Integrability of Symplectic structures A symplectic structure on even dimensional manifold is a non-degenerate closed two form and I understood integrability of symplectic structure is closedness as a differential 2-form which comes from involutivity of symplectic vector fields by Frobenius theorem. However, in my calculation of Lie derivative of semi symplectic form $ \omega$ which means merely non-degenerate 2-form with Lie bracket $[X, Y]$ of two symplectic vector fields $X$ and $Y$ is zero without d closed condition.
My question is that did I misunderstand of the notion of integrability of symplectic structures in the sense of Frobenius, or mistake in the following calculation?
Assume that $0=\mathcal{L}_X\omega, \ 0= \mathcal{L}_Y\omega$, since $X$ and $Y$ are symplectic.
We now compute $\mathcal{L}_{[X, Y]}\omega$ using a formula $\mathcal{L}_{[X, Y]}=\mathcal{L}_X \mathcal{L}_Y -\mathcal{L}_Y\mathcal{L}_X$.
\begin{align}
\mathcal{L}_{[X, Y]}\omega & = (\mathcal{L}_X \mathcal{L}_Y -\mathcal{L}_Y\mathcal{L}_X)\omega \\
& = 0. \\
\end{align}
Now $\mathcal{L}_{[X, Y]}\omega$ vanished, it implies that $[X, Y]$ is also symplectic without using d-closed condition.
How should I use d-closed condition to confirm integrability of symplectic structures?
 A: So the notion of Integrability on a Symplectic manifold is by definition the following:
A symplectic structure on a manifold M is a differential $2$-form $\omega$ satisfying two conditions:

*

*$\omega$ is non-degenerate, i.e. for each $p \in M$ and tangent vector $\tilde{u}$ based at $p$, if $\omega_p(\tilde{u},\tilde{v}) = 0$ for all tangent vectors $\tilde{v}$ based at $p$, then $\tilde{u}$ is the zero vector;

*$\omega$ is closed, i.e. the exterior derivative of $\omega$ is zero, i.e. $\mathrm{d}\omega = 0$.

Condition (2) is often noted as the integrability condition.
A: The two defining conditions of a symplectic manifold $(M,\omega)$ are

*

*$\omega$ is non-degenerate, and

*$\omega$ is closed.

The first one ensures that $X\mapsto i_X\omega$ is an isomorphism $T_pM\to T^*_pM$ for all $p\in M$. Therefore for any smooth function $f:M\to \mathbb R$ we have one and only one vector field, denoted $X_f$ and called Hamiltonian vector field, such that $i_{X_f}\omega=d f$. The second condition then ensures the following two things.

*

*The Hamiltonian vector fields are symplectic. Indeed: $$Lie_{X_f}\omega = i_{X_f}d\omega+\underbrace{d i_{X_f}\omega}_{d^2f=0}= i_{X_f}d\omega$$
so that $d\omega =0$ makes also the remaining term vanish. Note also that conversely, if $Lie_X\omega=0$ then $d\omega=0$ ensures that $d i_X \omega = 0$ and therefore, at least locally, you can "integrate" $i_X\omega=d f$, i.e. you can find an Hamiltonian function $f$ such that $X=X_f$.


*(Maybe more related to "integrability in the sense of Frobenius" or "involutivity of vector fields".) The Hamiltonian vector fields form a Lie subalgebra in the Lie algebra of all vector fields, or in other terms, the commutator of two Hamiltonian vector fields is an Hamiltonian vector field. To see it:
$$
di_{[X_f,X_g]}\omega =Lie_{[X_f,X_g]}\omega-i_{[X_f,X_g]}d\omega
$$
and if $d\omega=0$ the second term vanishes and the first term vanishes by the previous point and the computation you have done in the statement of the question.
