Related to 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

- The most common bracket system is the Poisson bracket system:
$$ \{f, g\} + \{g, f\} = 0$$

$$ \{ \{f, g\}, h\} + \{ \{g, h\}, f\} + \{ \{ h, f\}, g\} = 0$$

- A Riemannian metric can also be turned into a bracket, which has only the equation
$$ \lfloor f, g \rfloor - \lfloor g, f \rfloor = 0$$

- 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):
$$ \sum_{\sigma \in \mathfrak{S}_n} sgn(\sigma) \prod_{i = 1}^n \{f_i, g_{\sigma(i)} \} = 0$$

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.

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.

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)?

Question 3: Are there any other properties that seem natural to describe bracket systems?

Question 4: As linked, this was originally about "degeneracy-generalizations"; is there an obvious way to fit in concepts of "nondegeneracy"?

Jacobi-Jordanalgebra as seen in the previous link. I am not familiar with the literature enough to answer your question fully, but I did find it interesting that there is a name for that structure. $\endgroup$ – Victoria M Mar 28 '19 at 17:26inner(or they are "Hamiltonian vector fields"); this resembles Poisson cohomology in degree 1, but is not quite the same. $\endgroup$ – Ricardo Buring Apr 2 '19 at 9:20