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.


1 Answer 1


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 $f^{-1}(0)$.

  • $\begingroup$ I have understand. Well, Are the begining of my solution above and your this addition enough to prove the problem? $\endgroup$
    – 1190
    Commented May 7, 2013 at 18:18
  • $\begingroup$ @B11: I think your solution is correct and complete. By the way, the geometrical meaning of $\omega|_{f^{-1}(0)}$ is the normal "vector" of the surface $f^{-1}(0)$. $\endgroup$
    – 23rd
    Commented May 7, 2013 at 18:23
  • $\begingroup$ Okay thank you:) I dont understand the last thing you said? How do I use here? $\endgroup$
    – 1190
    Commented May 7, 2013 at 18:24
  • $\begingroup$ @B11: To solve your problem, you do not need to use it. It is just a comment for helping you to understand the motivation in defining $\omega$. $\endgroup$
    – 23rd
    Commented May 7, 2013 at 18:27
  • $\begingroup$ Okay thank you:)) I have learnt this topic for a week. So I dont have an enough knowledge. So I was confused and asked this. Thank you for comment and solution. :)) $\endgroup$
    – 1190
    Commented May 7, 2013 at 18:30

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