# Introducing division strategically in operads to accommodate formulas like the Baker-Campbell-Hausdorff formula

I am not as familiar with operad terminology as I'd like to be, so I might be missing some well-known term in the area. If so, I'd appreciate any pointers to the correct terms.

Consider the following two objects:

1. The free associative operad with one binary operation over $\mathbb{Q}$. What I mean is that we are working with $\mathbb{Q}$-vector spaces, and consider the operad where the generating set is an operation of arity 2 and the relation is the associativity relation.
2. The free associative operad with one binary operation over $\mathbb{Z}$. What I mean is that we are working with $\mathbb{Z}$-modules (aka abelian groups), and consider the operad where the generating set is an operation of arity 2 and the relation is the associativity relation.

(2) can be viewed as a $\mathbb{Z}$-suboperad of (1). I am interested in defining an intermediate $\mathbb{Z}$-suboperad of (1). The goal is to allow for a limited amount of division based on the degree.

The intermediate suboperad I want to define is the suboperad of (1) generated by the following infinite set of operations: for each prime number $p$, consider:

$$(x_1,x_2,\dots,x_p) \mapsto \frac{1}{p}(x_1x_2 \dots x_p)$$

The representations of the operad (1) correspond to associative $\mathbb{Q}$ algebras and the representations of the operad (2) correspond to associative $\mathbb{Z}$-algebras (also known as associative rings). The representations of the intermediate operad defined above are associative rings with additional "multiply them all and divide by $p$" operations associated to strings of length $p$ for every prime $p$.

The intermediate object I have defined above is closely related to the nature of denominators that appear in the Baker-Campbell-Hausdorff formula. If we perform a similar construction for the Lie operad instead of the associativity operad (UPDATE: There is a slight issue, in that the assumption of the alternating condition rather than skew symmetry is a non-operadic identity in so far as it involves a repeated variable, and this could be an issue if we are working over a base where 2 is non-invertible), then the representations of that operad will be the "Lie algebras" in a suitable generalization of the Lie correspondence or Lazard correspondence (there are some messy details I am not getting into here).

[As a quick illustration, note that the Baker-Campbell-Hausdorff formula has a denominator of 12 for its degree three terms. To make sense of a denominator of 12, note that we can separately make sense of denominators of 3 and 4 for length three Lie products: the denominator of 3 makes sense directly setting $p = 3$, and the denominator of 4 makes sense by consider $\frac{1}{2}[x,\frac{1}{2}[y,z]]$. We can combine these using the Euclidean algorithm to get a denominator of $1/12$ (in this case, $1/12 = 1/3 - 1/4$). There are general bounds on the denominators that appear in the Baker-Campbell-Hausdorff formula that guarantee that the denominators can always be achieved.]

Additionally, the general theme behind this sort of operad seems well-suited to situations where we want to introduce divisibility by certain primes but only in high degrees dependent on the prime. This might make it suitable to the study of things like the representation theory of symmetric groups, though I might be barking up the wrong tree here.

I'd like to know if operads of this sort have been studied before, and if so, I'd like any pointers to the appropriate terms and references. If not, I would appreciate any views on whether this is useful as an intermediate framework between the extremes of $\mathbb{Z}$-operads and $\mathbb{Q}$-operads.

If $C$ is an operad and $A$ is a $C$-algebra then you have maps $(C(n)\otimes A^{\otimes n})_{S_n}\to A$, where the domain is the module of coinvariants for the action of $S_n$ on $C(n)\otimes A^{\otimes n}$. The construction $x\mapsto\sum_{\sigma\in S_n}\sigma.x$ gives a map from the coinvariants to the invariants, which is an isomorphism if everything is a vector space over $\mathbb{Q}$. You can define a DP-$C$-algebra to be an object $A$ with maps $(C(n)\otimes A^{\otimes n})^{S_n}\to A$ subject to certain axioms. If you do this with the commutative operad, I think you are supposed to get divided power algebras. You might need to do a nonunital version to get the details straight. Anyway, this is certainly something intermediate between the integral and rational cases. I have never tried to work with it myself but I have heard Clark Barwick talk about it so you could try looking in his papers.

• Thanks. I had originally thought that my construction was like divided power algebras, and it resembles it in some ways, but it didn't seem to fit precisely enough for the setting I was in. I'll check out Barwick's papers. – Vipul Naik Jun 27 '13 at 20:23

Nice question! I've enjoyed thinking of this operad you define. I don't know of any reference where this sort of operad has been considered. Let me just record here some properties I have found out. I will work in the context of nonsymmetric operads, unlike Neil.

For $k$ a commutative ring, let $\mathtt{Ass}_{k}$ be the associative operad in the category of $k$-modules. This operad is given by $\mathtt{Ass}_{k}(n)=k$, $n>0$, and $\mathtt{Ass}_{k}(0)=0$. Composition laws are given by multiplication in $k$. This operad has a presentation with a single generator $\mu_2$ in degree $2$ and two relations: $\mu_2\circ_1\mu_2=\mu_2\circ_2\mu_2$. Notice that this operad is not free in any sense. I will write $\mu_n$ for the $(n-1)^{\text{th}}$ power of $\mu_2$. Due to the defining relation, we need not specify in which way this power is taken.

