What other monoidal structures exist on the category of sets?

Question

I know of the following monoidal structures over $\mathbf{Set}$ (taken from here):

• The Cartesian product $\otimes=\times$ (categorical product)
• The disjoint union $\otimes=+$ (categorial coproduct)
• $A\otimes B = A\times S\times B + A + B$ for some given set $S$
• $A\otimes B = (A\times S\times B\times S)^* \times (A + A\times S \times B) + (B\times S\times A\times S)^* \times (B + B\times S \times A)$ where $*$ is the Kleene star (free monoid construction)
• If $A=\emptyset$ then $A\otimes B=B$, if $B=\emptyset$ then $A\otimes B=A$, otherwise $A\otimes B=I$

I want to know if there are any more that are known or, conversely, if there are any demonstrations that additional examples don't exist.

Also, what if we restrict to, say, braided or symmetric monoidal categories, or traced monoidal categories (or some example), are there any results/examples regarding the existence/non-existence of monoidal structures over $\mathbf{Set}$ in these cases?

Answer 1

Plausibly we can construct examples from the power series expansion of formal group laws. We can try to write down such examples by writing down nice functions $q(x)$ and hoping that the formal group law

$$q^{-1} (q(x) + q(y)) = \sum s_{i, j} x^i y^j$$

has positive integer coefficients so it can be upgraded to an "analytic bifunctor"

$$\text{Set} \times \text{Set} \ni (A, B) \mapsto \bigsqcup_{i, j} S_{i, j} \times A^i \times B^j \in \text{Set}$$

which is a candidate to be a monoidal structure. For example,

• the disjoint union corresponds to choosing $q(x) = x$,
• the cartesian product corresponds to choosing $q(x) = \log x$,
• the third example corresponds to choosing $q(x) = \log (sx + 1)$.

The fourth example should also arise in this way with $q(x)$ related to the arctangent; the relevant formal group law is $\frac{x + y + 2s xy}{1 - s^2 xy}$ which resembles the tangent addition formula $\frac{x + y}{1 - xy}$. Speaking of which, this should give another family of examples (with $q(x) = \arctan sx$), namely

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B).$$

Edit: Apparently we are already pretty close to exhausting all of the formal group laws that can be expressed as rational functions! See this blog post. It seems plausible, but I haven't checked, that

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B + T \times A \times B)$$

works.

Edit #2: Hmm. I can only find an associator in the case already mentioned in the problem, where $T = S + S$. In this case, as mentioned in the link, $A \otimes B$ can be interpreted as the set of all words which start with an element of $A$ or $B$ and then alternate between $A$ and $B$, with the alternations delimited by elements of $S$, and ending with an element of $A$ or $B$. Then $(A \otimes B) \otimes C$ and $A \otimes (B \otimes C)$ both describe the corresponding construction where we don't repeat among the choices $A, B, C$ consecutively.

Answer 2

The following family of examples can be extracted from Lenart and Ray, Some applications of incidence Hopf algebras to formal group theory and algebraic topology (Postscript). Let $B_2$, $B_3$, $B_4$, … be any sequence of sets. An element of $X \otimes Y$ is a plane tree $T$, where every non-leaf has at least $2$ children, where every leaf is labeled with an element of $X \sqcup Y$, and every non-leaf with $i$ children is labeled with an element of $B_i$, subject to the condition:

If $v$ is a non-leaf but all the children of $v$ are leaves, then the children of $v$ do not come entirely from $X$ nor entirely from $Y$.

To see associativity, note that both $(X \otimes Y) \otimes Z$ and $X \otimes (Y \otimes Z)$ can be canonically bisected with plane trees whose non-leaves are labeled with $X \sqcup Y \sqcup Z$, subject to the analogous conditions. (The analogue of the boxed condition is that the children of $v$ do not come entirely from $X$, nor entirely from $Y$, nor entirely from $Z$.)

I think the logarithm of the corresponding group law is $t - \sum_{i \geq 2} |B_i| t^i$. Since this method can only categorify group laws whose logarithm has integer coefficients, this clearly cannot include many of the most obvious examples.

Answer 3

Here is another possible source of tensor products on $\mathbf{Set}$: transferring tensor products from categories $\mathbf{Set}$ is coreflective in. Unfortunately I have no good example.

Suppose $(\mathbf{C},\otimes,I)$ is a monoidal category with a coreflection from $\mathbf{Set}$; that is, there are functors $F\colon \mathbf{Set} \to \mathbf{C}$ and $G \colon \mathbf{C} \to \mathbf{Set}$ such that $F$ is left adjoint to $G$ with isomorphic unit $\eta_X \colon X \to GF(X)$. Moreover, suppose that $FG \colon \mathbf{C} \to \mathbf{C}$ is monoidal. Then $X \otimes Y := G(FX \otimes FY)$ is a monoidal product on $\mathbf{Set}$. (See e.g. Theorem 5 here.)

For example, these conditions are automatically fulfilled if $\mathbf{C}$ is the category of topological spaces and $G$ is the forgetful functor. However, a quick browse gives results classifying the tensor products on $\mathbf{Top}$ (*), so this is probably not a good source of examples. Are there others?

(*) That are canonical - meaning they just give the tensor product $\times$ on $\mathbf{Set}$ - and closed; but we are precisely looking for noncanonical ones...

Answer 4

So this is a partial answer, using the ideas of Qiaochu Yuan's answer. Whilst this question is obviously of general interest, it turns out that for the specific purposes I have in mind only products where the unit is the singleton are useful (since, as Tom Goodwillie points out, the only possible units in $\mathbf{Set}$ are the unit and the singleton). I leave it here in the hope that it may be useful to other people.

This answer then says that the only monoidal structure over $\mathbf{Set}$ that has the singleton as the unit which can be derived from Qiaochu Yuan's construction is the cartesian product. This is because there are no possible functions $q(x):\mathbb{C}\rightarrow \mathbb{C}$ such that $$q^{-1}(q(x)+q(y)) = \sum_{i,j}s_{i,j}x^iy^j$$ for integer coefficients and that $q(1)=0$ (which would give the tensor unit as the singleton set).

For such a $q$ to exist, it would have to have $q^{-1}(0)=1$ and therefore $$1 = \sum_{i,j}s_{i,j}$$ so that $s_{\alpha,\beta}=1$ for some specific pair $(\alpha,\beta)$ and all other $s_{i\neq\alpha,j\neq\beta}=0$ since they're all positive integers. Moreover, $\alpha=\beta$ since the LHS of the first displayed equation above is symmetric under $x\leftrightarrow y$ interchange. So we have $$\log f(x)f(y) = \log f((xy)^\alpha)$$ where $\log f = q$. Requiring $f$ to be analytic and expanding them as power series, it is easily seen that the only possible choice is $\alpha=1$, which is simply $q(x)=\log x$ and gives rise to the cartesian product, as noted by Qiaochu Yuan.

The same argument and conclusion is found when using $$q^{-1}(q(x)q(y)) = \sum_{i,j}s_{i,j}x^i y^j$$ rather than the above formal group law.

What this means is that if it can be proved that the above construction gives every possible monoidal structure on $\mathbf{Set}$ then all except the cartesian product have the empty set as their unit.