This question goes on where this question ended. Given is an non-empty $(n-1)$-dimensional ($n\ge2$) differentiable submanifold $X\subset\mathbb{R}^n$ such that there exists an open $U\subset\mathbb{R}^n$ with $X\subset U$ and a differentiable function $f:U\to\mathbb{R}$ such that $X=f^{-1}(0)$ and $Df(p)\neq0$ for all $p\in X$. Then $X$ is (as shown in the other question) the topological boundary in $U$ of $A:=\{p\in U\,|\,f(p)<0\}$.

I have to show that $X$ is an orientable manifold. That is, there is an differentiable atlas $\mathcal{A}$ such that $\det(D(h_i\circ h_j^{-1})(h_j(p))>0$ for all $p\in U_i\cap U_j$, for all charts $(U_i,h_,V_i),(U_j,h_j,V_j)\in\mathcal{A}$. Alternatively, there exists a $(n-1)$-form on $X$ which vanishes nowhere.

Now intuitively I feel like this is clear. Let $V\subset X$ be a connected component; I think that it is the case that $X$ is actually the smooth boundary of $A$, since if it wasn't, then $V$ (or some other connected component) would have some sort of kink somewhere, which would contradict it being a submanifold; as $A$ is open in $\mathbb{R}^n$ (since $U$ is open and $A$ is open in $U$), $A$ is orientable (restricting the standard orientation on $\mathbb{R}^n$ to $A$), and an orientation on $A$ induces an orientation on $\partial A=X$. Can this be turned into a formal argument?

Alternatively, how would one approach a proof using the two definitions of orientation given above? As the gradient $Df(p)$ is non-zero everywhere on $X$, and $X$ is the level set of $f$ at $C=0$, the gradient would be an ideal candidate to provide the outward pointing normal vector needed.

Any help is much appreciated. This question has kept me busy for some time now.

  • $\begingroup$ It would be helpful if you could explain what $A$ is without referring to some answer to some other question. $\endgroup$ – Thomas May 10 '16 at 19:03
  • $\begingroup$ You are right, I have edited. It wasn't in the answer of some other question, but in the question itself, though. $\endgroup$ – B. Pasternak May 10 '16 at 19:04

The following $n-1$ form $\omega$ on $X$ is nowhere zero and thus orients $X$:

For $x\in X$ and $v_1,v_2,\cdots,v_{n-1}\in T_x(X)=(\operatorname {grad}f(x) )^\perp\subset T_x(\mathbb R^n)= \mathbb R^n$ define $$\omega_x(v_1,v_2,\cdots,v_{n-1})=\det (v_1,\cdots,v_{n-1},\operatorname {grad}f(x) ).$$


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