Why this special method introduce a nowhere vanishing form on $\mathbb{S^1}$?

I'm studying "An Introduction to Manifolds, Loring W. Tu". In page 216 of it, Loring W. Tu introduces a method to find a nowhere vanishing form on $$\mathbb{S}^1$$ as follows:

To find a nowhere-vanishing 1-form on $$\mathbb{S}^1$$, we take the exterior derivative of both sides of the equation $$x^2 + y^2 = 1.$$ Using the antiderivation property of $$d$$, we get $$2xdx+2ydy=0.(*)$$ Of course, this equation is valid only at a point $$(x,y) \in \mathbb{S}^1.$$ Let $$U_x =\{(x,y)∈\mathbb{S}^1|x \not=0\}$$ and $$U_y =\{(x,y)∈\mathbb{S}^1 |y\not=0\}.$$ By $$(*)$$, on $$U_x \bigcap U_y$$: $$\frac{dy}{x} = -\frac{dx}{y}$$ Define a 1-form $$\omega$$ on $$\mathbb{S}^1$$ by $$\omega =\begin{cases} \frac{dy}{x}& on \space U_x \\ -\frac{dx}{y} & on \space U_y\\ \end{cases}$$

I know this form is well defined and smooth, but I have two questions:

$$1$$-Why this method makes a "nowhere vanishing form"?

$$2$$-Suppose $$f:\mathbb{R}^3\rightarrow \mathbb{R}$$ be smooth and $$f^{-1}(0)$$ is a regular level set.How can I generalize this method for manifold $$f^{-1}(0)$$ and find a nowhere-zero 2 form?

• For (2), you can't, because it's not true. The sphere is a level set, and has no nowhere-vanishing 1-form. Jan 10, 2020 at 7:09
• @SteveD I want to find a 2-form for sphere by generalized method. Jan 10, 2020 at 7:15
• It works for smooth curves $f(x,y)=0$. Jan 10, 2020 at 7:17
• @reuns So , Is there any other method to find? Jan 10, 2020 at 7:21

The $$1$$-form Tu defines is nowhere vanishing for the following reason. Consider $$dx/y$$ as a $$1$$-form on the subset of the circle where $$y\ne 0$$. It will vanish (as a $$1$$-form on $$\Bbb S^1$$) at the point $$(a,b)\in\Bbb S^1$$ only if the tangent space at $$(a,b)$$ is spanned by $$\partial/\partial y$$. But the equation $$x\,dx+y\,dy=0$$ on $$\Bbb S^1$$ tells you that this precisely when $$y=0$$, i.e., at points $$(\pm 1,0)$$, but we have restricted precisely to the complement of that set.
This method will generalize to hypersurfaces in arbitrary dimension, as long as you take the preimage of a regular value. For a surface $$S$$ given by $$f(x,y,z)=0$$, since $$df\ne 0$$ at every point of $$S$$, it follows that some partial derivative must be nonzero at every point of $$S$$. By analogy with what Tu gave for the circle, note that because $$df|_S = 0$$, we have $$\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz = 0$$ as a $$1$$-form on $$S$$. It follows that $$\frac{dy\wedge dz}{\partial f/\partial x} = \frac{dz\wedge dx}{\partial f/\partial y} = \frac{dx\wedge dy}{\partial f/\partial z}$$ whenever the expressions make sense. And at each point, (at least) one of them must be defined. You can check as I indicated for the circle case that the $$2$$-form is nowhere-vanishing on $$S$$. (For example, $$dy\wedge dz$$ vanishes on $$S$$ only at a point where $$\partial/\partial x$$ is tangent to $$S$$, and this happens precisely when $$\partial f/\partial x = 0$$.)