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

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?

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.

• Thanks. It seems that we are implicitly using $\Lambda^{k}(T_p^\ast M) \cong (\Lambda^{k}(T_p M))^\ast \cong Alt^{k}(T_p M,\mathbb{R})$, and thus for a point $p\in M$, it is enough to say what $\omega(p) (v_1,\ldots,v_{n-1})$ is for a basis $v_1,\ldots,v_{n-1}$ of $T_pM$. Why do I need that $v_1,\ldots,v_{n-1},v$ is a basis of $T_p\mathbb{R}^n$; can't I define $\omega(p)(v_1,\ldots,v_{n-1},v):=dV(v_1,\ldots,v_{n-1},v)$ for some random $v$? And how does it make sense to write $\nabla f \not\in T_p M$? – user363120 Mar 9 '17 at 14:16
• Moreover, you consider $v_i$ as an element of $T_p\mathbb{R}^n$. By this you mean $\iota_p(v_i)$, where $\iota_p \colon T_pM \to T_p\mathbb{R}^n$ is induced by the inclusion? Why can we then say that $v_1,\ldots,v_{n-1}$ remain linearly independent in $T_p\mathbb{R}^n$ (just wondering because it is not generally true that the induced map on the tangent spaces in injective if the map itself is injective)? Sorry for all the questions. – user363120 Mar 9 '17 at 14:23
• @user363120 you are looking at this from a very technical perspective. But you need to understand the geometry in order to write down a proof. It is true because you have a globally defined nonzero normal vector which allows you to reduce an $n-$ form on the ambient space to an $n-1$-form you can restrict to $T_p M$ If you use a random $v$ then you ignore the information that $M$ is the $0$ set of a $f$. $f$ is defined on the ambient space, so a priori it's gradient is not tangent to $M$, so it's not in $T_p M$ In fact it is normal to $T_p$, as already mentioned. – Thomas Mar 9 '17 at 14:51
• @Thomas Yes, I do look at this from a technical perspective. Before building up geometric intuition I want to make sure that I get things formally correct. So you are saying that in this case $\iota_p \colon T_p M \to T_p \mathbb{R}^n$ is injective? Is there any particular reason for normalizing $\nabla f$? – user363120 Mar 9 '17 at 15:59
• @user363120 Yes, this inclusion is injective in this case (because such manifolds are embedded hypersurfaces). You don't have to normalize the gradient. I'm used to do this since it makes some calculations easier, but you only need to know it's nowhere zero. You will get the same form multiplied by a nonzero scalar function. – Thomas Mar 9 '17 at 16:03

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).

• Thanks for your reply. Would you mind explaining your notation? For me a differential $k$-form $\omega$ on $M$ is an assignment $p\mapsto \omega(p)\in \Lambda^{k} (T^{\ast}_p M)$. – user363120 Mar 9 '17 at 13:36
• @user363120 For $p\in M$ choose $v_i \in T_p M$ and define, in that point, $\omega$ as in my answer. Note that $dV(p)\in \Lambda^n ((T_p\mathbb{R}^n)^*)$ – Thomas Mar 9 '17 at 13:48