nLab tensor product of commutative monoids

Idea

For $A$ and $B$ two commutative monoids, their tensor product $A \otimes B$ is a new commutative monoid such that a monoid homomorphism $A \otimes B \to C$ is equivalently a bilinear map out of $A$ and $B$.

Another way to think of it is that commutative monoid maps $A \otimes B \to C$ are in natural bijection with commutative monoid maps $A \to [B, C]$, where $[B, C]$ is the set of commutative monoid maps $B \to C$ provided with the pointwise-defined commutative monoid structure.

Definition

Definition

Let CMon be the collection of commutative monoids, regarded as a multicategory whose multimorphisms are the multilinear maps $A_1, \cdots, A_n \to B$.

The tensor product $A, B \mapsto A \otimes B$ in this multicategory is the tensor product of commutative monoids.

Equivalently this means explicitly:

Definition

For $A, B$ two commutative monoids, their tensor product of commutative monoids is the commutative monoid $A \otimes B$ which is the quotient of the free commutative monoid on the product of their underlying sets $A \times B$ by the relations

• $(a_1,b)+(a_2,b)\sim (a_1+a_2,b)$

• $(a,b_1)+(a,b_2)\sim (a,b_1+b_2)$

• $(0,b)\sim 0$

• $(a,0)\sim 0$

for all $a, a_1, a_2 \in A$ and $b, b_1, b_2 \in B$.

Remark

By definition of the free construction and the quotient there is a canonical function of the underlying sets

$$p_{A,B} \;\colon\; U(A) \times U(B) \overset{}{\longrightarrow} U(A \otimes B)$$

(where $U \colon CMon \to Set$ is the forgetful functor).

On elements this sends $(a,b)$ to the equivalence class that it represents under the above equivalence relations.

The following relates the tensor product to bilinear functions. It is a definition or a proposition depending on whether one takes the notion of bilinear function to be defined before or after that of tensor product of commutative monoids.

Definition/Proposition

A function of underlying sets $f : A \times B \to C$ is a bilinear function precisely if it factors by the morphism of through a monoid homomorphism $\phi : A \otimes B \to C$ out of the tensor product:

$$f : A \times B \stackrel{\otimes}{\to} A \otimes B \stackrel{\phi}{\to} C \,.$$

Properties

Symmetric monoidal closed category structure

Proposition

Equipped with the tensor product $\otimes$ of def. and the exchange map $\sigma_{A, B}: A\otimes B \to B \otimes A$ generated by $\sigma_{A, B}(a, b) = (b, a)$, CMon becomes a symmetric monoidal category.

The unit object in $(CMon, \otimes)$ is the additive monoid of natural numbers $\mathbb{N}$.

Proof

To see that $\mathbb{N}$ is the unit object, consider for any commutative monoid $A$ the map

$$A \otimes \mathbb{N} \to A$$

which sends for $n \in \mathbb{N}$

$$(a, n) \mapsto n \cdot a \coloneqq \underbrace{a + a + \cdots + a}_{n\;summands} \,.$$

Due to the quotient relation defining the tensor product, the element on the left is also equal to

$$(a, n) = (a, \underbrace{1 + 1 \cdots + 1}_{n\; summands}) = \underbrace{ (a,1) + (a,1) + \cdots + (a,1) }_{n\; summands} \,.$$

This shows that $A \otimes \mathbb{N} \to A$ is in fact an isomorphism.

Showing that $\sigma_{A, B}$ is natural in $A, B$ is trivial, so $\sigma$ is a braiding. $\sigma^2$ is identity, so it gives CMon a symmetric monoidal structure.

Notice the symmetry or braiding provides a natural isomorphism $A \otimes - \cong - \otimes A$.