We pull $\mathtt{Ass}_{\mathbb Q}$ back to the category of operads of abelian groups. As you point out, there is a morphism (inclusion) of operads $\mathtt{Ass}_{\mathbb Z}\subset \mathtt{Ass}_{\mathbb Q}$ given aritywise by the usual inclusion $\mathbb Z\subset\mathbb Q$. Let me denote your operad by $\mathtt{O}$, $\mathtt{Ass}_{\mathbb Z}\subset\mathtt{O}\subset \mathtt{Ass}_{\mathbb Q}$, generated by $\frac{1}{n}\mu_n$, $n\geq 2$.

The first interesting question is, what abelian group is $\mathbb Z\subset\mathtt{O}(n)\subset\mathbb Q$, $n>0$? Obviously, $\mathtt{O}(0)=0$. Notice also that $\mathtt{O}(1)=\mathtt{Ass}_{\mathbb Z}(1)=\mathbb Z$. Moreover, $\mathbb Z\subset\mathtt{O}(2)=\left(\frac{1}{2}\right)\subset\mathbb Q$ is the abelian subgroup generated by $\frac{1}{2}$. The description of $\mathtt O$ in terms of generators shows that $\mathtt O(3)$ contains $\frac{1}{3}\mu_3$, $\left(\frac{1}{2}\mu_2\right)^2=\frac{1}{4}\mu_3$ and their integer multiples, so it is the subgroup generated by $\left(\frac{1}{3},\frac{1}{4}\right)=\left(\frac{1}{12}\right)$. Recall that for $a,b\in\mathbb Z$, $\left(\frac{1}{a},\frac{1}{b}\right)=\left(\frac{1}{\text{least common multiple of }a\text{ and }b}\right)$. Reasoning in this way, one can easily check that $\mathtt{O}(n)=\left(\frac{1}{a_n}\right)$, where $a_n$ is the least common multiple of the numbers $p_1\cdots p_r$, $r\geq 1$, such that $p_1+\cdots+p_r-r+1=n$ and $p_i\geq 2$. I haven't found any easy way of computing these numbers. I've calculated them for small values of $n$, $n=1,2,3,4,5,6,7$, and the outcome is $$a_n=1,2,12, 24, 720, 1440, 60480.$$

At a first glance, there seems to be some relation between these numbers and the denominators in the Baker-Campbell-Hausdorff formula, but I cannot precise what. Looking this sequence up in The On-Line Encyclopedia of Integer Sequences® produces only one outcome: http://oeis.org/search?q=2%2C12%2C24%2C720&language=english&go=Search I haven't checked much, but I don't see any immediate link between the description of the sequence therein and $\{a_n\}_{n\geq 1}$.

Anyway, $\mathtt O$ can be presented as an operad of abelian groups in the following way: a sequence of generators $\{m_n\}_{n\geq 2}$, which correspond to $m_n=\frac{1}{a_n}\mu_n$, and relations $m_p\circ_im_q=\frac{a_{p+q-1}}{a_p\cdot a_q}m_{p+q-1}$, $p,q\geq 2$, $1\leq i\leq p$. Notice that he coefficient $\frac{a_{p+q-1}}{a_p\cdot a_q}$ is always an integer. I don't think this operad can be presented with finitely many generators. An $\mathtt O$-algebra is therefore an abelian group $B$ together with $n$-ary operations, $(b_1,\dots,b_n)\mapsto m_n(b_1,\dots,b_n)\in B$ such that

$$m_p(b_1,\dots,b_{i-1},m_q(b_i,\dots,b_{q+i-1}),b_{q+i},\dots,b_{p+q-1})=\frac{a_{p+q-1}}{a_p\cdot a_q}m_{p+q-1}(b_1,\dots,b_{p+q-1}).$$

I don't know if any of this can be of help for you. If you happen to find any sense to the $a_n$'s, please let me know. I'm curious about it!

• Thanks, a very useful response! I appreciate it. I think the sequence description you give is the same as that in OEIS (you have to click through to the description of the other sequence). The relation with the Baker-Campbell-Hausdorff formula (I think) is this: all the denominators in the degree n part of the BCH formula must divide the n^{th} term of the sequence. Some of the denominators may well be smaller, and I'm not sure if the bound is tight (i.e., whether the a_n of the sequence must exactly be the lcm of the denominators). – Vipul Naik Jul 10 '13 at 18:15
• UPDATE: The bound seems to be tight at least up to n = 5. See groupprops.subwiki.org/wiki/… for instance – Vipul Naik Jul 10 '13 at 18:18
• An alternate description that might be easier for explicitly computing the terms of the sequence is that the exponent on any prime $p$ is the greatest integer of $(n - 1)/(p - 1)$. I think one can work this out from your description by choosing as many of the primes as possible to be $p$. – Vipul Naik Jul 10 '13 at 18:21
• A colleague has implemented a maple code computing the terms of the sequence $a_n$. You can email me if you need it. – Fernando Muro Jul 13 '13 at 16:13
• Just in case, some detailed tables for BCH up to degree 20 seem to be available here: ehu.es/ccwmuura/bch.html - one can try to compare the data to what this operad predicts. – Vladimir Dotsenko Sep 12 '13 at 13:55