Prove that restricting a vector field $X$ over $\mathbb R^2$ to $S^1$ gives a vector field over $S^1$

Question

I find the following problem a very interesting and very natural question to ask, but I could not prove or disprove it.

$Question$: Let $X$ be a vector field on $\mathbb{R^2}$. If we restrict this vector field $X$ on $S^1 \subset \mathbb{R^2}$ then it is a vector field over $S^1$.

MY attempt:

Since $X$ is a vector field over $\mathbb{R^2}$. then,

$X=a\frac{\partial}{\partial x} +b\frac{\partial}{\partial y}$.

That is for any $p\in \mathbb{R^2}$,

$X(p)=a(p)\frac{\partial}{\partial x}|_p +b(p)\frac{\partial}{\partial y}|_p$.

Now consider the restriction map on $S^1$ that is,

$X(p)=a(p)\frac{\partial}{\partial x}|_p +b(p)\frac{\partial}{\partial y}|_p$ ; where $p\in S^1$

Now (let's try to) show $a(p)\frac{\partial}{\partial x}|_p +b(p)\frac{\partial}{\partial y}|_p\in T_pS^1 = span \{\frac{\partial}{\partial x}|_p\}$

It implies that

$a(p)\frac{\partial}{\partial x}|_p +b(p)\frac{\partial}{\partial y}|_p= c(p) \frac{\partial}{\partial x}|_p$

That is,

$b(p)\frac{\partial}{\partial y}|_p= (c(p)-a(p)) \frac{\partial}{\partial x}|_p$

I am not sure how to proceed any further?

Thanks for any insight!!

Answer 1

1. The tangential space of $\mathbb{S}^1$ in $p = (x,y)$ is not $\mathrm{span}\left\{\left.\frac{\partial}{\partial x}\right|_p\right\}$ but $\mathrm{span}\left\{y \left.\frac{\partial}{\partial x} \right|_p- x \left.\frac{\partial}{\partial y}\right|_p\right\}$, at least if you use the standard coordinates on $\mathbb{R}^2$
2. Over every point $p$ the vectors $\left.\frac{\partial}{\partial x}\right|_p$ and $\left.\frac{\partial}{\partial y}\right|_p$ are linearly independent, so for example the constant vectorfield $X = \frac{\partial}{\partial x}$ on $\mathbb{R}^2$ will not restrict to a vectorfield on $\mathbb{S}^1$ because $\left.\frac{\partial}{\partial x}\right|_{(1,0)} \notin \mathrm{span}\left\{\left.\frac{\partial}{\partial y}\right|_{(1,0)}\right\} = T_{(1,0)} \mathbb{S}^1$.