9

Question. How do we call a pair of two binary operations $m,m' : X \times X \rightrightarrows X$ on a set such that

  1. $m(x,m(y,z)) = m(m(x,y),z)$ ($m$ is associative)
  2. $m'(x,m'(y,z)) = m'(m'(x,y),z)$ ($m'$ is associative)
  3. $m(x,m'(y,z)) = m'(m(x,y),z)$
  4. $m'(x,m(y,z)) = m(m'(x,y),z)$

Maybe we can say "$m$ associates with $m'$" for (3)? But I haven't found this terminology in the literature. Clearly, (4) is symmetric to (3), and an answer to (3) will probably be sufficient. It can also be depicted via a commutative diagram.

$$\require{AMScd}\begin{CD} X \times X \times X @>{X \times m'}>> X \times X \\ @V{m \times X}VV @VV{m}V \\ X \times X @>>{m'}> X \end{CD}$$

Of course the same definitions work in any monoidal category (not necessarily symmetric).

The question A "group" with two binary operations that inter-associate is somewhat related, but it has the additional assumption of a common identity element, which I don't have, and which makes the question kind of trivial. It also suggests the terminology "inter-associate", which couldn't locate in the literature (well, it seems to be used in chemistry for something else).

Notice that $m$ and $m'$ do not necessarily commute, meaning $m(m'(x,y),m'(x',y')) = m'(m(x,x'),m(y,y'))$ does not hold. In fact, the operands $x,y,z$ above always stay in their order.

Background. I was wondering if it's possible to classify the symmetric monoidal structures on the category of sets and found this MO thread. It mentions the very interesting symmetric monoidal structure $$X \otimes_S Y := X \!\times\! S \!\times\! Y + X + Y,$$ where $S$ is any set. The unit object is the empty set $0$. To understand this better, I wanted to determine the monoids with respect to this monoidal structure. For $S = 1$ we just get semigroups (which is kind of funny, semigroups obtain a "neutral element" in this abstract sense, the unique map $0 \to X$). For $S = 2$ we exactly get what I wrote above: two associative operations which "associate with each other". We don't have a neutral element in the classical sense. For a general set $S$, a monoid structure on a set $X$ with respect to $\otimes_S$ consists of binary operations

$$(m_s : X \times X \to X)_{s \in S}$$

such that for all $x,y,z \in X$ and $s,s' \in S$ we have

$$m_s(x,m_{s'}(y,z)) = m_{s'}(m_s(x,y),z))$$

Question. Where can I find more about this interesting symmetric monoidal structure?

Bill Dubuque
  • 282,220
  • Isn't 3. just the (left) distributive property of $m$ over $m'$? https://en.wikipedia.org/wiki/Distributive_property – Jochen Dec 25 '24 at 15:26
  • 2
    @Jochen no, 3 is $x * (y + z) = (x * y) + z$, not $(x * y) + (x * z)$. – Naïm Camille Favier Dec 25 '24 at 15:30
  • There is the notion of a bimodule over a pair of monads, which looks quite similar. Maybe you can get something out of applying the notion of bimodule over a pair of monads to the monad on $\mathbf{Set}$ induced by the forgetul-free adjunction between sets and semigroups? – Lukas Heger Dec 25 '24 at 16:25
  • Thanks @NaïmFavier for pointing this out and for the better readable formulation. – Jochen Dec 25 '24 at 20:58
  • I don't think this answers the actual question, but the associativity here is the same as the linear distributivity in a linearly distributive category. As mentioned in Weakly Distributive Categories, if both operations $\otimes$ and $\oplus$ have units (unlike your situation, I think), then the linear distributors are isomorphisms if and only if $A \oplus B$ is naturally isomorphic to $A \otimes D \otimes B$ for some invertible (with respect to $\otimes$) object $D$. – S.C. Dec 26 '24 at 12:13
  • 1
    I think in German they are called "Doppelalgebren", so I would guess it is "double-algebra" then. The only place I have seen them is in the diploma thesis of Birgit Richter. I recently played a bit with the operads representing $n$-fold algebras (monoids for $\otimes_S$ with $S={1,...,n}$) for reasons unrelated to your question... – Jonas Linssen Dec 28 '24 at 13:47
  • @JonasLinssen Thanks! Can you please post this as an answer? Also with the title and maybe link of her thesis? – Martin Brandenburg Dec 28 '24 at 14:33
  • Sure. It is pure coincidence that I happened to work with them three weeks ago. My motivation was to have some more interesting $\mathsf{Set}$-enriched operads to play around with. Now I am left wondering, in which way this is related to your goal of classifying monoidal structures on $\mathsf{Set}$, knowing that certain multicolored operads are precisely monoidal categories... – Jonas Linssen Dec 28 '24 at 18:38
  • @JonasLinssen I don't think a classification is possible (after reading the MO discussion), and then I was basically just happy enough with these specific examples. – Martin Brandenburg Dec 28 '24 at 21:43
  • It may be helpful to note that this notion forms part of the definition of a "matched pair of semigroups" (or simply "matched semigroups") that appears in the study of bicrossed products. It may be that the literature on matched (semi)groups therefore has relevant terminology, though I'm not familiar enough with that literature to know. – varkor Dec 29 '24 at 09:14
  • 1
    Regarding your second question, I asked the same question for the special case $S = 1$ and so far received no answers. It would be nice to have a reference for this family of monoidal structures and its properties; I found it mentioned in several places online, but could not find a reference in the literature. – varkor Dec 29 '24 at 09:16
  • 1
    A slightly tangential comment: it might be interesting to understand what the categories enriched in this symmetric monoidal category look like. Do they correspond to anything known? – varkor Dec 29 '24 at 09:20
  • @JonasLinssen: I would be interested to hear more about your operadic approach to double algebras. It seems potentially relevant to some questions I had when I was thinking about the topic. Perhaps we could all have a discussion on the Category Theory Zulip? – varkor Dec 29 '24 at 09:40
  • @varkor unfortunately I don't really think I have anything valuable to share and surprisingly I didn't know about these monoidal structures before I stumbled upon this post here. Nevertheless I would be open and interested in a discussion, after I finished my holiday- and conference travels end of January... – Jonas Linssen Dec 29 '24 at 12:59
  • Completely by coincidence, I found an answer to Question 2, which I have written up as an answer here. – varkor Dec 29 '24 at 20:32

