# Why is a monoid with closed symmetric monoidal module category commutative?

Given a symmetric monoidal category and a monoid object A in it, one can form the category of modules over this monoid object, i.e. objects are $A \otimes M \rightarrow M$ satisfying the natural properties analogous to modules over a ring and morphisms respecting this. The following seems to be true and I would like to know why:

If the category of modules has a closed symmetric monoidal structure with A as unit object, then A is a commutative monoid.

This is how I read the statement right after Proposition 2.3.4 in Hovey/Shipley/Smith's paper "Symmetric Spectra" and it would give an excellent motivation for introducing symmetric spectra...

• I find this a really strange statement. Are the monoidal structure and the symmetry on A-Mod really not required to bear any kind of relationship to the monoidal structure and symmetry on your original category? Feb 10, 2010 at 11:01
• Tom Leinster - one does have to require that the natural right action of the ambient category is compatible with the monoidal structure on the category of modules, but that's it. Feb 10, 2010 at 11:43
• OK, let's see if I have this right. Tyler's "no" and Clark's "yes" are both correct. Tyler's "no" answers the question posed. Clark's "yes" answers the same question but under the assumption of a (sensible) extra hypothesis. Right? Feb 11, 2010 at 1:23
• That is how I understood it - Tyler's newly edited answer gives now an easy counterexample to my original statement (the former one used things I didn't know). And Clark's proof works and tells me what I wanted to know. Feb 11, 2010 at 19:45
• This seems to be the old result of Kock on commutative monads (the monad here being $(A\otimes−)$). What is true is that TFAE: 1. Mod(A) is closed (resp monoidal) and the free A-module adjunction $F\dashv U:\mathbf{Mod}(A)\to\mathcal{V}$ is a closed adjunction (resp. monoidal adjunction). 2. A is commutative. Jul 9, 2011 at 16:56

(I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!)

The situation is even better than that! Suppose we are given an $E_1$-algebra $A$ of a presentable symmetric monoidal $\infty$-category $\mathcal{C}$.

Call an $E_n$-monoidal structures on the $\infty$-category $\mathbf{Mod}(A)$ of left $A$-modules allowable if $A$ is the unit and the right action of $\mathcal{C}$ on $\mathbf{Mod}(A)$ is compatible with the $E_n$ monoidal structure, so that $\mathbf{Mod}(A)$ is an $E_n$-$\mathcal{C}$-algebra. Then the space of allowable $E_n$-monoidal structures is equivalent to the space of $E_{n+1}$-algebra structures on $A$ itself, compatible with the extant $E_1$ structure on $A$. (This is even true when $n=0$, if one takes an $E_0$-monoidal category to mean a category with a distinguished object.) The object $A$, regarded as the unit $A$-module, admits an $E_n$-algebra structure that is suitably compatible with the $E_1$ structure an $A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.]

Let's sketch a proof of this claim in the case Peter mentions. Suppose $A$ is a monoid in a presentable symmetric monoidal category $(\mathbf{C},\otimes)$. Suppose $\mathbf{Mod}(A)$ admits a monoidal structure (not even a priori symmetric!) in which $A$, regarded as a left $A$-module, is the unit. I claim that $A$ is a commutative monoid. Consider the monoid object $\mathrm{End}(A)$ of endomorphisms of $A$ as a left $A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give $\mathrm{End}(A)$ the structure of a commutative monoid object. The multiplication on $A$ yields an isomorphism of monoids $A\simeq\mathrm{End}(A)$.

In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If $X$ admits magma structures $\circ$ and $\star$ with the same unit (Below, Tom Leinster points out that I only have to assume that each has a unit, and it will follow that the units are the same. He's right, of course.) with the property that

$$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$

for any $a,b,c,d\in X$, then (1) the magma structures $\circ$ and $\star$ coincide; (2) the product $\circ$ is associative; and (3) the product $\circ$ is commutative. That is, a unital magma in unital magmas is a commutative monoid.

