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...

  • $\begingroup$ 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? $\endgroup$ Feb 10, 2010 at 11:01
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    $\begingroup$ 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. $\endgroup$ Feb 10, 2010 at 11:43
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    $\begingroup$ 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? $\endgroup$ Feb 11, 2010 at 1:23
  • $\begingroup$ 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. $\endgroup$ Feb 11, 2010 at 19:45
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    $\begingroup$ 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. $\endgroup$ Jul 9, 2011 at 16:56

4 Answers 4


(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.

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    $\begingroup$ 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. $\endgroup$ Feb 10, 2010 at 6:02
  • $\begingroup$ @Clark: I just fixed a little display error that was happening in the LaTeX. I hope you don't mind. $\endgroup$ Feb 10, 2010 at 11:50
  • $\begingroup$ Harry: Oh, no, of course not! I appreciate it! $\endgroup$ Feb 10, 2010 at 12:04
  • $\begingroup$ 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)? $\endgroup$ Feb 10, 2010 at 13:58
  • $\begingroup$ @ 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'). $\endgroup$ 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.

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    $\begingroup$ 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 :-) ? $\endgroup$ Feb 10, 2010 at 6:44
  • $\begingroup$ 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. $\endgroup$ Feb 10, 2010 at 11:40
  • $\begingroup$ @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. $\endgroup$ Feb 10, 2010 at 13:57
  • $\begingroup$ @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? $\endgroup$ Feb 10, 2010 at 14:05
  • $\begingroup$ 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 ? $\endgroup$ 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:


and the references on it:



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