# Orientation on level set manifold

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.

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.