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Let $H_1,H_2\in\mathcal{C}^1(\mathbb{R}^3;\mathbb{R})$ be two scalar fields. Consider a trajectory $\vec{x}(t)\in\mathbb{R}^3$ such that, for all observable $f\in\mathcal{C}^1(\mathbb{R}^3;\mathbb{R})$,

$$\frac{\mathrm{d}}{\mathrm{d}t}f(x)=\det\big(\nabla f,\nabla H_1, \nabla H_2\big)=\frac{\partial(f,H_1,H_2)}{\partial\vec{x}}.$$

This dynamical system recalls a Hamiltonian system with hamiltonian $H$ on the phase space $\lbrace(x,p)\in\mathbb{R}^2\rbrace$ such that for all observable $f\in\mathcal{C}^1(\mathbb{R}^2;\mathbb{R})$:

$$\frac{\mathrm{d}}{\mathrm{d}t}f(x)=\det\big(\nabla f,\nabla H\big)=\frac{\partial(f,H)}{\partial(x,p)}=\frac{\partial f}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial x}=\big\lbrace f,H\big\rbrace,$$

the Poisson bracket. Hence I would like to say that my dynamical system is a kind of "multi-hamiltonian" system. Is there any reference in which this kind of generalisation is studied?

Edit: it can be generalised to a system with $d-1$ scalar fields $(H_i)$ on $\mathbb{R}^d$ satisfying:

$$\frac{\mathrm{d}}{\mathrm{d}t}f(x)=\det\big(\nabla f,\nabla H_1,... \nabla H_{d-1}\big)=\frac{\partial(f,H_1,...,H_{d-1})}{\partial\vec{x}}.$$

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  • $\begingroup$ How does your proposal behave under changes of coordinates? I ask to see if we can figure out if there is any further geometric meaning to it. $\endgroup$ Nov 16, 2020 at 1:12
  • $\begingroup$ A set of new coordinates $\vec{y}\in\mathbb{R}^3$ satisfying $\partial\vec{y}/\partial \vec{x}=1$ preserves the equation of evolution (due to the fact that $\partial(f,H_1,H_2)/\partial\vec{x}=\partial(f,H_1,H_2)/\partial{\vec{y}}\times\partial\vec{y}/\partial\vec{x}$). This is analogous to a conanical change of coordinates for a usual Hamiltonian system. $\endgroup$
    – G. Panel
    Nov 16, 2020 at 1:32
  • $\begingroup$ In particular, the orders of the $x$ dimensions seems tied to the orders of the $H$'s, so it might be better to think of it as something like $(x, p_{1}, p_{2})$ rather than $\vec x$ $\endgroup$ Nov 16, 2020 at 2:07
  • $\begingroup$ My first question would be, " can you reduce this down to a normal Hamiltonian system by eliminating the third variable"? $\endgroup$ Nov 16, 2020 at 2:09
  • $\begingroup$ If we assume that $\partial(x_1,x_2,H_2)/\partial\vec{x}=1$, then $\partial(f,H_1,H_2)/\partial\vec{x}=\partial(f,H_1,H_2)/\partial(x_1,x_2,H_2)=\partial(f,H_1)/\partial(x_1,x_2)$ (after development toward the third column), and the system is then hamilonian in this specific case. $\endgroup$
    – G. Panel
    Nov 16, 2020 at 3:02

1 Answer 1

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Answered in PhysicsOverflow; reposting here for convenience:

Intriguingly enough, a quick search for "multi Hamiltonian physics" does not give any meaningful result; the mechanics described by the OP is Nambu mechanics, and is linked to nonassociative algebras appearing in e.g. M-theory.

In contrast to the construction by the OP, Nambu generalizes the Poisson bracket (rather than using a higher-dimensional matrix determinant) and writes the equations of motion $$\frac{df}{dt}=\{f,H_1,H_2,…,H_n\}$$

See https://arxiv.org/abs/hep-th/0212267 | https://arxiv.org/abs/1903.05673

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