I have a question reading a proof of the following theorem.
Theorem. Let $M$ be an oriented smooth manifold, and suppose $S\subset M$ us a regular level set of a smooth function $f:M\to \Bbb R$. Then $S$ is orientable.
Proof. Choose a Riemannian metric $g$ on $M$, and let $N=\text{grad}f|_S$. The hypotheses imply that $N$ is nowhere tangent vector field along $S$, so the result follows.
I know that $N$ is normal to $S$ at each point of $S$, but how do we know that $N$ does not vanish at each point of $S$? If $(U,x^1,\dots,x^n)$ is a chart of $M$ near a point in $S$, then $g$ can be written as $g=g_{ij}dx^i \otimes dx^j$ where $(g_{ij})$ is a positive definite real symmetric matrix. Then $\text{grad}f|_S$ equals $g^{ij} \dfrac{\partial f}{\partial x^i} \dfrac{\partial }{\partial x^j}$ where $(g^{ij})=(g_{ij})^{-1}$. Since $S$ is a regular level set, some $\dfrac{\partial f}{\partial x^i}$ does not vanish at each point of $S$, but how do we know that $g^{ij} \dfrac{\partial f}{\partial x^i}$ does not vanish for some $j$, at each point of $S$?
Best Answer
Arguing by contradiction, if
$$\tag{1} g^{ij} \frac{\partial f}{\partial x^i} = 0$$
for all $j$, then for each $k$, multiply the above by $g_{jk}$ and sum over $j$,
$$\sum_{j=1}^ng_{jk} g^{ij} \frac{\partial f}{\partial x^i} = 0\Rightarrow \delta_{ik} \frac{\partial f}{\partial x^i} = 0.$$
Thus $\frac{\partial f}{\partial x^k} = 0$ for all $k$, which is a contradiction.
(This is just linear algebra: (1) holds for all $j$ if and only if $$ g^{-1} \nabla_0 f = 0, $$ where $\nabla_0 f = \left( \frac{\partial f}{\partial x^1} , \cdots, \frac{\partial f}{\partial x^n}\right)^t$. To show that $\nabla_0 f$ is nonzero we, of course, multiply $g$ to the left).