# The Moyal action of a planar vector field

Let $X=P\frac{\partial}{\partial x}+Q\frac{\partial}{\partial y}$ be a polynomial vector field on $\mathbb{R}^{2}$. Consider the following (Moyal) operator on $\mathbb{C}[x,y]$:

$\tilde{D}_{X}(f)=f_{x}*P+f_{y}*Q$ or $\tilde{D}_{X}(f)=P*f_{x}+Q*f_{y}$ where $*$ is the Moyal product. The later has the potential of being an elliptic operator since it is of even order.

This is a noncommutative version of the standard derivation $D_{X}(f)=f_{x}P+f_{y}Q$:

I have two questions:

1) Motivated by the Leibniz formula for $D_{X}$, I ask that: what is a natural formula for $\tilde{D}_{X}(f*g)=?$

2)Can the number of limit cycles of $X$ be bounded by the codimension of the range of $\tilde{D}_{X}$?

In particular is the following statement, true?

• For a limit cycle $\gamma$ of a vector field $X$ on the plane and for an smooth function $f$ on $\mathbb{R}^{2}$, there is a point $p$ on $\gamma$ such that $\tilde{D}_{X}(f)(p)=0$.

A (commutative) motivation for the second question is the question in the following post

Note The * statement is a Moyal version of a key lemma to prove the commutative version of the second question. By commutative version I mean we replace the Moyal product by the usual product.

• Since the derivatives $\partial_x$ and $\partial_y$ are Moyal product derivations it is not too difficult to write down an explicit formula for 1). It will be a derivation formula plus correction terms taking into account $g\star P\ne P\star g$ (and the like for $Q$). – Nicola Ciccoli Jun 10 '18 at 9:09
• @NicolaCiccoli Thanks for your comment. I try to use your comment to derive the formula. – Ali Taghavi Jun 10 '18 at 13:18