Related to [Why symplectic geometry gives Poisson geometry](https://mathoverflow.net/questions/255034/why-symplectic-geometry-gives-poisson-geometry) by coming at it from the other side.


This isn't as fully formalized as it probably should be, but I think enough of the idea is there to ask the question.


Given an algebra $A$, a bracket on $A$ is a multilinear map $A^{\otimes n} \rightarrow A$ that is linear and Leibniz in each component. Geometrically, the idea is that a bracket is a way of generating vector fields on $\text{Spec}(A)$.


A bracket system is then a system of equations using (unspecified) brackets, rearrangement, and algebra operations. An algebra and a set of brackets satisfy a bracket system when the bracket system is true for any choice of elements of the algebra.



> 
> Examples
> 
> 
> 1. The most common bracket system is the Poisson bracket system:
> 
> 
> $$ \{f, g\} + \{g, f\} = 0$$
> 
> 
> $$ \{ \{f, g\}, h\} + \{ \{g, h\}, f\} + \{ \{ h, f\}, g\} = 0$$
> 
> 
> 2. A Riemannian metric can also be turned into a bracket, which has only the equation
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> $$ \lfloor f, g \rfloor - \lfloor g, f \rfloor = 0$$
> 
> 
> 3. Any bracket system can add a "dimension condition" on one (or any) input to a bracket, given by (using Poisson bracket notation, though this can work with any bracket):
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> $$ \sum\_{\sigma \in \mathfrak{S}\_n} sgn(\sigma) \prod\_{i = 1}^n \{f\_i, g\_{\sigma(i)} \} = 0$$
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> 
> Geometrically, the idea is that the vector fields generated should pointwise generate a vector space with dimension at most $n$.
> 
> 
> 


Call a bracket system "product-complete" if, whenever $(A, \{ \}\_A)$ and $(B, \{ \}\_B)$ satisfy the bracket system, then $(A \otimes B, \{ \}\_{A \otimes B})$ also satisfies it, where $\{ \}\_{A \otimes B}$ is defined to be $\{ \}\_A$ whenever all arguments can be expressed in the form $a \otimes 1$, as $\{ \}\_B$ whenever all arguments can be expressed in the form $1 \otimes b$, and as $0$ whenever arguments are "mixed" (i.e. when one argument can be expressed in the form $a \otimes 1$ and another in the form $1 \otimes b$).



> 
> Both the Poisson bracket and the Riemann bracket are product-complete.
> 
> 
> Either one with an added dimension condition, however, will not be product-complete. Geometrically, this is because the dimension of the vector space should add.
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> 
> 


This is the part I think might be harder to formalize: call a bracket system "Lie-complete" if, whenever the bracket system can be used to find a derivation, that derivation comes a bracket where the input for the derivation is explicitly an argument (with some expression using the other terms). Geometrically, the idea is that if a vector field can be calculated using a bracket system, then it should "explicitly already be" a bracket.



> 
> The Poisson bracket is Lie-complete. For example, the map $(f, g, h) \rightarrow \{\{f, g\}, h\} - \{f, \{g, h\}\}$ is a derivation on $g$, and the Jacobi identity shows that it is equal to $-\{\{h, f\}, g\}$, a bracket where $g$ is explicitly one of the arguments.
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> 
> The Riemann bracket is not Lie-complete. For example, the map $(f, g, h) \rightarrow \lfloor \lfloor f, g\rfloor, h\rfloor - \lfloor f, \lfloor g, h\rfloor \rfloor$ is a derivation on $g$, and there is no expression $M$ in terms of $f, h$ such that the above expression is equal to $\lfloor M(f, h), g\rfloor$.
> 
> 
> 


Finally for the questions.



> 
> Question 1: Is there a system to naturally formalize this idea?
> Answer: this is naturally formalized using operads.
> 
> 
> Question 2: The Poisson bracket system is Lie-complete and product-complete. Is there another Lie-complete and product-complete system that's not either 1) trivial (no brackets or only unary brackets), or 2) the Poisson bracket (with possible additional unary brackets)?
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> 
> Question 3: Are there any other properties that seem natural to describe bracket systems?
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> Question 4: As linked, this was originally about "degeneracy-generalizations"; is there an obvious way to fit in concepts of "nondegeneracy"?
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> 
> 



Bracket systems (generalization of Poisson brackets)
I'm no expert on operads but it seems that


* a "bracket system" can be formalized as an [operad](https://en.wikipedia.org/wiki/Operad); see e.g. [the Poisson operad here](https://mathoverflow.net/a/261448/61197),
* 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](https://ncatlab.org/nlab/show/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).


