# No where vanishing exact $1$-form on compact manifold.

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?

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.

Exactness means $$\omega = df$$ for some smooth $$f:M\rightarrow \mathbb{R}$$. Because $$M$$ is compact, there is a $$p\in M$$ for which $$f(p)$$ is an absolute maximum.

We claim that $$d_p f = 0$$, so that $$\omega = df$$ is not non-vanishing. To show this, we pick $$v\in T_p M$$ and want to show that $$(d_p f) v = 0$$. To that end, let $$\gamma:(-\epsilon, \epsilon)\rightarrow M$$ be a smooth curve with $$\gamma(0) = p$$ and $$\gamma'(0) = v$$.

We want to show that $$(d_p f) v = 0$$, or, said another way, that $$\frac{d}{dt}|_{t=0} f(\gamma(t)) = 0$$. By definition of derivative, we need to show that $$\lim_{h\rightarrow 0} \frac{f(\gamma(h)) - f(\gamma(0))}{h} = 0.$$

First, the numerator is $$f(\gamma(h) - f(p) \leq 0$$ since $$f(p)$$ is a maximum of $$f$$. It follows that $$\lim_{h\rightarrow 0^+} \frac{f(\gamma(h))- f(p)}{h} \leq 0$$ and that $$\lim_{h\rightarrow 0^-} \frac{ f(\gamma(h)) - f(p)}{h} \geq 0$$.

By assumption, $$\lim_{h\rightarrow 0} \frac{f(\gamma(h)) - f(p)}{h}$$ exists, so the left and right hand limits must match. Since one is non-negative and the other is non-positive, the conclusion is that $$\lim_{h\rightarrow 0} \frac{f(\gamma(h)) - f(p)}{h} = d_p f v = 0$$. $$\square$$

$$\$$

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.)

On the compact manifold with boundary $$[0,1]$$, the differential $$1$$-form $$dx$$ is nowhere vanishing and exact. It is the exterior derivative of $$x$$.

Consider the cylinder, $$S^1 \times [0, 1]$$, parameterized by $$\theta, t$$.

The everywhere nonzero form $$dt$$ is exact (it's the differential of the function $$t$$ that gives the "height" coordinate).

So the answer is "no".

• See @JasonDeVito's answer. – John Hughes Nov 15 '18 at 2:47