• For what it's worth, in Eckmann--Hilton you don't need to assume that the two operations have the same unit: that comes for free. Feb 10, 2010 at 6:02
• @Clark: I just fixed a little display error that was happening in the LaTeX. I hope you don't mind. Feb 10, 2010 at 11:50
• Harry: Oh, no, of course not! I appreciate it! Feb 10, 2010 at 12:04
• For the Eckmann-Hilton argument, what should the two operations be in my case? The module structure morphism and the monoid multiplicaion? I already know that those coincide, but why should they satisfy the interchange law (which then is equivalent to being commutative)? Feb 10, 2010 at 13:58
• @ Peter: Because the monoidal structure on A-modules is a functor from A-mod x A-mod to A-mod. Hence it commutes with composition. Specifically, if we apply the functor to (f,g) and (f',g') and then compose to get (f * g)(f' * g'), that is the same as first composing to get (ff', gg') and then tensoring to get (ff') * (gg'). Feb 10, 2010 at 14:33

I would second Tyler's answer. For example consider symmetric tensor category $Rep(G)$ of finite dimensional complex representations of $G$ where $G={\mathbb Z}/2 \times {\mathbb Z}/2$. Let $V$ be the 2-dimensional irreducible projective representation of $G$; then $A=V\otimes V^*$ is a non-commutative algebra in $Rep(G)$. It is easy to see that any $A-$module in $Rep(G)$ is a direct sum of several copies of $A$; in other words the category of $A-$modules is equivalent to the category of finite dimensional complex vector spaces. Thus the category of $A-$modules has an obvious structure of symmetric tensor category (of course $A$ is a unit for this structure).

I believe that the answer to the question as stated is: No, $A$ does not have to be commutative. EDIT: The original answer I posted was overcomplicated and it didn't have $A$ as a unit for the tensor product.

Let our abelian category be graded rational vector spaces under graded tensor product, with symmetry isomorphism given by $\alpha \otimes \beta \mapsto \beta \otimes \alpha$ (rather than the standard homological algebra sign convention). Let $A$ be the graded exterior algebra over $\mathbb{Q}$ with generators $x$ and $y$ in degree $1$. In particular it is a monoid but not a commutative monoid in this "no-sign-convention" symmetric monoidal category, but is a commutative monoid in the "sign-convention" symmetric monoidal category.

However, the category of left $A$-modules makes no reference to the symmetry isomorphism, and hence the "sign-convention" symmetric monoidal structure on left $A$-modules gives us a symmetric monoidal structure, with $A$ as unit, on the category of left $A$-modules in the "no-sign-convention" symmetric monoidal category.

Implicit in the question is perhaps the assumption that the symmetric monoidal structure on $A$-modules have as its chosen "twist" isomorphism something determined by the twist isomorphism in the underlying symmetric monoidal category, and this is simply not the case here.

• Tyler, two questions. (1) You need A to be a monoid in your monoidal category, i.e. a Q-algebra; yet A as you define it doesn't have scalar multiplication over Q. Did you mean it to be the Q-algebra (not the ring) given by those generators and relations? (2) Would you like to rewrite your final sentence :-) ? Feb 10, 2010 at 6:44
• Hi Tyler! I think the question of whether the monoidal structure on Mod(A) is symmetric is a red herring. The key point is that A, viewed as a left A-module is required to be the unit, and one needs for the right action of the ambient category to be compatible with the new tensor product. Feb 10, 2010 at 11:40
• @Tom: Thanks, I've edited on both counts. No matter how many times I vow not to write something late at night, I never learn my lesson. Feb 10, 2010 at 13:57
• @Clark: Sorry, I'm a little slow this morning - to which point are you objecting? That the monoidal structure I want to compose isn't compatible enough with the monoid structure of A? Feb 10, 2010 at 14:05
• So, (1) Is it still a counterexample if you take the Q-algebra with the presentation you. (2) Do you have an idea what is the explicit assupmtion on the symmetry on A-mod which would make the statement true ? Feb 10, 2010 at 14:33

Given a monoid $A$ in a symmetrical monoidal category, the monoidal category of $A$-modules has (I think) two interpretation.

1) implicitly they suppose $A$ commutative , then the left $A$-module and right $A$-modules and $(A, A)$-bimodules are identified.

2) It is the category of the $(A, A)$-bimodules. .

Then the question become: there exist a noncommutative monoid $A$ such that the $(A, A)$-bimodules category is symmetrical?

The answer is YES, I asked this question in MO (but it has been moved), see:

https://math.stackexchange.com/questions/433078/a-example-of-a-monoidal-non-symmetric-category-of-r-bimodules

and the references on it: