# Action of the center of an enriched category on a module as an enriched functor

If $M$ is a left-module over a ring $R$, then $M$ carries an action of $Z(R)$, the center of $R$, since it is a subring of $R$, by restriction of scalars. Since $Z(R)$ is commutative we may view this as a left or right action. Similarly, if $M$ is a right $R$-module and $N$ is a left $R$-module, while one usually defines $M\otimes_R N$ as just an abelian with no action, it actually carries an action of $Z(R)$.

Categorifying this picture horizontally, let $C$ be any category enriched over $\mathcal{V}$, let $M\colon C^\text{op}\to \mathcal{V}$ and $N\colon C\to\mathcal{V}$ be contravariant and covariant $\mathcal{V}$-enriched functors, which we view as right- and left-modules, respectively. Then we define the center to be $Z(C)=\hom(1_C,1_C),$ the endonatural transformations of the identity functor on $C$. We can define $\mathcal{V}$-natural transformations as elements $I\to\hom(x,x)$, for $x\in C$ and where $I$ is the unit of the monoidal structure on $\mathcal{V}$. In this case, $Z(C)$ is a set and it is easy to see how elements act on modules. As Martin points out in the comments, for $\alpha$ any natural transformation in $Z(C)$, $M\alpha$ is a natural transformation $M\to M$.

However it is more natural for $Z(C)$ to retain the structure of an object of $\mathcal{V}$, (the center of a ring is a ring, for example). Hence we should define $Z(C)$ to be the end $\int_{x\in C}\hom_C(x,x)$. This is an object of $\mathcal{V}$, hence is not a set, and we may not speak of its elements. We might consider the set of generalized elements, morphisms $I\to\hom_C(x,x)$, which are the $\mathcal{V}$-natural transformations. But these do not constitute an object of $\mathcal{V}$, and hence do not give an action on $M$ (which, remember, is defined as a $\mathcal{V}$-enriched functor).

So how do we construct this action as an enriched functor, from the functor $M$? How should we think about this generally? Like any endomorphism hom-object, $Z(C)$ must be a monoid object in $\mathcal{V}$. Does a functor out of a category somehow induce a functor on the delooping of this monoid of endomorphisms of the identity functor on the category?

• Surely the first thing to do is to determine how $Z (\mathcal{C})$ acts on $M$? – Zhen Lin Mar 3 '14 at 10:08
• @ZhenLin: Yes, that is surely the thing to do. But it's not clear in the enriched case, is it? – ziggurism Mar 6 '14 at 1:12

I think I was confused by the module language. I was looking for a functor from $Z(C)$ to $\mathcal{V}$. But an action of a category $C$ on a module $M$ is just morphisms $C(a,b)\otimes M(a)\to M(b)$. So an action isn't a functor, although a functor is a thing which always has an action. And then we can loosen the notion of action to be any collection $\{C_a\}_{a\in C}$ of $\mathcal{V}$-objects indexed by objects of $C$ with morphisms $C_a\otimes M(a)\to M(a)$, and the center $Z(C)$ has an action of this form by virtue of the components of the end $\int_C\hom(1,1)\to\hom(a,a)\to\hom(Ma,Ma).$ Basically the center acts because it maps to the hom-objects of the category, so we have something like "restriction of scalars" of the module $M$. And this of course agrees with the obvious action by multiplication in the set-enriched case.