Hilbert 16th problem asks for a uniform upper bound $H(n)$ for the number of limit cycles of a polynomial vector field of degree $n$ on the plane. Here is an updated proof of the finitness part of the Hilbert 16th problem:

http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=8352&option_lang=eng

This problem is open even for $n=2$.

In this question, by a quadratic vector field we mean a polynomial vector field $$P(x,y)\partial_x+Q(x,y) \partial_y\;\;\;(V)$$ where $P,Q\in \mathbb{R}[x,y]$ are polynomials of degree 2 with $P(0,0)=Q(0,0)=0$.

So the question is that

**(Hilbert 16th problem for n=2) Is $H(2)$ a finite number?**

As a possible approach to this question we would like to look at limit cycles of a vector field as closed geodesics. Namely we would like to put a Riemannian metric on the (regular part of the ) phase space such that the trajectories of the vector field would be unparametrized geodesics. By regular part we mean the complement of singularities of the vector field. Thus each limit cycle would be a closed geodesic. Then using Gauss Bonnete theorem we try to count the number of closed geodesics. So the sign of the Gaussian curvature play a very important role. But the first problem is that, regardless of the curvature sign, is there a Riemannian metric on the regular part of the phase space such that the trajectories of the vector field would be geodesic.

This example shows that, for a polynomial vector field of arbitrary degree such metric does not necessarily existed. But for quadratic vector field the situation is different as follows. In the following $C$ is the algebraic curve $$yP(x,y)-xQ(x,y)=0\;\;\;\;\;(C)$$

Observation:There is a Riemannian metric on $\mathbb{R}^2 \setminus C$ such that all solutions ofQuadratic vector field$(V)$ are geodesics. Furthermore every limit cycle of $(V) $ which surround the origin can not intersect the curve $C$

The proof of this observation is based on Proposition 6.7 and 6.8 as follows:

So we are sure that, on the complement of the above algebraic curve $C$, we have a Riemanian metric such that all solutions of $(V)$ are geodesics.

The methods of the proof of the Propositions 6.7 and 6.8 in the book Geometry of foliation suggest that we choose a Riemannian metric whose orthonormal base is the following:

$$\left\{\frac{x^2+y^2}{yP(x,y)-xQ(x,y)}V,\ \frac{1}{x^2+y^2}W\right\}.$$

where $V=P\partial_x +Q\partial_y$ is the quadratic vector field as in $(V)$ and $W$ is the radial vector field $W=x\partial_x+y\partial_y$. In fact $W$, as it is required in the proof of the above two propositions, lies in the kernel of $1\_$ form $\psi=\frac{1}{x^2+y^2}(ydx-xdy)$. This $1\_$ form is very essential to apply propositions 6.7 and 6.8 in the above reference. In fact this is a closed form which is identically equal to $1$ on the first vector of the above orthonormal frame.That is $\psi(\frac{x^2+y^2}{yP-xQ}(V))=1$.

I am really indebted to Ben McKay who suggested the $1\_$form $d\theta$ as a required $1\_$ form for possible satisfactions of proposition 6.8.

On the other hand the method of the proof of the Proposition 6.7 shows that, on the complement of
$C$, all trajectories of $V$ are geodesics for the metric arising from the above orthonormal frame. **Moreover we are free to rescale the second vector of the frame, arbitrarily**.

Now the main problem is that we control the sign of the curvature of such metric to make facilities to count the number of closed geodesics. In fact we use the Gauss Bonnete theorem.

Call the curvature of this metric $\kappa$.

Question: When a quadratic vector field $V$ does not have a center on the plane, is the curve $$\{(x,y)\mid \kappa(x,y)=0\}$$ transverse to $V$?

If not, what appropriate rescaling of the the second vector $ \frac{1}{x^2+y^2} W$ of the orthonormal frame would give a positive answer?

Notes:

1)A center is a singularity which is surrounded by a band of closed orbits. For quadratic vector fields they are classified at this paper.

2)**Of course existence of a positive answer to this question implies that $H(2)$ is finite.** Because the number of limit cycles of $V$ which suround the origin can be uniformly bounded since $\kappa=0$ is an algebraic curve so in every connected component of this algebraic curve, we have at most on limit cycle surrounding origin.
On the other hand there are at most 2 nest of limit cycles. So this would imply that $H(2)<\infty$

The motivation for this post is mentioned in this answer and this post.

However the initial motivation is mentioned in page 3, item 5 of this arxiv note.