Todd Trimble Notes on operads and the Lie operad

Contents

The plug-in product

Let $FB$ be the monoidal groupoid of finite sets and bijections, where the monoidal product is obtained by restricting the coproduct on the category of finite sets. Let $F, G: FB \to V$ be species valued in a bicomplete symmetric monoidal closed category $V$. The plug-in product $F \star G$ is defined by the formula

where $S/T$ denotes the pushout of $T \to 1$ along the inclusion $T \subseteq S$. This includes the degenerate case where $T = 0$; in this case $S/0$ is the result of freely adjoining a point to $S$.

The set $S/T$ has $T/T$ as distinguished basepoint, and if $U$ is complementary to $T$, we have a natural bijection $S/T \cong U + \{T/T\}$. It follows that

where $F' \otimes G$ refers to the usual Day convolution product, induced from the tensor product on $FB$. Since differentiation of species,

is cocontinuous, and since Day convolution is cocontinuous in each of its arguments, we see that

is separately cocontinuous in each of its arguments $F$, $G$. We remark below on the fundamental implications of this observation.

For $V = Set$, a structure of species $F \star G$ is given by three data:

• A tree obtained by grafting the root of a sprout (aka corolla) with leaf set $\tau$ to a leaf of another sprout with leaf set $\sigma$;
• An element of $F[\sigma]$;
• An element of $G[\tau]$.

Lax monoidal structure of plug-in product

We note that the plug-in poduct is not associative up to isomorphism, but there is a lax associativity. Specifically, we calculate

so that there is a noninvertible associativity (an inclusion)

which is natural in each of its arguments $F$, $G$, and $H$, and which satisfies an evident pentagon coherence condition. Additionally, there is a lax unit, defined as the species $X$ for which $X[S]$ is the monoidal unit of $V$ if $card(S) = 1$, else $X[S]$ is initial, for which we have evident natural maps

The first map $\lambda_F$ is invertible, but the second is not: its component at a finite set $S$ is the codiagonal

However, the standard unit coherence conditions for monoidal categories hold. We may call such a structure, relaxing the condition of invertibility of $\alpha$ and $\rho$ but retaining the naturality and coherence conditions, a lax monoidal category.

In the sequel, $F^{\star n}$ will denote the iterated plug-in product defined recursively by

so that all parentheses in $F^{\star n}$ are to the left. We have a partial coherence theorem:

For $F$ a $Set$-valued species, an element of $F^{\star n}$ is described by data as follows:

1. A structure of tree constructed from $n$ sprouts in a prescribed plugging order;

2. An element of $F[\sigma]$ for each sprout $\sigma$ used in the construction of the tree in 1.

Thus we expect some connection between the “geometric series” $\sum_n F^{\star n}$ and the free operad on $F$. The only difference between the two is that the standard construction of free operads via trees is in no way dependent on plugging order. Thus, to obtain the free operad, we have to mod out by isomorphisms on $F^{\star n}$ induced by “inessential” differences in plugging order. This will be described in detail in the section on trees and hierarchies.

Interlude: the importance of separate cocontinuity of plug-in

The are various well-traveled roads to operads and free operads. For example, there is a well-known concise definition of operad, as monoid in the category of species with respect to a monoidal product known as the substitution product (denoted $\circ$). It may therefore be wondered what is the point of introducing another approach, based on the somewhat outré “lax monoidal product” we have named the plug-in product.

A key point for us is that the substitution product is only partially adapted to developing the analogy between bar/cobar constructions for algebras and bar/cobar constructions for operads, and that the plug-in product is actually better adapted. On the substitution product side, we have the analogy

• An algebra $A$ is a monoid in the monoidal category $(Vect, \otimes)$.

• An operad $M$ is a monoid in the monoidal category $(Species, \circ)$.

Now the bar construction of an algebra is based on a simplicial resolution of $A$ whose components $A^{\otimes n}$ are the graded components of the free (tensor) algebra $\sum_n A^{\otimes n}$. (There are in fact several versions of bar resolutions; another is based on the tensor algebra of an augmentation ideal.) At this point however the analogy on the side of operads begins to break down, because the free operad on (the underlying species) $M$ is not $\sum_n M^{\circ n}$.

Analyzing the reasons for failure of the analogy, we recall the following classical result:

The substitution product $F \circ G$ does not distribute over countable coproducts. It is true that the functor $- \circ G$ preserves countable products, and this partial result can be exploited to yield simplicial resolutions (as we recount below in our summary of Joyal’s approach to the Lie species), but the bar constructions on operads (as in the work on Koszul operads initiated by Ginzburg and Kapranov) are not of this character.

At this point we recall our earlier observation that $\star$ does distribute over countable coproducts (and over colimits generally). We wish to pursue the very rough analogy:

As is obvious from the phrasing (and as already noted), the analogy is still somewhat imperfect, but the imperfections are easier to smooth over than is the case with the earlier analogy based on the substitution product. The analogy is actually worth pursuing for its own sake, as it lays bare other connections between algebras and operads which as far as I know have attracted little notice in the literature.

Trees and hierarchies

It is interesting to resolve $F^{\star n}$ into a sum of monomials, each of which is a tensor product of derivatives of $F$. For example, we have

There are in fact $(n-1)!$ monomial terms, each a tensor product of $n$ derivative terms, and each corresponding to a function $f: [2, n] \to [1, n]$ such that $f(k) \lt k$ for all $k \in [2, n]$. We call such an $f$ a hierarchy. In detail, if the $S^{th}$ derivative of a species $F$ is defined by $F^{(S)}[T] = F[S + T]$, we have

as may be proved by an easy induction.