3 Answers3

4

As it turns out my advisor Prof. Dr. Birgit Richter has studied this kind of algebraic structure in her diploma thesis (last link on the webpage above) under the German name Doppelalgebra (chapter 3), which translates to double algebra. I do not know, whether this name is standard, or whether this algebraic gadget has been more thoroughly studied by someone before.

When playing around with the obvious generalization to $n$ multiplications, I could not find a name for it in the literature. In my private notes I used $n$-fold (nonunital) algebra, but I do think one can improve on that terminology...

Jonas Linssen
  • 12,065
  • Thanks again! It is not too surprising that the term "double algebra" has already been used for something else, at least occasionally. Namely, in https://doi.org/10.1016/0021-8693(85)90043-2 on p. 134 a "double algebra" is defined (a specific construction associated to a Peano space; I am not familiar with it). https://arxiv.org/pdf/1611.01992 defines "double algebras" via a certain linear map $V \otimes V \to V \otimes V$. We also have the Drinfeld double (algebra) https://ncatlab.org/nlab/show/Drinfel%27d+double, and maybe more. So I guess that any usage requires some explanation. – Martin Brandenburg Dec 28 '24 at 22:08
  • PS: I am aware that you didn't answer my second question, but this is my fault: only one question per thread (need to remember this). PPS: Grüße nach Hamburg! – Martin Brandenburg Dec 28 '24 at 22:20
  • I guess not answering the second question was one reason I first wrote this as a comment ;) I don't really like the name double algebra either and I am not surprised it is overused already. PPS: Danke, werden ausgerichtet :) – Jonas Linssen Dec 29 '24 at 12:46
2

This is a particular case (for $S=\{1,2\}$) of the general algebraic structure that I studied together with Sergey Shadrin and Bruno Vallette for $S$ being any topological space (in this case, one can also take homology and have interesting algebraic structures) in https://arxiv.org/abs/1510.03261, Section 3.1.1. I think we called the corresponding operad $As_S$, and the corresponding algebras $S$-associative algebras.

  • Thank you! Can you please explain a bit more the connection between the symmetric monoidal structure and the operad $\mathrm{As}_S$? I don't see it right now. – Martin Brandenburg Dec 29 '24 at 15:20
  • @MartinBrandenburg An algebra (i.e. representation) for this operad is the same as a monoid in your monoidal category. – Pedro Dec 29 '24 at 16:54
1

It has been asked in the comments how $(\mathbf{Set},\otimes_S)$-enriched categories can be described, which is a natural extension of the description of the $(\mathbf{Set},\otimes_S)$-monoids (=one-object enriched categories). Let me answer this here just by unwinding the definitions. Afterwards, we will see that it can be simplified a lot.

A $(\mathbf{Set},\otimes_S)$-enriched category consists of

  1. a collection of objects,
  2. for every pair of objects $X,Y$ a set $\mathrm{Hom}(X,Y)$,
  3. for every $s \in S$ and for every triplet of objects $X,Y,Z$ a map $$\circ_s : \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,Z) \to \mathrm{Hom}(Y,Z)$$
  4. for every triplet of objects $X,Y,Z$ a map $$\rho_{X,Y,Z} : \mathrm{Hom}(X,Y) \to \mathrm{Hom}(X,Z)$$
  5. for every triplet of objects $X,Y,Z$ a map $$\lambda_{X,Y,Z} : \mathrm{Hom}(Y,Z) \to \mathrm{Hom}(X,Z)$$

subject to the following axioms:

