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Stream: learning: questions

Topic: Tensor product of (arbitrary) monoids

Emily (Nov 01 2021 at 03:02):

There's a number of different tensor products for monoids (with counterparts for groups):

• We have the usual tensor product of commutative monoids $\otimes_{\mathbb{N}}$, which gives $\mathsf{CMon}$ a symmetric monoidal structure. Then the monoids in $\mathsf{CMon}$ with respect to $\otimes_{\mathbb{N}}$ are semirings.
• Then there's a tensor/copower of monoids by sets, $(X,M)\mapsto X\odot M$, explained in Example 4.3 here. This tensor induces "left" and "right" tensor products $\lhd$ and $\rhd$ on $\mathsf{Mon}$, given by $M\lhd N=|N|\odot M$, characterised by a bijection of the form

$$\left\{\text{morphisms of monoids $A\lhd B\to C$}\right\} \cong \left\{\text{left-bilinear morphisms $A\times B\to C$}\right\},$$

(here left-bilinear means $f(ab,c)=f(a,c)f(b,c)$ and $f(1_A,b)=1_C$.). The tensor products $\lhd$ and $\rhd$ give $\mathsf{Mon}$ a skew-monoidal structure, and the monoids with respect to each of these are left and right near-semirings.

• Finally, there's a "two-sided" tensor product of (not necessarily commutative) monoids, explained here, and characterised instead by a bijection involving two-sided bilinear maps:

$$\left\{\text{morphisms of monoids $A\otimes B\to C$}\right\} \cong \left\{\text{bilinear morphisms $A\times B\to C$}\right\}.$$

Does this tensor product give $\mathsf{Mon}$ some kind of lax monoidal structure?

We do have natural non-invertible maps $A\to\mathbb{N}\otimes A$ and $A\to A\otimes\mathbb{N}$ as $\mathbb{N}\otimes A:=\mathrm{N}(A)\cong A/\{(ab)^n=a^nb^n\}_{n\in\mathbb{N}}$, though I'm not sure there's also a natural (lax) associator map $(A\otimes B)\otimes C\to A\otimes(B\otimes C)$...

(Incidentally, there's also a paper of Grillet, The Tensor Product of Semigroups, on this tensor product, which in particular proves in Prop. 3.7 that $A\otimes B\cong\mathrm{N}(A)\otimes\mathrm{N}(B)$.)

Graham Manuell (Nov 02 2021 at 17:40):

This isn't an answer to your question, but I thought I'd just mention something else that could also be called a 'tensor product of monoids'. If $M$ and $N$ are monoids in a monoidal category, then a monoid structure on $M \otimes N$ can be given by a distributive law (viewing the monoids as monads in the corresponding one-object bicategory). For monoids in Set this gives the Zappa–Szép product of the monoids. For a braided monoidal category the braiding gives a canonical choice of distributive law (and with this we would obtain the usual product in the case of a cartesian monoidal category).