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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 $\nabla_X X=0$?

Ali Taghavi
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1 Answers1

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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.

S.Surace
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  • Thank you very much for your answer. I will learn its details. BTW is it a metric connection? – Ali Taghavi Jun 25 '18 at 08:25
  • Wouldn't metricity contradict the non-geodesibility of the vector field? – S.Surace Jun 25 '18 at 10:43
  • Yes I am sorry for my stupid comment. – Ali Taghavi Jun 25 '18 at 11:48
  • I don't think it is a stupid question. When you try to solve the equations $\nabla_{\rho}g_{\mu\nu}=0$ for the unknown metric, using the connection given above, a quick calculation seems to suggest (but I may have made a mistake, so I would suggest to try for yourself) that you can find a nontrivial solution in some finite disk, but not the entire punctured plane (the metric seems to become singular on the boundary of said disk). I did not study the proof of the non-geodesibility that you linked to, so I cannot vouch for its accuracy. – S.Surace Jun 25 '18 at 12:25