when expressed in the language of the symplectic form $\omega$, the Jacobi identity turns out to be equivalent to its closedness
$$
{f,{g, h}}+{h,{f, g}}+{g,{h, f}}=0 \quad \Leftrightarrow \quad d \omega=0
$$
when expressed in the language of the Poisson tensor $\mathcal{P}$, the Jacobi identity turns out to be equivalent to its invariance with respect to an arbitrary Hamiltonian field ${ }^{250}$
$$
{f,{g, h}}+{h,{f, g}}+{g,{h, f}}=0 \quad \Leftrightarrow \quad \mathcal{L}{\zeta \uparrow} \mathcal{P}=0, \psi \in \mathcal{F}(M) $$ and in components this gives the differential identity $$ \mathcal{L}{\zeta_{f}} \mathcal{P}=0 \text { for all } f \in \mathcal{F}(M) \quad \Leftrightarrow \quad \mathcal{P}^{r[i} \mathcal{P}_{, \gamma}^{j k]}=0
$$
So this is the identity (mentioned above) which each Poisson tensor is to satisfy.
MATH3968 COURSE NOTES ：
The coordinates $\left(q^{a}, p_{a}\right)$, in which the symplectic form $\omega$ looks like this and whose existence is guaranteed by the Darboux theorem, are called the canonical coordinates in canonical coordinates the corresponding Poisson tensor is
$$
\mathcal{P} \equiv-\omega^{-1}=\frac{\partial}{\partial p_{a}} \wedge \frac{\partial}{\partial q^{a}}:=\frac{\partial}{\partial p_{a}} \otimes \frac{\partial}{\partial q^{a}}-\frac{\partial}{\partial q^{a}} \otimes \frac{\partial}{\partial p_{a}}
$$
in canonical coordinates the Hamiltonian field is
$$
\zeta_{f}=\left(\partial_{j} f\right) \mathcal{P}^{j /} \partial_{i} \equiv \frac{\partial f}{\partial p_{a}} \frac{\partial}{\partial q^{a}}-\frac{\partial f}{\partial q^{a}} \frac{\partial}{\partial p_{a}}
$$