4
$\begingroup$

Whenever $R$ is a commutative ring, write $R[x^{(n)}]$ for the set of all $p \in R[x]$ such that $p$ is a monic polynomial of degree $n$. Then $R[x^{(n)}]$ is not closed under sums, nor does it contain $0$. Nonetheless, it becomes an $R$-affine space in a natural way. Furthermore, for all natural numbers $n$ and $m$, there is a product function

$$C^{n,m} : R[x^{(n)}] \times R[x^{(m)}] \rightarrow R[x^{(n+m)}]$$

defined by $C^{n,m}(p,q) = pq.$ This holds irrespective of whether or not the commutative ring $R$ is integral. Furthermore, the $C$ family of functions satisfies analogues of all the usual axioms for a commutative monoid: for example, commutativity becomes:

$$C^{m,n}(p,q) = C^{n,m}(q,p)$$

Hence the monic polynomials in $R[x]$ form a multisorted algebraic structure in a natural way. The most elegant way to describe such algebraic structures uses multicategories.

To any additively-denoted commutative monoid $N$, we can assign a symmetric multicategory $N^{sym}$ whose objects are the elements of $N$, presented as follows.

Generators. The following collection of multiarrows, as $n$ and $m$ and vary over $N$.

$$C^{n,m} : n,m \rightarrow n+m.$$

Relations. The obvious analogues of the usual commutative monoid axioms.

(Not 100% sure this description of $N^{sym}$ works, but you get the general idea.)

Anyway, the upshot is that:

  • $\{0\}^{sym}$ is the symmetric operad whose algebras in $\mathbf{Set}$ are commutative monoids.
  • $\mathbb{N}^{sym}$ is the symmetric multicategory whose algebras in the symmetric multicategory of $R$-affine spaces are precisely the kinds of things we were trying to describe.

Question. Has this construction, which builds a symmetric multicategory from a commutative monoid, been described or studied anywhere, and if so, where? And if not, is it nonetheless a special case of a familiar construction?

I am also interested in terminology for some or all of the following concepts:

  • The construction $N \mapsto N^{sym}$ that takes commutative monoids to symmetric multicategories.
  • The symmetric multicategory $\mathbb{N}^{sym}$
  • The algebras of $\mathbb{N}^{sym}$ in the symmetric multicategory of $R$-affine spaces.
  • The functor that maps a commutative ring $R$ to the collection (viewed as an object of the above category) of monic polynomials in $R[x].$
$\endgroup$

2 Answers 2

4
$\begingroup$

Let me work instead in the language of symmetric monoidal categories. There is a forgetful functor from categories to sets sending a category to its set of objects, and this functor has a left adjoint sending a set to the "discrete category" on that set, with no non-identity morphisms.

This left adjoint is symmetric monoidal with respect to the cartesian product, and so it preserves both monoids and commutative monoids; in particular, it sends monoids in sets to monoidal categories and commutative monoids in sets to symmetric monoidal categories. The outputs of these constructions are distinguished by the fact that their underlying categories are discrete.

I don't have a name for this construction, but then again I also don't have a name for the construction which, given a group, equips that group with the discrete topology.

$\endgroup$
3
$\begingroup$

Every (commutative) monoid can be regarded as a (symmetric) monoidal category as in Qiaochu's answer.

There is an adjunction between the categories of multicategories and monoidal categories, see for example:

C. Hermida, Representable multicategories, Advances in Mathematics 151 2 (2000) 164-225.

The left adjoint in this adjunction is usually called the "free monoidal category functor", and the right adjoint is called the "underlying multicategory functor".

So if you want a name, you can call your object the "underlying (symmetric) multicategory of a discrete (symmetric) monoidal category".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.