$M=\{x: f(x)=0\}$ is orientable -- non-vanishing form

Question

Let $f\colon \mathbb{R}^n \to \mathbb{R}$ be a $\mathscr{C}^{\infty}-$ function and let $M=\{x\in \mathbb{R}^n : f(x)=0\}$. Suppose that $df(p)\neq 0$ for all $p\in M$. Then $M$ is an orientable manifold (i.e. there exists a non-vanishing top form).

Let $x_1,\ldots,x_n$ denote the standard coordinates on $\mathbb{R}^n$. Let $p\in M$ and suppose that $\frac{\partial{f}}{\partial{x_i}}(p) \neq 0$. Let $\Psi_i(x_1,\ldots,x_n)=(x_1,\ldots, \hat{i},\ldots,x_n)$ denote the projection. By the implicit function theorem, $\Psi_i$ is locally around $p$ invertible, let $\varphi_i(x_1,\ldots,\hat{i},\ldots,x_n)$ denote the $i$-th component function of $\Psi_i^{-1}$. These charts are compatible and thus $M$ is a manifold. Let $dy_1,\ldots,\widehat{dy_i},\ldots,dy_n$ denote the basis of $T_p^\ast M$ induced by $\Psi_i$.

For proving that $M$ is orientable, it is suggested to look at $\omega_i=(-1)^{i} \frac{1}{\frac{\partial{f}}{\partial{x_i}}} dx_1\wedge \ldots \wedge \hat{i} \wedge \ldots \wedge x_n$. The inclusion $\iota \colon M\hookrightarrow \mathbb{R}^n$ is differentiable and thus we may consider $\eta_i:=\iota^{\ast}(\omega_i)$. Clearly, $\iota^{\ast}(dx_k)=dy_k$ and therefore $$ \eta_i(p)=(-1)^{i} \frac{1}{\frac{\partial{f}}{\partial{x_i}}(p)} dy_1 \wedge \ldots \wedge \widehat{dy_i} \wedge \ldots \wedge dy_n. $$ It remains to prove that this is well-defined, i.e. if $\frac{\partial{f}}{\partial{x_j}}(p) \neq 0$, then $\eta_i(p)=\eta_j(p)$. Let $dz_1,\ldots, \widehat{dz_j},\ldots, dz_n$ denote the basis of $T_p^\ast M$ induced by $\Psi_j$. For $k\neq i,j$, we have $dy_k=\iota^\ast (dx_k)=dz_k$.

$\textbf{The problem}$ is to compute $dz_i$ in terms of $dy_1,\ldots,\widehat{dy_i},\ldots,dy_n$.
The coefficient $a_{\ell}$ in the unique expression $dz_i=\sum_{\ell \neq i} a_\ell dy_\ell$ is given by $$ a_{\ell} = \frac{\partial{(\Psi_i)_{\ell}}}{\partial{z_i}}=\frac{\partial{(\Psi_i \circ \Psi_j^{-1})_{\ell}}}{\partial{x_i}}. $$

Now, $$\Psi_i \circ \Psi_j^{-1}(x_1,\ldots,\widehat{x_j},\ldots,x_n)=(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{j-1},\varphi_j(x_1,\ldots,\widehat{x_j},\ldots,x_n),x_{j+1},\ldots,x_n)$$

and thus $a_{\ell}=0$ for $\ell \neq j$. And for $\ell = j$, $$ a_j=\frac{\partial{\varphi_j(x_1,\ldots,\widehat{x_j},\ldots,x_n)}}{\partial{x_i}}.$$

$\textbf{Question}$ At this point I am stuck. How can I proceed?

Answer 1

To explain Thomas's answer a bit more, he chooses a basis $v_1,...,v_{n-1}$ for $T_pM$ which wlog we can assume is the positive orientation. Since $\nabla f \not = 0$ and $\nabla f \not \in T_pM$, we can can normalize it and then $v_1,...,v_{n-1}, v:=\nabla f/\|\nabla f\|$ is a basis for $T_p\mathbb{R}^3$ which induces the positive orientation i.e if $(U, x^1,...,x^n)$ is a chart on $M$ (n-manifold) then $dx^1 \wedge \cdots \wedge dx^n$ gives the positive orientation i.e:

$$\alpha:=\underbrace{dx^1 \wedge \cdots \wedge dx^n}_{dV}(v_1,...,v_{n-1},v) >0$$

Now then, just define $\omega(v_1,...,v_{n-1},v) = \alpha$ and this gives a non-vanishing top form.

Answer 2

I would not try to do that in a coordinate system. You will have to patch them together and show that you can consistently keep the orientation, which is a complicated task.

Hint: $\frac{\nabla{f}}{|\nabla f|}$ is a nonvanishing unit normal vector field defined on all of $M$. Define the $n$-form $\omega$ for $v_1, \ldots, v_{n-1}$ on $TM$ by

$$\omega(v_1, \ldots, v_{n-1}):= dV(v_1, \ldots, v_{n-1},\frac{\nabla{f}}{|\nabla f|} )$$ where $dV$ is the volume form of the ambient space (which may be more or less the same as what you have written down, but it's defined without using coordinate patches).