Try to find a nowhere vanishing vector field $F$ on $S^1$ such that $F(\vec{x})$ is tangent to $S^1$ at $\vec{x}$

Question

The definition of a tangent vector field in my book is the following:

This is from "Munkres-Analysis on Manifolds"

I want to define a vector field $V$ on a unit circle $S(1)$ that nowhere vanishes and $V(\vec{x})$ is tangent to $S(1)$ at $\vec{x}=(x_1,x_2)$ to give myself an example here.

I think all I need to do is just to define a tangent vector field $F$ on this unit circle, that is $$\begin{align*} F:& S^1 \to \mathbb{R}^2 \times \mathbb{R}^2 \\ &\vec{x} \to (\vec{x},f(\vec{x})) \end{align*}$$ where $$\begin{align*} f:S^1 \to \mathbb{R}^2 \end{align*}$$ First I have to make sure that $$\begin{align*} (x,f(x))\in {\scr T_x}(S^1) \end{align*}$$ I first consider the coordinate chart: $$\begin{align*} \alpha: &[0, 2\pi) \to S^1-\{0,1\} \\ \\\alpha & :t \to (\cos t, \sin t)=(x_1,x_2) \end{align*}$$ all vectors in ${\scr T_x}(S^1)$ are $$\begin{align*} (\vec{x};v)=(\alpha(t);D \alpha(t) \cdot u)=(\vec{x}; (-\sin t, \cos t)^T \cdot u)=(\vec{x},(-x_2,x_1)^T u) \end{align*}$$ where $v$ is an vector in $\mathbb{R}^2$ and $u$ is a vector in $\mathbb{R}$ , so it is just a scaler.

Then I just define $$\begin{align*} F:\vec{x} \to (\vec{x},(-x_2,x_1)^T) \end{align*}$$ where $\vec{x}=(x_1,x_2)$ to be the tangent vector field. and $f:(x_1,x_2) \to (-x_2,x_1)$

$F$ nowhere vanishes on $S^1-\{1,0\}$ , and by checking $x_1=\cos t, \ x_2=\sin t$ one can find this is true. Also, for each $\vec x$, $(-x_2,x_1)$ is indeed in the tangent space of $\vec x$ on $S^1$.

Thus I think $F(\vec{x})$ indeed in the tangent space of $\vec{x}$ on $S^1$. Thus, I can consider another coordinate chart on $S^1$, so I can construct a tangent vector field on $S^1$

I'm wondering if my thought and attempt are right. Any help on this? Thanks.

Answer 1

It is essentially correct, but there are a few inaccuracies.

1. The map $\alpha : [0,2\pi) \to S^1 \setminus \{p_0\}$ with $p_0 = (0,1)$ is not well-defined because $\alpha(0) = p_0$. Moreover, it cannot be a coordinate patch because $[0,2\pi)$ is not open in $\mathbb R$.
   This problem can easily be remedied by considering $\alpha : (0,2\pi) \to S^1 \setminus \{p_0\}$.

2. Your approach prodced a tangent vector field on $S^1 \setminus \{p_0\}$, but you missed to consider the point $p_0$.
   This can be done by considering the coordinate patch $\beta : (-\pi,\pi) \to S^1 \setminus \{p_1\}$ with $p_1 = (0,-1)$.

What you have done for the points $\xi \in S^1 \setminus \{p_0\}$ is to determine the tangent space $T_\xi S^1$. This is $$T_\xi S^1 = \{\eta \in \mathbb R^2 \mid \eta \cdot \xi = 0\} . \tag{1}$$ Here $\cdot$ denotes the standard scalar product on $\mathbb R^2$. As indicated in 2., $(1)$ also holds for $\xi = p_0$.

Clearly, if $\xi = (x_1,x_2)$, then $T_\xi S^1$ is spanned by $(-x_2,x_1)$.

Thus your tangent vector field is given by $$F : S^1 \to \mathbb R^2 \times \mathbb R^2, F(\xi) = (\xi,t(\xi))$$ with $t(x_1,x_2) = (-x_2,x_1)$.