# A curvature description for center condition for quadratic vector field

We consider the quadratic vector field $V$ $$\begin{cases} x'=P(x,y)\\ y'=Q(x,y) \end {cases}\;\;\;\;(V)$$

where $P,Q \in \mathbb{R}[x,y]$ are polynomials of degree $2$ with $P(0,0)=Q(0,0)=0$.

We denote by $(W)$ the linear radial vector field $$W=x\partial_x+y\partial_y\;\;\;\;(W)$$

Consider the $1\_$form $$\psi=\frac{1}{x^2+y^2}(ydx-xdy)$$

Let $C$ be the algebraic curve $C=\{(x,y)\mid yP(x,y)-xQ(x,y)=0\}$

We consider the Riemannian metric on $\mathbb{R}^2 \setminus C$ whose orthonormal frame is the following:$$\{V/\psi(V), W/(x^2+y^2)\}$$

For such orthonormal frame, the determinant of the corresponding tensor metric $$\begin{pmatrix}E&F\\F&G \end{pmatrix}$$ is identically $1$, that is $EG-F^2=1$. Furthermore the curvature is zero for the linear center $V=y\partial_x-x\partial_y$.

Question: Is it true to say that $V$ has a center at origin if and only if the Gaussian curvature of the above metric is zero?

Note that a center is a singularity which is surrounded by a band of closed orbits. for quadratic vector fields they are classified at this paper.

The motivation for this post is mentioned in this answer

Remark: The initial motivation is mentioned in page 3, item 5 of this arxiv note.

The quadratic system $$y\partial_x -(x+x^2)\partial_y$$ has a center at the origin. But the curvature of the metric associated to the frame described in the question is not identically zero. I computed it and substituted point $$x=1,y=0$$ at the curvature, the result was non zero.
Remark: This enable us to construct a non flat metric on the cylinder such that the cylinder is foliated by closed geodescis with the same length, a non flat analogy of the standard flat structure of the cylinder: We count a deleted neighborhood of the origin as a topological cylinder. Now $$d\theta.X=1$$ implies that all closed geodesics have the same lenght $$2\pi.$$ Here $$X$$ is $$X=V/\psi(V)$$ with notations as in the question.