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)



"Left Brace Module"
$\newcommand\P{\mathtt{P}}\newcommand{\M}{\mathtt{M}}$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*.


