A vector space associated with a vector field on a symplectic manifold

Question

$\DeclareMathOperator\Div{Div}$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.

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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".

Answer 1

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".