  1. Right neutrality: $\rho_{X,Y,Y} = \mathrm{id}$
  2. Left neutrality: $\lambda_{Y,Y,Z} = \mathrm{id}$
  3. Coherence of $\rho$ with $\rho$: $$\begin{array}{ccc} \mathrm{Hom}(X,Y) & \xrightarrow{\rho} & \mathrm{Hom}(X,U) \\ \hspace{2cm} {\scriptstyle \rho} \searrow & & \hspace{-2cm} \nearrow {\scriptstyle \rho} \\ & \mathrm{Hom}(X,Z) & \end{array}$$
  4. Coherence of $\lambda$ with $\lambda$: $$\begin{array}{ccc} \mathrm{Hom}(Z,U) & \xrightarrow{\lambda} & \mathrm{Hom}(X,U) \\ \hspace{2cm} {\scriptstyle \lambda} \searrow & & \hspace{-2cm} \nearrow {\scriptstyle \lambda} \\ & \mathrm{Hom}(Y,Z) & \end{array}$$
  5. Coherence of $\lambda$ with $\rho$: $$\require{AMScd}\begin{CD} \mathrm{Hom}(Y,Z) @>{\rho}>> \mathrm{Hom}(Y,U) \\ @V{\lambda}VV @VV{\lambda}V \\ \mathrm{Hom}(X,Z) @>>{\rho}> \mathrm{Hom}(X,U) \end{CD}$$
  6. Coherence of $\rho$ with $\circ_s$ for all $s \in S$: $$\begin{CD} \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,Z) @>{\mathrm{id} \times \rho}>> \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,U) \\ @V{\circ_s}VV @VV{\circ_s}V \\ \mathrm{Hom}(X,Z) @>>{\rho}> \mathrm{Hom}(X,U) \end{CD}$$
  7. Coherence of $\lambda$ with $\circ_s$ for all $s \in S$: $$\begin{CD} \mathrm{Hom}(Y,Z) \times \mathrm{Hom}(Z,U) @>{\lambda \times \mathrm{id}}>> \mathrm{Hom}(X,Z) \times \mathrm{Hom}(Z,U) \\ @V{\circ_s}VV @VV{\circ_s}V \\ \mathrm{Hom}(Y,U) @>>{\lambda}> \mathrm{Hom}(X,U) \end{CD}$$
  8. Coherence of $\circ_s,\lambda,\rho$ for all $s \in S$: $$\begin{CD} \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Z,U) @>{\mathrm{id} \times \lambda}>> \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,U) \\ @V{\rho \times \mathrm{id}}VV @VV{\circ_s}V \\ \mathrm{Hom}(X,Z) \times \mathrm{Hom}(Z,U) @>>{\circ_s}> \mathrm{Hom}(X,U) \end{CD}$$
  9. Coherence of $\circ_s$ with $\circ_{s'}$ for all $s,s' \in S$: $$\begin{CD} \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,Z) \times \mathrm{Hom}(Z,U) @>{\mathrm{id} \times \circ_{s'}}>> \mathrm{Hom}(X,Y) \times \mathrm{Hom}(Y,U) \\ @V{\circ_s \times \mathrm{id}}VV @VV{\circ_s}V \\ \mathrm{Hom}(X,Z) \times \mathrm{Hom}(Z,U) @>>{\circ_{s'}}> \mathrm{Hom}(X,U) \end{CD}$$

It follows from 9. in particular that each $\circ_s$ is associative, i.e. we have a bunch of semicategory structures. The axiom 9. is a natural way of saying that they are compatible with each other.

Axioms 1 and 3 imply that each $\rho$ is an isomorphism. Likewise, axioms 2 and 4 imply that each $\lambda$ is an isomorphism. So we are left with $\mathrm{Hom}(X,X)$, which becomes a $(\mathbf{Set},\otimes_S)$-monoid because of axiom 9, but they are all isomorphic because of axioms 6 and 7. I guess that this implies that the category of small $(\mathbf{Set},\otimes_S)$-enriched categories with at least one object is equivalent to $\mathbf{Set}_{\neq \emptyset} \times \mathrm{Mon}(\mathbf{Set},\otimes_S)$.

  • 1
    It's curious to see how much the definition simplifies! (Although a little disappointing it doesn't turn out to be something more interesting.) It reminds me of enrichment in the disjoint union of sets considered in this Zulip thread. Presumably there's a more general picture here... – varkor Dec 29 '24 at 15:21
  • I wonder whether it's true that for every monoidal structure arising from a formal group (monoid?) law, the hom-sets in every enriched category are isomorphic. Then the interesting question would be whether it is possible to characterise the monoids in such monoidal structures. Extrapolating from the example in your question, it seems likely that it is possible to characterise monoids as objects equipped with algebraic structure for each summand of the formal group law, subject to an evident compatibility laws between the structures for each pair of summands. – varkor Dec 29 '24 at 15:45
  • 1
    It seems fun to be able to give (at least sufficient) conditions on an algebraic theory in this way such that its category of models in Set (or a distributive monoidal category) is equivalent to a category of monoids for some monoidal structure on Set. – varkor Dec 29 '24 at 15:55