Involutivity of $TS^{\perp} \subseteq TS$, where $S$ is a coisotropic submanifold of a symplectic manifold $(M,\omega)$

Question

Let $(M, \omega)$ be a symplectic manifold and $S$ be a coisotropic submanifold of $M$.

$TS^{\perp}$ is defined by $$TS^{\perp} = \bigcup_{p \in S} (T_p S)^{\perp}.$$ Consider a bundle homomorphism $\widehat {\iota^* \omega} : TS \rightarrow T^*S$ defined by $\widehat {\iota^* \omega}(X) =\iota^* \omega (X, \cdot)$, where $\iota : S \hookrightarrow M$.

Then $\text{ker} \widehat {\iota^* \omega} = TS^{\perp}$ and $TS^{\perp}$ is a smooth subbundle of $TS$ since $\widehat {\iota^* \omega}$ is of constant rank.

I am expecting that $TS^{\perp}$ is involutive.

To check this , let $X, Y$ be two local sections of $TS^{\perp}$.

I want to show $[X,Y]$ is also a local section of $TS^{\perp}$.

I am stuck on this point.

How can I solve this. Thanks.

Answer 1

You could use the formula $[\iota_X, \mathcal L_Y]\omega = \iota_{[X,Y]}\omega$. By the assumption that $\iota_X \omega = 0$, the left side is $\iota_X \mathcal L_Y\omega$; expanding $\mathcal L_Y \omega$ using Cartan's magic formula, we obtain $\iota_X \iota_Y d\omega + \iota_X d\iota_Y\omega = 0$, because $d\omega = 0$ and $\iota_Y \omega = 0$. Thus $\iota_{[X,Y]}\omega = 0$ as well.