# “Left Brace Module”

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)

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.