Is there a connection $\nabla$ for which this particular non geodesible vector field $X$ satisfy $\nabla_X X=0$?

Question

Let $X$ be the following vector field on $\mathbb{R}^2\setminus \{0\}$ \begin{align} x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\ y' &= y\,(1-x^2-y^2)(x^2+y^2-3) + x\,(2-x^2-y^2). \end{align}

It is well known that there is no a Riemannian metric on the punctured plane such that the orbits of the above vector field would be unparametrized geodesics. The proof is based on existence of opposite orientation of two consecutive nested limit cycles in the phase portrait of the vector field.

Now we reduce the above geodesibility requirement to the following question;(Is this an obvious reduced question)?

Is there a (torsion free or metric or arbitrary) connection $\nabla$ on the punctured plane with $\nabla_X X=0$?

Answer 1

Your vector field can be written in polar coordinates as $$X=X^r\partial_r+X^{\varphi}\partial_{\varphi}=r(4r^2-r^4-3)\partial_r+(2-r^2)\partial_{\varphi},$$ which exhibits the rotational symmetry. The condition $\nabla_X X=0$ implies that \begin{align} X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+(\Gamma^r_{r\varphi}+\Gamma^r_{\varphi r}) X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\ X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+(\Gamma^{\varphi}_{r\varphi}+\Gamma^{\varphi}_{\varphi r}) X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0. \end{align} For a connection with vanishing torsion, we have $\Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}$ and $\Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}$, which leads to the simplification \begin{align} X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+2\Gamma^r_{r\varphi} X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\ X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+2\Gamma^{\varphi}_{r\varphi} X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0. \end{align} Since the vector field has rotational symmetry, I will look for a connection with coefficients that depend only on $r$. By making a fourth-order polynomial-in-$r$ ansatz for all Christoffel symbols, $$\Gamma^i_{jk}=\sum_{m=0}^4\Gamma^i_{jkm}r^m,$$ one can find the following solution: $$\Gamma^r_{rr}=-\frac{r}{2}, \quad \Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}=\frac{1}{4}(3-12r^2+r^4), \quad \Gamma^r_{\varphi\varphi}=0,$$ $$\Gamma^{\varphi}_{rr}=2, \quad \Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}=r(2-r^2), \quad \Gamma^{\varphi}_{\varphi\varphi}=0.$$

It's worth noting that anything lower than a fourth-order polynomial will not produce any solution. There does not seem to be any other connection with polynomial coefficients.