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)

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)

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.

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)