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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?

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  • $\begingroup$ by "d-closed" are you referring to $\mathrm{d} \omega=0$? $\endgroup$
    – user284001
    Sep 29 at 11:24
  • $\begingroup$ Yes, I am considering about the meaning of the definition of symplectic form $d \omega=0$. But I notice that I forgot to write the non-degenerate in the definition. $\endgroup$ Sep 29 at 11:34
  • $\begingroup$ What does it mean that a symplectic structure is "integrable"? A symplectic structure according to the most common definition (en.wikipedia.org/wiki/Symplectic_manifold) is always given by a closed nondegenerate 2-form...On the other hand, integrability in Frobenius sense is something that makes sense for 1-forms only (en.wikipedia.org/wiki/…) $\endgroup$
    – Giulio
    Sep 29 at 11:46
  • $\begingroup$ I think that the notion of integrability of symplectic manifold is $d \omega =0$and it also has a property about integrable distributions mentioned in Frobenius theorem like a NEWLANDER-NIRENBERG THEOREM which is actually involutive condition of complex geometry. Is it wrong? $\endgroup$ Sep 29 at 13:39
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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:

  1. $\omega$ 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;
  2. $\omega$ 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.

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  • $\begingroup$ Is there any other interpretation about integrability of symplectic structures? I think the flows of symplectic vector fields induce symplectic isomorphisms so $d \omega=0$ should has some integrability role in symplectic geometry. $\endgroup$ Sep 29 at 13:48
  • $\begingroup$ Yes, there are, for example one can define 'integrability' in Symplectic manifolds since they are very closely related to complex manifolds. A weak hint of this fact is that they both have to be even dimensional! In general , symplectic manifolds will have almost complex structures, and the vanishing of the Nijenhius tensor will give the integrability of the almost complex structure. $\endgroup$
    – user284001
    Sep 29 at 13:49
  • $\begingroup$ In fact, I knew NEWLANDER-NIRENBERG THEOREM which indicates equivalence between involutive and integrability conditions. I want to understand symplectic structures in the context of the smooth distributions. I think it must be related to the d-closedness. $\endgroup$ Sep 29 at 14:55
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The two defining conditions of a symplectic manifold $(M,\omega)$ are

  1. $\omega$ is non-degenerate, and
  2. $\omega$ 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.

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  • $\begingroup$ Thanks. Do you know the reasons about why Hamiltonian vector fields are more "natural" than symplectic vector fields in the context of the integrability. I still feel symplectic vector fields should have integrability related to $d \omega=0$ than Hemiltonian vector fields since symplectic isomorphism $\varphi_t \colon M \rightarrow M$ induce symplectic vector fields $\frac{d}{d t}\varphi_t = X$. $\endgroup$ Sep 30 at 6:35
  • $\begingroup$ I hardly see from geometry whether Hamiltonian vector fields are more natural than symplectic ones, because at least locally they are the same thing (indeed, thanks to $d\omega =0$). The major relevance of Hamiltonian vector fields stems from Classical Mechanics (in particular you may look at relations between Liouville theorem on integrable systems and Frobenius theorem on distributions). The condition $d\omega = 0$ also plays a major role in the Darboux theorem (en.wikipedia.org/wiki/Darboux%27s_theorem) which is widely regarded as a generalization of Frobenius theorem. $\endgroup$
    – Giulio
    Sep 30 at 7:22

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