Let $M$ be an orientable smooth manifold of dimension $n$ and $f:M\to \mathbb{R}$ smooth with $0$ as regular value. I want to show that the level set $f^{-1}(0)$ is orientable (we know it to be an embedded manifold of dimension $n-1$).
I sketched a proof using differential forms and would like a review of it. I would also be interested in other proofs involving (or not) differential forms. Thanks in advance.
Here's my attempt:
Let $d$ denote the exterior derivative of differential forms.
Let $\varphi=(\varphi^1,\cdots,\varphi^n)$ denote a chart map of $M$.
As $f$ is a $0$-form, $df$ is a $1$-form and since $f*_p$ is surjective when $p\in f^{-1}(0)$ we have that $df=\sum\limits_{k=1}^n \frac{\partial f\circ \varphi^{-1}}{\partial x^k}d\varphi^{k}$ is non zero in an open neighborhood of $f^{-1}(0)$, which is an embedded orientable submanifold of $M$ that I shall call $N$.
Claim: Since $df$ is non zero in $N$ and $N$ is orientable, I can obtain a basis $\{\omega_1,\cdots, \omega_{n-1},df\}$ for $\Omega^1N $ (the space of differentiable 1-forms on $N$) such that $\eta=\omega_1\wedge\cdots\wedge \omega_{n-1}\wedge df$ is nowhere $0$ on $N$. i.e. $\eta_p\neq 0$ for all $p\in N$.
Let $i:f^{-1}(0)\hookrightarrow N$ be the inclusion map.
Since $\omega_1\wedge\cdots\wedge \omega_{n-1}$ is an $n-1$-form on $N$, the pull-back
$$\rho = i^*(\omega_1\wedge\cdots\wedge \omega_{n-1})=i^*\omega_1\wedge\cdots\wedge i^*\omega_{n-1}=(\omega_1\circ i)\wedge\cdots\wedge (\omega_{n-1}\circ i)$$
is an $n-1$-form on $f^{-1}(0)$ and is nowhere vanishing, for if $\rho_q=0$ then $\eta_{i(q)}=0$
Thus $\rho$ is a volume form (nowhere vanishing) on $f^{-1}(0)$ which we know to induce an orientation on $f^{-1}(0)$.
To prove my claim, I first extend $\{df\}$ to a basis using that every vector space has one (or should I treat $\Omega^1 N$ as a free $C^\infty$-module?).
Then I re-order the basis so that the volume form $\eta$ induces the same orientation as that of $N$.
To prove $\eta$ is a volume form I take for each $p\in N$ a basis $\{v_1,\cdots,v_n\}$ of $T_pN$ such that $\{\omega_1,\cdots,\omega_{n-1},df\}$ is it's dual basis. i.e. $\omega_{k}(v_j)=\delta_{kj}$ and $df(v_j)=\delta_{nj}$ then $\eta_p(v_1,\cdots,v_n)=1$ and thus non zero.
Best Answer
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.