Let $A$ be an algebra over the brace tree operad and $M$ a module over some base ring. Is there a good notion of a "left brace module" over a brace algebra? I do not think the definition of a module over an algebra over an operad is the definition I'm looking for.

It should take any brace tree with white vertices labelled 1,...,k and send it to a linear map (thinking of the first k-1 labels as corresponding to $A$ and the last label corresponding to $M$.) $\psi_k^T:A^{\otimes k-1} \otimes M \to M$

and agree with the composition of the algebra over an operad structure of $A$.

For example, the linear tree corresponding to the sequence 121 should be a map $\psi_2^T:A\otimes M\to M$ which is a homotopy between the left action 12 and the right action 21. (ie. $M$ is a bi-module where the left and right action are homotopic)

...in the sense that

$$ [d,\psi_2^T](a\otimes m)= a\cdot m-m\cdot a $$

Note: brace trees and sequences are in bijective correspondence; for any brace tree we can produce a sequence by "following the tree clockwise" and recording the vertices we pass. The tree can be recovered from this sequence.

Comment: Is there a general (good, working) definition of a left module over an algebra over an operad? I have always been told that the "correct" definition of a module in the context of operads is the two-sided version. (ie. a module over an algebra over the $A_\infty$ operad is automatically meant to mean a bi-module)

  • $\begingroup$ The usual notion of a module over an algebra over an operad sounds pretty much what you're looking for. Any particular reason to think that it's not? For example, did you check if the action of $\psi^{T}_2$ in this case coincides with what you want? $\endgroup$ Apr 6, 2019 at 19:36

1 Answer 1


$\newcommand\P{\mathtt{P}}\newcommand{\M}{\mathtt{M}}$In general, the structure of a "module over an algebra over an operad" (a mouthful) is encoded by a moperad (module + operad).

If $\P = \{\P(n)\}_{n \ge 0}$ is an operad, then a $\P$-moperad is a monoid in the category of right $\P$-modules. Concretely, this is a symmetric sequence $\M = \{\M(n)\}_{n \ge 0}$ equipped with two kinds of structure maps: $$\circ : \M(n) \otimes \M(n') \to \M(n+n')$$ $$\circ_i : \M(n) \otimes \P(n') \to \M(n+n'-1)$$ satisfying appropriate unitality/associativity/equivariance conditions. The canonical example of a $\P$-moperad is the shifted module $\P[1] = \{\P(n+1)\}_{n \ge 0}$. The product $\circ : \P(n+1) \otimes \P(n'+1) \to \P(n+n'+1)$ is given by inserting at the first coordinate, while the right module structure $\circ_i : \P(n+1) \otimes \P(n') \to \P(n+n'-1+1)$ is really $\circ_{i+1}$. You can view a $\P$-moperad as a special kind of bicolored operad: for outputs in the first color you have $\P$; for outputs in the second color, you have $\M$ if there is exactly one input of the second color, and nothing otherwise.

Now suppose that $A$ is a $\P$-algebra. An $\M$-module over the $\P$-algebra $A$ is an object $N$ equipped with structure maps $\M(n) \otimes A^{\otimes n} \otimes N \to N$ satisfying the obvious axioms. (Draw trees.) The usual notion of "module over $A$" is obtained precisely if you set $\M = \P[1]$.

With the moperad technology, you can define "left" modules. For example, if $\newcommand{\Ass}{\mathtt{Ass}}\P = \Ass$ governs associative algebras, you can define the moperad $\Ass_L$ such that an $\Ass_L$-modules over $A$ is exactly a left $A$-module. If $\Ass(n) = \Sigma_n$, then $\Ass_L(n) \subset \Sigma_{n+1}$ is given by permutations fixing the last input. I believe you can do a similar thing for the braces operad and you get left modules.

A reference for all this would be Willwacher's paper The Homotopy Braces Formality Morphism. There is also Horel's paper Operads, Modules and Topological Field Theories.

  • $\begingroup$ How do you define left modules for algebras over general operads? $\endgroup$ Apr 6, 2019 at 19:32
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    $\begingroup$ @YonathanHarpaz You don't, actually. It's a case by case definition. You need a corresponding moperad. There is one for associative algebras, Lie algebras etc. But I don't see how it would go for general operads. In the general case, the only moperad is $\mathtt{P}[1]$ which gives standard modules over an algebra over an operad. $\endgroup$ Apr 7, 2019 at 2:14

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