Understanding Vector Fields

Question

I want to construct a smooth vector field on $\Bbb S^1 \subset \Bbb R^2$ with the standard smooth structure.

Can I just take a smooth vector field on $\Bbb R^2$, say $$V(x,y) = x \frac{\partial}{\partial x} + (1-y) \frac{\partial }{\partial y},$$ and restrict it to $\Bbb S^1$? How could I then show that this is a smooth vector field, and what would I need to do to determine its zeroes?

$\textbf{Edit}$

After Tsemo's remark, I've decided to consider the possibility of restricting the following vector field to $\Bbb S^1$.

$$V(x,y) = (y-y^2)\frac{\partial}{\partial x} + (xy-x)\frac{\partial}{\partial y}.$$

Answer 1

If X is tangent to $S^1$, for every $x\in S^1$, $X(x)$ is orthogonal to x. To see this, just calculate $\|{d\over{dt}}\phi_t(x)\|_{t=0}$.

$V(0,-1)=(0,-2)$ is not orthogonal to $(0,-1)$.

Answer 2

In calculus, we may take any path $\gamma\colon[a,b]\longrightarrow\mathbb{R}^d$ and define a vector field $\mathbf{F}$ on it by giving a formula $\mathbf{F}\colon\mathbb{R}^d\longrightarrow\mathbb{R}^d$; this allows us to compute things such as work done by a force along a trajectory, a very important concept in mechanics.

In particular, going the other way, if you are handed a force field $\mathbf{F}$, then it's restriction to the path $\gamma$ is exactly how we find the work done by that particular field over the trajectory. Another example, when $\gamma$ is a 2D surface rather than path $\gamma\colon[a,b]\times[c,d]\rightarrow\mathbb{R}^d$ would be calculating the flux of some flow across a surface, a concept that is at the very core of all fluid dynamics.

In short, we most certainly must be able to restrict vector fields in $\mathbb{R}^d$ to lower-dimensional manifolds to make meaningful computations.

In the context of manifolds however, by definition a vector space is a projection from the tangent space onto the manifold, so we add the further restriction that the formula $\mathbf{F}$ give a tangent vector along the restriction, but the same idea applies, and of course is possible.