The symmetric monoidal structure is naturally seen as adjoint to a closed structure, where the internal hom $[B, C]$ of commutative monoids, whose underlying set consists of commutative monoid maps $f: B \to C$, is defined in pointwise fashion:

$$(f + g)(b) \coloneqq f(b) + g(b), \qquad 0_{[B, C]}(b) \coloneqq 0_C.$$

That $f+g$ so defined is a homomorphism uses commutativity in the monoid $C$:

$$(f + g)(b + b') \coloneqq f(b + b') + g(b + b') = f(b) + f(b') + g(b) + g(b') = f(b) + g(b) + f(b') + g(b') = (f + g)(b) + (f + g)(b').$$

Proposition

For every B, there is an adjunction $- \otimes B \dashv [B, -]$,

$$CMon(A \otimes B, C) \cong CMon(A, [B, C]),$$

i.e., maps $A \to [B, C]$ are in natural bijection with maps $A \otimes B \to C$.

Proof

Given $f: A \to [B, C]$, define $f^\vee: A \otimes B \to C$ by $f^\vee(a \otimes b) = f(a)(b)$. The linearity of $f^\vee$ in the second argument $b$ follows from the fact that $f(a): B \to C$ is a linear map. The linearity in the first argument follows from the fact that $f$ is itself linear, and the pointwise-defined structure of $[B, C]$:

$$f^\vee((a + a') \otimes b) = f(a + a')(b) = (f(a) + f(a'))(b) \coloneqq f(a)(b) + f(a')(b) = f^\vee(a \otimes b) + f^\vee(a' \otimes b).$$

In the other direction, given $g: A \otimes B \to C$, define $g^\wedge: A \to [B, C]$ by $g^\wedge(a)(b) = g(a \otimes b)$. The linearity of $g^\wedge$ follows from an argument similar to the one in the paragraph above. The fact that the assignments $f \mapsto f^\vee$ and $g \mapsto g^\wedge$ are mutually inverse is obvious.

Since $- \otimes A$ and its isomorph $A \otimes -$ are thus left adjoint functors, and since left adjoints preserve coproducts, we also have the following result.

Proposition

The tensor product of commutative monoids distributes over the biproduct of commutative monoids

$$A \otimes \oplus_{s \in S} B_s \simeq \oplus_{s \in S} ( A \otimes B_s ) \,.$$

Monoids

Proposition

A monoid in $(CMon, \otimes)$ is equivalently a rig.

Proof

Let $(A, \cdot)$ be a monoid in $(CMon, \otimes)$. The fact that the multiplication

$$\cdot : A \otimes A \to A$$

is bilinear means by the above that for all $a_1, a_2, b \in A$ we have

$$(a_1 + a_2) \cdot b = a_1 \cdot b + a_2 \cdot b$$

$$b \cdot (a_1 + a_2) = b \cdot a_1 + b \cdot a_2 \,.$$

$$0 \cdot b = 0 \,.$$

$$b \cdot 0 = 0 \,.$$

This is precisely the distributivity law and absorption law of the rig.

In $(\infty,1)$-category theory

The assignment $\mathcal{C} \mapsto \mathrm{CMon}(\mathcal{C})$ satisfies base change for presentable (∞,1)-categories, i.e.

$$\mathcal{C} \otimes \mathrm{CMon}(\mathcal{S}) \simeq \mathrm{CMon}(\mathcal{S}).$$

In particular, $\mathrm{CMon}(S)$ is a mode; this implies that for $\mathcal{C}$ a presentably symmetric monoidal(∞,1)-category, $\mathrm{CMon}(\mathcal{C})$ possesses a unique symmetric monoidal structure subject to the condition that the free functor

$$\mathcal{C} \rightarrow \mathrm{CMon}(\mathcal{C})$$

is symmetric monoidal. This is due to Gepner-Groth-Nikolaus

Related concepts

• tensor product of abelian groups

• tensor product of chain complexes

• tensor product of modules

• tensor product of algebras over a commutative monad

References

• David Gepner, Moritz Groth, Thomas Nikolaus, Universality of multiplicative infinite loop space machines, (2013) (arXiv:1305.4550)