5
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ For (2), you can't, because it's not true. The sphere is a level set, and has no nowhere-vanishing 1-form. $\endgroup$
    – Steve D
    Jan 10, 2020 at 7:09
  • $\begingroup$ @SteveD I want to find a 2-form for sphere by generalized method. $\endgroup$ Jan 10, 2020 at 7:15
  • $\begingroup$ It works for smooth curves $f(x,y)=0$. $\endgroup$
    – reuns
    Jan 10, 2020 at 7:17
  • $\begingroup$ @reuns So , Is there any other method to find? $\endgroup$ Jan 10, 2020 at 7:21

1 Answer 1

9
$\begingroup$

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 $\partial f/\partial x = 0$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .