Let  $(M,\omega)$  be a $2n$ dimensional symplectic  manifold and  $X$ is  a smooth vector  field on $M$. Consider the following subvector  space of $\chi^{\infty}(M)$: $$S(X)=\{Y\in \chi^{\infty}(M)\mid  X.\omega(X,Y)=nDiv(X)\omega(X,Y)\}$$.Here $Div$ is the  divergence  correspond to the  volum form $\omega^{n}$


This  vector  space  contains the Lie algebra $C(X)=\{Y\in \chi^{\infty}(M)\mid [X,Y]=0\}$.  It also contains the  Lie  algebra $M(X)=\{fX\mid f\in C^{\infty}(M)\}$

>Is $S(X)$ a Li subalgebra of $\chi^{\infty}(M)$?

> If the  answer is no, is the Lie  algebra  generated by $S(X)$ equal to the lie  algebra generated by $C(X)$  and $M(X)$?

>Motivated by the fact that "triviality of  centralizer of  a vector  field is somehow a  generic  situation"  is it true to say that for  a generic  vector  field $X$  we have $S(X)=M(X)$?