Definition
Given a symplectic manifold $(M,\omega)$, the Poisson bracket on $C^\infty(M)$, the space of smooth functions on $M$, is a binary operation $\{\cdot, \cdot\} : C^\infty(M) \times C^\infty(M) \to C^\infty(M)$ defined as follows: for any two functions $f, g \in C^\infty(M)$, the Poisson bracket is given by
$$ \{f, g\}(p) = \omega(X_f, X_g)(p), $$where $X_f$ and $X_g$ are the Hamiltonian vector fields associated to $f$ and $g$ respectively, and $p$ is a point in $M$.
$\blacksquare$
In the canonical coordinates of a classical Hamiltonian system is written as
$$ \{f, g\}=\sum_{i=1}^N\left(\frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i}\right) $$The Poisson brackets of the canonical coordinates are
$$ \begin{array}{l} \left\{q_i, q_j\right\}=0 \\ \left\{p_i, p_j\right\}=0 \\ \left\{q_i, p_j\right\}=\delta_{i j} \end{array} $$which are reminiscent of canonical commutation relations.
The Poisson bracket induced by a symplectic form satisfies the following properties:
These properties make the Poisson bracket a Lie bracket, that is, it makes $(C^\infty(M), \{\cdot, \cdot\})$ into a Lie algebra. Also is a Poisson algebra. If we abstract these properties we can define a Poisson manifold.
For every $f$ we have that $\{-,f\}$ is a differential operator, so it is a vector field. Indeed, if the Poisson bracket is coming from a symplectic form, it is $\{-,f\}\equiv X_f$ since for any smooth function $g$
$$ \{g,f\}=\omega(X_g,X_f)=dg(X_f)=X_f(g) $$by the definition of the Hamiltonian vector field $X_g$.
On the other hand, observe that we can rewrite Jacobi identity in this way:
$$ \{f,\{g, h\}\}=\{\{h, f\}, g\}-\{h,\{f, g\}\} $$Leaving an slot instead of $f$ we get
$$ \{-,\{g,h\}\}=\{-X_h(-),g\}-\{h,X_g(-)\}= $$ $$ =-X_g(X_h(-))+X_h(X_g(-)) $$expression that can be rewritten as
$$ X_{\{g,h\}}=[X_h,X_g]=X_{\omega(X_g,X_h)} $$It can be shown (see @olver86 page 393 section "The structure functions") that in local coordinates $(x^1,\ldots,x^m)$ the Poisson bracket can be expressed:
$$ \{F, H\}=\sum_{i=1}^m \sum_{j=1}^m\left\{x^i, x^j\right\} \frac{\partial F}{\partial x^i} \frac{\partial H}{\partial x^j} $$The functions $\left\{x^i, x^j\right\}$ are called the structure functions of the Poisson bracket in this coordinates, and they can be arranged into a skew-symmetric matrix denoted $J(x)$. If we denote by $\nabla F$ the usual gradient we have that
$$ \{F,H\}=\nabla F^TJ \nabla H $$Example. In the natural example given in Poisson manifold the matrix is
$$ J=\left(\begin{array}{rrr} 0 & -I & 0 \\ I & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) $$in the $(p,q,z)$-coordinates.
$\blacksquare$
I think this matrix is a kind of degenerate symplectic form.
In this context, since
$$ X_H=\{-, H\}=\sum_{i=1}^m \sum_{j=1}^m J^{ij}(x) \frac{\partial H}{\partial x^j}\frac{\partial }{\partial x^i} $$Hamiltonian equations for a function $H$ are given by
$$ \frac{dx}{dt}=J(x)\nabla H(x). $$________________________________________
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Author of the notes: Antonio J. Pan-Collantes
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