Intuition behind the fact that the figure eight in plane is not a single integral curve of a vector field

Question

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.

Answer 1

Following up on the remark in the comments, the fact that there is no way to "tell the particle where to go" means that there is no well-defined tangent space at $(0,0)$. For, suppose we had a curve $\gamma$ on some open interval $0\in I$ of $\mathbb R$. Let $S=\gamma(I)$ and $\gamma(0)=0.$ Then, if $0\in U\subseteq \mathbb R^2$ be an open neighborhood, $S\cap U$ is open in $S$. Take $U$ sufficiently small so that $(S\cap U)\setminus \{0\}$ has four connected components. Then, $S\cap U$ cannot be homeomorphic to any open interval $J$ because $J\setminus \{0\}$ has two connected components.

Another way to see this which only uses the fact that

$T_0S= \{v\in T_0\mathbb R^2:v(f)=0\ \text{for all}\ f\in C^{\infty}(\mathbb R^2)\ \text{such that}\ f_{|S}=0\}$ and a quick calculation of derivatives is to note that

$\gamma_1(t) = (\sin 2t, \sin t))$ and $\gamma_2(t) = (-\sin 2t, \sin t)$ both have their images in $S$ and give rise to derivations $\gamma_1'(0)=(2,1)$ and $\gamma_2'(0)=(-2,1)\ \textit{in}\ T_0\mathbb R^2.$

Now, suppose $f\in C^{\infty}(\mathbb R^2)$ such that $f_{|S}=0.$ Then, for

$i=1,2,\ \gamma_i'(0)(f)=\displaystyle\left(\frac{d}{dt}\right)_{t=0}\left(f\circ \gamma_i(t)\right)=0$.

But now since $\gamma_1'(0)$ and $\gamma'_2(0)$ cannot both be in $T_0S$, we have that

$T_0S\neq \{v\in T_0\mathbb R^2:v(f)=0\ \text{for all}\ f\in C^{\infty}(\mathbb R^2)\ \text{such that}\ f_{|S}=0\}.$

Answer 2

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 $(0,0)$.