classical mechanics - Is there any relation between Poisson Brackets and the Jacobian Matrix?

The Poisson brackets for $u,v$ can be written as,

$$ \frac{\partial u}{\partial q} \frac{\partial v}{\partial p} - \frac{\partial u}{\partial p}\frac{\partial v}{\partial q}. $$

We can write this as determinant of this matrix

$$ \begin{bmatrix} \frac{\partial u}{\partial q} & \frac{\partial u}{\partial p} \\\frac{\partial v}{\partial q} & \frac{\partial v}{\partial p} \end{bmatrix} $$

Which is the Jacobian matrix.

Is there any relation between them?

Answer

1. 
   

   For a 2-dimensional phase space, they are the same.

   
   


2. 
   

   More generally, for a $2n$-dimensional symplectic manifold $(M;\omega)$ with symplectic two-form $$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI}, $$ the Poisson bracket is given by $$ \tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} \frac{\partial g}{\partial z^J}, \qquad \pi^{IJ} ~:=~(\omega^{-1})^{IJ} . $$ See also e.g. this and this Phys.SE posts.

   
   


3. 
   

   The canonical volume form on $(M;\omega)$ $$\tag{3} \Omega~=~\omega^n~=~\rho~ dz^1\wedge \ldots \wedge dz^{2n},$$ is the $n$'th exterior power of the symplectic two-form $\omega$, with volume density $$\tag{4} \rho~=~{\rm Pf}(\omega_{IJ}) $$ given by the Pfaffian, which is a square root of the determinant. This is closely related to Liouville's theorem.