Enriched monoidal categories - MathOverflow

Enriched monoidal categories

2011-01-11T18:50:11Z

This question is about (not necessarily symmetric) monoidal categories enriched over a symmetric monoidal category $\mathcal{V}$. Assume that $\mathcal{V}$ is closed. You may also assume that $\mathcal{V}$ is (co)complete if you wish.

If $k$ is a commutative ring, a $k$ algebra can be defined in two ways. Either as a $k$-module $R$ together with morphisms $k\rightarrow R$ and $R\otimes_{k}R\rightarrow R$ satisfying the well-known laws, or as a ring homomorphism to the center $k\rightarrow Z(R)$.

Let's see what happens in the categorical context.

The tensor product of $\mathcal{V}$-enriched categories can be straightforwardly defined, see Kelly's book. Then one can define what a monoidal $\mathcal{V}$-category is by reproducing the classical definition in the enriched context.

Assume now that $\mathcal{C}$ is an ordinary monoidal category. I believe that the braided center $Z(\mathcal{C})$ as defined by Joyal and Street is a well known construction. Suppose that we have a strong braided monoidal functor $z : \mathcal{V}\rightarrow Z(\mathcal{C})$ such that the functor $z(-)\otimes Y : \mathcal{V}\rightarrow \mathcal{C}$ has a right adjoint ${Hom}_{\mathcal{C}}(Y,-) : \mathcal{C}\rightarrow\mathcal{V}$ for any object $Y$ in $\mathcal{C}$. The counit is an evaluation morphism in $\mathcal{C}$,

$ev: z( {Hom}_{\mathcal{C}}(Y,Z))\otimes Y\longrightarrow Z$

One can define composition morphisms in $\mathcal{V}$

${Hom}(Y,Z)\otimes {Hom}(X,Y)\longrightarrow {Hom}_{\mathcal{C}}(X,Z) $

as the adjoint of

$z({Hom}(Y,Z)\otimes {Hom}(X,Y))\otimes X \cong z({Hom}(Y,Z))\otimes z({Hom}(X,Y))\otimes X \stackrel{id \otimes ev}\longrightarrow z({Hom}(Y,Z))\otimes Y \stackrel{ev}\longrightarrow Z $

I think it's pretty obvious that $\mathcal{C}$ becomes $\mathcal{V}$-enriched in this way. Moreover, one can also enrich the tensor product in $\mathcal{C}$ in a similar way.

Do you guys agree? Do you know of any reference where this is checked with some detail? Is it even more obvious than I think?

Any comment is welcome!

Answer by Theo Johnson-Freyd for Enriched monoidal categories

2011-01-11T22:45:37Z

My "answer" is somewhere between "answer" and "comment", but is much too long to fit in the comment box, so I'll put it here. The point, though, is that I don't really answer the original question. I'm pretty sure that yes, your construction does turn any monoidal category $\mathcal C$ with a sufficiently nice functor $\mathcal V \to Z(\mathcal C)$ into an enriched category. I haven't checked the details, but they don't look hard. I don't know a good reference.

However, you will need more than just enriched monoidal categories in order to build an equivalence between the words "Monoidal category $\mathcal C$ enriched over $\mathcal V$", and "Monoidal category $\mathcal C$ with a braided monoidal functor $\mathcal V \to Z(\mathcal C)$". For example, let $\mathcal C$ be the category of finite-dimensional vector spaces, and $\mathcal V$ the category of all vector spaces, both with their usual tensor structures. Then $\mathcal C$ is certainly enriched over $\mathcal V$, but the corresponding functor $\mathcal V \to Z(\mathcal C)$ does not exist, because $\mathcal V$ is so much bigger than $\mathcal C$.

I do know of at least one situation where something like this should work. It might be broadly known, but I don't think so; we (re?)construct it in current joint-work-in-progress with A. Chirvasitu.

We propose that the correct notion of "2-abelian group" is locally presentable category. The correct 2-category with objects the 2-abelian groups is the one where a 1-morphism $A \to B$ consists of an adjoint pair, a left adjoint $f: A\to B$ and its right adjoint $A \leftarrow B$ (and all the extra stuff). The 2-morphisms are natural transformations of adjunctions. Since a left adjoint determines its right adjoint up to unique isomorphism, in fact we set the 1-morphisms to be precisely those functors that are left adjoints. (Recall that a functor between locally presentable categories is a left adjoint iff it is cocontinuous, and a right adjoint iff it is continuous and commutes with $\kappa$-filtered colimits for sufficiently large cardinals $\kappa$; so the 1,2-opposite category is the one whose 1-morphisms are the continuous $\kappa$-filtered-colimit-preserving functors.)

