mathbf{KL}(\mathbb{T}_L)(A,-)$ commutes with the filtered colimit $B \to 2\cdot B \to \dots \to \omega \cdot B \to \dots \to \kappa \cdot $ for every infinite $\kappa$, where $B$ is any object of $\mathbf{KL}(\mathbb{T}_L)$ and $\alpha \cdot B$ is the $\alpha$-fold coproduct of $B$ with itself.

Putting these together, you can pass from $\mathbf{KL}(\mathbb{T}_L)$ to its finite rank objects to their reflexive coequalizer completion. Of course, you could also just construct $\mathbb{T}_L-\mathbf{Alg}$ from $\mathbf{KL}(\mathbb{T}_L)$ (I think it's also a reflexive coequalizer completion) and then just take the finitely presentable objects in the resulting category.

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  • $\begingroup$ Thanks! Just a clarification, $L$ (the Lawvere theory) is actually $(\mathbf{Kl}(\mathbb{T}_L)_f)^{\text{op}}$ always, right? (the category opposite to the full subcategory of finitely presentable objects of the Kleisli category). I think I am getting confused over that too now.. $\endgroup$ – MmOmT Aug 1 '16 at 15:03
  • $\begingroup$ Right. It's kind of confusing. Personally, I tend to think of $L^\mathrm{op} = \mathbf{Kl}(\mathbb{T}_L)$ as the "primary object". $\endgroup$ – tcamps Aug 1 '16 at 17:57
  • $\begingroup$ $L^{\text{op}}$ is the whole of Kleisli, not its finitely presentable objects? $\endgroup$ – MmOmT Aug 2 '16 at 13:25
  • $\begingroup$ Oh, that was silly of me, of course not. $L^\mathrm{op}$ is just the finitely presentable objects of the Kleisli category. $\endgroup$ – tcamps Aug 2 '16 at 13:35
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No, there's no relation. The Kleisli category is the category of free algebras, for instance, of all free groups. It has nothing to do with finiteness.

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