Each $f \in Hier(n)$ endows the set $[1, n]$ with a rooted tree structure. A rooted tree structure on a set $S$ is precisely a function $f: S \to S$ with a fixed point $r$ (the root) such that for every $s \in S$, some finite iterate of $f$ takes $s$ to $r$, i.e., $f^{(k)}(s) = r$ for some $k$. (The function $f$ simply moves a vertex one step closer to the root along the unique path to the root.) Given a hierarchy $f: [2, n] \to [1, n]$, the corresponding rooted tree structure on $[1, n]$ is the extension of $f$ which fixes the point $1$. The well-ordering of the vertices $1 \lt \ldots \lt n$ serves to prescribe a plugging order on sprouts, starting from the sprout at the root $1$, plugging the sprout rooted at $2$ to a leaf of the initial sprout, and so on.

Somewhat more generally, given any finite well-ordered set $S = \{s_1 \lt \ldots \lt s_n\}$, a hierarchy on $S$ is a function $f: S - \{s_1\} \to S$ such that $f(s_i) \lt s_i$ for $1 \lt i \leq n$, and this also gives rise to a rooted tree structure. A rooted tree whose vertices are equipped with a well-ordering and whose tree structure arises from a hierarchy function will be called a hierarchical tree. The root $r$ is of course the initial element $s_1$.

Now by removing the root of a finite rooted tree $(S, f, r)$, we obtain a forest of finite rooted trees $S/s$ whose roots $s$ range over $f^{-1}(r)$. An isomorphism $\Phi$ of rooted trees $(S, f, r) \to (S', f', r')$ may be described recursively as consisting of

1. A bijection $\phi: f^{-1}(r) \to (f')^{-1}(r')$;

2. Isomorphisms of rooted trees $\Phi_s: S/s \to S'/\phi(s)$ where $s \in f^{-1}(r)$.

If $(S, f)$ is hierarchical, then the subtrees $S/s$ inherit hierarchical tree structures from $S$ by restricting both the well-ordering and $f: S \to S$ to a function $f|: S/s \to S/s$. Thus isomorphisms of hierarchical trees $S$ (defined to be isomorphisms of the underlying rooted trees) can also be described recursively.

If $(S = \{s_1 \lt \ldots \lt s_n\}, f)$ is a hierarchical tree and $F$ is a species, we define a species

Observe that there is a well-defined isomorphism

by permuting tensor factors. (The order of tensor factors on either side is fixed by the well-ordering of vertices.)

Now if $\Phi: (S, f) \to (S', f')$ is an isomorphism of hierarchical trees, we define by recursion a species isomorphism to be the composite of

with factor-permuting isomorphisms

on either side. The result is an isomorphism $F_\Phi: F_{S, f} \to F_{S', f'}$.

There is a groupoid consisting of hierarchical trees and isomorphisms between them. Let $Hier_n$ be the full subgroupoid whose objects are hierarchical tree structures on $[1, n]$. (The isomorphisms in $Hier_n$ can be thought of as permutations on $[1, n]$ that relabel vertices to give a new plugging order.) Given a $V$-species $F$, there is a functor

which defines a diagram in the category of species. The colimit of this diagram is a certain quotient of

which we will denote as $F^{\star n}/Hier_n$. By taking this quotient, we eliminate dependence on a specified hierarchy or plugging order.

The constructions are made in view of the following result.

This will be explored further in the next section.

Operads

The notion of (permutative) $V$-valued operad may be formulated in terms of the plug-in product. Specifically, if $M$ is an operad, there is a natural map

and whose unit is again $u: X \to M$. The structure of an operad $M$ may be described entirely in terms of $m: M \star M \to M$, $u: X \to M$, and the notion of operad may be recast as appropriate $\star$-monoids (with a slight extra twist).

The map $\psi$ is not hard to describe: the plug-in product $F \star G$ may be alternatively described by the formula

where $\widehat{G}_j$ is a tensor product of $k$ factors

all factors being $X$ except for a single instance of $G$ as the $j^{th}$ tensor factor. If there is in addition a map $i: X \to G$ (for example, a unit map $u: X \to M$ of an operad), there is an induced map

and under these circumstances there is an induced map

• A multiplication $m: F \star F \to F$;

• A unit $u: X \to F$

These must satisfy the following axioms:

1. $$\array{ F \star (F \star F) & \stackrel{1 \star m}{\to} & F \star F & \\ \alpha \downarrow & & & \searrow m \\ (F \star F) \star F & \underset{m \star 1}{\to} F \star F & \underset{m}{\to} & F }$$

   commutes;

2. The two composites $F'' \otimes F^{\otimes 2} \to F$ named in

   $$F'' \otimes (F \otimes F) \stackrel{\overset{\sigma}{\to}}{\underset{1}{\to}} F'' \otimes (F \otimes F) \stackrel{i}{\to} (F \star F) \star F \stackrel{m(m \star 1)}{\to} F$$

   commute. Here $i$ denotes the inclusion complementary to $\alpha: F \star (F \star F) \to (F \star F) \star F$, and $\sigma$ is the involution

   $$\sigma_1 \otimes \sigma_2: F'' \otimes F^{\otimes 2} \to F'' \otimes F^{\otimes 2}$$

   where $\sigma: F'' \to F''$ is the natural involution on the second derivative

   $$(-)'' = (- + 2)*: V^{FB} \to V^{FB}$$

   induced by the nonidentity involution $2 \to 2$, and $\sigma_2$ is the symmetry isomorphism $F^{\otimes 2} \to F^{\otimes 2}$.

3. (To be added)