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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

  1. The most common bracket system is the Poisson bracket system:

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

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

  1. A Riemannian metric can also be turned into a bracket, which has only the equation

$$ \lfloor f, g \rfloor - \lfloor g, f \rfloor = 0$$

  1. 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"?

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  • $\begingroup$ This article may be of interest to you: homepage.univie.ac.at/Dietrich.Burde/papers/… . The properties of the bracket system in example 1 are the same identities in the definition of a Jacobi-Jordan algebra 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 at 17:26
  • $\begingroup$ @VictoriaM While the article is interesting - thank you! - I don't think it addresses the major points I'm aiming for. The brackets are meant to be additional structure on top of a multiplication structure, and are supposed to be Leibniz with that multiplication. On a side note, the identities above are those of a Lie algebra (as said on the second page: "In the case of Lie algebras, the product is not commutative but anti-commutative.") or a Poisson algebra (if you make it Leibniz). This is meant to have a geometric focus, and that paper seems to be purely algebraic. Thanks for the idea! $\endgroup$ – user44191 Mar 29 at 2:21
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    $\begingroup$ Probably you can formalize it in terms of operads (see e.g. the Poisson operad here). The "Lie-complete" condition in the Poisson case can be expressed by saying that some derivations (those defined in terms of brackets) are inner (or they are "Hamiltonian vector fields"); this resembles Poisson cohomology in degree 1, but is not quite the same. $\endgroup$ – Ricardo Buring Apr 2 at 9:20
  • $\begingroup$ @RicardoBuring I think product-completeness should match up to Hopf. So question 2 comes down to something like "Is there a system of Leibniz, Hopf operads such that all derivations are inner?", right? (Leibniz here meaning following the Leibniz rule in all coordinates) $\endgroup$ – user44191 Apr 2 at 16:01
  • $\begingroup$ @RicardoBuring You're the closest to having an answer (having certainly answered one of the questions), so I'd like to award you the bounty if no one comes up with anything else in the meantime. Would you mind writing an answer expanding on (or possibly repeating) your comment? $\endgroup$ – user44191 Apr 4 at 0:53
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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).

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    $\begingroup$ You make a good point I hadn't considered for what I mean by Lie-complete; I think what I was really looking for was for any derivation to be expressible with only one bracket around the relevant function ($h$ in your example). This should be possible with the Poisson bracket because any derivation should be possible to simplify with the Jacobi identity. $\endgroup$ – user44191 Apr 4 at 15:09
  • $\begingroup$ That may be true, but I don't know how to prove it. $\endgroup$ – Ricardo Buring Apr 6 at 12:58
  • $\begingroup$ Note $f_1\{g_1,h\} + f_2\{g_2,h\}$ doesn't simplify in general, so you need to allow sums. $\endgroup$ – Ricardo Buring Apr 6 at 14:52

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