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Let $X$ be a real analytic vector field on $S^{2}$ which satisfies:

  1. $X$ has a finite number of singularities on $S^{2}$

  2. The equator is invariant under flow of $X$

3.$g_{*}X=\pm X$ where $g$ is the antipodal map $g(x,y,z)=(-x,-y,-z)$

Can we say that $X$ is topological equivalent to the Poincare compactification of a polynomial vector field on $\mathbb{R}^{2}$?

This is motivated by the holomorphic situation:Every one dimensional singular holomorphic foliation of $\mathbb{C}P^{2}$ arise from a polynomial vector field(or polynomial one form) on $\mathbb{C}^{2}$. See "Dynamics of singular holomorphic foliations by curves" by Saeid Zakeri, Contem. Math $\;\;\;$ The first version at http://arxiv.org/pdf/math/9809099.pdf

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    $\begingroup$ I don't quite see why you speak about topological conjugacy in that case. Wouldn't you want something stronger, like orbital equivalence as foliated analytic spaces ? The topological setting destroys far too structure in my opinion, you take the risk of losing rigidity arising from analyticity.In the complex case, there is no such requirement, since all the usual charts on $\mathbb P_2(\mathbb C)$ are birationnally equivalent. This should also be the case in the real setting because stereographic projection is algebraic. $\endgroup$
    – Loïc Teyssier
    Commented May 18, 2015 at 7:43
  • $\begingroup$ @LoïcTeyssier Could you please more explain why in this question, orbital equivalent is more appropriate than topological conjugacy? Moreover by orbital equivalent do you mean a homeomorphism send leaves to leaves regardless of parametrization? In this case, is not this weaker that topological conjugccy(rather than stronger)? $\endgroup$
    – Ali Taghavi
    Commented May 18, 2015 at 17:41
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    $\begingroup$ By «orbital equivalence» above I meant to speak of analytical orbital equivalence (=conjugacy), as opposed to topological orbital equivalence (=conjugacy) you speak of. I'm not sure topological equivalence is really what you want, but I can't speak in your stead. $\endgroup$
    – Loïc Teyssier
    Commented May 18, 2015 at 19:58
  • $\begingroup$ @LoïcTeyssier so my question in the post is weaker than what you proposed.analytic conjugacy implies topological one. am I missing some things? $\endgroup$
    – Ali Taghavi
    Commented May 19, 2015 at 17:33
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    $\begingroup$ No, you're not missing anything. I just wanted to point out that since your motivation (foliations in $\mathbb P_2(\mathbb C)$) was not dealing with topological conjugacy, maybe you wanted a less general question. That's all… $\endgroup$
    – Loïc Teyssier
    Commented May 19, 2015 at 18:26

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