TITLE: Intuition behind the fact that the figure eight in plane is not a single integral curve of a vector field
QUESTION [7 upvotes]: Consider the figure eight in the plane, with concrete parametrization, for instance: $$f:(-\pi, \pi) \to \mathbb{R}^2, f(t) = (\sin(2t), \sin(t)).$$

We know that the image of $f$ is an immersed submanifold of $\mathbb{R}^2$, but it is not embedded in $\mathbb{R}^2$. My question is: why can't the image of $f$ be a single integral curve of (smooth) a vector field $X$ on $\mathbb{R}^2$? What is the intuition behind it, and how would a formal proof look like?

I assume that this is because of the self-intersection point, since then, the vector field would have to "split" its direction, but is this the full intuition behind it?
I also think that the figure eight should be the union of three integral curves of a vector field.

REPLY [2 votes]: The Cassini ovals form a family of curves that contains your lemniscate as a special individuum (by the way: your parametrization produces a "vertical" lemniscate). Here is a sketch of the family:

You can orient all occurring loops counterclockwise and then define a smooth vector field ${\bf v}$ in ${\mathbb R}^2$ having these curves as integral curves. This ${\bf v}$ has three zeros at $(0,0)$ and $(\pm c,0)$. Each half of the lemniscate is a "full" integral curve, since the moving particle takes time $-\infty<t<\infty$ to make the loop from $(0,0)$ to $(0,0)$.