Suppose $f(x_1,...,x_{n+1})$ is a$ C^∞$ function on $\Bbb R^{n+1}$ with $0$ as a regular value. Show that the zero set of $f$ is an orientable submanifold of $\Bbb R_{n+1}$. In particular, the unit n-sphere $S_n$ in $\Bbb R^{n+1}$ is orientable.
I think that By the regular level set theorem, if $0$ is a regular value of a $C^∞$ function $f(x_1,...x_{n+1})$ on $\Bbb R^{n+1}$, then the zero set $f^{−1}(0)$ is a $C^∞$ manifold. And I guess that I need to apply a theorem. The theorem is following..
THM: A manifold M of dimension n is orientable if and only if there exists a $C^∞$ nowhere-vanishing n-form on M.
But I dont know how to apply this theorem. Please help me show how to apply this theorem to my question explicitly and instructively if my solution way is correct.
Thank you for help.
Let $$\omega=\sum_{i=1}^{n+1}(-1)^{i-1}\frac{\partial f}{\partial x_i}dx_1\wedge\cdots \wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots \wedge dx_{n+1}$$ Since $0$ is a regular value of $f$, $$df\wedge \omega= \sum_{i=1}^{n+1}\left(\frac{\partial f}{\partial x_i}\right)^2dx_1\wedge\cdots\wedge dx_{n+1}$$ is nowhere vanishing on $f^{-1}(0)$. It implies that the restriction of $\omega$ on $f^{-1}(0)$ is nowhere vanishing, so it induces an orientation of $f^{-1}(0)$.