If two vector fields on a one dimensional manifold commute then they are linearly dependent Let $X,Y$ be smooth vector fields on the unit circle $M = S^1$ such that $[X,Y] = 0$, i.e. (treating tangent vectors as derivations) $X(x)(Yf) = Y(x)(Xf)$ for all $x\in M$. Assuming that $X(x)\ne 0$ for all $x\in M$ I want to show that there is a $c\in \Bbb R$ such that $Y = cX$.
My attempt: for all $x\in M$ we have $X(x), Y(x) \in T_x M$ which is one dimensional, and since $X(x) \ne 0$ there is a $c(x)\in \Bbb R$ such that $Y(x) = c(x) X(x)$. Assuming this $c: M\to \Bbb R$ is smooth (I don't know how to show this) we get that $Y(x)(Xf) = X(x)(Yf) = X(x)(cXf) = X(x)(c) (Xf)(x) + c(x) X(x)(Xf) = X(x)(c)X(x)(f) + Y(x)(Xf)$, so $X(x)(c) X(x)(f) = 0$. This holds for all smooth $f: M\to \Bbb R$, so since $X(x)\ne 0$ we get $X(x)(c) = 0$. I'm not sure how to proceed or if this is even correct.
 A: Well, essentially you have written the proof. Only two details remain:

*

*The function $c:M\to \mathbb R$ is smooth. This is because both vector fields are smooth, so in a local chart $(U,t)$ they can be written as $X(t)=f(t)\partial_t$ and $Y(t)=g(t)\partial_t$, with $f,g$ smooth and $f(t)\neq 0$. Therefore $c(t)=g(t)/f(t)$, which is smooth and well defined.


*In the same local chart, since $X(x)(c)=0$, you have $f(t)\partial_t (c(t))=0$ and since $f(t)\neq 0$ you have $0=\partial_t(c(t))=c'(t)$. Son $c(t)$ is locally constant. But $S^1$ in connected, so $c$ is constant.
I am abusing of the notation writing in the local chart, sorry.
A: Recall that on the unit circle, there is a nowhere vanishing vector field usually denoted by $\partial_{\theta}$ when $S^1$ is seen embedded in $\Bbb C$, or $\partial_t$ when it is thought of as $\Bbb R / \Bbb Z$.
For any two vector fields $X$ and $Y$, there then exist two functions $f$ and $g$ such that $X = f\partial_{\theta}$ and $Y = g \partial_{\theta}$.
We have
\begin{align}
[X,Y] &= XY-YX\\
&= f\partial_{\theta}(g\partial_{\theta}) - g\partial_{\theta}(f\partial_{\theta})\\
&= f(\partial_{\theta}g) \partial_{\theta} + fg \partial_{\theta}^2 - g(\partial_{\theta}f)\partial_{\theta} - gf\partial_{\theta}^2\\
&= \left(f(\partial_{\theta}g) - g(\partial_{\theta}f)\right)\partial_{\theta}.
\end{align}
Thus, if $[X,Y]=0$, then
$
f \partial_{\theta}g - g \partial_{\theta}f = 0.
$
Assume that $X$ is non vanishing.
Then $f\neq 0$ everywhere.
Hence,
$$
\partial_{\theta} \left(\frac{g}{f}\right) = \frac{g\partial_{\theta}f - f \partial_{\theta}g}{f^2}=0,
$$
This in turns implies that $\frac{g}{f}$ is constant, that is, $g=\lambda f$ for some $\lambda\in \Bbb R$.

Here is an overkill solution.
Since $X$ is nowhere vanishing, it trivializes the tangent bundle of $S^1$.
Hence, there exists a smooth function $h$ such that $Y=hX$.
But then, $[X,Y] = [X,hX] = (Xh)X$.
It follows that $[X,Y]=0$ implies $Xh=0$.
That is, $h$ is constant along the integral curves of $X$.
But there is one and only one integral curve of $X$, which is (a particular parametrisation of) $S^1$.
Finally, $h$ is constant on $S^1$, and the result is proven.
