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I have a $C^{1}$ real valued function $f$ defined on a connected manifold $M$, it doesn't have critical points, lets assume that $f^{-1}(0)$ is a (compact) connected submanifold of $M$, does that imply that every level set will be connected?

Miguel
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1 Answers1

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No. Consider $f\colon \mathbb R^2\smallsetminus\{0\}\to\mathbb R$ given by $f(x,y) = x$.

EDIT: Sorry, I missed the fact that you said $f^{-1}(0)$ is compact, so this is not a counterexample.

It's still not true as stated. To get a counterexample, you can take $M=\mathbb R^2 \smallsetminus \{(0,0),\, (2,0),\, (-2,0)\}$, and $f(x,y) = x^2 + y^2 -1$. Then $f^{-1}(0)$ is the unit circle, which is compact and connected, but $f^{-1}(3)$ is a disjoint union of two arcs.

Jack Lee
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  • thanks, but what if we consider M complete? – Miguel Apr 05 '15 at 18:51
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    Your question doesn't have anything to do with a metric on $M$. You can always put a complete metric on $M$ without affecting either the hypothesis or conclusion. For example, $\mathbb R^2\smallsetminus {0}$ is diffeomorphic to $\mathbb R\times \mathbb S^1$, and the product metric on the latter is complete. However, I seem to have missed the "(compact)" in your hypothesis, so my counterexample doesn't work. I'll edit it. – Jack Lee Apr 05 '15 at 18:56
  • Thanks again for the edit, so I think I must look under what conditions on $M$ the answer would be possitive. – Miguel Apr 05 '15 at 20:26
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    @Miguel: The answer is positive if every level set of $f$ is compact. In fact, in that case, all level sets are diffeomorphic to each other. This can be proved by lifting the vector fields $\partial/\partial x^i$ to $M$, and using the flows of the lifted vector fields to show that every compact level set has a neighborhood in which all level sets are diffeomorphic. This allows you to show that the subset of $f(M)$ consisting of points whose preimages are diffeomorphic to $f^{-1}(0)$ is both open and closed in $f(M)$, hence all of $f(M)$. – Jack Lee Apr 05 '15 at 20:59
  • Actually I was going to prove that all the level sets are compact from the fact that they are connected, because I have a vector field whose flow take level sets to level sets (hypeotesis). – Miguel Apr 05 '15 at 21:13