A vector space associated with a vector field on a symplectic manifold $\DeclareMathOperator\Div{Div}$Edit:  The  correct formulation of the  vector  space   $S(X)$ which is  defined in this  question is the  following:$$S(X)=\{Y\in \chi^{\infty}(M)\mid  X.\omega(X,Y)=(1/n)\Div(X)\omega(X,Y)\}.$$ This  mistake (typos)had  been occurred in remark 6, page 7 of Taghavi - On periodic solutions of Liénard equations.

Let  $(M,\omega)$  be a $2n$ dimensional symplectic  manifold and  $X$  a smooth vector  field on $M$. Consider the following subvector  space of $\chi^{\infty}(M)$: $$S(X)=\{Y\in \chi^{\infty}(M)\mid  X.\omega(X,Y)=n\Div(X)\omega(X,Y)\}.$$ Here $\Div$ is the  divergence  corresponding to the  volume 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 $M(X)=\{fX\mid f\in C^{\infty}(M)\}$.
Note that, according to the above definition of $S(X)$,  the inclusion $C(X)\subset S(X)$  sensitively depends on the scalar $n$.  If we replace $n$ by another scalar, this inclusion is no longer true. (Nevertheless the inclusion $M(X)\subset S(X)$ is not sensitive to this scalar, that is, it is valid  for every other scalar.)
Questions:

What other interesting Lie  algebras are  contained in $S(X)$?
Is $S(X)$ a Lie subalgebra of $\chi^{\infty}(M)$? If the answer is yes, what are some interesting ideals of $S(X)$?
  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 usual dynamical  question "Is the triviality of  centralizer a  generic situation?",  we  ask: Is it true to say that for  a generic  vector  field $X$  we have $S(X)=M(X)$?

Note: At the international workshop on dynamical system in ICTP, Italy, 2001, I heard from a specialist of dynamical system that "the centralizer problem has various aspects both in discrete and continuous dynamics, but I think that the symplectic version of this problem is interesting and unknown". So this my post is  a try for a possible symplectization of "centralizer problem".
 A: For  $n>1$, and the  standard  symplectic  structure  $\omega=\sum dx_i\wedge dy_i$ of $\mathbb{R}^{2n}=\{(x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n)\}$  and for the  vector  field $X=\partial/\partial_{x_1}$ it  is   easy to observe  that the following  vector  space  is  not a  Lie  algebra, since $Div(\partial/\partial_{x_1})=0$
$$S_{\lambda}(X)=\left\{Y\in \chi^{\infty}(\mathbb{R}^{2n})\mid X.\omega(X,Y)=\lambda Div(X)\omega(X,Y)\right \}$$
But for  $n=1$ and $\lambda=1$  it is always a  Lie  algebra.  In fact we have  the  following  obvious  fact:
Obvious Fact: Let  $(M,\omega)$  be  a  $2$-  dimensional symplectic  manifold(i.e: $\omega$ is  a  volume form on $M$)  and  $X$ is a  vector  field  on $M$. Then the    vector  space $$S(X)=\left\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)= Div(X)\omega(X,Y)\right \}$$ is  a  Lie  algebra. Moreover it   contains the  centralizer $C(X) $
Proof:    We apply the  well known formula  $$d\alpha(X,Y)=X.\alpha(Y)-Y.\alpha(X)-\alpha([X,Y])$$ to $\alpha=i_X(\omega)$.
So  we  conclude that the $S(X)$ in the  Obvious Fact is equal to $\{Y\in \chi^{\infty}(M)\mid \omega(X,[X,Y])=0\}$. The later is obviously a Lie  algebra containing the centralizer $C(X).$
Remark: For a  symplectic  manifold  $N$ of  arbitrary dimension $2n$ it can be  shown that the centralizer $C(X)$ of a  vector field $X$ is  contained in the following vector  space:
$$\left\{Y\in \chi^{\infty}(N)\mid X.\omega(X,Y)=(1/n) Div(X)\omega(X,Y)\right \}$$
So in the question of this  post one  should replace $n$ by $1/n$.
Proof  of  Remark:
Assume that $[X,Y]=0$. We prove that $X.\omega(X,Y)=(1/n)Div X\omega(X,Y)$. But we need only to prove this   formula  at all points $p\in N$ with $\omega(X(p),Y(p))\neq 0$. For  any  such  a point $p$, there exist locally a $2$  dimensional symplectic manifold  $M$ containing $p$ such that $X,Y$ are tangent to $M$. Now  we apply the  Obvious Fact  above to $M$. We have  $X.\omega(X,Y)=Div_{\omega}X.\omega(X,Y) $, where $Div_{\omega} X$ is  the  divergence of $X$ as  a  vector  field  on $M$ with the  volume form $\omega$. On the  other hand  $Div X=(1/n)Div_{\omega} X$ where $Div X$ is  the  divergence of  $X$ as a vector  field  on the  whole  manifold  $N$ with volume form $\omega^n$.This  completes  the  proof  of  "Remark".
