7
$\begingroup$

In Linear logic, monads and the lambda calculus (DRAFT), Proposition 3.0.2 says that the Eilenberg-Moore category for a commutative monad has the structure of a symmetric monoidal closed category. My question is: How do you construct the tensor product for the EM-category? When I followed the reference to Kiegher's paper "Symmetric monoidal closed categories generated by commutative adjoint monads", that paper only seems to give the construction for monads of the form A ⊗ -. I don't see how to generalize the construction to arbitrary commutative monads.

$\endgroup$
12
$\begingroup$

Caveat: this construction only works if your category of algebras has coequalizers of reflexive pairs, i.e. coequalizers of parallel pairs of arrows with a common right inverse.

Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure.

A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps.

This goes back to Anders Kock, see his papers "Closed categories generated by commutative monads" (J. Austr. Math. Soc. 12:405-424, 1975), and "Monads on symmetric monoidal closed categories" (Archiv der Mathematik, 21:1-10, 1970).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.