Two questions about commutative theories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:43:55Z http://mathoverflow.net/feeds/question/113919 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113919/two-questions-about-commutative-theories Two questions about commutative theories Martin Brandenburg 2012-11-20T08:05:29Z 2012-11-22T11:17:03Z <p>Let $\mathcal{T}$ be a commutative algebraic theory (for example sets, abelian groups, commutative monoids, but not groups etc.). References include the <a href="http://ncatlab.org/nlab/show/commutative+algebraic+theory" rel="nofollow">nlab</a> and Borceux' <em>Handbook of Categorical Algebra 2</em>, section 3.10. Then $\mathsf{Mod}(\mathcal{T})$ is a monoidal category with internal homs.</p> <p><strong>Question 1.</strong> (<strong>Answered:</strong> Yes) Can we find a property of concrete categories which holds for $\mathsf{Mod}(\mathcal{T})$ if and only if $\mathcal{T}$ is commutative? In other words, does commutativity of an algebraic category not depend on the presentation? </p> <p><strong>Question 2.</strong> (<strong>Answered:</strong> No) Let $\mathcal{T}$ be a commutative algebraic theory and $C=\mathsf{Mod}(\mathcal{T})$. Assume that $X \in C$ is a Co-$C$-algebra, i.e. we have a factorization of $\hom(X,-) : C \to \mathsf{Set}$ over $C$. Does this have to coincide with the usual factorization? This is well-known to be true in the examples I have mentioned above, for example for every abelian group $A$ there is only one natural abelian group structure on the hom-sets $\mathrm{hom}(A,B)$. This should be all well-known, but I don't know a reference.</p> http://mathoverflow.net/questions/113919/two-questions-about-commutative-theories/113946#113946 Answer by Todd Trimble for Two questions about commutative theories Todd Trimble 2012-11-20T14:50:01Z 2012-11-21T14:12:37Z <p>As for question 1: commutativity doesn't depend on the presentation of $T$. If $M = Mod(T)$ and $U: M \to Set$ is the forgetful functor, then commutativity can be formulated as saying that the monad $Ran_U U = U \circ Ran_U 1_M$ is commutative (or monoidal) in the sense of the nLab article <a href="http://ncatlab.org/nlab/show/monoidal+monad" rel="nofollow">here</a>. Perhaps the most interesting aspect of this is that commutativity is a <em>property</em>, not an extra structure on a monad (where the structure of a strength constraint on an endofunctor on $Set$ is canonically given because every such endofunctor is canonically $Set$-enriched). These observations also lift to the enriched setting, provided of course that the functors involved are given as enriched functors (with respect to a base of enrichment $V$). </p> <p>(Note: $Ran_U 1_M$, which invariably exists, is just the left adjoint $F$ of $U$ if $U$ has a left adjoint. Some related discussion on the codensity monad of a general functor $U: M \to Set$ can be found in <a href="http://golem.ph.utexas.edu/category/2012/09/where_do_monads_come_from.html" rel="nofollow">this post</a> by Tom Leinster.) </p> <p><b>Edit:</b> I had responded to question 2 earlier, but I am now editing that response out as it is superseded by Martin's second comment below, which makes the situation quite clear. Apologies for the noise. </p>