# 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".

• If there is an error in the arXiv paper, then why not update to fix it? – LSpice Sep 17 '18 at 16:00
• @LSpice For the moment I do not access to the Latex file of that paper but I can write a letter to the journal which published that paper and inform them of this error. Any way I thank you for your attention to my post and for your suggestion. – Ali Taghavi Sep 17 '18 at 21:00
• You can download the source from the arXiv: arxiv.org/e-print/math/0409594 . (At least for me, it comes without an extension; but it is a tar'd gzip'd file, and can be opened as such.) – LSpice Sep 17 '18 at 23:30

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".