In classical Hamiltonian mechanics evolution of any observable (scalar function on a manifold in hand) is given as
$$\frac{dA}{dt} = \{A,H\}+\frac{\partial A}{\partial t}$$
So Poisson bracket is a binary, skew-symmetric operation
$$\{f,g\} = – \{f,g\}$$
which is bilinear
$$\{\alpha f+ \beta g,h\} = \alpha \{f, g\}+ \beta \{g,h\}$$
satisfies Leibniz rule:
$$\{fg,h\} = f\{g,h\} + g\{f,h\}$$
and Jacobi identity:
$$\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0$$
How to physically interpret these properties in classical mechanics? What physical characteristic each of them is connected to? For example I suppose the anticommutativity has to do something with energy conservation since because of it $\{H,H\} = 0$.
Best Answer
Let's assume no explicit time dependence and that our Poisson bracket $\{,\}$ - I prefer curly brackets so square ones $[,]$ can be used to denote the commutator of vector fields - is non-singular, ie there's a corresponding symplectic product $\omega$.
The time derivative $$ \frac{\mathrm d}{\mathrm dt}=\{\,\cdot\,,H\} $$ is actually the Lie derivative with respect to the Hamiltonian vector field $X_H$ given by $$ X_H\rfloor\omega \equiv \mathrm dH $$ in disguise as can be seen by $$ \{f,H\} \equiv \omega(X_f,X_H)=(X_f\rfloor\omega)(X_H)=\mathrm df(X_H)=\mathcal{L}_{X_H}f $$ As $\mathcal{L}_{X_H}$ is a linear differential operator respecting the Leibniz rule, so is $\{\,\cdot\,,H\}$.
Antisymmetry translates to $$ \mathcal{L}_{X_f}g = -\mathcal{L}_{X_g}f $$ ie the change of $g$ with respect to the Hamiltonian flow induced by $f$ is the negative of the change in $f$ with respect to the Hamiltonian flow induced by $g$.
Rewriting the Jacobi identity as $$ \{f ,\{g,h\}\} = \{\{f,g\},h\} - \{\{f,h\},g\} $$ we see that $$ \mathcal{L}_{X_{\{g,h\}}}f=\left(\mathcal{L}_{X_h}\mathcal{L}_{X_g} - \mathcal{L}_{X_g}\mathcal{L}_{X_h}\right)f = \mathcal{L}_{[X_h,X_g]}f $$ ie $f\mapsto X_f$ is a Lie-algebra homomorphism.