The exact statement of the problem is:

If $M$ is compact and $0$ is a regular value of $f:M\to\mathbb R,$ then there is a neighborhood $U$ of $0\in\mathbb R$ such that $f^{-1}(U)$ is diffeomorphic to $f^{-1}(0)\times U$ by a diffeomorphism $\phi:f^{-1}(0)\times U\to f^{-1}(U)$ with $f(\phi(p,t))=t.$

I want to find a vector field $X$ on a neighborhood of $f^{-1}(0)$ which can be pushed forward to $d/dt$ on $\mathbb R.$ From there it seems like I could then find a coordinate neighborhood $x$ about each point in $f^{-1}(0)$ such that in this neighborhood $X$ looks like $\partial/\partial x^1,$ use a compactness argument, and somehow get the proof out of this. However, I don't know how to go about defining such a vector field $X.$ I'm also wondering whether this is true when $M$ is not compact, but I don't exactly know how I would go about proving this.

Even just a hint would be helpful.

  • 2
    $\begingroup$ Choose a Riemannian metric on $M$ and use the gradient field. $\endgroup$ – Neal Mar 12 at 19:49
  • $\begingroup$ @Neal I don't have any results about Riemannian metrics. A solution using vector fields is what I think I'm supposed to be doing. $\endgroup$ – D. Brogan Mar 12 at 22:03
  • $\begingroup$ You can always get a Riemannian metric on $M$ with a partition of unity. If you're not comfortable with that, take the gradient of $f$ on each open set in an oriented submanifold atlas of $f^{-1}(0)$ and use compactness + a partition of unity. $\endgroup$ – Neal Mar 12 at 22:45

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