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

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

2010-02-09T21:13:01Z

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...

Answer by Clark Barwick for Why is a monoid with closed symmetric monoidal module category commutative?

2010-02-10T01:43:33Z

(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 given an $E_1$ -algebra $A$ of a presentable symmetric monoidal ∞-category $\mathcal{C}$ .

Call an $E_n$ -monoidal structures on the ∞-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 is 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.

Answer by Tyler Lawson for Why is a monoid with closed symmetric monoidal module category commutative?

2010-02-10T06:26:19Z

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.

Answer by Victor Ostrik for Why is a monoid with closed symmetric monoidal module category commutative?

2010-02-10T17:17:16Z

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).