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I'm no expert on this but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

You may also be interested in the notion of algebra over an operad, more tangentially the notion of "convolution Lie algebra" associated to an operad, and operads defined by using graphs (as in graph complexes).

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But consider for example $f\{g,h\}$. It is a derivation with respect to $h$, but it is impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). (For example on $\mathbb{R}^2$ with $\{x,y\} = x$ it is impossible to solve $y\{x,-\} = \{Q,-\}$.) The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions (in general).

I'm no expert on operads but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

You may also be interested in the notion of algebra over an operad, more tangentially the notion of "convolution Lie algebra" associated to an operad, and operads defined by using graphs (as in graph complexes).

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But consider for example $f\{g,h\}$. It is a derivation with respect to $h$, but it is impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). (For example on $\mathbb{R}^2$ with $\{x,y\} = x$ it is impossible to solve $y\{x,-\} = \{Q,-\}$.) The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions (in general).

I'm no expert on this but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

You may also be interested in the notion of algebra over an operad, and more tangentially the notion of "convolution Lie algebra" associated to an operad.

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But consider for example $f\{g,h\}$. It is a derivation with respect to $h$, but it is impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions.

I'm no expert on this but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

You may also be interested in the notion of algebra over an operad, more tangentially the notion of "convolution Lie algebra" associated to an operad, and operads defined by using graphs (as in graph complexes).

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But consider for example $f\{g,h\}$. It is a derivation with respect to $h$, but it is impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). (For example on $\mathbb{R}^2$ with $\{x,y\} = x$ it is impossible to solve $y\{x,-\} = \{Q,-\}$.) The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions (in general).

I'm no expert on this but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But how about e.g. $f\{g,h\}$? It is a derivation with respect to $h$, but it seems impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions.

You may also be interested in the notion of algebra over an operad, and more tangentially the notion of "convolution Lie algebra" associated to an operad.

I'm no expert on this but it seems that

• a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here,

• the "product-complete" condition could be related to having a Hopf operad (see previous link),

• the "Lie-complete" condition says derivations are inner (I don't know how to phrase this using operads).

You may also be interested in the notion of algebra over an operad, and more tangentially the notion of "convolution Lie algebra" associated to an operad.

Note that in the Poisson case, a derivation being inner means that it is a Hamiltonian vector field. You claim that every derivation defined by a bracket expression (with numerical coefficients?) is inner. I understand the example with the Jacobi identity, but why should it be true in general? A necessary condition for a vector field to be Hamiltonian is to vanish on Casimirs of the Poisson structure; this is clearly satisfied. But consider for example $f\{g,h\}$. It is a derivation with respect to $h$, but it is impossible to solve $f\{g,h\} = \{Q,h\}$ for $Q$ in general (using only the axioms) if $f$ is not a Casimir (if $f$ is a Casmir then $Q = fg$ works). The space of Hamiltonian vector fields is a module over the space of Casimirs, not over the space of all functions.