I found several answers on the following question :
Does there exists a no where vanishing exact $1$-form on a compact manifold without boundary?
All answer says that certainly not. But I cannot understand use of the fact that Boundary of manifold is empty . These answers emphasis on the fact that manifold is compact.
Is it still true that for a compact manifold with boundary there does not exist a no where vanishing exact $1$-form?
Best Answer
The other two examples provide examples showing that you need to assume your manifold is boundaryless. I want to show where the "usual" proof in the boundaryless case breaks down. So, here is the usual proof.
$ \ $
So, where does this break down if $p\in \partial M$? Well, tangent vectors on the boundary are defined differently: you only need $\gamma$ to have a domain which is $(-\epsilon, 0]$ or $[0,\epsilon)$.
If we're in the case that the domain of $\gamma$ is $[0,\infty)$, then the one sided limit $\lim_{h\rightarrow 0^-} \frac{f(\gamma(h)) - f(p)}{h}$ no longer makes sense: $\gamma(h)$ doesn't make sense for negative $h$, so $f(\gamma(h))$ is meaningless.
Likewise, if the domain of $\gamma$ is $(-\epsilon, 0]$, then $\lim_{h\rightarrow 0^+} \frac{f(\gamma(h)) - f(p)}{h}$ no longer makes sense.
Thus, in either case, we lose one of the two inequalities. Without both, the proof no longer works to force $(d_p f) v = 0$. (And the other two answers show there is no way to "fix" the proof to handle the case where $p$ is a boundary point.)