# Is there an operad that codifies groupoids?

maybe this question is trivial and, then this is the reason I've never seen this written.

The motivation is to define internal $\infty$-groupoids (that are preferably) Kan fibrant and to see if Kan fibrancy is really a necessary ontological assumption. As I understand there are non-Kan fibrant models of smooth $\infty$-groupoids given by $\infty$-stacks over cartesian spaces (http://ncatlab.org/nlab/show/smooth+infinity-groupoid), however there are Kan fibrant versions for the smooth case too (http://ncatlab.org/nlab/show/Kan-fibrant+simplicial+manifold)

So the questions are more or less:

1) Given a (closed) monoidal (model) category $\mathcal{V}$, is there an operad (that I will call $\text{Grpd}$) such that algebras over $\text{Grpd}$ are exactly the internal groupoids?

2)If the answer to 1) is yes. Let $\text{Grpd}_{\infty}$ be the resolutions of $\text{Grpd}$. What's a $\text{Grpd}_{\infty}$-algebra?

EDIT: Apparently, there's a misunderstanding regarding the notion of operad as noticed by Todd Trimble in the comments (I didn't know about this notation issues). I'm assuming the definition of operad given from page 23 to 30 of http://www.cs.ox.ac.uk/people/bob.coecke/andrei.pdf

• Even in Cartesian monoidal categories, groups and groupoids cannot be encoded as algebras over operads. There is no way to encode the existence of inverses. – Gijs Heuts Oct 1 '15 at 17:31
• @user40276: the problem is that the defining property of inverses, namely that $a^{-1} a = a a^{-1} = 1$, requires that the same variable appear twice in an equation. Operads can't encode these sorts of equational laws, although Lawvere theories can. – Qiaochu Yuan Oct 1 '15 at 18:07
• (Operads in Qiaochu's last comment referring to the default notion of permutative operad = monoid with respect to plethystic monoidal structure on the category $Set^\mathbb{P}$ where $\mathbb{P}$ is the category of finite permutations. One could stretch the meaning of operad so that a Lawvere theory is a "cartesian operad". But then that stretch should be made clear in the question!) – Todd Trimble Oct 1 '15 at 18:52
• @user40276: that's a pretty nonstandard use of the term "operad" as far as I can tell. In any case, for going up to groupoids there's an additional problem even for writing down a Lawvere theory. First, it has to be two-sorted, since groupoids have both an object of objects and an object of morphisms. Second, and much more importantly, composition isn't a morphism from some product to some other product; instead it involves a pullback. So you need something more sophisticated than a Lawvere theory, namely (I think?) a finite limit sketch. And at this point I see no reason to expect... – Qiaochu Yuan Oct 1 '15 at 19:17
• @QiaochuYuan: it is not true that operads cannot encode relations with multiple occurrences of a given variable. For instance flexible algebras, that are nonassociative algebras wherein product satisfies $(xy)x = x(yx)$, can be encoded by a symmetric operad (see mathoverflow.net/questions/128547/…? for instance). – Samuele Giraudo Oct 1 '15 at 21:16

The question (under the relaxed notion of operad in the original post) will be settled negatively if we show that the category $\mathrm{Grpd}$ (internal groupoids in $\mathrm{Set}$) isn't monadic over $\mathrm{Set}$ for any choice of underlying functor $U: \mathrm{Grpd} \to \mathrm{Set}$. (Note: I am using $\mathrm{Grpd}$ to denote the category, not the putative operad of the OP.)

Categories that are monadic over $\mathrm{Set}$ enjoy strong exactness conditions: they are Barr-exact. In particular, a category that is monadic over $\mathrm{Set}$ in some way is a regular category, and I claim that $\mathrm{Grpd}$ is not regular. In fact I will give an example of a regular epi in $\mathrm{Grpd}$ that is not stable under pullback.

Let's take a connected groupoid $G$ with two objects $0, 1$ whose automorphism group at either object is say $\mathbb{Z}$ (it won't matter much what the automorphism group is, but to be definite let's take this one). Let $A$ be the coproduct of two copies of $G$, and let $B$ be the quotient of $A$ obtained by identifying $1$ of the first copy with $0$ of the second. Then $B$ is a connected groupoid with three objects $0, 1, 2$ and with automorphism group the free group $\mathbb{Z} \ast \mathbb{Z}$, with the object $1$ being the identification of the two aforementioned objects of $A$.

Let $C$ be the full subgroupoid of $B$ on the objects $0, 2$, with inclusion $i: C \to B$. Consider what you get by pulling back the quotient map $p$ along $i$:

$$\begin{array}{ccc} P &\to & A\\ q \downarrow & & \downarrow p \\ C & \stackrel{i}{\to} & B \end{array}$$

The domain $P$ sits inside $A$ as the full subcategory whose objects are $0$ of the first copy of $G$ and $1$ of the second copy. There is no way $q: P \to C$ can be a quotient map, since the images of the two automorphism groups of $P$ under $q$ are disjoint (hence miss the arrows connecting $0$ to $2$ in $C$.

Edit: Nor is $\mathrm{Grpd}$ described by a multisorted Lawvere theory. For example, the category of algebras of a $2$-sorted Lawvere theory is monadic over $\mathrm{Set}/2$, and such categories are again regular. One result along these lines is that if $\mathbf{D}$ is finitely complete and finitely cocomplete and coequalizers in $\mathbf{D}$ split, and $U: \mathbf{C} \to \mathbf{D}$ is monadic, then $\mathbf{C}$ is regular; see for example Proposition 4.6 here. This applies in particular to any $\mathbf{D} = \mathrm{Set}/X$.

• Regarding the edit: is there an argument for Barr-exactness using only finite limits etc.? The only one I'm aware of uses the fact that $\mathbf{Set}$ has intersections of arbitrary collections of subobjects. – Zhen Lin Oct 3 '15 at 14:29
• @ZhenLin Is this an example of what you mean? If $D$ is Barr-exact and coequalizers split in $D$, then for any monad $T$ the category of algebras $D^T$ is Barr-exact. I think it's just an easy argument using the fact that under the hypotheses, the monadic functor $U: D^T \to D$ preserves and reflects finite limits and coequalizers. – Todd Trimble Oct 3 '15 at 15:55
• Ah, sorry, I was thinking of Barr-exactness + existence of coequalisers, which is what you need arbitrary intersections for. (Take the smallest congruence containing the relation corresponding to the parallel pair etc.) – Zhen Lin Oct 3 '15 at 16:03