The category of 2-abelian groups is symmetric monoidal with the "tensor product" defined in the obvious way: it is straightforward to prove that for any two 2-abelian groups $A,B$, there is a 2-abelian group representing the category-valued 2-functor $\hom(A,\hom(B,-))$. (Note that for 2-abelian groups $B,C$, the category $\hom(B,C) = \operatorname{cocontinuous}(B \to C)$ is again a 2-abelian group.) Thus there is a natural notion of "2-ring" and "commutative 2-ring" and the morphisms between them, their modules, etc.

In any case, in this setting given a "commutative 2-ring" $A$ (i.e. a locally presentable category with a symmetric monoidal structure that is cocontinuous in each variable), there is an equivalence of 2-categories between: {commutative monoid objects in $A\text{-mod}$}, and {commutative 2-rings with a distinguished morphism from $A$}. (Note that these 2-categories are somewhat subtle. For example, just like {symmetric monoidal categories} is not full in {monoidal categories}, similarly "module of a commutative 2-ring" requires more data than "module of a 2-ring that happens to be commutative"; also the tensor structure in $A\text{-mod}$ requires some work.)

We don't consider non-commutative 2-algebra, so I'd have to go over the arguments again to see about relaxing the commutativity constraints; but it should work.

Answer by Sean Tilson for Enriched monoidal categories

2011-01-12T01:48:57Z

I am not really familiar with this stuff, but one possible reference came to mind, Fresse talks about this a bit. Here is the book he wrote, it might be a helpful reference, or maybe the bibliography will be helpful: http://math.univ-lille1.fr/~fresse/OperadModuleFunctors.pdf

Maybe Fresse will even show up to answer?

Answer by Steve Lack for Enriched monoidal categories

2011-01-13T22:56:28Z

There is a theorem in category theory, generally regarded as folklore, which says that for a symmetric monoidal closed category $V$, the following structures are equivalent:

a category $C$ with an action $V\times C\to C$ of the monoidal category $V$ on $C$, which we may write as $(v,c)\mapsto v*c$, for which $-*c:V\to C$ has a right adjoint for each $c\in C$ (here the action amounts to a strong monoidal functor $V\to [C,C]$.
a $V$-category $C$ for which the $V$-functor $C(c,-):C\to V$ has a left adjoint for each $c\in C$. (such a $V$-category is said to be "tensored'' or "copowered'')

You can see this, for example, in the appendix to this paper.

In your case, unless I've misunderstood, the centre $Z(C)$ plays little role. The point is that your functor $z:V\to C$ induces an action via $v*c=z(v)\otimes c$, and $-*c$ has a right adjoint by assumption, so you get the $V$-enrichment.

(There is an analogous characterization of $V$-categories $C$ which are cotensored/powered: this means that each $C(-,d):C^{op}\to V$ has a left adjoint.)

Answer by Fernando Muro for Enriched monoidal categories

2011-01-13T23:53:46Z

Let me clarify the role of the center in my question. The braided functor to the center is needed to extend the ordinary monoidal category structure on $\mathcal{C}$ to a monoidal $\mathcal{V}$-category structure.

A strong braided monoidal functor $z\colon \mathcal{V}\rightarrow z(\mathcal{C})$ is the same as a strong monoidal functor $z\colon \mathcal{V}\rightarrow \mathcal{C}$ together with natural isomorphisms $\zeta(X,Y) : z(X)\otimes Y \cong Y\otimes z(X)$ satisfying some coherence laws. We need these isomorphisms to define the $\mathcal{V}$-enrichment of the tensor product in $\mathcal{C}$: $$\otimes : Hom_{\mathcal{C}}(W,X)\otimes Hom_{\mathcal{C}}(Y,Z)\longrightarrow Hom_{\mathcal{C}}(W\otimes Y, X\otimes Z).$$ This morphism must be the adjoint of:

$\qquad z(Hom_{\mathcal{C}}(W,X)\otimes Hom_{\mathcal{C}}(Y,Z))\otimes W\otimes Y$

$\cong z(Hom_{\mathcal{C}}(W,X))\otimes z(Hom_{\mathcal{C}}(Y,Z))\otimes W\otimes Y$

$\stackrel{\zeta}\cong z(Hom_{\mathcal{C}}(W,X))\otimes W\otimes z(Hom_{\mathcal{C}}(Y,Z))\otimes Y\stackrel{ev \otimes ev}\longrightarrow X \otimes Z.$