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Let $(M,\omega)$ be a quasi-symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the pseudo-Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediately by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

We introduce also a sort of Jacobiator $J:C^{\infty}(M)\times C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ by defining $J(f,g,h)=\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. This is a trilinear antisymmetric map which measure how much the Jacobi identity for the bracket is not satisfied.

I outline two different approaches.

1)$d\omega=0$ implies the Jacobi identity for $\{\cdot,\cdot\}$.

The Jacobi identity for the pseudo-Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\textrm{Lie}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

2)$(d\omega)(X_f,X_g,X_h)=J(f,g,h)$, for any $f$,$g$,and $h$ smooth functions on $M$.

By Palais' expression of exterior derivative through Lie derivatives, we can express $(d\omega)(X_f,X_g,X_h)$ as the sum of two terms obtained respectively by $\mathcal{L}(X_f)(\omega(X_g,X_h)$ and by $\omega(X_f,[X_g,X_h])$ summing over the ciclic permutations of $(f,g,h)$.

Now $\mathcal{L}(X_f)(\omega(X_g,X_h)=-\{f,\{g,h\}\}$ and $\omega(X_f,[X_g,X_h])=\{g,\{h,f\}\}+\{h,\{f,g\}\}$, and so we get the thesis.

Let $(M,\omega)$ be a quasi-symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\mathcal{X}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\mathcal{X}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the pseudo-Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediately by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

We introduce also a sort of Jacobiator $J:C^{\infty}(M)\times C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ by defining $J(f,g,h)=\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. This is a trilinear antisymmetric map which measure how much the Jacobi identity for the bracket is not satisfied.

I outline two different approaches.

1)$d\omega=0$ implies the Jacobi identity for $\{\cdot,\cdot\}$.

The Jacobi identity for the pseudo-Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\mathcal{X}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

2)$(d\omega)(X_f,X_g,X_h)=J(f,g,h)$, for any $f$,$g$,and $h$ smooth functions on $M$.

By Palais' expression of exterior derivative through Lie derivatives, we can express $(d\omega)(X_f,X_g,X_h)$ as the sum of two terms obtained respectively by $\mathcal{L}(X_f)(\omega(X_g,X_h)$ and by $\omega(X_f,[X_g,X_h])$ summing over the ciclic permutations of $(f,g,h)$.

Now $\mathcal{L}(X_f)(\omega(X_g,X_h)=-\{f,\{g,h\}\}$ and $\omega(X_f,[X_g,X_h])=\{g,\{h,f\}\}+\{h,\{f,g\}\}$, and so we get the thesis.

Let $(M,\omega)$ be a symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediatly by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

The Jacobi identity for the Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\textrm{Lie}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

Probably, at this time and after Jose's answer, my answer could be not useful for you. But anyway I tried to find a coordinate-free proof of the Jacobi identity for the Poisson bracket over a symplectic manifold, in order to test my own comprehension. And now I try to write it here as an exercise in exposition and hoping to be useful to casual readers.

Let $(M,\omega)$ be a quasi-symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the pseudo-Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediately by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

We introduce also a sort of Jacobiator $J:C^{\infty}(M)\times C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ by defining $J(f,g,h)=\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. This is a trilinear antisymmetric map which measure how much the Jacobi identity for the bracket is not satisfied.

I outline two different approaches.

1)$d\omega=0$ implies the Jacobi identity for $\{\cdot,\cdot\}$.

The Jacobi identity for the pseudo-Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\textrm{Lie}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

2)$(d\omega)(X_f,X_g,X_h)=J(f,g,h)$, for any $f$,$g$,and $h$ smooth functions on $M$.

By Palais' expression of exterior derivative through Lie derivatives, we can express $(d\omega)(X_f,X_g,X_h)$ as the sum of two terms obtained respectively by $\mathcal{L}(X_f)(\omega(X_g,X_h)$ and by $\omega(X_f,[X_g,X_h])$ summing over the ciclic permutations of $(f,g,h)$.

Now $\mathcal{L}(X_f)(\omega(X_g,X_h)=-\{f,\{g,h\}\}$ and $\omega(X_f,[X_g,X_h])=\{g,\{h,f\}\}+\{h,\{f,g\}\}$, and so we get the thesis.

Let $(M,\omega)$ be a symplectic manifold.

Being $\omega$ a non-degenerate $2-\textrm{form}$ on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}-\textrm{linear}$.

We introduce the Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediatly by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}-\textrm{bilinear}$ map.

The Jacobi identity for the Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in\textrm{Lie}(M)$ is a homomorphism of $\mathbb{R}-\textrm{algebras}$.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement: if $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function. Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$. (Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

Probably, at this time and after Jose's answer, my answer could be not useful for you. But anyway I tried to find a coordinate-free proof of the Jacobi identity for the Poisson bracket over a symplectic manifold, in order to test my own comprehension. And now I try to write it here as an exercise in exposition and hoping to be useful to casual readers.

Let $(M,\omega)$ be a symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediatly by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

The Jacobi identity for the Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\textrm{Lie}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

Probably, at this time and after Jose's answer, my answer could be not useful for you. But anyway I tried to find a coordinate-free proof of the Jacobi identity for the Poisson bracket over a symplectic manifold, in order to test my own comprehension. And now I try to write it here as an exercise in exposition and hoping to be useful to casual readers.