I do not quite understand why the orientation of a manifold is defined to be a smooth nowhere vanishing differential form. For example, $S^1$, the unit circle, is an orientable manifold. Intuitively, I would like to pick counter-clockwise as my positive direction, but how does this correspond to a differential form? I just cannot associate vectors that define the orientation in the usual sense with differential forms. Can someone explain this to me a little bit and (hopefully) give some examples? Thanks a lot in advance!
Find the orientation of a manifold
differential-formsmanifoldsorientation
Related Solutions
Suppose $p \in U_1\cap U_2$ and consider $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p \in \bigwedge^nT_p^*M$. While $\phi_1^*(\operatorname{vol})_p$ and $\phi_2^*(\operatorname{vol})_p$ are non-zero elements of $\bigwedge^nT_p^*M$, they could be negative multiples of each other, in which case $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p$ could be zero.
For example, consider the manifold $S^1$. Let $U_1 = S^1\setminus\{1\}$, $\phi_1 : U_1 \to (0, 2\pi)$, $e^{i\theta}\mapsto \theta \bmod{2\pi}$, and $U_2 = S^1\setminus\{-1\}$, $\phi_2 : U_2 \to (-\pi, \pi)$, $e^{i\theta} \mapsto \pi - \theta \bmod{2\pi}$. Then $\phi_1^*(dx) = d\theta$ and $\phi_2^*(dx) = d(\pi - \theta) = -d\theta$ so $\rho_1(p)\phi_1^*(dx)_p + \rho_2(p)\phi_2^*(dx)_p = (\rho_1(p) - \rho_2(p))d\theta$ which is zero whenever $\rho_1(p) = \rho_2(p) = \frac{1}{2}$.
The idea that one construct a no-where vanishing $n-1$ form on $f^{-1}(0)$ using $df$ is of course correct. However, some of your argument is wrong.
In particular, sometimes it is not possible to find $\omega_1, \cdots, \omega_{n-1}$ such that $\omega_1\wedge \cdots \wedge \omega_{n-1}\wedge df$ is a volume form. This can be done locally, but generally not globally.
For example, take $M = \mathbb R^3$ and $f(x) = |x|^2-1$. So $f^{-1}(0)$ is the unit sphere, and $df$ is non-zero on $N = \mathbb R^3 \setminus \{\vec 0\}$. However, let $\omega$ be any differential one form on $N$. Then write
$$\omega(x) = \nu(x) + a(x) df(x),$$ where $\nu(x)$ is perpendicular to $df(x)$: that is $\nu = \sum \nu_i dx^i$ satisfies, $\nu_1(x_1) x_1 + \nu_2(x) x_2 + \nu_3(x)x_3 = 0$. Thus we can think of $\nu$ as a vector fields on the unit sphere (when restricted to the unit sphere), which must have a zero (at $x_0$) by the hairy ball theorem. Hence
$$ \omega(x_0)\wedge df (x_0) = a(x_0) df(x_0) \wedge df(x_0) = 0$$
This implies that no volume form on $N$ can be written as $\omega_1\wedge \omega_2 \wedge df$ for some one-forms $\omega_1, \omega_2$.
Back to your question. One way to show that $f^{-1}(0)$ is orientable is to show that there is a no-where vanishing normal vector along $f^{-1}(0)$. This can be done by giving $M$ a Riemannian metric and consider $\vec n = \nabla f$. Then $\alpha = \iota_{\nabla f} V$ is a $n-1$-nonvanishing form on $f^{-1}(0)$, where $V$ is a volume form on $M$.
For a slightly more general statement, see here.
Best Answer
Picking the counter-clockwise direction means that with each point of $S^1$, you associate a tangential vector pointing in that direction. This smells like a global section through the tangent bundle, whereas differential forms are sections through the co-tangential bundle - why the extra "co"? Well, you will get difficulties when trying to generalize this to $S^2$: The generalization would amount to picking a basis (that we want to declare as positively oriented) of the two-dimensional tangential space in each point and such that this choice depends smoothly on the point. But you can't do this, says the Hairy Ball Theorem.
Actually, instead of picking an example of a positively oriented base of the tangential space for every point, we should look for a method that can decide whether a base is positively or negatively oriented. That is, for $n$ vector fields $X_1,\ldots, X_n$, we want to compute some $\alpha(X_1,\ldots, X_n)$ such that the result is positive for positively oriented bases, negative for negatively oriented bases (and zero in case of linear dependnce). As each $X_i$ is in $TM$, this $\alpha$ is a map $TM^n\to \Bbb R$. To make life easy, we of course want $\alpha$ to be linear in each component (i.e., multi-linear). Moreover, swapping $X_i\leftrightarrow X_j$ shall result in a sign change; in particular, if $X_i=X_j$ with $i\ne j$, then the result shall be zero. This means that $\alpha$ is an alternating multi-linear form $\bigwedge^nTM\to \Bbb R$, in other words, a global section through the $n$th outer power of the co-tangential bundle - also known as a differential $n$-form.