Tensor product of monoids and arbitrary algebraic structures

Question. Do you know a specific example which demonstrates that the tensor product of monoids (as defined below) is not associative?

Let $C$ be the category of algebraic structures of a fixed type, and let us denote by $|~|$ the underlying functor $C \to \mathsf{Set}$. For $M,N \in C$ we have a functor $\mathrm{BiHom}(M,N;-) : C \to \mathsf{Set}$ which sends an object $K \in C$ to the set of bihomomorphisms $M \times N \to K$, i.e. maps $|M| \times |N| \to |K|$ which are homomorphisms in each variable when the other one is fixed. Then one can show as usual that $\mathrm{BiHom}(M,N;-)$ is representable and call the universal bihomomorphism $M \times N \to M \otimes N$ the tensor product of $M,N$. This is a straight forward generalization of the well-known case $C=\mathsf{Mod}(R)$ for a commutative ring $R$.

Actually, this is a special case of a more general tensor product in concrete categories, studied in the paper "Tensor products and bimorphisms", Canad. Math. Bull. 19 (1976) 385-401, by B. Banaschewski and E. Nelson.

Here are some examples: For $C=\mathsf{Set}$, the tensor product equals the usual cartesian product. This is also true for $C=\mathsf{Set}_*$. For $C=\mathsf{Grp}$, we get $G \otimes H \cong G^{\mathsf{ab}} \otimes_{\mathbb{Z}} H^{\mathsf{ab}}$, using the Eckmann-Hilton argument. (This differs from the "tensor product of groups" studied in the literature). The case $C=\mathsf{CMon}$ is very similar to the well-known case $C=\mathsf{Ab}$ and is spelled out here; namely, we have internal homs and therefore a hom-tensor-adjunction. The same is true for $C=\mathsf{Mod}(\Lambda)$ for a commutative algebraic monad $\Lambda$, see here, Section 5.3.

Note that the tensor product is commutative, and that it commutes with filtered colimits in each variable. However, the case $C=\mathsf{Grp}$ shows that it does not have to commute with coproducts. In particular, tensoring with some object is no left adjoint. Also, the free object on one generator is not a unit in general:

Let us consider $C=\mathsf{Mon}$. Then, we have

$\mathbb{N} \otimes M = M / \{ (mn)^p = m^p n^p \}_{m,n \in M, p \in \mathbb{N}}$

The usual proof of the associativity of the tensor product breaks down: There is a map $\beta : M \times (N \otimes K) \to (M \otimes N) \otimes K$ mapping $(m, n \otimes k) \mapsto (m \otimes n) \otimes k$, which is a homomorphism in the second variable. But what about the first variable? The equation $\beta(mm',t) = \beta(m,t) \beta(m',t)$ is clear if $t \in N \otimes K$ is a pure tensor. But for $t=(n \otimes k) (n' \otimes k')$ we end up with the unlikely equation

$((m \otimes n) \otimes k) ((m' \otimes n) \otimes k) ((m \otimes n') \otimes k') ((m' \otimes n') \otimes k')$ $=((m \otimes n) \otimes k) ((m \otimes n') \otimes k') ((m' \otimes n) \otimes k) ((m' \otimes n') \otimes k')$

Solution 1:

Claim. The tensor product of monoids is not associative: We have $\mathbb{N} \otimes (\mathbb{N}^2 \otimes \mathbb{N}^2) \not\cong (\mathbb{N} \otimes \mathbb{N}^2) \otimes \mathbb{N}^2$.

First, let us look in general what $\mathbb{N} \otimes -$ does. This was already mentioned without proof by the OP. If $M$ is a monoid and $\langle u \rangle = \mathbb{N}$ is the free monoid with one generator, then $\mathbb{N} \otimes M$ is generated by $\{u \otimes m : m \in M\}$ subject to the relation that $u^p \otimes -$ is a homomorphism for all $p \geq 0$, which means that $u \otimes -$ is a homomorphism and $u \otimes (m_1 m_2)^p = u \otimes m_1^p m_2^p$ holds for all $m_1,m_2 \in M$. Thus, we have an isomorphism $\mathbb{N} \otimes M \cong M / \bigl((m_1 m_2)^p = m_1^p m_2^p\bigr)$.

Definition. Let us call a monoid power-homomorphic if for each each $p \geq 0$ the $p$th power map $m \mapsto m^p$ is a homomorphism. Thus, $\mathbb{N} \otimes M$ is the universal power-homomorphic quotient of $M$. Then $M$ is power-homomorphic iff $M \cong \mathbb{N} \otimes M$ iff the canonical homomorphism $M \to \mathbb{N} \otimes M$ is an isomorphism.

Now let us dismantle the explicit construction of the tensor product of monoids.

Definition. Let $M,N$ be two monoids. Let $F$ denote the free monoid on the set $|M| \times |N|$. For $w,w' \in F$ we define $w \sim_1 w'$ when we can write $w$ and $w'$ in one of the following forms:

• $w = A (m,nn') B$ and $w' = A (m,n) (m,n') B$
• $w = A (mm',n) B$ and $w' = A (m,n) (m',n) B$
• $w = A (m,1) B$ and $w' = AB$
• $w = A (1,n) B$ and $w' = A B$

Here, $A,B \in F$, $m,m' \in M$ and $n,n' \in N$. Let $\sim_2$ denote the symmetric closure of $\sim_1$, so that $w \sim_2 w'$ holds iff $w \sim_1 w'$ or $w' \sim_1 w$. Finally, let $\sim_3$ denote the transitive closure of $\sim_2$, so that $w \sim_3 w'$ holds iff there is a chain $w = w_0 \sim_2 w_1 \sim_2 \dotsc \sim_2 w_n = w'$. Observe that $\sim_3$ is a congruence relation on $F$, in fact the smallest one which makes $(m,n)$ a homomorphism in both variables. Thus, $M \otimes N = F/{\sim_3}$.

Definition. Let $M$ be a monoid. We call $m \in M$ strongly irreducible if $m \neq 1$ and if whenever $m = ab$, then $a=1$ or $b=1$.

Basically, I introduce this strong property because I do not want to bother about units.

Lemma. Let $M,N$ be two monoids with the property $xy=1 \implies x=y=1$. Let $x_1,\dotsc,x_r \in M$ (resp. $y_1,\dotsc,y_r$) be a sequence of strongly irreducible elements in $M$ (resp. $N$), such that $x_i \neq x_{i+1}$ (resp. $y_i \neq y_{i+1}$) for $1 \leq i < r$. Consider the element $w := (x_1,y_1) \cdots (x_r,y_r)$ in the free monoid $F$ on the set $|M| \times |N|$. Then the $\sim_3$-equivalence class of $w$ consists precisely of those elements of the form $p_0 (x_1,y_1) p_1 \cdots (x_r,y_r) p_r$, where each $p_i$ is a (possibly empty) product of elements of the form $(1,n)$ or $(m,1)$.

Before we give the proof, let us demonstrate how the claim follows:

Proof of the claim. Since $\mathbb{N}^2$ is power-homomorphic (even commutative), we have $\mathbb{N} \otimes \mathbb{N}^2 \cong \mathbb{N}^2$. Thus, we need to prove that $\mathbb{N} \otimes (\mathbb{N}^2 \otimes \mathbb{N}^2) \not\cong \mathbb{N}^2 \otimes \mathbb{N}^2$, i.e. that $\mathbb{N}^2 \otimes \mathbb{N}^2$ is not power-homomorphic. We denote the canonical generators of the first tensor factor by $\mathbb{N}^2 = \langle x_1,x_2 : x_2 x_1 = x_1 x_2 \rangle$ and of the second tensor factor by $\mathbb{N}^2 = \langle y_2,y_2 : y_1 y_2 = y_2 y_1 \rangle$. Notice that these generators are strongly irreducible. The Lemma implies that in the free monoid on $\mathbb{N}^2 \times \mathbb{N}^2$ the element $(x_1,y_1) (x_2,y_2) (x_1,y_1) (x_2,y_2)$ is not equivalent to $(x_1,y_1) (x_1,y_1) (x_2,y_2) (x_2,y_2)$. This means that in $\mathbb{N}^2 \otimes \mathbb{N}^2$ we have $((x_1 \otimes y_1) (x_2 \otimes y_2))^2 \neq (x_1 \otimes y_1)^2 (x_2 \otimes y_2)^2$. Thus, $\mathbb{N}^2 \otimes \mathbb{N}^2$ is not power-homomorphic. $\checkmark$

Now we prove the Lemma.

Proof of the Lemma. It is clear that every element of the form $p_0 (x_1,y_1) p_1 \cdots (x_r,y_r) p_r$ is $\sim_3$-equivalent to $w$. Conversely, let $w''$ be $\sim_3$-equivalent to $w$. There is a chain of $\sim_2$-equivalences from $w$ to $w''$. By induction on its length, we see that there is some $w' = p_0 (x_1,y_1) p_1 \cdots (x_n,y_n) p_n$ such that $w' \sim_2 w''$. We have 2 cases, each one having 4 subcases:

Case 1: $w' \sim_1 w''$.

Case 1a: $w' = A (m,nn') B$, $w'' = A (m,n) (m,n') B$: Then $(m,nn')$ is either some $(x_i,y_i)$ or one of the pairs $(m,1)$ in some $p_i$, or one of the pairs $(1,n'')$ in some $p_i$. In the first case, we have $n=1$ or $n'=1$ since $y_i$ is strongly irreducible, hence $w'' = A (x_i,1) (x_i,y_i) B$ or $w'' = A (x_i,y_i) (x_i,1) B$. We deduce that $w''$ has the desired form, because it results from $w'$ by adding the factor $(x_i,1)$ somewhere. In the second case, we have $nn'=1$, hence $n=n'=1$, hence $w'' = A(m,1)(m,1)B$ has the desired form, because it results from $w'$ by adding the factor $(m,1)$ somewhere. In the third case, we have $m=1$, hence $w'' = A(1,n) (1,n') B$ has the desired form, because it results $w'$ by expanding $(1,nn')$ in $p_i$ to $(1,n) (1,n')$.

