Why is a monoid with closed symmetric monoidal module category commutative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:07:59Z http://mathoverflow.net/feeds/question/14808 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14808/why-is-a-monoid-with-closed-symmetric-monoidal-module-category-commutative Why is a monoid with closed symmetric monoidal module category commutative? Peter Arndt 2010-02-09T21:13:01Z 2010-02-10T17:17:16Z <p>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:</p> <p>If the category of modules has a closed symmetric monoidal structure with A as unit object, then A is a <em>commutative</em> monoid.</p> <p>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...</p> http://mathoverflow.net/questions/14808/why-is-a-monoid-with-closed-symmetric-monoidal-module-category-commutative/14840#14840 Answer by Clark Barwick for Why is a monoid with closed symmetric monoidal module category commutative? Clark Barwick 2010-02-10T01:43:33Z 2010-02-10T11:52:06Z <p>(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!)</p> <p>The situation is even better than that! Suppose given an <code>$E_1$</code>-algebra $A$ of a presentable symmetric monoidal &infin;-category <code>$\mathcal{C}$</code>. </p> <p>Call an <code>$E_n$</code>-monoidal structures on the &infin;-category <code>$\mathbf{Mod}(A)$</code> of left $A$-modules <em>allowable</em> if <code>$A$</code> 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.]</p> <p>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 $A\simeq\mathrm{End}(A)$.</p> <p>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 <em>a</em> unit, and it will <em>follow</em> that the units are the same. He's right, of course.) with the property that</p> <p>$$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$</p> <p>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.</p> http://mathoverflow.net/questions/14808/why-is-a-monoid-with-closed-symmetric-monoidal-module-category-commutative/14864#14864 Answer by Tyler Lawson for Why is a monoid with closed symmetric monoidal module category commutative? Tyler Lawson 2010-02-10T06:26:19Z 2010-02-10T14:58:46Z <p>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.</p> <p>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 <i>is</i> a commutative monoid in the "sign-convention" symmetric monoidal category.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/14808/why-is-a-monoid-with-closed-symmetric-monoidal-module-category-commutative/14919#14919 Answer by Victor Ostrik for Why is a monoid with closed symmetric monoidal module category commutative? Victor Ostrik 2010-02-10T17:17:16Z 2010-02-10T17:17:16Z <p>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).</p>