# Vector Fields – Tangent to a Meridian

dynamical systemsmanifoldsordinary differential equationsVector Fields

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 $$\mathbb S^2$$.

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.

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.