I'm trying to understand a little bit about operads. I think I understand that monoids are algebras over the associative operad in sets, but can groups be realised as algebras over some operad? In other words, can we require the existence of inverses in the structure of the operad? Similarly one could ask the same question about (skew-)fields.
Are groups algebras over an operad?
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0Now that Zhen Lin added the tag, no need to keep the request for someone to do it in the post! :-) – 2012-04-21
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0can't argue with that – 2012-04-21
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0Obviously you are not sitting in room 12A. :-) – 2012-04-21
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0had to look that one up :P – 2012-04-21
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0Two categorical machines you can use to get groups are Lawvere theories (http://en.wikipedia.org/wiki/Lawvere_theory) and monads (http://en.wikipedia.org/wiki/Monad_(category_theory)). – 2012-05-09
3 Answers
No, there is no operad whose algebras are groups. Since there are many variants of operads a more precise answer is that if one considers what are known as (either symmetric or non-symmetric) coloured operads then there is no operad $P$ such that morphisms $P\to \bf Set$, e.g., $P$-algebras, correspond to groups.
In general, structures that can be captured by symmetric operads are those that can be defined by a first order equational theory where the equations do not repeat arguments (for non-symmetric operads one needs to further demand that the order in which arguments appear on each side of an equation is the same). The common presentation of the theory of monoids is such and indeed there is a corresponding operad. The common presentation of the theory of groups is not of this form (because of the axiom for existence of inverses). This however does not prove that no operad can describe groups since it does not show that no other (super clever) presentation of groups can exist which is of the desired form.
It can be shown that the category of algebras in $Set$ for an operad $P$ has certain properties that are not present in the category $Grp$ of groups. This does establish that no operad exists that describes groups.
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2Having learned a bit about operads now, I would quite like to know what these properties of algebras over an operad are! – 2013-03-11
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0I'll need to do some digging to find the precise answer. From what I remember categories of operad algebras have rather strong decomposability properties of morphisms (I know that's not saying much). – 2013-03-11
Groups are not algebras for an operad. (Here, by ‘operad’ I mean the non-symmetric monochromatic version, but the same proof goes through for the symmetric coloured version as well.) Indeed, observe that the category of operads has a terminal object, namely the operad for monoids, and so if $\mathcal{P}$ is an operad, then there must be a functor $\textbf{Mon} \to \textbf{Alg}_{\mathcal{P}}$ that commutes with the underlying set functor. Of course, there is no such functor $\textbf{Mon} \to \textbf{Grp}$.
Indeed, suppose $F : \textbf{Mon} \to \textbf{Grp}$ is a functor that commutes with the underlying set functor. By general nonsense, $F$ must preserve all limits, so $F$ must carry internal monoids in $\textbf{Mon}$ to internal monoids in $\textbf{Grp}$. However, the Eckmann–Hilton argument says that an internal monoid in $\textbf{Mon}$ is the same thing as a commutative monoid, and an internal monoid in $\textbf{Grp}$ is the same thing as an abelian group. So if there were such a functor $F$, every commutative monoid would automatically be an abelian group with the same binary operation and unit. This is clearly absurd: $\mathbb{N}$ is a commutative monoid but not an abelian group.
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0Thanks for posting; this was helpful. – 2016-04-10
As far as I know, the answer is "no". The point is that the axioms of an operad must contain no repeated variables (think the associativity or commutativity law, which are written $(xy)z = x(yz)$ and $xy = yz$, or the Jacobi identity $[[x,y],z] + [[y,z],x] + [[z,x],y] = 0.$
On the hand, the axioms for a group include the axiom $x x^{-1} = 1$, which involves the same variable $x$ twice.
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1that's a good point. maybe more flexible objects, like modular operads or something, might remedy this shortcoming. (well, perhaps not modular operads) – 2012-04-21
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0@donkeykong: Wha? There's a very simple way to encode the group axioms categorically, if that is what you are inclined to do: take the category of all finitely-generated free groups. – 2012-04-21
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0@donkeykong: Dear Donkeykong, In the language of universal algebra (or at least, the language I learned many years ago), one could take the *clone* generated by the group operations. I've never quite understood (either mathematically or sociologically) the relationship between the universal algebra viewpoint and the operadic viewpoint. Regards, – 2012-04-21
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0@Zhen Lin: Dear Zhen Lin, I don't quite understand your comment. Regards, – 2012-04-21
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0@ZL: I don't understand what free groups have to do with it. the problem is not defining group-objects, that's easy. the problem is to realise groups as algebras over some operad in sets. (actually, it's not a problem, I don't think anyone should care) – 2012-04-21
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0@donkeykong: P.S. A quick look at wikipedia suggests that my use of the term *clone* is not quite correct. In any case, we can certainly look at all the $n$-ary operation generated by the group operations (I think this *is* an example of a clone), and then consider the axioms these satisfy. This collection of operations satisfying these axioms is an example of something (unfortunately, my universal algebra is too rusty to remember the correct name for that something!). – 2012-04-21
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0@ME: I'm not sure I want to go and fetch what clone means, as I'm completely ignorant about universal algebra and was planning to persist in this state :) but thank you, your answers and comments are always excellent. – 2012-04-21
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1@donkykong: Dear Donkeykong, In case you change your mind, googling "clone universal algebra" should bring up plenty of references, and avoid too many bad sci-fi links. And thanks for your kind words. Best wishes, – 2012-04-21
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0would taking inverses be an extra 1-ary operation? perhaps in the easier case of abelian groups where it is a morphism it's easier? – 2012-04-21
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0@donkeykong: Yes, there are three group operations: the $0$-ary operation of the identity, the $1$-ary operation of inverses, and the $2$-ary operation of group multiplication. The philosophy of both clones and operads is that you should look at *all* the $n$-ary operations that you can generate from these, since all of them have equal status. I once saw Drinfeld describe it this way: most of us (who don't think operadically) think about an algebraic structure by singling out a particular set of operations and the axioms they satisfy, rather than thinking about all the operations they ... – 2012-04-21
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0... generate and all the coincidences between them. His next remark was that to operadically inclined people, this makes us seems as primitive as someone who wants to consider a particular set of generators and the relations they satisfy, but refusing to consider the group that they generate. I always found this a rather compelling comment in favour of the operadic view-point, although not compelling enough (as my subsequent mathematical development shows) to actually adopt an operadic view-point myself. Regards, – 2012-04-21
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0thanks, that's a nice quote. – 2012-04-21
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1@MattE: The opposite of the category of all finitely-generated free groups precisely encodes all the $n$-ary operations expressible in the signature of the theory of groups, modulo the equations of the theory of groups. If we pass to the skeleton $\mathcal{T}$ of this category then we obtain the Lawvere theory of groups. A group object in a category $\mathcal{C}$ with finite products is a finite-product-preserving functor $\mathcal{T} \to \mathcal{C}$. All this was known before operads were invented. – 2012-04-21
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1@ZL: but can this be interpreted operadically? Or is this already an operadic definition? (in the same sense that an operad should be a functor from the category of trees, or something of that sort) – 2012-04-21
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0@donkeykong: I don't know – I haven't studied operads! My point was that you haven't told us your motivation. If you are looking for a way to describe groups as "algebras" over a category, then there's a simple way to do that. MattE's observation shows that it is unlikely that the notion of internal group can be generalised to the non-cartesian setting, so I doubt there's an operadic way of looking at it. (Monoidal categories do not usually come with a "duplication" operation.) – 2012-04-21