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I am reading the book Geometry theory of dynamical systems of Palis and de Melo, and I came across the following vector field $X(x,y,z)=(-xz,-yz,x^2+y^2)$. The book mentions that this field is tangent to the meridians of $\mathbb S^2$. I have graphed it and this is true, but how can I prove it? Any idea how to start? When it's a field in the plane it's easy, but now that the field is defined on a surface I can't see how to do it. All help is welcome. The real goal is to find the $\alpha-$limit and $\omega-$limit set of a point $p$ in $\mathbb S^2$.

enter image description here

What I don't understand well is because the stereographic projection guarantees me tangentiality, for example in the following image I can think of a vector field not tangent to a meridian, from which I extract a vector, and project it onto the plane, this projection would give us vectors with the same properties as the $\pi(X)$ projection.

enter image description here

Zaragosa
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2 Answers2

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Here's a bit more geometric approach. If you look at the meridian $y=0$, the vector field tangent to it is $(-z,0,x)$ — just think about the tangent vector to the unit circle in the plane being orthogonal to the position vector. Now all we have to do is make this vector field invariant under rotation about the $z$-axis. So we replace $x$ with $r = \sqrt{x^2+y^2}$, and $(1,0,0)$ (at the points of $y=0$) with the radial vector field $\frac1r(x,y,0)$. Thus, we end up with the vector field $\frac1r(-zx,-zy,r^2)$. Now rescale.

Ted Shifrin
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  • Thanks for the answer, I managed to understand the construction. But if I just wanted to know if this field is tangent to a family of curves such as meridians, is there a way just to check them? Or do I necessarily have to prove that the solution of ODE $x'=X(x)$ gives me exactly the meridians? – Zaragosa Apr 17 '22 at 07:09
  • Nah. Just check tangency, no integration required. If the curve is given parametrically, just find its velocity vectors. If it’s given implicitly (maybe in coordinates) as level curves of $f$, then apply $df$ to,the given vector field and see if you get $0$. – Ted Shifrin Apr 17 '22 at 20:41
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Hint: To what objects on the plane does the stereographic projection (say the one defined everywhere but the north pole) transform meridians and $X$? Note that indeed for the purposes of dynamics up to orbit equivalence the flat picture is sufficient (and the explicit expression for $X$ is not needed).

Alp Uzman
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  • Interesting, I have not been able to understand everything you mention but I will make the effort. My doubts arise when you say that the explicit form of $X$ is not entirely necessary, but putting another field would not necessarily make it tangent to the meridians. I would be infinitely grateful if you could complete your hint a little more. – Zaragosa Apr 16 '22 at 05:23
  • @Zaragosa "the full expression is not needed" means that you just need to look at the direction of $X$ and don't care about its norm, since integral lines of two proportional vector fields have the same support (only the parametrization is changed) – Didier Apr 16 '22 at 06:20
  • @Zaragosa What I meant is along the lines of Didier's comment. Of course there are many other vector fields on the sphere that are not necessarily along meridians. But for instance the fact that $X$ is a polynomial vector field is not important for the question of limit sets, or the topological picture. – Alp Uzman Apr 16 '22 at 14:50
  • Further, one can start with say $Y:(x,y)\mapsto (2x,2y)$ on the plane and redefine $X$ to be whatever vector field the stereographic projection (or its inverse) transforms $Y$ into (with an added singularity at the north pole). – Alp Uzman Apr 16 '22 at 14:53
  • @AlpUzman I'm not sure if I understood correctly, I made the stereographic projection of $X$, and I managed to obtain the following $\displaystyle \pi(X(x,y,z))=(\frac{-x}{z},\frac{-y}{z})$. But I still don't see how this can help me. Could you complete your answer please? Or at least give me the steps to get there? I just can't understand well. In my mind it is: parameterize any meridian and then obtain its vector, its tangent field, but I don't know... I'm confused. – Zaragosa Apr 16 '22 at 17:02
  • If you used the s.p. blowing up the north pole you would want to see a vector field on the plane made up of vectors emanating from the origin and shooting to infinity (your vector field is pointing toward the origin instead). Disregarding this matter of north/south pole, note that if you fix $x$ and $y$, varying $z$ will not change the direction. Thus points flowing according to $\pi(X)$ can not turn around the origin and can only come closer or further along rays. – Alp Uzman Apr 16 '22 at 17:14
  • @AlpUzman Hi, I modified my question by adding my doubt (I couldn't put it as a comment since I needed to show the graph). Could you please give a full answer? I'm asking you earnestly because I'm completely entangled. – Zaragosa Apr 16 '22 at 20:03
  • @Zaragosa I believe Prof. Shifrin's neat argument makes it clear enough (although of course it works because of the symmetry of the given vector field). The issue with your s.p. argument seems to be that you are applying $\pi$ to the vector as well, but the derivative of $\pi$ ought to be applied. Put differently, the $X$ you drew in your addendum is not a tangent vector; given any vector on the plane there will be a unique vector tangent to the sphere compatible with s.p. – Alp Uzman Apr 16 '22 at 23:45
  • That is why I had suggested earlier that it might be better to first verify that meridians are taken to rays, and then to apply the derivative of the inverse of s.p. to a vector field along rays. In any event though the calculation is somewhat messy (which is not a major issue from the perspective of orbit equivalence). – Alp Uzman Apr 16 '22 at 23:48
  • Thanks for your hint, now that I do the math I see things a little clearer. – Zaragosa Apr 17 '22 at 07:09