Case 1b: $w' = A (mm',n) B$, $w'' = A (m,n) (m',n) B$: Analogous to 1a.

Case 1c: $w' = A (m,1) B$, $w'' = AB$: Then $(m,1)$ is either some $(x_i,y_i)$, which is not possible since $y_i \neq 1$, or $(m,1)$ is some pair in some $p_i$. Then $w'' = AB$ has the desired form, because it resulsts from $w'$ by removing a factor of $p_i$.

Case 1d: $w' = A (1,n) B$, $w'' = AB$: Analogous to 1c.

Case 2: $w'' \sim_1 w'$.

Case 2a: $w'' = A (m,nn') B$, $w' = A (m,n) (m,n') B$: Then either $(m,n)=(x_i,y_i)$ and hence $(m,n')$ is the first factor of $p_i$, or $(m,n)$ is some factor of some $p_i$ and hence $(m,n')$ is the next factor, or even $(x_{i+1},y_{i+1})$. If the first is true, then $m=x_i$, $n=y_i$ and $n'=1$ (since $m \neq 1$). Then $w'' = A (x_i,y_i) B$ results from $w'$ by removing the first factor of $p_i$, hence has the desired form. If the second is true, and both $(m,n)$, $(m,n')$ are factors of $p_i$, then $m=1$ and hence $(m,nn') = (1,nn')$ is a allowed factor, or $n=1$ and hence $(m,nn') = (m,n')$ is an allowed factor as well. Now if $(m,n')=(x_{i+1},y_{i+1})$, then $m \neq 1$ and thus $n=1$, so that $w'' = A (x_{i+1},y_{i+1}) B$ results from $w'$ by removing the last factor of $p_i$, which thus has the desired form.

Case 2b: $w'' = A (mm',n) B$, $w' = A (m,n) (m',n) B$: Analogous to 2a.

Case 2c: $w'' = A (m,1) B$, $w' = AB$. This means that we insert some $(m,1)$ somewhere in one of the $p_i$ in $w'$, so that $w''$ has the desired form.

Case 2d: $w'' = A (1,n) B$, $w' = AB$: Analogous to 2c.

This finishes the proof. $\checkmark$

Of course, it would be nice to have a better understanding of $\mathbb{N}^2 \otimes \mathbb{N}^2$ which then directly gives the desired inequality of squares, but I wasn't able to find such a proof. Although you can write down a presentation of $\mathbb{N}^2 \otimes \mathbb{N}^2$ with 4 generators and an infinite set of relations, it is not clear a priori how to exclude unexpected relations which might follow from these relations, i.e. it is not clear a priori how to describe the generated congruence relation. But perhaps someone finds a better way.

Solution 2:

This does not exactly answer your question, but it should be pointed out that in some situations such as groups, Lie algebras, ... one wants to consider other kinds of tensor products in which the key notion is that of a biderivation. An example of this is the commutator map $[\; ,\; ]: M \times N \to G$ where $M,N$ are normal subgroups of the group $G$. See a bibliography on this nonabelian tensor product with 120 items.

Solution 3:

EDIT. This doesn't work: see the comments.

I'm not sure what you're thinking of when you say "usual proof of the associativity", but the one I have in mind doesn't use commutativity.

Define a multihomomorphism of algebras to be a function of finitely many variables that is a homomorphism in each variable separately, and then consider "trihomomorphisms" (terhomomorphisms?) $f : A \times B \times C \to D$. It is clear that we get a unique bihomomorphism $g : A \times (B \otimes C) \to D$ such that $g(a, b \otimes c) = f(a, b, c)$ – just think of $f$ as an $A$-indexed family of bihomomorphisms – and we also have a unique bihomomorphism $h : (A \otimes B) \times C \to D$ such that $h(a \otimes b, c) = f(a, b, c)$. Conversely, any bihomomorphism $A \times (B \otimes C) \to D$ or $(A \otimes B) \times C \to D$ gives rise to a unique trihomomorphism. Thus, we have natural bijections $$\textrm{Multi}(A \otimes B, C; D) \cong \textrm{Multi}(A, B, C; D) \cong \textrm{Multi}(A, B \otimes C; D)$$ and so the Yoneda lemma implies $$(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$$ as required. The same argument using "quadrihomomorphisms" (quaterhomomorphisms?) should be enough to verify the hexagon axiom.

It's not so clear to me how to make this argument work in the general setting of strong monads over a symmetric monoidal closed category... but it probably can be done, since Kock [1971] proved it for commutative monads.

As for literature – Borceux mentions it very, very briefly in [Handbook of categorical algebra, Vol. 2, §3.10].