In this post we would like to give a possible new approach to the second part of the [Hilbert 16th problem](https://en.wikipedia.org/wiki/Hilbert%27s_sixteenth_problem)
First we give a short introduction:
A quadratic system is a polynomial vector field on the plane in the form $$(X_{\alpha})\;\;\;\;\;\;\;\;\;\;\begin{cases}\dot x=P_{\alpha}(x,y)\\ \dot y=Q_{\alpha}(x,y) \end{cases}$$
where $P_{\alpha},Q_{\alpha}$ are polynomials of degree $2$ which are parametrized by a $10$ tuple parameter $\alpha$ of coefficients.
A **center** is a singularity of this vector field which is surrounded by a band of closed orbits.
All quadratic vector fields with center are classified as follows:
They correspond to a finite number of algebraic conditions in $\alpha$,
see ["Integrability of plane quadratic vector fields" Expos. Math(1990)3-25.](https://www.researchgate.net/publication/246484433_Integrability_of_plane_quadratic_vector_fields).
We denote these algebraic conditions by $Cent(\alpha)=0$
**Question:**
>Are there a family of (polynomial) Riemannian metrics $g_{\alpha}$ on the punctured plane(after removing singularities) with Gaussian curvature $\kappa_{\alpha}$ with the following properties:
>The solutions of $X_{\alpha}$, after a possible reparametrization, are geodesics of $g_{\alpha}$.Moreover $\kappa_{\alpha}$, is not zero except at a finite number of algebraic curves transverse to $X_{\alpha}$. Moreover $\kappa_{\alpha}$ is identically zero if $Cent(\alpha)=0$? The question, in particular its last part is well behaved and consistent since a quadratic system can not have a center and a limit cycle, simultaneously.
If the answer would be yes, [then $H(2)$, the maximum number of limit cycles of a quadratic system,](https://en.wikipedia.org/wiki/Hilbert%27s_sixteenth_problem) would be finite.
This question is already discussed at the comment-conversation of the following post:
http://mathoverflow.net/questions/160945/limit-cycles-as-closed-geodesicsin-negatively-curved-space